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GMAT Work Rate: Understanding Work Rate Problems
Getting to Know GMAT Work Rate Problems
Table of Contents
Key Takeaways:• GMAT work rate questions revolve around the basic formula: Work = Rate × Time.
• Combining work rates means adding individual rates to get total efficiency in joint tasks.
• Time and rate are inversely related; as rate increases, time to complete a task drops.
• These problems mimic real-world efficiency tests in roles like consulting or operations.
• Regular practice with word problems improves accuracy and speeds up GMAT math solving.
Work rate problems are a core part of the GMAT's quantitative section, designed to test your understanding of efficiency and time management. These problems involve calculating how long it takes for one or more workers (or machines) to complete a task, given their rates of work. Mastering the work rate formula, Work = Rate × Time, is crucial for solving these questions accurately and efficiently. This article will guide you through the essential concepts, formulas, and strategies to tackle GMAT work rate problems with confidence, helping you score higher on this challenging section.
Preparing for the GMAT Work Rate Problems
Preparing for the GMAT requires a well-structured approach. Utilizing the right study materials can significantly enhance your chances of achieving a high score. Here are some effective resources:
| Resource Type | Examples |
--- |
| Books | The Official GMAT Guide, GMAT For Dummies |
| Online Courses | Magoosh, Manhattan Prep |
| Practice Tests | Official GMAT Practice Exams, GMATPrep Software |
| Forums and Communities | GMAT Club, Reddit GMAT |
These resources cater to different learning styles, whether you prefer reading, interactive learning, or discussing with peers. Many of these platforms also provide video tutorials and forums where students can ask questions and share tips.
Creating a Study Plan
A tailored study plan is essential for effective GMAT preparation. Here are key steps to create one:
An effective study plan not only improves your knowledge but also builds your confidence as test day approaches. Staying organized and consistent in your preparation will lead to better outcomes.
GMAT Work Rate Problems
Work rate problems are a common type of question on the GMAT, particularly in the Quantitative section. These problems require you to calculate the amount of work done by one or more people, machines, or systems over a specific period. The key to solving these problems lies in understanding the relationship between work, rate, and time.
The basic formula for work rate problems is:
Work = Rate × Time
Where:
For example, if a person can complete a task in 4 hours, their work rate is 1 task / 4 hours = 0.25 tasks per hour.
Key Concepts to Remember
Strategies for Solving Work Rate Problems
The best way to excel in work rate problems is through practice. Familiarizing yourself with different types of work rate questions can help you identify patterns and apply the correct formulas. Here are some strategies to consider:
Additional Resources for Work Rate Problems
When preparing for work rate problems, it’s beneficial to use specific study materials focused on this topic. Here are some recommended resources:
| Resource Type | Examples |
--- |
| Books | GMAT Quantitative Strategy Guide, 5 lb. Book of GRE Practice Problems |
| Online Courses | Khan Academy GMAT Prep, GMAT Ninja |
| Practice Questions | GMAT Official Guide, Veritas Prep |
| YouTube Channels | GMAT Prep Now, Magoosh GMAT |
These materials provide targeted practice and explanations, helping you grasp the concepts of work rate problems more effectively.
Please refer GMAT Work Rate Problems for detailed analysis of GMAT work Rate problems
Common Work Rate Problem Scenarios
Scenario 1: Two Workers Completing a Task Together
One of the most common scenarios involves two workers completing a task together. Understanding how to calculate the combined work rate is essential for solving these problems.
For instance, consider Worker A, who can complete a task in 4 hours, and Worker B, who can finish the same task in 6 hours. To find their combined work rate, you use the formula:
Combined Rate = 1/Time taken by A + 1/Time taken by B = 1/4 + 1/6
To add these fractions, find a common denominator, which in this case is 12:
Combined Rate = 3/12 + 2/12 = 5/12
This means together, they can complete 5/12 of the task in one hour. To find out how long it takes them to finish the entire task, you can take the reciprocal of their combined rate:
Time to complete the task = 12/5 = 2.4 hours or 2 hours and 24 minutes
Scenario 2: Work Done in Portions
Another common scenario involves a task that is completed in portions, often over different time intervals. For example, let’s say a worker completes 60% of a task in the first 3 hours and the remaining 40% in the next 2 hours.
First, calculate the work rate for each portion:
In this scenario, both portions have the same work rate. Understanding how to break down the task into segments helps clarify the overall time taken and can make solving these problems more manageable.
Please refer GMAT Rate Problems for detailed analysis of GMAT Work Rate
Tips for Mastering Work Rate Problems
1. Practice with Various Problem Types
Different types of work rate problems can appear on the GMAT, from simple two-worker scenarios to more complex problems involving multiple workers and variable rates. Practicing a wide range of problems will help you build a strong foundation.
2. Use Practice Tests
Simulating test conditions with practice tests is an effective way to improve your time management skills. Many resources offer full-length practice exams that closely resemble the GMAT, allowing you to gauge your readiness and familiarize yourself with the types of work rate problems you might encounter.
3. Analyze Your Performance
After taking practice tests, spend time reviewing the questions you got wrong. Understanding the reasoning behind each question, especially work rate problems, will help reinforce your knowledge and improve your skills for the actual exam.
4. Study with a Group
Collaborating with fellow test-takers can provide different perspectives on solving work rate problems. Group study sessions can help clarify concepts, share strategies, and provide motivation as you prepare for the GMAT.
5. Keep Resources Handy
Having reliable resources at your disposal, such as formula sheets, online calculators, and study guides, can enhance your study sessions. They can serve as quick references while solving practice problems.
Related Blogs
Conclusion
Work rate problems play a significant role in the GMAT Quantitative section. Mastering these problems not only helps you perform better on the exam but also equips you with essential skills for real-life situations where teamwork and project completion are necessary. By understanding the principles of work rates, you can approach these questions with confidence and efficiency.
Abhyank Srinet, the founder of MiM-Essay, is a globally recognized expert in study abroad and admission consulting. His passion is helping students navigate the complex world of admissions and achieve their academic dreams. Abhyank earned a Master's degree in Management from ESCP Europe, where he developed his skills in data-driven marketing strategies, driving growth in some of the most competitive industries.
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Contents
1 Theorem
2 Proof 1
3 Proof 2
4 Proof 3
5 Proof 4
6 Proof 5
7 Source of Name
Theorem
∫∞0 sin x x d x=π 2∫0∞sinx x d x=π 2
Proof 1
By Fubini's Theorem:
∫∞0(∫∞0 e−x y sin x d y)d x=∫∞0(∫∞0 e−x y sin x d x)d y∫0∞(∫0∞e−x y sinx d y)d x=∫0∞(∫0∞e−x y sinx d x)d y
Then:
∫∞0 e−x y sin x d y∫0∞e−x y sinx d y==[−e−x y sin x x]∞0[−e−x y sinx x]0∞Primitive of e a x e a x
==sin x x sinx x
and:
∫∞0 e−x y sin x d x∫0∞e−x y sinx d x==[−e−x y(y sin x+cos x)y 2+1]∞0[−e−x y(y sinx+cosx)y 2+1]0∞Primitive of e a x sin b x e a x sinb x
==1 y 2+1 1 y 2+1
Hence:
∫∞0 sin x x d x∫0∞sinx x d x==∫∞0 1 y 2+1 d y∫0∞1 y 2+1 d y
==[arctan y]∞0[arctany]0∞Primitive of 1 x 2+a 2 1 x 2+a 2
==π 2 π 2 as lim y→∞arctan y=π 2 lim y→∞arctany=π 2
■◼
Proof 2
∫∞0 sin x x d x∫0∞sinx x d x is convergent as an improper integral.
Indeed, for all n∈N n∈N:
∫2 π n 0 sin x x d x∫0 2 π n sinx x d x==∑k=0 2 n−1∫π(k+1)π k sin x x d x∑k=0 2 n−1∫π k π(k+1)sinx x d x
==∑k=0 2 n−1(−1)k∫π 0 sin x x+π k d x∑k=0 2 n−1(−1)k∫0 π sinx x+π k d x
==∑k=0 2 n−1(−1)k π k∫π 0 sin x 1+x π k d x∑k=0 2 n−1(−1)k π k∫0 π sinx 1+x π k d x
But:
∣∣∣∫π 0 sin x 1+x π k d x−2∣∣∣|∫0 π sinx 1+x π k d x−2|≤≤∫π 0 sin x∣∣∣1 1+x π k−1∣∣∣d x∫0 π sinx|1 1+x π k−1|d x
==1 k π∫π 0 x sin x d x 1 k π∫0 π x sinx d x
so that:
∫π 0 sin x 1+x k π d x→k→∞2∫0 π sinx 1+x k π d x→k→∞2
Hence:
∫2 π n 0 sin x x d x=∑k=0 n−1 1 2 π k∫π 0 sin x 1+x 2 π k d x−1 π(2 k+1)∫π 0 sin x 1+x π(2 k+1)d x∫0 2 π n sinx x d x=∑k=0 n−1 1 2 π k∫0 π sinx 1+x 2 π k d x−1 π(2 k+1)∫0 π sinx 1+x π(2 k+1)d x
can be expressed as a series whose general term is equivalent to:
2 π×1 2 k(2 k+1)2 π×1 2 k(2 k+1)
which is the term of an absolutely convergent series.
By Modulus of Sine of x Less Than or Equal To Absolute Value of x:
∣∣∣e−α x sin x x∣∣∣≤e−α x|e−α x sinx x|≤e−α x
From Laplace Transform of Real Power:
∫∞0 e−α x d x=1 α∫0∞e−α x d x=1 α
Hence by Comparison Test for Improper Integral:
∫∞0 e−α x sin x x d x∫0∞e−α x sinx x d x
converges whenever α>0 α>0.
So, we can define a real functionI:(0..∞)→R I:(0..∞)→R by:
I(α)=∫∞0 e−α x sin x x d x I(α)=∫0∞e−α x sinx x d x
for each α∈(0..∞)α∈(0..∞).
Using Improper Integral of Partial Derivative on segments included in (0..∞)(0..∞):
I′(α)I′(α)==∂∂α∫∞0 e−α x sin x x d x∂∂α∫0∞e−α x sinx x d x
==∫∞0∂∂α e−α x sin x x d x∫0∞∂∂α e−α x sinx x d xLeibniz's Integral Rule
==−∫∞0 e−α x sin x d x−∫0∞e−α x sinx d xDerivative of Exponential Function
==[−e−α x(−α sin x+cos x)(−α)2+1]∞0[−e−α x(−α sinx+cosx)(−α)2+1]0∞Primitive of e α x sin b x e α x sinb x
==−1 α 2+1−1 α 2+1
Therefore, by Derivative of Arctangent Function
I(α)=−arctan α+K I(α)=−arctanα+K
for some K∈R K∈R.
We also have:
|I(α)||I(α)|==∣∣∣∫∞0 e−α x sin x x d x∣∣∣|∫0∞e−α x sinx x d x|
≤≤∫∞0∣∣∣e−α x sin x x∣∣∣d x∫0∞|e−α x sinx x|d xTriangle Inequality for Definite Integrals
≤≤1 α 1 α
so:
lim α→∞|I(α)|=0 lim α→∞|I(α)|=0
That is:
lim α→∞I(α)=0 lim α→∞I(α)=0
Therefore:
I(α)=π 2−arctan α I(α)=π 2−arctanα
since arctan α→α→∞π 2 arctanα→α→∞π 2.
Note that we have:
I(α)→α→0 π 2 I(α)→α→0 π 2
We now need to show that:
I(α)→α→0∫∞0 sin x x d x I(α)→α→0∫0∞sinx x d x
Observe for this purpose that:
I(α)I(α)==∫∞0 sin 2 x x e−2 α x d x∫0∞sin2 x x e−2 α x d x change of variable x↦2 x x↦2 x
==2∫∞0 sin x x e−2 α x cos x d x 2∫0∞sinx x e−2 α x cosx d x
==2[sin 2 x x e−2 α x]∞0−2∫∞0 sin x e−2 α x(−α sin x x+cos x x−sin x x 2)d x 2[sin 2x x e−2 α x]0∞−2∫0∞sinx e−2 α x(−α sinx x+cosx x−sinx x 2)d xintegration by Parts for improper integral
==2 α∫∞0 sin 2 x x e−2 α x d x+2∫∞0(sin x x)2 e−2 α x d x−I(α)2 α∫0∞sin 2x x e−2 α x d x+2∫0∞(sinx x)2 e−2 α x d x−I(α)by Continuous Real Function/Examples, sin x x→1 sinx x→1 in 0 0
where all the improper integrals appearing here are convergent by Comparison Test for Improper Integral, as used above for defining I(α)I(α).
Therefore:
I(α)=α∫∞0 sin 2 x x e−2 α x d x+∫∞0(sin x x)2 e−2 α x d x I(α)=α∫0∞sin 2x x e−2 α x d x+∫0∞(sinx x)2 e−2 α x d x
We also have:
∫∞0 sin x x d x∫0∞sinx x d x==2∫∞0 sin x x cos x d x 2∫0∞sinx x cosx d x by change of variable x↦2 x x↦2 x
==2[sin 2 x x]∞0−2∫α 0 sin x x cos x d x+2∫∞0(sin x x)2 d x 2[sin 2x x]0∞−2∫0 α sinx x cosx d x+2∫0∞(sinx x)2 d x
==−∫∞0 sin x x d x+2∫∞0(sin x x)2 d x−∫0∞sinx x d x+2∫0∞(sinx x)2 d x
where the improper integrals on the right hand side are convergent because the first one identifies with ∫∞0 sin x x d x∫0∞sinx x d x and the second one because sin 2 x x 2 sin 2x x 2 is integrable on (0..∞)(0..∞), since it has a finite limit at 0 0 and is smaller than 1 x 2 1 x 2 at ∞∞.
Hence:
∫∞0 sin x x d x=∫∞0(sin x x)2 d x∫0∞sinx x d x=∫0∞(sinx x)2 d x
where the second integral is absolutely convergent.
Moreover:
α∫∞0 sin 2 x x e−2 α x d x α∫0∞sin 2x x e−2 α x d x==α∫∞0 sin 2 x α x e−2 x d x α∫0∞sin 2x α x e−2 x d x
≤≤α⎛⎝∫α 0 x 2 α 2 x d x+∫1 α 1 x d x+∫∞1 e−2 x d x⎞⎠α(∫0 α x 2 α 2 x d x+∫α 1 1 x d x+∫1∞e−2 x d x)
==α(1 2−ln α+1 2 e 2)α(1 2−lnα+1 2 e 2)
→α→0→α→0 0 0
whenever α≤1 α≤1.
Also:
∫∞0(sin x x)2 e−2 α x d x→α→0∫∞0(sin x x)2 d x∫0∞(sinx x)2 e−2 α x d x→α→0∫0∞(sinx x)2 d x
This is because, for any positive R R and α α:
∫∞0(sin x x)2(1−e−2 α x)d x∫0∞(sinx x)2(1−e−2 α x)d x==∫1 α√0(sin x x)2(1−e−2 α x)d x+∫∞1 α√(sin x x)2(1−e−2 α x)d x∫0 1 α(sinx x)2(1−e−2 α x)d x+∫1 α∞(sinx x)2(1−e−2 α x)d x
≤≤(1−e−2 α√)∫∞0(sin x x)2 d x+∫∞1 α√(sin x x)2 d x(1−e−2 α)∫0∞(sinx x)2 d x+∫1 α∞(sinx x)2 d x
→α→0→α→0 0 0
because sin 2 x x 2 sin 2x x 2 is integrable on (0..∞)(0..∞).
Finally, we have:
I(α)I(α)==α∫∞0 sin 2 x x e−2 α x d x+∫∞0(sin x x)2 e−2 α x d x α∫0∞sin 2x x e−2 α x d x+∫0∞(sinx x)2 e−2 α x d x
→α→0→α→0∫∞0(sin x x)2 d x∫0∞(sinx x)2 d x
==∫∞0 sin x x d x∫0∞sinx x d x
as well as:
I(α)I(α)==π 2−arctan α π 2−arctanα
→α→0→α→0 π 2 π 2
So that, by uniqueness of limits:
∫∞0 sin x x d x=π 2∫0∞sinx x d x=π 2
■◼
Proof 3
Let:
f(x)=⎧⎩⎨e i x−1 x i x≠0 x=0 f(x)={e i x−1 x x≠0 i x=0
We have, by Euler's Formula, for x∈R x∈R:
I m(f(x))={sin x x 1 x≠0 x=0 I m(f(x))={sinx x x≠0 1 x=0
So:
I m(∫∞0 e i x−1 x d x)=∫∞0 sin x x d x I m(∫0∞e i x−1 x d x)=∫0∞sinx x d x
Let C R C R be the arc of the circle of radiusR R centred at the origin connecting R R and −R−Ranticlockwise.
Let Γ R=C R∪[−R..R]Γ R=C R∪[−R..R].
Then, by Contour Integral of Concatenation of Contours:
∮Γ R e i x−1 x d x=∫C R e i x−1 x d x+∫R−R e i x−1 x d x∮Γ R e i x−1 x d x=∫C R e i x−1 x d x+∫−R R e i x−1 x d x
From Linear Combination of Contour Integrals, we write:
∮Γ R e i x−1 x d x=∫C R e i x x d x−∫C R d x x+∫R−R e i x−1 x d x∮Γ R e i x−1 x d x=∫C R e i x x d x−∫C R d x x+∫−R R e i x−1 x d x
Note that f f is holomorphic inside our contour.
It then follows from the Cauchy-Goursat Theorem, that:
∮Γ R e i x−1 x d x=0∮Γ R e i x−1 x d x=0
We also have:
∣∣∣∫C R e i x x d x∣∣∣|∫C R e i x x d x|≤≤π max 0≤θ≤π∣∣∣1 R e i θ∣∣∣π max 0≤θ≤π|1 R e i θ|Jordan's Lemma
==π R π R
→→0 0 as R→∞R→∞
Therefore:
lim R→∞∫C R d x x=lim R→∞∫R−R e i x−1 x d x=∫∞−∞e i x−1 x d x lim R→∞∫C R d x x=lim R→∞∫−R R e i x−1 x d x=∫−∞∞e i x−1 x d x
Evaluating the integral on the left hand side:
∫C R d x x∫C R d x x==∫π 0 i R e i θ R e i θ d θ∫0 π i R e i θ R e i θ d θ Definition of Complex Contour Integral
==i∫π 0 d θ i∫0 π d θ
==π i π i
So:
∫∞−∞e i x−1 x d x=π i∫−∞∞e i x−1 x d x=π i
Taking the imaginary part:
∫∞−∞sin x x d x=π∫−∞∞sinx x d x=π
From Definite Integral of Even Function:
∫∞−∞sin x x d x=2∫∞0 sin x x d x∫−∞∞sinx x d x=2∫0∞sinx x d x
Hence:
∫∞0 sin x x d x=π 2∫0∞sinx x d x=π 2
■◼
Proof 4
From Integral to Infinity of Function over Argument:
∫∞0 f(x)x=∫→∞0 F(u)d u∫0∞f(x)x=∫0→∞F(u)d u
for a real functionf f and its Laplace transformL{f}=F L{f}=F, provided they exist.
Let f(x):=sin x f(x):=sinx.
Then from Laplace Transform of Sine:
L{f(x)}=1 s 2+1 L{f(x)}=1 s 2+1
Hence:
∫∞0 sin x x d x∫0∞sinx x d x==∫→∞0 d u u 2+1∫0→∞d u u 2+1
==[arctan u]∞0[arctanu]0∞Primitive of 1 x 2+a 2 1 x 2+a 2
==π 2 π 2
■◼
Proof 5
Let M∈R>0 M∈R>0.
Define a real functionI M:R→R I M:R→R by:
I M(α):=∫M 0 sin x x e−α x d x I M(α):=∫0 M sinx x e−α x d x
Then, for α>0 α>0:
|I M(α)||I M(α)|≤≤∫M 0∣∣∣sin x x e−α x∣∣∣d x∫0 M|sinx x e−α x|d xAbsolute Value of Definite Integral
≤≤∫M 0 e−α x d x∫0 M e−α x d xSine Inequality|sin x|≤|x||sinx|≤|x|
==[e−α x−α]M 0[e−α x−α]0 MPrimitive of e a x e a x
==1 α−e−α M α 1 α−e−α M α
(1):(1):≤≤1 α 1 α
On the other hand:
I′M(α)I M′(α)==∫M 0∂∂α(sin x x e−α x)d x∫0 M∂∂α(sinx x e−α x)d xDefinite Integral of Partial Derivative
==∫M 0−sin x e−α x d x∫0 M−sinx e−α x d xPrimitive of e a x e a x
==[−e−α x(−α sin x+cos x)(−α)2+1]M 0[−e−α x(−α sinx+cosx)(−α)2+1]0 MPrimitive of e α x sin b x e α x sinb x
(2):(2):==−1 α 2+1+cos M e−α M α 2+1+sin M α e−α M α 2+1−1 α 2+1+cosM e−α M α 2+1+sinM α e−α M α 2+1
Thus:
I M(A)−I M(0)I M(A)−I M(0)==∫A 0 I′M(α)d α∫0 A I M′(α)d αFundamental Theorem of Calculus
==−∫A 0 d α α 2+1+cos M∫A 0 e−α M α 2+1 d α+sin M∫A 0 α e−α M α 2+1 d α−∫0 A d α α 2+1+cosM∫0 A e−α M α 2+1 d α+sinM∫0 A α e−α M α 2+1 d α by (2)(2) and Linear Combination of Integrals
Thus:
∣∣∣I M(A)−I M(0)+∫A 0 d α α 2+1∣∣∣|I M(A)−I M(0)+∫0 A d α α 2+1|==∣∣∣cos M∫A 0 e−α M α 2+1 d α+sin M∫A 0 α e−α M α 2+1 d α∣∣∣|cosM∫0 A e−α M α 2+1 d α+sinM∫0 A α e−α M α 2+1 d α|
≤≤∣∣∣cos M∫A 0 e−α M α 2+1 d α∣∣∣+∣∣∣sin M∫A 0 α e−α M α 2+1 d α∣∣∣|cosM∫0 A e−α M α 2+1 d α|+|sinM∫0 A α e−α M α 2+1 d α|Triangle Inequality for Real Numbers
≤≤∫A 0 e−α M α 2+1 d α+∫A 0 α e−α M α 2+1 d α∫0 A e−α M α 2+1 d α+∫0 A α e−α M α 2+1 d α
≤≤2∫A 0 e−α M d α 2∫0 A e−α M d α
(3):(3):≤≤2 M 2 M similarly to (1)(1)
Therefore:
∣∣I M(0)−π 2∣∣|I M(0)−π 2|==∣∣∣∣(I M(A)−I M(A))+I M(0)+(−∫A 0 d α α 2+1+∫A 0 d α α 2+1)−π 2∣∣∣∣|(I M(A)−I M(A))+I M(0)+(−∫0 A d α α 2+1+∫0 A d α α 2+1)−π 2|adding zero
==∣∣∣∣I M(A)−(I M(A)−I M(0)+∫A 0 d α α 2+1)+∫A 0 d α α 2+1−π 2∣∣∣∣|I M(A)−(I M(A)−I M(0)+∫0 A d α α 2+1)+∫0 A d α α 2+1−π 2|rearranging
≤≤|I M(A)|+∣∣∣I M(A)−I M(0)+∫A 0 d α α 2+1∣∣∣+∣∣∣∫A 0 d α α 2+1−π 2∣∣∣|I M(A)|+|I M(A)−I M(0)+∫0 A d α α 2+1|+|∫0 A d α α 2+1−π 2|Triangle Inequality for Real Numbers
≤≤1 A+2 M+∣∣∣∫A 0 d α α 2+1−π 2∣∣∣1 A+2 M+|∫0 A d α α 2+1−π 2|by (1)(1) and (3)(3)
→→2 M 2 M as A→+∞A→+∞ by Definite Integral to Infinity of 1 x 2+a 2 1 x 2+a 2
As:
I M(0)=∫M 0 sin x x d x I M(0)=∫0 M sinx x d x
we have shown:
∀M∈R>0:∣∣∣∫M 0 sin x x d x−π 2∣∣∣≤2 M∀M∈R>0:|∫0 M sinx x d x−π 2|≤2 M
In particular:
∫∞0 sin x x d x∫0∞sinx x d x==lim M→+∞∫M 0 sin x x d x lim M→+∞∫0 M sinx x d x Definition of Improper Integral
==π 2 π 2
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14303 | https://www.fda.gov/files/drugs/published/N201-277S008-Gadobutrol-Clinical-PREA.pdf | CLINICAL REVIEW Application Type Application Numbers Priority or Standard Submit Dates PDUFA Goal Date Reviewer Name(s) Review Completion Date Established Name Trade Name Drug Class Applicant Formulation Dosing Regimen Indication(s) Pediatric Supplemental New Drug Application NDA 201277 s008 Priority Redacted text June 30, 2014 December 30, 2014 Anthony Fotenos, MD, PhD Gadobutrol Gadavist Diagnostic magnetic resonance imaging gadolinium-based contrast agent Bayer HealthCare Pharmaceuticals 1.0 M (604.72 g/L) solution 0.1 mL/kg IV, followed by 5-10 mL saline flush, at rate of 1 mL/1 2 seconds For intravenous use in diagnostic magnetic resonance imaging to detect and visualize areas with disrupted blood brain barrier and/or abnormal vascularity of the central nervous system. Intended Population(s) Patients younger than age 2 years Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) Table of Contents 1 RECOMMENDATIONS/RISK BENEFIT ASSESSMENT.............................................................................04 1.1 Recommendation on Regulatory Action.......................................................................................................... 04 1.2 Risk Benefit Assessment.................................................................................................................................. 04 1.3 Recommendations for Postmarket Risk Management Activities.................................................................... 04 1.4 Recommendations for Postmarket Studies/Clinical Trials.............................................................................. 04 2 INTRODUCTION AND REGULATORY BACKGROUND........................................................................... 05 2.1 Product Information ........................................................................................................................................ 07 2.2 Tables of Currently Available Treatments for Proposed Indications...............................................................08 2.3 Availability of Proposed Active Ingredient in the United States..................................................................... 08 2.4 Important Safety Issues With Consideration to Related Drugs....................................................................... 08 2.5 Summary of Presubmission Regulatory Activity Related to Submission........................................................10 3 ETHICS AND GOOD CLINICAL PRACTICES............................................................................................. 11 3.1 Submission Quality and Integrity.....................................................................................................................11 3.2 Compliance with Good Clinical Practices....................................................................................................... 11 3.3 Financial Disclosures....................................................................................................................................... 11 4 SIGNIFICANT EFFICACY/SAFETY ISSUES RELATED TO OTHER REVIEW DISCIPLINES..........11 4.1 Chemistry Manufacturing and Controls........................................................................................................... 11 4.3 Preclinical Pharmacology/Toxicology............................................................................................................. 11 4.4 Clinical Pharmacology..................................................................................................................................... 12 4.4.1 Mechanism of Action.............................................................................................................................. 12 4.4.2 Pharmacodynamics.................................................................................................................................. 12 4.4.3 Pharmacokinetics..................................................................................................................................... 12 5 SOURCES OF CLINICAL DATA......................................................................................................................13 5.1 Tables of Studies/Clinical Trials...................................................................................................................... 13 5.2 Review Strategy............................................................................................................................................... 13 6 REVIEW OF EFFICACY....................................................................................................................................14 6.1 Indication..........................................................................................................................................................14 6.1.1 Methods..................................................................................................................................................14 6.1.2/3 Demographics/Subject Disposition.....................................................................................................20 6.1.4 Analysis of Primary Endpoints.............................................................................................................. 22 6.1.5 Analysis of Secondary Endpoints 25 7 REVIEW OF SAFETY........................................................................................................................................ 25 7.1 Methods............................................................................................................................................................ 25 7.1.1 Studies/Clinical Trials Used to Evaluate Safety...................................................................................... 25 7.1.2 Categorization of Adverse Events........................................................................................................... 25 7.2 Adequacy of Safety Assessments.....................................................................................................................26 7.2.1 Overall Exposure at Appropriate Doses/Durations and Demographics of Target Populations...............26 7.2.4 Routine Clinical Testing.......................................................................................................................... 26 7.3 Major Safety Results........................................................................................................................................ 27 7.3.1 Deaths...................................................................................................................................................... 27 7.3.2 Nonfatal Serious Adverse Events............................................................................................................ 27 7.3.3 Dropouts and/or Discontinuations........................................................................................................... 27 7.3.4 Significant Adverse Events......................................................................................................................27 7.4 Supportive Safety Results................................................................................................................................ 28 8 POSTMARKET EXPERIENCE.........................................................................................................................28 9 APPENDICES....................................................................................................................................................... 29 9.1 Literature Review/References.......................................................................................................................... 29 9.2 Labeling Recommendations............................................................................................................................. 30 9.3 Advisory Committee Meeting.......................................................................................................................... 31 2 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) Table of Tables Table 1: Reviewer’s tabulation of currently available contrast agents for MRI of the CNS.....................................08 Table 2: Reviewer’s tabulation summarizing Gadavist’s regulatory history............................................................ 10 Table 3: Reviewer’s tabulation of study reports submitted with application............................................................ 13 Table 4: Sponsor’s minimum GFR requirements for patients 0-23 months..............................................................15 Table 5: Sponsor’s schedule of procedures............................................................................................................... 20 Table 6: Sponsor’s tabular demographic summary................................................................................................... 21 Table 7: Sponsor’s comprehensive summary of measured PK values by age-group for pediatric patients..............24 Table 8: Sponsor’s tabulated summary of all ADRs identified in ≥ 0.1% of study subjects.................................... 28 Table of Figures Figure 1 : Postnatal development of glomerular filtration in relation to normalization method............................... 06 Figure 2: Sponsor’s graph of measured plasma gadolinium (Gd) concentration in relation to time post injection.. 22 Figure 3: Sponsor’s graph of area-under-the-curve (AUC) for Gadavist in relation to age...................................... 23 3 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) 1 Recommendations/Risk Benefit Assessment 1.1 Recommendation on Regulatory Action We recommend approval of this eighth supplementary application to NDA 201277; specifically, that the central nervous system indication for Gadavist 0.1 mmol/kg be extended to include all pediatric patients (including term neonates). 1.2 Risk Benefit Assessment Our recommendation is primarily based on integration of the sponsor’s submitted preclinical, pharmacokinetic, and safety data; published accounts of postnatal renal development; and post-marketing experience with gadolinium-based contrast agents in pediatric patients younger than 24 months. In total, this evidence identifies no compelling reason why the favorable risk/benefit balance established as the basis of approval for Gadavist 0.1 mmol/kg in adults and older pediatric patients should not apply to pediatric patients younger than age 2 years. 1.3 Recommendations for Postmarket Risk Management Activities The human kidney achieves its full complement of approximately one million nephrons by 36 weeks gestation, and it is reasonable to hypothesize that renally attributable safety risks associated with Gadavist may be higher in premature compared to term infants. It is also reasonable to hypothesize that regulatory approval for usage of Gadavist in term infants may lead to an increase in off-label usage in premature infants under uncommon circumstances when MR imaging with contrast is contemplated in this age group. In future quarterly safety updates and annual reports, in order to assess potential need for labeling clarification and/or additional study, we propose to request comment by the sponsor on any observed change in Gadavist usage amongst premature infants. 1.4 Recommendations for Postmarket Studies/Clinical Trials We recommend updating to “fulfilled” the status of post-marketing requirement 1743-2. Specifically, study 91741 entitled, “Open-label, multicenter, pharmacokinetic, and safety study in children (term newborn infant to 23 months of age) undergoing a contrast-enhanced MRI with an intravenous injection of 0.1 mmol/kg BW gadobutrol 1.0 M (Gadovist 1.0)” fulfills requirement 1743-2. PH-36304 entitled, “Extended single dose toxicity study in neonatal (4 days old) rats” and PH-36683 entitled, “Repeated dose toxicity study in neonatal/juvenile rats” was found to fulfill requirement 1743-1 on November 9, 2012. 4 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) 2 Introduction and Regulatory Background On June 30, 2014, sponsor Bayer HealthCare Pharmaceuticals submitted a supplemental new drug application to expand its FDA-approved central nervous system (CNS) indication for Gadavist to include pediatric patients younger than age 2 years (hereafter also referred to as “young pediatric patients” and encompassing individuals born at 38 weeks gestation or older and alive less than 24 months per ICH E 11 “term newborn” and “infant/toddler” definitions). This document provides a primary clinical review of the sponsor’s application. If approved for young pediatric patients, Gadavist will be the first gadolinium (Gd3+)-based contrast agent (GBCA) specifically marketable to this age group. The main reason to question use of GBCAs in young pediatric patients depends on understanding human postnatal renal development. This is because the most important safety risks associated with GBCAs can be divided conceptually into three categories, some of which depend on renal function: A) hypersensitivity reactions observed in a timeframe of minutes to hours post exposure; B) nephrogenic systemic fibrosis (NSF) observed in a timeframe of days to months; and C) the theoretical possibility of not-yet-observed long-term manifestations of gadolinium toxicity. Since renal impairment is the primary determinant of category B and likely also associated with category C, the direction and magnitude of renal function in young pediatric subjects compared to adult/older pediatric subjects is central to understanding relative risk and the overall risk/benefit balance for Gadavist in young pediatric patients. Interestingly, discussion of postnatal glomerular development (the element of renal function most relevant for Gadavist) varies across the literature. For example, in a review on the, “Ontogeny of drug elimination by the human kidney,” Chen and colleagues write, “Presently, many believe that the clearance rate increases gradually after infancy, reaching functional adult levels in preschool years. This generally accepted belief is grossly erroneous; in fact, during pre school years the clearance rate of many drugs greatly exceeds adult rates” (Pediatr Nephrol 2006:21:160-168). Figure 1 illustrates the problem: 5 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) Figure 1 shows how the answer to basic developmental questions, such as whether postnatal glomerular filtration rate (GFR) in infants is markedly (bottom panel) or mildly (middle panel) decreased compared to maximally functional levels, and whether maximally functional levels occur in pre-schoolers (middle panel), depends on the normalization method used to quantify GFR. Indeed, mean normal GFR is 26 mL/min/1.73m2 for term newborns (UpToDate accessed 9/26/2014) and the bottom panel in Figure 1 shows how almost all normal pediatric subjects have GFR’s < 60 mL/min/1.73m2 during their first two months of life. Do normal pediatric subjects share the same risks as adult subjects with moderate (G3, defined as GFR 30-59 mL/min/1.73m2) and severe (G4, defined as GFR 15-29 mL/min/1.73m2) kidney disease? The dependence of ontogenical models on normalization method, published criticisms of body surface area (BSA) normalization (for example, Nephrol Dial Transplant 2009 24: 3593-3596 and Clin Physiol Func Imaging 2007 27:135-137), and the absence of dialysis from any list of normal developmental milestones all strongly suggest the answer is, “No.” The clinical significance of an individual having <60 mL/min/1.73m2 GFR is not the same in young pediatric compared to adult subjects. Reviewer’s summary comment: For any new GBCA indication in young pediatric subjects, the use of GFR cutoffs expressed in units normalized for BSA should be expressed in alternative non-normalized units/percentiles (eg, Figure 1 top panel) and/or qualitatively contextualized to avoid the misimpression that GFR < 60 mL/min/1.73m2 has the same clinical significance in young pediatric compared to adult subjects. 2.1 Product Information Gadavist belongs to the GBCA pharmaceutical class. Molecules of this class act as microscopic magnets to reduce local relaxation times, key physical tissue properties measured by magnetic resonance imaging (MRI). The contrast agent’s registered trade name is Gadavist in the United States and Gadovist elsewhere. The active ingredient is gadobutrol. Other names include Gd D03A-butriol, ZK 135079, and BAY 86-4875. Gadavist was originally approved by the FDA as a new molecular entity on March 14, 2011. There are currently nine GBCAs approved for marketing in the United States, listed in order of original FDA approval, oldest first: Magnevist (1988), Prohance (1992), Omniscan (1993), Optimark (1999), Multihance (2004), Eovist (2008), Ablavar (2008), Gadavist (2011), and Dotarem (2013). GBCAs may be classified according to whether they are ionic or non-ionic; linear or macrocyclic; non-, weakly, or strongly protein-binding; and FDA-labeled as relatively higher or lower NSF-risk. Gadavist is macrocyclic, non-ionic, non-protein-binding, and relatively lower NSF-risk. Gadavist is the only GBCA formulated at a high (1.0 M) concentration. Gadavist is currently indicated at a 1.0 M 1 ml/kg IV dose for use with MRI in adults and pediatric patients 2 years of age and older to detect and visualize areas with disrupted blood brain barrier (BBB) and/or abnormal vascularity of the central nervous system. The sponsor now proposes to expand this indication to include all pediatric patients (including term neonates), using the same dosing regimen. 7 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) 2.2 Tables of Currently Available Treatments for Proposed Indications No GBCA is currently FDA-approved for pediatric patients younger than age 2 years. Table 1 shows the pediatric approval of the GBCAs available for MRI of the CNS. Table 1: Reviewer’s tabulation of currently available contrast agents for MRI of the CNS Proprietary Name Nonproprietary Name Approved for Pediatric Use Magnevist gadopentetate dimeglumine Yes (age ≥ 2y) Prohance gadoteridol Yes (age ≥ 2y) Omniscan gadodiamide Yes (age ≥ 2y) Optimark gadoversetamide No Multihance gadobenate dimeglumine Yes (age ≥ 2y) Dotarem gadoterate meglumine Yes (age ≥ 2y) Dotarem is approved for use in pediatric patients younger than age 2 years outside the United States. At a meeting of the FDA Center for Drug Evaluation and Research Medical Imaging Drugs Advisory Committee on February 14, 2013, focused on whether initial marketing approval for Dotarem should include pediatric patients younger than age 2 years, it was estimated that on the order of 20,000 patients per year in this age range currently receive contrast for MR imaging off-label in the United States. Based on survey results of pediatric radiology division leaders from 2010, it was estimated that the majority of pediatric patients receive relatively higher-NSF risk agents (particularly Magnevist). Almost half of respondents also indicated that off-label doses higher than 0.1 mmol/kg were prescribed (Pediatr Radiol 2011 41:1271-1283). 2.3 Availability of Proposed Active Ingredient in the United States Gadavist is the only marketed drug in the United States that contains gadobutrol. 2.4 Important Safety Issues With Consideration to Related Drugs The package insert for Gadavist (and other GBCAs Multihance, Prohance, Eovist, Ablavar, and Dotarem associated with NSF at a relatively lower rate) carry a boxed warning quoted in the following indented text: WARNING: NEPHROGENIC SYSTEMIC FIBROSIS (NSF) Gadolinium-based contrast agents (GBCAs) increase the risk for NSF among patients with impaired elimination of the drugs. Avoid use of GBCAs in these patients unless the diagnostic information is essential and not available with non-contrasted MRI or other modalities. • The risk for NSF appears highest among patients with: o Chronic, severe kidney disease (GFR < 30 mL/min/1.73m2), or o Acute kidney injury. • Screen patients for acute kidney injury and other conditions that may reduce renal function. For patients at risk for chronically reduced renal function (for example, 8 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) age > 60 years, hypertension or diabetes), estimate the glomerular filtration rate (GFR) through laboratory testing. The package insert for GBCAs associated with NSF at a relatively higher rate (Magnevist, Omniscan, and Optimark) carry the following boxed warning (difference highlighted in italics): WARNING: NEPHROGENIC SYSTEMIC FIBROSIS (NSF) Gadolinium-based contrast agents (GBCAs) increase the risk for NSF among patients with impaired elimination of the drugs. Avoid use of GBCAs in these patients unless the diagnostic information is essential and not available with non-contrasted MRI or other modalities. • Do not administer [GBCA] to patients with: o Chronic, severe kidney disease (GFR < 30 mL/min/1.73m2), or o Acute kidney injury. • Screen patients for acute kidney injury and other conditions that may reduce renal function. For patients at risk for chronically reduced renal function (for example, age > 60 years, hypertension or diabetes), estimate the glomerular filtration rate (GFR) through laboratory testing. 9 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) 2.5 Summary of Presubmission Regulatory Activity Related to Submission Table 2 provides a timeline of the regulatory history surrounding pediatric use of Gadavist, based on the sponsor’s summary and review of referenced regulatory database submissions. Table 2: Reviewer’s tabulation summarizing Gadavist’s regulatory history Date Application Description 07/15/1998 IND 56410 Original IND activation by Berlex 12/29/2003 IND 56410 IND reactivated by Berlex 04/04/2007 IND 56410 Bayer HealthCare Pharmaceuticals acquires Berlex 08/27/2007 IND 56410 End of Phase 2 agreement on design of phase II/III study 310788 of pediatric patients ages 2 to 17 years. 01/27/2011 NDA 201277 Clinical review of study 310788 02/15/2011 NDA 201277 Nonclinical review of neonatal rat study report PH-36304 03/14/2011 NDA 201277 Marketing approval for Gadavist’s CNS indication in patients older than age 2 years, including pediatric post-marketing requirements (PMRs) for animal (1743-1) and human pharmacokinetic (1743-2) studies in patients younger than age 2 years, per the Pediatric Research Equity Act (PREA; 21 U.S.C. 355c) 04/08/2011 IND 56410 PMR 1743-2: design of pharmacokinetic/efficacy study 91741 07/27/2011 IND 56410 Clinical pharmacology agreement on study 91741 design 09/12/2012 NDA 201277 PMR 1743-1: preclinical review of neonatal rat repeat-dose study report PH-36683 11/09/2012 NDA 201277 PMR 1743-1 fulfillment letter 04/15/2014 NDA 201277 Pre-sNDA meeting response and meeting cancellation 06/30/2014 NDA 201277 Application for pediatric sNDA and fulfillment of PMR 1743-2 submitted The FDA’s approval letter of March 14, 2011 specifically described the sponsor’s required pediatric assessments, quoted in the indented text, as follows: • 1743-1. You must provide additional nonclinical (animal) data to support the safety of your product in the 0-23 month pediatric age group. These nonclinical data should be obtained from newborn to juvenile animals that model pediatric patients in this age group. The study will examine the safety of the product in newborn and neonatal animals, following a single dose and limited repeated dose administrations. o Final Protocol Submission: May, 2011 o Study/Trial Completion: January, 2012 o Final Report Submission: June, 2012 • 1743-2. Your study will examine patients 0-23 months of age who are referred for an MRI exam with contrast. A sufficient number of subjects will be studied to adequately characterize the pharmacokinetics of the product in this age group. At least 40 patients 10 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) will be evaluated in this study, and the study must include a sufficient number of subjects to adequately support the efficacy of Gadavist for central nervous system MRI. o Final Protocol Submission: July, 2012 o Study/Trial Completion: March, 2014 o Final Report Submission: January, 2015 3 Ethics and Good Clinical Practices 3.1 Submission Quality and Integrity Based on filing review August 12, 2014, the sponsor’s application was found to be sufficiently complete to permit substantive review. The sponsor was notified August 26, 2014. 3.2 Compliance with Good Clinical Practices The sponsor reports no deviation from the ethical principles detailed in the Declaration of Helsinki or specific ethical considerations and provisions for pediatric patients, as detailed in the International Conference on Harmonization document on clinical investigation of medicinal products in the pediatric population (E11). 3.3 Financial Disclosures The sponsor reports adequate collection of financial disclosure forms and no disclosable information from all investigators and sub-investigators who enrolled study subjects. 4 Significant Efficacy/Safety Issues Related to Other Review Disciplines 4.1 Chemistry Manufacturing and Controls A Chemistry Manufacturing and Control (CMC) supplement to modify the syringe tip closure for Gadavist prefilled syringes was reviewed on 7/31/2013 and approved. A CMC supplement adding a 2 mL vial for patients less than 44 lbs was reviewed on 12/11/2013 and approved. 4.3 Preclinical Pharmacology/Toxicology In PH-36304 and PH-36683, the sponsor reports three main positive findings in pediatric rats ages 4 to 24 days: 1) reversible renal tubular vacuolation (an iodine- and gadolinium-contrast class-wide phenomenon); 2) incompletely reversible atrophic clear cell tubules; and 3) pediatric-specific, dose-dependent gadolinium brain deposition associated with transiently increased microglia. Based on preclinical review, the estimated no-observed-adverse-effect level was 4.9 times the maximum human dose. 11 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) 4.4 Clinical Pharmacology 4.4.1 Mechanism of Action Gadolinium can carry up to 7 unpaired electrons and thus forms a strong internal, induced magnetic field in the presence of an externally applied magnetic field. This magnetism forms the basis for Gadavist’s mechanism of action as an imaging contrast agent in the 0.1 to 1 nm range of each of gadolinium atom. Specifically, Gadavist increases the variation in nuclear relaxation times imaged by MRI devices, a property referred to as relaxivity. 4.4.2 Pharmacodynamics Gadavist is physiologically inert. Thus, from the perspective of an image reader, injected Gadavist highlights the hematological system. Its localization follows the course of blood from injection site to vascular circulation/extracellular space to urinary excretion. The localization of Gadavist in a single static image thus depends on the timing of image acquisition relative to bolus injection. The distribution of the Gadavist in the brain also provides diagnostic information regarding pathological disruption of the blood brain barrier. 4.4.3 Pharmacokinetics Gadavist is non-metabolized and renally excreted with a clearance rate comparable to inulin. The mean terminal half-life for plasma clearance is 1.8 hours. The only new human trial data submitted with this pediatric application comes from study 91741, in which 44 pediatric patients younger than age 2 years were injected with Gadavist. The primary endpoint is pharmacokinetic. The rationale for study 91741, consistent with PMR 1743 2, is that safety and efficacy have already been established for Gadavist in adult and older pediatric patients and can be extrapolated to younger pediatric patients if the primary pharmacokinetic and secondary imaging endpoints in younger pediatric patients do not substantially differ. Our clinical perspective on the primary pharmacokinetic endpoints is thus deferred for discussion of efficacy in section 6, below. 12 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) 5 Sources of Clinical Data 5.1 Tables of Studies/Clinical Trials The sponsor’s application reported on two pre-clinical studies and one clinical study, summarized in Table 3 below. Table 3: Reviewer’s tabulation of study reports submitted with application Report Identifier Study Title Associated PMR PH-36304 Extended single dose toxicity study in neonatal (4 days old) rats 1743-1 PH-36683 Repeated dose toxicity study in neonatal/juvenile rats 1743-1 PH-37807 Study 91741: Exploratory population pharmacokinetic analysis of gadobutrol (Gadovist) in the Phase 1 study in pediatric patients aged 0-23 months 1743-2 PH-37277 Study 91741: Open-label, multicenter, pharmacokinetic, and safety study in children (term newborn infant to 23 months of age) undergoing a contrast-enhanced MRI with an intravenous injection of 0.1 mmol/kg BW gadobutrol 1.0 M (Gadovist 1.0) 1743-2 The sponsor also submitted PH-37523, a report on post-marketing safety study 14823 focused on adult and older pediatric subjects and thus essentially outside the scope specific to evaluation of Gadavist usage in pediatric patients younger than age 2 years. 5.2 Review Strategy Our review focused on safety data and the clinical relevance of primary pharmacokinetic data from study 91741, as well as the sponsor’s summary of postmarketing experience and independent literature. Efficacy endpoints based on diagnostic imaging interpretation from study 91741 were neither designed nor statistically powered to guide regulatory decision-making on an independent basis. They were mainly interpreted as low-quality evidence speaking to the validity of extrapolating from higher quality evidence in adults. Sections of the review template only relevant to the original NDA were omitted. 13 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) Table 4: Sponsor’s minimum GFR requirements for young pediatric patients (from PH-37277 Table 7-4) COPYRIGHT MATERIAL WITHHELD As the source for this table, the sponsor cites Pediatr Nephrol 1991: 5:5 11, which in turn cites a data book from 1974 without providing information on the sample or method of derivation. Nevertheless, these numbers appear grossly concordant with Figure 1 (bottom) and other independent sources (eg, Eur J Nucl Med Mol Imaging 2006:33:1477 1682). Study participation involved four days of subject-investigator interactions: the first, for consent and history; the second and longest, for dose injection, MR imaging, and pharmacokinetic blood sampling at 15-60 min, 2-4 hours, and 6-8 hours post-injection, with exact sampling times within each time-block randomized per subject; the third, for assessing short-term adverse effects; and the last, seven days post-injection, for final safety assessment over the phone. At four weeks, investigators completed a form indicating “final diagnosis” presumably on the basis of available clinical information at the time. For study 91741, clinically relevant endpoints based on image interpretation were secondary. The data consisted of investigator survey item selections on questions referring to image technical adequacy, post-drug image contrast quality, lesion detection, lesion/vessel contrast, lesion/vessel border delineation, lesion/vessel morphology, and lesion/vessel diagnosis. The methods for experimentally controlling image interpretation and survey item selection permitted considerable variability and bias. For example, MR acquisition sequences and sequence parameters could vary between subjects and study sites, with only a minimum set of sequences pre-specified per anatomical site. Investigators were all board-certified radiologists, responsible for completing separate semi-quantitative pre-drug and combined pre- and post-drug (hereinafter referred to as “paired”) case report forms at the same time they were responsible for generating narrative radiology reports for patient care, requiring integration of pre- and post-drug images. Study investigators were aware of the purpose of the study, the identity of A vs A+B experimental comparators (available contemporaneously), and the role each comparator played in the study design when completing case report forms (also available contemporaneously). Only a single investigator interpreted each subject’s images. 15 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) For reference, the sponsor’s complete eligibility criteria; schedule of procedures; and image interpretation survey questions/items are quoted in the indented blocks and Table 5, below: Inclusion Criteria 1. Aged < 2 years (term newborn infants to toddlers 23 months of age inclusive) at the time of gadobutrol injection 2. Scheduled to undergo routine gadolinium-enhanced MRI of any body region 3. Signed and dated informed consent of the legal representative(s) 4. Able to comply with the study procedures such as availability at the study center for 8 hours after the MRI examination for PK blood sampling and for safety assessments at 24 ± 4 hours after the administration of gadobutrol 5. Their legal representative(s) was/were able to provide contact information for a follow-up telephone call at 7 ± 1 days post-injection Exclusion Criteria 1. Clinically unstable subjects, e.g. subjects in whom fluctuations in safety parameters was observed during the study period due to underlying disease and/or treatment regimens such as polytrauma subjects 2. Subjects who had a change in chemotherapy ≤ 48 hours prior to and up to 24 hours after gadobutrol injection 3. Any planned intervention during the study and up to 24 hours after gadobutrol injection (excluding lumbar puncture) 4. Subjects who received or were planned to receive any investigational product within 48 hours before gadobutrol injection or during study participation 5. Subjects who received or were planned to receive any other contrast agent within 48 hours prior to gadobutrol injection or up to 24 hours after gadobutrol injection 6. Subjects with contraindication for MRI such as iron metal implants (e.g. aneurysm clips) 7. History of anaphylactoid or anaphylactic reaction to any allergen including drugs and contrast agents 8. Severe inborn or acquired heart rhythm anomalies 9. Congenital long QT syndrome or family history of congenital long QT syndrome 10. Any concomitant medication known to prolong the QT interval 11. Congenital heart defect or higher degree heart block 12. Uncorrected hypokalemia 13. Subject with known and clinically relevant deviations of available clinical laboratory parameters from reference ranges (e.g. more than 3 times upper limit of reference range), in particular with regard to liver/renal function and blood coagulation 14. Subject with renal insufficiency of any intensity, i.e. eGFR < 80% of age adjusted normal value calculated based on the Schwartz formula 15. Acute renal failure of any intensity, either due to hepato-renal syndrome or occurring in the peri-operative liver transplantation period 16. Previous participation in this study 17. Close affiliation with the study center, e.g. a close relative of the investigator or his/ her designee 18. Pediatric subject in institutionalized care (most vulnerable population) 16 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) Image interpretation survey questions/items • Anatomical area evaluated (select 1 of 9, pre/paired) o brain o spine o head/neck o heart o thorax o abdomen o pelvic area o retroperitoneal area o musculoskeletal • Basic technical adequacy for diagnosis (select 1 of 4, pre + paired) o region visualized with artifacts compromising quality and interpretability of images o only partial evaluation of images possible, region not covered adequately anatomically o region visualized with artifacts, partially compromising image quality but evaluation and diagnosis still possible o region clearly visualized, excellent quality • Assessment of contrast quality (select 1 of 5, post) o none (eg, in case of non-enhancing vessel) o poor o moderate o good o excellent • Presence of pathology (select 1 of 2, pre + paired) o yes o no • Degree of contrast-enhancement in lesion or vessel (select 1 of 4, pre + paired) o no, lesion or vessel is not enhanced o moderate, lesion or vessel is weakly enhanced o good, lesion or vessel is clearly enhanced o excellent, lesion or vessel is clearly and brightly enhanced • Border delineation of lesion or vessel (select 1 of 4, pre + paired) o none, no or unclear delineation of the boundary between the lesion or vessel and the surrounding tissue o moderate, some aspects of border delineation covered o good, almost clear delineation, but not complete on relevant slices o excellent, clear and complete delineation • Visualization of lesion-internal morphology (lesion characterization) or homogeneity of vessel enhancement (select 1 of 3, pre + paired) o poor, the structure and internal morphology of the lesion or vessel is poorly visible 17 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) o moderate, the structure and internal morphology of the lesion or vessel is visible but sufficient information cannot be obtained o good, the structure and internal morphology of the lesion or vessel is sufficiently visible for diagnostic purposes • Diagnosis (select 1 of 58, pre + paired) and 4-week “final” diagnosis o brain - malignant brain lesion o brain - benign brain lesion o brain - brain malformation o brain - brain stoke (infarction / hemorrhage) o brain - CNS inflammation o brain - demyelinating disease of the brain o spine - malignant spinal cord lesion o spine - benign spinal cord lesion o spine - spinal cord malformation o spine - spinal chord trauma o spine - demyelinating disease of the spine o head / neck - malignant head / neck lesion o head / neck - benign head / neck lesion o head / neck - head / neck malformation o head / neck - head / neck inflammation o cardiac / heart - congenital heart malformation o cardiac / heart - cardiac vasculature malformation o cardiac / heart - myocarditis o cardiac / heart - cardiac ischemia o chest / thorax - malignant lung lesion o chest / thorax - benign lung lesion o chest / thorax - lung inflammation o abdomen - liver - malignant liver lesion o abdomen - liver - benign liver lesion o abdomen - liver - biliary tract malformation o abdomen - spleen - malignant splenic lesion o abdomen - spleen - benign splenic lesion o retroperitoneal - kidney - malignant renal lesion o retroperitoneal - kidney - benign renal lesion o retroperitoneal - kidney - renal malformation o retroperitoneal - adrenal gland - malignant adrenal gland lesion o retroperitoneal - adrenal gland - benign adrenal gland lesion o pancreas - pancreatic malformation o pancreas - pancreatitis o pancreas - malignant pancreatic lesion o pancreas - benign pancreatic lesion o pelvic area – reproductive organs - malignant ovarian lesion o pelvic area – reproductive organs - benign ovarian lesion o pelvic area – reproductive organs - ovarian torsion o pelvic area – reproductive organs - malignant testes lesion 18 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) o pelvic area – reproductive organs - benign testes lesion o pelvic area – reproductive organs - testes torsion o pelvic area – reproductive organs - genital malformation o lower urinary tract - malignant bladder lesion o lower urinary tract - benign bladder lesion o lower urinary tract - urinary bladder malformation o musculoskeletal - malignant bone lesion o musculoskeletal - benign bone lesion o musculoskeletal - malignant soft tissue lesion o musculoskeletal - benign soft tissue lesion o lymphatic system - malignant lymph node lesion o lymphatic system - benign lymph node lesion o MRA / vessels - vascular malformation o MRA / vessels - vascular stenosis o metastases - metastatic lesion o not assessable o no lesion / abnormality (normal) o other – specify • Additional diagnostic gain by the contrast-enhanced image set (select 1 of 3, post) o initial diagnosis unchanged o initial diagnosis changed – improved, ie more specific o initial diagnosis changed – new diagnosis • Confidence in diagnosis (select 1 of 3, pre + post) o very confident o confident o not confident 19 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) Table 5: Sponsor’s schedule of procedures (from final study protocol Table 7-1) Table. Unfortunately, we cannot provide alternative text. Persons with disabilities having problems accessing this page may call (301) 796-3634 for assistance. Reviewer’s summary comments. 1) We note that the sponsor’s eligibility criteria allow for non-CNS indications, meaning the overlap between adult/established (CNS) and pediatric/extrapolated (CNS and non-CNS) study populations is only partial. However, in view of the primary pharmacokinetic endpoint, tradeoffs involved in pediatric research, and the circulatory mechanism of GBCAs, we do not believe these population differences substantially limit the validity of extrapolation. 2) Routine clinical practice will likely not reflect sponsor’s operational definition of renal insufficiency for study exclusion, reflecting ongoing controversy/challenge in the field around how best to quantify postnatal glomerular function. 3) Methodological deficiencies substantially limit our ability to interpret the sponsor’s secondary efficacy endpoints related to image interpretation. 6.1.2/3 Demographics/Subject Disposition The sponsor reports nested subject groups of size n=47 (referred), n=44 (injected), and n=43 (injected per-protocol). Of the n=47 group, the sponsor problematically reports one subject (SID 10020008, age 18.4 months) was excluded for being older than 24 months. Two subjects (age 7 and 10.4 months) were excluded for having eGFR values < 80% normal. All in the n=44 group were injected with Gadavist. Of the n=44 group, one subject (age 21 months) should have been injected with a dose of 1.1 mL, but instead received a dose of 0.11 mL. The n=43 group was used for the primary pharmacokinetic analysis of efficacy and the n=44 group was used for the safety and secondary efficacy analysis. Members of the n=44 group ranged in ages from 0.2 to 23 months; assuming a 30-day month, the youngest subjects were 6-day-olds. By design, the age distribution was non-uniform, with 11 subjects 2 months or younger (and 9 younger than 2 months). Nine investigators recruited 20 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) subjects across three countries, mostly Germany (32), but also the United States (8) and Canada (4). Forty of the 44 subjects (91%) were white. Thirty-one (70%) were referred for MR imaging of the brain, head/neck, or spine, consistent with the approved CNS indication; the remainder were referred for body (13) or musculoskeletal (1) imaging. Two (age 7 and 10 months) had eGFR values below the inclusion criteria cut-offs shown in Table 3; nevertheless, these two subjects were included in the safety and efficacy analyses. Overall, of the 47 subjects referred for MRI, 4 (9%) had eGFR values below the sponsor’s 80%-normal cutoff. For reference, Table 6 provides the sponsor’s tabular demographic summary of the n=44 group. Table 6: Sponsor’s tabular demographic summary (from PH-37277 Table 8-3) Table. Unfortunately, we cannot provide alternative text. Persons with disabilities having problems accessing this page may call (301) 796-3634 for assistance. Reviewer’s summary comment: No major deficiency is identified involving demographics or subject disposition in study 91741. 21 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) 6.1.4 Analysis of Primary Endpoints The primary endpoints of the sponsor’s application are pharmacokinetic. Accordingly, Figure 2 shows the sponsor’s graph of measured plasma gadolinium concentration in relation to time post injection. Figure 2: Sponsor’s graph of measured plasma gadolinium (Gd) concentration in relation to time post injection (from PH-37807 Figure 11-1) Each line represents an individual subject and connects three measured points derived from blood samples drawn per subject between 0-2, 2-4, and 4-6 hours post injection. Figure 2 makes clear that the measured initial gadolinium concentrations of two subjects (age 4 and 6 months) were extremely high relative to the rest of the n=44 group. In at least one of these subjects, in response to an information request, the sponsor confirmed that only a single catheter was in place for both Gadavist injection and blood sampling, in support of a methodological (not physiological) explanation. 22 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) These two outlying measurements were thus excluded from the pharmacokinetic conversion of individual gadolinium plasma concentrations into the pivotal result to inform dosing: the sponsor’s graph of area-under-the-curve (AUC) for Gadavist in relation to age, shown in Figure 3. Figure 3: Sponsor’s graph of area-under-the-curve (AUC) for Gadavist in relation to age (from PH-37807 Figure 11-5) Each point represents a measurement-derived estimate of an individual’s total exposure to Gadavist, integrating from injection to complete excretion. Figure 3 suggests that Gadavist exposure is mildly (~30%), not markedly, elevated between months 0-2 compared to months 2-23, a plateau-like period of minimal change. Indeed, placing the plateau-like period of Gadavist clearance between ages 2 to 23 months in the context of previously reviewed pharmacokinetic data from pediatric subjects ages 2 to 17 years (study 31078) supports the concept of a super-normal period for glomerular filtration during pre school years (as previously discussed, see Figure 1B). For example, the sponsor reports median AUC is 1070 µmolh/L for ages younger than 2 months (n=9), 751 µmolh/L for ages 2 to 23 23 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) months (n=34), and 1167 µmolh/L for ages 12 to 17 years (n=46). In units of hours for half-life, the corresponding median age-group values are t1/2= 2.6, 1.5, and 1.7, respectively. For reference, Table 7 shows the sponsor’s complete tabulation of measured PK values by age-group for pediatric patients ages 0 to 17 years. Table 7: Sponsor’s comprehensive summary of measured PK values by age-group for pediatric patients (from PH-37807 Table 11-4) Table. Unfortunately, we cannot provide alternative text. Persons with disabilities having problems accessing this page may call (301) 796-3634 for assistance. Parenthesis contain 5th through 95th percentile ranges; central values represent medians. In contrast to this pharmacokinetic evidence supporting a non-linear developmental model for Gadavist clearance/exposure, estimates for plasma concentration at 20 (C20) and 30 (C30) minutes suggest a mildly up-sloping developmental trajectory. These plasma concentration values are important because they represent a clinically reasonable surrogate for the magnitude of image contrast added by drug injection. For example, the sponsor reports C20 estimates of 313 µmol/L for ages younger than 2 months, 341 µmol/L for ages 2 to 23 months, and 518-523 µmol/L for ages 12 to 17 years. These figures suggest the magnitude of image contrast added by drug injection may be somewhat less in young pediatric patients compared to adult and older pediatric patients, despite overlap between age groups in terms of percentile range. Together, the combined pharmacokinetic data suggest that the Gadavist dose of 0.1 mL/kg in young pediatric patients represents a reasonable compromise between two clinically competing interests: one, in terms of safety/AUC, potentially favoring a lower dose and the other, in terms of efficacy/concentration, potentially favoring a higher dose. Reviewer’s summary comment: Sponsor’s primary pharmacokinetic endpoints support extrapolation of benefit that is similar or reduced in young pediatric patients compared to the benefit at developmental maturity. Sponsor’s primary pharmacokinetic endpoints support extrapolation of risk that is similar or mildly increased in pediatric patients younger than 2 months and similar or mildly decreased in pediatric patients ages 2 to 23 months compared to the risk at developmental maturity. In any case, the clinical meaningfulness of any true risk differences is likely less than the range spanned for currently indicated uses in older patients with eGFR > 60 mL/min/1.73m2. We conclude that the extrapolated balance of benefit-to-risk remains acceptable and optimal or near-optimal for Gadavist 0.1 mL/kg in young pediatric patients, favoring extension of the approved CNS indication to this age group. 24 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) 6.1.5 Analysis of Secondary Endpoints We have little confidence making independent inferences about efficacy for Gadavist in young pediatric patients on the basis of the sponsor’s reported secondary imaging interpretation endpoints, given the methodological deficiencies discussed above in section 6.1.1. The reported results we find least problematic pertain to investigators’ multiple-choice selections of descriptions for imaging border delineation and morphology, briefly summarized below. Investigators selected the best imaging descriptor “Excellent, clear and complete delineation” under the heading “Border delineation of lesions/vessels” on pre-contrast images for 24 (55%) of 44 of subjects. In reference to the paired pre- and post-drug images, investigators selected the same best descriptor for 42 (95%) of 44 subjects. Investigators selected the best imaging descriptor, “Good, the structure and internal morphology of the lesion or vessel is sufficiently visible for diagnostic purposes” under the heading “Visualization of lesion-internal morphology (lesion characterization) or homogeneity of vessel enhancement” on pre-contrast images for 27 (62%) of 44 subjects. In reference to the paired pre-and post-drug images, investigators selected the same best descriptor for 43 (98%) of 44 subjects. Reviewer’s summary comment: In the context of more definitive adult efficacy and pharmacokinetic surrogate (ie, plasma concentration) data, sponsor’s limited secondary endpoints based on imaging interpretation are not inconsistent, favoring extrapolation of a diagnostic benefit in terms of improved lesion visualization for Gadavist 0.1 mL/kg to young pediatric patients. 7 Review of Safety 7.1 Methods 7.1.1 Studies/Clinical Trials Used to Evaluate Safety As of February 26, 2014, the sponsor reports that 6330 subjects have been exposed to Gadavist across 41 phase II-IV studies; this number is increased from the previously reviewed sample size of 5748. Of these 6330 study subjects, 138 were pediatric patients ages 2 to 17 years (study 36304, previously reviewed) and 44 were pediatric patients younger than 24 months (study 91741). 7.1.2 Categorization of Adverse Events The sponsor reports that adverse events (AEs) were categorized according to MedDRA version 16.1. Only events that were new starting between the period of injection and 72 hours (treatment 25 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) emergent adverse events, TEAEs) were categorized as AEs; later events were described separately. The sponsor categorized adverse drug events (ADRs) using a lower suspicion for causality for serious compared to non-serious AEs. 7.2 Adequacy of Safety Assessments 7.2.1 Overall Exposure at Appropriate Doses/Durations and Demographics of Target Populations The demographics of the target population of central interest, young pediatric patients, is the same for safety as the n=44 study 91741 group reviewed for efficacy in section 6.1.2/3, above. Each subject was exposed to a single dose of 1.0 M Gadavist 0.1 mL/kg IV. The sponsor reports the mean volume of study drug administered was 0.75 mL. All subjects completed the full schedule of planned safety evaluations. For comparison, whereas over 90% of the n=44 group of young pediatric patients were white, the sponsor reports that the overall n=6330 safety group was somewhat more diverse: 60% Caucasian, 30% Asian, 5.5% Hispanic, and 1.5% Black. 7.2.4 Routine Clinical Testing For study 91741, the sponsor reports that safety testing included the following: • Continuous assessment for adverse events • At baseline visit ≤ 24 hours pre-injection o Vital signs o Medical history and physical exam o Blood sampling • General chemistry (Na, K, Cl, BUN) • Hematology (Hct/Hgb, platelets, RBC, WBC) o Medication check • At visit for injection/MRI o Vital signs o Blood sampling • Creatinine/eGFR o Pulse oximetry and cardiac rhythm monitoring o Medication check • At 24-hour follow-up visit o Vital signs o Physical exam o Blood sampling • General chemistry (Na, K, Cl, BUN) • Hematology (Hct/Hgb, platelets, RBC, WBC) o Medication check 26 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) • At 7-day telephone contact o Medication check Reviewer summary comments. 1) The design for routine laboratory testing did not include pre-and post-drug creatinine, lowering sensitivity in the event of asymptomatic Gadavist-induced kidney injury; 2) The delay between GBCA dosing and NSF ranges from days to years (median ~60 days), meaning post-marketing data is most sensitive for detection of NSF. 7.3 Major Safety Results 7.3.1 Deaths The sponsor reports no deaths associated with study 91741 or any other study involving pediatric subjects. 7.3.2 Nonfatal Serious Adverse Events The sponsor reports nonfatal serious adverse events (infected cyst and sedation-related respiratory failure) involving two subjects in study 91741, neither of which were attributed to Gadavist. After review of sponsor’s narrative event descriptions, we agree with the sponsor’s causal attribution. 7.3.3 Dropouts and/or Discontinuations The sponsor reports no dropouts or discontinuations associated with adverse effects in study 91741. 7.3.4 Significant Adverse Events For study 91741, the sponsor reports that one subject (age 23 months) experienced an episode of emesis during the injection/MR imaging visit. Event severity was categorized as mild and might have been caused by Gadavist or the routine sedation procedure required for MRI in this age group. For comparison to the overall safety profile for Gadavist suggested by sponsor’s analysis of the n=6330 group, 3.7% experience ADRs; nausea is the second most common (1.2%). Table 8 shows the sponsor’s tabulated summary of all ADRs identified in ≥ 0.1% of subjects. 27 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) Table 8: Sponsor’s tabulated summary of all ADRs identified in ≥ 0.1% of study subjects (from 2.7.4 Summary of Clinical Safety Table 2-6) Table. Unfortunately, we cannot provide alternative text. Persons with disabilities having problems accessing this page may call (301) 796-3634 for assistance. Reviewer’s summary comment. The safety data reported for Gadavist use in 44 young pediatric patients are in line with the currently labeled risks derived from larger studies in adult and older pediatric patients. 7.4 Supportive Safety Results For study 91741, the sponsor reports no common or unexpected pre-to-post-drug change in measured laboratory parameters, vital signs, or cardiac rhythm, including no evidence for post-drug elevation in BUN. 8 Postmarket Experience Between initial marketing in Switzerland in 1999 and February 26, 2014, the sponsor estimates Redacted text that more than patients have received injections of Gadavist across the 104 countries in which the drug is approved and the 69 in which it is marketed. Market surveys suggest Redacted text approximately doses have been administered off-label to young pediatric patients. The sponsor reports that 18 deaths have occurred as a result of hyperensitivity/anaphlactoid reactions to Gadavist. No deaths have been attributed to the drug for patients younger than age 31 years. The sponsor reports awareness of NSF in 11 individuals who were also exposed to Gadavist. No cases have been reported since 2009 and none have involved pediatric patients. The sponsor attributes likely causality to Gadavist (as opposed to other confounding GBCAs) in three cases. 28 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) The accumulated prevalence of NSF from any cause worldwide is approximately 500, including ~10 in pediatric patients, all age 6 years or older (Pediatr Nephrol 2014 29:1927-37). In its sNDA, the sponsor included report PH-37523 on its phase IV study 14823 entitled, “GARDIAN, Gadovist in Routine Diagnostic MRI – Administration in Non-selected patients.” In this study, investigators from 17 countries completed forms on 23,708 patients who received MR imaging with Gadavist from Aug 7, 2010 through March 11, 2014. Investigators provided information on patient demographics, medical conditions including eGFR and dialysis, dosing and MRI indication, contrast quality, and adverse events observed during the period of MRI visit. The study protocol called for 90-day phone follow-up of patients with eGFR ≤ 60 mL/min/1.73m2 (n = 131); however, 71% were lost to follow-up. The sponsor classified patients younger than age 18 years as pediatric patients (n=1,142); however, only 4 were less than age 2 years, meaning there was essentially no overlap in the scope of the GARDIAN study and the sponsor’s supplementary application to extend the CNS indication for Gadavist to include pediatric patients younger than age 2 years. The sponsor reports that adverse drug reactions were observed in 170 patients (0.7%). These were fully concordant with currently labeled safety risks. Adverse reactions were similar between adult and pediatric patients age 2 to 18 years, except no serious adverse event was reported in the pediatric age range. A published prospective study from Canada of safety and efficacy for Gadavist 0.1 mL/kg in 60 pediatric patients younger than age 2 years (including infants born prematurely) involved subject follow-up over a 4-month period (Magn Res Ins 2013 3:1-12). Exclusion criteria included “renal impairment” without specification of eGFR cutoffs. Zero adverse drug events were identified, adding to evidence of Gadavist safety in young pediatric patients. Using clinical, pathological, or follow-up imaging studies to establish the diagnostic reference standard for 57 subject-lesions, expected and observed accumulation of contrast was concordant in 24 of 24 “enhancing” lesions and 33 of 33 “non-enhancing” lesions, adding to evidence of Gadavist efficacy in young pediatric patients. Reviewer’s summary comment. Despite extensive post-marketing exposure to GBCAs, no case of NSF has ever been identified in pediatric patients younger than age 6 years. 9 Appendices 9.1 Literature Review/References Bhargava R. and Noga M. (2013) Safety and efficacy of gadobutrol-enhanced MRI in patients aged under 2 years - a single center observational study. Magn Res Ins 3:1–12. Chen N, et al. (2006) Ontogeny of drug elimination by the human kidney. Pediatr Nephrol 21:160-168. 29 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) Delanaye P, et al. (2009) Errors induced by indexing glomerular filtration rate for body surface area: reductio ad absurdum. Nephrol Dial Transplant 24: 3593-3596. Douville P, et al. (2009) Impact of age on glomerular filtration estimates. Nephrol Dial Transplant 24: 97-103. FDA (2013) Minutes for the February 14 Meeting of the Medical Imaging Drugs Advisory Committee, DrugsAdvisoryCommittee/ucm334176.htm (accessed 10/1/2014). Hayton WL. (2002) Maturation and growth of renal function: dosing renally cleared drugs in children. AAPS PharmSci 2:1-7. Heaf, JG. (2007) The origin of the 1.73 m2 body surface area normalization:problems and implications. Clin Phsyiol Funct Imaging 27:135-137. Heilbron DC, et al. (1991). Expressing glomerular filtration rate in children. Pediatr Nephrol 5: 5-11. Piepsz A, et al. (2006) Revisiting normal 51Cr-ethylenediaminetetraacetic acid clearance values in children. Eur J Nucl Med Mol Imaging 33:1477-1482. Piepsz A, et al. (2008) Escaping the correction for body surface area when calculating glomerular filtration rate in children. Eur J Nucl Med Mol Imaging 35:1669-1672. Ringer, SA (2010). Acute renal failure in the neonate. Neoreviews 11:243-251. Rhodin MM, et al. (2009) Human renal function maturation: a quantitative description using weight and postmenstrual age. Pediatr Nephrol 24: 67-76. Roosmarijin FW, et al. (2012) Maturation of the glomerular filtration rate in neonates, as reflected by amikacin clearance. Clin Pharmacokinet 2012:51 105-117. Weller A, et al. (2014). Gadolinium and nephrogenic systemic fibrosis: an update. Pediatr Nephrol (2014) 29:1927–1937. Thomsen HS, et al. (2014) Contrast media classification and terminology. In Thomsen HS and Webb JW (eds), Contrast Media. Springer (Berlin, Germany). 9.2 Labeling Recommendations • Section 5 Warning and Precautions o Add following sentence to 5.1 Nephrogenic Systemic Fibrosis 30 Reference ID: 3655693 Clinical Review Anthony Fotenos, MD, PhD NDA 201277s8 Gadavist (gadobutrol) o No case of NSF has been identified in pediatric patients age <= 6 years [see Use in Specific Populations (8.4)] • Section 8 Use in Specific Populations o Add the following sentence to 8.4 Pediatric Use • The safety and effectiveness of Gadavist have been established in pediatric patients born at or later than 38 weeks gestation based on imaging and pharmacokinetic data in 138 patients ages 2 to 17 years and 44 patients age younger than 2 years and extrapolation from adult data o Add the following sentence to 8.4 Pediatric Use • The safety and effectiveness of Gadavist have not been established in premature infants. o Add the following paragraph to 8.4 Pediatric Use • NSF risk. • No case of NSF associated with Gadavist or any other GBCA has been identified in pediatric patients younger than age 6 years. Pharmacokinetic studies suggest that clearance of Gadavist is similar in pediatric patients and adults, including pediatric patients younger than age 2 years. No increased risk factor for NSF has been identified in juvenile animal studies. Normal estimated GFR (eGFR) is around 30 mL/min/1.73m2 at birth and increases to mature levels around 1 year of age, reflecting growth in both glomerular function and relative body surface area. Clinical studies in pediatric patients younger than age 1 year have been conducted in patients with the following minimum eGFR: 31 mL/min/1.73m2 (ages 2 to 7 days), 38 mL/min/1.73m2 (ages 8 to 28 days), 62 mL/min/1.73m2 (ages 1 to 6 months), and 83 mL/min/1.73m2 (ages 6 to 12 months). • Section 14 Clinical Studies o Add the following paragraph to 14.1 MRI of the CNS • Pediatric patients • One study of 44 pediatric patients younger than age 2 years and another study of 138 pediatric patients ages 2 to 18 years supported extrapolation of adult CNS efficacy findings. For example, comparing pre-contrast vs paired pre- and post-contrast images, investigators selected the best of four descriptors under the heading “Visualization of lesion-internal morphology (lesion characterization) or homogeneity of vessel enhancement” in 27/44 (62%) vs 43/44 (98%) of patients younger than age 2 years and in 108/138 (78%) vs 111/138 (80%) of patients ages 2 to 18 years. 9.3 Advisory Committee Meeting No advisory committee meeting was held for this pediatric supplement.. 31 Reference ID: 3655693 ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------This is a representation of an electronic record that was signed electronically and this page is the manifestation of the electronic signature. /s/ ANTHONY F FOTENOS 11/07/2014 BRENDA Q YE 11/07/2014 I agree with Dr. Fotenos' assessment. Reference ID: 3655693 |
14304 | https://physics.nist.gov/PhysRefData/Star/Text/ESTAR.html | Published Time: Sat, 01 Jul 2023 18:57:35 GMT
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14305 | https://www.sciencedirect.com/science/article/abs/pii/S0379073815002984 | A new model for the estimation of time of death from vitreous potassium levels corrected for age and temperature - ScienceDirect
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Forensic Science International
Volume 254, September 2015, Pages 158-166
A new model for the estimation of time of death from vitreous potassium levels corrected for age and temperature
Author links open overlay panel B.Zilg a, S.Bernard b, K.Alkass a, S.Berg c, H.Druid a
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•The postmortem increase in vitreous potassium is not linear but asymptotic.
•The postmortem increase in vitreous potassium is dependent on decedent age.
•The postmortem increase in vitreous potassium is dependent on ambient temperature.
•We have developed a web application for the estimation of the postmortem interval.
Abstract
Analysis of potassium concentration in the vitreous fluid of the eye is frequently used by forensic pathologists to estimate the postmortem interval (PMI), particularly when other methods commonly used in the early phase of an investigation can no longer be applied. The postmortem rise in vitreous potassium has been recognized for several decades and is readily explained by a diffusion of potassium from surrounding cells into the vitreous fluid. However, there is no consensus regarding the mathematical equation that best describes this increase. The existing models assume a linear increase, but different slopes and starting points have been proposed. In this study, vitreous potassium levels, and a number of factors that may influence these levels, were examined in 462 cases with known postmortem intervals that ranged from 2 h to 17 days. We found that the postmortem rise in potassium followed a non-linear curve and that decedent age and ambient temperature influenced the variability by 16% and 5%, respectively. A long duration of agony and a high alcohol level at the time of death contributed less than 1% variability, and evaluation of additional possible factors revealed no detectable impact on the rise of vitreous potassium. Two equations were subsequently generated, one that represents the best fit of the potassium concentrations alone, and a second that represents potassium concentrations with correction for decedent age and/or ambient temperature. The former was associated with narrow confidence intervals in the early postmortem phase, but the intervals gradually increased with longer PMIs. For the latter equation, the confidence intervals were reduced at all PMIs. Therefore, the model that best describes the observed postmortem rise in vitreous potassium levels includes potassium concentration, decedent age, and ambient temperature. Furthermore, the precision of these equations, particularly for long PMIs, is expected to gradually improve by adjusting the constants as more reference data are added over time. A web application that facilitates this calculation process and allows for such future modifications has been developed.
Introduction
In forensic medicine, the determination of the postmortem interval (PMI) is commonly requested by the police. Therefore, the development of a straightforward and reliable method for estimating PMI has been an active area of research for a long time. Despite the development of new technologies, the most commonly used method for determining PMI is still the rectal body temperature. However, this method can only be applied before the decedent's body has equilibrated with the temperature of the ambient air, which typically occurs within 24 h of death. Other methods for determining PMI include an evaluation of skeletal muscle responses upon mechanical or electrical stimulation and an evaluation of pupil reactions in response to chemical stimulation. However, even these methods can only be applied during the early postmortem period. Thus, for longer PMIs, the forensic pathologist must rely on other methods, most of which often are considered inaccurate.
One method that may be applicable to a wide range of postmortem intervals is the analysis of vitreous potassium concentration in the eyeball. Under physiological conditions, intracellular levels of potassium are high (∼150 mmol/L) in all cells of the body . Upon death, active membrane transport, as well as selective membrane permeability, is disrupted, thereby leading to the gradual leakage of potassium out of the cells and into the surrounding extracellular fluid. The vitreous body contains very few cells and its potassium level is similar to any other extracellular fluid . After death, a rise in vitreous potassium due to leakage from surrounding retinal and choroidal cells takes place. Since the 1960s, this rise in potassium levels has been used to estimate the PMI , , , , , , , , , , , , , , , , , . However, despite extensive research regarding this method, there is still no agreement regarding the most accurate equation to describe this rise and to estimate the PMI (see Table 1 and Fig. 1).
The equations that have previously been proposed for estimating PMI are based on an approximation of the diffusion that occurs from the surrounding cells into the vitreous according to a linear model. This may be a reasonable model for a limited period following death. However, it is not appropriate for longer postmortem intervals when vitreous potassium concentration may be the most important means to estimate PMI. Therefore, for the present study, it is hypothesized that the rise in potassium displays a non-linear, asymptotic curve and approaches a maximum that is a function of the sum of the potassium ions present in the intra- and extracellular compartments, and their relative volumes. It is further hypothesized that this curve can be right- or left-shifted, and up- and down-shifted, and can have different slope characteristics due to influence of various factors. Regarding the latter, several factors have been proposed:
-ambient temperature , , , ,
-duration of the terminal episode ,
-renal failure with increased K+ levels, reflected by elevated urea concentration
-alcohol level at the time of death ,
-sampling method and instrumentation used for analysis , , .
In this study, we have focused specifically on the changes in postmortem vitreous potassium levels. There are other reports that have evaluated other biochemical parameters, including hypoxanthine , but the drawback is that such analyses are more time consuming, and typically require more sophisticated analytical procedures, and thus, the results cannot be immediately reported to the police. Moreover, since hypoxanthine also may represent a marker of hypoxia , , the antemortem state of the deceased and circumstances surrounding their death need to be clarified before interpreting the results.
For this study, the potassium levels of more than 3000 vitreous samples that were consistently collected over a three-year period at the Department of Forensic Medicine in Stockholm, Sweden, were compiled. Using these data, the aim of this study was to generate a more appropriate equation for calculating the PMI from vitreous potassium values by taking into account factors that may affect the postmortem diffusion process in the vitreous. Many possible influencing factors were evaluated, but in particular, ambient temperature, decedent age, the presence of alcohol, and agony were investigated to determine whether these factors influence the increase in potassium concentration in the vitreous.
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Section snippets
Study design
Between January 2003 and June 2006, vitreous fluid samples were consistently collected from all deceased subjects admitted to the Department of Forensic Medicine (Stockholm, Sweden) as soon as possible following the arrival of the bodies at the morgue. Briefly, 0.2 mL vitreous fluid was aspirated from the center of each eye using a 1 mL syringe equipped with an 18-gauge needle. The samples from each decedent were pooled in the same syringe and were not pre-treated or subjected to dilution,
General observations
For the samples examined, the rise in vitreous potassium levels was not linear. Therefore, a non-linear model was developed to better fit these data, see Fig. 2. Given that there are very few cells in the vitreous fluid compared to the number of cells in the retina, it was assumed that the rise in vitreous potassium is largely dependent on degradation of the retinal cells, and perhaps also to some extent, the cells in the choroid layer. Upon death, intracellular potassium ions diffuse from the
Discussion
We present a new model for estimating the postmortem interval, that in contrast to previous studies describes a non-linear relationship between vitreous potassium and PMI. This provides a better estimation of the PMI, especially in longer postmortem intervals (at least up to 5 days). In contrast to previous equations, our model provides confidence intervals for the estimation of PMI. Our study is based on a larger material than most previously published studies and provides potassium data for
Conclusion
In conclusion, we have found that the postmortem rise in vitreous potassium is non-linear and that decedent age and the ambient temperature significantly influence the rise. We have developed a new equation for the estimation of the postmortem interval that includes decedent age and ambient temperature and that provides confidence intervals. To facilitate the calculation, we have developed a web application that rapidly provides a PMI estimate with confidence intervals based on these equations.
Acknowledgements
We thank the staff at the Department of Forensic Medicine in Stockholm. These studies were supported in part by the Swedish National Board of Forensic Medicine and the Swedish Medical Society.
Recommended articles
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Cited by (86)
Vitreous humor endogenous compounds analysis for post-mortem forensic investigation
2020, Forensic Science International Show abstract The chemical and biochemical analysis of bodily fluids after death is an important thanatochemical approach to assess the cause and time since death. Vitreous humor (VH) has been used as a biofluid for forensic purposes since the 1960s. Due to its established relevance in toxicology, a literature review highlighting the use of VH with an emphasis on endogenous compounds has not yet been undertaken. VH is a chemically complex aqueous solution of carbohydrates, proteins, electrolytes and other small molecules present in living organisms; this biofluid is useful tool for its isolated environment, preserved from bacterial contamination, decomposition, autolysis, and metabolic reactions. The post-mortem analysis of VH provides an important tool for the estimation of the post-mortem interval (PMI), which can be helpful in determining the cause of death. Consequently, the present review evaluates the recent chemical and biochemical advances with particular importance on the endogenous compounds present at the time of death and their modification over time, which are valuable for the PMI prediction and to identify the cause of death.
### Metabolomics in the study of retinal health and disease
2019, Progress in Retinal and Eye Research Show abstract Metabolomics is the qualitative and quantitative assessment of the metabolites (small molecules < 1.5 kDa) in body fluids. The metabolites are the downstream of the genetic transcription and translation processes and also downstream of the interactions with environmental exposures; thus, they are thought to closely relate to the phenotype, especially for multifactorial diseases. In the last decade, metabolomics has been increasingly used to identify biomarkers in disease, and it is currently recognized as a very powerful tool with great potential for clinical translation. The metabolome and the associated pathways also help improve our understanding of the pathophysiology and mechanisms of disease. While there has been increasing interest and research in metabolomics of the eye, the application of metabolomics to retinal diseases has been limited, even though these are leading causes of blindness. In this manuscript, we perform a comprehensive summary of the tools and knowledge required to perform a metabolomics study, and we highlight essential statistical methods for rigorous study design and data analysis. We review available protocols, summarize the best approaches, and address the current unmet need for information on collection and processing of tissues and biofluids that can be used for metabolomics of retinal diseases. Additionally, we critically analyze recent work in this field, both in animal models and in human clinical disease, including diabetic retinopathy and age-related macular degeneration. Finally, we identify opportunities for future research applying metabolomics to improve our current assessment and understanding of mechanisms of vitreoretinal diseases, and to hence improve patient assessment and care.
### A reliable method for estimating the postmortem interval from the biochemistry of the vitreous humor, temperature and body weight
2019, Forensic Science International Citation Excerpt :
In those cases, the most commonly used techniques are high performance liquid chromatography (HPLC) [10,15,18–20] and liquid chromatography coupled to tandem mass spectrometry (LC-MSMS) [21,22] and for K+ the usual method involves using an indirect potentiometry-specific electrode [19,23–24]. Despite the numerous studies done in this field, there is still no consensus as to what formula or method is most accurate to estimate PMI from increased potassium and hypoxanthine in VH . The concentrations of both substances increase as PMI increases [14–18]. Show abstract The estimation of the time elapsed since death is of paramount importance in the field of forensic sciences and criminal investigation, owing, among other factors, to the possible legal repercussions. Over the past few years various formulae have been developed to calculate this interval using a combination of different statistical methods and the concentrations of substances found in the vitreous humor. Corrective factors, such as ambient temperature, cause of death or age, which can modify the concentration of these substances and therefore the estimation of the postmortem interval, have been incorporated into models. In this paper five simple and reliable models to estimate PMI based the on the analysis of potassium, hypoxanthine and urea in the vitreous humor are presented. Corrective factors, such as body weight, rectal temperature and ambient temperature, which can influence the estimation of this interval have been incorporated into the formulae. Finally, the R 2 and the mean squared error have been calculated for each model in order to select the best of the five. A free software program which calculates the PMI from the model and parameters used is available from the authors. It provides quick and reliable results as well as the error committed and R 2 for each case.
### Estimation of time since death by vitreous humor hypoxanthine, potassium, and ambient temperature
2016, Forensic Science International Show abstract Measurement of vitreous humor potassium (K+) has since the 1960s been recognized as an adjunct for estimation of time since death. In 1991 we introduced hypoxanthine (Hx) as a new marker. Furthermore we demonstrated that time since death estimation was more accurate when ambient temperature was included in the calculations, both for K+ and for Hx. In this paper we present a refined method. The subjects consist of 132 cases with known time of death and ambient temperature. One sample from each subject was used in the calculations. Vitreous humor Hx levels were available in all subjects, while K+ was measured in 106 of the subjects, due to insufficient volume of vitreous humor. Linear regression analysis was applied to model the correlation between vitreous humor Hx and K+, taking the interactions with temperature into consideration. The diagrams published in 1991, which also included ambient temperature, estimated median time since death with range between the 10th and 90th percentile, whereas the linear regression analysis presented in this paper estimates mean time since death with a corresponding 95% interval of confidence. We conclude that time since death may be estimated with relatively high precision applying vitreous humor Hx and K+ concentrations combined with ambient temperature.
### Methods for determining time of death
2016, Forensic Science Medicine and Pathology
### Postmortem muscle protein degradation in humans as a tool for PMI delimitation
2016, International Journal of Legal Medicine
View all citing articles on Scopus
View full text
Copyright © 2015 Elsevier Ireland Ltd. All rights reserved.
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14306 | https://blog.powerscore.com/lsat/logic-game-types-and-frequency-of-appearance/ | Logic Game Types and Frequency of Appearance - PowerScore Test Prep
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May 19, 2021
Logic Game Types and Frequency of Appearance
Anyone preparing for the LSAT is well aware of the unique difficulty presented by the Logic Games section. What is occasionally overlooked is that certain concepts are far more critical to success than others. That is, specific game types and ideas routinely appear and regularly serve as the basis for the entire section. Other outlier notions are tested so infrequently that they deserve far less attention.
Basic Game Types & Frequency of Appearance
To help you best prioritize your studies, we’ve put together a classification count 1 for the released logic games from June 2007 through the May 2020 LSAT. If you’re counting, that’s a total of 160 games. Review the comparison of frequencies from 2013 to the present relative to 2007-2020 to discover the more recent trends and tendencies. The final piece is a breakdown of advanced features such as Templates and Numerical Distributions.
Use these lists to guide your efforts as you anticipate what’s most likely to appear on test day.
| Game Type | # of Games Jun'07-May'20 | # of Games Jun'13-May'20 | Frequency Percentage 2007-2020 | Frequency Percentage 2013-2020 | Percentage Change |
--- --- --- |
| Linear (Basic & Advanced) | 79 | 38 | 49.4% | 43.2% | -6.2% |
| Grouping | 52 | 30 | 32.5% | 34.1% | 1.6% |
| Grouping/Linear Combo | 14 | 10 | 8.8% | 11.4% | 2.6% |
| Pure Sequencing | 10 | 5 | 6.3% | 5.7% | -0.6% |
| Pattern | 3 | 3 | 1.9% | 3.4% | 1.5% |
| Circular² | 1 | 1 | 0.6% | 1.1% | 0.5% |
| Mapping³ | 1 | 1 | 0.6% | 1.1% | 0.5% |
A quick glance at this table reveals that Linear and Grouping games dominate. 91% of the games appear as either Grouping, Linear, or Grouping/Linear Combination. The total is 97% if you consider that Pattern games are also Linear in nature. So anyone learning games would obviously be well-served to gear their prep towards Linear and Grouping concepts. These occur on every test and are by far the most frequently appearing game types.
You can also see that strict Linear games are being tested somewhat less frequently these days, in favor of Grouping and, in particular, Grouping/Linear Combination games. This is not surprising as the combined grouping and linear elements generally make for a more challenging construction than games featuring only linear ideas.
Further, given the reappearance of Pattern games—one in each year from 2014-2016. It would be wise to review the three most recent examples and ensure that you’re comfortable with that type on the off chance another one is presented on your test. Ditto for Circular and Mapping, although you’ll have to go back to the 90s and early 2000s to find instances of those. A full list is provided here.
Linear Games & Advanced Concepts
Within certain game types, further analysis is helpful. Linear games can be divided into two types—Basic and Advanced. While the two appear nearly as often as one another, Basic Linear have been slightly more common over the last decade.
| Linear Game Type | # of Games Jun'07-May'20 | #of Games Jun'13-May'20 | Percentage of All Games 2007-2020 | Percentage of All Games 2013-2020 | Percentage Change |
--- --- --- |
| Basic Linear | 44 | 21 | 27.5% | 23.9% | -3.6% |
| Advanced Linear | 35 | 17 | 21.9% | 19.3% | -2.6% |
Some of the more advanced game concepts and techniques specifically addressed in the PowerScore LSAT courses also appear quite frequently.
| Concept/Technique Type | # of Games Jun'07-May'20 | #of Games Jun'13-May'20 | Percentage of All Games 2007-2020 | Percentage of All Games 2013-2020 | Percentage Change |
--- --- --- |
| Identify the Templates | 41 | 25 | 28.4% | 28.4% | 2.8% |
| Idenfity the Possibilities | 3 | 1 | 1.9% | 1.1% | -0.7% |
| Numerical Distributions | 33 | 16 | 20.6% | 18.2% | -2.4% |
Over 30% of games require an advanced approach such as Identify the Templates or Identify the Possibilities. Over 20% of games feature a challenging element such as a Numerical Distribution. These high rates of appearance put an unprepared test taker at a severe disadvantage. So again, it is imperative that you learn how to handle these higher-level concepts before test day!
A Final Note
The takeaway here is that there are no guarantees as to exactly what you’ll encounter on your LSAT. But you can reasonably anticipate the concepts most likely to appear and thus structure your prep accordingly! In other words, the LSAT is not a collection of random ideas. The test protocols dictate that certain concepts must be tested regularly. An understanding of those patterns allows you to direct your studies in the direction that will bring the greatest reward.
Master the fundamentals of linearity and grouping, as well as the more complex techniques of templates and distributions. Then you’ll be ready for anything the test makers throw at you!
Questions or comments on the LG section and its most recent trends? Let us know below or via email at lsat@powerscore.com!
1 The nomenclature above is based on PowerScore’s classification system outlined in our courses and books and helps students specifically identify the critical features of each game. However, the discussion here uses only a few very basic classification terms. For a detailed discussion of the entire Unified Games Theory™ classification system, please refer to the Logic Games Bible.
2 The game counts are only for released LSATs, and thus do not include the Circular game presented on the February 2014 test. That game also reappeared on the July 2018 exam.
3 Note that we classified game 4 from September 2016 (the infamous Computer Virus game) as Linear/Mapping. I’ve counted it as Linear here since that element was far more impactful than any Mapping features.
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14307 | https://benvitalenum3ers.wordpress.com/2016/04/02/alphametic-puzzles-part-2/?ak_action=reject_mobile | Alphametic puzzles — Part 2 | Fun With Num3ers
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Alphametic puzzles — Part 2
Posted onApril 2, 2016bybenvitalis
(1)
are positive integers in arithmetic sequence such that
,
and
Find
(2)
Substitute each of the letters with a different decimal digit from 0 to 9, to satisfy the following system of simultaneous modular alphametic equations:
(3)
Solve in base-10
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Equation: 5^{x} + 7^{y} + 11^{z} = 0 (mod 13)July 14, 2016 In "Number Puzzles"
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← Pythagorean triangle – repdigit &pandigital
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6 Responses to Alphametic puzzles — Part 2
pipo says: April 2, 2016 at 11:36 pm Puzzle one:
(x, y, z) = (8, 9, 10)
Puzzle two:
581 ≡ 1 mod 145
581 ≡ 1 mod 290
290 ≡ 0 mod 145
Puzzle three has many solutions, like:
2437
59168+
61605 Reply
benvitalissays: April 3, 2016 at 1:34 am
Reply
pipo says: April 3, 2016 at 2:27 am Oops you are right Reply
David @InfinitelyManic says: April 3, 2016 at 6:16 am I see Pipo already posted a solution to #1
8 9 10 271 217
However if you want to know how to ACT like a CAT then:
2 9 10 271 721
ACT = 271
CAT = 721 Reply
4. pipo says: April 3, 2016 at 8:46 am And if you want to know how she ACTS like two CATS then:
7 12 17: 1385 and 3185
ACTS = 1385
CATS = 3185 Reply
5. David @InfinitelyManic says: April 3, 2016 at 9:58 pm When PUPILS SLIPUP and hate math. I did not check for conditions in 2).
100^3 -73^3 = 610983 = PUPILS
73^3 -1^3 = 389016 = SLIPUP Reply
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14308 | https://mathworld.wolfram.com/GraphComplement.html | Graph Complement
Download
Wolfram Notebook
The complement of a graph , sometimes called the edge-complement (Gross and Yellen 2006,
p. 86), is the graph , sometimes denoted or (e.g., Clark and Entringer 1983), with the same vertex
set but whose edge set consists of the edges not
present in
(i.e., the complement of the edge set of with respect to all possible edges on the vertex
set of ).
The graph sum on a -node graph is therefore the complete graph , as illustrated above.
The adjacency matrix of the graph complement of the graph with adjacency
matrix
is given by
| |
| |
(Ellis-Monaghan and Merino 2008), where is the unit matrix and is the identity
matrix.
A graph complement can be computed in the Wolfram
Language by the command GraphComplement[g].
See also
Complement, Complete Graph, Cycle Complement Graph, Graph
Sum, Path Complement Graph, Rook
Complement Graph, Self-Complementary
Graph, Wheel Complement Graph
Explore with Wolfram|Alpha
More things to try:
graph complement
(2+3i)(5-i)
det {{a, b, c}, {d, e, f}, {g, h, j}}
References
Clark, L. and Entringer, R. "Smallest Maximally Nonhamiltonian Graphs." Periodica Math. Hungarica 14, 57-68, 1983.Ellis-Monaghan,
J. A. and Merino, C. "Graph Polynomials and Their Applications II: Interrelations
and Interpretations." 28 Jun 2008.
J. T. and Yellen, J. Graph
Theory and Its Applications, 2nd ed. Boca Raton, FL: CRC Press, 2006.Skiena,
S. "The Complement of a Graph." §3.2.3 in Implementing
Discrete Mathematics: Combinatorics and Graph Theory with Mathematica. Reading,
MA: Addison-Wesley, p. 93, 1990.
Referenced on Wolfram|Alpha
Graph Complement
Cite this as:
Weisstein, Eric W. "Graph Complement."
From MathWorld--A Wolfram Resource.
Subject classifications
Find out if you already have access to Wolfram tech through your organization |
14309 | https://www.theexpertta.com/lab-files/261/7-Phys261-Exp5-ConicalPendulum-Spr%202025.pdf | 47 Experiment V The Conical Pendulum I.
Preparing for Lab The purpose of this experiment is to test a model of the conical pendulum, which is an interesting and unique example of centripetal motion. To prepare for this lab before your session starts, read through the Physical Theory section below; for further reference, see section 6.2 on centripetal motion and sections 16.2 – 16.8 on oscillations in your textbook. If you wish to review it, a video walkthrough of a similar experimental setup is available here. Finally, you must complete the Pre-Lab questions on Expert TA before your lab starts. Equipment: Centripetal motion apparatus with fixed optical gate Removable 1-cm x 1-cm grid with green screen on reverse 3 m measuring tape Motor controller Lab Pro Interface Logitech web camera Excel spreadsheet template Logger Pro template for Conical Pendulum II.
Physical Theory Review of Circular Motion A body is accelerating whenever the magnitude or the direction of its velocity vector v is changing. When an object moves in a circular path at a constant speed v, the velocity is continuously changing in direction despite the constant magnitude. This type of acceleration is called centripetal acceleration. The magnitude of a centripetal acceleration is determined by the size of the circle (via radius r) as well as the speed of the motion 2 c dv v a dt r (1) where v is the linear speed; since the motion is circular and periodic, the speed can be understood as Spring 2025 v2 48 2 r v (2) where r 2 is the circumference of the circle and is the period, i.e. the time it takes to complete one revolution. Substituting Equation (2) into (1) gives: 2 2 4 r ac (3) The direction of the centripetal acceleration vector c a is always towards the center of the circle the object moves in, which is always changing; this can be complicated to state in a Cartesian coordinate system, but in terms of the polar unit vector r , it can be described generally as ˆ ( ) c c a a r (4) where r points from the center of the circle towards the mass, so that - r points inward. As is the case with any acceleration, according to Newton's 2nd law, a is the result of the presence of a net force on the object; net i i F F ma In the case of centripetal motion, that net force must also continuously point toward the center of the circle; we call this net force a centripetal force ˆ c c F ma r (5) Relationship between the period and the radius r in a conical pendulum In this experiment a spherical mass m is suspended from a light string and is driven to swing in a horizontal circular path. This arrangement is called a conical pendulum. Figure 2(a) shows the geometry of the mass when it is moving, while Figure 2(b) shows the geometry when it is Figure 2. (a) Geometry of the pendulum at an instant when the ball is smoothly revolving about the central axis on the string of length L. (b) Geometry of the pendulum when the ball is stationary; note the offset distance r0 from the center of rotation. (c) Free body diagram for the mass m during motion. (a) (b) (c) 49 hanging stationary in equilibrium. L is the length of the string, the angle the string makes with the vertical, and r is the radius of the circular orbit. In order to promote stable circular motion, the string is attached to the rotor at a small offset distance r0 from the center. Let y be the vertical coordinate of the bob when the pendulum is revolving, and y0 be the vertical coordinate of the center of the mass when the mass is at rest and 0. When the mass is revolving steadily, it is in a dynamic equilibrium, with the centripetal acceleration constant in magnitude and always pointing toward the center of the path; since the circular path is in the horizontal place, ac is in the horizontal direction only, while ay = 0. Figure 2(c) shows the free-body diagram of the dynamic equilibrium; the only forces on the mass are due to gravity and the tension T in the string. Examining the diagram, one can see that in the vertical direction Newton’s second law gives mg T cos (6) while in the horizontal direction sin c T ma (7) Our conical pendulum apparatus is designed to easily allow for measuring the period of the motion and the radius of the circular path r (actually the diameter). Therefore we would like to use the 2nd law equations (6) and (7) with the centripetal acceleration magnitude expression in equation (3) to derive a model equation for (r). Dividing Equation (7) by Equation (6) gives tan g ac (8) With a little effort, one can find from Figure 2(a) that 0 2 2 0 tan r r L r r (9) Substituting equation (3) for ac and equation (9) for tan into equation (8), one can re-arrange to find an expression for period in terms of orbit radius r and string length L o o r r g r r L r 2 2 2 (10) This is clearly a complex equation, but notice a few important characteristics: A little simplification shows that the dimensions under the radical are still L/g, which is the same as for a simple pendulum harmonic oscillator Also like the simple pendulum, the period does not depend on the mass of the bob. The expression would be much simpler if the small offset length r0 was zero, but must be non-zero so that the motor can apply torque to the mass to create a stable orbit and bring the system into dynamic equilibrium. The minimum radius of the orbit is r = r0 at which the period goes to infinity, and the maximum radius of the orbit is r = r0 + L at which the period would go to zero (equivalent to infinite frequency). 50 III. Experiment SAFETY WARNING! This apparatus has exposed rotating gears in the rotor above the pendulum. These rotating components can cause injury through entanglement with hair or clothing. Stay clear of the gears at all times. Remove or secure any loose fitting sleeves or obtrusive jewelry, and tie back any hair from your face. Part A: Getting started and Setting up the Camera (1) Open the Excel spreadsheet template for Lab 5 found in the Lab Templates folder on your lab station computer. (2) Fill in your name and your lab partner’s name, and choose your lab section number. Failure to provide the correct section may result in grading problems with your report; please ask your TA if you aren’t sure of your section number. (3) In this lab, we will use the web cam for data collection, so you won’t take the usual selfie today. To enable quantitative analysis of data images, the camera needs to be in the same location for all of the images you acquire. You should find it is near the right edge of the work desk, about 70-75cm to the right of the front edge of the apparatus (see Figure 1). If it is not at least 65-70cm away, rearrange the set up so you can move it back to this distance. Figure 1. Photograph of the centripetal motion apparatus, motor controller and camera. 51 Adjust the camera height so that it is at or just below the top of the pendulum string. Then go to the computer desktop and click on the Logitech Camera icon. In the app you should see a live video feed. (4) Slide the large board with the black and white screen in the holder on the apparatus base directly under the pendulum, nearer to the front (camera end). Carefully adjust the height, angle, and alignment of the camera until you see: the pendulum support at the top of the picture the red bob is centered horizontally in the grid the grid is perpendicular to the camera and centered on the display NOTE: You will have to manually adjust the height and angle of the camera to get the desired orientation. If the zoom seems off, adjust the camera position on the desk slightly until the preview image looks similar Figure 2. (5) Check that the camera is set with the autofocus off. If needed, adjust the focus of the camera to ensure that the grid is fairly sharp. (6) Close the Camera software, and in your spreadsheet, click “Grid Background” to capture your image like the image below in Figure 2. The macro uses a MatLab script to take the picture and import it into your template. If you get an error message, click “End” and try again. If it still does not work, notify your TA. Once you are satisfied with your grid background image, do not touch or move the camera for the rest of the lab, or the size scale on your images will not be accurate. Figure 2. Sample picture of pendulum centered in background grid. The top of the pendulum support is visible at the top of the screen. The bob is horizontally centered. The grid lines are fairly square. 52 If you need to revisit the Camera controls at any point, close them again before taking any images in Excel, as Matlab and the Camera app can only control the camera one at a time. (7) Measure the distance H from the camera to the center of the pendulum rotor, and record the value in your template in cm. For this lab, all your length measurements will be in cm. (8) Measure the length L of the pendulum from the top of the string to the middle of the bob; record L and an estimate of the uncertainty L in your spreadsheet. (9) Notice that offset length r0 (see Figure 2) is given in the template. (10) Before moving on to part B, remove the grid board, turn it around so that the green side is facing the camera, and slide it in the holder on the base at the back of the apparatus, farthest from the camera. This background will serve as a “green screen” on which to overlay the grid images for measuring the pendulum path diameter. Part B: Measuring the Period and obtaining images of the motion The goal of this part of the experiment is to measure the period of the bob in motion at eight rotation frequencies and collect an image of each case. (1) Make sure that the green screen is in place and everything is out of the way of the pendulum. (2) On your computer desktop click on the “Logger Pro Templates” folder and open the “Conical Pendulum” file. (3) Turn the motor controller power on and adjust the potentiometer on the small box (Figure 3) to set the motor rotation frequency to 3000 rpm. Note that 3000 rpm (50Hz) will NOT be the rotation rate of the pendulum because the large gear wheel at the top of the apparatus reduces the rate by a factor of about 27, so the period of the bob rotation will be only about 1.8 Hz. Figure 3. Motor Speed Control; use the knob on the small attached box, not on the motor. 53 (4) The rotating gear will of course turn as fast as it is being driven, but it will take several minutes for the bob to “catch up” with the gear and reach dynamic equilibrium at this drive frequency. More importantly, It is essential that your bob reach the same rotation rate as the gear before you take the picture, in order for your data to be even usable later. The only way to see if the rate is correct is to watch the “keyhole” in the gear (located in the same direction as the string offset) and the bob itself, both at the same time, carefully and for several periods, until you can visually confirm the motions are synchronized. There will be several “fake equilibria” during which the bob will semi-stably rotate in resonance with the gear but NOT in unison with it, at rates like 5-to-3 and 3-to-2. Hence this is a rather difficult task that can be very taxing. The best solution would be to leave the apparatus alone and running for a “very long time” to ensure synchronization. That obviously is not feasible given the time constraints. So we suggest the following strategy. (5) While you are waiting at first, click "Collect" in Logger Pro to start the collection of time data from the optical gate mounted in with the gear; this will measure the rotation period of the gear. The data should appear as points on the graph and as numbers in the three columns on the left of the screen. (6) The collection of timing data should finish after 15 s. From the statistics pop-up window, enter the values of the mean period τavg, the standard deviation Δτ, and the number of periods N into the 3000rpm section of part B section of your spreadsheet. (7) When you believe the pendulum motion has stabilized, click the button labelled “3000 rpm” in your spreadsheet. The macro will use MatLab to take a series of pictures and stack the images together to create a single “blurred” image of the trajectory of the bob, with fairly sharp outer limits during which the bob is moving primarily perpendicular to the image plane. Do NOT stop the bob rotation now or until after all images are collected. Note also that the routine has overlaid your background grid image, which now provides a faithful 1cm x 1cm grid in the plane cutting the pendulum cone in half; you can therefore use this grid to measure the diameter of the orbit just by counting the boxes (from bob center on the left to bob center on the right). (8) To check that your pendulum has actually synced with the rotation of the gear, we will proceed with the first measurement of the orbit of the bob and check that its value is reasonable. This will avoid potential problems with imaginary values in your fit later! Carefully examine the output image for 3000rpm, and use the 1cm x 1cm grid to find the apparent diameter D’ of the orbits, simply by counting the boxes (to at least the nearest half-box). Make sure to measure from the center of the ball on the extreme left side of the orbit to the center of the ball on the right (see Figure 4 below). If it is difficult to determine where the center of the mass is due to the blur, estimate the radius of the bob and approximate where 54 the center would be. This box-counting can be quite tedious… it is best recommended to enlarge the image when counting to manage it, be sure to return the image to precisely the same size afterward. If your bob was synced with the gear rotation, the diameter of the path D’ should be not much smaller than double your string length, 2L. For example, if your string length is 16cm, then D’ should be around 30cm. The angle is quite large (> 60o) so this is reasonable. If your value is close to this expected range, then record the value in the table above the image and proceed to step 9. If your value is considerably less than expected (e.g. < 29cm for a 16cm string), then your bob had not yet synced with the gear. By this point it has likely been long enough for sync to occur, so delete your initial image and retake the picture. Re-check the value of D’ for the new image in the same manner, and repeat this process until sync is evident. (9) Repeat steps (3) – (7) for the frequencies specified in your template. If you adjust the frequency downward VERY slowly, it will sync with the gear rather quickly… in only about 30 seconds or so. At the lowest frequencies, the bob may never stabilize… if you find yourself waiting more than a couple of minutes for sync, then just proceed anyway. Figure 4. Image for a mid-range frequency with the cone geometry and the pertinent diameter D’ superimposed. The number of boxes gives the length of D’ (in real life) in cm. 55 IV. Analysis In this part you will analyze the videos you took of the motion of the pendulum bob and compare your results to the theory. (1) Carefully examine each remaining output image using the grid to find the apparent diameter D’ of each of the orbits. Plug your values for D’ for each orbit into the designated area of your spreadsheet in part B. (2) Click the “Sort Data” button in the analysis section of the template. This macro will organize the data you collected into a table and put all of your images into one stacked comprehensive picture. Note this macro is extremely demanding for Excel and may fail for any number of reasons, if the image does not appear after a few tries, just notify your TA and proceed without it. (3) Calculate the uncertainties in the average period N avg and the apparent radii of the orbits 2 / ' ' D r in the appropriate columns. (4) Given that the camera is a distance H from the screen, there will be some parallax in your measurement of the apparent radius r’ of the orbit. The “Correct for Parallax” macro will adjust your measured values of r’ accordingly based on how far away the camera is from the pendulum and record the true values for r in the next column. (5) Make a scatter plot of your eight measurements with on the y-axis and r on the x-axis. Label both axes, and add units and a chart title. As usual plot your data as points without any line connecting them. Finally, add error bars in the y-axis direction using your values for avg. (6) Once all of your data is filled in and error bars are added, click “Fit Data” to run the least-squares regression macro and find best fit values for L, r0, and g. Compare the fit values to the expected values. If you think you made a mistake in recording your data or uncertainties, you can go back and fix it, but make sure you rerun the macro, as it does not auto-update. IMPORTANT: Since the equation for contains several square roots of differences, it is feasible for the fit to result in imaginary results if the data is bad enough. If you are getting “#NUM!” in one of more cells for your fit, it means your D’ values were recorded for un-synced pendula. Consult your TA for how to proceed. (7) Once you are satisfied with your fit results, add the fit vs r data to your plot as a 2nd set of points. Make sure the plot has a legend that clearly labels each set of points (i.e. data and fit). (8) Input equation (10) to calculate the expected theoretical value fortheory as a function of the radius r in the designated area of your spreadsheet. Notice we have included a large number of radius values so that you can get a smooth curve for the theory. For this calculation you 56 will need to use your measured value for L, the given value for r0, and g = 980.1 cm/s2. Add the theory to your plot of vs r and reformat the theory points to be a line without markers. Final Question 1: Based on your fit to and plot of τ vs r, do your results agree with Eqn (10)? Give a statistical argument. Final Question 2: Are your fitting parameter values for L and r0 consistent with the values found in part A? Comment on possible causes of any discrepancies. Final Question 3: Calculate the percent difference between your fit value for g and the known value of g = 980.1 cm/s2. V. Finishing Up Before Leaving the Lab (1) Record your answers to the Final Questions in your Lab spreadsheet. Yes or no questions should be justified or explained adequately. (2) Check over your spreadsheet to make sure that you have completed everything, and that you have not missed any steps or left red feedback messages unaddressed. The automatic feedback system on the template has limited ability to detect problems, so check carefully, and consult the TA if you think your work is incorrect. (3) Save your spreadsheet using the provided button and submit your spreadsheet on ELMS before you leave. Both partners should do this. (4) Log out of ELMS when you are done, but do NOT log out on the computer, just leave it at the desktop. Each student needs to submit a copy of their spreadsheet to their own account on ELMS before leaving the lab. |
14310 | https://www.thoughtco.com/multiplication-rule-for-independent-events-3126602 | Multiplication Rule for Independent Events
Key Takeaways
It is important to know how to calculate the probability of an event. Certain types of events in probability are called independent. When we have a pair of independent events, sometimes we may ask, "What is the probability that both of these events occur?" In this situation, we can simply multiply our two probabilities together.
We will see how to utilize the multiplication rule for independent events. After we have gone over the basics, we will see the details of a couple of calculations.
Definition of Independent Events
We begin with a definition of independent events. In probability, two events are independent if the outcome of one event does not influence the outcome of the second event.
A good example of a pair of independent events is when we roll a die and then flip a coin. The number showing on the die has no effect on the coin that was tossed. Therefore these two events are independent.
An example of a pair of events that are not independent would be the gender of each baby in a set of twins. If the twins are identical, then both of them will be male, or both of them would be female.
Statement of the Multiplication Rule
The multiplication rule for independent events relates the probabilities of two events to the probability that they both occur. In order to use the rule, we need to have the probabilities of each of the independent events. Given these events, the multiplication rule states the probability that both events occur is found by multiplying the probabilities of each event.
Formula for the Multiplication Rule
The multiplication rule is much easier to state and to work with when we use mathematical notation.
Denote events A and B and the probabilities of each by P(A) and P(B). If A and Bare independent events, then:
Some versions of this formula use even more symbols. Instead of the word "and" we can instead use the intersection symbol: ∩. Sometimes this formula is used as the definition of independent events. Events are independent if and only if P(A and B) = P(A) x P(B).
Example #1 of the Use of the Multiplication Rule
We will see how to use the multiplication rule by looking at a few examples. First suppose that we roll a six sided die and then flip a coin. These two events are independent. The probability of rolling a 1 is 1/6. The probability of a head is 1/2. The probability of rolling a 1 and getting a head is 1/6 x 1/2 = 1/12.
If we were inclined to be skeptical about this result, this example is small enough that all of the outcomes could be listed: {(1, H), (2, H), (3, H), (4, H), (5, H), (6, H), (1, T), (2, T), (3, T), (4, T), (5, T), (6, T)}. We see that there are twelve outcomes, all of which are equally likely to occur. Therefore the probability of 1 and a head is 1/12. The multiplication rule was much more efficient because it did not require us to list our the entire sample space.
Example #2 of the Use of the Multiplication Rule
For the second example, suppose that we draw a card from a standard deck, replace this card, shuffle the deck and then draw again. We then ask what is the probability that both cards are kings. Since we have drawn with replacement, these events are independent and the multiplication rule applies.
The probability of drawing a king for the first card is 1/13. The probability for drawing a king on the second draw is 1/13. The reason for this is that we are replacing the king that we drew from the first time. Since these events are independent, we use the multiplication rule to see that the probability of drawing two kings is given by the following product 1/13 x 1/13 = 1/169.
If we did not replace the king, then we would have a different situation in which the events would not be independent. The probability of drawing a king on the second card would be influenced by the result of the first card.
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14311 | https://www.youtube.com/watch?v=AuGiOuxeawE | Pure Maths EQ - Recurrence Relation/Periodic Sequence/Periodic Series/Summation
Zeeshan Zamurred
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Posted: 9 Jun 2023
In this video I cover an exam question on periodic sequence and evaluating a summation involving a periodic series.
The topics include:
General sequence
Sigma notation
Summation
Periodic sequence
Periodic series
Order of a periodic sequence
Order of a periodic series
Evaluating a summation
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Transcript:
welcome back in this YouTube video I'm looking at pure mathematics exam question on recurrence relation periodic sequence periodic series here is the exam question a sequence A1 A2 A3 and so on is defined by a n is equal cos squared in brachia and pi over 3. find the exact values of part A Part 1 A1 part 2 A2 part 3 A3 please pause the video attempt part a part one part two part three once you've got your complete solution then play the video let's have a look at the solution to parte part one right so we want to work out A1 we need to substitute n equal 1 into this formula okay we know that cos squared is the same as writing square bracket to the power 2 cosine inside so we've got our angle to be 1 pi over 3 which is pi over 3. okay we substitute n equal one right provided that my calculator is on radiant mode if I put this into my calculator I get a quota in the same way we can actually work out A2 and A3 so part two A2 equal square bracket cos 2 pi over 3 to the power 2 this will equal 1 over 4. part three A3 equals square bracket cos 3 pi over 3 to the power two if I put this into my calculator this is equal to 1. this completes party of the question let's move on Part B hence so we have to refer back to our answering party find the exact value of the sum from n equal 1 to 50 or what we have inside the special brackets n plus cos squared in bracket and pi over three you must make your method clear please pause the video have a go at Part B once you've got your complete solution then play the video let's have a look at the solution to Part B of the exam question right so we've got Part B now from part A we know that A1 is equal to a quarter A2 is equal to a quarter A3 is equal to one I can continue with this process and work out A4 A5 A6 the reason why I'm going to do this is because that maybe I've got a periodic sequence I need to check this before I proceed forward with calculating the exact value of this summation so if I use my calculator I get A4 is equal to a quarter A5 is equal to a quarter and A6 is equal to one so precisely we have a periodic sequence so we have repetitions in cycles of three this implies that the order will equal three of the periodic sequence that order three is very important in helping us calculate the exact value of this summation now let's have a look at that summation so we've got the sum from n equal 1 to 50. so special bracket n Plus cos squared and bracket n pi over three okay we can open the sum into two parts so we've got the sum from n equal 1 to 50 of n plus the sum from n equal 1 to 50 of cos squared n pi over 3 in Brackets let's talk about evaluating this sum over here so the sum from n equal 1 to 50 of n if I substitute n equal 1 I get one so the first term of the series is one plus if I substitute n equal to I get two second term of the series is two so we've got one plus two plus three plus dot dot dot plus the final term if I substitute n equal 50 I get 50. right so this over here is the sum of the first 50 terms of the series over here we notice that we have a common difference each time we're adding one so it is an arithmetic series we're going to calculate s50 using the formula s and equal a half n in bracket a plus L we can use this formula because we know the last term of the series so the s50 is equal to one over two multiplied by n which is the number of terms we're looking at 50 terms multiplied by in bracket a the first term of the arithmetic series is one plus L the last term of the arithmetic series is 50. okay so I can put this into my calculator and this gives me 1275. okay so I've got 1275 plus this particular summation right let's have a look at evaluating this sum now so we've got the sum from n equal 1 to 50 of cos squared in bracket and pi over 3. now we know that this over here ladies and gents is technically a n so I'm going to replace this with a n so I've got the sum from n equal 1 to 50 of a n I'm going to open up the summation okay so if I open up the summation this is equivalent to A1 Plus A2 plus A3 plus A4 plus A5 plus A6 plus dot dot dot plus a 49 plus the last term which is a 50. okay so this is equal to I can replace these terms with the numerical values a quarter plus a cool R plus one a quarter plus a quarter plus one plus dot dot dot Plus a49 plus a 50. so I can put a bracket around the sum of these three terms put a bracket around the sum of these three terms now we need to determine how many repetitions are there of these brackets over here to do this I take the number of terms which is 50 and I divide by the order of the periodic Series so the order is free so 50 divided by three this gives me 16.66 dot dot dot I'm going to round down so if I round down I'm looking at 60. so 16 repetitions so 16 lots of a quarter plus a quarter plus one how many terms will this cover so it will be 16 times 3 48 terms covered so 48 terms are covered this implies that repetition would start again at a 49 so a49 must equal a quarter and so A50 is also equal a quarter right so I can simplify this particular summation so this will equal 16 lots of a quarter plus a quarter plus one plus the a49 so the a49 is a quarter as explained plus the A50 which is also a quarter as explained right so now I can put this into my calculator and if I do this I get 49 over 2. right so I can go back over here I've got 1 275 plus the value of the summation here the value of this summation is 49 over 2. and ladies and gents I am getting closer to my final answer so my final answer will be 2599 over 2. that is the exact value of this beautiful summation how beautiful is that amazing okay so that completes Part B of the exam question and this teaching video pure mathematics exam question on recurrence relation periodic sequence periodic series if you found this teaching video useful please don't forget to subscribe leave a like leave a comment turn on your notification Bell so that you receive notifications every time I post a new teaching video |
14312 | https://online.stat.psu.edu/stat500/ | STAT 500 | Applied Statistics
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STAT 500 | Applied Statistics
Department of Statistics
About this course
STAT 500 | Applied Statistics
About this course
Lessons
0: Overview
1 Collecting and Summarizing Data
2 Probability
3 Probability Distributions
4 Sampling Distributions
5 Confidence Intervals
6 Hypothesis Testing
7 Comparing Two Population Parameters
8 Chi-Square Test for Independence
9 Linear Regression Foundations
10 Introduction to ANOVA
11 Introduction to Nonparametric Tests and Bootstrap
12 Summary and Review
Contents
About this course
Currently enrolled?
Course Overview
Lessons
Categories
Bayes
Bias
Confidence interval
Descriptive statistics
Empirical rule
Estimation
Expected value
Inference
Minitab
Parameter
Population
Population mean
Population proportion
Probability
Probability distributions
Random variables
Sample
Sample means
Sample proportions
Sample size
Sampling distributions
Set notation
Types of studes
Types of variables
z-scores
About this course
Welcome to the course notes for STAT 500: Applied Statistics. These notes are designed and developed by Penn State’s Department of Statistics and offered as open educational resources. These notes are free to use under Creative Commons license CC BY-NC 4.0.
This course is part of the Online Master of Applied Statistics program offered by Penn State’s World Campus.
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Currently enrolled?
If you are a current student in this course, please see Canvas for your syllabus, assignments, lesson videos, and communication from your instructor.
Course Overview
This course provides an introduction to the basic concepts of probability, common distributions, statistical methods, and data analysis. It is intended for graduate students who have one undergraduate statistics course and who wish to review the fundamentals before taking additional 500 level statistics courses. This course is cohort-based, which means that there is an established start and end date, and that you will interact with other students throughout the course.
Check out the STAT 500 Online Sample Syllabus.
Course Goals
Appreciate and understand the role of statistics in your field.
Develop an ability to apply appropriate statistical methods to summarize and analyze data for some of the more routine experimental settings.
Make sense of data and be able to report the results in appropriate tables or statistical terms for inclusion in your thesis or paper.
Interpret results from various computer packages (Minitab, SPSS, SAS) and be able to use Minitab to perform appropriate statistical techniques.
Lessons
##### 0: Overview Minitab Population Parameter Sample Population 6/25/25
##### Collecting and Summarizing Data Bias Types of studes Types of variables Descriptive statistics 8/25/25
##### Probability Probability Bayes Set notation 7/30/25
##### Probability Distributions Random variables Expected value Probability distributions z-scores Empirical rule 6/20/25
##### Sampling Distributions Sampling distributions Sample means Sample proportions 6/19/25
##### Confidence Intervals Estimation Inference Confidence interval Population mean Population proportion Sample size 8/25/25
##### Hypothesis Testing 7/19/25
##### Comparing Two Population Parameters 7/19/25
##### Chi-Square Test for Independence 7/19/25
##### Linear Regression Foundations 7/20/25
##### Introduction to ANOVA 7/28/25
##### Introduction to Nonparametric Tests and Bootstrap 6/19/25
##### Summary and Review 7/28/25
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14313 | https://www.starship.org.nz/guidelines/tachyarrhythmia-in-infants-and-children | Donate
Starship clinical guidelines
Principles of management
Interpretation of ECG
Supra-ventricular tachycardia (SVT) management algorithm
SVT - acute management
SVT Ongoing Care
Ventricular Tachycardia (VT) management algorithm
Ventricular Tachycardia (VT) - acute management
References
Document Control
Principles of management
Rapid heart rate in children may be due to:
Sinus tachycardia
Supraventricular tachycardia (SVT) - post cardiac surgery junctional ectopic tachycardia (JET) is common
Ventricular tachycardia (VT)
The following guidelines give practical suggestions for the diagnosis and management of these problems. Relevant history, physical examination and investigations are important however effective intervention must not be unduly delayed. Early consultation with a Paediatric Cardiologist is desirable and is mandatory for patients with complex congenital heart disease.
Recommended approach:
Basic “ABC” assessment. Is the child in shock?
12 lead ECG. Is there a tachyarrhythmia? Broad or narrow complex? P-waves?
Institute appropriate management (see SVT/VT algorithm).
Consider underlying causes / precipitating factors.
Review diagnosis if management unsuccessful.
Repeat 12 lead ECG.
Arrange appropriate ongoing care and follow-up.
Interpretation of ECG
Determine heart rate. One useful method is to count the number of RR intervals in six large squares then multiply by 50 - see chart below for normal rates.
Is the tachycardia wide or narrow complex? - see chart below for normal durations.
Can you identify P waves and what is their relationship to the QRS complex? They may be absent, hidden or inverted. There may be AV dissociation.
Assume a broad complex tachycardia is due to VT rather than SVT with aberrant conduction unless there is clear evidence it is not VT.
Separating sinus tachycardia from SVT can be difficult. Some clues:
Sinus tachycardia: child appears sick/underlying systemic illness, normal heart rate variability, heart rate usually <220 bpm and normal P wave axis.
Atrial tachycardia: 'fixed' heart rate usually >220 bpm, relatively well looking child, abnormal P wave axis.
| | | |
---
| Normal heart rates | | |
| Age (years) | Heart rate (bpm) | Tachyarrhythmia should be considered if a newborn or infant has a heart rate >200 beats/minute, a toddler >180 beats/minute or a school-aged child >160 beats/minute. Sepsis is an important differential diagnosis which may be difficult to recognise and is potentially more dangerous than a stable tachyarrythmia. Other contributing factors to tachycardia include pain, anxiety, fever Cardiac Monitors are not accurate for heart rates above 200 |
| <1 | 110 - 160 |
| 1-2 | 100 - 150 |
| 2-5 | 95 - 140 |
| 5-12 | 80 - 120 |
| >12 | 60 - 100 |
Normal QRS Duration: Average (and upper limits) for age:
| | | | | | | | | |
--- --- --- ---
| | 0-1mo | 1-6mo | 6-12mo | 1-3yr | 3-8yr | 8-12yr | 12-16yr | Adult |
| Seconds | 0.05 (0.07) | 0.05 (0.07) | 0.05 (0.07) | 0.06 (0.07) | 0.07 (0.08) | 0.07 (0.09) | 0.07 (0.10) | 0.08 (0.10) |
Modified from Guntheroth WG Pediatric electrocardiography, Philadelphia 1965, Saunders
Supra-ventricular tachycardia (SVT) management algorithm
(APLS, 2021)
Supraventricular tachycardias (SVT) are common in infancy and childhood, with an incidence between 1:250 and1:1,000. In most cases it is due to a re-entrant mechanism and usually occurs in otherwise normal children.
30-40% of children presenting with new-onset SVT do so in the first few weeks after birth. SVT can often be tolerated for many hours but because infants cannot verbally communicate their perceptions of tachycardia early signs may be subtle until they develop haemodynamic compromise. They may have poor feeding, tachypnea, pallor, sweating, lethargy and irritability. Later they can develop signs of congestive heart failure. In rare instances SVT may have been present for days in a young infant.
Older children usually perceive tachycardia (palpitations) and may feel ill, breathless, sweaty, nauseated, and dizzy or have orthostatic hypotension. Onset often occurs at rest, while they are sitting, reading, watching TV or in the 5-10 minutes after exercise. SVT rarely occurs in sleep or causes syncope. Sudden death from SVT has been reported but is rare.
SVT - acute management
Vagal manoeuvres should be tried first but not cause undue delay. The technique is based on age:
Neonates and infants: Facial immersion in ice water. This technique must not be used for infants in circulatory shock. The baby is attached to a cardiac monitor, arms are wrapped in a towel, and the whole face is immersed in ice water slurry for five seconds. It is unnecessary to occlude the nostrils. This technique is safe and 90% effective in terminating a reentrant tachycardia. Explain carefully to the parents what you are doing. The baby will not drown!
Toddlers: Ice cold facecloth to the face. Older infants resist being dipped into water as above, but this technique is almost as effective. Eyeball pressure is no longer recommended. Unilateral carotid sinus massage can be useful in older children, but it is often difficult to perform.
School-aged children: Valsalva technique; ask the child to blow on their thumb after full inspiration for 10-15 seconds. Demonstrate the technique and have the child copy you. There should be no air escape and the child should be seen to strain (“playing the trumpet silently”).
A Handstand is also a useful option with adult assistance, if the child is well enough to perform a handstand.
Consider tilting child head down (or passive leg raising in older child) or to facilitate other vagal manouevers.
2. Adenosine: This is the drug of first choice for narrow complex tachycardia. This is both therapeutic and sometimes diagnostic. Technique is important due to a very short half-life.
Give through a large vein (antecubital fossa) using a three way tap close to the cannula.
Always attempt to record a rhythm strip during adenosine administration as the response may be helpful in the diagnosis of the underlying tachycardia even if the tachycardia does not terminate (can use the defibrillator rhythm strip for this purpose).
Give adenosine on one port by rapid IV push
.
First dose of adenosine is 100 micrograms/kg followed by 200 micrograms/kg then 300 micrograms/kg (max 12 mg).
Followed immediately with a rapid push of sodium chloride 0.9% through the other port.
Sodium chloride 0.9% recommended flush volume
5 mL < 3 kgs
10 mL < 20 kgs
20 mL > 20 kgs
Older children complain of transient flushing, chest pain and nausea. Infants will frequently cry. The drug works by temporarily blocking the AV node and on ECG you frequently see a brief pause before sinus rhythm returns.
The main cause of adenosine “non-response” is incorrect administration technique.
If the technique is correct and the child does not revert to sinus rhythm after the maximum dosage, you should review your initial diagnosis and contact cardiology for further management (is it atrial tachycardia/atrial flutter, JET, fascicular VT).
3. Amiodarone: This is the safest antiarrhythmic in the post-operative situation, or where ventricular function is compromised, although some negative inotropic effect is present when given intravenously, particularly if infused rapidly. Amiodarone is useful for atrial tachycardias, particularly atrial flutter, which will not respond to adenosine, and resistant reentrant circuits as well as "stable" ventricular tachycardia.
There is limited data available for the use of intravenous Amiodarone in children but the following doses have been routinely used:
IV: 25 microgram/kg/min for 4 hr then 5-15 microgram/kg/min (max 1200 mg/24 hr).
More rapid administration of the loading dose can be used but there is a significant risk of hypotension. In children who are unstable, electrical cardioversion is a safer option than a more rapid infusion of Amiodarone.
See the Starship Amiodarone for use in paediatric cardiology guideline
Verapamil is contraindicated in children under 1 year of age as it is associated with profound hypotension. Fatalities have been reported.
4. DC cardioversion: Can be used if there is no response to adenosine and/or there is evidence of shock. Consider calling a Paediatric Code Blue in this circumstance unless sufficient senior support is present..If the patient is conscious they should have intravenous sedation/anaesthesia prior to DC cardioversion. Give a synchronous DC shock at 0.5-1 J/kg. Further shocks at 2 J/kg may be necessary.
SVT Ongoing Care
A 12 lead ECG must always be performed once sinus rhythm has been achieved. Ongoing care should be discussed with the senior clinician responsible for the child. Discuss with the on call Paediatric Cardiologist before starting prophylactic anti-arrhythmic medication.
Wolff-Parkinson-White syndrome - accounts for approximately 25% of SVT in children. They have a tendency to atrial fibrillation with rapid AV conduction. This predisposes them to ventricular tachycardia and sudden death, approximately 1% risk per ten years. They should not be given Digoxin and should be referred to the Paediatric Cardiology service for an electrophysiology test/ablation.
Infants – Most SVT that has its onset in infancy will resolve by 12 months of age and long-term treatment might not be necessary, however tachycardia may recur in late childhood. Infants are more likely to be compromised by SVT and are usually placed on prophylactic antiarrhythmic therapy until they are over the age of one year. Options include a beta-blocker (Propranolol or Atenolol), Sotalol and Flecainide. Sotalol and Flecainide should always be commenced after discussion with a Paediatric Cardiologist and in hospital due to their pro-arrhythmic effects.
School Age Children - In this age group, SVT is generally a nuisance rather than life-threatening and prophylactic antiarrhythmic agents might not be required. Persistence is however more common and careful follow-up should be arranged. The incidence of underlying congenital heart disease is up to 20%, with congenitally corrected transposition (L-TGA) and Ebstein's anomaly being the most common.
Admission: A study of ED presentations with new SVT found the majority of early recurrences and all adverse outcomes were identified if all patients under 3 months were admitted and the older patients were monitored in ED for a minimum of 90 minutes, with a 24 hour review following discharge. Infants being started on antiarrhythmic medications always require admission.
Follow up: All children with SVT should have Paediatric Cardiology follow-up. A copy of the 12-lead ECG in SVT, the rhythm strip during reversion to sinus rhythm, and the 12-lead ECG in sinus rhythm should be sent with the referral letter.
Ventricular Tachycardia (VT) management algorithm
(APLS, 2021)
Ventricular Tachycardia (VT) - acute management
The differential diagnosis of broad complex tachycardia includes VT, any SVT/atrial tachycardia or intra-atrial reentrant tachycardia with either aberrant conduction (or rate-related bundle branch block) or underlying bundle branch block, antidromic reentrant tachycardia (down the accessory pathway and up the AV node in WPW) and pre-excited atrial fibrillation. The QRS may be broad at rest due to congenital or operated heart disease, so it is important to check if the patient has a baseline ECG and what is the QRS morphology. However, particularly in the absence of this information, assume broad complex tachycardia is due to VT rather than SVT with aberration. Initial management does not need a precise diagnosis and it is important not to delay a safe therapeutic intervention as the rhythm often deteriorates into pulseless VT or VF.
Consideration of possible underlying cause (congenital heart disease, cardiac surgery, electrolyte imbalance, pro-arrhythmic medications and poisoning) and early consultation with a Paediatric Cardiologist is recommended.
Amiodarone: This is the safest antiarrhythmic in the post-operative situation, or when ventricular function is compromised. Some negative inotropic effect is present when given intravenously, particularly at fast infusion rates. Amiodarone is useful for atrial tachycardias and atrial flutter, which will not respond to Adenosine, resistant reentrant circuits and “stable" ventricular tachycardia.
There are few data available for the use of intravenous Amiodarone in children but the following doses have been routinely used:
IV: 25 microgram/kg/min for 4 hr then 5-15 microgram/kg/min (max 1200 mg/24 hr)
More rapid administration of the loading dose can be used but there is a significant risk of hypotension. In children who are unstable, electrical cardioversion is a safer option than a more rapid infusion of Amiodarone
Lignocaine: stabilises the “irritable” ventricle and reduces recurrence of VT after cardioversion. Give 0.5-1.0mg/kg initial bolus followed by infusion of 10-50mg/kg/min
DC cardioversion: first line if haemodynamically unstable. Can also be used in drug toxicity. Conscious patients should have intravenous sedation/anaesthesia prior to DC cardioversion. Give synchronous DC shock at 1 J/kg. For a second shock use 2J/kg. Use non-synchronous shocks if the defibrillator cannot 'track' QRS complexes.
Note: The only VT unlikely to respond to Amiodarone is fascicular ("verapamil sensitive") VT. This is has a characteristic pattern of RBBB with a superior QRS axis. If suspected, verapamil infusion should be administered after consultation with a Paediatric Cardiologist.
References
APLS. (2021). Advanced paediatric life support algorithms. Retrieved 14 September 2021, APLS,
Appelboam. A., Reuben, A., Mann, C., Gagg, J., Ewings, P., Barton, A., … Benger, J. (2015) Postural modification to the standard valsalva manoeuvre for emergency treatment of supraventricular tachycardias (REVERT): A randomised controlled trial. The Lancet , 2015; 386: (10005)
Perry JC, Garson A Jr. Supraventricular tachycardia due to Wolff-Parkinson-White syndrome in children: early disappearance and late recurrence. J Am Coll Cardiol 1990; 16:1215.
Ko JK, Deal BJ, Strasburger JF, Benson DW Jr. Supraventricular tachycardia mechanisms and their age distribution in pediatric patients. Am J Cardiol 1992; 69:1028.
Erickson LC, Cocalis MW, The acute management of paroxysmal supraventricular tachycardia. Pediatr Rev, 1993; 14:273-274.
BET 2: Ice immersion, other vagal manoeuvres or Adenosine for SVT in children. Emerg Med J. 2017 Jan:34(1):58-60.
Figa FH et al, Clinical efficacy and safety of intravenous Amiodarone in infants and children. American Journal of Cardiology 74:573-577, 1994.
Kothari DS, Skinner JR. Neonatal tachycardias: an update. Arch Dis Child Fetal Neonatal Ed. 2006 Mar; 91(2):F136-44.
Losek J, Endom E, Dietrich A, Adenosine and pediatric supraventricular tachycardia in the emergency department: multicentre study and review, Annals Emerg Med, 1999; 33(2):185-191.
Prystowsky E, Inpatient versus outpatient initiation of antiarrhythmic drug therapy for patients with supraventricular tachycardia. Clin Cardiol 1994; 17 (suppl II)
Pudpud AA, Linares MY, Greenberg B, Is hospitalisation necessary for treatment of SVT? Predictive variables for recurrence & negative outcome, Am J Emerg Med,1999; 17:512-516
Wu MH, Chang YC, Lin JL, Probability of supraventricular tachycardia recurrence in pediatric patients, Cardiology 1994;85:284-289
Document Control
Clinical Guideline
Starship Child Health
2 year(s)
Luciana Marcondes
Marion Hamer |
14314 | https://courses.grainger.illinois.edu/ece110/sp2022/content/labs/Experiments/C_RC_TimeConstant_Falstad.pdf | Equation Reference: ECE210 textbook, page 97, Analog Signals and Systems by Kudeki and Munson. The RC Time Constant (Falstad) Outline 3, 2, 1…Blast Off! Typically, the path for charging a capacitor and the path for discharging the capacitor is current-limited by a resistor, the value of which can be changed to alter the rate at which charging or discharging can occur. This timing mechanism is exploited in the construction of oscillators that create controlled voltage swings at a given frequency, systems (filters) that respond favorably only to signals that change at particular rates/frequencies, and also electronic timing to delay a response. The calculation of the “time constant” and, even better, “rise time” of many of these devices is very simple, requiring a simple multiplication! Let’s learn more… Prerequisites • Familiarity with capacitors and their function as energy-storage devices by charge separation. • Familiarity with the Falstad circuit simulator through a short tutorial or willingness to self-learn this simple tool. Parts Needed • A computer with an Internet browser. Access to Learning Objectives • You will be able to provide a definition and formula of the RC time constant as well as the 10%-to-90% rise time. • You will be able to predict how a change in the capacitor or resistor values will alter the time constant of a circuit. Capacitor Charging and Discharging When a capacitor is charged by a constant (DC) voltage supply of 𝑉𝑉 𝐷𝐷𝐷𝐷, the time-domain voltage across the capacitor is given as 𝑣𝑣(𝑡𝑡) = (𝑉𝑉 𝑖𝑖−𝑉𝑉 𝐷𝐷𝐷𝐷)൬𝑒𝑒−t 𝑅𝑅𝑅𝑅൰+ 𝑉𝑉 𝐷𝐷𝐷𝐷 where 𝐶𝐶 is the capacitance being charged, 𝑉𝑉 𝑖𝑖 is the initial voltage on the capacitor at time 𝑡𝑡= 0, and 𝑅𝑅 is the series resistance within the charging path. If the initial voltage is 0 𝑉𝑉 and the voltage supply is 9 𝑉𝑉, then this equation simply becomes 𝑣𝑣(𝑡𝑡) = ൬1 −𝑒𝑒−t 𝑅𝑅𝑅𝑅൰𝑉𝑉 𝐷𝐷𝐷𝐷 Notes: “on the order of”: In engineering, this may also be an expression meaning on the same “order of magnitude as” something else. Order of magnitude is usually expressed in powers of 10. For example, if 𝜏𝜏= 5 𝑠𝑠, then most of the time it takes to charge is surely contained in the range [0.5 𝑠𝑠, 50 𝑠𝑠]. The voltage across the capacitor would be asymptotically approaching 9 𝑉𝑉, although most of the charging occurs in a short time span on the order of the time constant which is the product 𝑅𝑅 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝐶𝐶. 𝜏𝜏= 𝑅𝑅𝑅𝑅 Figure 1: The waveform 𝑣𝑣(𝑡𝑡) across a capacitor while charging to 𝑉𝑉 𝐷𝐷𝐷𝐷. When a capacitor is being discharged to ground voltage (0 V), the time-domain voltage across the capacitor is given by 𝑣𝑣(𝑡𝑡) = 𝑉𝑉 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑒𝑒−𝑡𝑡 𝑅𝑅𝑅𝑅 Figure 2: The waveform 𝑣𝑣(𝑡𝑡) across a capacitor while discharging to ground voltage. 𝑣𝑣(𝑡𝑡) 𝑣𝑣(𝑡𝑡) Notes: (a) (b) Figure 3: Charging (a) and discharging (b) schematics for capacitor. The arrow shows the direction positive-valued current will flow as the capacitor charges and discharges, respectively. The time constant, while a very simple equation, is neither easy to remember nor easy to measure. Technically, it is the time an uncharged capacitor would take to charge to 63.8% of its final voltage value. A better metric would be the 10%-to-90% rise time, which, for a capacitor charging from 0 to 𝑉𝑉 𝐷𝐷𝐷𝐷, it would be the time between the point the voltage reaches 10% of 𝑉𝑉 𝐷𝐷𝐷𝐷 until it reaches 90% of 𝑉𝑉 𝐷𝐷𝐷𝐷. Given a value for the capacitor, 𝐶𝐶, and the resistor, 𝑅𝑅, for Figure 3 (a), the capacitor will have a rise time given by 𝑡𝑡𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟≈2.2 𝑅𝑅𝑅𝑅 Figure 4: The 10%-to-90% rise time. Notes: Falstad To create a short investigation into time constants, let’s build a circuit in Falstad’s circuit simulator at www.falstad.com/circuit. Start with CircuitsBlank Circuit. Use the keyboard shortcut v to draw a voltage source. Edit the value to 10 𝑉𝑉 (so that finding the 10%-to-90% rise time will not take any math!). Remember that the space bar can allow you to quickly drag an element if it doesn’t lie where you want it. Use c to add a capacitor. Set the value of the capacitor to 100 𝜇𝜇𝜇𝜇 (it is fine to leave “Trapezoidal Approximation” checked and 1m as the initial voltage). Use S (capitalized) to add a single-pole-double-throw switch, r to add two resistors (10 Ω and 100 Ω), and w to complete the wiring as shown in Figure 5. Figure 5: Falstad circuit simulation for studying timing circuits. On the left, the switch is charging the capacitor, on the right, the switch is discharging the capacitor. The switch will now allow us to charge the capacitor through a 10 Ohm resistor and then, clicking on the switch, disconnect the charging circuit and connecting the discharging circuit (through the 100 Ohm resistor). Considering time constants, the capacitor is unchanged, but the resistance in each path differs by a factor of 10. Warning: You cannot connect a wire from, say, the bottom of the battery to the bottom of the 100 Ω resistor without connecting first at the bottom of the capacitor. The simulator will not automatically make that connection! Notes: The following questions should be discussed as a team. One teammate will record the answers and enter a team submission on behalf of all participating team members. Consider 𝑡𝑡𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟= 2.2 𝑅𝑅𝑅𝑅. Use the equation to estimate the rise time of the voltage across the capacitor to be during the charging cycle. Which resistance do you use in this calculation? Capacitors also discharge with a 90%-to-10% “fall time” given by the same calculation. 𝑡𝑡𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓≈2.2 𝑅𝑅𝑅𝑅 Use the equation to estimate the fall time of the voltage across the capacitor to be during the discharging cycle. Which resistance do you use in this calculation? In your circuit simulation, toggle the switch several times and observe the current flow as the capacitor charges and discharges as evidenced by the moving dots. To get a time display of voltage and current, hover over the capacitor, right-click, and select View in Scope. Again, toggle the switch to make observations of the charging and discharging cycles. Press the RUN/Stop button to freeze the oscilloscope (scope) view with at least one full charging cycle and one full discharging cycle. Use your mouse to move a cursor over the plot. For a charging cycle, record the time the voltage reaches 1 volt (10%) and the time it reaches 9 volts (90%). Record these values and take the difference to find 𝑡𝑡𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟. The measurements will not be too precise, so don’t worry too much about precision. Notes: Repeat for a discharging cycle to find 𝑡𝑡𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓. Record the two observed times as well as the calculation. Without simulating it yet, Use the equation to estimate rise and fall times if the capacitance is increased to 1000 𝜇𝜇𝜇𝜇. Discuss both the charging (rise) and discharging (fall) times. Use simulation to confirm your predictions and comment here in a few sentences on the accuracy or failure of any predictions. Your answers to the previous question will not be counted against you as long as you can identify here what you did incorrectly! |
14315 | https://www.scribd.com/document/859789784/Acceleration-IGCSE-Physics-Revision-Notes | Acceleration - IGCSE Physics Revision Notes | PDF | Acceleration | Velocity
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14316 | https://www.freemathhelp.com/forum/threads/points-on-a-lattice.115519/ | New posts
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Points on a Lattice
Thread starter rakam
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rakam
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#1
I got this question somewhere:Hanna moves in a lattice where every point can be represented by a pair of integers. She moves from point A to point B and then takes a turn 90 degrees right and starts moving till she reaches the first point on the lattice. Find what's the point she would reach? In essence the problem boils down to finding the first point where the perpendicular to a line will intersect.I did find one solution on stack overflow but can't understand it. Can someone please explain me the following solution: Let (dx,dy) = (Bx,By)-(Ax,Ay), the vector from point A to point B. We can rotate this by 90 degrees to give (dy,-dx). After hanna turns right at B, she will head out along that rotated vector toward (Bx+dy,By-dx) Since she is moving in a straight line, her vector from B will follow (t.dy,-t.dx), and will hit another lattice point when both of those components are integers, i.e... She will hit another lattice point at: (Bx + dy/GCD(|dx|,|dy|), By - dx/GCD(|dx|,|dy|) )
Dr.Peterson
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#2
Please tell us which parts you do understand, and which you don't. In particular, how much do you know about vectors? I would actually rather guide you through figuring out an answer based on what you already know, than teach you topics you don't know just to understand this person's work. That's a major disadvantage of trying to solve a problem by searching for ready-made answers, rather than thinking for yourself, and is why we don't give ready-made answers. Also, I would recommend that you start with a drawing of an example. Try this one:
R
rakam
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Thanks Dr. Peterson for the reply.So I do understand the following: Let (dx,dy) = (Bx,By)-(Ax,Ay), the vector from point A to point B. We can rotate this by 90 degrees to give (dy,-dx). After hanna turns right at B, she will head out along that rotated vector toward (Bx+dy,By-dx) Since she is moving in a straight line, her vector from B will follow (t.dy,-t.dx), and will hit another lattice point when both of those components are integers, i.e... The part that I don't understand is: She will hit another lattice point at: (Bx + dy/GCD(|dx|,|dy|), By - dx/GCD(|dx|,|dy|) ) That is how did we get the 1/GCD(|dx|,|dy|) component.
Dr.Peterson
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#4
Look at my example, where (dx,dy) is u=(2, 6), so (dy,-dx) is (6, -2). Here, GCD(2, 6) = 2, which means we can divide both components by 2 and still get a pair of integers. The resulting vector, v=(3, -1) takes us to the first lattice point we come to, which is B+v = (4,7) + (3,-1) = (7,6), my point C.This is much like reducing a fraction to lowest terms by dividing terms by the GCD.
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rakam
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#5
Ah.. got itThank you so much for explanation. Really appreciate your help
R
rahuls
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#6
Thanks Dr. Peterson. I have one question though.How did you come at resulting vector v=(3, -1) ? I came at vector resulting vector v = (3,4) as following:GCD (6,2 ) => 2 (BX, BY)=> 4 , 7(dx, dy) => 6 , -2(Bx + dy/GCD(|dx|,|dy|), By - dx/GCD(|dx|,|dy|) ) (4 - 2/2, 7 - 6/2) ==> (3,4)
Dr.Peterson
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#7
rahuls said:
How did you come at resulting vector v=(3, -1) ?I came at vector resulting vector v = (3,4) as following: GCD (6,2 ) => 2 (BX, BY)=> 4 , 7 (dx, dy) => 6 , -2 (Bx + dy/GCD(|dx|,|dy|), By - dx/GCD(|dx|,|dy|) ) (4 - 2/2, 7 - 6/2) ==> (3,4)
Click to expand...
Divide (6, -2) by 2 and you get the VECTOR v = (3, -1). Right? I also showed how the POINT C = (4,7) + (3, -1) = (7, 6). That's what the formula is supposed to produce: C, not v. You also wrongly took (dx,dy) to be my rotated vector (6, -2). But in the formula, (dx,dy) is the unrotated vector u = AB = (2, 6). Putting that in the formula, you get C = (4 + 6/2, 7 - 2/2) = (7, 6) which is C in my picture (again, not v).
R
rahuls
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#8
Dr.Peterson said:
Divide (6, -2) by 2 and you get the VECTOR v = (3, -1). Right?I also showed how the POINT C = (4,7) + (3, -1) = (7, 6). That's what the formula is supposed to produce: C, not v.You also wrongly took (dx,dy) to be my rotated vector (6, -2). But in the formula, (dx,dy) is the unrotated vector u = AB = (2, 6). Putting that in the formula, you get C = (4 + 6/2, 7 - 2/2) = (7, 6) which is C in my picture (again, not v).
Click to expand...
Thank you. Got it.
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On balance
By Fraser Scott2020-04-22T11:27:00+01:00
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Get students to leave behind troublesome techniques when solving algebraic equations
Source: © Science Photo Library
The balance model can help students with algebra
Algebra is the branch of maths where letters, and other symbols, are used to represent numbers in order to construct equations and formulas. We find it many times in our chemistry curriculums: from solving PVT calculations, interpreting chemical properties graphically, through to investigating chemical kinetics.
Fundamental issues
Students find various aspects of algebra tricky. Sometimes the symbols represent physical quantities, sometimes pure numbers. This is even more challenging in chemistry because we have lots of additional uses of symbols of our own. In addition, the meaning of an algebraic expression is often not understood, ie the relationship between the various quantities.
The equals sign can also be an issue. Students often think that ‘=’ is an invitation to perform a calculation, rather than it being a symbol of equivalence.
Trouble understanding the fundamentals of algebra can mean students prefer alternative methods to solving equations when they encounter them in subjects other than maths. For example, the use of ‘magic triangles’ to solve a whole range of chemistry problems.
While ‘magic triangles’ can work in some situations, we should avoid them and instead help our students to understand the fundamental algebra. We can do this by:
Modelling our own algebraic thinking, to make algebraic ideas explicit.
Emphasising when symbols represent physical quantities, and when they do not.
Asking students to describe in words what they understand by the relationships represented in symbols.
Even if you can help your students to move away from the use of magic triangles or the like and get them to approach equations more algebraically, it is important that they are doing so correctly.
Download this
Download this student worksheet as MS Word or pdf. It starts with some simple, context-free algebraic manipulations and then introduces relevant chemistry contexts. The questions that ask for a specific algebraic operation to be performed should tease out common misconceptions often possessed by students, particularly if they use the transition model detailed below.
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Download this
This student worksheet starts with some simple, context-free algebraic manipulations, and then introduces relevant chemistry contexts. The questions that ask for a specific algebraic operation to be performed should tease out common misconceptions often possessed by students, particularly if they use the transition model: rsc.li/3eHJmAR
Let’s change the subject
Most algebraic operations have one goal, changing the subject of the equation to a different variable. To see how your students tackle this sort of algebraic thinking, you could give them this simple diagnostic question:
Ask your students to explicitly show all working and thinking – this will quickly let you know how they approach their algebra, and if they are confident in their understanding of the underlying mathematics. Also, for this specific question, it will tease out if they are prone to using ‘magic triangles’, and if so, you can encourage them to approach the problem algebraically.
If you do persuade your students to use algebra, there is a very common approach that can lead to significant issues, and it is important that you know what it is so you can look out for it. First the correct method …
The balance model
This method recognises that algebraic manipulation of an equation must preserve the equality of both sides of the equation, ie whatever operation you do to one side you must do to the other side.
Let’s consider the following equation where we are asked to change the subject to x, ie to have x by itself on one side of the equation and everything else on the other side:
There is more than one way to approach the task, but if we recognise that the standard convention is to have the subject on the left, we will begin by adding 3x to both sides. We add 3x because it is the inverse operation of the ‘subtract 3x’ term that we want to remove from the right-hand side. Applying this operation leaves us here:
Next, we can subtract 18 from both sides of the equation. Again, we do this because we are performing the inverse operation of the ‘add 18’ term. After this operation we have:
The last task is to obtain an x term, rather than a 3x term, and we do this by dividing both sides by 3. This is the inverse operation of multiplying by 3. As there are multiple terms on one side of the equation, it is important to remember that all terms are being divided by 3, not just a single term. Finally, we are left with an equation with x as the subject:
The transition model (the ‘change side change sign’ rule)
Here, we select a problematic term and ‘move’ it to the other side of the equation, and in doing so we change the sign so that the term is now the inverse operation. You may recognise this as the ‘change side change sign’ rule, and you may even have been taught it yourself. Please note that the transition model is an undesirable approach for many reasons, detailed below, and it is only explained here to help you identify if your students are using it!
Let’s do the same example as above but approach it using the transition model:
First, we will move ‘subtract 3x’ from the right-hand side to the left side of the equation, where it becomes ‘add 3x’:
Next, we will move ‘add 18’ to the right side of the equation, which will give ‘subtract 18’. Following this, we are left with:
Finally, we move the ‘multiply by 3’ in front of x, to the other side, where it becomes ‘divide by 3’, and gives us the final answer as before.
On the surface, it may seem almost identical to the balance model, but the transition model can lead to significant issues for students.
Balance model vs transition model
The most obvious problem with the transition model is that it is a ‘magic’ rule that isn’t readily understood – why does the sign change when the term moves sides? Whereas, the balance model is the fundamental framework on which the rules of algebra are built, ie for both sides of an equation to remain equal, the same operation must be performed to both sides.
A pragmatic argument could be put forward for a ‘magic’ rule as it might help a student carry out a task that they might otherwise not be able to do, ie they make the situation easier, allowing the student to get the correct answer without them necessarily understanding the underlying theory. However, in the case of the transition model, it can be the cause of some of common errors encountered in algebra.
Consider the final step in the transition model example above. When moving the ‘multiply by 3’ across to the other side, where exactly does the resultant ‘divide by 3’ go? Instead of dividing all terms by 3, many students will only divide one of the terms. This is a consequence of how they think about the process of ‘moving’ the operation from one side to another.
Similarly, consider this example where a student decides to deal with the 3 in front of the x at this stage:
Using the transition model, a student can move the ‘multiply by 3’ to the other side, and even if they remember to divide both terms on the right by 3, the rule doesn’t allow them to correctly divide the second term, the 12, on the left hand side by 3. The transition model will always get this sort of algebraic operation wrong.
Moreover, it is not easy to extend the transition model to more challenging algebraic situations involving exponentials, logarithms or trigonometry – all of which are routinely encountered in chemistry! However, these are no trouble for the balance model.
If you spot your students using the transition model, you should encourage them to start using the balance model. To help with this, you could explicitly show them examples, like the above, where the transition model doesn’t work.
It could be that this is the method that your students have been taught. If so, perhaps a chat with your colleagues in the maths department will help. However, sometimes students simply develop this type of approach on their own.
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Sending data from Python to R to perform statistical test using rpy2
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I want to use the Fisher's exact test functionality (specifically, the MC simulation functionality) of R with an interface to Python. I'm trying to do that using rpy2, but it's more difficult than I thought.
I can get an interface to the Fisher's test method using the following code:
coffeescript
import rpy2.robjects as robjects
fisher = robjects.r['fisher.test']
However, how do I pass a 2xN matrix to the function and retrieve the p-value?
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asked Apr 27, 2016 at 19:42
pirpir
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Consider importing R's stats package and run the Fisher Test as a Python function. Do note, the result object is <class 'rpy2.robjects.vectors.ListVector'> and hence must be converted to a Python dictionary as shown below.
```python
import rpy2
from rpy2.robjects.numpy2ri import numpy2ri
from rpy2.robjects.packages import importr
import numpy as np
cont = np.reshape(np.arange(0,4), (2,2))
statspackage = importr('stats', robject_translations={'format_perc': '_format_perc'})
result = statspackage.fisher_test(numpy2ri(cont), simulate_p_value = True, B = 100)
DEPRECATED CONVERSION
import pandas.rpy.common as com
pyresultdict = com.convert_robj(result)
for k, v in pyresultdict.items():
print(k, v)
data.name ['structure(c(0L, 2L, 1L, 3L), .Dim = c(2L, 2L))']
p.value [1.0]
estimate odds ratio 0.0
dtype: float64
null.value odds ratio 1.0
dtype: float64
conf.int [0.0, 77.90626902008512]
alternative ['two.sided']
method ["Fisher's Exact Test for Count Data"]
```
Another note, you may receive a warning about the deprecation of com.convert_to_r_dataframe and com.convert_robj(rdf) which should be replaced with pandas2ri.pandas2ri() and pandas2ri as suggested here. However, the conversion on my end does not work for the ListVector object. Ideally, above conversion would be replaced with below:
```coffeescript
CURRENT CONVERSION
from rpy2.robjects import pandas2ri
pandas2ri.activate()
pyresultdict = pandas2ri.ri2py(result)
for k, v in pyresultdict.items():
print(k, v)
```
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edited Apr 28, 2016 at 15:53
answered Apr 28, 2016 at 15:43
ParfaitParfait
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pir
pirOver a year ago
Thanks, looks like a less hacky approach. Is there any direct benefit of this approach?
2016-04-28T20:30:54.753Z+00:00
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Parfait
ParfaitOver a year ago
You work in the Python object model and avoid passing R script lines for each task.
2016-04-28T21:37:10.593Z+00:00
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lgautier
lgautierOver a year ago
@Parfait I have the import of the R package working with simply statspackage = importr('stats') (using R-3.3.0-beta)
2016-04-29T01:18:37.407Z+00:00
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ParfaitOver a year ago
There was a bug in earlier version with naming conflict of format_perc so ignore if works for you. It might help future readers.
2016-04-29T02:10:31.913Z+00:00
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ParfaitOver a year ago
Curious @lgautier if you are same BitBucket user, did the bottom conversion work in converting the R vectors.ListVector object to Python dictionary? I posted an issue but was immediately closed as pandas does not maintain rpy2.
2016-04-29T13:06:57.97Z+00:00
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Here is one way to do this:
```coffeescript
import rpy2.robjects as robjects
from rpy2.robjects import r
from rpy2.robjects.numpy2ri import numpy2ri
from rpy2.robjects.packages import importr
import numpy as np
cont = np.reshape(np.arange(0,4), (2,2))
print cont
r_cont = numpy2ri(cont)
r.assign("cont", r_cont)
r("res <- fisher.test(cont, simulate.p.value = TRUE, B = 100)")
r_result = r("res")
p_value = r_result
print r_result
print p_value
```
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answered Apr 28, 2016 at 14:18
pirpir
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14321 | https://physics.stackexchange.com/questions/477106/inconsistency-in-spring-compression-under-falling-weight | newtonian mechanics - Inconsistency in spring compression under falling weight - Physics Stack Exchange
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Inconsistency in spring compression under falling weight
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I think this is a simple enough question but I am not able to see the consistency in two different math of the same problem. Here is the problem,
One has to determine the compression of the spring. h′h′ in the figure is the final height of the top of the spring after the mass has come to rest and h h is the initial height. So I applied the conservation of energy equation by saying that,
m g(H+h)=m g h′+1 2 k(h−h′)2 m g(H+h)=m g h′+1 2 k(h−h′)2
This can be rearranged to get,
1 2 k(h−h′)2−m g(h−h′)−m g H=0 1 2 k(h−h′)2−m g(h−h′)−m g H=0
which is a quadratic equation in the compression term (h−h′)(h−h′) and can be solved easily with quadratic formula. But what if you used Hooke's law to determine the compression? In which case,
F=W=k(h−h′)F=W=k(h−h′)
and then
(h−h′)=m g k(h−h′)=m g k
which will give a different answer. If these two ways were consistent then substituting the later relation in the former energy equation, should give an identity but it does not.
I am missing some very obvious piece of the puzzle here but I can't figure out what that piece is.
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edited May 1, 2019 at 6:03
ShazShaz
asked May 1, 2019 at 5:32
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1 Related How much work can a force on a spring do? Why are two methods wrong?Farcher –Farcher 2019-05-01 06:06:46 +00:00 Commented May 1, 2019 at 6:06
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There are two things wrong with the last equation.
First, it doesn’t account for the loss of potential energy m g H m g H of the falling mass.
Second, even if the weight were released on top of the spring instead of dropping it from a height H above the spring, the equation wouldn’t account for the change in potential energy that occurs during the compression. The correct equation in this case is obtained from the first equation by setting H=0 H=0, which is the equivalent of releasing the weight on top of the spring. When you do that you will get a maximum displacement of
(h−h′)=2 m g k(h−h′)=2 m g k
This is twice the displacement in your last equation and will result in oscillation of the mass about the displacement (h−h′)=m g k(h−h′)=m g k.
The displacement in your last equation would apply if instead of releasing the weight on top of the spring, you were to support the mass on the spring and slowly reduce your upward force against gravity bringing the mass to rests at the maximum displacement of (h−h′)=m g k(h−h′)=m g k.
Hope this helps
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edited May 1, 2019 at 14:10
answered May 1, 2019 at 8:26
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In your second method you are incorrectly assuming that at maximum compression the system is in equilibrium where the forces cancel out. This is not the case.
Determine the spring force from the displacement found in your first method. You will see that it is larger than the object's weight.
If you want to go ahead and use forces, accelerations, etc. then you need to first determine the velocity with which the object hits the spring:
v 0=2 g H−−−−√v 0=2 g H
And then use Newton's second law to determine the acceleration as a function of position (down is positive, y=0 y=0 is at the top of the uncompressed spring ):
a=F m=g−k m y a=F m=g−k m y
This gives you a second order differential equation you can solve using initial conditions of y(0)=0 y(0)=0 and v(0)=v 0 v(0)=v 0 to get y(t)y(t). From there you can find the maximum compression of finding y(T)y(T), where T T is a time such that v(T)=0 v(T)=0. I will leave this to you should you choose to do so.
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edited May 1, 2019 at 14:42
answered May 1, 2019 at 14:31
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14322 | https://colorado.pressbooks.pub/introorbitalmechanics/front-matter/introduction/ | Primary Navigation
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Introduction
1. Chapter 1 - Orbital Basics
2. Chapter 2 - Orbit Geometry
3. Chapter 3 - The Classical Orbital Elements (COEs)
4. Chapter 4 - Preliminary Orbit Determination
5. Chapter 5 - Launch Windows and Time
6. Chapter 6 - Kepler's Prediction Problem
7. Chapter 7 - Manuevering
8. Chapter 8 - Rendezvous
9. Chapter 9 - Ground Tracks
10. Chapter 10 - Orbital Perturbations
11. Chapter 11 - Interplanetary Travel
Appendix
Introduction to Orbital Mechanics
Welcome to the fascinating world of Astrodynamics! In this course, you will learn about orbital mechanics, maneuvers, and their basic applications. Have you ever wondered why we launch spacecraft from specific locations and during certain time windows? What is a geosynchronous orbit, and what is needed to get satellites there? How do spacecraft get to Mars and other interplanetary destinations? In this book, you will find the answers to these questions and others. You will learn about satellite orbits and why specific ones are chosen as well as how to change satellite orbits. It should be a great learning experience and will be helpful in your chosen career field.
For an introduction to the author and additional information about orbital mechanics, see Video 1. For information about the primary editor, see Video 2.
Video 1: Instructor information
Video 2: Editor Information
Expected Outcomes:
This book is divided into roughly twelve parts, the last two of which are optional for an introductory course.
See the Schedule for a detailed list of topics and schedule by lesson.
These topics cover enough fundamental information to give you insight into orbital mechanics, regardless of whether you are a student, professional, or educator. It is intended for people who have a basic background in mathematics and physics; however, an in-depth understanding of those topics is not required.
At this time, I would like you to pause and think about the reasons why we want to go into space. What types of space missions do we have currently? Some people argue that space systems are a waste of time, money, and valuable resources. What would you say to counter that claim?
OVERVIEW
Space systems give us a global perspective and serve as the ultimate high ground (Sellers and Astore). What types of orbits do we use for space missions? You probably know we use weather satellites and other earth observing missions from relatively low orbits, lower than about 500 km (310 miles). But do you know how these ideas got started?
Manned balloon flights were recorded as early as the 1700s in France. By the 1790s, the French military had formed balloon units for observation purposes. The Union army used tethered, gas-filled balloons to observe enemy troops and to direct artillery fire during the Civil War (Figure 1). The Confederate soldiers would try to bring them down, but the balloons would float just beyond the range of enemy guns (Military History Now, 2015). Today, we regularly use Google Maps and other satellite-based imagery to get as global or close a perspective as we need.
An advantage of space systems is that we can see a LONG WAY, unobscured by Earth’s atmosphere. The James Webb Space Telescope (JWST) took this photo of the Carina Nebula, one of our galaxy’s largest star-forming regions (Figure 2). It is at a distance of 7500 light years and has a diameter of over 200 light years. The JWST was launched in 2021 into Earth-Sun Lagrange point 2 (L2). It orbits the Sun at about 1.5 million kilometers (km), or 932,000 miles, away from the Earth. Lagrange points in space are locations at which three bodies can orbit each other yet stay in the same position relative to each other (Webb/NASA). These are very stable orbits, and you will find out more about why L2 is a fitting orbit for the JWST later this course (NASA, 2023). The telescope is the largest, most powerful, and most complex one to date. It is an infrared observatory that has longer wavelength coverage and better sensitivity than previous telescopes.
Its predecessors, the Hubble Space Telescope (HST) and the Kepler Space Telescope (KST), were the pathfinders for the JWST. The HST was launched in 1990 into a low earth orbit (LEO) at an altitude of 538 km and at an inclination, or orbital tilt, of 28.5 degrees. In 2009, the KST was launched from Cape Canaveral into an orbit around the Sun. In July 2015, the KST discovered the planet Kepler-452b, which is about 1,400 light years away from Earth. Kepler-452b is the first near-Earth-sized world to be found in the habitable, or Goldilocks zone, of a star that is similar to our Sun (Figure 3). The Goldilocks zone is a region around a star where temperatures are ideal for water to pool on its surface (Sawe, 2017). The Kepler Space Telescope ran out of fuel needed for operations and was retired in 2018 after finding more than 2,600 exoplanets (Johnson, 2015). Overall, over 5500 exoplanets have been discovered by our telescope assets (NASA, 2021).
Another advantage of space systems is the free-fall environment of space. It does not just make astronauts queasy and give them chicken legs and fat faces! It has many advantages that have led to advances in materials science, food growth, and biological research. But what is meant by free-fall?
Is it zero gravity? No! It simply means there are no contact forces on the body. In the case of satellites in orbit, the satellite is falling toward the earth because of gravity, but it is continually missing the earth because of its forward speed and the earth’s curvature. We often refer to this as microgravity. You will understand more about this topic in chapter 1.
In microgravity, crystal structures tend to grow larger and with fewer impurities than they do on the surface. Shown in Figure 4 on the left is an example of protein crystals grown in microgravity. Compare them to those grown on Earth on the right (Choi, 2016).
The flatworm shown in Figure 5 was originally an amputated flatworm fragment sent to the International Space Station in 2017. If you look closely, it has two heads with two separate sets of eyes, which is an interesting phenomenon that occurs due to free-fall (Fredette, Smith, Sable, & Kasap).
Space has a vast availability of resources. Not only do we have a practically unlimited supply of solar power, but there are also minerals and other extraterrestrial materials out there. We know the moon alone has an ample supply of ice at the poles, rich regolith, and abundant Helium-3 (He-3), which is scarce on Earth. He-3 has the highest energy to mass ratio of any substance found in nature. When combined with deuterium, or heavy hydrogen often found in Earth’s oceans, it can be used for fusion reactors. It is possible to get highly efficient and powerful engines from a He-3/deuterium combination, up to 55 times more efficient than any chemical rockets in use today (Javier, 2019).
There are any number of resources available on other planets. Just take the red planet, Mars, for example, shown in Figure 6. Its color alone indicates an abundant supply of iron. Other ores also exist on Mars including nickel, copper, titanium, platinum, palladium, and chromium.
Mars also features Olympus Mons, which is the tallest mountain on the planet, and the largest in our solar system. Valles Marineras Canyon is Mars’ version of Earth’s Grand Canyon, but a whopping 4000 km (2500 miles) long. It is large enough to stretch the distance across the United States from coast to coast. We can only imagine what resources this and other features on Mars may contain, not to mention the resources on other planets (Redd, 2017).
Space is an adventure for the human race – exciting yet challenging. It is human nature to want to explore the unknown, push the limits of what we can accomplish, do what nobody else has done before, and make new discoveries. It is a unique challenge.
I can only image the excitement the world felt when Neil Armstrong and Buzz Aldrin took their first steps on the moon. We have had plenty of discoveries throughout the years that have both intrigued and piqued man’s curiosity. One of them, aptly named the Mars Curiosity Rover, did just that (Figure 7). Early in its mission, in Gale Crater, Curiosity found chemical and mineral evidence of past habitable environments on Mars. We are learning so much about Mars’ climate and geology from the chemistry and structure of the rocks and soil Curiosity’s 10 science instruments and 17 cameras have found (Mars Science Lab, 2019).
There are many, many other discoveries that have been made just in our galaxy. The Cassini mission visited Saturn and its moons beginning in 2004, New Horizons visited Pluto in 2015, and Juno explored Jupiter starting in 2016. The Dawn mission explored some of the largest asteroids in our solar system: Vesta in 2011, and Ceres in 2015. Data gathered was in both visible and infrared wavelengths, and we found that both planet-like worlds evolved very differently (Figure 8). OSIRIS-REx was the first US mission to collect a sample from an asteroid, Bennu, which was returned to earth on Sept 24, 2023. NASA is always interested in exploring other worlds to expand our understanding of the solar system (NASA Dawn, 2020).
So now that you have some background on the reasons for space missions, let us discuss some types of space missions. Pause for a minute to think about some types of space missions you have heard about. We will discuss these in the next section.
For a recap of this section, see Video 3.
Video 3: Course Introduction
TYPES OF SPACE MISSIONS
Space systems are widely used in our daily lives. We rely on space systems to help us navigate around on Earth. We also use space assets for weather monitoring, television, communication, and to develop many technologies that we use every day. NASA has a website that highlights space technologies that benefit life on Earth in the form of commercial products. NASA has profiled more than 2,000 spinoffs since 1976 – there is more space in your life than you might think! (NASA, 2020).
First up is space-based communication. One set of satellites NASA uses is TDRSS, or the Tracking and Relay Data Satellite System (Figure 9). It is a network of satellites and ground stations that allow spacecraft to communicate with the ground as needed. As the mission manager for a satellite launched on the Titan IV launch vehicle, I used TDRSS to communicate closely with the ground during the launch phase. TDRSS has provided critical communication to thousands of satellites.
The Mobile User Objective System (MUOS) is an ultra-high frequency (UHF) satellite communication system developed for the US Navy (Figure 10). MUOS satellite radio terminal users can connect around the globe into the Global Information Grid as well as into the Defense Switching Network. MUOS provides voice, video, and mission data on a high-speed Internet protocol-based system.
The Wideband Global Satcom System, WGS, is a US satellite constellation that provides military communication around the globe (Figure 11). The system was initiated in 2001 as a replacement for the Defense Satellite Communication System, DSCS, which operated from the 1980s into the 2000s (WGS, 2020). Although DSCS performed extremely well for many years, WGS delivers newer and greater capabilities and provides the primary communication for the Department of Defense. It operates alongside the Air Force’s Advanced Extremely High Frequency, AEHF, and the Navy’s Mobile User Objective System satellite communication systems. In fact, the US Space Force’s first launch was the AEHF-6 satellite on a United Launch Alliance Atlas V rocket in March 2020 (Blastoff!, 2020).
The Advanced Extremely High Frequency (AEHF) program is a constellation of communication satellites operated by the US Space Force (Figure 12). They are used to relay security communication for the US, British Armed Forces, the Canadian Armed Forces, the Netherlands Armed Forces, and the Australian Defence Force. The AEHR system is a joint service communications system that provides survivable, global, secure, protected, and jam resistant communications for high priority military ground, sea, and air assets.
Most satellite communication systems operate from geostationary or geosynchronous orbits. GEO orbits are at an altitude of about 35,800 km (22,250 miles). Satellites at that altitude with an inclination, or orbital tilt of 0º are called geostationary. Those at an inclination other than zero are called geosynchronous. You will learn more about these orbits and why they are useful for communication systems a little later.
Remote sensing is an important type of space mission. Shown here are only unclassified satellite systems, but remote sensing is a lot of what people think of as “spy satellites” that can be used to monitor troop movement and other land-based activities. Some of the more well-known systems are the Landsat and SPOT satellites.
Landsat is a US satellite system that shows Earth from space (Figure 13). Since the first Landsat was launched in 1972, the satellite systems have collected data on forests, urban areas, and fresh water. Its data is freely available and has helped us to better understand environmental changes, manage agricultural activities, allocate water resources, and respond to natural disasters. Landsat 8, shown below, is in a near-polar orbit called sun-synchronous. It repeats its orbital pattern every sixteen days (Jenner, 2015).
The French SPOT, Satellite pour l’Observation de la Terre, satellites have been providing high resolution optical imagery since 1986 (Figure 14). The system consists of five satellites and has revealed the surface of the Earth in enough detail that it has led to new applications in mapping, vegetation monitoring, land use and land cover, and the impacts of natural disasters. Shown here is a photo of Brisbane, Australia taken from the SPOT-6 satellite (SPOT-6, 2017).
SBIRS, or the Space-Based Infrared Radar System, is a US space surveillance system that provides key capabilities in the areas of missile warning, missile defense, and defensive and offensive space characterization. It is operated by the US Space Force and consists of satellites in geostationary orbits and highly elliptical orbits. It can identify a missile’s trajectory and provide real-time information to develop an appropriate defense strategy. It has sensitive infrared sensors that enable identification of missile types and has extremely precise prediction accuracy (SBIRS, 2020).
The Sentinel-3 satellite was launched to measure ocean surface topography, ocean and land surface temperatures, and ocean and land surface colors. It supports ocean forecasting systems, environmental monitoring, and climate monitoring. It is jointly operated by the European Space Agency, ESA, and EUMETSAT, the European operational satellite agency responsible for monitoring weather, climate, and the environment (Navigation, 2020).
Aiding navigation is a useful space mission. Probably the most well-known space system on this mission is the Global Positioning System, or GPS. Not only is it used to find our way around, but it can also be used for tracking, mapping, and timing for precision agriculture, transportation, and defense purposes. But, GPS is not the only space navigation system. The Russians’ Global Navigation Satellite System, or GLONASS, uses 24 satellites to provide worldwide services. These satellites are located in medium, circular orbits at an altitude of approximately 19,100 km (14,430 miles), and at an orbital inclination, or tilt of 64.8°. Again, you will find out more about these types of orbits later in the course (Blau, 2017).
Galileo is Europe’s global navigation system. It is interoperable with GPS and GLONASS, and the complete constellation will consist of twenty-four operational satellites in medium Earth orbit at 23,222 km altitude and an inclination of 56° (What is Galileo, 2020).
We have sent numerous missions to explore our solar system and beyond. The Cassini mission to Saturn made numerous discoveries, including the extraordinary weather patterns at the north pole. The hexagon shape comes from a six-sided jet stream that spans about 30,000 km (20,000 miles) across (Figure 15). It has winds of about 322 km/hr (200 miles/hour) and has a massive, rotating storm at the center. There is no other known weather feature like this in our solar system (NASA Cassini, 2013).
The Voyager spacecraft, launched in the 1970s, are still out there and sending back data. Voyager 1 is in interstellar space, the space between the stars, while Voyager 2 is still exploring the outer layer of the solar bubble. These missions are continuing to characterize the outer solar system environment and search for our heliopause boundary, the outer limits of the Sun’s magnetic field and outward flow of the solar wind. It is believed the Voyagers have enough electrical power and thrust fuel to keep their current science instruments on until at least 2025. Perhaps the Voyagers are destined to wander the Milky Way eternally! (Voyager, 2020).
There are many future space missions underway. Although it is impossible to predict what will happen, the human condition will continue to be improved by exploration of our solar system. A society that fails to invest in its future may have no future in the end (Sellers, 2020). For a recap of this section, see Video 4.
Video 4: Course Introduction Part 2 – Types of Space Missions Video
COURSE OVERVIEW
Now that we know some of the reasons for space systems and have reviewed some typical missions, let us take a look at what space is. Have you thought about where space begins? Well, a person who reaches 93 km (58 miles) altitude earns his or her astronaut wings. But technically, space begins at the altitude where an object will remain in orbit, at least temporarily (Figure 16).
It was a 1950s engineer and physicist, Theodore von Kármán, who defined the Kármán Line, the boundary between the sciences of aeronautics and astronautics. Kármán calculated where this boundary would be. He concluded that space must begin where the atmosphere becomes too thin to support conventional airplane flight. Based on his calculations, this boundary was around 100 km (62 miles). At this altitude, a vehicle would have to travel at velocities about 8 km/s (17,900 miles per hour) in order to generate enough lift to stay aloft. At that velocity, lift would become irrelevant and the vehicle would be in orbit (Sellers & Astore).
Then, how do we get vehicles further out? There is a great program to explore this called Systems Tool Kit, or STK. As an educator or student, you can download your own free version at the website linked here ( This STK simulation shows one way we move satellites from smaller to larger orbits (Video 5). This maneuver is called a Hohmann transfer/combined plane change maneuver. You will learn more about maneuvering later in this course.
Video 5: Combined Plane Change
To give you a better idea of the size of relatively near space, we start at the earth, which has a diameter of 12,800 km (7850 miles). Another number you will become very familiar with in this course is the radius of the Earth, which is about an average of 6,378 km (3963 miles). Low Earth orbit, or LEO, is very close. If the Earth were the size of a peach, LEO would be about the height of the peach fuzz! It ends at an altitude of about 2,000 km (1200 miles). But where does it begin? You might guess 100 km, the Kármán Line, but we are going to use 200 km (120 miles). Below that, drag will have a major effect on the orbit. You will learn more about drag and other perturbations in this course.
Where is GEO? Geostationary orbits have inclinations equal to zero degrees, or no tilt relative to the equator. Geosynchronous orbits have inclinations other than zero degrees. All GEO orbits have periods equal to 24 hours. The orbits are at an altitude of approximately 35,780 km (22,300 miles). This allows satellites located on the equator to stay above the same position on Earth at all times. The speed of the satellite is exactly that of the earth’s rotation. To get a better sense of how far away geo orbits are, consider the earth. A geo satellite will be at an altitude of about three Earth diameters! These are excellent orbits for communication satellites.
Then, of course, further out is the moon at 384,000 km (238,600 miles), or at thirty times Earth’s diameter away. The sun is at the center of it all and is about 150 million km (93 million miles) from the earth.
The first few lectures will be on some of the fundamental laws that govern satellite orbits: Newton’s Laws, his Universal Law of Gravitation, and Kepler’s three Laws. They will help explain why satellites travel faster at periapsis, or perigee if around the earth. Perigee is the closest point in an orbit. Satellite travel slower at apoapsis, or apogee if around the earth (Figure 17).
Have you ever wondered how we get something into space? Perhaps you have seen a rocket launch, live or on video. You notice how the rocket takes off straight up, but what happens after that? You will see it turn over to a nearly horizontal position. This gives it the horizontal velocity the spacecraft needs to get enough speed around the earth to stay in orbit briefly. Recall this was about 8 km/sec (17,900 miles per hour).
Next, we will study the classical orbital elements (Figure 18). These are:
a, Semimajor Axis, which describes the size of the orbit
e, Eccentricity, which describes the shape of the orbit
i, Inclination, the orbital tilt
Ω, Right Ascension of Ascending Node, or the swivel of an orbit
ω, Argument of Perigee, or the orientation of an elliptical orbit
ν, True Anomaly, which tells you where the satellite is located in its orbit
.
These laws, when combined, make up the two-body equation of motion for satellites. This equation and its solution will be critical to understanding orbits. Earth orbits will be referenced to the geocentric equatorial coordinate system. The video at the end if this section will help explain some of these termsWe will also look at preliminary orbit determination and ground tracks. Ground tracks are the paths a satellite makes on the surface of a planet directly below it, or the projection of a satellite’s orbit on the surface of the earth.
I had seen various types of ground tracks before but did not know until I began studying astrodynamics that you could get nearly round shaped ground tracks. These types of ground tracks are not from circular orbits! You will learn more about different types of ground tracks in this block (Figures 19 and 20).
Now that you will know the basics of orbital motion and how to describe orbits, how can you predict where a satellite will be in the future so you can communicate with it? There are two types of Kepler’s problems. The first type is to find how long it takes to get from one position in space to another. This is not a difficult problem for circular orbits as the satellites in them are travelling at the same speed all the time, but what about satellites in elliptical orbits? They are constantly slowing down near apogee and speeding up near perigee. The second type is Kepler’s prediction problem. When given an initial position and time of flight, where will the satellite be in the future (Figure 21)?
The next block will be about maneuvering satellites around in orbit. We will cover some common maneuvering techniques. First, we will consider the Hohmann transfer, which is used to change an orbit’s size (Figure 22).
Plane changes can be used to change the tilt or the swivel of the orbit. Combined plane changes combine the two to change both the size and the tilt of an orbit (Figure 23).
The International Space Station often receives shipments of supplies and has astronauts coming and going. Have you ever thought about how these rendezvous missions work? We will consider two types of rendezvous scenarios: co-planar and co-orbital. Co-planar rendezvous occurs when two satellites are at different altitudes and need to rendezvous. Co-orbital rendezvous occurs when you have two satellites in the same orbit. It then becomes an important issue of timing the maneuvers correctly, so the two satellites meet up together at the same time! We will also consider some more complex maneuvers (Figure 24).
We just discussed how to predict where a satellite will be in the future. That is really only possible in an ideal world. In reality, satellite orbits are subject to disturbances, or perturbations. We will consider two main ones, geopotential and drag effects. The geopotential effect, often called the J2 effect, describes the earth’s oblateness. The earth is not perfectly symmetrical. This effect will cause a torque on an orbit, which will cause the orbit to move.
Drag will reduce the size and affect the shape of an orbit and can eventually cause satellites to de-orbit. Below is the Russian MIR space station, which reentered Earth’s atmosphere in 2001 (Figure 25). Fortunately, most of its pieces disintegrated in the atmosphere and the pieces that did not land in the ocean.
A course in astrodynamics would not be complete without at least a brief discussion of interplanetary travel. In this block, we break the multi-body interplanetary transfer problem into a series of two body problems that we will study, enabling each section to be solved for and then “patched” together. Hence, this technique is called the patched conic approach. Finally, we will take a look at gravitational assists, which are often used in interplanetary travel. For example, the Cassini mission flew by Venus twice, Earth, and Jupiter on the way to Saturn. As it passed by each planet, it got a velocity boost, enabling Cassini to reach its destination more efficiently.
If time allows, a few additional topics are recommended, which are provided in the last few lessons. Launch windows and time will give you an introduction to when and where to launch. The launch velocities lesson will show you how to determine the velocity boost needed to get there. Finally, the last two lessons will consider some more advanced astrodynamics topics. For a video overview of the book and course, see video 6.
Video 6: Course Introduction Part 3 – Course Overview
SLIDES
Instructor Information
Course Introduction Pt 1 – Overview
Course Introduction Pt 2 – Types of Space Missions
Course Introduction Pt 3 – Course Overview
VIDEOS REFERENCES
Instructor Information
Course Introduction Pt 1 – Overview
Mice aboard the International Space Station. (2019, April 11).
Course Introduction Pt 2 – Types of Space Missions
Observing Weather From Space. (2020).
Ray, J. (2014, December 18). Newest GPS satellite goes active.
SBIRS – Spacecraft & Satellites. (2020).
SPOT-6 Satellite Image of Brisbane, Australia. (2017).
SPOT 7. (2020).
Voyager – The Interstellar Mission. (2020).
WGS – Wideband Global Satcom – Spacecraft & Satellites. (2020).
What is Galileo? (2020).
Wu, K. (2019, December 12). The European Space Agency Is Sending a Robot to Hug Junk Out of Space.
BOOK REFERENCES
Instructor Information
Doupe, C., Swenson, E., George, L, and Black, J. Finite Element Model Tuning with 3D Mode Shapes from FalconSAT-5. (2009, May). 50th AIAA/ASME/ASCE/AHS/ASC Structures, Structural Dynamics and Materials Conference.
Dunbar, B. (2012, October 15). A Long and Winding Road: Cassini Celebrates 15 Years.
George, Lynnane E. (2002). Active Vibration Control of a Flexible Base Manipulator. Doctoral dissertation, Georgia Institute of Technology.
George, L. E., & Kos, L. D. (1998). Interplanetary mission design handbook: Earth-to-Mars mission opportunities and Mars-to-Earth return opportunities 2009-2024. Huntsville, Ala.: National Aeronautics and Space Administration, Marshall Space Flight Center.
George, L.E. (1991). Wind Profile Data Gap Study. (ATM NO: 91-(6530-06)-10. The Aerospace Corporation.
Herbert, K. (2020). Falconsat-3.
Pike, J. (n.d.). Space.
Progressive Management. (2014). NASA Space Technology Report: Pogo in Rockets and Launch Vehicles.
Kepler-452B: A second earth?. The Science Explorer. (2015, September 8).
NASA. (2021, March 22). Exoplanet catalog. NASA.
USAF, Kirtland AFB. Air Force Research Laboratory’s Roll-Out Solar Array (ROSA) (af.mil)
Wilke, P., Johnson, C., Grosserode, P., & Scuilli, D. (2000, March). Whole-Spacecraft Vibration Isolation for Broadband Attenuation.
Course Introduction Pt 1 – Overview
Aeolus. (n.d.).
APOD: 2015 February 8 – Carina Nebula Dust Pillar. (n.d.).
Choi, Charles. (2019, April 15). The First Detailed Study Of How Mice Behave In Space Reveals Strange, Coordinated Zooming.
Choi, C. (2016, August 17). Space May Be the Best Place to Grow Bone Formation Protein Crystals.
Dvorsky, George. (2019, April 12). The First Detailed Study Of How Mice Behave In Space Reveals Strange, Coordinated Zooming. from
Early U.S. rocket and space launch failures and explosion. (2007, June 28).
Fredette, K., Smith, A., Sable, J., & Kasap, H. (n.d.). Space Worms?
Google. (2020).
History – Remembering First Landing on Moon – 20 July 1969. (2014, July 20).
In Depth. (2019, July 15).
James Webb Space Telescope – Webb/NASA. (2023).
Johnson, M. (2015, July 25). A Spin Around An Exoplanet Most Like Earth.
Joint Chiefs of Staff. (2023).
Mars Science Laboratory. (2019, September 07).
Military History Now. (2012, July 05). The First Air Forces – A Century of Balloons at War.
NASA Dawn. (2020, November 04).
NASA. (2023, September 14). NASA’s Webb takes star-filled portrait of pillars of creation. NASA.
NASA Jet Propulsion Laboratory Blog. (2014, May 23).
Redd, N. (2017, December 09). Valles Marineris: Facts About the Grand Canyon of Mars.
Ten Years Aboard the International Space Station. (2010, November 02).
Sawe, B. (2017, December 12). What is the Goldilocks Zone?
Sellers, J. J., & Astore, W. J. (2020). Understanding space: An introduction to astronautics. New York: McGraw-Hill.
StarTrek’s Monologue. (2008, January 25).
Javier, Yanes. (2019, April 01). Helium-3: Lunar Gold Fever.
Course Introduction Pt 2 – Types of Space Missions
Blau, P. (2017, September 23). Glonass Satellite Blasts Off on Soyuz to Replenish Russian Navigation Constellation.
Dunbar, B. (2015, May 01). Tracking and Data Relay Satellite (TDRS) Fleet.
Flybys. (2019, March 20).
Jenner, L. (2015, April 01). Landsat Overview.
MUOS. (2020).
NASA. (2020).
NASA’s Cassini Spacecraft Obtains Best Views of Saturn Hexagon. (2013, December 04).
Navigation. (2020).
Observing Weather From Space. (2020).
Ray, J. (2014, December 18). Newest GPS satellite goes active.
SBIRS – Spacecraft & Satellites. (2020).
SPOT-6 Satellite Image of Brisbane, Australia. (2017).
SPOT 7. (2020).
Voyager – The Interstellar Mission. (2020).
WGS – Wideband Global Satcom – Spacecraft & Satellites. (2020).
What is Galileo? (2020).
Wu, K. (2019, December 12). The European Space Agency Is Sending a Robot to Hug Junk Out of Space.
Course Introduction Pt 3 – Course Overview
AGI, Free Trial Options. (2021).
Convert Keplerian Orbital Elements to a State Vector. (n.d.).
Deorbit of Russia’s Mir Space Station was 10 Years Ago Today. (2011, March 23).
Orbital Mechanics 101, page 1. (n.d.).
Sellers, J. J., & Astore, W. J. (2021) Understanding space: An introduction to astronautics. New York: McGraw-Hill.
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Zheng et al. BMC Infectious Diseases (2024) 24:414
BMC Infectious Diseases
Correspondence:
Ying Zhu
ahfyww@126.com
Full list of author information is available at the end of the article
Abstract
Background Lobar pneumonia caused by Mycoplasma pneumoniae is a relatively difficult-to-treat pneumonia in
children. The time of radiographic resolution after treatment is variable, a long recovery time can result in several
negative effects, and it has attracted our attention. Therefore, exploring factors associated with delayed radiographic
resolution will help to identify these children at an early stage and prepare for early intervention.
Methods The data of 339 children with lobar pneumonia caused by Mycoplasma pneumoniae were collected from
the Department of Pediatrics of Fu Yang People’s Hospital, China from January 2021 to June 2022. After discharge,
the children were regularly followed up in the outpatient department and on the WeChat platform for > 8 weeks.
According to whether pulmonary imaging (chest radiography or plain chest computed tomography) returned to
normal within 8 weeks, the children were divided into the delayed recovery group (DRG) ( n = 69) and the normal
recovery group (NRG) ( n = 270). The children’s general information, laboratory examination findings, bronchoscopy
results, and imaging findings were retrospectively analyzed. Single-factor analysis was performed to identify the risk
factors for delayed radiographic resolution of lobar pneumonia caused by Mycoplasma pneumoniae , and the factors
with statistically significant differences underwent multiple-factor logistic regression analysis. Receiver operating
characteristic (ROC) analysis was then performed to calculate the cutoff value of early predictive indicators of delayed
radiographic resolution.
Results Single-factor analysis showed that the following were significantly greater in the DRG than NRG: total
fever duration, the hospitalization time, C-reactive protein (CRP) level, lactate dehydrogenase (LDH) level, D-dimer
level, pulmonary lesions involving two or more lobes, a large amount of pleural effusion, the time to interventional
bronchoscopy, and mucus plugs formation. Multivariate logistic regression analysis showed that the hospitalization
time, CRP level, LDH level, pulmonary lesions involving two or more lobes, and a large amount of pleural effusion
were independent risk factors for delayed radiographic resolution of lobar pneumonia caused by Mycoplasma
pneumoniae . The cutoff values on the receiver operating characteristic curve were a hospitalization time of ≥ 10.5
days, CRP level of ≥ 25.92 mg/L, and LDH level of ≥ 378 U/L.
Early predictors of delayed radiographic
resolution of lobar pneumonia caused
by Mycoplasma pneumoniae in children:
a retrospective study in China
Yu Zheng 1, Guoshun Mao 1, Hongchen Dai 1, Guitao Li 1, Liying Liu 1, Xiaying Chen 1 and Ying Zhu 1 Page 2 of 8 Zheng et al. BMC Infectious Diseases (2024) 24:414
Introduction
Lobar pneumonia is a common type of Mycoplasma
pneumoniae pneumonia (MPP) in children, and it is
typically a mild condition that responds well to macro -
lide antibiotics (MAs). A follow-up study revealed that
in approximately 70% of children’s pulmonary imag -
ing changes can return to normal within 8 weeks when
antibiotics and glucocorticoids are used appropriately [ 1,
2]. In some children, however, the time of radiographic
resolution significantly exceeds 8 weeks. When the clini -
cal condition is accompanied by necrotizing pneumonia,
the time of radiographic resolution can reach 6 months
, significantly impacting the patient’s quality of life.
Furthermore, children with long-term lung lesions may
develop persistent fever, impaired lung function, and
worsening of the disease [ 4, 5].
This study was performed to identify predictors that
can be used for early prediction of delayed radiographic
resolution of lobar pneumonia caused by Mycoplasma
pneumoniae (M. pneumoniae ) in children, thus facilitat -
ing early interventions to prevent serious complications
such as necrotizing pneumonia, bronchiectasis, and
bronchial occlusion.
Patients and methods
Patients
We retrospectively analyzed the data of 339 children with
lobar pneumonia caused by M. pneumoniae who were
admitted to the Department of Pediatrics of Fu Yang
People’s Hospital from 1 January 2021 to 30 June 2022.
The children were divided into a normal recovery group
(NRG) ( n = 270) and a delayed recovery group (DRG)
(n = 69) ( Fig. 1 ). The research protocol was approved by
the Ethical Committee of Fu Yang People’s Hospital (ethi -
cal code: 2018 − 167).
Inclusion and exclusion criteria
The inclusion criteria were (1) chest radiography or
lung computed tomography findings consistent with the
imaging characteristics of lobar pneumonia and (2) sat -
isfaction of the diagnostic criteria for MPP in the Expert
Consensus on Laboratory Diagnostics and Clinical Prac -
tice of Mycoplasma pneumoniae Infection in Children in
China (i.e., manifestations of pneumonia and/or imaging
changes combined with an MP antibody titer of ≥ 1:160)
.
The exclusion criteria were (1) infection with other
pathogens, (2) immunodeficiency, (3) lack of bronchos -
copy with bronchoalveolar lavage, (4) incomplete medi -
cal records, and (5) loss to follow-up.
Data collection
The following data were retrospectively collected from
the children’s medical records. (1) Clinical data: age, sex,
total fever duration, peak body temperature, hospital -
ization time, season of onset, and time to interventional
bronchoscopy ( The “time” refers to the interval from
the onset of illness to undergoing bronchoscopy with
bronchoalveolar lavage, it is not a treatment time.). (2)
Laboratory examination findings: including white blood
cell count (WBC), neutrophil ratio (NE%), lymphocyte
ratio (LYM%), C-reactive protein (CRP) level, D-dimer
level, lactate dehydrogenase (LDH) level, alkaline phos -
phatase (ALP) level, and immunoglobulins A, M, and
G (IgA, IgM, IgG). (3) Other conditions: glucocorticoid
resistance, a large amount of pleural effusion, atelecta -
sis, lung lesions involving two or more lobes, and mucus
plugs formation as shown by bronchoscopy. A retrospec -
tive analysis was performed to understand the relation -
ship between the above factors and delayed radiographic
resolution of lobar pneumonia caused by M. pneumoniae
in children.
Definitions
(1) According to previous reports, the time of radio -
graphic resolution in most patients with lobar pneumo -
nia caused by M. pneumoniae is 1 to 2 months [ 7–9].
Therefore, we defined delayed radiographic resolution
as lack of recovery on chest radiographs or lung com -
puted tomography scans from admission to 8 weeks of
treatment. Radiologic follow-up was performed every 4
weeks.
(2) MAs non-responsiveness was defined as a persis -
tent fever in a child with MPP after 72 h of regular MAs
treatment, with either no improvement in clinical signs
and pulmonary imaging findings or the development of
further aggravation [ 10 ].
(3) Glucocorticoid resistance was defined as a persis -
tent fever for > 72 h after intravenous administration of
methylprednisolone at 1 to 2 mg/kg per day [ 11 , 12 ].
Conclusion If patients with lobar pneumonia caused by Mycoplasma pneumoniae have a hospitalization time of
≥ 10.5 days, CRP level of ≥ 25.92 mg/L, and LDH level ≥ 378 U/L, the time of radiographic resolution is highly likely to
exceed 8 weeks. Pediatricians must maintain a high level of vigilance for these factors, control the infection as early as
possible, strengthen airway management, and follow up closely to avoid complications and sequelae of Mycoplasma
pneumoniae pneumonia.
Keywords Children, Mycoplasma pneumoniae , Lobar pneumonia, Imaging Page 3 of 8 Zheng et al. BMC Infectious Diseases (2024) 24:414
(4) A large amount of pleural effusion was defined as
a total pleural fluid volume of ≥ 500 mL and a low to
medium volume of pleural effusion was defined as a total
pleural fluid volume of < 500 mL.
Bronchoscopy with bronchoalveolar lavage
All children in this study received routine anti-infection
therapy and nebulization for 3 days after admission. If
the children still had persistent fever, persistent cough,
and no significant improvement in pulmonary signs,
bronchoscopy with bronchoalveolar lavage was rec -
ommended. The alveolar lavage fluid was reserved for
pathogenic nucleic acid detection after obtaining written
informed consent from the child’s family. The equipment
used for bronchoscopy with alveolar lavage is PENTAX
fiber bronchoscopy (device model: EB-1170 K).
Patients were discharged when the rales disappeared,
and the body temperature was normal for ≥ 3 consecu -
tive days after completing routine anti-infective and
anti-inflammatory treatment and bronchoscopy with
bronchoalveolar lavage.
Statistical analysis
The data were analyzed using SPSS 25.0 software (IBM
Corp., Armonk, NY, USA). Measurement data conform -
ing to a normal distribution are presented as mean ± stan -
dard deviation, and data with a skewed distribution are
presented as median with interquartile range (Q25–
Q75). Comparisons between groups were performed
with the t-test or Mann–Whitney U test [ 13 ]. Enumer -
ated data are expressed as rates, and the χ 2 test was used
for comparison between groups. We conducted a single-
factor analysis to identify the risk factors for delayed
radiographic resolution. We then conducted a multivari -
ate logistic regression analysis of statistically significant
factors to summarize the high-risk factors. Finally, the
relevant factors were analyzed using a receiver operat -
ing characteristics (ROC) curve, and the early predictors
of delayed radiographic resolution were summarized.
Two-sided P˗values of < 0.05 were considered statistically
significant.
Fig. 1 Flow chart of patient selection Page 4 of 8 Zheng et al. BMC Infectious Diseases (2024) 24:414
Results
Clinical features
General information
A total 339 children with lobar pneumonia caused by M.
pneumoniae were divided into 270 cases with NRG and
69 cases with DRG. The mean age at onset was 7.17 ±
2.37 years in the NRG and 6.86 ± 1.50 years in the DRG.
The male: female ratio was approximately 1.4:1.0 (NRG
vs. DRG). There was no significant difference in sex, sea -
son of onset, or age between the two groups. Total fever
duration and hospitalization time were significantly lon -
ger in the DRG than NRG ( P < 0.05). The time to inter -
ventional bronchoscopy was significantly longer in the
DRG than NRG (11.06 ± 3.53 vs. 9.03 ± 2.76 days, respec -
tively) (Table 1).
NRG, normal recovery group; DRG, delayed recovery
group.
Laboratory test results
Comparison of the laboratory examination results
between the two groups showed that the CRP, D ˗dimer,
and LDH levels were significantly higher in the DRG than
NRG ( P < 0.05). There were no significant differences in
the white blood cell count, neutrophil ratio, lymphocyte
ratio, alkaline phosphatase level, or immunoglobulin lev -
els ( P > 0.05) (Table 2).
Other conditions
There were statistically significant differences in gluco -
corticoid resistance, a large amount of pleural effusion,
atelectasis, pulmonary lesions involving two or more
lung lobes, and mucus plugs formation between the NRG
and DRG ( P < 0.05). However, the incidence of MAs non-
responsiveness was not significantly different between
the two groups (Table 3).
Multivariate logistic regression analysis
The statistically significant factors in the above-described
univariate analysis were taken as independent variables,
and whether delayed recovery had occurred was taken
as the dependent variable (no = 0, yes = 1); a binary logis -
tic regression analysis was then performed. The final
results showed that the hospitalization time [odds ratio
(OR) = 1.584, 95% confidence interval (CI) = 1.082–2.321,
P = 0.018], LDH level (OR = 1.022, 95% CI = 1.008–1.036,
P = 0.002), CRP level (OR = 1.079, 95% CI = 1.045–1.114,
P < 0.001), pulmonary lesions involving two or more lung
lobes (OR = 8.997, 95% CI = 1.698–47.682, P = 0.010), and
a large amount of pleural effusion (OR = 11.568, 95%
CI = 1.767–75.728, P = 0.011) were independent risk fac -
tors for delayed radiographic resolution of lobar pneu -
monia caused by M. pneumoniae in children ( P < 0.05)
(Table 4).
ROC curve regression analysis
The cutoff values for maximum sensitivity and specific -
ity of predictive indicators were determined according
to Youden’s method. The results showed that the cutoff
values for predictive indicators were a hospitalization
duration of ≥ 10.5 days, CRP level of ≥ 25.92 mg/L, and
LDH level of ≥ 378 U/L. The corresponding area under
the curve was 0.835, 0.944, and 0.917; the sensitivity was
0.667, 0.913, and 0.812; and the specificity was 0.844,
0.844, and 0.900, respectively (Fig. 2; Table 5).
Table 1 Comparison of clinical data between NRG and DRG
Characteristics NRG
(n=270)
DRG
(n=69)
P˗value
Gender (male/female) 159/111 36/33 0.098
Onset seasons (spring and
summer/autumn and winter)
138/132 29/40 0.178
Age (years) 7.17 ± 2.37 6.86 ± 1.50 0.179
Total fever duration (days) 6.34 ± 2.63 11.13 ± 4.37 <0.001
Peak body temperature ( ℃)39.15 ± 0.73 39.29 ± 0.64 0.114
Hospitalization time (days) 8.36 ± 2.26 12.70 ± 4.07 <0.001
Time to interventional bron -
choscopy (days)
9.03 ± 2.76 11.06 ± 3.53 <0.001
Table 2 Comparison of laboratory examination results between
NRG and DRG
Characteristics NRG
(n=270)
DRG
(n=69)
P˗value
WBC (×10 9/L) 7.27 ± 2.24 7.72 ± 1.75 0.121
NE% 61.74 ± 11.14 64.79 ± 12.68 0.068
LYM% 29.20 ± 9.55 27.47 ± 10.82 0.223
CRP (mg/L) 10.61(5.21, 17.03) 57.18(35.11, 98.09) <0.001
D-D (mg/L) 0.85(0.41, 2.21) 3.18(2.09, 5.53) <0.001
ALP (U/L) 174.22 ± 52.42 163.12 ± 37.31 0.099
LDH (U/L) 307.99 ± 58.77 515.00 ± 215.48 <0.001
IgA (g/L) 1.37 ± 0.55 1.45 ± 0.70 0.418
IgM (g/L) 1.47 ± 0.52 1.55 ± 0.74 0.251
IgG (g/L) 8.85 ± 1.88 9.00 ± 1.97 0.569
NRG, normal recovery group; DRG, delayed recovery group; WBC, white blood
cell count; NE%, neutrophil ratio; LYM%, lymphocyte ratio; CRP, C-reactive
protein; D-D, D-dimer; LDH, lactate dehydrogenase; ALP, alkaline phosphatase;
IgA, immunoglobulin A; IgM, immunoglobulin M; IgG, immunoglobulin G
Table 3 Comparison of other conditions between NRG and DRG
Characteristics NRG
(n=270)
DRG
(n=69)
P˗value
MAs non-responsiveness [ n(%)] 105(38.9) 35(50.7) 0.075
Glucocorticoid resistance [ n(%)] 27(10.0) 40(58.0) <0.001
A large amount of pleural effusion
[n(%)]
20(7.4) 28(40.6) <0.001
Atelectasis [ n(%)] 54(20.0) 29(42.0) <0.001
Pulmonary lesions involving two or
more lung lobes [ n(%)]
54(20.0) 50(72.5) <0.001
Mucus plugs [ n(%)] 59(21.9) 38(55.1) <0.001
NRG, normal recovery group; DRG, delayed recovery group; MAs, macrolide
antibiotics
Page 5 of 8 Zheng et al. BMC Infectious Diseases (2024) 24:414
Prognosis
Comparing the prognosis of two groups of children, it
was found that the patients of DGR had an increased
risk of adverse prognosis ( P < 0.05), such as necrotizing
pneumonia, bronchiectasis, obliterative bronchiolitis and
recurrent wheezing. The top two are recurrent wheezing
and necrotizing pneumonia. They account for 47.8% and
36.2%, respectively (Table 6).
Discussion
Lobar pneumonia caused by M. pneumoniae in children
often presents as pulmonary lobe or segmental consolida -
tion (37%) and slightly less commonly as, infiltration near
the hilum or around the bronchus (27%), local reticularis
nodule infiltration (21%), and patchy infiltration (15%) [ 14 ].
Unilateral pulmonary lobe involvement is more common
than bilateral involvement. Imaging examination is one of
the main bases for clinical severity and prognosis assess -
ment. In this study, we found that children in the DRG
were more likely to have a worsened prognosis. Delayed
radiographic resolution may lead to necrotizing pneumo -
nia, bronchiolitis obliterans, bronchiectasis, and recurrent
wheezing, their incidence increases in DRG significantly.
After the above-mentioned situations occur, almost all
patients will experience worsening pulmonary function
and decreased exercise tolerance. Therefore, in order to
Table 4 Multivariate logistic regression analysis of delayed
radiographic resolution of lobar pneumonia caused by
Mycoplasma pneumoniae in children
Variables B SE Wald P-value Odds
ratio 95%
confidence
interval (CI)
Hospitalization
time
0.460 0.195 5.583 0.018 1.584(1.082–
2.321)
LDH 0.022 0.007 9.648 0.002 1.022(1.008–
1.036)
CRP 0.076 0.016 21.734 < 0.001 1.079(1.045–
1.114)
Pulmonary le -
sions involving
two or more
lung lobes
2.197 0.851 6.666 0.010 8.997(1.698–
47.682)
A large amount
of pleural
effusion
2.448 0.959 6.522 0.011 11.568(1.767–
75.728)
Total fever
duration
0.183 0.157 1.369 0.242 1.201(0.884–
1.632)
Glucocorticoid
resistance
1.476 0.917 2.592 0.107 4.375(0.725–
26.381)
D-D -0.180 0.101 3.154 0.076 0.835(0.685–
1.019)
Mucus plugs -0.686 0.874 0.616 0.432 0.503(0.091–
2.793)
Time to
interventional
bronchoscopy
0.128 0.105 1.484 0.223 1.137(0.925–
1.398)
Atelectasis 0.902 0.831 1.179 0.278 2.465(0.484–
12.560)
Hospitalization time, CRP level, LDH level, D-D level, total fever duration and
time to interventional bronchoscopy are measurement data. Pulmonary
lesions involving two or more lung lobes, a large amount of pleural effusion,
glucocorticoid resistance, mucus plugs and atelectasis are enumerated data
(0 =no, 1 =yes)
NRG, normal recovery group; DRG, delayed recovery group; LDH, lactate
dehydrogenase; CRP, C-reactive protein; D-D, D-dimer; SE, standard error
Table 5 Receiver operating characteristic threshold values of
predictive indexes of delayed radiographic recovery in children
with lobar pneumonia induced by Mycoplasma pneumoniae
Mycoplasma pneumoniae
Variables Cutoff
value
The sensitivity The specificity AUC
Hospitalization
time (days)
10.50 0.667 0.844 0.835
CRP (mg/L) 25.92 0.913 0.844 0.944
LDH (U/L) 378 0.812 0.900 0.917
CRP, C-reactive protein; LDH, lactate dehydrogenase; AUC, area under the curve.
Table 6 Comparison of prognosis between NRG and DRG of
discharged children
Characteristics NRG
(n = 270)
DRG
(n = 69)
P˗value
Necrotizing pneumonia [ n(%)] 5(1.9) 25(36.2) < 0.001
Bronchiectasis [ n(%)] 2(0.7) 10(14.5) < 0.001
Bronchiolitis obliterans [ n(%)] 5(1.9) 12(17.4) < 0.001
Recurrent wheezing [ n(%)] 11(4.1) 33(47.8) < 0.001
Fig. 2 Receiver operating characteristic curve of predictive value of hos -
pital stay, CRP level, and LDH level on delayed radiographic resolution in
children with lobar pneumonia induced by Mycoplasma pneumoniae CRP,
C-reactive protein; LDH, lactate dehydrogenase Page 6 of 8 Zheng et al. BMC Infectious Diseases (2024) 24:414
strengthen post-discharge management, early identification
of these children is necessary.
This study involved 339 children with MPP who showed
lobar changes on lung imaging. The children’s clinical fea -
tures, laboratory examination findings, and bronchoscopic
findings were retrospectively analyzed, and the risk fac -
tors for delayed radiographic resolution were summa -
rized. Univariate analysis showed statistically significant
differences between the NRG and DRG in the CRP level,
LDH level, total fever duration, hospitalization time, glu -
cocorticoid resistance, a large amount of pleural effu -
sion, pulmonary lesions involving two or more lung lobes,
and atelectasis ( P < 0.05). Further analysis using a multi -
variate logistic regression model and ROC curve showed
that a of hospitalization time of ≥ 10.5 days, CRP level of
≥ 25.92 mg/L, LDH level of ≥ 378 U/L, pulmonary lesions
involving two or more lung lobes, and a large amount of
pleural effusion were risk factors for delayed radiographic
resolution of lobar pneumonia caused by M. pneumoniae in
children. Radiographic resolution of pneumonia is affected
by multiple factors, which are generally divided into two
categories: factors related to the disease severity and fac -
tors related to the treatment effect. The more risk factors are
combined in the same children, the greater the probability
of delayed radiographic recovery.
More severe clinical features and higher laboratory indi -
cators may be associated with delayed radiographic recov -
ery. Lobar pneumonia caused by M. pneumoniae occurs
mostly in children aged >5 years, because their immune
function is stronger, their autoimmune reaction after M.
pneumoniae infection is stronger, and lung tissue damage is
more serious than in children aged < 5 years. Zhao et al. [ 15 ]
found that age had an impact on the prognosis of children
with MPP, and the critical age threshold was 4.5 years. The
mean age of the children in the NRG and DRG in this study
was 7.17 ± 2.37 and 6.86 ± 1.50 years respectively, both were
above the threshold of 4.5 years, and there was not find any
impact of age on the prognosis of the two groups of patients.
Patients in the DRG had a longer hospital stay (12.70 vs.
8.36 days) and a longer fever duration (11.13 vs. 6.34 days)
than those in the NRG. Moreover, the levels of CRP (57.18
vs. 10.61 mg/L), LDH (515.00 vs. 307.99 U/L), and D-dimer
(3.18 vs. 0.85 mg/L) were higher in the DRG than NRG.
These findings suggest that the pulmonary inflammatory
reaction or tissue damage was more severe in the DRG [ 16 ].
An elevated CRP level is an early indicator of lung tissue
necrosis, indicating a systemic inflammatory reaction sec -
ondary to severe lung infection. Additionally, studies have
suggested that the incidence of pleural effusion, myocardial
injury, and liver injury is remarkably increased in children
who have MPP with elevated indices [ 17 , 18 ]. Such findings
are consistent with the slower radiographic resolution seen
in the children in the DRG of our study. The multivariate
logistic regression analysis showed that the CRP and LDH
levels were independent risk factors for delayed radio -
graphic resolution, and the ROC curve analysis showed that
the CRP and LDH levels (area under the curve = 0.944 and
0.917, respectively) had high predictive value for delayed
radiographic resolution.
Several serious complications can lead to delayed radio -
graphic resolution of pneumonia. The most common com -
plications of M. pneumoniae -induced lobar pneumonia
include pleural effusion, atelectasis, and mucus plugs for -
mation, which seriously affect the recovery of lung imag -
ing. Kim et al. [ 19 ] showed that children who had MPP
with pleural effusion had more severe pneumonia lesions
and poorer responses to treatment, resulting in prolonged
radiographic resolution of lung abnormalities. In other stud -
ies, a large amount of pleural effusion was closely associated
with the occurrence of necrotic pneumonia [ 2, 8, 20 ]. The
imaging changes of necrotic pneumonia were character -
ized by the emergence of voids after destruction of the lung
parenchymal structure, and the radiographic resolution of
necrotic pneumonia was thus slower than that of common
pneumonia. Undoubtedly, children with extensive pulmo -
nary inflammatory lesions had a longer recovery time. In
this study, 70% (235/339) of the children had a single pul -
monary lobe lesion, and only 30% of the children had lung
lesions involving more than two pulmonary lobes. It indi -
cates that lobar pneumonia caused by M. pneumoniae is
usually characterized by involvement of a single pulmonary
lobe. Notably, however, 72.5% of the children in the DRG
had pulmonary lesions involving two or more lobes, sug -
gesting that once inflammatory lesions involve multiple
lobes, the recovery period is prolonged. The formation of
mucus plugs is an important manifestation of progressive
airway mucosal damage, which can lead to irreversible air -
way dysventilation and atelectasis. Zhang et al. [ 21 ] found
that children with mucus plugs formation were more likely
to have intrapulmonary complications such as pleural effu -
sion and extrapulmonary complications, making treatment
more difficult and inflammation more persistent.
Factors related to the treatment effect have slightly dif -
ferent effects on radiographic resolution of pneumonia.
Such factors include MAs non-responsiveness, glucocor -
ticoid resistance, and the time to interventional bronchos -
copy. Effective antimicrobial and anti-inflammatory therapy
helps control disease and shorten the acute phase of dis -
ease. MAs therapy is still the first choice for the treatment
of M. pneumoniae infection, MAs inhibit protein synthe -
sis to achieve an anti-infection effect by binding to domain
II and/or V of 23 S rRNA in the ribosomal subunit of 50 S
bacteria. In recent years, however, the rate of resistance
to MAs has shown an increasing trend. Wang et al. [ 22 ]
found that 90.94% (1386/1524) of specimens from 1524
children with MPP were resistant to MAs. Gene mutation
in the 23 S rRNA domain V of M. pneumoniae is the main
mechanism of MAs resistance, and A2063 G/C, A2064 Page 7 of 8 Zheng et al. BMC Infectious Diseases (2024) 24:414
G/C, and C2617 G/A have been shown to be mutation sites
of MAs resistance [ 23 , 24 ]. In the study, the rate of MAs
non-responsiveness was 38.9% and 50.7% in the NRG and
DRG, respectively, with no statistically significant difference
between the two groups. This finding indicates that MAs
non-responsiveness did not affect the radiographic reso -
lution in children with MPP, which is consistent with the
results of reported by Deng et al. [ 25 ]. Additionally, Chen
et al. [ 26 ] stated that MAs resistance may prolong fever
duration and treatment time, but the chest X-ray examina -
tion findings and laboratory test indicators did not change
accordingly. Glucocorticoids are used as immunomodula -
tors to suppress overactive host immune responses, reduce
the intensity of local inflammation, and promote recov -
ery from disease. Patients with glucocorticoid resistance
often have more severe clinical manifestations and imag -
ing abnormalities. A strong cellular immune response will
lead to severe ciliary dysfunction, reduce airway immune
function, and destroy ciliary mucus clearance ability, these
changes will result in large-scale infiltration, atelectasis, and
mucus plugs formation, seriously affecting the recovery of
lung inflammation. Bronchoscopy with bronchoalveolar
lavage is a rapid and effective treatment for relieving clini -
cal symptoms. Many studies have shown that bronchos -
copy with bronchoalveolar lavage is beneficial for removing
mucus plugs and has significant advantages for patients
with pneumonia characterized by persistent atelectasis [ 27 ,
28 ]. Disease remission has been shown to become greatly
accelerated after intervention of bronchoalveolar lavage
[29 ]. The study also showed that the time to interventional
bronchoscopy was shorter in NRG than DRG, and early
treatment with bronchoalveolar lavage was conducive to the
improvement of imaging.
No single clinical or biochemical indicator can accu -
rately predict whether delayed radiographic resolution
of pulmonary abnormalities will occur. We performed a
multiple logistic regression analysis and ROC curve analy -
sis to eliminate the confounding among study factors and
objectively examine the clinical factors influencing delayed
radiographic resolution in children with lobar pneumonia
caused by M. pneumoniae . However, this study also had
some limitations; for example, it is a single-center study,
some patients were lost to follow-up, and some patients
were excluded because of incomplete data or lack of bron -
choscopy with bronchoalveolar lavage, which may have
biased the results. The sample size should be increased for
in further studies. We followed up on the pulmonary func -
tion of some patients in the DRG and found that they had
varying degrees of decreased pulmonary function. However,
pulmonary function tests could not be conducted on young
patients ( < 6 years old) who were unable to cooperate, pre -
venting an assessment of their pulmonary function. This
article also has the shortcoming of not making a compre -
hensive comparison of pulmonary function.
Conclusion
In the study, we found that delayed radiographic resolution
was mainly associated with the development of complica -
tions and high levels of inflammatory markers. Patients with
lung lesions involving two or more lobes, a large amount of
pleural effusion, high levels of inflammatory markers (CRP
level of ≥ 25.92 mg/L, LDH level of ≥ 378 U/L), or a long
hospital stay (a hospitalization time of ≥ 10.5 days) had lon -
ger radiographic resolution. These five factors are closely
related to radiographic resolution, and pediatrician should
pay attention to them in our clinical work. These find -
ings may help to identify these patients early in the course
of their disease and enhance patient management after
discharge.
Abbreviations
DRG Delayed recovery group
NRG Normal recovery group
ROC Receiver operating characteristic
AUC Area under the curve
CI Confidence interval
OR Odds ratio
MPP Mycoplasma Pneumoniae Pneumonia
MAs Macrolide antibiotics
WBC White blood cell count
NE% Neutrophil ratio
LYM% Lymphocyte ratio
CRP C-reactive protein
D-D D-dimer
LDH Lactate dehydrogenase
ALP Alkaline phosphatase
IgA Immunoglobulin A
IgM Immunoglobulin M
IgG Immunoglobulin G
Acknowledgements
The author thanks all patients enrolled in this study.
Author contributions
Y. Z. and Y. Z. were responsible for the conception and design of the paper,
statistical processing, analysis and interpretation of the results, and the final
writing of the paper. H. D. and G. L. were responsible for bronchoscopic
treatment and follow-up observation of patients, and carried out the
implementation and feasibility analysis of the study. L. L. and X. C. are
responsible for data collection and collation. Y. Z. and G. M. are responsible
for the quality control and proofreading of the article, as well as the overall
supervision and management of the article. All authors read and approved
the final manuscript.
Funding
This work was supported by Fu Yang Self-funded Science and Technology Plan
Project (FK202081038).
Data availability
Underlying research data and materials can be accessed by contacting the
corresponding author.
Declarations
Ethics approval and consent to participate
The performed study’s protocol was following the principles of the declaration
of Helsinki and approved by the Ethical Committee of Fu Yang People’s
Hospital (2018 −167). The informed consent was waived by the Ethical
Committee of Fu Yang People’s Hospital due to retrospective nature of study.
Page 8 of 8 Zheng et al. BMC Infectious Diseases (2024) 24:414
Consent for publication
Not applicable.
Competing interests
The authors declare no competing interests.
Author details
1
Department of Pediatrics, Fu Yang People’s Hospital, No.501, Sanqing
Road, Yingzhou District, 236000 Fuyang, Anhui Province, China
Received: 1 September 2023 / Accepted: 3 April 2024
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Conical pendulum question
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Asked 1 year, 8 months ago
Modified1 year, 7 months ago
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Hi, I was wondering why you get different answers when using different approaches to this question.
Using the method provided, I get the same answer but if I use trigonometry and do T=35 g cos(30)T=35 g cos(30), I get 396.06285 N 396.06285 N (using g=9.8 m s g=9.8 m s).
Contrarily, if I use the pythagorean theorem, I found F n e t F n e t with m v 2 r=196.23333 N m v 2 r=196.23333 N. Then, T=196.2333 2+(35 g)2−−−−−−−−−−−−−−−√=395.16647 N T=196.2333 2+(35 g)2=395.16647 N (again with g=9.8 m s g=9.8 m s).
Thus, I am confused why different answers are obtained with each of the methods, even after compensating for the difference in the true value of the acceleration due to gravity.
homework-and-exercises
newtonian-mechanics
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edited Jan 31, 2024 at 10:16
user220866
asked Jan 31, 2024 at 9:42
Vansh KumarVansh Kumar
23 3 3 bronze badges
6
1 Those answers are actually all very close to each other. The differences should be coming from the finite precision with which we state all the variables. For example, if you assume that the rope is exactly 3m long and the radius of the circle is exactly 1.5m, then the velocity is not exactly 2.9m/s, but only a little different.naturallyInconsistent –naturallyInconsistent 2024-01-31 10:13:37 +00:00 Commented Jan 31, 2024 at 10:13
So in an exam situation, if the question had required 3 s.f., which of them am I meant to use? I guess the method they used would be best as there is no imprecision of the acceleration due to gravity being rounded?Vansh Kumar –Vansh Kumar 2024-01-31 10:18:40 +00:00 Commented Jan 31, 2024 at 10:18
1 That's simple. If the question wanted 3sf, it would have given you data at least to 3sf. They have given you only 2sf here, and so the answer key is also only up to 2sf.naturallyInconsistent –naturallyInconsistent 2024-01-31 10:20:06 +00:00 Commented Jan 31, 2024 at 10:20
I don't follow. If I were to add 3 zeros after each value, I would get the same results as above but each answer would still not be correct to 5 s.f.Vansh Kumar –Vansh Kumar 2024-01-31 10:47:12 +00:00 Commented Jan 31, 2024 at 10:47
1 If you added 3 zeroes to 3.0m and 1.5m, then the velocity is 2.9147m/s and cannot be reconciled with 2.9m/s.naturallyInconsistent –naturallyInconsistent 2024-01-31 12:27:48 +00:00 Commented Jan 31, 2024 at 12:27
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Here's the thing, the data in the question is just so ever slightly wrong. Let's assume everything as is given except for the length of the rope. We don't know that it is 3m long. And since we don't know this length we also don't know the angle that the rope makes with the vertical. Let us assume this as θ θ.
Writing the equations using NLM we get:
T s i n(θ)=m v 2 r T s i n(θ)=m v 2 r
and
T c o s(θ)=m g T c o s(θ)=m g
taking their ratios gives us
t a n(θ)=v 2 r g t a n(θ)=v 2 r g
Putting the values we get t a n(θ)=0.5715256541 t a n(θ)=0.5715256541 (using g = 9.81)
This gives θ=29.7490743360347∘θ=29.7490743360347∘ (and not30∘30∘)
It's because of this slight inconsistency that the final answers vary. (As you might have noticed, you used θ θ in method 1 but not in method 2.)
Coming to the final conclusion, this really shouldn't matter much since the answers have a very small difference and in an exam scenario, it would be advisable to follow the method provided in the original question since the crux of the question is to test your knowledge of centripetal acceleration.
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answered Feb 16, 2024 at 13:39
ConundrumaniaConundrumania
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14325 | https://pmc.ncbi.nlm.nih.gov/articles/PMC2654706/ | Differential Assessment and Management of Asthma vs Chronic Obstructive Pulmonary Disease - PMC
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Differential Assessment and Management of Asthma vs Chronic Obstructive Pulmonary Disease
Barbara P Yawn
Barbara P Yawn, MD, MSc, FAAFP
1 Olmstead Medical Center Rochester, Minnesota
Find articles by Barbara P Yawn
1
Author information
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1 Olmstead Medical Center Rochester, Minnesota
Disclosure: Barbara P. Yawn, MD, MS, MSc, FAAFP, has disclosed that she serves as a consultant to Boehringer Ingelheim, Pfizer, AstraZeneca, and Schering-Plough related to COPD.
Collection date 2009.
©2009 Medscape
PMC Copyright notice
PMCID: PMC2654706 PMID: 19295941
Abstract
Context
Asthma and chronic obstructive pulmonary disease (COPD) are prevalent respiratory conditions with overlapping disease characteristics. Differentiation between asthma and COPD is important because several aspects of the guideline-recommended management strategies differ for these conditions. This review identifies the epidemiologic, etiologic, and clinical distinctions of these diseases to assist physicians and other clinicians in differentiating between asthma and COPD. Key components of the guideline-recommended management approaches for these conditions are also reviewed.
Search Strategies
Relevant articles were found by searching the MEDLINE database for “asthma” and “chronic obstructive pulmonary disease OR COPD” in association with the terms “diagnostic criteria” and “differential diagnosis”. Recent statistical summaries (meta-analyses), reviews, and consensus-type documents were also included.
Synthesis
A review of relevant articles found that, although asthma and COPD may occur simultaneously, differences between these diseases are frequently recognized in terms of age at onset, prevalence in relation to age and sex, potential for reversibility of airway obstruction, pathophysiology, and typical symptom presentation. A thorough clinical history in conjunction with lung function testing usually aids in diagnostic distinction and choice of therapeutic interventions. Radiologic imaging and inflammatory marker testing may also aid in the differentiation of these conditions. Over time, disease progression often differs between asthma and COPD.
Conclusions
Although overlaps exist in the disease characteristics of asthma and COPD, careful history, physical examination, and lung function testing often reveal information that facilitates distinction between these diseases, allowing physicians and other clinicians to better tailor their therapy.
A Snapshot of Obstructive Lung Disease
Asthma and chronic obstructive pulmonary disease (COPD) are characterized by a reduced rate of pulmonary airflow resulting from increased inflammation. Airway obstruction is typically fully or nearly fully reversible in patients with asthma, whereas COPD is characterized by airway obstruction that is not fully reversible. Patients with either disease may develop acute exacerbations characterized by increased inflammation of the airway and worsening airway obstruction.2,
Asthma most often presents at a younger age as recurrent episodes of increased airway obstruction that may have varying frequency and intensity, which then become recognized as a chronic pattern of reversible airway obstruction. In a subset of patients with long-term disease, reversibility of airway obstruction is diminished (due to airway remodeling), and a disease pattern similar to COPD may ensue. When the onset of asthma occurs in the sixth or seventh decade of life, recognition is more difficult because symptoms may be similar to those of cardiac disease and COPD, and patients often accept problems as secondary to aging rather than disease.
COPD is a progressive disease of declining lung function observed mainly in older adults with a history of cigarette smoking. COPD includes emphysema and chronic bronchitis, and identification of the clinical signs of these different conditions often occur simultaneously.6, The overlap between asthma and COPD has been recognized and its treatment delineated (ie, manage overlapping COPD and asthma as “asthma”). However, the combination can complicate diagnosis and treatment.6,
Although asthma and COPD share some degree of clinical overlap, they are frequently viewed as separate diseases. The most difficult group of patients in which to distinguish the 2 conditions are those who are current or former smokers with a 15- to 20-pack-year history. In smokers as young as 30 to 40 years, clinical and pathologic manifestations resembling early-stage COPD may be present, which is suggestive of a mixed or intermediary condition. Many older patients, either with recently diagnosed asthma or long-term asthma, have decreased lung function beyond that expected with aging and are at increased risk for fixed airflow obstruction compared with younger patients.11– Long-term follow-up studies show that up to 23% of patients with a history of asthma develop the COPD hallmark of fixed airflow obstruction.15,
Epidemiology
The rates for asthma prevalence have been consistently higher in children than in adults. The overall prevalence in patients younger than 18 years was 8.9% in the United States in 2005, which translates to approximately 6.5 million children. Asthma is more common in boys than in girls; estimates for the prevalence of asthma by sex in patients younger than 18 years in the United States in 2004 were approximately 10% for boys and 7% for girls. During adolescence, asthma incidence increases for girls and decreases for boys, resulting in a gender reversal in prevalence among adults. Estimates from 2004 show that approximately 5.1 million men (5%) and 9.2 million women (8%) had asthma.
In 2002, COPD was the fourth leading cause of death in the United States among people 65 years or older. An estimated 20% of smokers may eventually develop COPD. In an investigation of 169.3 million adults in the United States, obstructive lung disease was found to be present in 12.5% of smokers, 9.4% of former smokers, and 5.8% of people who had never smoked. Genetics also plays a role in the development of COPD. Currently, alpha 1-antitrypsin deficiency is the only genetic change that has been directly associated with COPD, but this disorder accounts for fewer than 5% of cases of COPD in the United States. Other genetic associations are likely to be discovered in the future that help explain why approximately 80% of long-term smokers do not develop COPD. Other causes of COPD include industrial or occupational exposure to air pollutants and exposure to wood or cooking fire smoke.
In 2006, an estimated 12.1 million adults in the United States were diagnosed with COPD. However, the overall prevalence of COPD has been estimated at 24 million, suggesting that many people with COPD remain unrecognized and undiagnosed. The prevalence of COPD is higher among whites than blacks and among women; rising rates of smoking have been accompanied by a rising prevalence of COPD – mortality from COPD is now higher for white women than for white men.
Although asthma is more prevalent than COPD, COPD imposes a heavier disease burden due to the greater number of hospitalizations, greater severity of exacerbations, uniform progression of the disease, and poorer overall prognosis. The estimated total cost of asthma in the United States in 2004 was $16.1 billion ($11.5 billion in direct medical expenditures, and $4.6 billion in indirect costs), whereas the total cost of COPD was more than twice as high at $37.2 billion ($20.9 billion in direct medical expenditures, and $16.3 billion in indirect costs).
Natural History
Asthma is most often associated with onset during childhood, prior atopic reactions, and a family history of atopy or asthma. The initial clinical presentation varies, most often with intermittent symptoms but sometimes with constant wheezing, cough, or shortness of breath. Wheezing on expiration is the classic symptom, but some patients present primarily with cough, especially at night. Symptoms typically increase with exposure to allergens and triggers, such as viral upper respiratory infections, pollen, dust mites, animal dander, environmental irritants (most commonly tobacco smoke), cold air, and perhaps physical exertion. In some cases, asthma symptoms diminish after childhood.
In contrast, COPD is essentially unknown in children and is rare in younger adults without a history of alpha 1-antitrypsin deficiency. After age 40, however, the prevalence of COPD increases substantially with aging, and the prevalence of patient-reported asthma declines slightly. In general, risk for COPD increases with pack-years of smoking or, less often, with ongoing occupational exposure to inhaled toxins or irritants. Although daily symptoms are present in only 27% of people with asthma, symptoms in COPD are more likely to be constant and progressive, reflecting the fact that airway obstruction in COPD is not due to the reversible airway constriction and inflammation of asthma but rather to structural changes and mechanical derangements with abnormal elastic recoil.
The nature and extent of the systematic components of COPD remain unclear. However, it is clear that COPD often coexists with cardiac disease, depression, and other chronic conditions. Some of the coexisting conditions may be unique disease processes secondary to long-term tobacco exposure or may be part of the same inflammatory process that affects the lungs in COPD.34– The importance of the coexisting conditions was highlighted recently in a review article in which lung cancer and cardiovascular disease were listed as the leading causes of death in patients with mild to moderate COPD.
Pathophysiology
Airway obstruction in asthma results from bronchial smooth muscle constriction, airway hyperreactivity to allergens and irritants, and inflammation accompanied by increased eosinophils and CD4+ lymphocytes. Both asthma and COPD are associated with inflammatory changes. However, the airway mucosa often appears markedly different between the 2 conditions, particularly in regard to the appearance of the surface epithelium and the reticular basement membrane (Figure).
Figure.
Open in a new tab
A, Airway mucosa (bronchial biopsy) from a patient with asthma, showing loss of surface epithelium and early basement membrane thickening. B, Airway mucosa (bronchial biopsy) from a heavy smoker with COPD with an FEV1 of 40% predicted, showing epithelial squamous metaplasia. Adapted with permission from Jeffery PK. Remodeling in asthma and chronic obstructive lung disease. Am J Respir Crit Care Med. 2001;164:S28-S38.
Airway obstruction in COPD is associated with cellular damage and mucus hypersecretion; inflammation becomes more prominent in exacerbations of severe disease. Cellular damage is a progressive process initiated by inflammatory changes induced by smoking and other environmental toxins to pulmonary tissues. Emphysema is characterized by loss of lung elastic recoil and loss of alveolar structure resulting from inflammatory cell-mediated damage to bronchioles, alveolar ducts, and alveoli. In chronic bronchitis, chronic cough and increased sputum production occur as a consequence of mucosal infiltration by inflammatory cells that lead to cellular damage to the airway mucosa. Most ominously, the cellular destruction and structural changes associated with COPD interfere with oxygenation and pulmonary circulation, which may cause an increased burden on the right side of the heart.
Clinical Assessment
The classic clinical presentation of asthma is a younger patient with recurrent episodes of wheezing and coughing that may be accompanied by chest tightening or breathlessness. Acute symptoms usually respond quickly to bronchodilators.1,29, In contrast, the classic presentation of COPD is an older smoker with progressively worsening shortness of breath and possible cough (with copious mucus in chronic bronchitis) that is often assumed to be a sign of aging and accompanied with decreasing physical activity; COPD may only be diagnosed after a moderate to severe exacerbation. In the classic presentation, reversing pulmonary obstruction after bronchodilator treatment does not return lung function to normal or nearly normal levels and is lower than that observed with asthma, although recent data show that considerable reversibility can be achieved by patients with moderate to severe COPD (stages II–IV). Furthermore, patients with COPD can experience day-to-day variability in both pre-43, and postbronchodilator43, forced expiratory volume in one second (FEV 1) reversibility.
When the clinical history of a patient who is suspected to have or is at high risk for obstructive airway disease is being taken, a proactive approach is warranted. Patients with mild symptoms, including those with early-stage COPD, may not volunteer information about symptoms that they do not consider serious and may ignore symptoms or regard them as a sign of aging. To facilitate differentiation of COPD from asthma, the patient history should include specific queries regarding the following:
Age at onset of symptoms – younger with asthma, older with COPD. Reporting of symptoms may require specific prompts, such as “What were you able to do 2 to 5 years ago that you cannot do now because of breathing problems?”
Any history of atopy – such conditions as allergy, eczema, and rhinitis are associated mainly with asthma; asthma is also suggested when symptoms are acutely triggered by exercise or exposure to cold air.
Characteristics of cough – more likely to be dry in asthma, productive in COPD.
History of smoking – pack-years likely to be lower in asthma than in COPD.
Symptom variability – daily changes in symptom frequency or intensity are common in asthma; COPD symptoms are less variable.
In addition, note that viral upper respiratory tract infection can result in exacerbations of either condition, and this factor is therefore of little help in distinguishing asthma from COPD. However, exacerbations associated with bacterial respiratory infection are more suggestive of COPD than asthma. The clinical assessment should also elicit evidence of comorbid conditions. For example, gastroesophageal reflux disease is present in patients with asthma at 3 times the rate of the general population, and the conditions exacerbate each other. Patients with COPD have an increased risk for right-side congestive heart failure, as previously mentioned, and are at increased risk for all types of cardiovascular disease; muscle wasting; weight loss; chronic hypoxia; and depression related to marked and progressive functional impairment, systemic effects of tobacco smoking, and perhaps to the systemic inflammation associated with COPD.47–
Lung Function Testing
Spirometry is the most commonly performed noninvasive test of lung function and is considered the most practical and reliable tool for establishing the presence and severity of obstructive airway diseases, including asthma and COPD.32, Standardized guidelines for obtaining accurate and interpretable spirometry tests are available (eg, the American Thoracic Society/European Respiratory Society Standardization of Lung Function Testing) and should be followed to ensure reproducible and clinically meaningful results. Commonly used spirometry measurements of relevance for the differentiation of asthma from COPD include the volume of air that can be forcibly exhaled in a single breath after a maximum inspiration (forced vital capacity [FVC]), the FEV 1 of this maneuver, and the ratio of these measurements (FEV 1/FVC). Much emphasis is placed on FEV 1 and FEV 1/FVC measures because abnormal levels of each (ie, FEV 1/FVC ratio <70% predicted is required for a diagnosis of COPD) are considered indicators of airflow obstruction.2, FEV 1 and FEV 1/FVC measurements must also be measured after 2 puffs of a short-acting bronchodilator to complete reversibility testing that is useful in differentiating between classical presentations of asthma (ie, often fully reversible airway obstruction) and COPD (ie, not fully reversible or irreversible airflow obstruction).2,Table 1 summarizes the criteria for interpretation of spirometry results in the diagnosis of these diseases on the basis of FEV 1 readings and as percentages of expected normal values for the FEV 1/FVC ratio.2, However, the frequent variability in reversibility over time43, and overlap of asthma and COPD (eg, fixed airflow obstruction in elderly patients with asthma11, and significant bronchodilator response in patients with COPD) limit the use of bronchodilator reversibility testing as the sole criterion for differentiating asthma from COPD. Although these measures are not perfect, they improve the ability to identify reversible and nonreversible airway obstruction compared with clinical impression alone.
Table 1.
Spirometry Essentials for Asthma and COPD
| | Adult Lung Function Parameters |
:---: |
| | |
| Condition and Severity | FEV 1 | FEV 1/FVC |
| Asthma |
| Intermittent | >80% predicted | Normal |
| Mild persistent | >80% predicted | Normal |
| Moderate persistent | 60%–79% predicted | Reduced ≤5% from normal |
| Severe persistent | <60% predicted | Reduced >5% from normal |
| COPD† |
| Mild | ≥80% predicted | <0.7 |
| Moderate | 50%–79% predicted | <0.7 |
| Severe | 30%–49% predicted | <0.7 |
| Very severe | <30% predicted or <50% predicted plus chronic respiratory failure | <0.7 |
Open in a new tab
COPD = chronic obstructive pulmonary disease; FEV 1=forced expiratory volume in 1 s; FVC=forced vital capacity
Normal FEV 1/FVC = 85% for patients aged 8 to 19 y; 80% for patients aged 20 to 39 y; 75% for patients aged 40 to 59 y; 70% for patients aged 60 to 80 y.
†
Postbronchodilator FEV 1 recommended for diagnosis and assessment.
Other less commonly used spirometry measures that may be helpful in differentiating between asthma and COPD include total lung capacity, residual volume, and functional residual capacity. These measures are reported to be lower in patients with asthma than in those with COPD,53, although none has been shown to be good predictors of asthma vs COPD. The diffusing capacity for carbon monoxide is another lung function measure that may assist in distinguishing asthma from COPD. This measure has been reported to be higher in patients with asthma than in those with COPD.53– Exhaled nitric oxide has also been shown to be a good predictor of asthma vs COPD in that levels greater than 16 parts per billion suggested a history of asthma, whereas exhaled nitric oxide levels below 16 ppb were suggestive of a history of COPD.Methacholine challenge testing may be used in some specialty settings but is seldom done in primary care. The appropriate use of bronchoprovocation testing, its exact meaning, and the best provocative agent to use remain controversial.56–
Hyperinflation is an elevation in resting functional residual capacity that results in dyspnea and limits exercise capacity.32, It is also associated with a reduction in inspiratory capacity. Hyperinflation is often observed in COPD patients at rest, with worsening during exercise or exacerbation. In patients with asthma, hyperinflation is generally restricted to periods of moderate to severe exacerbations, although a few patients may maintain elevated residual volumes at rest.
Other Diagnostic Modalities
When the diagnosis remains unclear after full clinical and lung function testing assessments, a number of other diagnostic modalities may be used. A chest radiograph can be performed and is often normal in asthma, whereas signs of hyperinflation, bronchial thickening, and increased basilar markings may be observed in COPD. Other useful diagnostic modalities include those based on differences in cellular inflammation profiles for asthma and COPD (Table 2). For example, eosinophil levels are higher and neutrophil levels are lower in bronchoalveolar lavage fluid and sputum samples from patients with asthma vs COPD. In some countries, such as Japan, as well as in some subspecialty offices, more advanced imaging studies are being used to diagnose COPD, especially in confusing or atypical cases.
Table 2.
Differences in Inflammation Profiles Between Asthma and COPD
| Asthma | COPD |
:---: |
| Cell types infiltrating airways |
| CD4+ lymphocytes | CD8+ lymphocytes |
| ↑ Ratio of activated CD4+/CD8+ | ↓ Ratio of activated CD4+/CD8+ |
| Eosinophils | Neutrophils |
| Mast cells | Macrophages |
| Neutrophils (severe asthma) | Eosinophils (exacerbation) |
| Upregulated cytokines/chemokines observed |
| IL-4, IL-5, IL-13 | IL-8, IL-1 |
| RANTES, eotaxins, MCP-1 | Leukotriene B4, interferon-γ |
Open in a new tab
COPD = chronic obstructive pulmonary disease; IL = interleukin; MCP-1 = monocyte chemoattractant protein-1; RANTES = regulated on activation, normal T-cell expressed and secreted
Adapted with permission from Mauad T, Dolhnikoff M. Pathologic similarities and differences between asthma and chronic obstructive pulmonary disease. Curr Opin Pulm Med. 2008;14:31–38.
Disease Management
The most important reason to accurately diagnose asthma vs COPD is probably because management approaches and goals for these conditions differ. However, some elements of management are similar – for example, both asthma and COPD involve lifestyle modification concomitant with pharmacotherapy tailored to the severity of the condition.2, Treatment of comorbid conditions is crucial in both disorders to differentiate symptoms and to ensure that appropriate treatment is being provided. Patient education also plays a key role in both conditions in optimizing compliance with lifestyle adjustment and pharmacotherapy.2,
Lifestyle Modification
Lifestyle modification begins with smoking cessation, regardless of the diagnosis. Asthma patients must be reminded that smoking exacerbates their condition. Continued smoking may accelerate loss of lung function in COPD patients, whereas smoking cessation can significantly reduce this decline. Thus, for the patient with obstructive airway disease, smoking cessation is a high priority.
Avoidance of environmental air pollution is helpful to minimize exacerbation risk in patients with asthma or COPD, whereas allergen avoidance is more effective for reducing exacerbation risk in patients with asthma. Although total avoidance may not be possible, any meaningful reduction in exposure is beneficial. Other avoidance measures, such as using high-efficiency particulate air filters in the home and changing the furnace/air conditioner filters every 3 months; cleaning the home thoroughly on a weekly basis; and remaining indoors when pollen or mold counts are highest, are recommended.
Another lifestyle modification that may be considered is weight normalization, which may benefit acute or chronic obstructive airway disease and any comorbid cardiovascular disease. Furthermore, exercise as part of a pulmonary rehabilitation program may be important for maintaining and improving health-related quality of life in patients with COPD.66– Aerobic exercises, such as walking, swimming, or bicycling, are important and have been shown to improve cognition and reduce depression. In addition, low-intensity weight training may decrease or delay muscle wasting and improve functional capacity.
Pulmonary rehabilitation that includes disease-specific education, exercise training, nutritional education, and social support can improve most aspects of COPD and its comorbid conditions, such as depression.66– This is the most comprehensive type of COPD therapy and addresses all aspects of the burden on individuals and families.
Pharmacotherapy
Because asthma is a chronic, eosinophil-based inflammatory disorder of the airways, reduction of airway inflammation is a fundamental goal of asthma pharmacotherapy. The types of pharmacotherapy and the doses used are based on assessment of asthma control as measured by symptom burden and effect of symptoms on daily life and numbers and severity of exacerbations and can include spirometry assessment.2, Inhaled corticosteroids (ICSs) reduce airway inflammation and are the cornerstone of pharmacotherapy for treatment of persistent asthma. These drugs do not elicit immediate relief during episodes of acute bronchospasm – their role is mainly to control symptoms by minimizing the inflammatory component of asthma on an ongoing basis. ICS therapy should be titrated up or down based on assessment of control.2, Thus, short-acting bronchodilators, such as short-acting beta 2-adrenergic agonists, are required in conjunction with ICS therapy to treat breakthrough symptoms. For patients whose asthma is not well controlled on ICS therapy alone, adding long-acting beta 2-adrenergic agonists may be considered. For patients aged 12 years or older whose asthma is not well controlled with low-dose ICS therapy, asthma treatment guidelines recommend adding a long-acting beta 2-adrenergic agonist to low-dose ICS treatment or increasing ICSs to a medium dose. Other types of anti-inflammatory therapy include leukotriene inhibitors or modifiers, cromoglycates, and theophylline.2, In both the National Heart Lung and Blood Institute/National Asthma Education and Prevention Program and Global Initiative for Asthma guidelines, ICSs are listed as the preferred therapy for asthma inflammation but other considerations, such as allergic disease, adherence, and acceptability, should be considered when individualizing therapy.2, Anti-IgE therapy is available in an injection form and should be considered in patients whose asthma is difficult to control and who have a confirmed allergic basis for their asthma. Few primary care physicians initiate anti-IgE therapy.2,71– Oral steroids are a major part of therapy for moderate to severe asthma exacerbations and are usually given in a short 5- to 10-day burst of the equivalent of 10 mg of prednisone daily. No tapering is required for this dosing, and no advantage has been found for use of intramuscular or intravenous corticosteroids in persons not experiencing respiratory failure.
Although anticholinergic drugs can be used as alternative reliever therapy in asthma patients who cannot tolerate short-acting beta 2-agonists, bronchodilatory response in asthma patients after anticholinergic treatment has been shown to be more variable than with beta 2-adrenergic agonist treatment. Furthermore, no long-acting anticholinergic medications are currently indicated for use in the United States for the treatment of asthma.
Treatment of COPD with bronchodilator medications is fundamental for symptomatic management of the disease. Selective inhaled beta 2-adrenergic agonists relax bronchial smooth muscle and may help improve mucociliary activity. Short-acting bronchodilators (eg, short-acting beta 2-agonists and short-acting anticholinergics) are recommended as short-term therapy for patients with COPD at any level of severity (stages I-IV). Long-acting bronchodilators (eg, anticholinergics, beta 2-adrenergic agonists, or methylxanthines) are more effective than short-acting bronchodilators,75– and COPD guidelines recommend treatment with one or more of these medications to control symptoms in patients with moderate to severe COPD (stages II-IV) who do not respond to as-needed short-acting beta 2-agonist therapy. Because COPD is typically associated with neutrophilic vs eosinophilic inflammation (the latter is more commonly observed in asthma), ICS treatment is not a first-line therapy for COPD and is reserved for use in combination with a long-acting bronchodilator in patients with severe to very severe COPD (stages III-IV) who have severe or frequent exacerbations. In COPD therapy, ICS is not titrated on the basis of control but rather on exacerbation rates. Usually the dose is only titrated upward or may be discontinued if it does not reduce exacerbation rates.5,8, Systematic corticosteroids are used to treat COPD exacerbations in short bursts of oral steroids usually lasting 10 days and in doses equivalent to 10 to 20 mg of prednisone per day.5, There is no role for daily oral steroid maintenance in COPD with the exception of very severe and late-stage disease in which any and all types of therapy may be tried.5,
Controlled oxygen support may be used in advanced cases of COPD.32, Current recommendations are for oxygen supplementation when oxygen saturation (SaO 2) is 88% or lower. This therapy is also recommended for use in asthma but only for management of exacerbations to maintain SaO 2 at 90% or greater. Recommendations for earlier use of oxygen supplementation cannot be made because it has not been sufficiently studied.
Conclusions
Although asthma and COPD often overlap in clinical features, they are distinct conditions whose differences influence both management and prognosis. Concurrent assessment of clinical history, current clinical symptoms, lung function tests, and other diagnostic modalities (eg, radiographic imagery and cellular markers of inflammation) can help facilitate an accurate diagnosis of asthma vs COPD. Establishing the correct diagnosis is the essential first step in developing an individualized disease management plan that will optimize quality of life and physical functioning for patients with asthma or COPD.
Acknowledgments
The author acknowledges Tracy J. Wetter, PhD, and Rhonda L. Croxton, PhD, from Complete Healthcare Communications, Inc., for providing medical writing support funded by AstraZeneca LP.
Footnotes
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References
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14326 | https://www.youtube.com/watch?v=47uESIjJSTg | The 4 Right Hand Rules of Electromagnetism ("Easiest explanation on entire YouTube!")
Iain Explains Signals, Systems, and Digital Comms
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Posted: 20 Dec 2021
Explains the 4 different "Right Hand Rules" of Electromagnetism, showing when they apply and what they tell us.
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Transcript:
so let's look at the four right hand rules of electromagnetism and it can be confusing because if you search for right hand rules you'll often just be told one of them now let's look at what the four of them are and how they relate so the first rule relates to having a current in a wire so here we have a wire and there's a current in that wire what that current does is it generates a magnetic field around the wire so if you put your thumb in the direction of the current then the curl your fingers around that will be the direction of the magnetic field that results from having a current in the wire okay so that one's quite straightforward what about the second rule so this also relates to having a wire with a current in it this rule though relates to what happens when you have another magnetic field acting on that wire so that wire will have its own field just like this one did if the current's in this direction its own field will be curled around just like it is on the left but let's imagine there's another magnetic field also acting on this wire and that's when this rule happens so this rule is with an open palm you put again your thumb in the direction of the current then now you face your open fingers in the direction of the magnetic field and your palm shows you the direction of the force that's acting on the positive particles in the positive charges in this wire so here in this case the current's going up the magnetic field let's say it's a magnetic field from this wire over here so this wire over here the magnetic field acting on this wire will be into the page because we've got this wire going up curling around and now coming down into the page on this second wire so this second wire if we use this rule now use a flat palm we put our thumb facing up and our fingers flat down into the page then it shows a force to the left on the positive particles now the positive particles in a wire are in the lattice structure the electrons are the negative particles they're flowing freely but the positive particles are trapped in the structure and the force will cause them then to move to the left so this wire will bend to the left towards this wire and this is the right hand rule that tells you about that of course if the current in this wire was in the other direction then you would have your thumb facing downwards still you would have the magnetic field into the page because that's coming from this wire over here and then the bending the force would be towards the right and this wire would bend out of course this wire also has a field acted on it from this wire here in exactly the same way so this wire here if we apply this rule as we said the the field from this wire is going around this way so on it's acting on this wire will be coming out of the page on this wire so let's apply this rule here then to find out the force so the field from this wire is acting on this wire we put our thumb in the direction of the current which is up the field from this wire on this wire is up now and we have our flat hand and the force is in so both of these wires with the way it's shown here will bend in so these are these first two right hand rules what about a third right hand rule well this relates to coils and it's you can see it's a curl rule again but the directions are flipped so instead of having the current in the direction of the thumb now we have the field in the direction of the thumb so that can be confusing but i'll show that actually they're equivalent but let's look at this for a moment if the if as shown here the current is going around this way and if you put your fingers in the direction of the current which is curling around this way then your thumb shows the direction of the magnetic field that results in the middle of the coil and that's a very important thing which is often overlooked that this field here is in the middle of the coil of course the field lines go up and then they come out on the outside and then they meet back up the bottom these magnetic field lines so on the outside it's in the other direction but this shows the direction of the magnetic field in the center of the coil now how is it that this one and this are the same and they look opposite well let's apply this rules to this scenario and this rule we have the current the thumb in the direction of the current so this is in this way the current is coming let's look at it specifically for the wire so on this part of the wire here the current is going from left to right so then our curled magnetic field is going this way in here and you can see hope you can visualize it on the front of this tube here the field will be going up behind the wire and then down and over in front of the wire and up behind the wire is in the center of the coil and as you move your hand around following the direction of the wire moving your thumb around you always have the maintained curls of your fingers so always on the inside of it's facing up and on the outside it's facing down and that's what this shows here this is the magnetic field on the inside of the coil so these two actually make sense and they match each other up of course when you look at them you can be confused but if you understand that this is the current in the wire of a specific wire then you can see that they do in fact match up now what's common to all of these three rules one thing is well the common thing is that they all have a current source so in each of these cases there's a current source supplying current to the wire and we are learning about either the induced magnetic field or the induced force so what about the fourth rule and that applies when there is a magnetic source and the current is the thing that's induced and so here's the scenario here you've got a magnet north magnet and a south magnet they are going to have magnetic field lines between those two magnets and here i've drawn a wire in between those two magnets now if we move that wire vertically up let's say out of the page vertically up then you can use this right hand rule and the thumb is in the direction of the motion the fingers open hand are in the direction of the magnetic field and then it this palm shows you the direction of the current in that wire if it's in a loop if it's not in a loop then there will be a voltage difference between the ends of that wire that would then result in a current if you put the wire in a loop so here's here's a wire let's look at it for this scenario here the field lines are from left to right the motion is vertically up and so the current is going in this direction here through the wire of course if you reverse the direction of the the magnets or if you reverse the direction of the motion and put the motion going down then if the motion is going down then your thumb is facing down and it's hard to show in the video but your thumb is facing down and the current would be in the other direction and this is often shown when we're looking at motors and generators where the electrical wire is in a loop that is rotating vertically and and up and down through a loop rotating in this magnetic field so this is the fourth uh right hand rule so i think hopefully if you see it this way it can categorize and can understand clearly three of them reply when there's a current source in the wire one of them applies when there's a magnetic source and the current is induced so if this video has helped you to understand the topic please give it a thumbs up it helps others to find the video subscribe to the channel for more videos and check out the webpage in the description below where you'll find a full categorized listing of all the videos on the channel |
14327 | https://homepages.gac.edu/~sskulrat/Courses/2013S-256/notes/2d.pdf | MCS-256: Discrete Calculus and Probability Handout #2d San Skulrattanakulchai Gustavus Adolphus College March 6, 2013 Sum Calculus Indefinite Sum (Antidifference) In integral calculus we learned about the indefinite integrals and the definite integrals of real-valued functions. The indefinite integral or the antiderivative of a function f is defined to be any function F such that DF = f. Antiderivatives are not unique. In fact, if F is an antiderivative of f, then any function G defined by G(x) = F(x) + C for all x, where C is any real constant, is also an antiderivative of f. So when we write Z f(x) dx = F(x) + C, what we actually mean is that the set of all antiderivatives of f is {F + C : C is a constant function} where F is some fixed antiderivative of f.
Let’s try to define for discrete calculus a concept parallel to the antiderivative. For any real-valued function f, define an indefinite sum or antidifference of f to be any function F such that ∆F = f. We also write F = P f to mean F is an indefinite sum of f.
Lemma. F and G are antidifferences of f if and only if G(x) = F(x) + C(x) for all x, where C is some function satisfying C(x + 1) = C(x) for all x, i.e., C is a periodic constant function with period 1.
2 MCS-256: Handout #2d Proof. . . .
In abused notation, one writes X f(x) δx = F(x) + C(x).
Note that in this notation, sigma does not mean taking the sum over a set of numbers like in the sigma notation we have been using. Following tradition, we’ll be using this abused notation, and omitting the periodic constant function, for the rest of this section.
The most basic formulas for antidifferences are Theorem.
X [f(x) + g(x)] δx = X f(x) δx + X g(x) δx X [f(x) −g(x)] δx = X f(x) δx − X g(x) δx X cf(x) δx = c X f(x) δx.
Proof. . . .
How do we find an antidifference of a given function f? In the continuous case, we do it by “inverting the theorems for derivatives”, e.g., from the theorem Dxn = nxn−1, we get R xn dx = xn+1 n+1 + C. We’ll do the same thing in the discrete case.
Here is a collection of theorems we get by inverting the theorems for antidifferences we already knew.
MCS-256: Handout #2d 3 Theorem.
X a δx = ax (a ∈R) X xm δx = xm+1 m + 1 (m ∈Z, m ̸= −1) X (ax + b)m δx = (ax + b)m+1 a(m + 1) (a ∈R, m ∈Z, a ̸= 0, m ̸= −1) X ax δx = ax a −1 (a ∈R, a ̸= 1) Proof. . . .
Notice that we still can’t do P x−1 δx = P 1 x+1 δx yet. The situation is similar to the con-tinuous calculus where inverting the formula for Dxm does not give us an antiderivative for 1 x. We had to find out as a separate fact that Z 1 x dx = ln |x|.
What about the function in discrete calculus that is similar to ln x? It turns out that the Harmonic Numbers Hn defined by Hn = 1 + 1 2 + 1 3 + · · · + 1 n is the discrete parallel to the natural log function.
Theorem.
X 1 x + 1 δx = Hx.
Proof. . . .
We also have the following theorem similar to “integration by parts” of continuous cal-culus.
4 MCS-256: Handout #2d Theorem (Summation by parts).
X f∆g = fg − X Eg∆f Proof. . . .
Definite Sum and the Fundamental Theorem The definite integral of a function f from the lower bound a to the upper bound b is written Z b a f(x) dx and it denotes the area bounded by the curves x = a, y = f(x), x = b, and y = 0.
(The precise definition is as the limit of a certain sum.) The Fundamental Theorem of Calculus states that if f is continuous on [a, b] and F is an antiderivative of f, then Z b a f(x) dx = F(x)|b a = F(b) −F(a).
Let’s define the definite sum of the function f from the lower bound a to the upper bound b, written Pb a f(x) δx, to be b X a f(x) δx := b−1 X x=a f(x).
Make sure you understand the difference in meanings of the two sigmas used in the above definition.
Theorem (Fundamental Theorem of Discrete Calculus). If F is the antidifference of f then b X a f(x) δx = F(x)|b a = F(b) −F(a).
Proof. . . .
MCS-256: Handout #2d 5 Exercises 1. Find the following antidifferences.
(a) P[x3 + 3x2 −5x1 + 4] δx (b) P(3 + 2x)5 δx (c) P(4 −3x)−3 δx (d) P x(x + 3)(x + 6) δx (e) P x−3(2x + 1) δx (f) P xnax δx (g) P xna−x δx for n = 1, 2, 3 2. Using sum calculus, prove that (a) a + (a + d) + (a + 2d) + · · · + [a + (n −1)d] = an + dn(n −1) 2 (b) a + ar + ar2 + · · · + arn−1 = a · rn −1 r −1 if r ̸= 1 (c) Pn k=1 k = n(n + 1) 2 (d) Pn k=1 k2 = n(n + 1)(2n + 1) 6 3. Evaluate the following finite sums: (a) 2 · 4 · 6 + 4 · 6 · 8 + · · · + 16 · 18 · 20.
(b) 1 1 · 2 + 1 2 · 3 + · · · + 1 20 · 21 (c) 1 1 · 2 · 3 + 1 2 · 3 · 4 + · · · + 1 20 · 21 · 22 (d) 2 · 5 · 8 + 5 · 8 · 11 + 8 · 11 · 14 + · · · + 20 · 23 · 26.
4. Evaluate the following finite sums: (a) Pn k=0 k2k (b) Pn k=0 k32k |
14328 | https://study.com/learn/lesson/least-common-denominator-definition-examples.html | Least Common Denominator | LCD Definition, Calculation & Examples - Lesson | Study.com
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Science Courses / General Studies Math: Help & Review Course
Least Common Denominator | LCD Definition, Calculation & Examples
Lesson
Transcript
Yuanxin (Amy) Yang Alcocer, Joshua White
Author Yuanxin (Amy) Yang Alcocer
Amy has a master's degree in secondary education and has been teaching math for over 9 years. Amy has worked with students at all levels from those with special needs to those that are gifted.
View bio
Instructor Joshua White
Josh has worked as a high school math teacher for seven years and has undergraduate degrees in Applied Mathematics (BS) & Economics/Physics (BA).
View bio
Understand what the least common denominator (LCD) is, what a common denominator is, & how to find LCD. Using examples, solve for the lowest common denominator. Updated: 11/21/2023
Table of Contents
What is a Common Denominator
Least Common Denominator Definition
Lesson Summary
Show
Frequently Asked Questions
What is the LCD of 9 and 12?
The term LCM is used when comparing two numbers. The term LCD is used when comparing two fractions.
Using the steps, the multiples of 9 and 12 are the following.
9: 9, 18, 27, 36, 45, ...
12: 12, 24, 36, 48
The lowest multiple they have in common is the 36. So 36 is the LCM of 9 and 12.
How do you find the LCD or LCM?
To find the LCD for fractions or the LCM between numbers, follow these steps.
Step 1: Find several multiples of each denominator.
Step 2: Identify the smallest multiple all the denominators have in common.
These steps work for fractions that use numbers and not for rational expressions.
What is the LCD of 4 and 8?
When referring to numbers and not fractions, the term LCM is used instead of LCD.
Using the steps, the multiples of 4 and 8 are the following.
4: 4, 8, 12, ...
8: 8, 16, 24, ...
The smallest multiples they have in common is 8. So 8 is the LCM of 4 and 8.
How do you find the LCD?
To find the LCD, follow these steps.
Step 1: Find several multiples of each denominator.
Step 2: Identify the smallest multiple all the denominators have in common.
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Table of Contents
What is a Common Denominator
Least Common Denominator Definition
Lesson Summary
Show
What is a Common Denominator ----------------------------
Fractions have numerators and denominators. The numerator is the top part of the fraction, and the denominator is the bottom part of the fraction.
Fractions have numerators and denominators
When adding or subtracting fractions, common denominators are needed. What is a common denominator? A common denominator means two or more fractions have the same denominator. Before two or more fractions can be added or subtracted, they all need common denominators.
For example, 1 4 and 1 12 cannot be added together unless both fractions possess a common denominator. One method of finding a common denominator is to multiply the two denominators together.
4∗12=48
To get to 48 in the denominator, the first fraction needs to be multiplied by a 12 12.
1 4×12 12=12 48
The second fraction needs to be multiplied by a 4 4.
1 12×4 4=4 48
The addition becomes this.
12 48+4 48=16 48
This problem is not considered complete until it needs to be simplified. The resulting fraction 16 48 can be simplified because both the 16 and the 48 are divisible by 16. Therefore, the fraction simplifies like this.
16 48=1 3
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0:03 LCD: Analogy & Definition
1:09 How to Find the LCD
3:16 Sample Problem #1
3:46 Sample Problem #2
4:16 Sample Problem #3
5:43 Lesson Summary
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Least Common Denominator Definition -----------------------------------
While the above produces a common denominator, it does not always produce the most easy-to-manage numbers. Another method uses the least common denominator as the common denominator. The least common denominator (LCD) definition is the smallest common denominator two or more fractions have. When the common denominator also happens to be the smallest common denominator, they are the same. Otherwise, the least common denominator will always be smaller than a common denominator.
Finding the least common denominator involves finding the least common multiple (LCM), the first and smallest multiple, of the denominators involved in the problem. For example, the problem of adding the fractions 1 4 and 1 12, has the least common denominator of 12. Using the least common denominator to add these fractions produces the following.
1 4+1 12=3 12+1 12=4 12=1 3
When the least common denominator is used, the numbers are smaller and more manageable.
How to Find the LCD
The steps for how to find the LCD are the following.
Step 1: Find several multiples of each denominator.
Step 2: Identify the lowest multiple that the denominators have in common.
The term LCD is used when referring to the denominators in fractions. The term LCM is used when referring to just the numbers. When looking for a denominator, use the term LCD. When looking for multiples, use the term LCM.
Using the addition of 1 4 and 1 12 as an example, finding the LCD for these two fractions looks like this.
Step 1: Find several multiples of each denominator.
Multiples of 4 are 4, 8, 12, 16, ....
Multiples of 12 are 12, 24,...
Finding the LCD involves finding the LCM
Step 2: Determine the smallest multiple all the denominators have in common.
The smallest of these that both have in common is the 12. Therefore, 12 is the LCD for the two fractions and the LCM for the numbers 4 and 12.
Example 1: What is the Least Common Denominator for 1 3 and 7 15?
Let's practice using the method to find the least common denominator for the fractions 1 3 and 7 15 and then adding them up.
Step 1: Find several multiples of each denominator.
The first step involves finding several multiples of each denominator.
The multiples of 3: 3, 6, 9, 12, 15, 18...
The multiples of 15: 15, 30, 45...
15 is the LCM making it the LCD
Step 2: Identify the smallest multiple all the denominators have in common.
The LCM they have in common is that 15. So that is the LCD.
Changing the fractions, so both fractions have 15 as the denominator produces the following.
1 3+7 15=5 15+7 15
With an LCD, the fractions can now be added.
5 15+7 15=12 15
The 12 15 cannot be simplified any further, so the sum of 1 3 and 7 15 is 12 15.
Example 2: Least Common Denominator for Fractions
Let's try a different problem.
3 15+6 9
Following the steps, this is what happens.
Step 1: Find several multiples of each denominator.
Finding multiples of each denominator gives this.
| 15 | 9 |
--- |
| 30 | 18 |
| 45 | 27 |
| 60 | 36 |
| 75 | 45 |
Step 2: Identify the smallest multiple all the denominators have in common.
The smallest multiple of 15 and 9 is 45. So that makes 45 the LCM of 15 and 9 and the LCD of the fractions 3/15 and 6/9.
Converting the fractions to the LCD and then proceeding with the addition gives this.
3 15+6 9=9 45+30 45=39 45=13 15
This problem required simplification to get to the 13 15. By using the LCD, the addition is more manageable. If the other method was used, the common denominator could have been 15 9 = 135. The 45 is easier to use than the 135.
Example 3: Lowest Common Denominator
Here's one more example to find the lowest common denominator and then the sum.
3 x+6 x 2 y
This problem includes rational expressions, fractions that also have polynomials in either the numerator or denominator or both. To sum up these fractions, the process is a bit different. The least common denominator is still desired. To find the LCD of rational expressions, each variable and coefficient is examined separately. If there are coefficients, then the LCM of the coefficients are found. For the variables, the variable with the highest exponent is kept.
Comparing the denominators here, the x and the x 2 have two variables with 1 as the coefficient. Since both coefficients are 1, the LCM of 1 and 1 is 1. Now the variables are compared one by one. The x has an x in the first fraction and an x 2 in the second fraction. The x 2 is the highest, so that is kept in the LCD. Next, there is no y in the first fraction, but there is one y in the second fraction, so a y is kept. This produces an LCD of x 2
Changing the fractions to the LCD gives the following result.
3 x+6 x 2 y=3 x y x 2 y+6 x 2 y
An xy is multiplied by the first fraction because the denominator needs another x and another y to turn it into the LCD. The second fraction already has the needed LCD. Going ahead with the addition produces this result.
3 x y x 2 y+6 x 2 y=(3 x y+6)x 2 y
It is perfectly fine to leave the answer in rational expression form with a polynomial in the numerator or denominator or both when working with rational expressions.
Lesson Summary
In review, a common denominator means two or more fractions have the same denominator. The least common denominator (LCD) definition is the smallest common denominator two or more fractions have. The least common denominator (LCD) definition is the smallest common denominator two or more fractions have. Rational expressions are fractions that also include polynomials in either the numerator or denominator or both.
The steps for finding the LCD for fractions with numbers are Step 1) Find several multiples of each denominator, and Step 2) Identify the smallest multiple all the denominators have in common. If the fractions are rational expressions, then the LCD is found by keeping the variables with the largest exponent and finding the LCM of the coefficients of the denominators.
To unlock this lesson you must be a Study.com Member.
Create your account
Least Common Denominator Definition -----------------------------------
While the above produces a common denominator, it does not always produce the most easy-to-manage numbers. Another method uses the least common denominator as the common denominator. The least common denominator (LCD) definition is the smallest common denominator two or more fractions have. When the common denominator also happens to be the smallest common denominator, they are the same. Otherwise, the least common denominator will always be smaller than a common denominator.
Finding the least common denominator involves finding the least common multiple (LCM), the first and smallest multiple, of the denominators involved in the problem. For example, the problem of adding the fractions 1 4 and 1 12, has the least common denominator of 12. Using the least common denominator to add these fractions produces the following.
1 4+1 12=3 12+1 12=4 12=1 3
When the least common denominator is used, the numbers are smaller and more manageable.
How to Find the LCD
The steps for how to find the LCD are the following.
Step 1: Find several multiples of each denominator.
Step 2: Identify the lowest multiple that the denominators have in common.
The term LCD is used when referring to the denominators in fractions. The term LCM is used when referring to just the numbers. When looking for a denominator, use the term LCD. When looking for multiples, use the term LCM.
Using the addition of 1 4 and 1 12 as an example, finding the LCD for these two fractions looks like this.
Step 1: Find several multiples of each denominator.
Multiples of 4 are 4, 8, 12, 16, ....
Multiples of 12 are 12, 24,...
Finding the LCD involves finding the LCM
Step 2: Determine the smallest multiple all the denominators have in common.
The smallest of these that both have in common is the 12. Therefore, 12 is the LCD for the two fractions and the LCM for the numbers 4 and 12.
Example 1: What is the Least Common Denominator for 1 3 and 7 15?
Let's practice using the method to find the least common denominator for the fractions 1 3 and 7 15 and then adding them up.
Step 1: Find several multiples of each denominator.
The first step involves finding several multiples of each denominator.
The multiples of 3: 3, 6, 9, 12, 15, 18...
The multiples of 15: 15, 30, 45...
15 is the LCM making it the LCD
Step 2: Identify the smallest multiple all the denominators have in common.
The LCM they have in common is that 15. So that is the LCD.
Changing the fractions, so both fractions have 15 as the denominator produces the following.
1 3+7 15=5 15+7 15
With an LCD, the fractions can now be added.
5 15+7 15=12 15
The 12 15 cannot be simplified any further, so the sum of 1 3 and 7 15 is 12 15.
Example 2: Least Common Denominator for Fractions
Let's try a different problem.
3 15+6 9
Following the steps, this is what happens.
Step 1: Find several multiples of each denominator.
Finding multiples of each denominator gives this.
| 15 | 9 |
--- |
| 30 | 18 |
| 45 | 27 |
| 60 | 36 |
| 75 | 45 |
Step 2: Identify the smallest multiple all the denominators have in common.
The smallest multiple of 15 and 9 is 45. So that makes 45 the LCM of 15 and 9 and the LCD of the fractions 3/15 and 6/9.
Converting the fractions to the LCD and then proceeding with the addition gives this.
3 15+6 9=9 45+30 45=39 45=13 15
This problem required simplification to get to the 13 15. By using the LCD, the addition is more manageable. If the other method was used, the common denominator could have been 15 9 = 135. The 45 is easier to use than the 135.
Example 3: Lowest Common Denominator
Here's one more example to find the lowest common denominator and then the sum.
3 x+6 x 2 y
This problem includes rational expressions, fractions that also have polynomials in either the numerator or denominator or both. To sum up these fractions, the process is a bit different. The least common denominator is still desired. To find the LCD of rational expressions, each variable and coefficient is examined separately. If there are coefficients, then the LCM of the coefficients are found. For the variables, the variable with the highest exponent is kept.
Comparing the denominators here, the x and the x 2 have two variables with 1 as the coefficient. Since both coefficients are 1, the LCM of 1 and 1 is 1. Now the variables are compared one by one. The x has an x in the first fraction and an x 2 in the second fraction. The x 2 is the highest, so that is kept in the LCD. Next, there is no y in the first fraction, but there is one y in the second fraction, so a y is kept. This produces an LCD of x 2
Changing the fractions to the LCD gives the following result.
3 x+6 x 2 y=3 x y x 2 y+6 x 2 y
An xy is multiplied by the first fraction because the denominator needs another x and another y to turn it into the LCD. The second fraction already has the needed LCD. Going ahead with the addition produces this result.
3 x y x 2 y+6 x 2 y=(3 x y+6)x 2 y
It is perfectly fine to leave the answer in rational expression form with a polynomial in the numerator or denominator or both when working with rational expressions.
Lesson Summary
In review, a common denominator means two or more fractions have the same denominator. The least common denominator (LCD) definition is the smallest common denominator two or more fractions have. The least common denominator (LCD) definition is the smallest common denominator two or more fractions have. Rational expressions are fractions that also include polynomials in either the numerator or denominator or both.
The steps for finding the LCD for fractions with numbers are Step 1) Find several multiples of each denominator, and Step 2) Identify the smallest multiple all the denominators have in common. If the fractions are rational expressions, then the LCD is found by keeping the variables with the largest exponent and finding the LCM of the coefficients of the denominators.
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Video Transcript
LCD: Analogy & Definition
Imagine that you and a group of friends are ordering some pizzas for dinner. However, each person wants a different topping. For example, if five people want pepperoni, you'll probably need to order more than one pepperoni pizza to make sure that each one has enough to eat. But if only one person wants sausage, one pizza with that topping will probably be sufficient.
The process you just went through, determining how much of an item from a group of items is needed, is similar to finding the least common denominator among a group of fractions in math. Let's see how.
The least common denominator (LCD), also known as the lowest common denominator, is the least common multiple for all the denominators in a group of fractions or rational expressions. In a group of numbers, the least common multiple is the smallest number that will divide evenly into all of the numbers in the group. Rational expressions are just fractions that have variables in their denominators. For example, the pizza order you placed was similar to an LCD in that the order was the smallest number of pizzas that allowed each person to have a sufficient amount of pizza with the toppings he or she desired.
The main reason to find the LCD in math is when you have a problem where you need to add or subtract fractions or rational expressions. Since the numerators of fractions can only be combined when the fractions have identical denominators, finding the LCD is something that needs to be done quite frequently.
How to Find the LCD
To find the LCD for two or more fractions, first determine the factors for each denominator by identifying the prime factors. For example, if a fraction has a denominator of 60, you might first factor it as 6 10. Here, 2 time 3 equals 6 and 2 time 5 equals 10, so 60 = 2 2 3 5.
Repeat this same process for all the denominators in your problem until you have the prime factors for each one. Then, find the LCD by multiplying the maximum number of each prime factor together.
For example, let's say you have 60 and 42 as the two denominators. You already know that 60 = 2 2 3 5. Factoring 42 gives you 42 = 2 3 7. Now, let's count up the factors. The number 2 appears at most twice in one of the factorizations, while 3, 5, and 7 appear just once in each of the factorizations. Therefore, the LCD will be 2 2 3 5 7 or 420.
There are a couple of important things to note here. The first is that you only need two 2s, not three, because two is the maximum number of times that a 2 appears in either of the factorizations. The second thing to note is that you could have just multiplied to get the common denominator: 60 42 = 2520. However, 2520 is not the least or smallest common denominator.
Although it's mathematically sound to use 2520 as the common denominator instead of 420, it will make the numbers in both of the numerators larger than they need to be and most likely require spending extra time simplifying the final answer for accuracy. With that in mind, let's take a look at some more examples.
Sample Problem #1
Find the LCD for 2/15 and 3/8.
Here, the two denominators are 15 and 8. 15 = 3 5 while 8 = 2 2 2. To find the LCD for 8 and 15, you will need three 2s, one 3, and one 5. Multiplying these numbers together gives 2 2 2 3 5 or 120. So, the LCD for 2/15 and 3/8 is 120.
Sample Problem #2
Find the LCD for 10/21, 1/4, and 5/12.
Here, the three common denominators are 21, 4, and 12. Let's factor them:
21 = 3 7
4 = 2 2
12 = 2 2 3
To determine the LCD, you'll need two 2s, one 3, and one 7. So, 2 2 3 7 = 84, which is the LCD for 10/21, 1/4, and 5/12.
Sample Problem #3
Find the LCD for the following expression:
These are not fractions but rather two rational expressions since each one of the denominators has both numbers and variables. However, the process to find the LCD is similar. If you factor the first denominator, x^2y^3z, you should get x x y y y z. And, if you factor the second denominator, you should get x y y y y z. Following the same process used in the previous two sample problems, you can determine the number of each variable needed in the LCD. For example, you'll need two x s, four y s, and one z. Therefore, the LCD will be x^2y^4z.
Note that when finding the LCD in problems with variables, you can use a shortcut. Rather than factoring each variable, take the highest power of each variable from among all the denominators to determine the LCD. For example, in this problem, the highest power of x is 2 (in the first expression), the highest power of y is 4 (in the second expression), and the highest power of z is 1 (the same in both expressions). This is an exact match for the LCD: x^2y^4z.
Lesson Summary
Finding the least common denominator (LCD), or the least common multiple for all the denominators in a group of fractions or rational expressions, is not all that different from compiling a large pizza order.
When working with fractions, use factoring to determine the prime factors for each denominator, and then multiply them based on the maximum number of times they appear. When working with variables, multiply the highest power of the individual variables together to find the LCD. Being able to determine the LCD among a group of fractions or expressions is an extremely important concept in math, since it is often required in order to add or subtract fractions and rational expressions.
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14329 | https://www.omnicalculator.com/statistics/99-confidence-interval | Last updated:
99% Confidence Interval Calculator
If you've got a data set and want to perform statistical calculations on it, our 99% confidence interval calculator is a good place to start. We've got you covered — not only if you want to know how to find this 99% confidence interval, but also to learn:
What is Z-score;
What does confidence interval even mean; and
How to find a margin of error for a 99% confidence interval.
What is a confidence interval?
A confidence interval is the range of values you expect your parameter to fall in if you repeat a test multiple times. Let's see an example that puts confidence intervals into real life.
Becky sells homemade muffins, and she wants to check the average weight of her baked goods. She found that 99% of her muffins weigh between 121 and 139 grams (4.27–4.9 oz), while one muffin came out of the oven at a whopping 160 grams (5.64 oz) — much bigger than expected!
The 99% confidence interval of Becky's muffins' weights is the range of 121 to 139 g. And so, when selling muffins, she can be 99% sure that any muffin she baked weighs between 121 and 139 g. But 1% of the time, she might accidentally produce a chonky muffin (or a tiny one!)
How do I find a 99% confidence interval?
Estimating confidence intervals is a bit of work, but we'll walk you through the process step by step.
First, you need to know:
n — the size of your sample (the number of measurements taken);
μ — the mean (average) of the measurements; and
σ — the standard deviation of the measurements.
Now, you can calculate standard error and margin of error with the formulas:
SE=nσME=SE×Z(0.99)
where:
SE — the standard error;
ME — the margin of error;
Z(0.99) — the z-score corresponding to the chosen confidence level (which you'll find in statistical tables).
Add and subtract the margin of error value from the mean to obtain your confidence interval. It is the range between these lower and upper bounds.
upper bound=μ+MElower bound=μ−ME
How to use 99% confidence interval calculator
Now that you know how complicated it is to count confidence intervals on your own, you know what's going on behind the scenes with our 99% confidence interval calculator! Here's how to use it:
Fill in the sample mean (x̅) in the first row.
Enter the standard deviation (s).
Enter the sample size (n).
Your confidence level is already filled in (99%), but keep in mind you can change it anytime.
The Z-score will update automatically as you decide on the confidence interval.
And that's it! At the bottom of the calculator, you'll see:
A chart describing your data;
The confidence interval range — with a lower and upper bound indicated; and
The margin of error.
Confidence interval tools
If you're interested in statistics, you might find those tools helpful:
Confidence interval calculator;
95% confidence interval calculator;
90% confidence interval calculator.
FAQs
What is the Z-score for a 99% confidence interval?
The z-score for a two-sided 99% confidence interval is 2.807, which is the 99.5-th quantile of the standard normal distribution N(0,1).
How to find the margin of error for a 99% confidence interval?
To find the margin of error for a 99% confidence interval:
Find Z(0.99) (the z-score for 99% confidence) in the statistical table.
Z(0.99) = 2.576
Calculate the standard error with the formula SE = σ/√n, where σ is the standard deviation and n is the sample size.
Multiply Z(0.99) by the standard error to obtain the margin of error, ME.
ME = Z(0.99) × SE
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14330 | https://byjus.com/chemistry/henderson-hasselbalch-equation-questions/ | The Henderson-Hasselbalch equation provides a relationship between the pH of acids (in aqueous solutions) and their pKa (acid dissociation constant). The pH of a buffer solution can be estimated with the help of this equation when the concentration of the acid and its conjugate base, or the base and the corresponding conjugate acid, are known.
| |
| Definition: The Henderson-Hasselbalch equation, pH = pKa + log([A–]/[HA]), can be used to calculate the pH of a buffer. The equilibrium concentrations of the conjugate acid-base pair used to create the buffer solution are denoted by [HA] and [A–] in this equation. When [HA] = [A–], the pH of the solution equals the pKa of the acid. |
Henderson Hasselbalch Equation Chemistry Questions with Solutions
Q1. The Henderson Hasselbalch equation explains the relationship between-
a.) pH and pOH
b.) pH and logKa
c.) pH and pKa
d.) pOH and pKa
Correct Answer– (c.) pH and pKa
Q2. A buffer solution contains 0.36 M sodium acetate (CH3COONa) and 0.45 M acetic acid (CH3COOH), pKa = 4.8. What is the pH of this buffer solution?
a.) 4.7
b.) 6.3
c.) 5.5
d.) 4.2
Correct Answer– (a.) 4.7
Q3. If the pH = pKa, the log of the ratio of dissociating acid and associated acid will be-
a.) equal
b.) zero
c.) greater than 1
d.) lesser than 3
Correct Answer– (b.) zero
Q4. When the pH equals the pKa, the acid will be:
a.) fully dissociated
b.) half dissociated
c.) partially dissociated
d.) None of the above
Correct Answer– (b.) half dissociated
Q5. Which of the following is true regarding the Henderson-Hasselbalch equation?
I. The pH of the solution is always greater than the pKa of the solution.
II. As the ratio of conjugate base to acid increases, the pH increases.
III. The hydrogen ion concentration can never equal the acid dissociation constant.
a.) I and II
b.) II only
c.) I only
d.) II and III
Correct Answer– (b.) II only
Q6.What are the four-variable in the Henderson Hasselbalch equation?
Answer. The Henderson Hasselbalch Equation is:
(\begin{array}{l}pH=pK_{a}+log\frac{\left [ A^{-} \right ]}{\left [ HA \right ]}\end{array} )
.
The four variables present in this equation are pH, pKa, [A–] and [HA].
On knowing any three values, the unknown value can be calculated.
Q7. Is the Henderson-Hasselbalch equation only applicable to acids?
Answer. Only if the consecutive pK values of a polybasic acid differ by at least 3 can the Henderson-Hasselbalch equation be applied.
Q8. Calculate the pH of a buffer that contains 0.7 M ammonia and 0.9 M ammonium chloride. (pKa = 9.248).
Answer. By using the Henderson Hasselbalch Equation is:
(\begin{array}{l}pH=pK_{a}+log\frac{\left [ A^{-} \right ]}{\left [ HA \right ]}\end{array} )
pH = 9.248 + log [0.9 / 0.7]
pH = 9.248 + log 1.29
pH = 9.248 + 0.11
pH = 9.358
Q9. How does the buffer concentration affect the buffer capacity?
Answer. A buffer’s ability to neutralise added acid or base is determined by the concentrations of HA and A– in the solution. The greater the concentrations for a given ratio of [HA] to [A– ], the greater the overall buffer capacity. When [HA] exceeds [A–], the capacity for added base exceeds that of acid.
Q10. What are the limitations of the Henderson Hasselbalch equation?
Answer. The limitations of the Henderson Hasselbalch equation are as follows:
The Henderson-Hasselbalch equation fails to predict accurate values for strong acids and strong bases because it assumes that the concentration of the acid and its conjugate base at chemical equilibrium will be the same as the formal concentration (the binding of protons to the base is neglected).
Since the Henderson-Hasselbalch equation does not account for water’s self-dissociation, it cannot provide accurate pH values for extremely dilute buffer solutions.
Q11. A solution of hydrofluoric acid has a concentration of 0.3M.
The Ka for HF is 7.2 × 10–4.
If sodium hydroxide is slowly added to this solution, what will the pH be at the half equivalence point?
Answer. By using the Henderson Hasselbalch Equation
(\begin{array}{l}pH=pK_{a}+log\frac{\left [ A^{-} \right ]}{\left [ HA \right ]}\end{array} )
We don’t need to worry about using the molar amounts of the acid and the base if we use the Henderson-Hasselbalch equation. The conjugate base concentration is equal to the weak acid concentration at the half equivalence point. That is, the equation can be simplified.
A simplified equation for the half equivalence point:
pH = pKa
Since the acid dissociation constant is 7.2 × 10–4, the pH will be:
pH = –log [7.2 × 10–4] = 3.14.
Hence, the pH at the half equivalence point is 3.14.
Q12. Calculate the pH of a solution containing 3.0 M hydrofluoric acid and 2.5 M fluoride.
(Ka for hydrofluoric acid is 6.76 × 10–4.)
Answer. To begin, convert the acid-dissociation constant into pKa. The Henderson-Hasselbalch equation can then be used to calculate the pH of this solution.
pKa = –log (Ka)
pKa = –log(6.76 × 10–4)
pKa = 3.17
By using the Henderson Hasselbalch Equation
(\begin{array}{l}pH=pK_{a}+log\frac{\left [ base \right ]}{\left [ acid \right ]}\end{array} )
pH = 3.17 + log [2.5/3]
pH = 3.09
Q13. What is the ratio of the concentration of acetic acid and acetate ions required to prepare a buffer with pH 5.20? The pKa of acetic acid is 4.76.
Answer. Given: pH of the buffer = 5.20
pKa of acetic acid = 4.76
By using the Henderson Hasselbalch Equation
(\begin{array}{l}pH=pK_{a}+log\frac{\left [ proton acceptor \right ]}{\left [proton donor \right ]}\end{array} )
(\begin{array}{l}log\frac{\left [ proton acceptor \right ]}{\left [proton donor \right ]}=pH-pKa\end{array} )
(\begin{array}{l}log\frac{\left [ proton acceptor \right ]}{\left [proton donor \right ]}=5.20-4.76\end{array} )
(\begin{array}{l}log\frac{\left [ proton acceptor \right ]}{\left [proton donor \right ]}=0.44\end{array} )
(\begin{array}{l}\frac{\left [ proton acceptor \right ]}{\left [proton donor \right ]}=antilog 0.44\end{array} )
=2.75
Thus, the ratio of the concentration of acetic acid and acetate ions required to prepare a buffer with pH 5.20 is 2.75.
Q14. What is the importance of Henderson – Hasselbalch equation?
Answer. For acid buffer solution the equation is:
(\begin{array}{l}pH=pK_{a}+log\frac{\left [ salt \right ]}{\left [ acid\right ]}\end{array} )
For basic buffer solution the equation is:
(\begin{array}{l}pH=pK_{a}+log\frac{\left [ salt \right ]}{\left [ base\right ]}\end{array} )
The Henderson Equation’s Importance:
1) The pH of a buffer solution can be calculated using Henderson’s equation.
2) Henderson’s equation can be used to calculate the salt: acid ratio required to produce a buffer solution with the desired pH value.
3) Given the pH and total buffer concentration, the concentration of acid and salt (or base and salt) can be calculated.
Q15. Derive Henderson – Hasselbalch equation.
Answer. Let us take an example of ionization of weak acid HA:
HA +H2O ⇋ H+ + A–
Acid dissociation constant is given as:
(\begin{array}{l}K_{a}=\frac{\left [ H^{+} \right ]\left [ A^{-} \right ]}{\left [ HA \right ]}\end{array} )
Taking negative log of the above equation
(\begin{array}{l}-logK_{a}=-log\frac{\left [ H^{+} \right ]\left [ A^{-} \right ]}{\left [ HA \right ]}\end{array} )
(\begin{array}{l}-logK_{a}=-log\left [ H^{+} \right ]-log\frac{\left [ A^{-} \right ]}{\left [ HA \right ]}\end{array} )
Since, –log [H+] = pH and –log Ka = pKa
The equation can be written as:
(\begin{array}{l}pKa=pH-log\frac{\left [ A^{-} \right ]}{\left [ HA \right ]}\end{array} )
On rearranging the equation:
(\begin{array}{l}pH=pK_{a}+log\frac{\left [ A^{-} \right ]}{\left [ HA \right ]}\end{array} )
Practise Questions on Henderson Hasselbalch Equation
Q1. A solution of acetic acid (pKa = 4.75) has a pH of 6.75. The ratio of acid to the conjugate base is ___.
a.) 100 CH3COO– to 1 CH3COOH
b.) 100 CH3COOH to 1 CH3COO–
c.) 1 CH3COOH to 100 CH3COO–
d.) 0.01 CH3COOH to 100 CH3COO–
Q2. You need to produce a buffer with a pH of 5.75. You have a solution with 30.0g of acetic acid (pKa=4.75). How many moles of sodium acetate must you add to achieve the desired pH?
a.) 0.3 mol
b.) 5 mol
c.) 0.5 mol
d.) 3 mol
Q3. Find [H+] in a solution of 1.0M pf HNO3 and 0.225 M NaNO2. The Ka for HNO2 is 7.4 × 10–4.
Q4. Calculate the pH of a buffer solution made from 0.20 M CH3COOH and 0.50 M CH3COO– that has an acid dissociation constant for CH3COOH of 1.8 x 10–5.
Q5. When is the Henderson-Hasselbalch Equation used?
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14331 | https://english.stackexchange.com/questions/430977/if-craven-means-contemptibly-lacking-in-courage-what-do-craven-desires-and | meaning - If craven means "contemptibly lacking in courage," what do "craven desires" and "craven idols" mean? - English Language & Usage Stack Exchange
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If craven means "contemptibly lacking in courage," what do "craven desires" and "craven idols" mean?
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I was recently reviewing the meaning of the word "craven," which included trying to find some good examples of its usage. More often than not, I came across the phrases, "craven desires" and "craven idols," neither of which seem to be using the definitions I find for "craven." I am assuming that the word "craven" in these instances is acting more like a participle of some sort for the verb "to crave," but, thus far, I have not found any dictionary entry that alludes to this. For phrases so common as "craven desires" and "craven idols" why do I not see some mention made of this?
Addendum:
Well, there's at least one decent answer in this thread that I will award the green checkmark to if no others surface, so I don't know how necessary this is at this point. However, since I felt my initial question was sufficient and sense that others will, too, I will elaborate a bit to explain why I am adding to my original question. First of all, I am really quite shocked that so many have never heard of the phrase "craven idol(s)" before. (See comments below.) I will wholeheartedly admit that I was unaware that it was a rock band and I did have to do some major revisions to my search parameters to finally get something that wasn't about that band. Nevertheless, in my defense, I had heard or seen the phrase "craven idols" so long ago that I couldn't even tell you when it became part of my memory bank. Not considering myself all that "well read," I was surprised to see that so many in this alleged erudite forum would be so dumbfounded by the phrase "craven idol." But to suffice the skepticism of those below, what follows are some examples of the written word and its use of the phrase "craven idol(s)." [I won't bother adding examples of "craven desires" because someone already took the time and trouble to do so (see the answers below)]. Examples of craven idol and/or craven idols follow:
"Saint Paul makes the law of nature the basis for his claim that God can legitimately punish those who violate its precepts and inclinations. He especially condemns the pagan practice of idol worship — of making God into the image of mortals, birds, and snakes. Saint Paul is not referring to people who find themselves ignorant about the true God. He addresses those whose personal wickedness compels them to reduce God to a craven idol."
Source:Praying with Saint Paul: Daily Reflections on the Letters of Saint Paul
"Garrett, damned to hell as he could well see, knew now, much to his horror, that Vanra's loyalty had been easily purchased. The bodyguard's honour was swayed by gold alone and, alas, Garratt's reliance on the mere honour of the soldiering brotherhood could never topple that craven idol of greed."
Source:Shadows Beneath the Watch Tower
"In some universities, lecturers fearful of sparking a suicide refuse to fail even the poorest students. "I often think of the implications for the student and the family before failing anyone," says one professor at Sogang University. But Korean students and their families, whose faith in the omnipotent powers of education is greater than in any other Confucian country in Asia, are worshipping a craven idol. If anyone takes too close a look at the system, they will be in for a rude awakening."
Source:Korea Economic Report, Vol. 8, p. 36
I see I still haven't added an example of "craven idols" yet, so here's one more. For the record, there were several I could have chosen. I suppose this is as good as any:
"These 'evil men who had been used to milking the villagers for years by parading bronze peacocks and other such craven idols among the villages and thus collecting large sums for their own use' had to be put down."
Source:The Well-protected Domains: Ideology and the Legitimation of Power in the Ottoman Empire, 1876-1909
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edited Feb 17, 2018 at 16:49
Edwin Ashworth
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asked Feb 13, 2018 at 20:25
Lisa BeckLisa Beck
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1 Can you quote and link to some examples? I only find those terms as the names of a blogger and a heavy metal band, respectively (plus one article about a town in England named Craven); in other words, as nonsensical strings of words chosen for sound, rather than because they actually mean anything.1006a –1006a 2018-02-13 20:38:20 +00:00 Commented Feb 13, 2018 at 20:38
5 Is no one going to mention the phrase that OP actually heard was gravenidols?Dan Bron –Dan Bron 2018-02-13 22:03:16 +00:00 Commented Feb 13, 2018 at 22:03
1 @1006a: on craven desire, use a minus sign (-bull, -pitbull, -Yorkshire) to eliminate false hits or use an NGram and follow those links.KarlG –KarlG 2018-02-13 22:50:33 +00:00 Commented Feb 13, 2018 at 22:50
1 @KarlG That's good advice—for the OP. If this is a real usage, it's the OP's responsibility to include evidence of that in the question.1006a –1006a 2018-02-13 23:05:42 +00:00 Commented Feb 13, 2018 at 23:05
2 It's really really stupid and small-minded that anyone should downvote a question that "now" has references and sources. I checked each link, and each one is valid. Don't like the question? Move along. Downvote Qs if they are low quality and do not share any research. This is not an example of a LQQ.Mari-Lou A –Mari-Lou A 2018-02-17 13:56:26 +00:00 Commented Feb 17, 2018 at 13:56
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The Desires
The standard dictionary definition of craven is "cowardly":
This could be illustrated through Ancient History in the conduct of Ephialtes, the "Judas Iscariot" of Greece, who in satisfying his craven desires, brought untold misery to his fellow patriots by his treachery... Henry Germanus Maeder, Educational Outcomes Unique to Ancient History, 1927
Ephialtes betrayed the united Greek forces to the Persians at Thermopylæ in hope of gain, his craven, i.e. cowardly desires. Greed in itself is not always cowardly, but accepting payment for betrayal, as Judas Iscariot, may certainly be deemed so.
In other usages, however, moral failings other than cowardice seem to be on the author's mind: craven desires are base, often debauched, and in the fiery rhetoric of Christian moralists, a grievous sin. For those wanting to perfume their otherwise wanting prose with a whiff of scandal, the word may equally serve.
They laugh and dance, indulge and spend, and find some satisfaction in pursuing thrill, excitement, danger, popular acclaim, and material security. They follow their selfish instincts or yield to their craven desires. Leslie F. Brandt, Jesus/Now, 1978.
Your religious training may have taught you that the body is a repository of base instincts and desires that must be repressed, ignored and squelched, lest you accede to your craven desires and spend eternity regretting the decision! Ramón Stevens, Whatever Happened to Divine Grace, 1988.
And now for someone apparently ready to chuck it all for (shudder!) one night of pleasure:
She entered the living room in a most charming negligee. She had never interested me sexually in all the years I had known her — somehow I had never thought of her in that way. She had always been on a pedestal above all such craven desires. Her seductive negligee plus the liquor was sufficient for me to discover charms never before unfolded. William Ward Smith, A Letter from my Father, 1976.
This isn't cowardice; it's sex, drugs and rock 'n' roll. How lexicographers have missed this usage is beyond me, but only Superman-sized conceptual leaps can get to cowardice from charming, seductive negligees.
Craven Desire in the Singular
Craven desires can just as easily be condemned in the singular, where the distance from the original meaning of "cowering, cowardly" is even more apparent. Occasionally, the familiar moral tone of the plural is still heard. Thus paraphrasing the Buddha:
the craven desire is the root of all suffering
And a Christian writer in an imaginary conversation with Hieronymus Bosch:
...the craven desire is for immediate gratification. It uses the other person, or thing, or even God, to meet this consummation of desire.
This suggests a meaning of inordinate, selfish desire, or, in a word, lust, which can be described negatively in countless scenarios without audible religious overtones. Thus a craven desire
for power, again here, for media attention,attention and power, money,to stay in power, for publicity, to be liked, to seem young, purely for commercial purposes (Hannah Montana), to be provocative (Rolling Stones), or to be on television (reality stars).
One writer turns to Norman Mailer for a critique of the media in the run-up to the Second Gulf War:
In an effort to analyze our wanton and exuberant Iraq war-lust, Mailer points to our craven desire for manipulated and televised displays of dominance as a major factor for the drive to the military invasion of Baghdad.
A journalist takes a look at the 1936 Berlin Olympics and
the Germans' craven desire to paint a sanitized picture of their society to a skeptical world.
While craven is never completely free of opprobrium, there are cases where the desire is neither overwhelming nor morally suspect, merely very strong. Thus a business has a craven desire to win over consumers or fill a niche market, and the director of the film Benjamin Button has a craven desire to elicit emotion.
One writer suggests that US senators' reluctance to use email comes from their craven desire to avoid record-keeping, while the Far Right in the US has a craven desire for academic acceptance.
Finally, a self-designated perfectionist has a "craven desire to do everything perfectly the first time" and a woman in India describes humorously her craven desire for a Michael Kors handbag.
No stretch of the imagination can inject cowardice into any of these usages of craven.
The Idols
Craven idol received no hits in a Google NGram; a collocation search only yielded results such as fear/fears, unambiguously in the sense of 'cowardly.' Since Craven Idol is also the name of a heavy metal band, googling becomes a frustrating exercise in using the minus sign. What non-metalic results appear, however, suggest that the band is not using a nonce word for its name.
A Time Magazinereview (31 Mar 1961) of a decidedly anti-modernist book by Robert Elliot Fitch, Dean of the Pacific School of Religion, is entitled The Craven Idol.
The Golden Calf of the Exodus saga is craven:
But there in that wilderness/ before the craven idol/ before the king demanding to be god/ Refusing to bow, to deny the creator/ honoring the words written so long ago/ Prepared to be burned, to submit to the furnace/ entering in, faithful to God. Raymond A. Foss, The First Commandment, 2009.
God was about to destroy the children of Israel because of their craven idol that [they] had made and also because of Aaron's sin. Erica Michael, Birthing His Glory Through Dance, Part 1, 2010.
Foss also alludes to the statue of the Roman emperor to which Christians were to pay homage or suffer punishment.
In The Invisible Dragon: Essays on Beauty, Revised and Expanded, 2012, author Dave Hickey describes a painting of Jesus, which
... thanks to oil glazing, seems an uncanny incarnation of Christ? Is it a picture, an icon, a craven idol, or something else?
A four-item catalogue such as this is the last place one would expect an affect-laden or polemic adjective, though idol is the only noun that gets any adjective at all. Whatever the intentions of the author, pausing in the middle of a list to comment on the "cowardly" nature of a cultic statue is not likely to be among them.
Almost all of these uses of craven, whether collocated with desires or idol, come from literature with an overtly religious stance, or certainly a moral one.
For the prophet Habakkuk (2.18), graven or molten images are made from human hands: they are "dumb," inert stone or metal incapable of speech. In 1 Cor. 12.2, they are merely "dumb idols." It would be difficult to imagine a writer of devotional literature ascribing any affect at all to an idol, even a negative one such as cowardice.
The Dilemma
Craven desires and craven idol seem to inhabit the same Christian moral and polemic universe, yet the usual definition of craven as "cowardly" is a poor fit for either. A meaning of "depraved, debauched" might work for desires, but not for idols. Try as I might, I can only see the use of craven for idols as a strange variant of graven, even by those intimate with the latter term from the Hebrew and Greek Scriptures.
The parallel of bronze peacocks and craven idols in the Ottoman history would suggest an adjective equally parallel to bronze, and none of the sources for craven idol would suffer in the least were graven used in its stead.
The use of the singular diverges just as strongly from dictionary entries, yet it's clear that the word craven is being used in ways other than that prescribed. Perhaps craven suggests to some speakers an excessive craving for most any desired object, which can be acknowledged but not necessarily condemned. The disparity between living language and lexicography for this word, however, remains an intriguing dilemma.
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edited Feb 18, 2018 at 20:11
answered Feb 13, 2018 at 22:12
KarlGKarlG
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Appreciate the time/effort you put into this answer. All things considered, you may walk away with the green checkmark. The best part of your answer, to me, however, is when you acknowledge that it is something of a mystery how lexicographers missed this usage. Perhaps they are too busy adding new words like "selfie-stick" to the dictionary. I'm holding out for someone who might be able to address that aspect a bit more thoroughly. (If that's you even better!) If not, you certainly have answered the question best. But considering the question has been put on hold, somewhat of a moot point.Lisa Beck –Lisa Beck 2018-02-17 13:04:33 +00:00 Commented Feb 17, 2018 at 13:04
1 @LisaBeck: Three users, including me, have voted to reopen the question. Only two more are required. I only knew this word in its "fire and brimstone" sense and was surprised that wasn't reflected in dictionaries.KarlG –KarlG 2018-02-17 13:27:45 +00:00 Commented Feb 17, 2018 at 13:27
There are many more hits on the internet for 'secret desires'. These may be collocations, but they are transparent and hardly more than adjective + noun strings.Edwin Ashworth –Edwin Ashworth 2018-02-17 13:35:34 +00:00 Commented Feb 17, 2018 at 13:35
@EdwinAshworth: The point is about the lexical disparity between dictionary definitions and actual usage, at least for me. Why should "many more hits for 'secret desires'" be more relevant here than more for "potato salad"?KarlG –KarlG 2018-02-17 13:47:44 +00:00 Commented Feb 17, 2018 at 13:47
1 @Lambie: I assumed it was so obvious I didn't need to mention it. In the traditions with which I'm most familiar (Lutheran, Catholic), that verse is still part of the First Commandment. But this is part of the dilemma, why would someone use the word craven in a poem about the First Commandment no matter where the verses are divided, when the word graven is right in front of his nose?KarlG –KarlG 2018-02-18 21:01:39 +00:00 Commented Feb 18, 2018 at 21:01
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Sorry, you're all out to lunch. The idols are graven (carved) not craven. Misheard, misspelled, misused....please do not ever write craven idols. The rock band's name was clearly a joke, which apparently nobody gets.. I feel sorry for the word craven, now used to mean corrupt although for centuries it had a plain meaning: cowardly, weak and slavish. And now this! Pls see Exodus 20 4-6 for the graven images.
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answered Jan 11, 2021 at 19:51
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1 My thoughts entirely. The use of "craven" idols is an eggcorn or simple ignorance or a misprint. Eggcorn: independent.co.uk/life-style/…Greybeard –Greybeard 2021-02-02 20:38:39 +00:00 Commented Feb 2, 2021 at 20:38
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They mean desires and idols that are craven (contemptibly lacking in courage; cowardly). Idols are not actually alive, so any "desires" they have can be described as craven. Also, Craven Idol is the name of a rock band; and pretty much anything goes when it comes to names; they don't have to "make sense" or be common phrases (examples: Smashing Pumpkins, Goo Goo Dolls, Mr. Mister, Limp Bizkit; your mileage may vary; and I am not knocking any of the named bands).
The same can be said of "craven desires," and the good old Guardian has made a pun on the phrase in its article:
Craven desires: why living in the Yorkshire Dales makes you happy
Here, Craven is apparently "a local government district of North Yorkshire." But the people of Craven, or perhaps Craven itself, can have desires. Perhaps this is a reference to the other Craven Desires that is on the internet, which seems to be some kind of controversial blogger.
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edited Feb 13, 2018 at 21:59
answered Feb 13, 2018 at 21:29
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14332 | https://pubmed.ncbi.nlm.nih.gov/17337027/ | The evolution of venom-conducting fangs: insights from developmental biology - PubMed
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. 2007 Jun 1;49(7):975-81.
doi: 10.1016/j.toxicon.2007.01.007. Epub 2007 Jan 26.
The evolution of venom-conducting fangs: insights from developmental biology
Kate Jackson1
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1 Department of Zoology, University of Toronto, 25 Harbord Street, Toronto, Ont., Canada M5S 3G5. k.jackson@utoronto.ca
PMID: 17337027
DOI: 10.1016/j.toxicon.2007.01.007
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The evolution of venom-conducting fangs: insights from developmental biology
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. 2007 Jun 1;49(7):975-81.
doi: 10.1016/j.toxicon.2007.01.007. Epub 2007 Jan 26.
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Kate Jackson1
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1 Department of Zoology, University of Toronto, 25 Harbord Street, Toronto, Ont., Canada M5S 3G5. k.jackson@utoronto.ca
PMID: 17337027
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The present study of the origin of the various types of fang represented among colubroid snakes (i.e., tubular, grooved, and ungrooved) attempts to reconcile the morphology of adult fangs with current phylogenetic hypotheses. Observations of growth series of developing tubular fangs were hypothesised to shed light on the evolutionary origin of fangs in snakes. While molecular phylogenies and evolutionary studies of venom proteins and of other anatomical components of the venom-delivery system reconstruct a consistent evolutionary scenario, the character of a tubular venom-conducting fang does not fit in this scenario. The present review offers a series of possible scenarios to resolve this anomaly. Of these, a new idea argues that a heterochronic mechanism (alteration of the timing of developmental events) may provide the answer that the ungrooved and grooved teeth of colubrid snakes evolved from an ancestral tubular fang by means of attachment of replacement tubular fangs to the maxilla at an earlier developmental stage than usual (precocial ankylosis).
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14333 | https://www.themathdoctors.org/necessary-and-sufficient-conditions-if-or-only-if/ | Necessary and Sufficient Conditions: If, or Only If? – The Math Doctors
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Necessary and Sufficient Conditions: If, or Only If?
May 14, 2018 January 10, 2024 / Geometry, Logic / Words / By Dave Peterson
Sometimes in math, we trip over words, especially when they are used in ways that differ from everyday usage, or when the associated grammar is complicated. This set of three answers from our archive, each of which is referred to by the next one, look at relationships among the ideas of “necessary and sufficient conditions”, and “if and only if”.
Does “necessity” mean “if” or “only if”?
First, from 1999, we have a question about the words “necessary” and “sufficient” in the statement of a theorem to be proved; such a statement is also called a “biconditional”, as we have conditions in both directions. The geometrical theorem here is simple, probably intended just to demonstrate the form of this sort of theorem. It happens to tie in to our recent discussions of inclusive definitions:
Parts of a Biconditional Statement
I came across the following arguments (in a book) involving the biconditional and the author's proof confused me. The author stated the following theorem:
A quadrilateral is a square if and only if it is both a rhombus and a rectangle,
and proved the theorem in two steps as follows:
Step 1 (the "if" part): Let Q be a quadrilateral which is both a rhombus and a rectangle. From the definition of a rhombus, all four sides are equal in length. From the definition of a rectangle, all four angles are right. A square is a four-sided figure in which all angles are right and all sides are equal. Therefore, Q is also a square.
Step 2 (the "only if" part): Let Q be a square. By definition, all four sides are equal, so Q is also a rhombus. Again by definition, all angles are right, so Q is a rectangle.
Until here I have followed the author's proof, but further he re-proves the same biconditional using "necessary" and "sufficient"; and there he wrote that:
step 1 (the "necessity" part): the same as "if" part in the last proof.
Step 2 (the "sufficiency" part): same as the "only if" part above.
I believe it's just the other way around: that is, "necessity" corresponds to "only if" and "sufficient" corresponds to "if."
Could anybody be of some help? Am I right? More explanation to make the idea clearer is very welcome.
Unfortunately, Abdellah didn’t quote the “necessary and sufficient” formulation of the theorem; there are two possibilities, and if the book itself didn’t state it explicitly, that may be the source of confusion.
My first thought was that the restatement of the theorem would most naturally be something like this, where the condition is the more complicated statement:
“A necessary and sufficient conditionfor a quadrilateral to be a square is that it is both a rhombus and a rectangle.”
If so, then Abdellah is arguing that the statement that being both a rhombus and a rectangle is necessary for being a square is equivalent to “A quadrilateral is a square only ifit is both a rhombus and a rectangle.” He would be right, though honestly it took me a while to convince myself of this, because the words are so convoluted!
On the other hand, it could be this, swapping the roles of the clauses to put “necessary and sufficient condition” where “if and only if” was:
“For a quadrilateral, being a square is anecessary and sufficient conditionfor it to be both a rhombus and a rectangle.”
If that is what they meant, the book is right!
Doctor Mike properly assumes that the book is correct (taking the first clause to be the “condition” for the second, as in my second version), and explains the meaning of the words:
Maybe it would help to actually write out the sentences, but let's use "S" to mean Square and "R+R" to mean Rhombus and Rectangle.
The first one is "S is a sufficient condition for R+R." This means that if you are given S, then you have "sufficient" (or "enough") information to prove R+R. That's exactly what was done for the "only if" part.
Next, look at "S is a necessary condition for R+R". This should be the other direction, so let's see why. If you focus on the first part, "S is a necessary condition," then you see we really are talking about S being a necessary and logical consequence of something, namely R+R. Assuming R+R and proving S is what was done above in the "if" step.
Here is a summary. The following 4 sentences mean the same.
S ----> R+R
S implies R+R
S only if R+R
S is sufficient for R+R
The other direction also has the 4 similar variants.
S <---- R+R
S is implied by R+R
S if R+R
S is necessary for R+R
He left out two forms that can help clarify (or confuse): S→R+R can be read as “if S, then R+R”, Similarly,“S if R+R” means the same thing as “if R+R, then S”. Often, putting the “if” first clarifies the meaning.
As I think about the use of “necessary” and “sufficient” in logic, I realize that these are, in a sense, two different senses of what we think of as a “condition” in everyday life, and part of the confusion may be the ambiguity of that ordinary usage.
When I look up “condition” in a dictionary, the relevant definition is “prerequisite”; in the Merriam-Webster dictionary, an example given is “Available oxygen is an essential condition for animal life”. A condition in this sense is something without which something wouldn’t happen – it will happen only if the condition is satisfied. That’s exactly what a “necessary condition” is. As Doctor Mike said, we can just as well think of this as a logical consequence: If someone is alive, then we know he must have oxygen. But this is not to say that life causes the oxygen! Nor is oxygen a sufficient condition for life; you need other things as well, such as food.
On the other hand, “condition” also means, in grammar, the “if clause” in a conditional sentence like “If A, then B”; in this sense a “condition” is only sufficient, giving a condition under which B will be true, but saying nothing about what happens if A is false. So when we connect the words “condition” and “if”, we tend to think of a sufficient condition (a fact from which we can conclude that something else is true), not a necessary condition (a fact that is required in order for something else to be true).
The same is true in logic: When we talk about a “conditional statement”, we mean A→B, or “If A, then B”, where again A is a sufficient, not necessary, condition for B. If A is true, we can be sure that B is true. (In logic, A is called the antecedent and B the consequent – the word “condition” is not used.)
When we use the words “necessary” or “sufficient” with “condition”, we are overriding these uses, and taking a “condition” merely as any statement, which has whichever relation we specify with the other statement.
Necessary or sufficient
Adeel asked about these words in 2002:
What do we mean by 'necessary condition' and 'sufficient condition' (and sometimes we call a condition both 'necessary and sufficient')? I am very much confused. Help!
I used an example (unlike Abdellah’s question above) in which only one part is true, which makes it a little easier to see the distinctions:
Let's look at the two statements (predicates), "X is a mammal" and "X is a dog". Call the first statement A, and the second B.
Now, A is a necessary condition for B, because A must be true in order for B to be true. B can only be true if A is true; if A is not true, then B can't be true. We can say this in several ways:
A is a necessary condition for B
A <== B (A is implied by B)
B ==> A (B implies A)
A if B (whenever B is true, A will be true)
B only if A (B is true only when A is true)
On the other hand, A is not a sufficient condition for B, which would mean that in order to know that B is true, it is enough to know that A is true. It is not enough to know that X is a mammal, because there are other mammals besides dogs. But if we reverse the two statements, we find that B is a sufficient condition for A: if we know that X is a dog, we know that it is a mammal. So these statements are equivalent:
A is a sufficient condition for B
B <== A (B is implied by A)
A ==> B (A implies B)
B if A (whenever A is true, B will be true)
A only if B (A is true only when B is true)
I needed a different example to illustrate “necessary and sufficient”:
Note that "necessary condition" and "sufficient condition" are opposites; "A is a necessary condition for B" means the same thing as "B is a sufficient condition for A".
Now, if A is a necessary AND sufficient condition for B, then the implication works both ways; it can be expressed as
A <==> B (A is equivalent to B)
A iff B (A if and only if B)
This means that if A is true, B must be true, and if B is true, A must be true. That is not the case in our example statements; but it would be true, for example, if A were "X is less than Y" and B were "Y is greater than X". These two statements mean the same thing; if one is true, then the other is true. So if we want to prove B, it is necessary for A to be true, and it is sufficient to prove that A is true.
What does “only if” mean?
I think some additional explanation is needed for the meaning of “only if”; but I couldn’t find a good discussion of this in the archive. What follows is an unarchived question from 2011:
The statement "A if and only if B" can be taken apart as "A if B" and "A only if B". Both parts need to be either true or false for the biconditional to be true. Shown with a truth table, easily proven.
Wanting to deeply understand the meaning of the parts, I ran into some struggle. Since I'm not native English speaking, probably I have some problems with the words...but I don't know if that is really the case.
"A only if B"
This is the easy one for me. It says, only if B is true, A is true. So, if B is true and A is true, the formula is true. And if B is true but A is false, the formula is false. And for the rest of possibilities, there is nothing we can say. So this is equivalent to "B --> A".
"A if B"
This is the difficult one for me. Since 'B' comes syntactically after the 'if', this looks for me like "B --> A". But this case is already given above. So how can I rephrase this in a way making sense in English? Or is this just some sloppiness of the definition and I have to get over it? Or did I not understand the real meaning? Can you explain on this? Thank you.
Panny actually got the “if” case right, but wrongly thinks that “only if” means the same thing. That’s easy to do! I replied, first about the language issue:
I think most native English speakers have trouble with this! In fact, they may have more trouble, because they are somewhat familiar with the phrases but have never stopped to think about exactly what they mean. When you think you understand something, you are in greater danger of fooling yourself!
Then, about the “only if” case:
This is actually the harder part; you got it wrong, as most of us do at first. Even I have to think about it a moment to be sure I'm explaining it correctly.
Take an example: "The road is wet only if it rains". What does that statement claim? It says that the only time the road will be wet is if it rains. (This is not really true, of course; snow may have melted on the road.) So if you see that the road is wet, it must be because it rained. Therefore, we can restate this as, "if the road is wet, then it has rained".
Do you see what this means? The statement "A only if B" turns out to mean "if A, then B", or "A -> B". (In my example, A is "the road is wet" and B is "it has rained".)
Again, it is saying that A is true ONLY if B is true, which means that if A is true, you know that B must be true!
Finally, the “if” case:
Here your reasoning is correct; "A if B" means "if B then A", which means "B -> A".
So (a) "A if B" and (b) "A only if B" mean
(a) "B -> A" and (b) "A -> B" respectively.
Another example
I will quote only parts of one last question and answer on this, because the example is from modular arithmetic, which not everyone will want to get into. Read the whole thing if you do!
Necessary and/or Sufficient Conditions with Modular Math
... identify which of the conditions below are "sufficient", "necessary", "necessary and sufficient" or none of these ...
I don't understand what is meant by sufficient, necessary, and sufficient and necessary.
Here is the generally useful part of the answer, which touches on the “only if” question as well:
A condition A is "necessary" for a result B if B is true ONLY if A is true; that is, A HAS TO be true in order for B to be true. We say that B implies A, or "B only if A".
A condition A is "sufficient" for a result B if B is true WHENEVER A is true; that is, A is ENOUGH to force B to be true. We say that A implies B, or "B if A".
A condition A is "necessary AND sufficient" for a result B if both of the above are valid; we say that A is true IF AND ONLY IF B is true. A and B are equivalent.
As an example, suppose I asked whether "x and y are both even" is a necessary and/or sufficient condition for the product xy to be even. Since IF x and y are both even, THEN xy is even, it is a sufficient condition; knowing they are both even is enough to be certain that the product is even. But xy will also be even if only one of the factors is even; so having both even is NOT necessary.
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14334 | https://phys.libretexts.org/Bookshelves/Electricity_and_Magnetism/Electricity_and_Magnetism_(Tatum)/17%3A_Magnetic_Dipole_Moment/17.02%3A_The_SI_Definition_of_Magnetic_Moment | Skip to main content
17.2: The SI Definition of Magnetic Moment
Last updated
: Mar 5, 2022
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17.1: Introduction to Magnetic Dipole Moments
17.3: The Magnetic Field on the Equator of a Magnet
Page ID
: 5523
Jeremy Tatum
University of Victoria
( \newcommand{\kernel}{\mathrm{null}\,})
If a magnet is placed in an external magnetic field B, it will experience a torque. The magnitude of the torque depends on the orientation of the magnet with respect to the magnetic field. There are two oppositely-directed orientations in which the magnet will experience the greatest torque, and the magnitude of the magnetic moment is defined as the maximum torque experienced by the magnet when placed in unit external magnetic field. The magnitude and direction of the torque is given by the equation
τ=p×B.(17.2.1)
The SI unit for magnetic moment is clearly N m T−1.
If an electric current I flows in a plane coil of area A(recall that area is a vector quantity – hence the boldface), the torque it will experience in a magnetic field is given by
τ=IA×B.(17.2.2)
This means that the magnetic moment of the coil is given by
p=IA.(17.2.3)
Thus the unit A m2 is also a correct SI unit for magnetic moment, though, unless the concept of “current in a coil” needs to be emphasized in a particular context, it is perhaps better to stick to N m T−1.
While “J T−1” is also formally dimensionally correct, it is perhaps better to restrict the unit “joule” to work or energy, and to use N m for torque. Although these are dimensionally similar, they are conceptually rather different. For this reason, the occasional practice seen in atomic physics of expressing magnetic moments in MeV T−1 is not entirely appropriate, however convenient it may sometimes seem to be in a field in which masses and momenta are often conveniently expressed in MeV/c2 and MeV/c.
It is clear that the unit “T M3”, often seen for “magnetic moment” is not dimensionally correct for magnetic moment as defined above, so that, whatever quantity is being expressed by the often-seen “T M3”, it is not the conventionally defined concept of magnetic moment.
The magnetization Mof a material is defined by the equation
B=μ0(H+M)(17.2.4)
Equations 17.2.2 and 17.2.4 for the definitions of magnetic moment and magnetization are consistent with the alternative concept of magnetization as “magnetic moment per unit volume”.
17.1: Introduction to Magnetic Dipole Moments
17.3: The Magnetic Field on the Equator of a Magnet |
14335 | http://www.mhtlab.uwaterloo.ca/courses/ece309/lectures/notes/S16_chap6_web.pdf | Convection Heat Transfer Reading Problems 12-1 →12-8 12-40, 12-49, 12-68, 12-70, 12-87, 12-98 13-1 →13-6 13-39, 13-47, 13-59 14-1 →14-4 14-18, 14-24, 14-45, 14-82 Introduction ↙ ↘ Newton’s Law of Cooling ˙ Qconv = ∆T Rconv = hA(Tw −T∞) ⇒ Rconv = 1 hA Typical Values of h (W/m2K) Natural gases: 3-20 Convection water: 60 - 900 Forced gases: 30 - 300 Convection oils: 60 - 1800 water: 100 -1500 Boiling water: 3000 - 105 Condensation steam: 3000 - 105 Controlling Factors Geometry: shape, size, aspect ratio and orientation Flow Type: forced, natural, laminar, turbulent, internal, external Boundary: isothermal (Tw = constant) or isoflux ( ˙ qw = constant) Fluid Type: viscous oil, water, gases or liquid metals Properties: all properties determined at film temperature Tf = (Tw + T∞)/2 Note: ρ and ν ∝1/Patm ⇒see Q12-40 density: ρ ((kg/m3) specific heat: Cp (J/kg · K) dynamic viscosity: µ, (N · s/m2) kinematic viscosity: ν ≡µ/ρ (m2/s) thermal conductivity: k, (W/m · K) thermal diffusivity: α, ≡k/(ρ · Cp) (m2/s) Prandtl number: P r, ≡ν/α (−−) volumetric compressibility: β, (1/K) 1 Forced Convection The simplest forced convection configuration to consider is the flow of mass and heat near a flat plate as shown below.
• flow forms thin layers that can slip past one another at different velocities • as Reynolds number increases the flow has a tendency to become more chaotic resulting in disordered motion known as turbulent flow – transition from laminar to turbulent is called the critical Reynolds number, Recr Recr = U∞xcr ν – for flow over a flat plate Recr ≈500, 000 – x < xcr the boundary layer is laminar; x > xcr the boundary layer is turbulent Boundary Layers Velocity Boundary Layer • the region of fluid flow over the plate where viscous effects dominate is called the velocity or hydrodynamic boundary layer • the velocity at the surface of the plate, y = 0, is set to zero, U@y=0 = 0 m/s because of the no slip condition at the wall 2 • the velocity of the fluid progressively increases away from the wall until we reach approxi-mately 0.99 U∞which is denoted as the δ, the velocity boundary layer thickness.
• the region beyond the velocity boundary layer is denoted as the inviscid flow region, where frictional effects are negligible and the velocity remains relatively constant at U∞ Thermal Boundary Layer • the thermal boundary layer is arbitrarily selected as the locus of points where T −Tw T∞−Tw = 0.99 • for low Prandtl number fluids the velocity boundary layer is fully contained within the ther-mal boundary layer • conversely, for high Prandtl number fluids the thermal boundary layer is contained within the velocity boundary layer Flow Over Plates 1. Laminar Boundary Layer Flow, Isothermal (UWT) All laminar formulations for Re < 500, 000. The local value of the Nusselt number is given as Nux = 0.332 Re1/2 x P r1/3 ⇒local, laminar, UWT, P r ≥0.6 An average value of the heat transfer coefficient for the full extent of the plate can be obtained by using the mean value theorem.
NuL = hLL kf = 0.664 Re1/2 L Pr1/3 ⇒average, laminar, UWT, P r ≥0.6 For low Prandtl numbers, i.e. liquid metals Nux = 0.565 Re1/2 x P r1/2 ⇒local, laminar, UWT, P r ≤0.6 2. Turbulent Boundary Layer Flow, Isothermal (UWT) All turbulent formulations for 500, 000 ≤Re ≤107. The local Nusselt number is given as Nux = 0.0296 Re0.8 x P r1/3 ⇒ local, turbulent, UWT, 0.6 < P r < 60 3 and the average Nusselt number is NuL = 0.037 Re0.8 L P r1/3 ⇒ average, turbulent, UWT, 0.6 < P r < 60 3. Combined Laminar and Turbulent Boundary Layer Flow, Isothermal (UWT) When (Tw −T∞) is constant hL = 1 L Z L 0 hdx = 1 L (Z xcr 0 hlam x dx + Z L xcr htur x dx ) NuL = hLL k = (0.037 Re0.8 L −871) P r1/3 ⇒ average, combined, UWT, 0.6 < P r < 60, 500, 000 ≤ReL ≤107 4. Laminar Boundary Layer Flow, Isoflux (UWF) Nux = 0.453 Re1/2 x P r1/3 ⇒local, laminar, UWF, P r ≥0.6 5. Turbulent Boundary Layer Flow, Isoflux (UWF) Nux = 0.0308 Re4/5 x P r1/3 ⇒local, turbulent, UWF, P r ≥0.6 Example 6-1: Hot engine oil with a bulk temperature of 60 ◦C flows over a horizontal, flat plate 5 m long with a wall temperature of 20 ◦C. If the fluid has a free stream velocity of 2 m/s, determine the heat transfer rate from the oil to the plate if the plate is assumed to be of unit width.
4 Flow Over Cylinders and Spheres 1. Boundary Layer Flow Over Circular Cylinders, Isothermal (UWT) The Churchill-Bernstein (1977) correlation for the average Nusselt number for long (L/D > 100) cylinders is NuD = 0.3 + 0.62 Re1/2 D P r1/3 [1 + (0.4/P r)2/3]1/4 1 + ReD 282, 000 !5/8 4/5 ⇒average, UWT, ReD < 107, 0 ≤P r ≤∞, ReD · P r > 0.2 All fluid properties are evaluated at Tf = (Tw + T∞)/2.
2. Boundary Layer Flow Over Non-Circular Cylinders, Isothermal (UWT) The empirical formulations of Zhukauskas and Jakob given in Table 12-3 are commonly used, where NuD ≈hD k = C Rem D P r1/3 ⇒see Table 12-3 for conditions 3. Boundary Layer Flow Over a Sphere, Isothermal (UWT) For flow over an isothermal sphere of diameter D, Whitaker recommends NuD = 2 + h 0.4 Re1/2 D + 0.06 Re2/3 D i P r0.4 µ∞ µs !1/4 ⇒ average, UWT, 0.7 ≤P r ≤380 3.5 < ReD < 80, 000 where the dynamic viscosity of the fluid in the bulk flow, µ∞is based on the free stream tem-perature, T∞and the dynamic viscosity of the fluid at the surface, µs, is based on the surface temperature, Ts. All other properties are based on T∞.
5 Example 6-2: An electric wire with a 1 mm diameter and a wall temperature of 325 K is cooled by air in cross flow with a free stream temperature of 275 K. Determine the air velocity required to maintain a steady heat loss per unit length of 70 W/m.
Internal Flow Lets consider fluid flow in a duct bounded by a wall that is at a different temperature than the fluid.
For simplicity we will examine a round tube of diameter D as shown below The Reynolds number is given as: ReD = UmD ν .
For flow in a tube: ReD < 2300 laminar flow 2300 < ReD < 4000 transition to turbulent flow ReD > 4000 turbulent flow 6 For engineering calculations, we typically assume that Recr ≈2300, therefore ReD ( < Recr laminar > Recr turbulent Hydrodynamic (Velocity) Boundary Layer • when the boundary layer grows to the tube radius, r, the boundary layers merge – this flow length is called the flow entrance length, Lh – 0 ≤x ≤Lh is the hydrodynamic entrance region – Lh < x ≤L is the fully developed region or hydrodynamically developed region • the hydrodynamic boundary layer thickness can be approximated as δ(x) ≈5x Umx ν !−1/2 = 5x √Rex • the hydrodynamic entry length can be approximated as Lh ≈0.05ReDD (laminar flow) Thermal Boundary Layer • a thermal entrance region develops from 0 ≤x ≤Lt • the thermal entry length can be approximated as Lt ≈0.05ReDP rD = P rLh (laminar flow) • for turbulent flow Lh ≈Lt ≈10D 7 Wall Boundary Conditions 1. Uniform Wall Heat Flux: The total heat transfer from the wall to the fluid stream can be deter-mined by performing an energy balance over the tube. If we assume steady flow conditions, ˙ min = ˙ mout = ˙ m then the energy balance becomes ˙ Q = ˙ qwA = ˙ m(hout −hin) = ˙ mCp(Tout −Tin) Since the wall flux ˙ qw is uniform, the local mean temperature is linear with x.
Tm,x = Tm,i + ˙ qwA ˙ mCp The surface temperature can be determined from Tw = Tm + ˙ qw h 2. Isothermal Wall: Using Newton’s law of cooling we can determine the average rate of heat transfer to or from a fluid flowing in a tube ˙ Q = hA (Tw −Tm) | {z } average ∆T From an energy balance over a control volume in the fluid, we can determine ˙ Q = ˙ mCpdTm Equating the two equations above we find ˙ mCpdTm = hA (Tw −Tm) | {z } average ∆T By isolating the temperature terms and inte-grating we obtain ln Tw −Tout Tw −Tin !
= −hA ˙ mCp Because of the exponential temperature decay within the tube, it is common to present the mean temperature from inlet to outlet as a log mean temperature difference where ∆Tln = Tout −Tin ln Tw −Tout Tw −Tin ! = Tout −Tin ln(∆Tout/∆Tin) ⇒ ˙ Q = hA∆Tln 8 1. Laminar Flow in Circular Tubes, Isothermal (UWT) and Isoflux (UWF) For laminar flow where ReD ≤2300 NuD = 3.66 ⇒fully developed, laminar, UWT, L > Lt & Lh NuD = 4.36 ⇒fully developed, laminar, UWF, L > Lt & Lh NuD = 1.86 ReDP rD L !1/3 µb µs !0.14 ⇒ developing laminar flow, UWT, P r > 0.5 L < Lh or L < Lt For non-circular tubes the hydraulic diameter, Dh = 4Ac/P can be used in conjunction with Table 13-1 to determine the Reynolds number and in turn the Nusselt number.
2. Turbulent Flow in Circular Tubes, Isothermal (UWT) and Isoflux (UWF) For turbulent flow where ReD ≥2300 the Dittus-Boelter equation (Eq. 13-68) can be used NuD = 0.023 Re0.8 D P rn ⇒ turbulent flow, UWT or UWF, 0.7 ≤P r ≤160 ReD > 2, 300 n = 0.4 heating n = 0.3 cooling For non-circular tubes, again we can use the hydraulic diameter, Dh = 4Ac/P to determine both the Reynolds and the Nusselt numbers.
In all cases the fluid properties are evaluated at the mean fluid temperature given as Tmean = 1 2 (Tm,in + Tm,out) except for µs which is evaluated at the wall temperature, Ts.
9 Natural Convection What Drives Natural Convection?
• fluids tend to expand when heated and contract when cooled at constant pressure • therefore a fluid layer adjacent to a surface will become lighter if heated and heavier if cooled by the surface • a lighter fluid will flow upward and a cooler fluid will flow downward • as the fluid sweeps the wall, heat transfer will occur in a similar manner to boundary layer flow however in this case the bulk fluid is sta-tionary as opposed to moving at a constant ve-locity in the case of forced convection In natural convection, the Grashof number is analogous to the Reynolds number.
Gr = buouancy force viscous force = gβ(Tw −T∞)L3 ν2 Natural Convection Over Surfaces • natural convection heat transfer depends on geometry and ori-entation • note that unlike forced convec-tion, the velocity at the edge of the boundary layer goes to zero • the velocity and temperature profiles within a boundary layer formed on a vertical plate in a stationary fluid looks as fol-lows: 10 Natural Convection Heat Transfer Correlations The general form of the Nusselt number for natural convection is as follows: Nu = f(Gr, P r) ≡CGrmP rn where Ra = Gr · P r • C depends on geometry, orientation, type of flow, boundary conditions and choice of char-acteristic length.
• m depends on type of flow (laminar or turbulent) • n depends on the type of fluid and type of flow • Table 14-1 should be used to find Nusselt number for various combinations of geometry and boundary conditions – for ideal gases β = 1/Tf, (1/K) – all fluid properties are evaluated at the film temperature, Tf = (Tw + T∞)/2.
1. Laminar Flow Over a Vertical Plate, Isothermal (UWT) The general form of the Nusselt number is given as NuL = hL kf = C gβ(Tw −T∞)L3 ν2 | {z } ≡Gr 1/4 ν α |{z} ≡P r 1/4 = C Gr1/4 L P r1/4 | {z } Ra1/4 where RaL = GrLP r = gβ(Tw −T∞)L3 αν 2. Laminar Flow Over a Long Horizontal Circular Cylinder, Isothermal (UWT) The general boundary layer correlation is NuD = hD kf = C gβ(Tw −T∞)D3 ν2 | {z } ≡Gr 1/4 ν α |{z} ≡P r 1/4 = C Gr1/4 D P r1/4 | {z } Ra1/4 D where RaD = GrDP r = gβ(Tw −T∞)L3 αν 11 Natural Convection From Plate Fin Heat Sinks The average Nusselt number for an isothermal plate fin heat sink with natural convection can be determined using NuS = hS kf = " 576 (RaSS/L)2 + 2.873 (RaSS/L)0.5 #−0.5 Two factors must be considered in the selection of the number of fins • more fins results in added surface area and re-duced boundary layer resistance, R ↓= 1 hA ↑ • more fins leads to a decrease fin spacing and a decrease in the heat transfer coefficient R ↑= 1 h ↓A A basic optimization of the fin spacing can be obtained as follows: ˙ Q = hA(Tw −T∞) where the fins are assumed to be isothermal and the surface area is 2nHL, with the area of the fin edges ignored.
For isothermal fins with t < S Sopt = 2.714 L Ra1/4 L !
with RaL = gβ(Tw −T∞)L3 ν2 P r The corresponding value of the heat transfer coefficient is h = 1.307kf/Sopt All fluid properties are evaluated at the film temperature.
12 Example 6-3: Find the optimum fin spacing, Sopt and the rate of heat transfer, ˙ Q for the following plate fin heat sink cooled by natural convection.
Given: W = 120 mm H = 24 mm L = 18 mm t = 1 mm Tw = 80 ◦C T∞ = 25 ◦C P∞ = 1 atm fluid = air Find: Sopt and the corresponding heat transfer, ˙ Q 13 |
14336 | https://www.health.com/adenomyosis-8700361 | What To Know About Adenomyosis
Jess Sims is a health writer specializing in women's and reproductive health, neurology. immunology, and genetics. She's been published in Women's Health, SELF, Harper'z Bazaar, Teen Vogue, Glamour UK, and many other amazing places.
LukaTDB / Getty Images
Adenomyosis is a chronic condition in which the endometrial tissue that lines the uterus grows into the muscular wall of the uterus. Symptoms include heavy menstrual bleeding, pelvic pain, painful intercourse, and infertility—though one in three people with the condition report having no symptoms.
Unlike endometriosis, where the tissue grows outside the uterus, adenomyosis occurs within the uterine wall. However, it's not uncommon for people to have both conditions.
There is no consensus on the diagnostic criteria of adenomyosis, so it is challenging to fully understand the condition's prevalence. According to some reports, 21-34% of people who visit the gynecologist (a healthcare provider who primarily treats health concerns related to female sex organs and hormones) show signs of the condition in sonographic imaging.
While most diagnoses occur in people aged 40-50, younger people with infertility are increasingly being diagnosed due to better imaging techniques.
Types of Adenomyosis
Currently, adenomyosis is not categorized into a single, agreed-upon typing system. Over the years, experts have attempted to classify the condition based on the location of the abnormal tissue or tissue growth progression. Some generally agreed-upon terms refer to the distribution or location of the abnormal tissue and help healthcare providers with prognosis and treatment.
It is important to note that the following subtypes of adenomyosis are not always associated with the severity of symptoms or tissue growth.
Diffuse Adenomyosis
Diffuse adenomyosis refers to the presence of abnormal tissue throughout many areas of the uterine wall, not limited to one location. It is associated with an enlarged, more rounded uterus. In research, people with this type of adenomyosis exhibited more symptoms.
Diffuse adenomyosis is more commonly seen in older people.
Focal Adenomyosis
When healthcare providers use the term focal, they are indicating that the abnormal tissue growth is localized to one specific area of the myometrium, which is the middle layer of the uterus where adenomyosis is commonly found.
In one study, people with focal adenomyosis were at higher risk for infertility and miscarriage, though those findings are not conclusive.
Uterine Adenomyomas
An adenomyoma is a subtype of focal adenomyosis that evolves into a larger nodular mass that thickens the surrounding myometrium. Adenomyomas can originate in the myometrium or endometrium (the inner layer of the uterus) and grow as polyps that become nodular masses.
Little is known about why these nodules form, but they are present in many people who have heavy uterine bleeding and infertility.
Cystic Adenomyosis
Cystic adenomyosis is rare and is characterized by small cysts filled with dark brown fluid. The cysts may be responsible for the chronic pelvic pain some people with adenomyosis experience, which is a less common symptom.
Adenomyosis Symptoms
Symptoms of adenomyosis include:
Clinical signs of adenomyosis also include what’s referred to as a “boggy” uterus, which is in itself a clinical diagnosis. When examining a healthy uterus, the muscular organ is resistant to touching or probing. In a boggy uterus, the organ is enlarged and tender, which can cause pain in the lower abdomen and back.
Another common sign of adenomyosis is infertility. However, because the majority of people diagnosed are in perimenopause or menopause, the exact prevalence is unclear.
Up to a third of people with adenomyosis don’t have any symptoms. Those with symptoms typically find that symptoms increase in severity around 40-50 years old. That does not mean the condition isn’t present in younger people–quite the opposite, as emerging research suggests younger people may just be underdiagnosed.
Causes
Adenomyosis occurs when abnormal tissue grows into the wall of the uterus where it does not belong, often leading to an enlarged uterus, among other signs and symptoms. Due to a lack of research, experts are unsure what causes adenomyosis.
There are two commonly accepted theories about the development of adenomyosis:
Risk Factors
The most common risk factors for adenomyosis are age, previous pregnancies and childbirth, and a history of uterine procedures.
About 70-80% of people who elect to undergo a hysterectomy (removal of the uterus) for the treatment of adenomyosis are in their 40s and 50s. That suggests that being older is a risk factor for adenomyosis. However, newer research is finding that the condition may actually start when the patient is younger, but the diagnosis doesn’t come until older age.
A large percentage of people with adenomyosis have had multiple pregnancies and childbirths. Some research also suggests that prior Cesarean sections are correlated with a higher incidence of adenomyosis, though those findings are inconclusive. These findings support the theory that injured or damaged endometrium is the cause of adenomyosis.
Diagnosis
Historically, signs of adenomyosis were only seen during examinations of post-hysterectomy uterine tissue. In recent years, improvements in imaging technology have made it easier for providers to identify and diagnose the condition.
Diagnosing adenomyosis starts with a healthcare provider’s evaluation and is confirmed through transvaginal ultrasound and pelvic magnetic resonance imaging (MRI). Some of the signs of adenomyosis providers are looking for include:
In focal adenomyosis, it is not always apparent whether the growths are endometrial tissue or fibroids using just transvaginal sonography. In that case, an MRI scan can differentiate between the two conditions for a definitive diagnosis.
Adenomyosis Treatment
The route you take for adenomyosis treatment may largely depend on whether you may want to have children. The only way to cure adenomyosis is through surgery, by removing your uterus with a hysterectomy. If you want to preserve your uterus, treatment would instead focus on controlling symptoms like heavy, painful periods through medications or other surgeries.
Often, healthcare providers try to treat adenomyosis with medication first and work up to surgery if needed.
Medication
Currently, there are no therapies specific to treating adenomyosis approved by the U.S. Food and Drug Administration (FDA). Instead, many people turn to over-the-counter (OTC) pain medications such as non-steroidal anti-inflammatory drugs (NSAIDs) to manage painful periods, chronic pelvic pain, or painful sex.
For ongoing treatment, a healthcare provider may prescribe hormonal contraception, with the ultimate goal of suppressing ovulation, thereby regressing the growth of endometrial tissue. These include:
Surgical Interventions
The most prevalent treatment for adenomyosis is a hysterectomy. One large study found that 80% of people with the condition opted to undergo this surgery, which is the only way to cure it.
If you don’t want a hysterectomy because you want to preserve your fertility and hormonal therapies have not helped symptoms, other surgical options may help. These less invasive procedures focus on reducing the abnormal tissue rather than removing the entire uterus.
One procedure is endometrial ablation, which destroys the lining of the uterus. It is usually offered to people who experience abnormal or heavy menstrual bleeding. This is not normally recommended because it can worsen pain. It might be used in addition to other treatments to reduce heavy bleeding, but know that it has the potential side effect of worsening pain.
Another procedure is myometrial excision or reduction. Surgical removal or reduction of the lesions may be possible in visible focal adenomyosis. However, the effectiveness of excision is less than 50%.
Other surgical options may be more invasive but still preserve fertility. These include the removal of uterine fibroids (myomectomy) and a partial removal of the uterus (partial hysterectomy). Still, the procedures can lead to complications with future pregnancies.
Prevention
There are currently no recognized methods to prevent the development of adenomyosis, as it develops somewhat randomly and does not have clear diagnostic criteria to help target preventative factors.
However, if you develop adenomyosis, you can try to prevent its progression. The only evidence-backed methods to prevent the condition's progression are hormonal medication or surgical interventions.
Related Conditions
Adenomyosis is closely related to uterine fibroids and endometriosis. The prevalence of co-occurring conditions is high—as many as 50% of people with adenomyosis have uterine fibroids and 25-75% of people with endometriosis also have adenomyosis.
The high number of people with these concurrent conditions may be attributed to the connection of abnormal tissue growth in and around the uterus.
One other common symptom between all three conditions is infertility, though there is minimal research into the rates of impaired fertility attributed to adenomyosis.
Living With Adenomyosis
Adenomyosis is not a fatal condition, and there is no evidence it negatively affects your life expectancy. However, it can be a challenging condition to manage.
Fertility concerns, medication management, and weighing the risks and benefits of invasive procedures are just a few of the factors to consider. You may need accommodations for some of the more disruptive symptoms, such as heavy or painful periods and chronic pelvic pain.
If possible, track your menstrual cycle so you can prepare ahead of your period. If the pain affects your quality of life, consider speaking with your healthcare provider about writing a note to your employer for work accommodations.
Adenomyosis often coexists with other gynecologic conditions, making treatment decisions more complicated. Luckily, there are ways to manage symptoms. Even some lifestyle changes can help, and emerging therapies are promising. Until a cure is found, symptom management remains the optimal mode of care.
Frequently Asked Questions
Adenomyosis requires lifelong treatment. Otherwise, there could be continued or worsened impacts on menstrual symptoms or fertility. Pregnancy outcomes may also be affected, such as a high risk of miscarriage.
There are no indications that any food should be avoided if you have adenomyosis.
There is no clinical evidence that adenomyosis causes weight gain. However, some of the medications used to manage adenomyosis symptoms, such as danazol, may cause weight gain.
Gunther R, Walker C. Adenomyosis. In: StatPearls. StatPearls Publishing; 2024.
Tellum T, Munro MG. Classifications of adenomyosis and correlation of phenotypes in imaging and histopathology to clinical outcomes: A review. Curr Obstet Gynecol Rep. 2022;11:1–11. doi:10.1007/s13669-021-00320-5
Rees CO. Adenomyosis: Diagnostic features and clinical impact. In: Eindhoven University of Technology. Eindhoven University of Technology; 2024.
Schrager S, Yogendran L, Marquez CM, Sadowski EA. Adenomyosis: Diagnosis and management. Am Fam Physician. 2022;105(1):33-38.
Latif S, Saridogan E. Uterine adenomyoma—what we know, and what we don’t know: A narrative review. Gynecol Pelvic Med. 2022;5. doi:10.21037/gpm-21-50
Breathnach F, Geary M. Uterine atony: Definition, prevention, nonsurgical management, and uterine tamponade. Semin Perinatol. 2009; 33:2:82-87. doi:10.1053/j.semperi.2008.12.001
Taran FA, Stewart EA, Brucker S. Adenomyosis: Epidemiology, risk factors, clinical phenotype and surgical and interventional alternatives to hysterectomy. Geburtshilfe Frauenheilkd. 2013;73(9):924-931. doi:10.1055/s-0033-1350840
Yu O, Schulze-Rath R, Grafton J, Hansen K, Scholes D, Reed SD. Adenomyosis incidence, prevalence and treatment: United States population-based study 2006-2015. Am J Obstet Gynecol. 2020;223(1):94.e1-94.e10. doi:10.1016/j.ajog.2020.01.016
Borghese G, Doglioli M, Orsini B, et al. Progression of adenomyosis: Rate and associated factors. Intl J Gynecology & Obste. 2024. doi:10.1002/ijgo.15572
Vannuccini S, Petraglia F. Recent advances in understanding and managing adenomyosis. F1000Res. 2019;8:283. doi:10.12688/f1000research.17242.1
Etrusco A, Barra F, Chiantera V, et al. Current medical therapy for adenomyosis: From bench to bedside. Drugs. 2023;83(17):1595-1611. doi:10.1007/s40265-023-01957-7
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The Art of Counting
Course Description
The subject of enumerative combinatorics deals with counting the number of elements of a finite set. For instance, the number of ways to write a positive integer n as a sum of positive integers, taking order into account, is 2n-1. We will be concerned primarily with bijective proofs, i.e., showing that two sets have …
The subject of enumerative combinatorics deals with counting the number of elements of a finite set. For instance, the number of ways to write a positive integer n as a sum of positive integers, taking order into account, is 2n-1. We will be concerned primarily with bijective proofs, i.e., showing that two sets have the same number of elements by exhibiting a bijection (one-to-one correspondence) between them. This is a subject which requires little mathematical background to reach the frontiers of current research. Students will therefore have the opportunity to do original research. It might be necessary to limit enrollment.
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The total number of ways to cover an m x n chessboard (and many other nonrectangular boards as well, such as the Young diagram of a partition) with disjoint snakes is a product of Fibonacci numbers. (Image adapted from Homework 7 (PDF).)
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Probabilities of Drawing Balls (Reference From HKIMO Only) Q
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There are 7 balls in a box including 5 black balls and 2 white balls. Also, 1 black ball will be added at 0:09 and then 1 black ball is added every 10 seconds ( 0 :19, 0 : 29, 0 : 39,...,0 : 59 ). At 25 seconds, 1 white ball will be added into the box. If we draw balls from the box one at a time without replacement, and it needs to wait 31 seconds after each draw, find the maximum possible value of the probability to draw at least one white ball within 1 minute
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There are 6 balls in a box including 3 black balls and 3 white balls. Also, 1 black ball will be added at 0:04 and then 1 black ball is added every 5 seconds (0 : 09, 0 :14, 0 :19,...,0 : 49, 0 : 54, 0 : 59 ). At 29 seconds, 1 white ball will be added into the box. If we draw balls from the box one at a time without replacement, and it needs to wait 21 seconds after each draw, find the maximum possible value of the probability to draw at least one white ball within 1 minute
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There are 6 balls in a box including 3 black ba lls and 3 w hite balls. Also, 1 black ball will be added at 0:02 and then 1 black ball is added every 5 seconds ( 0 : 07, 0 :12, 0 :17, 0 : 22,
…..
,0 : 47, 0 : 52, 0 : 57 ). At 28 and 58 seconds, 1 white ball will be added into the box. If we draw balls from the box one at a time without replacement, and it needs to wait 31 seconds after each draw, find the maximum possible value of the probability to draw at least one white ball within 1 minute.
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There are 7 balls in a box including 2 black balls and 5 white balls. Also, 1 black ball will be added at 0:01 and then 1 black ball is added every 3 seconds ( 0 : 04, 0 : 07, 0 :10, 0 :13,
….
,0 : 52, 0 : 55, 0 : 58 ). At 11,26, 41 and 56 seconds, 1 white ball will be added into the box. If we draw balls from the box one at a time without replacement, and it needs to wait 16 seconds after each draw, find the maximum possible value of the probability to draw at least one white ball within 1 minute.
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MUAC Interpretation Using Shakir's Tape
The tape is color-coded to indicate nutritional status:
Green Zone (≥12.5 cm): Normal nutritional status
Yellow Zone (11.5–12.5 cm): Moderate malnutrition
Red Zone (<11.5 cm): Severe acute malnutrition
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14340 | https://www.cnblogs.com/HuZihu/p/10142737.html | 机器学习---最小二乘线性回归模型的5个基本假设（Machine Learning Least Squares Linear Regression Assumptions）
在之前的文章《机器学习---线性回归（Machine Learning Linear Regression）》中说到，使用最小二乘回归模型需要满足一些假设条件。但是这些假设条件却往往是人们容易忽略的地方。如果不考虑模型的适用情况，就只会得到错误的模型。下面来看一下，使用最小二乘回归模型需要满足哪些假设，以及如果不满足这些假设条件会产生怎样的后果。
最小二乘回归模型的5个基本假设：
自变量（X）和因变量（y）线性相关
自变量（X）之间相互独立
误差项（ε）之间相互独立
误差项（ε）呈正态分布，期望为0，方差为定值
自变量（X）和误差项（ε）之间相互独立
第一个假设：自变量（X）和因变量（y）线性相关
线性相关（linearly dependent）是最基本的假设。如果自变量和因变量之间没有关系或者是非线性关系，那么就无法使用线性回归模型进行预测，或者无法预测出准确的结果。
第二个假设：自变量（X）之间相互独立
如果我们发现本应相互独立的自变量出现了一定程度（甚至高度）的相关性，那么我们就无法知道自变量和因变量之间的真正关系，这称之为共线性（collinearity）。当共线性出现的时候，变量之间的联动关系会导致我们估计的参数的标准差变大，置信区间变宽，由此来看，参数的估计值会变得不稳定，对参数的假设检验也会变得不准确。
（注：两个特征之间相互关联被称之为共线性，但是也有可能三个或更多的特征之间相互关联，即使这些特征两两之间并没有很高的关联，这被称之为多重共线性（multicollinearity））
第三个假设：误差项（ε）之间相互独立
随机误差项的各期望值之间存在着相关关系，称随机误差项之间存在自相关性（autocorrelation）。自相关性通常出现在时间序列里，后一项依赖于前一项；也可能出现在有偏差的样本里，比如样本搜集自同一个家庭的成员。当自相关性出现的时候，预测值的标准差往往比真实的小，进而会导致置信区间变窄，同时，较低的标准差会导致p值较小，这会让我们得到错误的假设检验结果。
第四个假设：误差项（ε）呈正态分布，期望为0，方差为定值
这里其实分为两个假设。第一个假设：误差项服从均值为0的正态分布。第二个假设：误差项的方差为定值（不变）。这两个假设是为了保证回归模型在小样本下能够顺利进行假设检验。正态分布假设仅在小样本的情况下需要，大样本的情况下则不需要，因为有中心极限定理做正态性的支撑。而方差齐性则保证最小二乘法估计出来的统计量具有最小的方差。如果违反了这个假设，置信区间会变宽，这称之为异方差性（heteroscedasticity）。当异方差性出现的时候，如果仍采用最小二乘法估计参数，会导致参数的t检验值被高估，可能造成本来不显著的某些参数变为显著，使假设检验失去意义。
第五个假设：自变量（X）和误差项（ε）之间相互独立
模型中一个或多个自变量与随机误差项存在相关关系，这称之为内生性（endogeneity）。内生性通常由于遗漏变量而导致的，因此是一个普遍存在的问题。内生性会导致模型参数估计不准确。
分类:
Machine Learning
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最小二乘, 线性回归, Least Squares, Linear Regression, Assumptions
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14342 | https://www.quora.com/What-is-the-Nernst-equation-and-its-application | What is the Nernst equation and its application? - Quora
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What is the Nernst equation and its application?
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The Nernst equation is a fundamental equation in electrochemistry that relates the concentration of ions in solution to the electrical potential (voltage) of an electrochemical cell. It provides a way to calculate the cell potential under non-standard conditions.
Nernst Equation
The general form of the Nernst equation is:
E=E∘−R T n F ln Q E=E∘−R T n F lnQ
Where:
E E = cell potential under non-standard conditions (in volts)
E∘E∘ = standard cell potential (in volts)
R R = universal gas constant (8.314 J/(mol·K))
T T = temperature in Kelvin
n n = number of moles of electrons transferred in the r
Continue Reading
The Nernst equation is a fundamental equation in electrochemistry that relates the concentration of ions in solution to the electrical potential (voltage) of an electrochemical cell. It provides a way to calculate the cell potential under non-standard conditions.
Nernst Equation
The general form of the Nernst equation is:
E=E∘−R T n F ln Q E=E∘−R T n F lnQ
Where:
E E = cell potential under non-standard conditions (in volts)
E∘E∘ = standard cell potential (in volts)
R R = universal gas constant (8.314 J/(mol·K))
T T = temperature in Kelvin
n n = number of moles of electrons transferred in the reaction
F F = Faraday's constant (approximately 96485 C/mol)
Q Q = reaction quotient, which is the ratio of the concentrations of the products to the reactants, each raised to the power of their stoichiometric coefficients.
Applications of the Nernst Equation
Determining Cell Potentials: The Nernst equation is used to calculate the potential of electrochemical cells under varying concentrations of reactants and products, allowing predictions of cell behavior in real-world conditions.
Electrode Potentials: It can be applied to individual half-cells to determine their electrode potentials based on ion concentrations, which is important in designing batteries and fuel cells.
pH Measurement: The Nernst equation is utilized in pH meters, where the potential of a glass electrode is related to the hydrogen ion concentration, allowing for accurate pH measurements.
Biological Systems: In biology, the Nernst equation helps to understand the electrochemical gradients across cell membranes, which are crucial for processes like nerve impulse transmission and muscle contraction.
Corrosion Studies: It is used to predict the corrosion potential of metals in various environments, aiding in the development of corrosion-resistant materials.
Thermodynamic Calculations: The Nernst equation links thermodynamics with electrochemistry, providing insights into the spontaneity of reactions based on cell potentials.
Example Calculation
For example, if you want to calculate the potential of a copper electrode in a solution where the concentration of Cu²⁺ is 0.01 M at 298 K, with a standard potential E∘E∘ of +0.34 V for the half-reaction:
Cu 2++2 e−↔Cu Cu 2++2 e−↔Cu
You would substitute into the Nernst equation as follows:
Identify n=2 n=2 (two electrons are transferred).
Calculate Q Q: Q=1[C u 2+]=1 0.01=100 Q=1[C u 2+]=1 0.01=100.
Use R=8.314 J/(mol⋅K)R=8.314 J/(mol·K), F=96485 C/mol F=96485 C/mol, and T=298 K T=298 K.
E=0.34−(8.314)(298)(2)(96485)ln(100)E=0.34−(8.314)(298)(2)(96485)ln(100)
Calculating the logarithmic term and substituting back will yield the potential E E.
This equation is critical for understanding and predicting the behavior of electrochemical systems across various fields.
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What are the applications of the Nernst equation?
What does the Nernst equation allow us to calculate?
How does the Nernst equation work practically?
What is the Nernst equation?
What are the concepts behind the Nernst equation?
Ajay Adarsh S
B.E in Mechanical Engineering, Sri Krishna College of Engineering and Technology, Tamil Nadu, India (SKCET) (Graduated 2021) · Author has 65 answers and 130.5K answer views
·4y
The Nernst equation is an equation that relates the reduction potential of an electrochemical reaction (half cell or full cell reaction) to the standard electrode potential, temperature, and activities such as concentrations of the chemical species undergoing reduction and oxidation.
Applications of Nernst Equation:
1.One of the major application of Nernst equation is in determining ion concentration.
It is also used to calculate the potential of an ion of charge “z” across a membrane.
It is used in oxygen and the aquatic environment.
It is also used in solubility products and potentio-met
Continue Reading
The Nernst equation is an equation that relates the reduction potential of an electrochemical reaction (half cell or full cell reaction) to the standard electrode potential, temperature, and activities such as concentrations of the chemical species undergoing reduction and oxidation.
Applications of Nernst Equation:
1.One of the major application of Nernst equation is in determining ion concentration.
It is also used to calculate the potential of an ion of charge “z” across a membrane.
It is used in oxygen and the aquatic environment.
It is also used in solubility products and potentio-metric titrations.
It is also used in pH measureme.
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M Srinath
Studied at RV College of Engineering (RVCE) (Graduated 2022)
·7y
Originally Answered: What is the Nernst equation? ·
Nernst equation is an important equation in electrochemistry in order to find cell voltage when conditions are not standard i.e 298K temperature,1 ATM pressure and 1 M concentration
The equation is as below
Ecell = E°cell- RT/nF(ln [product]/[reactant])
Ecell is the electrode potential
E°cell is the standard electrode potential
R- universal Gas constant
T-temperature
n-number of electrons transferred in the redox reaction
F -Faraday constant (96500 C)
[Reactant]-reactant concentration
[Product]-product concentration
However we can also use to find the cell potential under standard conditions
In that case
Continue Reading
Nernst equation is an important equation in electrochemistry in order to find cell voltage when conditions are not standard i.e 298K temperature,1 ATM pressure and 1 M concentration
The equation is as below
Ecell = E°cell- RT/nF(ln [product]/[reactant])
Ecell is the electrode potential
E°cell is the standard electrode potential
R- universal Gas constant
T-temperature
n-number of electrons transferred in the redox reaction
F -Faraday constant (96500 C)
[Reactant]-reactant concentration
[Product]-product concentration
However we can also use to find the cell potential under standard conditions
In that case RT/nF ln becomes 0.059/n log
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Venkat Vuddagiri
Software Engineer 1 at NCR (company) (2022–present)
·7y
Originally Answered: What is the Nernst equation? ·
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Which is a good affordable wireless laser printer that prints both sides of paper automatically?
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Toppr
Learning App for Class 5-12 (Official Account) · Author has 576 answers and 449.3K answer views
·4y
Originally Answered: What is the Nernst equation? ·
Nernst Equation:
Nernst equation is a general equation that relates the Gibbs free energy and cell potential in electrochemistry. It is very helpful in determining cell potential, equilibrium constant etc.
It takes into account the values of standard electrode potentials, temperature, activity and the reaction quotient for the calculation of cell potential. For any cell reaction, Gibbs free energy can be related to standard electrode potential as:
ΔG =-nFE
Where, ΔG= Gibbs free energy, n = number of electrons transferred in the reaction, F = Faradays constant (96,500 C/mol) and E= cell potential.
Continue Reading
Nernst Equation:
Nernst equation is a general equation that relates the Gibbs free energy and cell potential in electrochemistry. It is very helpful in determining cell potential, equilibrium constant etc.
It takes into account the values of standard electrode potentials, temperature, activity and the reaction quotient for the calculation of cell potential. For any cell reaction, Gibbs free energy can be related to standard electrode potential as:
ΔG =-nFE
Where, ΔG= Gibbs free energy, n = number of electrons transferred in the reaction, F = Faradays constant (96,500 C/mol) and E= cell potential. Under standard conditions, the above equation can be given as,
ΔGo =-nFEo
According to the theory of thermodynamics, Gibbs free energy under general conditions can be related to Gibbs free energy under the standard condition and the reaction quotient as:
ΔG=ΔGo + RT lnQ
Where, Q= reaction quotient, R= universal gas constant and T= temperature in Kelvin. Incorporating the value of ΔG and ΔGo, from the first two equations, we get:
-nFE = -nFE0 + RT lnQ
E = E0 – (RT/nF) lnQ
Converting natural log to log10, the above equation is known as the Nernst equation. Here, it relates the reaction quotient and the cell potential. Special cases of Nernst equation:
E = Eo − (2.303RT/nF) log10Q
At standard temperature, T= 298K:
E = Eo − (0.0592V/n) log10Q
At standard temperature T = 298 K, the 2.303RTF, term equals 0.0592 V.
Under Equilibrium Condition:
As the redox reaction in the cell proceeds, the concentration of reactants decreases while the concentration of products increases. This goes on until equilibrium is achieved. At equilibrium, ΔG = 0. Hence, cell potential, E = 0. Thus, the Nernst equation can be modified to:
E0 – (2.303RT/nF) log10Keq = 0
E0 = (2.303RT/nF) log10Keq
Where, Keq = equilibrium constant and F= faradays constant. Thus, the above equation gives us a relation between standard electrode potential of the cell in which the reaction is taking place and the equilibrium constant.
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Why do we need Nernst equation?
What does the Nernst equation tell us?
What does a negative Nernst potential mean?
How nernst equation works?
What is the relationship between the Nernst equation and the equilibrium constant?
Vineet Verma
B.Sc from Veer Bahadur Singh Purvanchal University (Graduated 2019) · Author has 60 answers and 31.2K answer views
·5y
Originally Answered: What are the applications of the Nernst equation? ·
1.One of the major application of Nernst equation is in determining ion concentration.
is also used to calculate the potential of an ion of charge “z” across a membrane.
is used in oxygen and the aquatic environment.
is also used in solubility products and potentiometric titrations.
is also used in pH measurements.
i hope it will help you
regards
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Vishnuthirtha Madaksira
Professor in Chemistry at Sri Guru Krupa Coaching Centre (2009–present) · Author has 10.6K answers and 2.7M answer views
·3y
Originally Answered: What are the applications of the Nernst equation? ·
Applications of the Nernst equation :
• The EMF of the cell can be calculated if given different concentrations of the redox reaction.
• The equilibrium constant of the reaction can be calculated.
• The pH of the solution can be calculated from the standard hydrogen electrode, given the H+ ion concentration.
etc…..
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9 2
Abhiram Kosuru
Journey beyond!! · Author has 204 answers and 552K answer views
·8y
Originally Answered: What is the Nernst equation? ·
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Meher Memon
6y
Originally Answered: What is Nernst equation? ·
In electrochemistry , the Nernst equation is an equation that relates the reduction potential of an electrochemical reaction (half-cell or full cell reaction) to the standard electrode potential, temperature, and activities (often approximated by concentrations) of the chemical species undergoing reduction and oxidation. It is the most important equation in the field of electrochemistry. It was named after Walther Nernst, a German physical chemist who formulated the equation.
When the cell potential equals zero, the reaction is at equilibrium.
Nernst Equation
Can be used to find the cell pote
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In electrochemistry , the Nernst equation is an equation that relates the reduction potential of an electrochemical reaction (half-cell or full cell reaction) to the standard electrode potential, temperature, and activities (often approximated by concentrations) of the chemical species undergoing reduction and oxidation. It is the most important equation in the field of electrochemistry. It was named after Walther Nernst, a German physical chemist who formulated the equation.
When the cell potential equals zero, the reaction is at equilibrium.
Nernst Equation
Can be used to find the cell potential at any moment in during a reaction or at conditions other than standard-state.
E = cell potential (V) under specific conditions
E°= cell potential at standard-state conditions
R = ideal gas constant = 8.314 J/mol-K
T = temperature (kelvin), which is generally 25°C (298 K)
n = number of moles of electrons transferred in the balanced equation
F = Faraday's constant, the charge on a mole of electrons = 95,484.56 C/mol
lnQc= the natural log of the reaction quotient at the moment in time
Sample Calculation
Calculate the cell potential for the following system:
Write the half-reactions with the half-cell potentials:
Multiply the reactions to get the lowest common multiple of electrons:
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9 1
Wonderslate Technologies
Support at Wonderslate (2015–present) · Author has 705 answers and 228.2K answer views
·3y
The Nernst equation is often used to calculate the cell potential of an electrochemical cell at any given temperature, pressure, and reactant concentration. The equation was introduced by a German chemist named Walther Hermann Nernst.
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9 1
Vikas Gupta
Btech in IT Engineering, K. J. Somaiya College of Engineering (Graduated 2022) · Author has 418 answers and 704K answer views
·7y
Originally Answered: What is the Nernst equation? ·
Nernst equation is a equation which relates reduction potential of a cell to the standard reduction potential of a cell, temperature and concentration of product and reactant
This equation is derived from Gibbs free energy equation
Equation :
E c e l l=E∘c e l l−R T n F L n Q E c e l l=E c e l l°−R T n F L n Q
where
E c e l l=E c e l l=reduction potential at temperature T T and at reaction quotient Q Q
E∘c e l l=E c e l l°=standard reduction potential
R=R=uni
Continue Reading
Nernst equation is a equation which relates reduction potential of a cell to the standard reduction potential of a cell, temperature and concentration of product and reactant
This equation is derived from Gibbs free energy equation
Equation :
E c e l l=E∘c e l l−R T n F L n Q E c e l l=E c e l l°−R T n F L n Q
where
E c e l l=E c e l l=reduction potential at temperature T T and at reaction quotient Q Q
E∘c e l l=E c e l l°=standard reduction potential
R=R=universal gas constant=[math]8.314JK^...
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Puneet20102010
1y
The Nernst equation formula establishes a relationship between the reaction quotient, electrochemical cell potential, temperature, and the standard cell potential
Upvote ·
Rod Glen
Master of Science in Analytical Chemistry, University of Hull (UK) (Graduated 1998) · Author has 3.7K answers and 1.1M answer views
·4y
Originally Answered: What is the Neinst equation? ·
Do you mean the Nernst equation? This is an equation that is used in electrochemistry to describe the behaviour of ion selective cells/probes. The most common of these are those used in pH meters.
See here for a fuller description
Nernst equation - Wikipedia General form with chemical activities edit When an oxidized species ( Ox ) accepts a number z of electrons ( e − ) to be converted in its reduced form ( Red ), the half-reaction is expressed as: Ox + ze − ⟶ Red {\displaystyle {\ce {Ox + ze- -> Red}}} The reaction quotient ( Q r ), also often called the ion activity product ( IAP ), is the ratio between the chemical activities ( a ) of the reduced form (the reductant , a Red ) and the oxidized form (the oxidant , a Ox ). The chemical activity of a dissolved species corresponds to its true thermodynamic concentration taking into account the electrical interactions between all ions present in solution at elevated concentrations. For a given dissolved species, its chemical activity (a) is the product of its activity coefficient (γ) by its molar (mol/L solution), or molal (mol/kg water), concentration (C): a = γ C. So, if the concentration ( C , also denoted here below with square brackets [ ]) of all the dissolved species of interest are sufficiently low and that their activity coefficients are close to unity, their chemical activities can be approximated by their concentrations as commonly done when simplifying, or idealizing, a reaction for didactic purposes: Q r = a Red a Ox = [ Red ] [ Ox ] {\displaystyle Q_{r}={\frac {a_{\text{Red}}}{a_{\text{Ox}}}}={\frac {[\operatorname {Red} ]}{[\operatorname {Ox} ]}}} At chemical equilibrium , the ratio Q r of the activity of the reaction product ( a Red ) by the reagent activity ( a Ox ) is equal to the equilibrium constant K of the half-reaction: K = a Red a Ox {\displaystyle K={\frac {a_{\text{Red}}}{a_{\text{Ox}}}}} The standard thermodynamics also says that the actual Gibbs free energy Δ G is related to the free energy change under standard state Δ G o by the relationship: Δ G = Δ G ⊖ + R T ln Q r {\displaystyle \Delta G=\Delta G^{\ominus }+RT\ln Q_{r}} where Q r is the reaction quotient and R is the universal ideal gas constant . The cell potential E associated with the electrochemical reaction is defined as the decrease in Gibbs free energy per coulomb of charge transferred, which leads to the relationship Δ G = − z F E . {\displaystyle \Delta G=-zFE.} The constant F (the Faraday constant ) is a unit conversion factor F = N A q , where N A is the Avogadro constant and q is the fundamental electron charge. This immediately leads to the Nernst equation, which for an electrochemical half-cell is E red = E red ⊖ − R T z F ln Q r = E red ⊖ − R T z F ln a Red a Ox . {\displaystyle E_{\text{red}}=E_{\text{red}}^{\ominus }-{\frac {RT}{zF}}\ln Q_{r}=E_{\text{red}}^{\ominus }-{\frac {RT}{zF}}\ln {\frac {a_{\text{Red}}}{a_{\text{Ox}}}}.} For a complete electrochemical reaction (full cell), the equation can be written as E cell = E cell ⊖ − R T z F ln Q r {\displaystyle E_{\text{cell}}=E_{\text{cell}}^{\ominus }-{\frac {RT}{zF}}\ln Q_{r}} where: E red is the half-cell reduction potential at the temperature of interest, E o red is the standard half-cell r
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14343 | https://www.albert.io/blog/solving-literal-equations/ | Skip to content
Solving Literal Equations: Explanation, Review, and Examples
Look at you go! You’ve now learned how to solve one-step equations, two-step equations, and multi-step equations.
Now it’s time to discuss solving literal equations.
In this lesson, we will define literal equations, examine examples of literal equations, as well as learn how to solve literal equations. Let’s literally get excited to begin!
What We Review
What is a literal equation?
Recall from our earlier study, that an equation is a mathematical sentence that uses an equal sign, =, to show that two expressions are equal. Unlike other equations you have already worked with, literal equations are equations primarily made up of letters and variables.
Many of the literal equations you have worked with in your life have been formulas. Although these equations will look different than our normal equations, they still follow the same rules of solving.
Examples of literal equations
Although the idea of equations including mostly letters may seem like a foreign topic, you have used literal equations many times in your life. Here are some of the examples of literal equations you have already worked with in your life:
Area of a Rectangle
A=b⋅h
Circumference of a Circle
C=π⋅D
Simple Interest Formula
I=p⋅r⋅t
Each letter (or variable) in the literal equation has a specific meaning and changes from problem to problem.
Return to the Table of Contents
How to solve literal equations
Solving literal equations follows the same rules as solving a one-step or two-step equation. The idea of “solving” a literal equation essentially means we are rearranging the letters (or variables) to isolate a new variable. A literal equation is “solved” when the variable of interest is alone on one side of the equation.
Explore Albert school licenses!
Similar to solving equations, we will use inverse operations to isolate the variable by itself. Here are examples of inverse operations:
Addition↔Subtraction
Multiplication↔Division
Here are a few examples of solving literal equations:
Example 1
Solve for h in the following literal equation:
A=b⋅h
Remember this formula? As stated above, this is the Area of a Rectangle. As noted previously, there are mainly letters and variables in Literal Equations. If this was a simple equation such as 10=2x, we would’ve simply divided both sides by 2 to achieve my final answer.
When “solving” Literal Equations, we follow the same rules as simple equations. Therefore, in order to solve for h in this equation, we need to isolate it by itself. Therefore, we will divide both sides by b.
bA=bb⋅h
Doing so will isolate h, giving us the answer:
h=bA
Example 2:
Although formulas are a common example of Literal Equations, not all Literal Equations are formulas. We can also rearrange and “solve” a Literal Equation for any variable. For Example:
Solve for m in the following equation:
| | |
--- |
| x=m+n | Original equation |
| x−n=m+n−n | Subtract n from both sides |
| x−n=m | m is now isolated |
| m=x−n | |
Even though there were no numbers in the equation, we have “solved” the Literal Equation for m.
Return to the Table of Contents
Multi-step literal equations
Example 1
Not all Literal Equations only take one step to solve. Here is an example of using multiple steps to solve Literal Equations.
Start practicing Algebra 1 on Albert now!
Solve for r in the following equation:
V=πr2h
Given is the formula for the Volume of a Cylinder. In order to solve for r, we must first get r2 by itself:
πhV=πhπr2h
Then we are left with:
πhV=r2
Then to solve for r, we must take the square root of both sides:
πhV=r2
Now we have r isolated by itself, giving us the new Literal Equation:
r=πhV
Example 2
Here is an example of a Literal Equation that is not a formula, but which we can still solve for a variable.
Solve for x in the following equation:
4(x+y)=P
There are two ways to approach this problem. The first method is to treat it as an equation and distribute the 4, then solve:
4x+4y=P
We can then subtract 4y from each side:
4x+4y−4y=P−4y
Then we will need to divide each side by 4:
44x=4P−4y
Finally, we need to simplify our equation:
x=4P−44y
x=4P−y
Now, we have finally solved for x in the Literal Equation. Let’s see how you can solve the equation, without having to simplify at the end.
In the other method, we can simply divide by 4 at the beginning to avoid using the Distributive Property. For example:
44(x+y)=4P
Then, we would simply have to subtract y from each side:
x+y−y=4P−y
Therefore, we end up with:
x=4P−y
Notice how the equation is already simplified, and no other steps are needed.
Return to the Table of Contents
Here’s a short video demonstrating how to solve literal equations:
Literal equations with fractions
Let’s work on some examples of literal equations involving fractions!
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Example 1
Many literal equations, and formulas, involve fractions in some sort. For example, here’s the formula for the Volume of a Sphere:
V=34πr3
Let’s say we are asked to solve for the radius, r. First, we must eliminate the fraction. Let’s do this by multiplying by the reciprocal.
| | |
--- |
| V=34πr3 | Original equation |
| 43⋅V=43⋅34πr3 | |
| 43⋅V=πr3 | |
| 4π3V=ππr3 | Divide each side by π |
| 4π3V=r3 | Simplify |
| 34π3V=3r3 | Cube Root both Sides |
| r=34π3V | |
Now that r is isolated, we have successfully solved for r.
Example 2
What happens if we simply want to rearrange an equation for the other variable?
For instance, solve the following equation for x:
y=4x−81
Notice how there are two variables in the equation and our ultimate goal is still to isolate x. Remember, we can eliminate all fractions in one move by multiplying all terms by the Least Common Denominator.
In this Literal Equation, the least common denominator is 8. Therefore, we will multiply each term by 8.
8⋅y=8⋅4x−8⋅81
This will give us an equation that no longer has fractions:
8y=2x−1
Then continue solving just as you would a normal equation:
| | |
--- |
| 8y=2x−1 | Original equation |
| 8y+1=2x−1+1 | Add 1 to each side |
| 8y+1=2x | Simplify |
| 28y+1=22x | Divide each side by 2 |
| 28y+21=x | Simplify |
| 4y+21=x | Simplify |
Now that we have isolated x by itself, we have correctly “solved” the Literal Equation.
Return to the Table of Contents
Keys to Remember: Solving Literal Equations
A Literal Equation is an equation that contains all letters (or variables) or an Equation that has multiple variables
Formulas, such as P=2L⋅2W, are common examples of Literal Equations
We solve Literal Equations by isolating a determined variable on one side of the equation
Solving Literal Equations follow the same rules as normal equations, so we must do Inverse Operations in order to solve
Remember, whatever you do to one side, you must do to the other!
To solve Literal Equations with fractions, you can multiply each term by the Least Common Denominator to eliminate the fractions
Return to the Table of Contents
Read these other helpful posts:
Solving One-Step Equations
Solving Two-Step Equations
Solving Multi-Step Equations
Forms of Linear Equations
View ALL Algebra 1 Review Guides
Return to the Table of Contents
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14344 | https://www.sciencedirect.com/science/article/pii/S0146000519300424 | Including ultrasound scans in antenatal care in low-resource settings: Considering the complementarity of obstetric ultrasound screening and maternity waiting homes in strengthening referral systems in low-resource, rural settings - ScienceDirect
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Abstract
Keywords
Introduction
The First Look Study and the WHO ultrasound recommendation
The First Look Study and descriptive study
Discussion
Conclusion
Ethics approval and consent to participate
Consent for publication
Availability of data and material
Competing interests
Funding
Authors’ contributions
Acknowledgments
References
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Seminars in Perinatology
Volume 43, Issue 5, August 2019, Pages 273-281
Including ultrasound scans in antenatal care in low-resource settings: Considering the complementarity of obstetric ultrasound screening and maternity waiting homes in strengthening referral systems in low-resource, rural settings
Author links open overlay panel David L.Swanson a, Holly L.Franklin b, Jonathan O.Swanson a, Robert L.Goldenberg c, Elizabeth M.McClure b, Waseem Mirza d, David Muyodi e, Lester Figueroa f, Nicole Goldsmith a, Nancy Kanaiza e, Farnaz Naqvi d, Irma Sayury Pineda f, Walter López-Gomez f, Dorothy Hamsumonde g, Victor Lokomba Bolamba h, Jamie E.Newman b, Elizabeth V.Fogleman b, Sarah Saleem d, Fabian Esamai e, Sherri Bucher j…Robert O.Nathan a
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Abstract
Recent World Health Organization (WHO) antenatal care recommendations include an ultrasound scan as a part of routine antenatal care. The First Look Study, referenced in the WHO recommendation, subsequently shows that the routine use of ultrasound during antenatal care in rural, low-income settings did not improve maternal, fetal or neonatal mortality, nor did it increase women's use of antenatal care or the rate of hospital births. This article reviews the First Look Study, reconsidering the assumptions upon which it was built in light of these results, a supplemental descriptive study of interviews with patients and sonographers that participated in the First Look study intervention, and a review of the literature. Two themes surface from this review. The first is that focused emphasis on building the pregnancy risk screening skills of rural primary health care personnel may not lead to adaptations in referral hospital processes that could benefit the patient accordingly. The second is that agency to improve the quality of patient reception at referral hospitals may need to be manufactured for obstetric ultrasound screening, or remote pregnancy risk screening more generally, to have the desired impact. Stemming from the literature, this article goes on to examine the potential for complementarity between obstetric ultrasound screening and another approach encouraged by the WHO, the maternity waiting home. Each approach may address existing shortcomings in how the other is currently understood. This paper concludes by proposing a path toward developing and testing such a hybrid approach.
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Keywords
Pregnancy risk screening
Maternity waiting home
Referral systems
Continuum of care
Task shifting
Midwifery
Introduction
As of 2016, the World Health Organization (WHO) recommends an ultrasound scan as a part of routine antenatal care (ANC). This recommendation is part of a comprehensive WHO guideline on routine ANC for pregnant women and adolescent girls.1 Specifically, one ultrasound scan before 24 weeks’ gestation is recommended for pregnant women to estimate gestational age, improve detection of fetal anomalies and multiple pregnancies, reduce induction of labor for post-term pregnancy, and improve a woman's pregnancy experience.
The development group for the WHO recommendations acknowledges that the use of early pregnancy ultrasound has not been shown to reduce perinatal mortality. The Global Network First Look Study (2013–2016), which was underway at the time of publication, is discussed among the considerations of the ultrasound recommendation. This multi-country cluster randomized trial, according to the WHO ANC guideline, “should contribute further evidence on the health effects, health care utilization and implementation-related information on ultrasound in rural low-resource settings.1 ”
Of these expectations, the First Look Study did provide further evidence on health effects and health care utilization when the results were published in 2018. The results showed, however, that the routine use of ultrasound during ANC did not improve maternal, fetal or neonatal mortality, or maternal near-miss. Moreover, ultrasound screening during ANC did not increase women's use of ANC or the rate of hospital births.2
The expectation that the study should generate ‘implementation related information’ on ultrasound in low-resource settings was also met. Elements of the study's implementation have been discussed in detail elsewhere, including the training of ultrasound naïve health personnel in obstetric ultrasound screening,3 the development of web-based quality assurance,4 and an evaluation of their combined effectiveness and accuracy.5 A subsequent case study on challenges of implementing the obstetric ultrasound screening for the First Look Study in the Democratic Republic of the Congo details some of the political, logistical, infrastructural and resource challenges acknowledged in the WHO guideline.6
This article reviews the intervention of the First Look Study, reconsidering the assumptions upon which it was built in light of the study's results, WHO recommendations, a supplemental study of interviews with patients and sonographers that participated in the First Look Study intervention, and a review of the literature. What surfaces is a potential for complementarity between obstetric ultrasound screening and another approach encouraged by the WHO, the maternity waiting home.7 Each approach may address existing shortcomings in how the other is currently understood. A hybrid approach might also strike a better balance between evaluation and implementation. This paper concludes by proposing a path toward developing and testing such a hybrid approach.
The First Look Study and the WHO ultrasound recommendation
The First Look Study was a cluster-randomized trial conducted in rural areas of the Democratic Republic of Congo (DRC), Guatemala, Kenya, Pakistan, and Zambia primarily to evaluate the impact of obstetric ultrasound screening at routine ANC visits on maternal, fetal and neonatal mortality and maternal near-miss. The study aimed, secondarily, at evaluating the intervention's effect on women's use of ANC and the rate of hospital births. The study also assessed the quality of the field sonographers’ ultrasound examination through review of a proportion of the examinations by expert sonographers using a web-based program.4 Finally, First Look investigators conducted a qualitative study to better understand the reasons for women's acceptance of referrals.
Clusters were predefined geographical areas with a health center, approximately 300–500 expected deliveries per year, and a Maternal Newborn Health Registry, an independent study which documented all pregnancies and their outcomes to 6 weeks post-delivery. Ultrasound units and training were introduced in intervention cluster health centers, as well as in those hospitals to which patients from intervention and control clusters were referred. Standard care was provided in the control clusters.
The trial was approved by all participating institutional review boards and ethic review committees; all women and sonographers who participated provided informed consent. The study design, training, and ethical approvals are described in detail elsewhere.3, 8
Task shifting
The recent WHO ANC guideline notes that antenatal ultrasound is a task which potentially can be shifted from trained sonographers and doctors at hospitals to skilled attendants in rural health care facilities.1 This approach was utilized in the First Look Study, where obstetric ultrasound screening during ANC visits was performed by rural health center personnel. We use the term ‘field sonographer’ throughout to encompass the skilled attendants – the nurses, midwives, medical and clinical officers – trained to use ultrasound at ANC in the First Look Study intervention clusters. Field sonographers were taught to assess gestational age, to identify high-risk pregnancies – including multiple gestations, malpresentation, placenta previa, intrauterine growth restriction, and some fetal anomalies – and when and how to refer patients to hospitals providing comprehensive emergency obstetric and neonatal care.
The WHO ANC guideline also emphasizes the importance of quality assurance, ongoing training, supervision and staff retention, all of which were reinforced by the First Look Study. The concept of ‘obstetric ultrasound screening’ employed in the study embodies the integration of task-shifting with oversight. In this approach, positive screening results require confirmation before consequent interventions are undertaken. Patients screening positively for high-risk pregnancies by field sonographers at intervention health centers, for example, were encouraged to refer to a hospital for a confirmatory ultrasound by hospital sonographers and physicians. This relationship, with the field sonographer acting as an extension of the hospital sonographer, also provided supervision. With the assistance of a web-based quality assurance process, involving the remote review of stored images of ultrasound scans, hospital sonographers could be aimed or could target where continued training of field sonographers in obstetric ultrasound screening was required.4 Hospital sonographers also trained replacement field sonographers in obstetric ultrasound screening as necessary. A strong indication of the effectiveness of these task-shifting and quality assurance processes is seen in the review of stored images using the web-based quality assurance process. The concordance between field sonographers and reviewers in the ultrasound diagnosis was 99.4%.5
The continuum of care
Through a broader lens, the First Look Study's intervention increased the pregnancy risk screening capacity of skilled attendants at rural, primary health care facilities in low-resource settings. The effectiveness of skilled attendants on improving healthy pregnancy outcomes in these settings, according to the WHO, is also dependent upon their role within a continuum of care.9 For the sake of clarity, ‘continuum of care’ in this article refers to the household-to-hospital continuum of care, spanning the home, community, health center, and hospital.10, 11 The continuum starts at home with the woman and her family, is followed by first level care that involves the provision of high-quality care, and – when complications occur – may require care at secondary or tertiary levels of the health system. The importance of a viable continuum of care in ensuring quality care at the time of birth remains a priority in recent literature.12
An awareness of the importance of this continuum of care is apparent in the First Look Study's protocol.8 Although the focus of the study's intervention was on training skilled attendants, other inputs were incorporated across the continuum. Skilled attendants were trained to be field sonographers, for example, and to encourage women screening positively for potential complications to refer to hospitals. Hospital sonographers were taught to confirm the findings of these screenings upon referral. In communities targeted for the intervention, sensitization activities were conducted to inform women and their families of the availability of ultrasound at their antenatal care clinics. In referral hospitals, training in the management of obstetric and neonatal care was provided to staff as necessary. Furthermore, guidance was offered to health system officials and hospital administrators on possible referral system enhancements. An outward manifestation of this guidance was the creation of referral algorithms – developed with input from the local health system and posted near each ultrasound – to help field sonographers determine the need for and the timing of referrals.
Increasing the capacity of the skilled attendant, however, remained the primary focus of the study's intervention. As such, the First Look Study depended in large part upon the skilled attendant playing a central role in the continuum of care. The training emphasized the importance of communicating with the patient during the scan, helping the patient understand what ultrasound can show, and incorporating these findings into a patient's birth-preparedness plan. These emphases correspond to the skilled attendant's ‘core skills and abilities’ spelled out by the WHO, including the capacity to “assess individual needs, give appropriate advice and guidance, calculate the expected date of delivery and perform specific screening tests.9” The skilled attendant's enhanced risk-assessment capabilities and effectiveness in using them to encourage patients with high-risk pregnancies to deliver at a hospital, in other words, were the fundamental means by which pregnancy outcomes were intended to improve.
This concept of positioning the skilled attendant at the center of the continuum of care is one that the WHO has strongly advocated. This concept is articulated in the 2004 joint statement of the WHO, the International Confederation of Midwives and the International Federation of Gynecology and Obstetrics, Making pregnancy safer: the critical role of the skilled attendant. The statement emphasizes that in childbearing, “women need a continuum of care to ensure the best possible health outcome for them and their newborns.” In doing so, it positions the “skilled attendant… at the center of this continuum of care.” It elaborates on this, stating, “At the primary health care level, she/he will need to work with other care providers in the community, such as traditional birth attendants and social workers. She/he will also need strong working links with health care providers at the secondary and tertiary levels of the health system.9 ”
The results of the First Look Study, showing no improvement in maternal, fetal and neonatal outcomes and no increase in the rate of hospital births, however, may open the possibility of revisiting the reasoning behind this approach. Perhaps, even with enhanced risk-assessment and improved birth-preparedness skills, the position of the skilled attendant – particularly one stationed in a rural health center in a low-recourse health system – lacks the agency to improve and maintain a continuum of care sufficient for improved outcomes.
The First Look Study and descriptive study
Communicating across the continuum of care
Because the First Look Study was conducted across 5 study sites in 5 different countries, numerous concerns arose; some unique, some ubiquitous.6 Approaches to addressing these concerns were shared across sites at the study's inception and throughout its course. Bound by the study's protocol and respectful of the autonomy of each country site within the Global Network, solutions implemented in one site were suggested to other sites. Many of these suggestions were aimed at improving the continuum of care through small measures which incorporated ultrasound-enhanced risk-screening.
An approach originating from the Kenya site involved the use of a black book that tracked those patients with potentially high-risk pregnancies. A patient, for example, with a fetus lying transversely, was recommended to return to the health center for a follow-up scan at 36 weeks to determine whether the condition persisted and referral was recommended. The black book contained the recommended date of the follow-up scan and a contact cellphone number of the patient, a family member, a community health worker or a neighbor, depending upon availability. A call was placed to encourage a follow-up visit by the patient once the recommended date passed. This simple approach aimed to strengthen the home-to-health center section of the continuum of care. The prevalence of cellphone technology makes this possible and would be enhanced by well-organized community health worker networks.
Across study sites, a similar system was suggested for the hospitals to which patients from intervention health centers were referred. Here hospital staff – often sonographers – were encouraged to receive calls or texts from field sonographers and record the screening results and timing of referrals. In the ideal, a hospital could then keep track of expected referrals, communicating back to health centers when an expected patient did not arrive by a given date. Personnel at the health center could then reach out across the continuum of care to the community and household to further encourage the patient to make her referral.
Where First Look Study sites encouraged field sonographers to communicate at the household and community levels, real and perceived hierarchies in health systems made field sonographers communication with referral hospitals substantially more problematic. This is reflected in interviews conducted as part of a descriptive study in the intervention clusters of the First Look study, described in detail elsewhere.13 The descriptive study consisted of structured interviews conducted near the end of the 18-month study at all five country sites. Individual structured interviews were conducted in each site with field sonographers, hospital sonographers, and patients recommended for referral during ANC ultrasound screenings.
In these structured interviews, 3 of the 38 field sonographers responded – when asked for “any other comments regarding the ultrasound referral” – with basic concerns about communication with referral hospitals. Following are their responses: “Need of a direct contact person at the referral centre;” “Proper link to referral site like phone numbers and the person receiving a client on the other side;” “Patients are not told who will attend them.” When asked for some of the reasons that women are not going to the referral visit, 8 of 38 field sonographers specified that “The referral hospital is not attentive to patients that we send.” Field sonographers’ responses were channeled to boxes checked by the interviewers, and multiple boxes could be checked, the last being “Other, specify.” Two responses specified: “Non-availability of the referral sonographer and the woman gave birth before the sonographer was available” and “Don't know which department to go to as where (sic) it was located.” Details of the survey are described elsewhere.13
Where the field sonographers could communicate with the hospital about referrals, they appear not to have had the means to ensure that their improved risk-assessment capacity was used effectively. Even where they had “strong working links” with hospital health care providers, as recommended by the WHO,9 these seemed insufficient to bring about the structural change necessary to accommodate their improved risk screening capacity. In Zambia, for example, the relationship between field sonographer and hospital sonographer was established through two weeks of initial training and was strengthened through continued training and communication over referrals. As a result, the hospital sonographer received referred patients directly, saving patients from ANC processing at the hospital, which could consume a day. The sonographer could not, however, facilitate the patient's interaction with the hospital further than the provision of a confirmatory scan. In a structured interview with hospital sonographers conducted concurrently with those mentioned above, one sonographer responded as follows when asked at the end of a structured survey interview if there is anything else he thought important to mention: “The rest of (the hospital) is not as responsive or say they have no bed space (or) more serious conditions to deal with.” Another respondent mentioned, “If the (maternal child health department) was better engaged,” similarly seeming to desire better connectivity across hospital departments.
Concerns surrounding the need for structural change in hospitals to accommodate the improved risk screening of the intervention surfaced in all study sites throughout the course of the study.6 This observation suggests that the study's emphasis on training skilled attendants, in line with WHO recommendations, with merely the provision of guidance to health system and hospital officials, may in retrospect have been insufficient to improve outcomes. The temporary nature of the study and the need to contain its intervention into a measurable, scalable entity may have limited the intervention's impact as well.6 The importance of making structural changes to referral systems, particularly along the health center-to-hospital section of the continuum of care, to accommodate the increased capacity in risk screening of ultrasound rose to prominence for some investigators during the course of the study. In light of the study's results, this aspect of the approach appears to warrant further scrutiny.
The complexity of referral
Obstetric referrals from rural primary health centers to hospitals in low-resource settings involve levels of complexity that can create barriers for rural women. This receives little to no attention in the literature concerned with ultrasound in ANC in low-resource settings. Moreover, the task-shifting nature of obstetric ultrasound screening, with the corresponding need for confirmation, may increase this complexity.
To better understand this complexity, consider, as an example, a woman from Lukolis, a rural community in Kenya. Upon receiving an ultrasound scan as a part of an antenatal visit at her local health center, this woman is told by the field sonographer that she has screened positively for twins. The field sonographer recommends the woman refer to a hospital in the city of Busia, 20 km away, for a confirmatory ultrasound, ideally initiating a process intending the woman to deliver there.
In the literature concerning ultrasound in antenatal care, a referral is often considered a kind of finality for the ultrasound intervention. The field sonographer discovers potentially complicated pregnancies and refers those patients for treatment at the referral hospital, theoretically leading to better maternal and neonatal outcomes. This is expressed in terms like: “Once diagnosed, patients with complicating conditions…would ideally be referred to a regional obstetric center where they would be managed appropriately.14 ”
The First Look Study protocol reflects the literature in this regard. The protocol emphasized that for obstetric ultrasound screening in primary health centers to have a chance at being effective: “Having a referral institution with staff trained to review ultrasound findings and manage complications is crucial.8 ” As to what happens to the patient between being referred at the primary health center and delivering at the hospital, the protocol is brief, limiting the extent of intervention to be targeted at this section of the continuum of care. In its discussion of referral and system enhancement the protocol says, “While this will not be a major trial component, we expect to hold several sessions with appropriate health system leaders and administrators to discuss integration of obstetric/neonatal care between the primary health clinics and referral hospitals.8 ” Again, this was because the focus was on building the capacity of the skilled attendant who, in the words of the WHO, is “pivotal in reducing maternal mortality and morbidity.9 ”
What appears to be overlooked here is the complexity of a referral from the patient's perspective. For the woman from Lukolis, Kenya, a referral means that she must pay for a ride on a minibus to the district hospital in Busia. Once she finds the hospital, she discovers that she must first attend a processing visit in the ANC department before she can schedule a confirmatory ultrasound scan in the radiology department. It takes most of the day for the ANC visit to be completed, and the confirmatory scan is scheduled for the following morning. She now either needs to travel back home and return to Busia Hospital by the morning or find a place to spend the night in Busia.
This complexity is reflected in the patients’ responses during interviews conducted as part of a descriptive study mentioned above.13 Individual structured interviews for patients were conducted in each country site with a convenience sample of women at 6 weeks post-delivery for whom referral was recommended during ANC ultrasound screenings in primary health centers. An additional interview was conducted with women who made the referral; another for those who did not. Of the 190 interviews conducted with women who did not make their referrals, 54 indicated that they attempted to visit the referral hospital. Table 1 compiles the reasons that women did not attend the referral visit. Of these responses, 25 pertain to the patient not receiving adequate attention at the hospital and 13 relate to the hospital being an intimidating and/or difficult place for a patient to find her way through.
Table 1. Reasons women did not attend the referral visit.
| Reason | Frequency |
--- |
| I was told to come back later | 14 |
| I did not know where to go in the hospital | 10 |
| I was not attended to on the day I visited | 5 |
| I was not comfortable being at the hospital | 3 |
| I was told to come back the following day and had nowhere to spend the night | 3 |
| I had an appointment, but was not attended to on that day | 3 |
| Multiple response | 10 |
| Missing | 6 |
Of the 135 women interviewed who did not attempt to visit the referral hospital, the barriers most often identified were cost (45% of interviewees), transportation (16%) and distance to the hospital (14%). Disapproval by the father (20%), other family members/neighbors (9%), and traditional healers or clergy/pastor (7%) were also among the barriers cited. Concerns about the hospital – ‘heard about bad experiences as the hospital’ (10%), ‘not comfortable going to the hospital’ (7%) – also surfaced in the responses. 10 interviewees (7%) who responded to an “Other, specify” option also cited concerns with the hospital, including, “hospital staff treats patients poorly” and “fear of the hospital.”
Continuing with our example, suppose the woman from Lukolis overcomes these initial barriers, traveling 20 km home for the night and returning to Busia District Hospital the next day. There the sonographer is able conduct an ultrasound exam confirming that she has twins. The woman is then told to go to the maternity ward. The nurses there inform her that she should deliver at the hospital; but, because of the limited availability of beds, she should present at the hospital only once labor begins. The woman now needs to find and afford direct transportation from her home in Lukolis or stay near the hospital in Busia in the days or weeks before her due date, allowing her to be present at the hospital soon after the onset of labor.
Instead of a simple visit to the hospital, the woman's referral now requires three visits to Busia and possibly accommodation in the city for days or weeks. Each of these steps increases costs and time away from a household that depends upon her. Each step represents an additional barrier which may keep the woman from delivering in a facility that provides the comprehensive emergency obstetric and neonatal care she may need.
The structured interviews conducted in conjunction with the First Look Study provide a glimpse into how women are affected by additional barriers created by having to travel to the referral hospital more than once. Of the 510 women that indicated that they attended a referral visit during their pregnancies when asked at 6 weeks post-delivery, 121 indicated that they did not deliver at a hospital. Table 2 details reasons women provided for why they did not deliver at a hospital. The additional costs in terms of time and money of returning to the hospital at delivery became prohibitive for many.
Table 2. Reasons why women that attended the referral visit did not deliver at a hospital.
| Reason | Frequency | Percent |
---
| Expense / lack of money | 21 | 19.63% |
| Time | 23 | 21.50% |
| No transportation | 10 | 9.35% |
| Distance to referral hospital | 13 | 12.15% |
| Other, specify | 40 | 37.38% |
| Frequency missing=14 |
Of those who responded “Other”, 9 responses related to the timing of deliveries in the vein of: “labour was sudden” or “delivered before time.” 9 responses referred to matters of choice like: “there was no need, the baby was fine” or “she was afraid to have a cesarean,” while another 4 indicated that delivering at the health center seemed sufficient with statements such as: “Midwife at the clinic was able to deliver the mother.” 5 responses refer to advice or a change in diagnosis such as: “In the hospital they told me that everything was fine” or “Baby changed position to normal.” 6 responses specified concerns with the hospital such as: “They do not take good care at the hospital” or, perhaps most notably, “I went to deliver at the hospital, but it took long for baby to come, I was sent back home. At arrival in my residence the labor worsen after 20 min I delivered my baby boy and girl.” The remaining responses indicated some misunderstanding of the question, of the response, or in the translation.
The series of interviews referenced here was conceived by the investigators at a point during the study when concerns about the referral processes across sites mounted and preconceptions of the barriers to referral began to predominate. What these interviews provide is some elucidation into the complexity of a referral as experienced by the patient. Through this lens, the barriers relating to money, time, distance and social and cultural constraints appear to remain formidable; but, they may be compounded by the poor quality of reception that referral hospitals offer patients. Moreover, by increasing the number of visits to a referral hospital a woman is asked to make during pregnancy, obstetric ultrasound screening may to some extent counteract its intended outcome of helping more women with high risk pregnancies deliver in hospitals.
Discussion
With the WHO ANC guideline mentioned at outset desiring the First Look Study to generate ‘implementation related information’, we turn here toward recommendations for future approaches of incorporating obstetric ultrasound screening into ANC provision in rural primary health care settings. In this regard, two themes emerge from this review. The first is that the focused emphasis that the First Look Study put on building the skills of the skilled attendant did not take into account the lack of agency rural primary health setting personnel have in building and improving the continuum of care in the direction of the referral hospital. The second is that a means of manufacturing such agency to improve the quality of patient reception at referral hospitals may be essential for obstetric ultrasound screening to have its desired impact.
Improvements on approaches to incorporating obstetric ultrasound screening into ANC, in this light, should have a dual emphasis on increasing the risk-screening capacity of health center personnel and improving the quality of reception of patients at referral hospitals. The concept of ‘streamlining’ patients who have screened positively for high-risk pregnancies, enabling them to bypass ANC processing, receive a confirmatory scan, and meet with an obstetrician or nurses in the maternity department for the purpose of developing birth-preparedness plans, all within a day's visit to the hospital, has been discussed elsewhere.6 Beyond streamlining, the need exists for improvements that address the barriers of transport, timing and cost for women without an immediate means of reaching the hospital at the onset of labor. In the ideal, what is needed is an entity within health systems that can build, improve and maintain continuums of care in a way less heavily reliant on health center personnel at their centers.
Maternity waiting homes
In 1996, the WHO published Maternity Waiting Homes: A review of experiences, endorsing the concept as low-cost means of bringing women closer to needed obstetric care, as part of a comprehensive package of essential obstetric services.7 The maternity waiting home (MWH) is loosely defined as a shelter located near a hospital or primary health center for pregnant women to reside for a period prior to delivery.15 MWHs range from simple shelters to facilities with beds, showers, and kitchens, managed by nurses, linked to the adjoining hospitals, and offering health-related courses to visiting women by day. With encouraging anecdotal evidence indicating that MWHs were successful in reducing maternal mortality, the WHO report states in 1996 that “little quantitative research has been conducted to prove their efficacy.7 ” Two decades later, a Cochrane review still found insufficient evidence to determine the effectiveness of MWH for improving maternal and neonatal outcomes.16
The concept of prenatal risk selection has played an important role in descriptions of the MWH.16, 17 Risk screening algorithms that include maternal age, parity, height, and obstetric history – sometimes factoring in the distance of the patient's home from the health facility – have been used to determine which patients are recommended to MWHs. These have tended to have relatively low positive predictive values because of the low-risk and high-prevalence of conditions targeted in the algorithms.18 A frequent concern that arises in the literature is whether these risk screening algorithms lead to an effective use of resources.16 On the other hand, much of prenatal care is devoted to screening for specific conditions not likely to be diagnosed by the algorithms described above or by ultrasound, such as preeclampsia. The discussion below is relevant to these conditions as well.
A combined intervention
Within the breadth of MWHs' definition may be found a complementary intervention to obstetric ultrasound screening in ANC. The WHO review emphasizes that the MWH “is not a stand-alone intervention, but rather serves to link communities with the health system in a continuum of care.7 ” Our findings point to the need for a strengthening of this continuum of care to make effective use of increased risk screening in remote health care settings. Determining more concise parameters and minimum requirements for what constitutes MWHs may help strengthen the continuum of care where it appears most needed.
It is worth noting that, by combining the MWH with the improved risk screening capacity of ANC ultrasound, as well as other screening tests, the result should constitute a needed improvement acknowledged throughout the literature on MWHs. With a 99.4% level of diagnostic accuracy of ultrasound screenings determined through web-based quality assurance in the First Look Study,5 and with the conditions discovered by ultrasound mostly low-prevalence, high-risk pregnancy complications,18 risk selection may become one of the combined intervention's strengths.
With regard to the themes that emerged from this review – keeping in mind scalability and measurability – we recommend some parameters for the MWH. Each of these recommendations reflect the experience of the First Look Study, the development of its ultrasound intervention, other conditions screened for during ANC, and aspects of existing MWHs described in the literature. We recommend that the following be evaluated:
1.An MWH connected to and in the vicinity of a referral hospital that provides continuous, quality comprehensive emergency obstetric and neonatal care.
2.Hospital staff engaged at an MWH to strengthen the link between communities and health systems, increasing connectivity with the referral hospital in particular.
3.An MWH tasked with managing referred patients’ relationships with the hospital, streamlining their interactions with its departments, to ensure sufficient agency exists along the continuum of care to advocate on their behalf.
4.Non-emergent, screened patients referred to an MWH, allowing the MWH to manage and track referrals and communicate with primary health centers for follow-up.
5.An MWH house and care for those screened patients required – while seeking hospital care – to spend the night away from home, as well as those with high-risk pregnancies nearing their due date or the appointed date for a planned cesarean section or induction.
Further parameters might be established through discussion with stakeholders and evaluators of MWHs, with emphasis placed on allowing for local adaptations to accommodate cultural norms and political environments.
Finally, one reason that the First Look Study of routine use of ultrasound during ANC may have failed to improve maternal, fetal or neonatal mortality could be that the conditions screened by ultrasound may not present enough risk, even if appropriately treated, to have influenced these outcomes.2 Some of the issues brought to light by the study and the accompanying structured interviews, however, likely pertain to remote obstetric risk screening more broadly. Patients screening positively for preeclampsia, for example, may benefit from improvements along the continuum of care that an MWH aimed at accommodating remote obstetric risk screening provides. Similarly, as improvements on risk screening are developed, this approach may provide a basis for their integration into rural, low-resource health settings.
Conclusion
With the existing WHO recommendation of one ultrasound scan before 24 weeks’ gestation for pregnant women, and in light of the First Look Study's and its supplementary study's results, this article considers a means of improving the impact of obstetric ultrasound screening in ANC. In line with recent findings, our recommendations aim to improve access to a range of maternal health services, which has been shown to be one of the two most important predictors of maternal mortality, along with per capita income.16, 19 By combining the interventions of obstetric ultrasound screening and screening for other high-risk conditions with a redefined MWH, the possibility exists for improving and maintaining the continuum of care in a way that focusing primarily on building the capacity of the primary health center personnel appears unable to do. At present, evidence that triaging mothers with high obstetric risk for hospital delivery or to an MWH may be associated with improved perinatal outcomes remains limited to low-quality observational studies, although it is recommended that risk screening tools be evaluated as an intervention in combination with access to obstetric care.18 In current analyses of MWHs, ultrasound has not been considered as a tool for risk screening in rural, low-resource settings. We recommend researching the development and use of the MWH as a means of capturing the benefits of the improved risk screening which ultrasound provides. To do this, we recommend first defining more sharply the parameters of the MWH with regard to its role in supporting remote obstetric risk screening. The refined MWH could then be piloted in different settings with existing ultrasound and other screening services to better understand the impact it can have on improving communication along the continuum of care and the quality of high-risk patient referrals. If evaluations of the piloted MWHs warrant, a study on whether obstetric ultrasound screening with MWHs positively impacts maternal and neonatal outcomes might further the understanding of ultrasound's value in rural, primary antenatal care in low-resource settings.
Ethics approval and consent to participate
The First Look Study and subsequent descriptive study were approved by all participating institutional review boards and ethic review committees; all women and sonographers who participated provided informed consent. The study design, training, and ethical approvals are described in detail here:3, 8
Consent for publication
Not applicable.
Availability of data and material
The data that support the findings of this study are available through the corresponding author upon reasonable request.
Competing interests
The authors declare that they have no competing interests.
Funding
The study was conducted by sites of the Eunice Kennedy Shriver National Institute of Child Health and Human Development (NICHD) Global Network for Women's and Children's Health Research through grants from NICHD to the sites and to RTI International, the data coordination center. Funding for the trial also was provided by the Bill and Melinda Gates Foundation: OPP1056057 to the sites with additional funding to RTI. General Electric provided the ultrasound units to the sites and in addition provided support to the Radiology Department of the University of Washington for the training component of the study. None of the funders were involved in the data analysis or writing any of the manuscripts using data from this study.
Authors’ contributions
DS, HF, JS, EM, RG, RN, JN, CB and WC were involved in drafting the manuscript or revising it critically for important intellectual content in line with their substantial contributions to conception, design and analysis and interpretation. WM, DM. LF, NG, NK, FN, ISP, WLG, DH, VLB, EF, SS, FE, SB, EL, AG, NFK, KMH, EC, MB, MM, AT and AL made substantial contributions to conception and design, or acquisition of data, or analysis and interpretation of data.
Acknowledgments
FIRST LOOK: Ultrasound Study Investigators and Lead Sonographers include Democratic Republic of Congo Site: Antoinette Tshefu, MD, Adrien Lokangaka, MD, and Victor Lokomba Bolamba, MD, University of Kinshasa School of Public Health; Carl Bose, MD; University of North Carolina at Chapel Hill; Zambia Site: Elwyn Chomba, MBChB, DCH, MRCP, Musaku Mwenechanya, MD, and Dorothy Hamsumonde, University Teaching Hospital, Lusaka, Zambia; Wally Carlo, MD, University of Alabama, Birmingham, Alabama; Guatemala Site: Ana Garces, MD, MPH, Lester Figueroa, MD, and Irma Sayury Pineda, MD, Instituto de Nutrición de Centroamérica y Panamá (INCAP), Guatemala City, Guatemala; Michael Hambidge, MD and Nancy Krebs, MD, University of Colorado Health Care System (UCHSC), Denver, Colorado; Pakistan Site: Omrana Pasha, MD, Sarah Saleem, MD, Farnaz Naqvi, MSc, and Waseem Mirza, MD, Aga Khan University, Karachi, Pakistan; Robert Goldenberg, MD, Department of Obstetrics and Gynecology, Columbia University, New York, NY; Kenya Site: Fabian Esamai, MBChB, MMed, PhD and David Muyodi, MD, and Nancy Kanaiza Moi University School of Medicine, Eldoret, Kenya, Ed Liechty, MD, Indiana University School of Medicine, Indianapolis, Indiana; and the Data Coordinating Center: Elizabeth McClure, PhD; Dennis Wallace, PhD and Holly Franklin, MPH, Research Triangle Institute, Durham NC. Eunice Kennedy Shriver National Institute of Child Health and Human Development (NICHD): Menachem Miodovnik, MD and Marion Koso-Thomas, MD, MPH; University of Washington: Robert Nathan, MD; Jonathan Swanson, MD and David Swanson, MIDP, Department of Radiology, University of Washington, Seattle, WA.
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Intention to use maternity waiting home and associated factors among pregnant women in Gamo Gofa zone, Southern Ethiopia, 2019
2021, Plos One Show abstract A maternity waiting home is a temporary residence in which pregnant women from remote areas wait for their childbirth. It is an approach targeted to advance access to emergency obstetric care services especially, in hard-to-reach areas to escalate institutional delivery to reduce complications that occur during childbirth. Apart from the availability of this service, the intention of pregnant women to utilize the existing service is very important to achieve its goals. Thus, this study aimed to assess the intention to use maternity waiting homes and associated factors among pregnant women. Community-based cross-sectional study was conducted among 605 pregnant women using a multistage sampling technique from March 10 to April 10, 2019, by using a structured questionnaire through a face-to-face interview. The collected data was entered into Epi-Data version 3.1 and analyzed using the SPSS version 24 statistical package. Logistic regression analysis was used to test the association. All variables at p-value < 0.25 in bivariate analysis were entered into multivariate analysis. Lastly, a significant association was declared at a P-value of < 0.05 with 95% CI. In this study, the intention to use maternity waiting homes was 295(48.8%, 95%CI: 47%-55%)). Occupation (government employee) (AOR:2.87,95%CI: 1.54–5.36), previous childbirth history (AOR:2.1,95%CI:1.22–3.57), past experience in maternity waiting home use AOR:4.35,95%CI:2.63–7.18), direct (AOR:1.57,95%CI:1.01–2.47) and indirect (AOR: 2.18, 1.38,3.44) subject norms and direct (AOR:3.00,95%CI:2.03–4.43), and indirect (AOR = 1.84,95%CI:1.25–2.71) perceived behavioral control of respondents were significantly associated variables with intention to use maternity waiting home. The magnitude of intention to use maternity waiting homes among pregnant women is low. Community disapproval, low self-efficacy, maternal employment, history of previous birth, and past experiences of MWHs utilization are predictors of intention to use MWHs, and intervention programs, such as health education, strengthening and integration of community in health system programs need to be provided.
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Current State of Health of Women and Children in Low and Middle Income Countries
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Edward A.Liechty
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4: Introduction to Organic Compounds
CHEM 118 Textbook
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4.1: Aliphatic Hydrocarbons
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Table of contents
1. Learning Objectives
2. Note
3. Alkanes
1. condensed structural formula
2. Example (\PageIndex{1})
1. Solutions
3. Exercise \(\PageIndex{1}\)/CHEM_118_Textbook/04:_Introduction_to_Organic_Compounds/4.1:_Aliphatic_Hydrocarbons#Exercise_.5C(.5CPageIndex.7B1.7D.5C))
Homologous Series
Note
Naming (IUPAC Nomenclature)
Note
Example (\PageIndex{2})
Solution
Exercise (\PageIndex{2})
Isomerism-Branched Chain Alkanes
Naming Branched Alkanes
Example (\PageIndex{3}):
Solution
Exercise (\PageIndex{3})
Answer
Example (\PageIndex{4}):
Solution
Exercise (\PageIndex{4})
Answer
Properties of Alkanes
Cyclic Hydrocarbons
Unsaturated Hydrocarbons: Alkenes and Alkynes
Alkenes
Note
Alkynes
Naming Alkenes and Alkynes
Example (\PageIndex{5})
Solution
Exercise (\PageIndex{5})
Example (\PageIndex{6})
Solution
Exercise (\PageIndex{6})
Properties of Alkenes and Alkynes
Geometric Isomers (Cis-Trans) Isomers
Summary
Contributors and Attributions
Learning Objectives
Define and identify alkanes, alkenes, alkynes, and cyclic hydrocarbons.
List some properties of hydrocarbons.
Identify and name simple (straight-chain) alkanes given formulas and write formulas for straight-chain alkanes given their names.
Define structural and geometric isomers.
Write condensed structural formulas and draw line-angle formulas given complete structural formulas for alkanes,alkenes, alkynes, and cyclic hydrocarbons.
The simplest organic compounds are those composed of only two elements: carbon and hydrogen. These compounds are called hydrocarbons. Hydrocarbons themselves are separated into two types: aliphatic hydrocarbons and aromatic hydrocarbons. Aliphatic hydrocarbons are hydrocarbons based on chains of C atoms. There are three types of aliphatic hydrocarbons. Alkanesare aliphatic hydrocarbons with only single covalent bonds. Alkenes are hydrocarbons that contain at least one C–C double bond, and alkynes are hydrocarbons that contain a C–C triple bond. Occasionally, we find an aliphatic hydrocarbon with a ring of C atoms; these hydrocarbons are called cycloalkanes(or cycloalkenes or cycloalkynes). Because alkanes have the maximum number of H atoms possible according to the rules of covalent bonds, alkanes are also referred to as saturated hydrocarbons.
Note
The word saturated has the same meaning for hydrocarbons as it does for the dietary fats and oils: the molecule has no carbon-to-carbon double bonds (C=C).
Alkanes
The simplest alkanes have their C atoms bonded in a straight chain; these are called normal alkanes. They are named according to the number of C atoms in the chain. The smallest alkane is methane:
Figure (\PageIndex{1}). Two dimensional model of methane (top); Three-dimensional representation of methane © Thinkstock. The methane molecule is three dimensional, with the H atoms in the positions of the four corners of a tetrahedron. (bottom)
The diagrams representing alkanes are called structural formulas because they show the structure of the molecule. As molecules get larger, structural formulas become more and more complex. One way around this is to use a condensed structural formula, which lists the formula of each C atom in the backbone of the molecule.
condensed structural formula
The condensed formulas show hydrogen atoms right next to the carbon atoms to which they are attached, as illustrated for butane:
The ultimate condensed formula is a line-angle formula (or line drawing) , in which carbon atoms are implied at the corners and ends of lines, and each carbon atom is understood to be attached to enough hydrogen atoms to give each carbon atom four bonds. For example, we can represent pentane (CH 3 CH 2 CH 2 CH 2 CH 3) and isopentane [(CH 3)2 CHCH 2 CH 3] as follows:
Table (\PageIndex{1}) Structural formulas for methane, ethane , and propane| Structural Formula | | | |
| Ball and Stick Model | | | |
| Space Filling model | | | |
| Name | Methane | Ethane | Propane |
| Condensed Structural Formula | CH 4 | CH 3 CH 3 | CH 3 CH 2 CH 3 |
| Molecular Formula | CH 4 | C 2 H 6 | C 3 H 8 |
| Line angle drawing | | | |
Example (\PageIndex{1})
Provide the molecular formula, the complete structural formula, the condensed structural formula and the line angle formula for a straight chain alkane with 8 carbon atoms.
Solutions
Example (\PageIndex{2}) writing formulas for a straight chain alkane with eight carbon atoms.| structural formula | Explanation | Answer |
---
| Molecular Formula | The general formula for an alkane is C n H 2 n+ 2 For an alkane with 8 carbon atoms, C 8 H(2 × 8) + 2 = C 8 H 18. | C 8 H 18. |
| Complete Structural Formula | Eight carbons first linked together and all remaining bonds to hydrogen to fulfill the octet rule. | C 1-C 2-C 3-C 4-C 5-C 6-C 7-C 8 |
| Condensed Structural Formula | The terminal carbon atoms (C 1 and C 8) are bonded to three hydrogen atoms each and the six middle carbon atoms are bonded to two hydrogen atoms each. The six middle carbon atoms are all bonded to two hydrogen atoms so the structural formula can be simplified further. | CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 CH 3(CH 2)6 CH 3 |
| Line Angle Drawing | Eight carbon atoms are implied at the corners and ends of lines, and each carbon atom is understood to be attached to enough hydrogen atoms to give each carbon atom four bonds. | |
Exercise (\PageIndex{1})
Provide the molecular formula, the complete structural formula, the condensed structural formula, and the line angle formula for a straight chain alkane with 7 carbon atoms.
Answermolecular formula: C 7 H 16 complete structural formula: condensed structural formula: CH 3(CH 2)5 CH 3 line angle drawing:
Homologous Series
Methane (CH 4), ethane (C 2 H 6), and propane (C 3 H 8) are the beginning of a series of compounds in which any two members in a sequence differ by one carbon atom and two hydrogen atoms—namely, a CH 2 unit. The first 10 members of this series are given in Table (\PageIndex{1}).
Table (\PageIndex{1}) The First 10 Straight-Chain Alkanes| Molecular Formula | Condensed Structural Formula | Name |
---
| CH 4 | CH 4 | methane |
| C 2 H 6 | CH 3 CH 3 | ethane |
| C 3 H 8 | CH 3 CH 2 CH 3 | propane |
| C 4 H 10 | CH 3 CH 2 CH 2 CH 3 | butane |
| C 5 H 12 | CH 3 CH 2 CH 2 CH 2 CH 3 | pentane |
| C 6 H 14 | CH 3(CH 2)4 CH 3 | hexane |
| C 7 H 16 | CH 3(CH 2)5 CH 3 | heptane |
| C 8 H 18 | CH 3(CH 2)6 CH 3 | octane |
| C 9 H 20 | CH 3(CH 2)7 CH 3 | nonane |
| C 10 H 22 | CH 3(CH 2)8 CH 3 | decane |
Note
Parentheses in condensed structural formulas indicate that the enclosed grouping of atoms is attached to the adjacent carbon atom.
Consider the series in Figure (\PageIndex{2}). The sequence starts with C 3 H 8, and a CH 2 unit is added in each step moving up the series. Any family of compounds in which adjacent members differ from each other by a definite factor (here a CH 2 group) is called a homologous series. The members of such a series, called homologs, have properties that vary in a regular and predictable manner. The principle of homology gives organization to organic chemistry in much the same way that the periodic table gives organization to inorganic chemistry. Instead of a bewildering array of individual carbon compounds, we can study a few members of a homologous series and from them deduce some of the properties of other compounds in the series.
Figure (\PageIndex{2}) Members of a homologous series.
The principle of homology allows us to write a general formula for alkanes: C n H 2 n+ 2. Using this formula, we can write a molecular formula for any alkane with a given number of carbon atoms. For example, an alkane with eight carbon atoms has the molecular formula C 8 H(2 × 8) + 2 = C 8 H 18.
Naming (IUPAC Nomenclature)
A systematic way of naming hydrocarbons and other organic compounds has been devised by the International Union of Pure and Applied Chemistry (IUPAC). These rules, used worldwide, are known as the IUPAC System of Nomenclature. A stem name in Table (\PageIndex{1}) indicates the number of carbon atoms in the longest continuous chain (LCC). Atoms or groups attached to this carbon chain, called substituents, are then named, with their positions indicated by numbers.
Table (\PageIndex{2})Stems That Indicate the Number of Carbon Atoms in Organic Molecules| Stem | Number of C Atoms |
--- |
| meth- | 1 |
| eth- | 2 |
| prop- | 3 |
| but- | 4 |
| pent- | 5 |
| hex- | 6 |
| hept- | 7 |
| oct- | 8 |
| non- | 9 |
| dec- | 10 |
Note
A continuous (unbranched) chain of carbon atoms is often called a straight chain even though the tetrahedral arrangement about each carbon gives it a zigzag shape. Straight-chain alkanes are sometimes called normal alkanes, and their names are given the prefix n-. For example, butane is called n-butane. We will not use that prefix here because it is not a part of the system established by the International Union of Pure and Applied Chemistry.
Example (\PageIndex{2})
What is the name of an eight carbon alkane?
Solution
The stem that refers to eight carbon atoms is octplus the suffix -anefor an alkane, so the answer is Octane.
Exercise (\PageIndex{2})
Provide the name for the following compounds:
a. seven carbon alkane
b. six carbon alkane
Answer
a. heptane
b. hexane
Isomerism-Branched Chain Alkanes
One of the interesting aspects of organic chemistry is that it is three-dimensional. A molecule can have a shape in space that may contribute to its properties. Molecules can differ in the way the atoms are arranged - the same combination of atoms can be assembled in more than one way. These compounds are known as isomers. Isomers are molecules with the same molecular formulas, but different arrangements of atoms. We will look at some isomer possibilities for alkanes and alkenes.
We can write the structure of butane (C 4 H 10) by stringing four carbon atoms in a row,
–C–C–C–C–
and then adding enough hydrogen atoms to give each carbon atom four bonds:
Figure (\PageIndex{2}) Complete structural formula of butane
The compound butane has this structure, but there is another way to put 4 carbon atoms and 10 hydrogen atoms together. Place 3 of the carbon atoms in a row and then branch the fourth one off the middle carbon atom:
Figure (\PageIndex{3}) Incomplete structural formula
Now we add enough hydrogen atoms to give each carbon four bonds.
There is a hydrocarbon that corresponds to this structure, which means that two different compounds have the same molecular formula: C 4 H 10. The two compounds have different properties—for example, one boils at −0.5°C; the other at −11.7°C. Different compounds having the same molecular formula are called isomers.
Figure (\PageIndex{4}) Butane and isobutane. The ball-and-stick models of these two compounds show them to be isomers; both have the molecular formula C4H10.
Notice that C 4 H 10 is depicted with a bent chain in Figure (\PageIndex{4}). The four-carbon chain may be bent in various ways because the groups can rotate freely about the C–C bonds. However, this rotation does not change the identity of the compound. It is important to realize that bending a chain does not change the identity of the compound; all of the following represent the same compound, butane:
The structure of isobutane shows a continuous chain of three carbon atoms only, with the fourth attached as a branch off the middle carbon atom of the continuous chain, which is different from the structures of butane (compare the two structures in Figure (\PageIndex{4}).
Unlike C 4 H 10, the compounds methane (CH 4), ethane (C 2 H 6), and propane (C 3 H 8) do not exist in isomeric forms because there is only one way to arrange the atoms in each formula so that each carbon atom has four bonds.
Next beyond C 4 H 10 in the homologous series is pentane. Each compound has the same molecular formula: C 5 H 12. The compound at the far left is pentane because it has all five carbon atoms in a continuous chain. The compound in the middle is isopentane; like isobutane, it has a one CH 3 branch off the second carbon atom of the continuous chain. The compound at the far right, discovered after the other two, was named neopentane (from the Greek neos, meaning “new”). Although all three have the same molecular formula, they have different properties, including boiling points: pentane, 36.1°C; isopentane, 27.7°C; and neopentane, 9.5°C.
Figure (\PageIndex{5}) complete structural formulas for pentane isopentane and neopentane
Naming Branched Alkanes
You have already learned the basics of nomenclature-the names of the first 10 normal hydrocarbons. Here, we will add some steps to the procedure so you can name branched hydrocarbons.
First, given the structure of an alkane, identify the longest continuous chain of C atoms. Note that the longest chain may not be drawn in a straight line. The longest chain determines the parent name of the hydrocarbon. For example, in the molecule
Figure (\PageIndex{6}) structural formula for hexane
the longest chain of carbons has six C atoms. Therefore, it will be named as a hexane. However, in the molecule
Figure (\PageIndex{7}) structural formula for heptane
the longest chain of C atoms is not six, but seven, as shown. So this molecule will be named as a heptane.
The next step is to identify the branches, or substituents,on the main chain. The names of the substituents, or alkyl groups, are derived from the names of the parent hydrocarbons; however, rather than having the ending -ane, the substituent name has the ending -yl. Table (\PageIndex{3}), lists the substituent names for the five smallest substituents.
Table(\PageIndex{3})Substituent Names.| Substituent Formula | Number of C Atoms | Name of Substituent |
---
| CH 3 | 1 | methyl- |
| CH 3 CH 2 | 2 | ethyl- |
| CH 3 CH 2 CH 2 | 3 | propyl- |
| CH 3 CH 2 CH 2 CH 2 | 4 | butyl- |
| CH 3 CH 2 CH 2 CH 2 CH 2 | 5 | pentyl- |
| and so forth | and so forth | and so forth |
In naming the branched hydrocarbon, the name of the substituent is combined with the parent name of the hydrocarbon without spaces. However, there is likely one more step. The longest chain of the hydrocarbon must be numbered, and the numerical position of the substituent must be included to account for possible isomers. As with double and triple bonds, the main chain is numbered to give the substituent the lowest possible number. For example, in this alkane
Figure (\PageIndex{7}) Structural formula for 3-methylpentane
the longest chain is five C atoms long, so it is a pentane. There is a one-carbon substituent on the third C atom, so there is a methyl group at position 3. We indicate the position using the number, which is followed by a hyphen, the substituent name, and the parent hydrocarbon name-in this case, 3-methylpentane. That name is specific to that particular hydrocarbon and no other molecule. Organic chemistry nomenclature is very specific!
It is common to write the structural formula of a hydrocarbon without the H atoms, for clarity. So we can also represent 3-methylpentane as
Figure (\PageIndex{8}) Structural formula for 3-methylpentane without the hydrogen atoms
where it is understood that any unwritten covalent bonds are bonds with H atoms. With this understanding, we recognize that the structural formula for 3-methylpentane refers to a molecule with the formula of C 6 H 14.
Example (\PageIndex{3}):
Name this molecule.
Solution
The longest continuous carbon chain has sevenC atoms, so this molecule will be named as a heptane. There is a two-carbon substituent on the main chain, which is an ethyl group. To give the substituent the lowest numbering, we number the chain from the right side and see that the substituent is on the third C atom. So this hydrocarbon is 3-ethylheptane.
Exercise (\PageIndex{3})
Name this molecule.
Answer
2-methylpentane
Branched hydrocarbons may have more than one substituent. If the substituents are different, then give each substituent a number (using the smallest possible numbers) and list the substituents in alphabetical order, with the numbers separated by hyphens and with no spaces in the name. So the molecule
Figure (\PageIndex{8}) 3-ethyl-2-methylpentane
is 3-ethyl-2-methylpentane.
If the substituents are the same, then use the name of the substituent only once, but use more than one number, separated by a comma. Also, put a numerical prefix before the substituent name that indicates the number of substituents of that type. The numerical prefixes are listed in Table (\PageIndex{4}). The number of the position values must agree with the numerical prefix before the substituent.
Table (\PageIndex{4}) Numerical Prefixes to Use for Multiple Substituents| Number of Same Substituent | Numerical Prefix |
--- |
| 2 | di- |
| 3 | tri- |
| 4 | tetra- |
| 5 | penta- |
| and so forth | and so forth |
Consider this molecule:
Figure (\PageIndex{9}) 2,3-dimethylbutane
The longest chain has four C atoms, so it is a butane. There are two substituents, each of which consists of a single C atom; they are methyl groups. The methyl groups are on the second and third C atoms in the chain (no matter which end the numbering starts from), so we would name this molecule 2,3-dimethylbutane. Note the comma between the numbers, the hyphen between the numbers and the substituent name, and the presence of the prefix di- before the methyl. Other molecules-even with larger numbers of substituents-can be named similarly.
Example (\PageIndex{4}):
Name this molecule.
Solution
The longest chain has seven C atoms, so we name this molecule as a heptane. We find two one-carbon substituents on the second C atom and a two-carbon substituent on the third C atom. So this molecule is named 3-ethyl-2,2-dimethylheptane.
Exercise (\PageIndex{4})
Name this molecule.
Answer
4,4,5-tripropyloctane
Properties of Alkanes
All alkanes are composed of carbon and hydrogen atoms, and have similar bonds, structures, and formulas; noncyclic alkanes all have a formula of C n H 2n+2. The number of carbon atoms present in an alkane has no limit. Greater numbers of atoms in the molecules will lead to stronger intermolecular attractions (dispersion forces) and correspondingly different physical properties of the molecules. Properties such as melting point and boiling point (Table (\PageIndex{5})) usually change smoothly and predictably as the number of carbon and hydrogen atoms in the molecules change.
Table (\PageIndex{5}) Properties of Some Alkanes| Alkane | Molecular Formula | Melting Point (°C) | Boiling Point (°C) | Phase at STP4 | Number of Structural Isomers |
--- --- --- |
| methane | CH 4 | –182.5 | –161.5 | gas | 1 |
| ethane | C 2 H 6 | –183.3 | –88.6 | gas | 1 |
| propane | C 3 H 8 | –187.7 | –42.1 | gas | 1 |
| butane | C 4 H 10 | –138.3 | –0.5 | gas | 2 |
| pentane | C 5 H 12 | –129.7 | 36.1 | liquid | 3 |
| hexane | C 6 H 14 | –95.3 | 68.7 | liquid | 5 |
| heptane | C 7 H 16 | –90.6 | 98.4 | liquid | 9 |
| octane | C 8 H 18 | –56.8 | 125.7 | liquid | 18 |
| nonane | C 9 H 20 | –53.6 | 150.8 | liquid | 35 |
| decane | C 10 H 22 | –29.7 | 174.0 | liquid | 75 |
| tetradecane | C 14 H 30 | 5.9 | 253.5 | solid | 1858 |
| octadecane | C 18 H 38 | 28.2 | 316.1 | solid | 60,523 |
Alkanes are relatively stable molecules, but heat or light will activate reactions that involve the breaking of C–H or C–C single bonds. Combustion is one such reaction:
[\ce{CH4}(g)+\ce{2O2}(g)⟶\ce{CO2}(g)+\ce{2H2O}(g) \nonumber ]
Alkanes burn in the presence of oxygen, a highly exothermic oxidation-reduction reaction that produces carbon dioxide and water. As a consequence, alkanes are excellent fuels. For example, methane, CH 4, is the principal component of natural gas. Butane, C 4 H 10, used in camping stoves and lighters is an alkane. Gasoline is a liquid mixture of continuous- and branched-chain alkanes, each containing from five to nine carbon atoms, plus various additives to improve its performance as a fuel. Kerosene, diesel oil, and fuel oil are primarily mixtures of alkanes with higher molecular masses. The main source of these liquid alkane fuels is crude oil, a complex mixture that is separated by fractional distillation. Fractional distillation takes advantage of differences in the boiling points of the components of the mixture (Figure (\PageIndex{10})). You may recall that boiling point is a function of intermolecular interactions, which was discussed in the chapter on solutions and colloids.
Figure (\PageIndex{10}) In a column for the fractional distillation of crude oil, oil heated to about 425 °C in the furnace vaporizes when it enters the base of the tower. The vapors rise through bubble caps in a series of trays in the tower. As the vapors gradually cool, fractions of higher, then of lower, boiling points condense to liquids and are drawn off. (credit left: modification of work by Luigi Chiesa)
In a substitution reaction, another typical reaction of alkanes, one or more of the alkane’s hydrogen atoms is replaced with a different atom or group of atoms. No carbon-carbon bonds are broken in these reactions, and the hybridization of the carbon atoms does not change. For example, the reaction between ethane and molecular chlorine depicted here is a substitution reaction:
Figure (\PageIndex{11}) the reaction between ethane and molecular chlorine
The C–Cl portion of the chloroethane molecule is an example of a functional group, the part or moiety of a molecule that imparts a specific chemical reactivity. The types of functional groups present in an organic molecule are major determinants of its chemical properties and are used as a means of classifying organic compounds as detailed in the remaining sections of this chapter.
Cyclic Hydrocarbons
A cyclic hydrocarbon is a hydrocarbon in which the carbon chain joins to itself in a ring. A cycloalkane is a cyclic hydrocarbon in which all of the carbon-carbon bonds are single bonds. Like other alkanes, cycloalkanes are saturated compounds. Cycloalkanes have the general formula (\ce{C_{n}H_{2n}}). The simplest cycloalkane shown below is cyclopropane, a three-carbon ring.
Figure (\PageIndex{11}) Cyclopropane the simplest cycloalkane.
The structural formulas of cyclic hydrocarbons can be represented in multiple ways, two of which are shown above. Each atom can be shown as in the structure on the left from the figure above. A convenient shorthand is to omit the element symbols and only show the shape, as in the triangle on the right. Carbon atoms are understood to be the vertices of the triangle.
The carbon atoms in cycloalkanes are still (sp^3) hybridized, with an ideal bond angle of (109.5^\text{o}). However, an examination of the cyclopropane structure shows that the triangular structure results in a (\ce{C-C-C}) bond angle of (60^\text{o}). This deviation from the ideal angle is called ring strain and makes cyclopropane a fairly unstable and reactive molecule. Ring strain is decreased for cyclobutane, with a bond angle of (90^\text{o}), but is still significant. Cyclopentane has a bond angle of about (108^\text{o}). This minimal ring strain for cyclopentane makes it a more stable compound.
Figure (\PageIndex{12}) Cyclobutane and Cyclopentane
Cyclohexane is a six-carbon cycloalkane shown below
Figure (\PageIndex{13}) Cyclohexane
All three of the depictions of cyclohexane are somewhat misleading because the molecule is not planar. In order to reduce the ring strain and attain a bond angle of approximately (109.5^\text{o}), the molecule is puckered. The puckering of the ring means that every other carbon atom is above and below the plane. The figure below shows two possibilities for the puckered cyclohexane molecule. Each of the structures is called a conformation. The conformation on the right is called the boat conformation, while the one on the left is called the chair conformation.
Figure (\PageIndex{14}) Chair (left) and boat (right) conformations for cyclohexane.
While both conformations reduce the ring strain compared to a planar molecule, the chair is preferred. This is because the chair conformation results in fewer repulsive interactions between the hydrogen atoms. However, interconversion readily occurs between the two conformations.
Larger cycloalkanes also exist, but are less common. Cyclic hydrocarbons may also be unsaturated. A cycloalkene is a cyclic hydrocarbon with at least one carbon-carbon double bond. A cycloalkyne is a cyclic hydrocarbon with at least one carbon-carbon triple bond. Shown below are the simplified structural formulas for cyclohexene and cyclooctyne.
Figure (\PageIndex{15}) Cyclohexene(left) and cyclooctyne(right)
Unsaturated Hydrocarbons: Alkenes and Alkynes
Alkenes
Organic compounds that contain one or more double or triple bonds between carbon atoms are described as unsaturated. Unsaturated hydrocarbons have less than the maximum number of H atoms possible. Unsaturated hydrocarbon molecules that contain one or more double bonds are called alkenes. Carbon atoms linked by a double bond are bound together by two bonds, one σ bond and one π bond. Double and triple bonds give rise to a different geometry around the carbon atom that participates in them, leading to important differences in molecular shape and properties. The differing geometries are responsible for the different properties of unsaturated versus saturated fats.
Note
The word un saturated has the same meaning for hydrocarbons as it does for the dietary fats and oils: the molecule has at least one carbon-to-carbon double (C=C) or triple bond (C=C).
Ethene, C 2 H 4, is the simplest alkene. Each carbon atom in ethene, commonly called ethylene, has a trigonal planar structure. The second member of the series is propene (propylene) (Figure (\PageIndex{16})); the butene isomers follow in the series. Four carbon atoms in the chain of butene allows for the formation of isomers based on the position of the double bond, as well as a new form of isomerism.
Structural formulas of Alkenes| Structural Formula | | | |
| Ball and Stick Model | | | |
| Space Filling model | | | |
| Name | ethene | propene | 1-butene |
| Condensed Structural Formula | CH 2 CH 2 | CH 2 CHCH 3 | CH 3 CH 2 CHCH 2 |
| Molecular Formula | C 2 H 2 | C 3 H 6 | C 4 H 8 |
| Line angle drawing | | | |
The name of an alkene is derived from the name of the alkane with the same number of carbon atoms. The presence of the double bond is signified by replacing the suffix -ane with the suffix -ene. The location of the double bond is identified by naming the smaller of the numbers of the carbon atoms participating in the double bond:
Figure (\PageIndex{16}) Four structural formulas and names are shown. The first shows two red C atoms connected by a red double bond illustrated with two parallel line segments. H atoms are bonded above and below to the left of the left-most C atom. Two more H atoms are similarly bonded to the right of the C atom on the right. Beneath this structure the name ethene and alternate name ethylene are shown. The second shows three C atoms bonded together with a red double bond between the red first and second C atoms moving left to right across the three-carbon chain. H atoms are bonded above and below to the left of the C atom to the left. A single H is bonded above the middle C atom. Three more H atoms are bonded above, below, and to the right of the third C atom. Beneath this structure the name propene and alternate name propylene is shown. The third shows four C atoms bonded together, numbered one through four moving left to right with a red double bond between the red first and second carbon in the chain. H atoms are bonded above and below to the left of the C atom to the left. A single H is bonded above the second C atom. H atoms are bonded above and below the third C atom. Three more H atoms are bonded above, below, and to the right of the fourth C atom. Beneath this structure the name 1 dash butene is shown. The fourth shows four C atoms bonded together, numbered one through four moving left to right with a red double bond between the red second and third C atoms in the chain. H atoms are bonded above, below, and to the left of the left-most C atom. A single H atom is bonded above the second C atom. A single H atom is bonded above the third C atom. Three more H atoms are bonded above, below, and to the right of the fourth C atom. Beneath this structure the name 2 dash butene is shown.
Alkynes
Hydrocarbon molecules with one or more triple bonds are called alkynes; they make up another series of unsaturated hydrocarbons. Two carbon atoms joined by a triple bond are bound together by one σ bond and two π bonds. The sp-hybridized carbons involved in the triple bond have bond angles of 180°, giving these types of bonds a linear, rod-like shape.
The simplest member of the alkyne series is ethyne, C 2 H 2, commonly called acetylene. The Lewis structure for ethyne, a linear molecule, is:
Figure (\PageIndex{17}) The structural formula and name for ethyne, also known as acetylene, are shown. In red, two C atoms are shown with a triple bond illustrated by three horizontal line segments between them. Shown in black at each end of the structure, a single H atom is bonded.
The IUPAC nomenclature for alkynes is similar to that for alkenes except that the suffix -yne is used to indicate a triple bond in the chain. For example, (\mathrm{CH_3CH_2C≡CH}) is called 1-butyne.
Naming Alkenes and Alkynes
Alkenes and alkynes are named in a similar fashion. The biggest difference is that when identifying the longest carbon chain, it must contain the C–C double or triple bond. Furthermore, when numbering the main chain, the double or triple bond gets the lowest possible number. This means that there may be longer or higher-numbered substituents than may be allowed if the molecule were an alkane. For example, this molecule
Figure (\PageIndex{18}) 2,4-dimethyl-3-heptene
is 2,4-dimethyl-3-heptene (note the number and the hyphens that indicate the position of the double bond).
Example (\PageIndex{5})
Name this molecule.
Solution
The longest chain that contains the C–C triple bond has six C atoms, so this is a hexyne molecule. The triple bond starts at the third C atom, so this is a 3-hexyne. Finally, there are two methyl groups on the chain; to give them the lowest possible number, we number the chain from the left side, giving the methyl groups the second position. So the name of this molecule is 2,2-dimethyl-3-hexyne.
Exercise (\PageIndex{5})
Name this molecule.
Answer
2,3,4-trimethyl-2-pentene
Example (\PageIndex{6})
Identify the molecule as an alkane, an alkene, or an alkyne.
Solution
The molecule has at least one carbon-to-carbon triple bond (C=C) so it is an alkyne.
The molecule has a one carbon-to-carbon double bond (C=C) so it is an alkene.
The molecule only has carbon-to-carbon single bonds so it is an alkane.
Exercise (\PageIndex{6})
Identify the molecule as an alkane, an alkene, or an alkyne.
Answer
1. alkane, 2. alkene, 3. alkyne
Properties of Alkenes and Alkynes
Ethylene (the common industrial name for ethene) is a basic raw material in the production of polyethylene and other important compounds. Over 135 million tons of ethylene were produced worldwide in 2010 for use in the polymer, petrochemical, and plastic industries. Ethylene is produced industrially in a process called cracking, in which the long hydrocarbon chains in a petroleum mixture are broken into smaller molecules.
Halogens can also react with alkenes and alkynes, but the reaction is different. In these cases, the halogen reacts with the C–C double or triple bond and inserts itself onto each C atom involved in the multiple bonds. This reaction is called an addition reaction. One example is
Figure (\PageIndex{19})[CH_{2}=CH_{2}+H_{2}\overset{metal\: catalyst}{\rightarrow} CH_{3}CH_{3} \nonumber ]
The reaction conditions are usually mild; in many cases, the halogen reacts spontaneously with an alkene or an alkyne.
Hydrogen can also be added across a multiple bond; this reaction is called a hydrogenation reaction. In this case, however, the reaction conditions may not be mild; high pressures of H 2 gas may be necessary. A platinum or palladium catalyst is usually employed to get the reaction to proceed at a reasonable pace:
[CH_{2}=CH_{2}+H_{2}\overset{metal\: catalyst}{\rightarrow} CH_{3}CH_{3} \nonumber ]
Geometric Isomers (Cis-Trans) Isomers
With a molecule such as 2-butene, a different type of isomerism called geometric isomerism can be observed. Geometric isomers are isomers in which the order of atom bonding is the same but the arrangement of atoms in space is different. The double bond in an alkene is not free to rotate because of the nature of the pi bond. Therefore, there are two different ways to construct the 2-butene molecule. The image below shows the two geometric isomers, called cis-2-butene and trans- 2-butene.
Figure (\PageIndex{20})\ cis-2-butene and trans- 2-butene.
The cis isomer has the two single hydrogen atoms on the same side of the molecule, while the trans isomer has them on opposite sides of the molecule. In both molecules, the bonding order of the atoms is the same. In order for geometric isomers to exist, there must be a rigid structure in the molecule to prevent free rotation around a bond. If the double bond in an alkene was capable of rotating, the two geometric isomers above would not exist. In addition, the two carbon atoms must each have two different groups attached in order for there to be geometric isomers. Propene has no geometric isomers because one of the carbon atoms has two single hydrogens bonded to it.
Figure (\PageIndex{21})\ Propene
Physical and chemical properties of geometric isomers are generally different. While cis-2-butene is a polar molecule, trans-2-butene is nonpolar. Heat or irradiation with light can be used to bring about the conversion of one geometric isomer to another. The input of energy must be large enough to break the pi bond between the two carbon atoms, which is weaker than the sigma bond. At that point,the now single bond is free to rotate and the isomers can interconvert.
As with alkenes, alkynes display structural isomerism beginning with 1-butyne and 2-butyne. However, there are no geometric isomers with alkynes because there is only one other group bonded to the carbon atoms that are involved in the triple bond.
Summary
Simple alkanes exist as a homologous series, in which adjacent members differ by a CH 2 unit.
Condensed chemical formulas show the hydrogen atoms (or other atoms or groups) right next to the carbon atoms to which they are attached.
Line-angle formulas imply a carbon atom at the corners and ends of lines. Each carbon atom is understood to be attached to enough hydrogen atoms to give each carbon atom four bonds.
Alkanes have both common names and systematic names, specified by IUPAC.
Names and structures of typical cyclic hydrocarbons are given.
Structural and geometric isomers are defined.
Examples of alkane and alkene isomers are given.
Contributors and Attributions
OpenSTAX
TextMap: Beginning Chemistry (Ball et al.)
Marisa Alviar-Agnew(Sacramento City College)
Henry Agnew(UC Davis)
4.1: Aliphatic Hydrocarbons is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
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The Physics Classroom » Physics Tutorial » Vibrations and Waves » Traveling Waves vs. Standing Waves
Vibrations and Waves - Lesson 4 - Standing Waves
Traveling Waves vs. Standing Waves
Traveling Waves vs. Standing Waves
Formation of Standing Waves
Nodes and Anti-nodes
Harmonics and Patterns
Mathematics of Standing Waves
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A mechanical wave is a disturbance that is created by a vibrating object and subsequently travels through a medium from one location to another, transporting energy as it moves. The mechanism by which a mechanical wave propagates itself through a medium involves particle interaction; one particle applies a push or pull on its adjacent neighbor, causing a displacement of that neighbor from the equilibrium or rest position. As a wave is observed traveling through a medium, a crest is seen moving along from particle to particle. This crest is followed by a trough that is in turn followed by the next crest. In fact, one would observe a distinct wave pattern (in the form of a sine wave) traveling through the medium. This sine wave pattern continues to move in uninterrupted fashion until it encounters another wave along the medium or until it encounters a boundary with another medium. This type of wave pattern that is seen traveling through a medium is sometimes referred to as a traveling wave.
Traveling waves are observed when a wave is not confined to a given space along the medium. The most commonly observed traveling wave is an ocean wave. If a wave is introduced into an elastic cord with its ends held 3 meters apart, it becomes confined in a small region. Such a wave has only 3 meters along which to travel. The wave will quickly reach the end of the cord, reflect and travel back in the opposite direction. Any reflected portion of the wave will then interfere with the portion of the wave incident towards the fixed end. This interference produces a new shape in the medium that seldom resembles the shape of a sine wave. Subsequently, a traveling wave (a repeating pattern that is observed to move through a medium in uninterrupted fashion) is not observed in the cord. Indeed there are traveling waves in the cord; it is just that they are not easily detectable because of their interference with each other. In such instances, rather than observing the pure shape of a sine wave pattern, a rather irregular and non-repeating pattern is produced in the cord that tends to change appearance over time. This irregular looking shape is the result of the interference of an incident sine wave pattern with a reflected sine wave pattern in a rather non-sequenced and untimely manner. Both the incident and reflected wave patterns continue their motion through the medium, meeting up with one another at different locations in different ways. For example, the middle of the cord might experience a crest meeting a half crest; then moments later, a crest meeting a quarter trough; then moments later, a three-quarters crest meeting a one-fifth trough, etc. This interference leads to a very irregular and non-repeating motion of the medium. The appearance of an actual wave pattern is difficult to detect amidst the irregular motions of the individual particles.
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What is a Standing Wave Pattern?
It is however possible to have a wave confined to a given space in a medium and still produce a regular wave pattern that is readily discernible amidst the motion of the medium. For instance, if an elastic rope is held end-to-end and vibrated at just the right frequency, a wave pattern would be produced that assumes the shape of a sine wave and is seen to change over time. The wave pattern is only produced when one end of the rope is vibrated at just the right frequency. When the proper frequency is used, the interference of the incident wave and the reflected wave occur in such a manner that there are specific points along the medium that appear to be standing still. Because the observed wave pattern is characterized by points that appear to be standing still, the pattern is often called a standing wave pattern. There are other points along the medium whose displacement changes over time, but in a regular manner. These points vibrate back and forth from a positive displacement to a negative displacement; the vibrations occur at regular time intervals such that the motion of the medium is regular and repeating. A pattern is readily observable.
The diagram at the right depicts a standing wave pattern in a medium. A snapshot of the medium over time is depicted using various colors. Note that point A on the medium moves from a maximum positive to a maximum negative displacement over time. The diagram only shows one-half cycle of the motion of the standing wave pattern. The motion would continue and persist, with point A returning to the same maximum positive displacement and then continuing its back-and-forth vibration between the up to the down position. Note that point B on the medium is a point that never moves. Point B is a point of no displacement. Such points are known as nodes and will be discussed in more detail later in this lesson. The standing wave pattern that is shown at the right is just one of many different patterns that could be produced within the rope. Other patterns will be discussed later in the lesson.
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Why just read about it and when you could be interacting with it? Interact - that's exactly what you do when you use one of The Physics Classroom's Interactives. We would like to suggest that you combine the reading of this page with the use of our Standing Wave Patterns Interactive. You can find it in the Physics Interactives section of our website. TheStanding Wave Patterns Interactiveprovides the learner an environment for exploring the formation of standing waves, standing wave patterns, and mathematical relationships for standing wave patterns.
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Formation of Standing Waves
Nodes and Anti-nodes
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Mathematics of Standing Waves
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Unlocking BRRETA: navigating broker duties and responsibilities in Georgia real estate transactions
3/18/24
By: Brian Goldberg and Zachary H. Waldrop
In Georgia, the Brokerage Relationships in Real Estate Transactions Act, commonly referred to as BRRETA, was enacted to govern the relationships between sellers, landlords, buyers, tenants, and real estate brokers and their affiliated licensees to the extent not governed by specific written agreements between and among the parties. In place of general fiduciary duties, the Act enumerates specific duties which real estate brokers must exercise with reasonable care. O.C.G.A. §§ 10-6A-2(a); see also O.C.G.A. § 10-6A-5; Atlanta Partners Realty, LLC v. Wohlgemuth, 365 Ga. App. 386, 394, 878 S.E.2d 615, 622 (2022), cert. denied (May 31, 2023) (citing Starks v. Carver, 360 Ga. App. 366, 368 (1), 861 S.E.2d 193 (2021)). An important ramification of the lack of a fiduciary duty is that, in the context of a fraud claim, a client cannot justifiably rely on a broker’s representations without exercising their own due diligence. See, e.g., Spies v. Deloach Brokerage, Inc., 169 F. Supp. 3d 1365, 1375-76 (S.D. Ga. 2016). A broker, however, still has statutorily mandated disclosure obligations under BRRETTA. For example, when a broker with an existing brokerage relationship with a customer or client enters into a new brokerage relationship with the customer or client, the broker must timely disclose this new relationship to everyone involved in the contemplated real estate transaction. O.C.G.A. § 10-6A-4(b).
Generally, a broker has a duty to:(1) perform the terms of the brokerage engagement; (2) promote the interests of the client, including by seeking a sale, purchase, or lease at the price and terms agreed to, timely presenting offers, disclosing to the client all material facts concerning the transaction of which the broker has actual knowledge, advising the seller to obtain expert advice on matters beyond the broker’s expertise, and timely accounting; (3) exercise reasonable skill and care; (4) comply with all requirements of BRRETA and other applicable statutes and regulations; and (5) keep confidential all information received that is made confidential by express request from the client. See O.C.G.A. §§ 10-6A-5(a), 10-6A-6(a), 10-6A-7(a), 10-6A-8(a).
A. Brokers engaged by sellers and landlords.
More specifically, for brokers engaged by sellers and landlords, BRRETA effectively provides the same duties and responsibilities for each. In these relationships, BRRETA additionally requires the broker to timely disclose to all parties with whom the broker is working all adverse material facts regarding the property’s physical conditions and physical conditions within one mile of the property that are actually known by the broker and cannot be discovered by a buyer’s or tenant’s reasonably diligent inspection. See O.C.G.A. §§ 10-6A-5(b)(1)-(2), 10-6A-6(b)(1)-(2). Notably, however, BRRETA expressly provides that a broker has no duty to seek to discover such adverse material facts. See O.C.G.A. §§ 10-6A-5(b)(2), 10-6A-6(b)(2). Moreover, while brokers are prohibited from providing false information to prospective buyers or tenants, a broker cannot be liable where it did not have actual knowledge that the information was false and discloses the information’s source to the buyer or tenant. See id. Further, no cause of action may arise against a broker for disclosing information in compliance with subsection (b) of O.C.G.A. §§ 10-6A-5 and 10-6A-6. Additionally, a broker engaged with a seller or landlord cannot be held liable for failure to disclose any matter other than those expressly enumerated in those subsections. See id. Nor can any violation of that subsection create liability for the broker without a finding of fraud perpetrated by the broker. See id. In other words, to be held liable, a broker must have actual knowledge of false information given or facts not disclosed and an intention to deceive another party. Importantly, a broker engaged by a seller or landlord may provide assistance to the buyer or tenant by performing ministerial acts, and performing such ministerial acts will not violate the broker’s engagement agreement with the seller or landlord or form a brokerage engagement with the buyer or tenant. See O.C.G.A. §§ 10-6A-5(c), 10-6A-6(c). Finally, a broker engaged by a seller or landlord may show alternative properties to prospective buyers or tenants without breaching any duty or obligation owed to the seller or landlord. See O.C.G.A. § 10-6A-5(d), 10-6A-6(d).
B. Brokers engaged by buyers and tenants.
The duties and responsibilities of brokers engaged by buyers and tenants are very similar to those for brokers engaged by sellers and landlords. The same prohibitions on providing false information exist, as do the protections against a broker’s liability. See O.C.G.A. §§ 10-6A-7(b), 10-6A-8(b). Brokers engaged by buyers or tenants may also similarly provide assistance by performing ministerial acts for the seller and landlord. See O.C.G.A. §§ 10-6A-7(c), 10-6A-8(c). These duties and responsibilities, however, do contain some differences from those for brokers engaged by sellers and landlords. For example, BRRETA requires that a broker engaged by a buyer must timely disclose to a prospective seller, with whom the broker is working as a customer and who is selling property which will be financed by loan assumption or the seller, all material adverse facts, of which the broker has actual knowledge, concerning the buyer’s financial ability to perform the terms of the sale. See O.C.G.A. § 10-6A-7(b).
In addition, where the transaction is residential, a broker must disclose the buyer’s intent to occupy the property as a principal residence. See id. Similarly, brokers engaged by tenants must timely disclose to a prospective landlord with whom the broker is working all adverse material facts, of which the broker possesses actual knowledge, concerning the tenant’s financial ability to perform the terms of the lease or letter of intent to lease or intent to occupy the property. See O.C.G.A. § 10-6A-8(b). Moreover, a broker engaged by a buyer or tenant does not breach any duty or obligation by showing properties in which the buyer or tenant is interested to other prospective buyers or tenants. See O.C.G.A. §§ 10-6A-7(d), 10-6A-8(d).
C. Other duties enumerated.
These code sections set forth the general duties and responsibilities for brokers under BRRETA. Other code sections set forth more specific mandates regarding the duration of engagements, disclosures to clients, dual agency, and ministerial acts, etc. See O.C.G.A. §§ 10-6A-9 (duration), 10-6A-10 (disclosures), 10-6A-12 (dual agency), 10-6A-14 (ministerial acts). It is imperative that brokers understand their duties to clients and other parties as, in real estate transactions, questions often arise regarding duties to disclose and conflicting relationships. Sometimes these questions may be difficult to answer. In that case, a broker should consult with an attorney to determine the proper course of action. Regardless, brokers should lean towards caution regarding disclosures and conflicts and never provide false information.
For more information, please contact Zachary Waldrop at zhwaldrop@fmglaw.com or Brian Goldberg at brian.goldberg@fmglaw.com.
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How to Find the Perimeter of Similar Figures
Oklahoma Academic Standards for Mathematics (OAS-M) - Geometry Skills Practice
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Example SolutionsPractice Questions
Steps for Finding the Perimeter of Similar Figures
Step 1: Assign a variable to the length of a side of figure 1, whose perimeter is given.
Step 2: Multiply the length of the side of figure 1 by the given scale factor to get an expression for the length of the corresponding side of figure 2, whose perimeter is unknown.
Step 3: Set up a proportion that sets the ratio of the corresponding side lengths of figure 1 and figure 2 equal to the ratio of the perimeters of figure 1 and figure 2. Solve the equation using cross multiplication and inverse operations as required to get the unknown perimeter of the similar figure.
Vocabulary, Definitions, and Formulas for Finding the Perimeter of Similar Figures
Similar Figures: Similar figures are figures that have the exact same shape but can differ in size. For example, a figurine that is an exact replica of the Taj Mahal at a gift shop is similar to the original Taj Mahal, as it has the exact same shape but is a different size.
Perimeter of Similar Figures: The perimeter of a figure is the total distance around the boundary of the figure. The ratio of the perimeters of two similar figures is equal to the ratio of their corresponding side's lengths. To further explain, let figures 1 and 2 be similar. Let the perimeters of figure 1 and 2 be P 1 and P 2, respectively. Also let two corresponding side lengths of figure 1 and figure 2 be s 1 and s 2, respectively. Then, we have the following:
P 1:P 2=s 1:s 2
This can also be written in fraction form as the proportion:
P 1 P 2=s 1 s 2
Scale Factor: A scale factor of two similar figures (figure 1 and figure 2) is the ratio of the length of a side of figure 1 to the length of the corresponding side of figure 2, or vice versa, depending on the context of the problem.
We will solve two examples using these steps, definitions, and vocabulary to get a better understanding of the concept of finding the perimeter of similar figures.
Example Problem 1- Finding the Perimeter of Similar Figures
The perimeter of rectangle A B C D is 36 m. Find the perimeter of rectangle P Q R O that is similar to A B C D and whose longest side is 2 times the length of the longest side of the rectangle A B C D.
Step 1: Assign a variable to the length of a side of figure 1, whose perimeter is given.
Let the length of side A D in rectangle A B C D be x.
Step 2: Multiply the length of the side of figure 1 by the given scale factor to get an expression for the length of the corresponding side of figure 2, whose perimeter is unknown.
The longest side length of rectangle P Q R O is 2 times the longest side length of the rectangle A B C D. Thus, the length of side Q P is 2 times x, or 2 x.
Step 3: Set up a proportion that sets the ratio of the corresponding side lengths of figure 1 and figure 2 equal to the ratio of the perimeters of figure 1 and figure 2. Solve the equation using cross multiplication and inverse operations as required to get the unknown perimeter of the similar figure.
Let the perimeter of rectangle A B C D be P 1, and let the perimeter of rectangle P Q R O be P 2. Also, let the length of side A D be s 1, and the length of side Q P be s 2.
We can now set up our ratio, plug in any known values, convert the ratio to a proportion involving fractions, and then solve the equation for the missing perimeter. This goes as follows:
P 1:P 2=s 1:s 2 36:P 2=x:2 x 36 P 2=x 2 x 36 P 2=x 2 x 36 P 2=1 2 36×2=P 2 72=P 2
The perimeter of rectangle P Q R O is 72 m.
Example Problem 2- Finding the Perimeter of Similar Figures
The perimeter of triangle F G H is 11 m. Find the perimeter of triangle W Y X that is similar to F G H and whose hypotenuse is three times the length of the hypotenuse of triangle F G H.
Step 1: Assign a variable to the length of a side of figure 1, whose perimeter is given.
Let the length of hypotenuse F H in triangle F G H be x.
Step 2: Multiply the length of the side of figure 1 by the given scale factor to get an expression for the length of the corresponding side of figure 2, whose perimeter is unknown.
The length of the hypotenuse W X in triangle W Y X will be 3 times x or 3 x.
Step 3: Set up a proportion that sets the ratio of the corresponding side lengths of figure 1 and figure 2 equal to the ratio of the perimeters of figure 1 and figure 2. Solve the equation using cross multiplication and inverse operations as required to get the unknown perimeter of the similar figure.
11:P 2=x:3 x 11 P 2=x 3 x 11 P 2=x 3 x 11 P 2=1 3 11×3=P 2 33=P 2
The perimeter of triangle W Y X is 33 m.
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Table of Contents
Steps for Finding the Perimeter of Similar Figures
Vocabulary, Definitions, and Formulas for Finding the Perimeter of Similar Figures
Example Problem 1- Finding the Perimeter of Similar Figures
Example Problem 2- Finding the Perimeter of Similar Figures
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14349 | https://onlinelibrary.wiley.com/doi/10.1111/jne.13423 | European Neuroendocrine Tumor Society (ENETS) 2024 guidance paper for the management of well‐differentiated small intestine neuroendocrine tumours - Lamarca - 2024 - Journal of Neuroendocrinology - Wiley Online Library
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Journal of Neuroendocrinology
Volume 36, Issue 9 e13423
CLINICAL GUIDELINE
Open Access
European Neuroendocrine Tumor Society (ENETS) 2024 guidance paper for the management of well-differentiated small intestine neuroendocrine tumours
Angela Lamarca,
Corresponding Author
Angela Lamarca
angela.lamarca@quironsalud.es
orcid.org/0000-0001-9696-6122
Department of Oncology – Onco Health Institute, Fundación Jiménez Díaz University Hospital, Madrid, Spain
Department of Medical Oncology, The Christie NHS Foundation, Manchester, Division of Cancer Sciences, University of Manchester, Manchester, UK
Correspondence
Angela Lamarca, Department of Oncology, OncoHealth Institute, Fundación Jiménez Díaz University Hospital, Avda Reyes Catolicos 2, Madrid 28040, Spain.
Email: angela.lamarca@quironsalud.es
Contribution: Conceptualization, Methodology, Visualization, Project administration, Supervision, Writing - review & editing, Writing - original draft, Validation
Search for more papers by this author
Detlef K. Bartsch,
Detlef K. Bartsch
Department of Visceral-, Thoracic- and Vascular Surgery, Philipps University Marburg, Marburg, Germany
Contribution: Methodology, Writing - original draft, Writing - review & editing
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Martyn Caplin,
Martyn Caplin
Neuroendocrine Tumour Unit, Royal Free Hospital, London, UK
Contribution: Methodology, Writing - original draft, Writing - review & editing
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Beata Kos-Kudla,
Beata Kos-Kudla
orcid.org/0000-0003-3755-1722
Department of Endocrinology and Neuroendocrine Tumors, ENETS Center of Excellence, Department of Pathophysiology and Endocrinology, Medical University of Silesia, Katowice, Poland
Contribution: Methodology, Writing - original draft, Writing - review & editing
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Andreas Kjaer,
Andreas Kjaer
Department of Clinical Physiology and Nuclear Medicine and Cluster for Molecular Imaging, Copenhagen University of Copenhagen–Rigshospitalet, Department of Biomedical Sciences, University of Copenhagen, Copenhagen, Denmark
Contribution: Methodology, Writing - original draft, Writing - review & editing
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Stefano Partelli,
Stefano Partelli
Pancreas Translational and Clinical Research Centre, Pancreatic and Transplant Surgery Unit, Vita-Salute San Raffaele University, Milan, Italy
Contribution: Methodology, Writing - original draft, Writing - review & editing
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Anja Rinke,
Anja Rinke
orcid.org/0000-0002-8156-2915
Department of Gastroenterology, University Hospital Marburg and Philipps University Marburg, Marburg, Germany
Contribution: Methodology, Writing - original draft, Writing - review & editing
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Eva Tiensuu Janson,
Eva Tiensuu Janson
Department of Medical Sciences, Endocrine Oncology Unit, Uppsala University, Uppsala, Sweden
Contribution: Methodology, Writing - original draft, Writing - review & editing, Supervision
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Christina Thirlwell,
Christina Thirlwell
Department of Medical Oncology, University of Exeter Medical School, Exeter, UK
Contribution: Methodology, Writing - review & editing
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Marie-Louise F. van Velthuysen,
Marie-Louise F. van Velthuysen
orcid.org/0000-0003-0435-9494
Department of Pathology, Erasmus MC University Medical Center, Erasmus University, Rotterdam, The Netherlands
Contribution: Methodology, Writing - original draft, Writing - review & editing
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Marie-Pierre Vullierme,
Marie-Pierre Vullierme
Department of Radiology, Paul Brousse University Hospital, AP-HP-University Paris Saclay, Villejuif, France
Contribution: Methodology, Writing - original draft, Writing - review & editing
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Marianne Pavel,
Marianne Pavel
Department of Medicine 1, Friedrich-Alexander University Erlangen-Nürnberg, ENETS Center of Excellence Erlangen, CCC Erlangen- EMN, and Deutsches Zentrum Immuntherapie (DZI), University Hospital Erlangen, Erlangen, Germany
Contribution: Methodology, Writing - original draft, Writing - review & editing, Supervision, Visualization, Project administration, Validation
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Angela Lamarca,
Corresponding Author
Angela Lamarca
angela.lamarca@quironsalud.es
orcid.org/0000-0001-9696-6122
Department of Oncology – Onco Health Institute, Fundación Jiménez Díaz University Hospital, Madrid, Spain
Department of Medical Oncology, The Christie NHS Foundation, Manchester, Division of Cancer Sciences, University of Manchester, Manchester, UK
Correspondence
Angela Lamarca, Department of Oncology, OncoHealth Institute, Fundación Jiménez Díaz University Hospital, Avda Reyes Catolicos 2, Madrid 28040, Spain.
Email: angela.lamarca@quironsalud.es
Contribution: Conceptualization, Methodology, Visualization, Project administration, Supervision, Writing - review & editing, Writing - original draft, Validation
Search for more papers by this author
Detlef K. Bartsch,
Detlef K. Bartsch
Department of Visceral-, Thoracic- and Vascular Surgery, Philipps University Marburg, Marburg, Germany
Contribution: Methodology, Writing - original draft, Writing - review & editing
Search for more papers by this author
Martyn Caplin,
Martyn Caplin
Neuroendocrine Tumour Unit, Royal Free Hospital, London, UK
Contribution: Methodology, Writing - original draft, Writing - review & editing
Search for more papers by this author
Beata Kos-Kudla,
Beata Kos-Kudla
orcid.org/0000-0003-3755-1722
Department of Endocrinology and Neuroendocrine Tumors, ENETS Center of Excellence, Department of Pathophysiology and Endocrinology, Medical University of Silesia, Katowice, Poland
Contribution: Methodology, Writing - original draft, Writing - review & editing
Search for more papers by this author
Andreas Kjaer,
Andreas Kjaer
Department of Clinical Physiology and Nuclear Medicine and Cluster for Molecular Imaging, Copenhagen University of Copenhagen–Rigshospitalet, Department of Biomedical Sciences, University of Copenhagen, Copenhagen, Denmark
Contribution: Methodology, Writing - original draft, Writing - review & editing
Search for more papers by this author
Stefano Partelli,
Stefano Partelli
Pancreas Translational and Clinical Research Centre, Pancreatic and Transplant Surgery Unit, Vita-Salute San Raffaele University, Milan, Italy
Contribution: Methodology, Writing - original draft, Writing - review & editing
Search for more papers by this author
Anja Rinke,
Anja Rinke
orcid.org/0000-0002-8156-2915
Department of Gastroenterology, University Hospital Marburg and Philipps University Marburg, Marburg, Germany
Contribution: Methodology, Writing - original draft, Writing - review & editing
Search for more papers by this author
Eva Tiensuu Janson,
Eva Tiensuu Janson
Department of Medical Sciences, Endocrine Oncology Unit, Uppsala University, Uppsala, Sweden
Contribution: Methodology, Writing - original draft, Writing - review & editing, Supervision
Search for more papers by this author
Christina Thirlwell,
Christina Thirlwell
Department of Medical Oncology, University of Exeter Medical School, Exeter, UK
Contribution: Methodology, Writing - review & editing
Search for more papers by this author
… See all authors
First published: 08 July 2024
Citations: 35
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Abstract
INTRODUCTION
1 WHICH FEATURES NEED TO BE TAKEN INTO CONSIDERATION FOR THE MANAGEMENT OF SI-NET?
2 IS THERE ANY ROLE OF CIRCULATING BIOMARKER MEASUREMENT IN SI-NET?
3 WHICH IS THE MOST SUITABLE DIAGNOSTIC AND STAGING WORK-UP FOR PATIENTS WITH SMALL INTESTINAL NET?
4 WHAT IS THE ROLE OF SURGERY IN LOCALISED SI-NET?
5 WHAT IS THE RECOMMENDED FIRST-LINE SYSTEMIC TREATMENT FOR ADVANCED DISEASE?
6 WHAT IS THE RECOMMENDED TREATMENT BEYOND FIRST-LINE SSA THERAPY?
7 WHAT IS THE ROLE OF LOCOREGIONAL AND ABLATIVE THERAPIES IN ADVANCED SI-NET?
8 WHICH IS THE BEST THERAPEUTIC STRATEGY IN PATIENTS WITH CARCINOID SYNDROME FOR SYNDROME AND TUMOUR GROWTH CONTROL?
9 WHAT ARE THE MOST RECENT DEVELOPMENTS FOR SI-NET?
10 WHAT IS THE RECOMMENDED FOLLOW-UP FOR SI-NET?
11 CONCLUSION
AUTHOR CONTRIBUTIONS
CONFLICT OF INTEREST STATEMENT
APPENDIX A
Open Research
REFERENCES
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Bluesky
Abstract
Both the incidence and prevalence of well-differentiated neuroendocrine tumours from the small intestine (Si-NET) are gradually increasing. Most patients have non-functioning tumours with subtle GI symptoms and tumours are often discovered incidentally by endoscopy or at advanced disease stages by imaging depicting mesenteric lymph node and /or liver metastases while around 30% of the patients present with symptoms of the carcinoid syndrome. Adequate biochemical assessment and staging including functional imaging is crucial for treatment-related decision-making that should take place in an expert multidisciplinary team setting. Preferably, patients should be referred to specialised ENETS Centres of Excellence or centres of high expertise in the field. This guidance paper provides the current evidence and best knowledge for the management of Si-NET grade (G) 1–3 following 10 key questions of practical relevance for the diagnostic and therapeutic decision making.
INTRODUCTION
Neuroendocrine neoplasms (NENs) arising from the small intestine (Si-NEN) are considered rare malignancies. Despite this, both the incidence and prevalence are gradually increasing.1 NENs are divided into well-differentiated neuroendocrine tumours (NET) and poorly differentiated neuroendocrine carcinomas (NEC) according to the most recent World Health Organization (WHO) 2022 classification. NEC are very rare among Si-NEN and covered in a separate ENETS Guidance paper2 Grading is an essential component of the WHO classification. Si-NETs are graded as follows: NET G1 (Ki-67 <3%; <2 mitotic count in 2 mm 2), NET G2 (Ki-67 3%–20%; mitotic count 2–20 in 2 mm 2), NET G3 (Ki-67 >20%; mitotic count >20 in 2 mm 2).3, 4
According to the most updated data from the Surveillance, Epidemiology, and End Results (SEER) programme that analysed 64,971 cases of NEN, there was an age-adjusted annual incidence rate increase by 6.4-fold from 1973 (1.09 per 100,000) to 2012 (6.98 per 100,000).1 This increase occurred across all NEN, for all sites, stages and grades. The commonest individual organ with the highest incidence is the lung (1.49 per 100,000 persons). Gastro-entero-pancreatic (GEP) NEN amount to 3.56 per 100,000 population and include a variety of primary tumours, the most common being Si-NEN (1.05 per 100,000 persons). Available data on incidences from national and regional cancer registries in North America, Western Europe and Japan, identified slight variations across countries with a trend toward slightly higher incidence rate for United States, Norway and Sweden compared to United Kingdom, Italy, Austria and France.5 However, these variations may derive from heterogeneity of data sources and means of registration rather than a true variation.5 In a more recent retrospective patient based study from the United Kingdom the incidence of SI NET was 1.46 per 100,000 in 2018 and equalled the incidence of lung NET.6 Si-NET is also one of the sites with most pronounced increase of prevalence. This is most likely due to the indolent disease course of Si-NET with long overall survival (OS) reported in the SEER database (median OS of 14 years for Si-NET diagnosed at localised stages and median OS of 8.6 years for Si-NET with distant metastatic disease).
Most patients have non-functioning tumours with subtle or no symptoms while around 30% of the patients may present with symptoms of the carcinoid syndrome (CS), that may lead to specific symptoms including diarrhoea, flushing, and more rarely bronchial obstruction induced by hypersecretion of serotonin and other bioactive substances from the tumour. Some patients may develop carcinoid heart disease (CHD) as a long-term sequela of CS. Comprehensive information on diagnosis and treatment of CS and CHD are provided in a separate guidance paper.7
This guidance paper provides the current evidence and best knowledge for the management of Si-NET (predominantly jejunal and ileal primaries) G1-3 following 10 key questions (Table1) of practical relevance for the diagnostic and therapeutic decision making. Grades of recommendations are provided based on criteria available in Appendix FigureA1.
TABLE 1. Ten major questions on diagnostic and therapeutic management of NET.
| Q. No. | Questions on diagnostic and therapeutic management of NET |
--- |
| Q1 | Which features need to be taken into consideration for the management of Si-NET? |
| Q2 | Is there any role of circulating biomarker measurement in Si-NET? |
| Q3 | Which is the most suitable diagnostic and staging work-up for patients with small intestinal NET? |
| Q4 | What is the role of surgery in localised SI-NET? |
| Q5 | What is the recommended first-line systemic treatment for advanced disease? |
| Q6 | What is the recommended treatment beyond first- line SSA therapy? |
| Q7 | What is the role of locoregional and ablative therapies in advanced Si-NET? |
| Q8 | Which is the best therapeutic strategy in patients with Carcinoid Syndrome and tumour growth control? |
| Q9 | What are the most recent developments for Si-NET? |
| Q10 | What is the recommended follow-up for Si-NET? |
1 WHICH FEATURES NEED TO BE TAKEN INTO CONSIDERATION FOR THE MANAGEMENT OF SI-NET?
Overall, diagnosis and management of Si-NETs requires a specialised multidisciplinary team (MDT) approach (Figures1 and 2). Si-NET are often diagnosed at an advanced stage of the disease either incidentally following abdominal imaging for another reason or due to unspecific symptoms associated with metastases, mesenteric fibrosis, or bowel complaints due to intestinal obstruction. Some patients present with specific symptoms of the carcinoid syndrome (CS) due to secretion of biologically active compounds such as serotonin and tachykinins or are discovered during surgery for intestinal obstruction or bleeding. Most Si-NET are indolent slow-growing, well -differentiated NET G1 or G2 and only rarely NET G3. Patients have an excellent 5- and 10-year survival, even in the metastasised setting.8, 9
FIGURE 1
Open in figure viewerPowerPoint
Diagnosis of Si-NET.
FIGURE 2
Open in figure viewerPowerPoint
Staging and work-up of Si-NET.
Risk factors for developing Si-NET remain poorly understood. The relationship of their development with smoking and gallbladder diseases has been described.10 There is emerging evidence of the presence of inherited variants of Si-NET, but the genetic background of this familial subgroup is still not identified.11 It should also be noted that Si-NET may coexist with other neoplasms, such as gastrointestinal adenocarcinoma or breast cancer.8, 9
From a molecular point of view, relatively few genetic alterations are found in Si-NET and the landscape is very different from that seen in Pan-NET.10 50%–78% of Si-NET show loss of chromosome 18, and about 10% of cases have mutations of the CDKN1B gene, however without any hotspot mutations in this gene.12, 13 Further research is needed to clarify whether genetic changes might give clues to new prognostic or predictive factors.
Assessment of symptoms with special focus on CS is crucial. CS can be the clinical sign leading to the diagnosis of Si-NET. Symptoms of CS occur when serotonin and other substances (tachykinins, prostaglandin and histamine) are secreted directly into the systemic circulation, therefore symptoms of the CS occur in general when NET are metastasised to the liver. CS not only significantly worsens the quality of life of Si-NET patients, but it is also associated with CHD in up to 30% of the patients and adversely affects overall survival.14, 15
It is important to note that patients suffering from CS are treated differently from patients with non-functioning Si-NET. Control of the CS has a priority before any other procedure to avoid carcinoid crisis7; the mainstay of treatment of CS is somatostatin analogues (SSA). The majority of patients however, have non-functioning NET although a transition into a functioning tumour in the course of the disease or with increasing tumour burden may occur. If the disease is not resectable, first-line treatment for most Si-NET patients is SSA. However, some patients with NET G1 and limited tumour burden may be appropriate for active surveillance.7, 16, 17
There are variable recommendations regarding the preparation of Si-NET patients for various medical procedures, including surgery, peptide receptor radionuclide therapy (PRRT) or locoregional therapies. It is recommended that in patients with CS, perioperative and periprocedural use of intravenous SSA is offered to patients regardless of the therapy with long-acting SSA.7, 16, 18, 19
A pathognomonic feature on conventional imaging is the typical mesenteric metastasis, reported in a large series of Si-NET patients in 64% of cases20 where radiating strands of soft tissue can be seen on CT-scan.21 At pathology, a sclerosing fibrosis around the tumour mass is often found encompassing vessels and nerves leading to multiple clinical problems including small intestinal ischaemia, venous congestion, ascites and bowel obstruction.
Recommendations:
Management of Si-NET requires a specialised multidisciplinary team (MDT) for the complexity of the disease for appropriate diagnostic and treatment decisions. Specialised centres with high level of expertise should treat most of these patients, especially in the advanced setting (RECOMMENDATION A-5). Assessment of symptoms with special focus on CS is crucial (RECOMMENDATION A-2b).
2 IS THERE ANY ROLE OF CIRCULATING BIOMARKER MEASUREMENT IN SI-NET?
Si-NET may produce and secrete polypeptides and amines, which play an important role for clinical symptoms related to the CS. These secretory compounds can also be used to guide diagnosis, follow-up after curative management and to monitor treatment response of Si-NET. Measuring such biomarkers at time of first diagnosis and as a baseline assessment at the time of starting a new treatment is recommended (Figure2).
Chromogranin A (CgA) is a polypeptide secreted by neuroendocrine cells and not specific for Si-NET. The level of CgA in untreated Si-NET patients can be correlated to tumour burden. Changes during the initial phase of treatment have in some studies been correlated to treatment outcome even though this is less well documented.22, 23 Risk of false positives should always be considered, especially in the absence of visible malignancy on imaging. CgA should not be used for screening of Si-NET, since levels are increased in several non-malignant diseases such as impaired liver and kidney function, atrophic gastritis and inflammatory bowel disease among others. Another drawback of this biomarker is the existence of different assays with different cut-off levels. It is therefore important to use the same assay over time to be able to compare results.
Serotonin is the main amine produced by Si-NET. Measurement of serotonin in whole blood is not recommended since platelets contain high levels, which influences the results. Instead, the breakdown product 5-hydroxyindoleacetic acid (5-HIAA) is used and should be monitored in Si-NET patients.24
5-HIAA has traditionally been measured in acidified urine in two 24 h collections (due to large variations between days) and under diet restriction since serotonin-rich food may increase the levels significantly. Recently, testing of plasma and serum 5-HIAA has been assessed and compared to urine 5-HIAA,7 and has become increasingly used in the clinic.25 However, impaired kidney function may influence serum 5-HIAA levels and should be taken into consideration when interpreting the results.
Tachykinins are small peptides secreted from Si-NET. Circulating levels correlate to flushing and bronchospasms. However, since only few laboratories can provide the assay, general measurement is not recommended.26
In patients with CHD, measurement of N-terminal pro-brain natriuretic peptide (NT-ProBNP) is helpful to assess the degree of heart failure and should be monitored regularly.7, 27, 28
The NETest® (Wren Laboratories) is an mRNA-based test that analyses 51 gene transcriptome. It may be used to monitor treatment, but similarly to CgA it is not specific for Si-NET and the usefulness in clinical practice is still under evaluation.29 This test is currently not widely available and its role in ‘real-world’ management as a biomarker remains to be determined.
In the few patients with Si-NET G3 it may be of interest to measure neuron specific enolase (NSE), lactate dehydrogenase (LDH) and aspartate aminotransferase (AST), since these biomarkers have shown to be of prognostic value in large cohort studies of G3 NEN patients.30, 31 However, there is very little specific data addressing their importance for Si-NET and more data is needed to determine their utility in this patient group.
Recommendations:
Circulating biomarkers can be used to guide diagnosis, follow-up after curative resection and to monitor treatment response of Si-NET (RECOMMENDATION A-2b).
CgA should not be used for screening of Si-NET (RECOMMENDATION B-2b), since levels are increased in several non-malignant diseases with significant number of false positive results.
5-HIAA, either measured in blood or 24 h urine, should be monitored in Si-NET patients with suspected CS and to assess treatment response in patients with CS (RECOMMENDATION A-2a).
In patients with CHD, measurement of NT-ProBNP is helpful to assess the degree of heart failure and should be monitored regularly (RECOMMENDATION A-2b).
3 WHICH IS THE MOST SUITABLE DIAGNOSTIC AND STAGING WORK-UP FOR PATIENTS WITH SMALL INTESTINAL NET?
3.1 Morphological imaging
Computed tomography (CT) with contrast enhancement is well established as the basic morphological imaging method for diagnosis and staging of patients with Si-NET and should be acquired as a triple-phase CT scan.32 As an alternative or in combination with CT, contrast enhanced magnetic resonance imaging (MRI) may also be used; for accurate assessment of liver metastases, this is the preferred imaging method. MRI is also preferable in young patients with long-term need of follow-up to reduce irradiation dose. In general, MRI has a better soft-tissue contrast and is superior to CT in detection of liver lesions as well as bone and brain metastases.32 Importantly, using conventional MRI small liver metastases are still missed,33 the addition of diffusion-weighted MRI (DWI) raised sensitivity to 80% compared to all imaging. Furthermore, using hepatospecific contrast agent (Gd EOB-DTPA) raised the sensitivity to 100% (compared to other nuclear medicine images with positron emission tomography (PET) such as DOTA-TOC imaging).34
For detection of peritoneal metastases there is no accepted reference standard. One recent meta-analysis including 24 articles shows that DWI provided the highest sensitivity for the detection of peritoneal metastases of gastrointestinal cancer.35 The most useful imaging tool to identify peritoneal carcinomatosis in NET is the combination of CT and somatostatin receptor (SST) imaging (SRI).36 Many of the imaging recommendations rely on availability.
In scenarios where the primary tumour cannot be identified with regular imaging, CT or MRI enterography may be performed for identification of an occult bowel primary.37 The use of capsule endoscopy is not generally recommended, but may be considered in very selected cases.
3.2 Somatostatin receptor imaging
SST expression, in particular subtype 2, is present in most Si-NET and may be visualised using SRI. SRI is nowadays almost entirely based on positron-emitting radioligands that bind to SST and may be detected by PET. Currently used PET tracers include 68 Ga-DOTATOC, 68 Ga-DOTATATE and 68 Ga-DOTANOC (here named 68 Ga-SSA-PET) that at large perform similarly.38 In addition, 64 Cu-DOTATATE has recently been approved and made available in the US and is used on a compassionate use basis in Europe. 64 Cu has a longer physical half-life (12.7 h vs. 1.1 h) and a shorter positron range compared with 68 Ga, making central labelling and distribution of 64 Cu-DOTATATE possible and the shorter position range leads to a better lesion detection rate.3968 Ga-SSA-PET should be performed as PET/CT or PET/MRI with diagnostic quality CT or MRI. The use of 68 Ga-SSA-PET in Si-NET serves a dual purpose. Firstly, as the majority of the tumours express SST, 68 Ga-SSA-PET is valuable for diagnostic work-up and staging with a high sensitivity and lesion detection rate.32 Secondly, non-invasive visualisation of SST expression may help determine whether patients are suitable candidates for SSA or SST-targeted PRRT with 177 Lu-DOTATATE or other SST targeting compounds. Finally, 68 Ga-SSA-PET may be used for therapy response monitoring.
3.3 18 F-Fluorodeoxyglucose (18 F-FDG) PET
FDG is a glucose analogue that is widely used in cancer imaging outside NET. It is taken up by the same transporters as glucose but not further metabolised and thus accumulates in cancer cells. For NET it has been shown that more aggressive phenotypes (e.g., higher Ki-67 index) have higher 18 F-FDG uptake and that 18 F-FDG uptake is highly prognostic across all NEN grades.40 In particular, 18 F-FDG PET is also highly prognostic for G1/G2 NET. A recent prospective study demonstrated that 39% (35 of 90) of Si-NET were 18 F-FDG positive.40 The same study revealed that across tumour origins, 37% (21 of 57) of G1 NET and 58% (48 of 83) of G2 NET were 18 F-FDG positive.40 Therefore, in addition to the use in high-grade tumours that are negative on 68 Ga-SSA-PET, there may be added value when using both 18 F-FDG and 68 Ga-SSA-PET more systematically in NET although prospective studies to document the value are warranted. In patients that are planned for PRRT, the additional use of 18 F-FDG PET to identify mismatched lesions, that is, lack of SST on lesions identified on 18 F-FDG PET may be used to identify patients that should be considered for other or additional treatment. Dual 18 F-FDG and SRI could be considered in G3 Si-NET and selected NET G2 patients (i.e., Ki-67 >10%).
3.4 18 F-DOPA PET – any role in Si-NET?
When comparing 18 F-DOPA with PET based SRI, studies have not been able to demonstrate any significant advantages. Accordingly, on a lesion basis several studies have shown better lesion detection performance of 68 Ga-SSA-PET when compared head-to-head with 18 F-DOPA.41-43 Therefore, although it cannot be ruled out that 18 F-DOPA may have value in single cases, especially in the rare events of SST negative Si-NET with low Ki-67, when also considering the availability of 18 F-FDG PET, we see no role for 18 F-DOPA in Si-NET in general.
A summary of the work up for Si-NETs is provided in Figure2.
Recommendations:
The initial staging investigation should always be performed using CT or MRI with contrast enhancement and CT should be acquired as a triple-phase scan (RECOMMENDATION A-2a).
MRI using contrast (RECOMENDATION A-2a), is the best method for accurate assessment of liver metastases and in young patients to reduce radiation dose.
Use of SRI with PET/CT or PET/MRI is recommended in the diagnostic work-up and staging (RECOMMENDATION B-2a), selection of patients for PRRT (RECOMMENDATION A-1b) and therapy monitoring in selected scenarios (RECOMMENDATION B-2b). In addition, 18 F-FDG PET may be valuable for risk stratification and prior to PRRT in patients with NET G2/ G3 (RECOMMENDATION B-3a). The role of 18 F-DOPA, if any, is limited in Si-NET (RECOMMENDATION C-3b).
4 WHAT IS THE ROLE OF SURGERY IN LOCALISED SI-NET?
All localised, resectable Si-NET should be operated, since this is the only chance for long-term cure given the high rate of lymph node metastases even in tumours <10 mm. A recent systematic review showed that the 90-day mortality after surgery for Si-NET is higher in low-volume centres compared to high-volume hospitals (4% vs. 1%).44 Frequently, patients affected by Si-NET are operated in an emergency setting due to intestinal occlusion or bleeding in centres with different levels of expertise, and as consequence with risk for inappropriate surgical management with regard to lymphadenectomy and resection of all NET that are often multiple in the small bowel. Therefore, it is crucial to refer these patients to dedicated, high-volume centres, to properly evaluate Si-NET surgery and to plan for an adequate reoperation with vessel- and bowel sparing surgery. In those rare cases of asymptomatic localised Si-NET that are incidentally discovered, surgery is always the treatment of choice. The extent of nodal involvement and the degree of mesenteric fibrosis (MF) represent the main determinants of surgical resectability. In particular, the possibility to perform a successful radical resection with adequate lymphadenectomy is strictly related to the localisation of superior mesenteric vessel encasement. The definition of surgical resectability should take especially into account the degree of arterial involvement by lymph-node metastases according to the classification of Ohrvall and colleagues.45 In particular, Si-NET with stage III nodal involvement might be difficult to resect, whereas Si-NET with stage 0, I, or II nodal involvement are generally considered resectable.46
An appropriate surgical resection of Si-NET has to meet specific criteria, including manual palpation of the entire small intestine and an adequate lymphadenectomy preserving as much small intestine as possible. An accurate manual palpation aims to detect multiple synchronous tumours. These tumours are found in 40%–60% of patients and are usually not detected by the preoperative imaging.45, 47 The aim of surgery is to perform a systematic lymphadenectomy avoiding a short bowel syndrome. For this reason, the surgical approach should not be a “pizza pie” approach (including a large intestinal resection), but a retrograde vessel-sparing lymphadenectomy.45, 47 It has been shown in a recent retrospective study that the resected bowel specimen is shortened by about half, when using this technique.47
Si-NET are frequently associated with MF that might cause bowel obstruction and bowel ischaemia. MF occurs more frequently in male patients than women and patients affected by MF have a significant poorer OS compared to those without MF.20 Surgery remains the cornerstone of treatment in patients who report symptoms related to MF such as abdominal pain or present with bowel obstruction. Nevertheless, the oncological benefits of surgical resection in patients with MF remain controversial. A large retrospective series found that patients with MF and metastatic Si-NET showed no improvement of survival rates after palliative surgery.20 There may however be a symptomatic benefit.
There are some conflicting results on the minimum number of lymph nodes to be retrieved during surgery but it seems that adequate lymphadenectomy should include at least 8 lymph nodes for examination.48 A recent multi-institutional series on surgical treatment of Si-NET of the terminal ileum, found that patients who underwent formal right hemicolectomy had similar long-term outcomes compared to those who underwent ileocaecectomy.49 Another single-centre study showed that right hemicolectomy is a positive prognostic factor regarding recurrence in localised Si-NET in the distal ileum.50 Figure3 summarises treatment planning for Si-NET.
FIGURE 3
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Treatment planning for Si-NET (part 1).
The number of metastatic lymph nodes is the main prognostic factor and determinant of disease recurrence after radical surgery.48 Patients with more than 4 metastatic nodes have a 3-year recurrence free survival of 80% compared to 90% and 93% in patients with 1–3 and 0 nodal metastases, respectively. 5-years recurrence rates of 30%–40% are reported in retrospective series (e.g., 48, 50). Since the role of adjuvant therapy, however, has yet not been explored, it is not indicated outside board approved studies.
A minimally-invasive approach for Si-NET resection has increasingly gained acceptance especially in high-volume centres.51 Laparoscopic approach showed similar results in terms of number of resected lymph nodes and long-term outcomes compared to open approach.51 Nevertheless, minimally-invasive surgery for Si-NET remains controversial, since an open approach has some proposed advantages such as manual palpation of the entire small intestine, a safer vascular control, especially when performing a vessel-sparing lymphadenectomy.50 Even when sometimes small SI-NEN are diagnosed by colonoscopy, endoscopic removal should not be performed. Resection of the primary tumour might not be complete or complicated (e.g., risk of perforation) and potentially metastatic lymph nodes will not be targeted.
The role of prophylactic cholecystectomy during resection for Si-NET is another controversial issue. Recently, it has been demonstrated that on-demand surgery can be considered non-inferior to the prophylactic cholecystectomy in patients affected by Si-NET.52
Following curative resection, there is no evidence supporting the use of adjuvant strategies for NET53 and current evidence supports follow-up only.
Recommendations:
All localised, resectable Si-NET should be operated on using organ-sparing techniques in high-volume centres (RECOMMENDATION-A-2b). The surgical gold standard includes an open approach with bimanual palpation of the entire small intestine and a vessel-sparing lymphadenectomy (at least >8 lymph nodes) aiming to limit the extent of small intestine resection (RECOMMENDATION A-2b). Adjuvant therapy following curative resection for Si-NET is not indicated. The oncological benefits of surgical resection in patients with MF remain controversial (RECOMMENDATION C-4).
5 WHAT IS THE RECOMMENDED FIRST-LINE SYSTEMIC TREATMENT FOR ADVANCED DISEASE?
SSA are well established as first line treatment in patients with CS7 and for antiproliferative effect (regardless of functionality) based on the results of two placebo-controlled phase 3 trials54, 55 (Figure4). The PROMID trial included therapy-naïve midgut NET patients, the majority being NET G1 and SST positive although these were not inclusion criteria. The CLARINET trial included non-functioning SST positive advanced GEP-NET with a proliferation rate up to a Ki-67 of 10%.56 Nearly all patients had a stable disease at the time of treatment start. These trials demonstrated significant prolongation of time to progression and progression free survival (PFS) compared to placebo, respectively, whereas a benefit on OS could not be proven, however the vast majority of patients crossed over to open label SSA upon progression on placebo.57 Objective remissions with SSA are rare. No clear superiority of one SSA over another has been shown.58 Quality of life is maintained or even improved during treatment.59 It is therefore widely accepted that SSA (either lanreotide autogel [AG] or octreotide long acting release [LAR]) is the preferred first-line treatment in the presence of advanced G1 or G2 (especially if Ki-67 below 10%) Si-NET suitable for systemic therapy. SSA are generally a well-tolerated treatment except for mostly transient gastrointestinal (GI) side effects. Patients may develop pancreatic insufficiency with steatorrhoea, in which case pancreatic enzyme replacement therapy should be provided.
FIGURE 4
Open in figure viewerPowerPoint
Treatment planning for Si-NET (part 2).
There are a few scenarios when SSA are not the preferred first-line treatment:
1. In grade 1, asymptomatic, low tumour burden and stable Si-NET a watch and wait strategy may be applied delaying start of treatment with SSA until progression occurs. The CLARINET study showed anti-proliferative activity of lanreotide AG in patients with progressive disease who crossed over from placebo.60
2. Around 10% of patients with well-differentiated Si-NET do not show expression of SST. In patients with tumour lesions less than 1 cm in size a SR-PET/CT should be performed (as generally recommended) to exclude false negative lesions. Although also receptor-independent mechanisms of SSA effects have been described, the antiproliferative effect is mediated via binding to SST.61 Therefore, in patients with SST negative disease other treatment strategies are needed.
IFN-α has been used for three decades in Si-NET patients for antisecretory as well as antiproliferative purposes.62 Prospective data are mainly available for combination treatment of IFN-α with octreotide in GEP-NET patients resulting in 6 months median time to progression in patients with progressive disease at baseline.63 However, in a large phase 3 study in patients with “carcinoids” including a mixed patient population of NET G1 and G2 patients with either progressive disease or other poor prognostic factors including around 36% midgut NET, a median PFS of 15.4 months was achieved with octreotide LAR and IFN-α.64 The objective response rate was lower with 4% as compared to older publications.64 Side effects have to be taken into consideration. The use of IFN-α is limited due to lack of availability.
Everolimus is approved for progressive non-functioning Si-NET based on the results of the RADIANT-4 trial. The percentage of patients who were treatment naïve (first-line) in this study is unclear. In patients with gastrointestinal primary the median PFS was 13.1 months in the everolimus arm compared to 5.4 months in the placebo-arm.65, 66 The response rate was 2% in patients treated with everolimus. Common toxicities include stomatitis, infections, diabetes, diarrhoea and pneumonitis, therefore careful follow up investigations and active side effect management is needed. Although everolimus is an approved therapy in progressive disease, locoregional therapies could be considered upfront.
The role of SSA in patients with NET G2 and proliferation rate above 10% is less clear, especially if a rapid progression is documented or if a high tumour burden is present.56 In these scenarios, more intense treatments are probably needed. Options of first-line systemic treatment in this setting may include everolimus, PRRT or systemic chemotherapy.
Everolimus may be utilised in this patient group in the presence of moderate tumour burden and moderate progression (based on the RADIANT-4 trial).65, 66
The use of upfront PRRT in selected SST positive NET patients with Ki-67 <10% is not an officially registered therapy but may be considered particularly in patients with high tumour burden if enrolment in a clinical trial is not feasible. It remains unclear which is the best treatment option in NET G2/ G3 patients with higher proliferation rate. The use of PRRT is supported by the NETTER-2 study results evaluating 177 Lu-DOTATATE versus high dose octreotide (double standard dose) in patients with higher proliferating tumours.67 In this phase III study, patients with newly diagnosed somatostatin receptor-positive NET G2 or G3 (Ki-67 ≥10% and ≤55%) advanced GEP-NETs (29.2% of small intestine origin) who had been randomised to the 177 Lu-DOTATATE arm showed an improved median PFS (22.8 vs. 8.5 months); stratified HR 0.276 (95% CI 0.182, 0.418; p< .0001) and overall response rate (43.0% vs. 9.3%). It is expected that patients represented in the NETTER-2 study are highly selected (i.e., patients who could have been randomised to high-dose SSA and therefore not in need of urgent reduction of tumour burden or other more aggressive therapies). The issue of the appropriate therapy in NET with Ki-67 from 15% to 55% is currently further investigated in the COMPOSE trial (PRRT vs. physicians choice, either everolimus or systemic chemotherapy).
The best chemotherapy regimen is unclear68 and could include temozolomide and capecitabine, 5-Fluorouracil-based schedules such as FOLFOX or combinations of capecitabine with SSA. A French phase II trial reported favourable results with capecitabine in combination with bevacizumab with a median PFS of 23.4 months.69 Altogether, data for use of chemotherapy in Si-NET is very limited and patients with high Ki-67 should preferentially be considered for clinical trials. Chemotherapy should only be used in highly selected patients, especially when high Ki-67, high tumour burden and/ or rapid tumour progression within 6 months or less is present.
NET G3 of small intestinal origin are rare and the best strategy in patients with advanced disease is unknown. Despite having more evidence for pancreatic than for Si-NET, most centres use chemotherapy with temozolomide in combination with capecitabine or FOLFOX as a first-line therapy.70, 71 This is based on the fact that adenocarcinoma-like and alkylating-based chemotherapies may be the most effective treatments in this setting, especially in terms of response rate and PFS, with etoposide-platinum chemotherapy showing poor efficacy.71 In carefully selected patients everolimus could be utilised. In addition, in selected patients with homogeneous SST expression PRRT may also be an option.72 The NETTER-2 study supports the use of PRRT in selected NET G3 (see section above).67 The COMPOSE (NCT04919226) study will provide more insight into the management of NET G3 with PRRT as compared to other therapies, such as everolimus, and chemotherapy (Temozolomide + Capecitabine or FOLFOX). Patients with NET G3 should preferentially be included in clinical trials.
Recommendations:
In the presence of advanced unresectable Si-NET G1 or G2 long-acting SSA (either lanreotide or octreotide) is the preferred first-line treatment (RECOMMENDATION A-1b). Watch and wait may be an option in selected patients (G1, asymptomatic, low tumour burden and/ or stable Si-NET) (RECOMMENDATION B-2b). In patients with advanced disease and SST negative lesions IFNα (RECOMMENDATION A-2b) or everolimus (RECOMMENDATION A-1b) can be used as first-line treatment if locoregional therapy is not an appropriate option. In patients with advanced intestinal NET G2 with Ki-67 >10% everolimus (RECOMMENDATION A-1b) and PRRT (RECOMMENDATION A-1b) are treatment options for selected patients. Chemotherapy has a minor role in highly selected patients (especially NET G3, and/or rapid tumour progression) (RECOMMENDATION C-4). Patients with NET G3 should be enrolled in clinical trials.
6 WHAT IS THE RECOMMENDED TREATMENT BEYOND FIRST-LINE SSA THERAPY?
When considering antiproliferative therapy in Si-NET pathological grade, SST expression, volume of disease, associated symptoms and comorbidities are key factors (Figure4). However, clinical trial data and predictive biomarkers are currently lacking to guide the sequence of systemic therapy. Overall, following progression on SSA, treatment with PRRT for patients with SST positive tumour lesions and everolimus for patients with SST negative lesions are the most likely scenarios. Third-line therapy following these are more difficult to state and may rely on prior treatment and specific characteristics at the time of progression, making discussion in a multidisciplinary team of huge relevance.73 The presence of CS may impact the choice of treatment for patients with Si-NET following SSA, this is further discussed in section 8.
6.1 PRRT
In the NETTER-1 phase 3 study patients with advanced progressive Si-NET were randomised to four cycles of 177 Lutetium (177 Lu)-DOTATATE (every 8 weeks) combined with Octreotide LAR 30 mg every 28 days or high dose Octreotide LAR 60 mg every 4 weeks. This study showed a significant benefit in terms of PFS for 177 Lu-DOTATATE combined with SSA compared to high-dose SSA with a hazard ratio 0.21 (p< .001).74 This was associated with a durable benefit (median PFS exceeding 28 months) and improved quality of life with the most significant benefit seen in global health but also overall physical and role function. Additionally there were symptomatic benefits for diarrhoea, pain and fatigue in terms of prolonged time to worsening of symptoms as compared to SSA.75 The final analysis of NETTER-1 performed 5 years after the last patient was enrolled, reported no additional long-term safety signals with 2% of patients developing myelodysplastic syndrome.76 Renal side effects may occur but are usually mild. PRRT is therefore considered the standard second-line treatment option in the presence of homogenous SST-positive disease. Major tumour shrinkage is, however, rarely achieved; the objective response rate in NETTER-1 was 18%. PRRT was followed by octreotide LAR at standard dose, however, it is not proven, that (in the absence of carcinoid syndrome) maintenance therapy with SSA is superior to just observation after application of PRRT, although some retrospective data suggest its additional value. Maintenance therapy with SSA should be given if patients have high tumour burden.
6.2 Rechallenge on PRRT
In view of the significant beneficial outcome of treatment with 177 Lu-DOTATATE and with limited other treatment options, there is a strong rationale for rechallenge (retreatment) with PRRT in case of persistent homogenous SST expression. The current consensus is that rechallenge PRRT may have a role for selected patients and that it could be delivered outside clinical trials if available. This could be considered especially if the disease was controlled for at least 9–12 months following the last cycle of PRRT. If utilised, rechallenge with PRRT would be delivering two more cycles of PRRT, and this could be repeated again in the case of progression after another >12 months and if the treatment was well tolerated.77 In a recent meta-analysis78 the pooled median PFS in patients initially treated with either 90 Y- or 177 Lu-PRRT who were given rechallenge treatment with 177 Lu-DOTATATE (5 studies, 272 patients) was 12.26 months and the patients who received only 177 Lu-DOTATATE had PFS 13.4 months. The median OS in two studies included in the meta-analysis for rechallenge PRRT was 26.8 months from the start of retreatment. The safety profile after rechallenge was similar to initial therapy with grade 3 or 4 haematological toxicity reported in 9% of patients and no patients reported for grade 3 or 4 renal toxicity. Thus rechallenge PRRT appears to be safe and demonstrates encouraging efficacy for selected patients. Prospective studies are ongoing to assess its value as compared to alternative treatment options such as everolimus.
6.3 High dose SSA
There are some retrospective and few prospective studies reporting data for tumour control with high dose SSA.56, 79, 80 Although PRRT with continuation of standard dose octreotide LAR was more effective as compared with double standard dose of octreotide LAR in midgut NET patients in the NETTER-1 trial, the median PFS was 8.4 months with octreotide 60 mg/month.74 The CLARINET forte trial evaluating the efficacy of lanreotide AG 120 mg every 14 days in patients with progressive Si-NET or Pan-NET following first-line standard-dose lanreotide AG treatment (120 mg every 28 days) reported a very similar median PFS of 8.3 months in the midgut cohort.81 In patients with Ki-67 ≤10% median PFS was 8.6 months. Median duration of response was 13.8 months. Based on these studies, above-label SSA is mainly an option in patients with slowly growing, low proliferative (Ki-67 ≤10%) and/or low tumour burden Si-NET where a meaningful delay to more toxic treatments can be reached. The control arm of the NETTER-2 study is challenging this assumption since a median PFS of 8.5 months was achieved with octreotide 60 mg/ month even in patients with Ki-67 above 10%. High dose SSA can also be used in patients with comorbidities such as severe renal insufficiency or impaired bone marrow function in whom PRRT is contraindicated or everolimus is not an option.
6.4 Everolimus
Everolimus given orally at 10 mg/day is EMA approved for G1/G2 intestinal, pancreatic and lung NET. Across all NET the overall response rate (ORR) was <10% and this should be taken into consideration when selecting systemic therapy, as those patients with significant symptoms from large volume disease may not benefit symptomatically. The RADIANT-4 trial investigated everolimus in 302 GI and lung NET patients. The median PFS was 11 months with everolimus and 3.9 months with placebo [hazard ratio (HR) 0.48], with PFS prolongation seen in the GI subgroup [HR 0.56].65 However, a heterogeneous response was identified on post hoc analysis with limited benefit identified in indolent Si-NET.66 Everolimus is currently recommended following PRRT in patients with SST expressing NET given the good tolerability of PRRT, favourable impact on quality of life and durable benefit although data from a prospective randomised trial is still lacking. However, in patients with absent or insufficient SST expression without CS, everolimus is the treatment of choice in progressive G1/G2 Si-NET. The ongoing phase III COMPETE trial (NCT03049189) will compare everolimus to PRRT with 177 Lu-edotreotide and will provide further evidence regarding adequate sequencing of treatment.
6.5 Tyrosine kinase inhibitors
Tyrosine kinase inhibitors (TKIs) are currently not approved in Si-NET in Europe but could be an option for patients with Si-NET with low or no SST expression or after failure to established standard therapies. Several phase II trials with TKIs including cabozantinib, lenvatinib, pazopanib and axitinib have demonstrated some efficacy in Si-NET.82-88 Axitinib and octreotide LAR demonstrated a prolongation of median PFS compared to placebo and octreotide LAR in central review of a randomised phase III trial in patients with extra-pancreatic NET (however failed to reach significant results in local radiological analysis what was the primary endpoint of the trial). The activity of cabozantinib has been confirmed in a randomised placebo-controlled phase III study, that showed prolongation of PFS with decisions regarding approval for this indication still pending.89 If accessible, a TKI may be considered (outside clinical trial setting) after failure of standard therapies (before chemotherapy) or in SST negative patients in case of intolerance to everolimus or IFN-α, or after failure of these therapies.
6.6 Chemotherapy
Efficacy of systemic chemotherapy in G1/G2 Si-NET is poor with ORR of 11.5% (range 5.8%–17.2%).68 However, in very selected cases of NET G2 with a higher Ki-67 (15%–20%) or in NET G3, in case of rapid clinical or radiological progression or where any aspects exclude other treatment options chemotherapy can be beneficial. When doing so, temozolomide and capecitabine or FOLFOX are the recommended schedules to use. Assessment of dihydropyrimidine dehydrogenase (DPD) testing to identify DPD deficiency (associated with poor metabolism of 5-FU or capecitabine) is recommended if these drugs are used; with dose adjustment if DPD deficiency is identified.
Recommendations:
The recommended second-line treatment in the presence of SST-positive Si-NET is PRRT (RECOMMENDATION A-1b), followed by everolimus (RECOMMENDATION A-1b). Rechallenge PPRT (RECOMMENDATION C-4) and high dose SSA (RECOMMENDATION B-2) may be considered in selected scenarios. TKI may be considered based on accessibility and reimbursement after failure to approved therapies or in SST negative NET (RECOMMENDATION C-2b). Chemotherapy is rarely recommended for Si-NET unless NET G2 with higher Ki-67 (15%–20%) or NET G3, with rapid tumour progression (RECOMMENDATION C-3).
7 WHAT IS THE ROLE OF LOCOREGIONAL AND ABLATIVE THERAPIES IN ADVANCED SI-NET?
For Si-NET patients with resectable G1/G2 liver metastases without extrahepatic tumour spread identified after thorough work-up (including MRI as well as functional imaging) and with no significant comorbidities, a curative surgical approach is the treatment of choice. This is supported by large series reporting favourable 5-year OS rates of 71%–74%90, 91 and 85% for all resected and R0 resected patients,92 respectively. For Si-NET, there is no established cut-off value for Ki-67 to exclude patients from surgery of resectable liver metastases. However, it has been shown in a retrospective study that the PFS is about half in patients with G3 liver metastases compared to G1/G2 metastases.93 Therefore, surgery for Si-NET G3 patients with potentially resectable single or few liver metastases should be discussed critically in the MDT. Furthermore, the recurrence rate after R0 resection for G1/G2 tumours, is high (about 80% in 5 years) and therefore a follow-up strategy is needed.90 The same is true for patients with resectable liver metastases of NET G1/G2 with unknown primary. In patients with not completely resectable liver metastases and/or unresectable extrahepatic metastatic disease debulking surgery might be considered after MDT discussion in case of functional NET (carcinoid syndrome) or local liver problems such as bile duct obstruction caused by the metastases.
For patients with G1/G2 Si-NET with predominant liver metastases who are not surgical candidates due to either performance status or extent of liver disease, liver-directed therapies such as radiofrequency ablation (RFA), microwave ablation (MWA), irreversible electroporation (IRE), transarterial embolization (TAE), or chemoembolization (TACE) and radioembolization (TARE), can be applied to provide local tumour control and improve symptoms of CS.94-96 The details of advantages and disadvantages of the distinct techniques are beyond the scope of this guidance paper and high-quality evidence supporting one treatment approach over another as well as for the optimal treatment sequencing is still very limited. Centres are encouraged to take into account the treatment options in which they have most experience on at the time of decision making. From a clinical point of view, percutaneous ablation techniques (RFA, MWA, IRE) are the treatments of choice in patients with oligometastatic disease (preferably maximum of 3–5 metastases and <3 cm in size) or oligoprogressive disease (1 or 2 metastases not responding to systemic treatment). It has been reported that in patients who underwent locoregional treatment for focal liver progression, the median time to new systemic therapy was 32 months.97 Larger size, hilar location, and the proximity to major bile ducts can increase the risk of complications, while proximity to large vessels can result in a cooling effect reducing its effectiveness. For patients with metastases >3 cm, aforementioned risk factors for ablation or with multifocal uni- or bilobular disease, transarterial therapies such as TAE, TACE or TARE are usually indicated. Few studies have compared different transarterial strategies. A recent retrospective study noted no difference between TACE and TARE regarding PFS and OS rates.98 An ongoing randomised controlled trial comparing TAE, TACE and drug-eluting beads (DEB)-TACE had to close the DEB-TACE arm due to increased severe complications.99 Based on the limited high-quality data, MDT decisions regarding liver-directed therapies are significantly influenced by institutional preferences and local experience of distinct techniques.20, 95
Palliative resection of the primary Si-NET in patients with advanced disease is also a matter of debate. Surgical resection or intestinal bypass are mandatory in the presence of primary-related complications such as intestinal obstruction, ischaemia or haemorrhage. The advantage of prophylactic surgery in asymptomatic patients affected by Si-NET is more controversial. Several retrospective studies have been performed.100 Most of these studies lack important information, such as the burden of metastatic disease, the cause of unresectable tumours, and the use and the type of associated systemic therapies. Moreover, palliative surgical resection has most probably been offered more frequently to patients with better performance status and less aggressive disease. Given this high-risk of bias, primary tumour resection is seemingly associated with a benefit in terms of survival compared to no resection.101, 102 However, there are at least three studies that have tried to remove selection bias by using propensity-score matching, and these studies show no benefit in terms of survival.20, 103, 104 Regarding timing of prophylactic surgery, the need of an upfront surgical approach in patients with asymptomatic stage IV Si-NET, has been challenged by a large retrospective study.103 In this study only 22% of patients in the surgery as needed group had to undergo surgery during follow-up. Delayed locoregional surgery (>6 months from diagnosis) in patients with advanced Si-NET is associated with similar overall and disease-specific survival compared to surgery performed within 6 months from initial diagnosis.103 Two recent retrospective studies105, 106 have shown that the combination of Si-NET resection and PRRT provides the best disease-specific survival for patients with diffuse stage IV disease, especially if the metastatic liver burden is less than 50%.106 At present the question is open and its firm answer requires a prospective controlled trial. Thus, benefit and risk of the primary tumour resection in asymptomatic patients with unresectable stage IV disease should be discussed with the patient and in the MDT. A rare consideration could be resection of the primary tumour to leave a patient with liver only disease and the consideration in selected individuals for orthotopic liver transplantation.
Cytoreduction for Si-NET with peritoneal carcinomatosis has been demonstrated to be effective with acceptable morbidity and mortality rates in selected patients.107 In a recent retrospective analysis of 98 patients, Si-NET patients without liver metastases and only localised peritoneal carcinomatosis in one abdominal quadrant with lesions <5 mm benefit the most. Their OS was 76 months compared to 32 months in patients with diffuse peritoneal carcinomatosis and lesions >5 mm.107 Thus, Si-NET patients with or without resectable liver metastases with limited peritoneal carcinomatosis are good candidates for cytoreductive surgery.84 The role of hyperthermic intraperitoneal chemotherapy (HIPEC) in addition to cytoreduction in patients with peritoneal carcinomatosis has been retrospectively explored in Si-NET.108 Based on these results it cannot be recommended for patients with Si-NET and peritoneal carcinomatosis at present.
Recommendations:
In patients with Si-NET G1/G2 stage IV disease and exclusive liver metastases a surgical approach is indicated, if R0 resection can be achieved (Recommendation B-4). In non-surgical candidates locoregional liver-directed therapies should be explored in predominant liver disease to control tumour growth, CS and oligometastatic progression (Recommendation B-3). Cytoreductive peritoneal surgery is recommended in patients with localised peritoneal carcinomatosis (Recommendation A-4). Benefit and risk of a primary tumour resection in asymptomatic patients with unresectable stage IV disease should be discussed critically with the patient and in the MDT.
8 WHICH IS THE BEST THERAPEUTIC STRATEGY IN PATIENTS WITH CARCINOID SYNDROME FOR SYNDROME AND TUMOUR GROWTH CONTROL?
Refractory CS (especially with worsening bowel movements (BM) or flushing) may be a challenging scenario faced by clinicians treating NET. Refractory CS has recently been defined by ENETS as “recurring or persisting CS symptoms and increasing or persistently high u5-HIAA levels despite the use of standard doses of SSA”. Refractory CS is classified as “non-aggressive or aggressive”, based on symptom burden together with disease status (stable or progressive), hepatic tumour burden, and/or the presence of CHD.7
In the presence of CS, the recommendations for antiproliferative therapy as outlined above (see Q6) may not be applicable, and alternative approaches (i.e., early liver directed therapies) may be considered. The requirements for treatment may also be different since patients may suffer from increasing symptoms but may not necessarily display obvious tumour growth. It is of crucial importance to determine the treatment goals first, either improvement of CS symptoms, tumour growth control or both.
8.1 Aiming for symptom control
In the scenario of refractory CS on SSA without evidence of disease progression, the aim of the treatment should focus on achieving an antisecretory effect.
High dose of SSA can be considered7 and has often been applied in clinical routine. It is supported by retrospective analyses but there is a lack of high quality data or evidence based randomised controlled trials.56 High dose SSA implies above labelled dose that can either be reached by shortening of the injection interval or by increasing the dose and maintaining the standard interval of 28 days. In retrospective trials symptomatic improvement for diarrhoea is reported in up to 79%109 and for flushing in up to 91% of patients.110 The only prospective randomised trial in patients with refractory CS compared an above label dose of octreotide LAR (40 mg) with pasireotide LAR 60 mg (a universal ligand to SST) every 28 days111 and reported symptom control in 27% and 21% of patients, respectively. Based on these data high dose of SSA can be used for symptom control in refractory CS.
Addition of short acting octreotide s.c. to depot preparations of SSA as rescue therapy for breakthrough symptoms may also be an option.112, 113 Pasireotide may be considered off-label in patients suffering from CS when other options have been exploited.
Telotristat ethyl, an oral inhibitor of the serotonin synthesis, significantly reduced diarrhoea and 5-HIAA levels in patients with refractory CS in two randomised placebo-controlled trials.107, 108 Flushing was not significantly improved as expected since it is related to secretion of other bioactive compounds than serotonin. The drug was generally well-tolerated, treatment related adverse events included nausea, abdominal pain and elevation of gamma-glutamyl transferase.114 Addition of telotristat ethyl to SSA is recommended in CS patients with refractory diarrhoea.
The use of liver directed therapies for poorly controlled CS should be considered upfront. In addition to debulking surgery, locoregional treatments can achieve symptomatic as well as tumour response. Evidence from prospective controlled trials is lacking. A recent review including more than 100 mainly retrospective studies and 5000 patients reported a symptomatic response in 55.2%.95 Outcome did not significantly vary depending on the type of embolization performed with a tendency for better symptomatic and tumour size response after radioembolization. However, radioembolization should be used with caution in patients with Si-NET given the risk of long-term sequelae including liver fibrosis and cirrhosis, particularly in low grade Si-NET who may have a long life expectancy exceeding 8 years. In general, bland embolization is usually preferred. In patients with persisting symptoms of CS despite SSA treatment, locoregional treatment of liver metastases should be discussed early in the MDT.
IFN-α is an approved antisecretory treatment for patients with CS and may alleviate not only diarrhoea but also flushing. Due to more side effects compared to SSA, it is usually used as an add-on option in patients with refractory CS or upfront in SST negative NET patients. Symptomatic response was reported in 50%–70% of patients.62, 115 The recommended dose is 3–5 MU/week, an off-label pegylated formulation once weekly may be better tolerated,116 but availability is limited due to discontinued production of some compounds.
PRRT is mainly used for its antiproliferative effects based on the results of the NETTER-1 trial and its labelled indication in progressive disease (see Q6).74 The proportion of patients suffering from CS symptoms on PRRT plus standard dose of octreotide versus high dose octreotide was the same. However, in NETTER-1 the median time to deterioration (TTD) of diarrhoea in a subset of patients with CS was significantly longer with PRRT and standard dose of octreotide LAR as compared to high-dose octreotide (p = .0107), whereas TTD for flushing was not significantly different when comparing treatment arms.75 A small retrospective monocentric study reported efficacy in refractory CS patients without documented disease progression; number of bowel movements as well as frequency of flushing episodes were reduced after PRRT with 177 Lu-DOTATATE.117 In highly symptomatic patients, however, symptom control is crucial before PRRT, and withdrawal of long-acting SSA prior to PRRT may lead to worsening of diarrhoea (eventually associated with renal insufficiency). Therefore, other options should be considered upfront for improvement of symptoms. Caution is also required in patients with CHD, where compensation of heart insufficiency may be required first and an experienced cardiologist should be consulted, (see separate guidance paper on CHD).7
Everolimus plays a minor, if any role in patients where syndrome control is the goal. However, in the RADIANT-2 study, everolimus plus octreotide LAR resulted in greater reductions of urinary 5-HIAA compared with placebo plus octreotide LAR, but the efficacy on symptom control was not reported in this trial.118 Reduction of 5-HIAA was sustainable and since reduction of 5-HIAA usually is associated with symptom improvement, an (off label) addition of everolimus may be considered in patients with CS, particularly if there is progressive disease and an antiproliferative therapy is warranted. A benefit on symptoms is also supported by a small retrospective study reporting symptom control in 7 of 10 patients with the addition of everolimus to SSA.119 Patients should be controlled for development of hyperglycaemia or diabetes when everolimus is combined with SSA.
8.2 Aiming for antiproliferative effect
In the presence of CS and evidence of disease progression, management may be complex since not all treatments with an antisecretory effect have proven anticancer effect (i.e., telotristat ethyl) and also because some treatments have not shown definitive activity in the presence of CS (i.e., everolimus).118
IFN-α, if available may be a preferred therapy in these patients because it also exerts antiproliferative effects. In addition to symptomatic response (see above) tumour control was reported with a median PFS of 15.4 months and a response rate of 4% when combined with SSA.64
The use of high-dose SSA is supported by the CLARINET forte study and the control arm of NETTER-1 study in patients with Ki-67 <10%. In patients with comorbidities, elderly patients or patients with indolent tumour growth, this may be an alternative approach when more aggressive therapies seem inappropriate or growth is slow over several years.
In the RADIANT-2 phase 3 trial, the primary endpoint PFS by central reading narrowly missed the prespecified threshold of significance (p = .026). Therefore, everolimus is not registered for patients with a history of CS. It may be considered in patients with significant growth when other options have been exploited (such as high dose SSA, locoregional therapies and PRRT).
PRRT is regarded as the standard antiproliferative second line treatment for metastatic and progressive SST positive Si-NET (see Q6). This treatment was more effective than high dose octreotide in patients regardless of the presence of carcinoid syndrome. PRRT is a good treatment option in patients with progressive disease especially if extrahepatic metastases are present. In liver predominant disease embolization treatment can induce fast symptomatic response and high radiological response rates (the objective response rate of pooled data in a recent review was 36.6%).95 In the presence of CS even a higher symptomatic response rate was seen (70%).7 PRRT can be recommended as preferred option although evidence from prospective controlled trials is not available.
Recommendations:
In patients with CS, a multidisciplinary approach is of crucial importance.
In patients with refractory CS and predominant diarrhoea, telotristat ethyl should be added to SSA (Recommendation A-1a). SSA dose escalation can be considered in patients with refractory CS (Recommendation A-3b) or alternatively addition of IFN-α (Recommendation B-2b). Locoregional and ablative treatments including debulking surgery, should be considered early for refractory CS (Recommendation B-3a). PRRT represents an option (Recommendation B-3b), especially if progressive metastatic disease is present. Everolimus may be considered in individual patients in combination with SSA if significant tumour growth is present after exploitation of other options (Recommendation C-4).
9 WHAT ARE THE MOST RECENT DEVELOPMENTS FOR SI-NET?
9.1 Molecular profiling and precision medicine
Despite molecular profiling being widely explored in Si-NET, precision medicine strategies in this patient population are scarce and remain investigational120 due to lack of specific mutations and rather presence of epigenetic alterations. Routine genomic profiling of tumour tissue is not recommended, but might be considered in patients with Si-NET (if available and reimbursed) when all treatment options have been exploited, but it is important to manage patients' expectations regarding the very low likelihood of the results impacting their treatment options.120
9.2 Latest developments on systemic treatments for Si-NET: targeting angiogenesis
Regarding new systemic therapy options for Si-NET, multiple clinical trials have been reported over the last few years related to antiangiogenic tyrosine-kinase inhibitors (TKIs) with promising results. However, only sunitinib is registered in Europe for Pan-NET so far.121 Si-NET patients were not included in the pivotal phase 3 clinical trial with sunitinib, and thus the field was unexplored for some time, until the recent years when a variety of TKIs have been studied in phase II and III clinical trials including Si-NET.
Some of the TKIs that have been explored in this setting include lenvatinib,84 surufatinib,87, 122 axitinib88, 123 and pazopanib.124 However, these compounds are not yet registered for NET in Europe. Cabozantinib is an emerging drug in this field. A phase II study with cabozantinib included a cohort of 41 heavily pretreated Si-NET patients.125 In this study the primary endpoint was ORR. Achieved ORR in the Si-NET cohort was 15%, with 26/41 patients having disease stabilisation. Median PFS was 31.4 months (95% CI 8.5-not reached). The CABINET phase III clinical trial data including 197 patients with extra-pancreatic NETs (129 randomised on cabozantinib and 68 on placebo) supports the activity of cabozantinib with improvement of PFS (HR 0.41; p-value <.0001).89 Data from the subgroup analysis in intestinal NET are still lacking.
Based on the study results, TKI preferably cabozantinib in view of the positive phase III trial but also lenvatinib (based on results from an uncontrolled phase 2 study with ORR 16.4% and mPFS 15.7 months in GI-NET patients) may be considered in patients in whom all treatment options have been exploited if available and reimbursed and no clinical trial available. TKI may also be a consideration in SST negative patients.
9.3 Immunotherapy
Immunotherapy approaches in Si-NET have been disappointing. PD-L1 expression within tumour cells has been described to be present in 12.8% of Si-NET.126 Monotherapy with checkpoint inhibitors (CPI) has not shown much activity, regardless of expression of PD-L1 or other potential biomarkers linked to immunotherapy activity.127-129 Combination strategies of dual CPI such as durvalumab and tremelimumab (DUNE study) have been explored but no objective responses have been reported in the Si-NET group neither.130 Despite this, the field is evolving and alternative combinations of immunotherapy and antiangiogenic compounds (i.e., atezolizumab and bevacizumab) look more promising (objective response rate of 15% in the extra-pancreatic NET cohort with durable benefit [median PFS 14.9 months]).131 Novel approaches such as CAR-T cells are in preclinical stage132; combination of CPI with TKIs are under development.
Outside the setting of mismatch repair deficiency, microsatellite instability or high TMB, that is rather seen in poorly differentiated NEC than in well differentiated NET, immunotherapy for Si-NET cannot be recommended based on the current evidence. It remains investigational and patients should be included in clinical trials.
Recommendations:
After failure to standard therapies, novel TKIs (RECOMMENDATION B-2b) could be considered if available, while immunotherapy has currently no place in the management of Si-NET, and strategies in this setting remain investigational (RECOMMENDATION C-2b). Patients diagnosed with Si-NET who failed available treatment options should be considered for participation in clinical trials.
10 WHAT IS THE RECOMMENDED FOLLOW-UP FOR SI-NET?
10.1 Imaging follow-up after surgery: curative setting
No prospective studies are available regarding the timing of follow-up and the imaging modalities to be utilised in patients who have had potentially curative resection for Si-NET. Follow-up should last for a long time (it may be required to be life-long) after resection since disease relapse can occur also 15 years after surgery.133, 134 MRI of the liver can be offered as the preferred imaging modality of follow-up for its accuracy in detecting liver metastases as the most frequent site of distant metastasis and the advantage of not exposing patients to ionising radiation. Staging intervals range between 3 and 12 months depending on extent of disease, grade, and length of follow-up. We recommend imaging 3–6 months after surgery, followed by 6 months intervals in NET G1, and closer intervals in NET G2. Recurrence most frequently occurs within the first 5 years after curative surgery. After 5 years of being tumour-free, staging intervals may be prolonged to 1 year.135 In patients with advanced disease (loco-regional LN involved, liver metastases), SRI is recommended after surgery within 3–6 months, and should be repeated in long-term follow up (after 3 years or if suspicious findings on conventional imaging or clinical examination).
10.2 Imaging follow-up: palliative setting
Frequency of follow-up during palliative treatment depend on clinical factors (more frequent follow-up in rapidly progressing tumours or G2 Si-NET) as well as the treatment administered. Follow-up should be performed in the form of CT or MRI with higher sensitivity for visualisation of liver metastases with MRI. After a first assessment at 3 months from starting a new treatment, imaging every 6 months for G1 Si-NET, treated with SSA may be acceptable. In contrast, patients on everolimus, IFN-α, TKI and chemotherapy require imaging every 3 months as established in oncological practice. Follow-up on patients receiving PRRT should be performed first after 3–4 months from the first cycle of PRRT, since around 20% do not respond to PRRT. The rare event of pseudoprogression should be kept in mind and SRI with PET/CT should be interpreted in the context of clinical and biochemical findings. Further re-evaluation intervals may range between 3 and 6 months depending on grade (3 months in higher NET G2 and NET G3). SRI may be recommended if clinically indicated or every 1–3 years depending on disease spread/sites (lung, bone involvement, peritoneal carcinomatosis, diffuse LN metastases).
Staging intervals with conventional imaging should be every 3 months in NET G3 irrespective of type of treatment. An 18 F-FDG-PET may be required to visualise bone metastases or other distant metastases if there is no uptake on SRI or if PRRT is a treatment option.
Bone MRI may be required if symptoms occur indicating bone lesions.
10.3 Biomarker follow-up
Assessment of circulating biomarkers such as CgA, and NSE (in NET G3) and 5-HIAA; see Question 2) at the time of first diagnosis and as a baseline assessment at the time of starting a new treatment is recommended. If elevated in these conditions, biomarker reassessment (every 3–6 months depending on the characteristics of the NET and the treatment being monitored) along with imaging could be of use, particularly to inform about the need to expand imaging to other regions than the abdomen. In long-term follow-up circulating biomarkers, if informative in the individual patient (elevated in patients with measurable disease on imaging) may be used alternating with imaging and prompting to additional diagnostic work-up in case of significant increase.
Recommendations:
All patients should be followed-up long-term after curative surgery (RECOMMENDATION A-4) preferably using MRI of the liver/abdomen (RECOMMENDATION C-5). Assessment of circulating biomarkers at time of first diagnosis, and as a baseline assessment at the time of starting a new treatment is recommended (RECOMMENDATION B-4). Radiological and biomarker follow-up during palliative treatment depends on clinical factors, grading, and the type of treatment being administered. (RECOMMENDATION B-5).
11 CONCLUSION
Management of Si-NET requires a multidisciplinary approach and referral to centres with adequate expertise. Accurate diagnosis and staging are crucial for proper treatment planning. Exploration of (epi)genetics, novel circulating biomarkers and improved radiology and nuclear medicine tools are ongoing areas of research in the field. Current management, especially in the setting of systemic therapies, relies on robust scientific evidence based on prospective clinical trials. Treatment options are still limited, but are expanding with novel TKIs and novel radiopeptides under development while immunotherapy has no role and novel approaches remain investigational. The most adequate sequence of therapies has still to be clarified. In the context of precision medicine the development of predictive biomarkers for tailored treatment planning is crucial.
AUTHOR CONTRIBUTIONS
Angela Lamarca: Conceptualisation; methodology; visualisation; project administration; supervision; writing – review and editing; writing – original draft; validation. Detlef K. Bartsch: Methodology; writing – original draft; writing – review and editing. Martyn Caplin: Methodology; writing – original draft; writing – review and editing. Beata Kos-Kudla: Methodology; writing – original draft; writing – review and editing. Andreas Kjaer: Methodology; writing – original draft; writing – review and editing. Stefano Partelli: Methodology; writing – original draft; writing – review and editing. Anja Rinke: Methodology; writing – original draft; writing – review and editing. Eva Tiensuu Janson: Methodology; writing – original draft; writing – review and editing; supervision. Christina Thirlwell: Methodology; writing – review and editing. Marie-Louise F. van Velthuysen: Methodology; writing – original draft; writing – review and editing. Marie-Pierre Vullierme: Methodology; writing – original draft; writing – review and editing. Marianne Pavel: Methodology; writing – original draft; writing – review and editing; supervision; visualisation; project administration; validation.
CONFLICT OF INTEREST STATEMENT
Dr Angela Lamarca has received travel and educational support from Ipsen, Pfizer, Bayer, AAA, SirtEx, Novartis, Mylan and Delcath; speaker honoraria from Merck, Pfizer, Ipsen, Incyte, AAA, QED, Servier, Astra Zeneca and EISAI; advisory and consultancy honoraria from EISAI, Nutricia Ipsen, QED, Roche, Servier, Boston Scientific, Albireo Pharma, AstraZeneca, Boehringer Ingelheim, GENFIT and TransThera Biosciences; she is member of the Knowledge Network and NETConnect Initiatives funded by Ipsen. Detlef K. Bartsch has received spekaer honoraria from AAA and Streamedup. Anja Rinke has received travel and educational support from Ipsen and Novartis, speaker honoraria from AAA, Advanz Pharma, Falk Foundation, Ipsen and Novartis and advisory honoraria from AAA, Advanz Pharma, Esteve, Ipsen and Novartis. Eva Tiensuu Janson has received speaker honoraria from Ipsen. Marie-Louise van Velthuysen has no conflict of interest. Marianne Pavel received speaker honoraria from AAA, IPSEN, Boehringer-Ingelheim, MSD, Lilly, Novartis, Serb, Streamedup and received fees for consultancy from Hutchmed, Riemser, AAA, IPSEN. B. Kos-Kudla: consulting and honoraria – Merck, IPSA, Ipsen, Novartis, Pfizer. Martyn Caplin received speaker honoraria/advisory board honoraria from AAA, Boehringer-Ingelheim, IPSEN, Novartis, Pfizer and research grants from AAA-Novartis and Ipsen. Andreas Kjaer is an inventor of and holds IPR on 64 Cu-DOTATATE. Marie-Pierre Vullierme is consultant for Guerbet International.
APPENDIX A
FIGURE A1
Open in figure viewerPowerPoint
Grade of recommendations. Grading according to published research. (“A” = Strong; “B” = Moderate; “C” = Low; “D” = Very low).
Open Research
PEER REVIEW
The peer review history for this article is available at
DATA AVAILABILITY STATEMENT
Data sharing not applicable to this article as no datasets were generated or analysed during the current study.
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Keywords
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Publication History
Issue Online: 07 September 2024
Version of Record online: 08 July 2024
Manuscript accepted: 02 June 2024
Manuscript revised: 28 May 2024
Manuscript received: 25 April 2024
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14351 | https://www.med.upenn.edu/eckenhofflab/assets/user-content/documents/3%20Anes%20Surg%20AD%20papers.pdf | Anesthetic, Surgery, and Perioperative Neurocognitive Disorders 2018 Recommendations for the nomenclature of cognitive change associated with anaesthesia and surgery-2018. Evered L, Silbert B, Knopman DS, Scott DA, DeKosky ST, Rasmussen LS, Oh ES, Crosby G, Berger M, Eckenhoff RG; Nomenclature Consensus Working Group. Br J Anaesth. 2018 Nov;121(5):1005-1012 Recommendations for the Nomenclature of Cognitive Change Associated with Anaesthesia and Surgery-2018. Evered L, Silbert B, Knopman DS, Scott DA, DeKosky ST, Rasmussen LS, Oh ES, Crosby G, Berger M, Eckenhoff RG; Nomenclature Consensus Working Group. Anesthesiology. 2018 Nov;129(5):872-879. Recommendations for the Nomenclature of Cognitive Change Associated With Anaesthesia and Surgery-2018. Evered L, Silbert B, Knopman DS, Scott DA, DeKosky ST, Rasmussen LS, Oh ES, Crosby G, Berger M, Eckenhoff RG; Nomenclature Consensus Working Group. Anesth Analg. 2018 Nov;127(5):1189-1195 Recommendations for the nomenclature of cognitive change associated with anaesthesia and surgery-2018. Evered L, Silbert B, Knopman DS, Scott DA, DeKosky ST, Rasmussen LS, Oh ES, Crosby G, Berger M, Eckenhoff RG; Nomenclature Consensus Working Group. Acta Anaesthesiol Scand. 2018 Nov;62(10):1473-1480. Recommendations for the nomenclature of cognitive change associated with anaesthesia and surgery-2018. Evered L, Silbert B, Knopman DS, Scott DA, DeKosky ST, Rasmussen LS, Oh ES, Crosby G, Berger M, Eckenhoff RG; Nomenclature Consensus Working Groupª. Can J Anaesth. 2018 Nov;65(11):1248-1257. Recommendations for the Nomenclature of Cognitive Change Associated with Anaesthesia and Surgery-20181. Evered L, Silbert B, Knopman DS, Scott DA, DeKosky ST, Rasmussen LS, Oh ES, Crosby G, Berger M, Eckenhoff RG; Nomenclature Consensus Working Group. J Alzheimers Dis. 2018;66(1):1-10. Best Practices for Postoperative Brain Health: Recommendations From the Fifth International Perioperative Neurotoxicity Working Group. Berger M, Schenning KJ, Brown CH 4th, Deiner SG, Whittington RA, Eckenhoff RG, Angst MS, Avramescu S, Bekker A, Brzezinski M, Crosby G, Culley DJ, Eckenhoff M, Eriksson LI, Evered L, Ibinson J, Kline RP, Kofke A, Ma D, Mathew JP, Maze M, Orser BA, Price CC, Scott DA, Silbert B, Su D, Terrando N, Wang DS, Wei H, Xie Z, Zuo Z; Perioperative Neurotoxicity Working Group. Anesth Analg. 2018 Oct 9. doi: 10.1213/ANE.0000000000003841. [Epub ahead of print] Between a ROCK and an IR Place. Pan JZ, Eckenhoff RG. Anesth Analg. 2018 Mar;126(3):750-751 One to rule them all? Terrando N, Eckenhoff RG. Br J Anaesth. 2018 Mar;120(3):428-430 Stroke 112: A Universal Stroke Awareness Program to Reduce Language and Response Barriers. Zhao J, Eckenhoff MF, Sun WZ, Liu R. Stroke. 2018 Jul;49(7):1766-1769. 2017 Perioperative cognitive disorders. Response to: Postoperative delirium portends descent to dementia. Evered L, Eckenhoff RG; International Perioperative Cognition Nomenclature Working Group. Br J Anaesth. 2017 Dec 1;119(6):1241. Effects of propofol and surgery on neuropathology and cognition in the 3xTgAD Alzheimer transgenic mouse model. Mardini F, Tang JX, Li JC, Arroliga MJ, Eckenhoff RG, Eckenhoff MF. Br J Anaesth. 2017 Sep 1;119(3):472-480 2016 From the Cover: Volatile Anesthetics Transiently Disrupt Neuronal Development in Neonatal Rats. Drobish JK, Gan ZS, Cornfeld AD, Eckenhoff MF. Toxicol Sci. 2016; 154(2):309-319. Absence of Neuropathology With Prolonged Isoflurane Sedation in Healthy Adult Rats. DeYoung TP, Li JC, Tang X, Ward CG, Dworkin BR, Eckenhoff MF, Kofke WA. J Neurosurg Anesthesiol. 2016 Sep 20. [Epub ahead of print] Fallacy…. Really? Eckenhoff RG, Hogan KJ, Evered L. Anesthesiology. 2016; 125(2):426-8. Mechanisms of the Immunological Effects of Volatile Anesthetics: A Review. Yuki K, Eckenhoff RG. Anesth Analg. 2016; 123(2):326-35. Neurocognitive Adverse Effects of Anesthesia in Adults and Children: Gaps in Knowledge. Ward CG, Eckenhoff RG. Drug Saf. 2016 Jul;39(7):613-26. 2015 Potential Adverse Effects of Anesthesia in Children. Ward CM, Hogan KJ, Eckenhoff RG. JAMA. 2015; 314(4):408-9. Functional outcomes after critical illness in the elderly. Neuman MD, Eckenhoff RG. Crit Care Med. 2015; 43(6):1340-1. Taxane modulation of anesthetic sensitivity in surgery for nonmetastatic breast cancer. Linganna RE, Levy WJ, Dmochowski IJ, Eckenhoff RG, Speck RM. J Clin Anesth. 2015; 27(6):481-5. Perioperative neurotoxicity in the elderly: summary of the 4th International Workshop. Terrando N, Eriksson LI, Eckenhoff RG. Anesth Analg. 2015; 120(3):649-52. Long-term dantrolene treatment reduced intraneuronal amyloid in aged Alzheimer triple transgenic mice. Wu Z, Yang B, Liu C, Liang G, Liu W, Pickup S, Meng Q, Tian Y, Li S, Eckenhoff MF, Wei H. Alzheimer Dis Assoc Disord. 2015; 29(3):184-91. 2014 General anesthetic and the risk of dementia in elderly patients: current insights. Hussain M, Berger M, Eckenhoff RG, Seitz DP. Clin Interv Aging. 2014; 9:1619-28. Anesthetic preconditioning inhibits isoflurane-mediated apoptosis in the developing rat brain. Peng J, Drobish JK, Liang G, Wu Z, Liu C, Joseph DJ, Abdou H, Eckenhoff MF, Wei H. Anesth Analg. 2014; 119(4):939-46. 2013 Anesthesia, surgery and neurodegeneration. Preface. Eckenhoff RG, Planel E. Prog Neuropsychopharmacol Biol Psychiatry. 2013; 47:121. Differential general anesthetic effects on microglial cytokine expression. Ye X, Lian Q, Eckenhoff MF, Eckenhoff RG, Pan JZ. PLoS One. 2013; 8(1):e52887. Modulation of murine Alzheimer pathogenesis and behavior by surgery. Tang JX, Mardini F, Janik LS, Garrity ST, Li RQ, Bachlani G, Eckenhoff RG, Eckenhoff MF. Ann Surg. 2013; 257(3):439-48. Anesthetic effects in Alzheimer transgenic mouse models. Tang JX, Eckenhoff MF. Prog Neuropsychopharmacol Biol Psychiatry. 2013; 47:167-71. 2012 Isoflurane binds and stabilizes a closed conformation of the leukocyte function-associated antigen-1. Yuki K, Bu W, Xi J, Sen M, Shimaoka M, Eckenhoff RG. FASEB J. 2012; 26(11):4408-17. Anesthesia, surgery, illness and Alzheimer's disease. Eckenhoff RG, Laudansky KF. Prog Neuropsychopharmacol Biol Psychiatry. 2013; 47:162-6. Dantrolene ameliorates cognitive decline and neuropathology in Alzheimer triple transgenic mice. Peng J, Liang G, Inan S, Wu Z, Joseph DJ, Meng Q, Peng Y, Eckenhoff MF, Wei H. Neurosci Lett. 2012; 516(2):274-9. Neurodevelopmental consequences of sub-clinical carbon monoxide exposure in newborn mice. Cheng Y, Thomas A, Mardini F, Bianchi SL, Tang JX, Peng J, Wei H, Eckenhoff MF, Eckenhoff RG, Levy RJ. PLoS One. 2012; 7(2):e32029. Postoperative cognitive decline: where art tau? Eckenhoff RG, Planel E. Anesthesiology. 2012; 116(4):751-2. 2011 Human Alzheimer and inflammation biomarkers after anesthesia and surgery. Tang JX, Baranov D, Hammond M, Shaw LM, Eckenhoff MF, Eckenhoff RG. Anesthesiology. 2011 Oct;115(4):727-32. Anesthesia in presymptomatic Alzheimer's disease: a study using the triple-transgenic mouse model. Tang JX, Mardini F, Caltagarone BM, Garrity ST, Li RQ, Bianchi SL, Gomes O, Laferla FM, Eckenhoff RG, Eckenhoff MF. Alzheimers Dement. 2011; 7(5):521-531.e1. Anesthetic modulation of neuroinflammation in Alzheimer's disease.Tang JX, Eckenhoff MF, Eckenhoff RG. Curr Opin Anaesthesiol. 2011; 24(4):389-94. The common inhaled anesthetic isoflurane increases aggregation of huntingtin and alters calcium homeostasis in a cell model of Huntington's disease. Wang Q, Liang G, Yang H, Wang S, Eckenhoff MF, Wei H. Toxicol Appl Pharmacol. 2011; 250(3):291-8. 2010 Acute anemia elicits cognitive dysfunction and evidence of cerebral cellular hypoxia in older rats with systemic hypertension. Li M, Bertout JA, Ratcliffe SJ, Eckenhoff MF, Simon MC, Floyd TF. Anesthesiology. 2010; 113(4):845-58. A smoking gun but still no victim. Eckenhoff MF, Eckenhoff RG. J Alzheimers Dis. 2010; 19(4):1259-61. Anesthetic-induced neurodegeneration mediated via inositol 1,4,5-trisphosphate receptors. Zhao Y, Liang G, Chen Q, Joseph DJ, Meng Q, Eckenhoff RG, Eckenhoff MF, Wei H. J Pharmacol Exp Ther. 2010; 333(1):14-22. Inhaled anesthetic potency in aged Alzheimer mice. Bianchi SL, Caltagarone BM, Laferla FM, Eckenhoff RG, Kelz MB. Anesth Analg. 2010; 110(2):427-30. Anesthesia and the old brain. Tang J, Eckenhoff MF, Eckenhoff RG. Anesth Analg. 2010; 110(2):421-6. 2009 Consensus statement: First International Workshop on Anesthetics and Alzheimer's disease. Baranov D, Bickler PE, Crosby GJ, Culley DJ, Eckenhoff MF, Eckenhoff RG, Hogan KJ, Jevtovic-Todorovic V, Palotás A, Perouansky M, Planel E, Silverstein JH, Wei H, Whittington RA, Xie Z, Zuo Z; First International Workshop on Anesthetics and Alzheimer's Disease.. Anesth Analg. 2009; 108(5):1627-30. 2008 A presenilin-1 mutation renders neurons vulnerable to isoflurane toxicity. Liang G, Wang Q, Li Y, Kang B, Eckenhoff MF, Eckenhoff RG, Wei H. Anesth Analg. 2008; 106(2):492-500. The common inhalational anesthetic isoflurane induces apoptosis via activation of inositol 1,4,5-trisphosphate receptors. Wei H, Liang G, Yang H, Wang Q, Hawkins B, Madesh M, Wang S, Eckenhoff RG. Anesthesiology. 2008; 108(2):251-60. Brain and behavior changes in 12-month-old Tg2576 and nontransgenic mice exposed to anesthetics. Bianchi SL, Tran T, Liu C, Lin S, Li Y, Keller JM, Eckenhoff RG, Eckenhoff MF. Neurobiol Aging. 2008; 29(7):1002-10. 2007 Inhaled anesthetic modulation of amyloid beta(1-40) assembly and growth. Carnini A, Lear JD, Eckenhoff RG. Curr Alzheimer Res. 2007; 4(3):233-41. 2006 Interactions of volatile anesthetics with neurodegenerative-disease-associated proteins. Carnini A, Eckenhoff MF, Eckenhoff RG. Anesthesiol Clin. 2006; 24(2):381-405. 2005 Isoflurane and sevoflurane affect cell survival and BCL-2/BAX ratio differently. Wei H, Kang B, Wei W, Liang G, Meng QC, Li Y, Eckenhoff RG. Brain Res. 2005; 1037(1-2):139-47. 2004 Inhaled anesthetic enhancement of amyloid-beta oligomerization and cytotoxicity. Eckenhoff RG, Johansson JS, Wei H, Carnini A, Kang B, Wei W, Pidikiti R, Keller JM, Eckenhoff MF. Anesthesiology. 2004; 101(3):703-9. |
14352 | https://artofproblemsolving.com/wiki/index.php/1974_IMO_Problems/Problem_4?srsltid=AfmBOopBQhyKkYywk4ePYDhf73RnjyqZTj5O2vvmy4JbSJ0UBof8ppSQ | Art of Problem Solving
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1974 IMO Problems/Problem 4
Problem
Consider decompositions of an chessboard into non-overlapping rectangles subject to the following conditions:
(i) Each rectangle has as many white squares as black squares.
(ii) If is the number of white squares in the -th rectangle, then
Find the maximum value of for which such a decomposition is possible. For this value of determine all possible sequences
Solution
Since each rectangle has the same number of black squares as white squares, . Clearly for to so so this forces . It is possible to decompose the board into rectangles, as we will show later. But first let us find all such sequences . Now . For a rectangle to have white squares, it will have an area of so it's dimensions are either or - neither of which would fit on a board. So .
If (which could fit as a rectangle) then . Then so . So are 6 numbers among 1-7. If is the number that is not equal to any , then so . Then . Such a decomposition is possible. Take a rectangle on the top left corner, where there are squares horizontally and vertically. Then directly below use a and a rectangle to cover the 3 rows below it. It's simple from there.
Similarly, you can find the other possibilities as or or . Tilings are not hard to find.
The above solution was posted and copyrighted by WakeUp. The original thread for this problem can be found here:
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1974 IMO (Problems) • Resources
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Problem 31•2•3•4•5•6Followed by
Problem 5
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Experimental Determination of Wavelength in Dielectric-Filled Periodic Structures (Correspondence) | IEEE Journals & Magazine | IEEE Xplore
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Journals & Magazines>IRE Transactions on Microwave...>Volume: 7 Issue: 4
Experimental Determination of Wavelength in Dielectric-Filled Periodic Structures (Correspondence)
Publisher: IEEE
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Let it be required to determine the guided wavelength in a dielectric-filled periodic structure, such as a corrugated wall or serrated waveguide. The accepted traveling p...Show More
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Let it be required to determine the guided wavelength in a dielectric-filled periodic structure, such as a corrugated wall or serrated waveguide. The accepted traveling probe technique requires a slot in the broad wall of the guide and a groove in the dielectric material. Even if the errors introduced by these modifications could be tolerated, other effects render this technique unsuitable, One of these is a surface-wave effect which resuIts in a measured wavelength higher than the one in the guide and lower than the free-space value. If the structure is dissipative, such as a serrated guide, more difficulties arise.
Published in:IRE Transactions on Microwave Theory and Techniques ( Volume: 7, Issue: 4, October 1959)
Page(s): 480 - 481
Date of Publication: 31 October 1959
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DOI:10.1109/TMTT.1959.1125427
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Wave Propagation in a Rectangular Waveguide Loaded with an H-Plane Dielectric Slab (Correspondence)
IEEE Transactions on Microwave Theory and Techniques
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An Exact Treatment of a Rectangular Waveguide Symmetrically Loaded with Resistively Coated Dielectric Slabs for Maximum Attenuation
1986 16th European Microwave Conference
Published: 1986
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14354 | https://academic.oup.com/rheumap/article/9/2/rkaf051/8156941 | Published Time: 2025-06-04
Imaging of gout: an atlas | Rheumatology Advances in Practice | Oxford Academic
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Volume 9 Issue 2 2025
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Imaging of gout: an atlas Open Access
Luqman Wali,
Luqman Wali
Radiology Department, Royal National Orthopaedic Hospital NHS Trust
, London,
UK
Correspondence to: Luqman Wali, Radiology Department, Royal National Orthopaedic Hospital NHS Trust, Brockley Hill, Stanmore, London HA7 4LP, UK. E-mail:l.wali@nhs.net
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Emma Rowbotham
Emma Rowbotham
Radiology Department, Leeds Teaching Hospitals NHS Trust
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Rheumatology Advances in Practice, Volume 9, Issue 2, 2025, rkaf051,
Published:
04 June 2025
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16 February 2025
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Luqman Wali, Emma Rowbotham, Imaging of gout: an atlas, Rheumatology Advances in Practice, Volume 9, Issue 2, 2025, rkaf051, Select Format Download citation Close
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Abstract
Gout is a common systemic disease defined by deposition of monosodium urate (MSU) crystals in articular and peri-articular structures, leading to recurrent bouts of inflammation. Imaging plays an important role in establishing the diagnosis when crystal aspiration is not feasible and the clinical diagnosis is uncertain. Each imaging modality has a unique role. Radiographs can demonstrate characteristic erosions and tophi in later stages of gout. Ultrasound has a major role in the diagnosis and assessment of gout. Dual-energy computed tomography (DECT) enables precise visualization of MSU deposits and can determine disease burden. MRI can assess for non-specific inflammatory and structural changes. Both ultrasound and DECT are emphasized as part of diagnostic algorithms and the role of imaging is expanding with more recent advancements and evidence. This review provides an imaging-centric overview of each modality and its evolving significance in gout.
gout, urate, MSU, imaging, radiology, ultrasound, CT, dual-energy CT, MRI, X-ray
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Lay Summary
What does this mean for patients?
Gout is a common condition marked by painful inflammation of the joints and tendons due to crystal build-up. The condition often requires imaging for diagnosis when direct crystal testing, i.e. taking a fluid sample from the joint using a needle, is not feasible or has not established the diagnosis. X-rays can show typical joint damage that occurs with long-term gout. Ultrasound can identify crystals within the joint but suffers from technical limitations when looking at deeper structures. Dual-energy computed tomography (DECT), a recent and advanced way of using CT imaging, can highlight clusters of crystals in and around joints. MRI shows more generic joint damage and inflammation. Recent guidelines for assessing gout include the use of both ultrasound and DECT due to their accuracy. This review explores each type of imaging and how it can help in the overall assessment of gout.
Key messages
Each imaging modality has advantages and disadvantages in imaging of gout.
DECT and ultrasound are important and can be used in the diagnosis of gout.
Imaging can assess for disease burden and monitor response to treatment.
Clinical background
Gout is a systemic disease defined by the deposition of monosodium urate (MSU) crystals and constitutes the most common form of inflammatory arthritis in the UK and across the world [1, 2]. The disease is more common in men and the prevalence is increasing due to higher rates of obesity and changing dietary habits .
Hyperuricaemia is a prerequisite to MSU crystal deposition but is often asymptomatic and does not necessarily lead to gout . Hyperuricaemia develops due to overproduction and decreased excretion of urate due to a combination of environmental and genetic factors . Likewise, the development of MSU crystals is multifactorial and deposition within joints can precipitate an acute immune-mediated inflammatory response . This is termed an acute gout flare and is typically self-limiting. The disease commonly affects the peripheral joints and adjacent soft tissue structures. Gout classically involves the first MTP joint but can be found in any joint, including the axial skeleton .
The intercritical period follows resolution of acute symptoms, during which the disease is clinically quiescent, although accumulated MSU crystals may remain within the affected joint or tissue . Persistent hyperuricaemia and recurrent flares can lead to features of chronic gout in the form of tophi, erosions and bony proliferation. Gouty tophi are stigmata of chronic uncontrolled disease and represent a granulomatous inflammatory response formed by a central core of MSU crystals surrounded by inflammatory and fibrovascular soft tissue .
The gold standard for diagnosis of gout is the aspiration of MSU crystals and confirmation of their presence by using polarizing light microscopy, which conveys 100% specificity for the disease . Serum urate and other biochemical markers serve as useful adjuncts but are not sufficient for a definitive diagnosis . Despite the high specificity of positive crystal aspiration, sensitivity can be lacking due to technical difficulty in aspirating small joints, timing of aspiration and sample analysis . In cases where aspiration is not possible, the results are negative or if there is ongoing clinical uncertainty, imaging plays an important role in establishing the diagnosis.
Radiographs
The classical radiographic appearance of gout is well-defined juxta-articular erosions with sclerotic margins and overhanging edges (Fig.1). These findings can take many years to manifest and the early radiographic appearance of gout can be subtle, with only peri-articular soft tissue swelling being present [11, 12]. With recurrent flares, progressive plain film soft tissue and osseous changes occur, reflected in the relatively low sensitivity (31%) but high specificity (93%) of radiographs as compared with clinical diagnosis .
Figure 1.
Classic radiographic appearances of gout. (A) A juxta-articular erosion (star) of the first MTP joint with well-defined sclerotic margins and preserved bone mineral density. Note the overlying soft tissue swelling (arrowheads) corresponding to a gouty tophus. (B) Erosion of the index finger proximal IP joint with an overhanging edge as a result of new bone proliferation (arrowhead). (C) Erosion of the distal IP joint of the second toe which, despite the advanced disease, maintains a well-defined sclerotic margin
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Tophi appear as dense peri-articular or intra-articular masses and are typically not visible until they are up to 5–10 mm in size. The presence of tophi reflects longstanding, recurrent gout. Tophi can become calcified over time, although this finding is variable. Suh et al. demonstrated that the radiographic size of tophi can correlate with serum urate levels, with a decrease in size after successful treatment.
Tophaceous gout typically results in punched-out, well-defined erosions that involve the margins of the joint. Progressive erosions can become intra-articular and central in location. The exact underlying mechanism is uncertain but is thought to relate to the pressure effects of tophi and granulomatous synovitis . An abundance of osteoclasts and paucity of osteoblasts at sites of tophi have been implicated on a cellular level [16, 17]. Erosions are often accompanied by ill-defined new bone proliferation, usually in the form of sclerosis and osteophytosis, although the exact mechanism for this is unclear . Unlike RA, peri-articular osteopenia is not a feature and joint spaces are well preserved until late in the disease process .
Radiographs can readily assess for chronic changes and multiple scoring methods have been described . The presence of at least one gout-related erosion in the hands or feet, defined as a cortical break with sclerotic margins and overhanging edges, forms part of the 2015 ACR/EULAR diagnostic criteria for gout . However, erosions can take years to develop, and while useful for screening of chronic gout-related changes, radiographs have a limited role in the acute setting. Instead, the updated 2018 EULAR guidelines suggest that ultrasound is a more useful modality in the acute assessment of gout .
Ultrasound
Similar to conventional radiography, ultrasound is readily available, low cost and well tolerated by patients. In addition, it lacks any form of ionizing radiation and can guide both therapeutic and diagnostic interventions, such as joint aspirations. Disadvantages include reliance on having an appropriate acoustic window to enable visualization of the target area and limited definition of deeper structures. Generic inflammatory and more specific gout-related features can be identified by ultrasound. Non-specific features include the presence of joint effusions, synovial hypertrophy and hyperaemia. Four key ultrasound features of gout were defined through a Delphi exercise by the OMERACT working group in 2015 : aggregates, the double contour sign, erosions and tophi (Figs 2 and 3).
Figure 2.
Ultrasound detection of MSU deposition in gout. (A) MTP joint effusion that contains hyperechoic foci of MSU aggregates and a subtle double contour sign (arrowheads). (B) MSU aggregates within a joint effusion and along the joint lining (arrowhead). (C) Large tendon sheath effusion containing tiny hyperechoic foci representing MSU aggregates and giving rise to the snowstorm appearance
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Figure 3.
Appearance of tophi and erosions on ultrasound. (A) Heterogeneous mass (arrowheads) with central hypoechogenicity and posterior acoustic shadowing, reflecting the typical appearance of a tophus. (B) Another typical tophus demonstrating low-grade internal vascularity. (C) Cortical discontinuity consistent with a juxta-articular erosion (arrowhead)
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As part of the disease process, MSU crystals can aggregate within joint fluid, synovium and hyaline cartilage as well as other extra-articular soft tissue structures such as tendons and bursae . Aggregates appear as tiny hyperechoic foci within these structures and, when floating in synovial fluid, give rise to the snowstorm appearance.
The double contour sign occurs when MSU crystals are deposited onto hyaline cartilage, resulting in a hyperechoic band that can be irregular or discontinuous, but is independent of the angle of insonation. The double contour sign is included in the 2015 ACR/EULAR diagnostic criteria for gout . This sign can also appear in calcium pyrophosphate (CPP) deposition disease. Dynamic manoeuvres can help distinguish between the two conditions, as MSU crystals are deposited onto cartilage and move with the underlying bone, whereas CPP crystals move in the opposite direction .
Tophi are heterogeneous, hyperechoic or hypoechoic circumscribed masses that may have small anechoic rims and variable posterior acoustic shadowing. The acoustic shadowing is likely due to the density of MSU crystals and the presence of some degree of calcification [24, 25]. Erosions are intra- or extra-articular discontinuities of the bone surface that must be visible in two perpendicular planes.
The intra-observer and interobserver reliability of these definitions was assessed by the OMERACT group using static images and in patients [22, 26]. Reliability ranged from moderate to excellent in static images, but aggregates were the least reliable, with a Cohen’s κ coefficient of 0.54 for intra-observer and 0.65 for interobserver reliability. Interobserver reliability was only 0.21 for aggregates in the patient group. Therefore, aggregates were redefined in 2021 by Christiansen et al. , with the overarching principle that aggregates can only be deemed to be present if there are other ultrasound features of gout. In addition, aggregates cannot be located within tophi. The new definition of aggregates improved reliability, which was found to be comparable to other ultrasound features. The 2021 revision also introduced a semiquantitative scoring system of 0 (normal), 1 (possibly present), 2 (definitely present but mild) or 3 (definitely present and severe).
Multiple studies have demonstrated that ultrasound can be useful in establishing the diagnosis of gout. Ogdie et al. demonstrated that the double contour sign, tophi and snowstorm appearance are highly specific features (>0.9) in aspiration-proven cases of gout. However, relatively low sensitivity was found in cases of early disease of <2-years duration. In 2021, Christiansen et al. compared all four OMERACT ultrasound features against reference standards of diagnosis and found that all features are highly sensitive (0.77–0.95) but differed in specificity. Tophi and the double contour sign both demonstrated greater specificity (0.88–0.98) compared with aggregates and erosions (0.32–0.59). Christiansen et al. employed an extensive scanning protocol that took ≈1 h to complete. However, a reduced scanning protocol involving both knee and first MTP joints, as well as the peroneal tendons, would have identified the double contour sign or tophi in all patients, without the need to scan other areas. Similar conclusions were drawn by Cipolletta et al. in a 2023 study assessing different scanning protocols and the impact this has on diagnostic accuracy. The authors conclude that due to the preferential involvement of the knee and MTP joints in gout, these joints should be included in routine scanning protocols even if asymptomatic.
The diagnostic potential of ultrasound was assessed in a meta-analysis by Lee and Song in 2018, including 938 patients with gout and 788 controls. The pooled sensitivity of the double contour sign, tophi and snowstorm appearance was 65.1%, 54.3% and 30.8%, respectively. The pooled specificity, however, was far greater, at 89.0%, 93.2% and 90.6%, respectively. A network meta-analysis by Stewart et al. compared the accuracy of ultrasound features vs DECT in patients with suspected gout. The pooled accuracy of DECT, tophi and the double contour sign were similar at 0.77, 0.76 and 0.75, respectively. The accuracy of aggregates was lower, at 0.63, and the included data were more varied.
Ultrasound features of gout can be serially assessed and can become less conspicuous with successful reduction in urate levels by the use of urate-lowering therapy (ULT). After starting ULT, patients in a trial conducted by Ebstein et al. demonstrated early disappearance of the double contour sign and an overall decrease in the size of tophi at 6 months. Lower serum urate levels also correlated with greater decreases in the size of tophi and disappearance of the double contour sign. Similarly, a study by Hammer et al. demonstrated a reduction in tophi size and presence of the double contour sign at 12 months.
Aside from its diagnostic potential in symptomatic patients, ultrasound can identify subclinical gout-related changes in asymptomatic hyperuricaemic individuals before these findings become clinically apparent . The severity of MSU deposition, in the form of aggregates, tophi or the double contour sign, predicts the likelihood of future gout flares, albeit in patients with a pre-existing diagnosis of gout . Furthermore, Cipolletta et al. showed that ultrasound can be used to predict successful clinical remission at 12 months, with a statistically significant correlation with baseline disease burden as assessed by ultrasound. Only 33.3% of patients with ultrasound evidence of disease at baseline achieved remission compared with 87.5% of those without any structural abnormalities (P< 0.01).
CT and DECT
Gouty erosions and tophi can be depicted by conventional CT as hyperdense (average HU 170) articular or periarticular masses with variable calcification and characteristic erosions in typical locations (Fig.4) [37–39]. However, conventional CT has now been largely superseded by DECT.
Figure 4.
Appearance of tophi on CT. (A) Sagittal and (B) axial CT images of a large tophus in the form of a lobulated hyperdense mass, circumferentially surrounding the middle finger IP joint (arrowhead)
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DECT can differentiate materials based on relative absorption and attenuation of X-rays at different energy levels. High atomic number materials, such as calcium, demonstrate greater attenuation differences at higher and lower energy levels compared with low atomic number materials, such as MSU crystals. This principle can enable specific tissue characterization and accurately separate deposited MSU crystals from surrounding soft tissues.
Distinct energy levels, typically 80kVp and 140kVp, can be produced using dual-source scanners or by single tube scanners, the latter using either sequential acquisition or rapid voltage switching. Patients can be scanned sequentially, once at 80kVp and then again at 140kVp, resulting in longer acquisition times and possible misregistration artefacts . Rapid kVp switching between high and low kVp using a single tube system is an alternative method that minimizes misregistration but can result in a higher overall radiation dose . Both methods are acceptable and the exact choice of system is at the discretion of the local imaging department.
DECT has a number of different clinical applications and various DECT post-processing algorithms are available to meet these requirements . In the context of gout, a two-material decomposition algorithm can be applied, separating MSU crystals from calcium and background soft tissue. Depending on the vendor and software package, gouty deposits can be reconstructed and displayed as colour-coded lesions overlayed onto two-dimensional or three-dimensional CT reformatted images (Fig.5). Preprocessed CT images are also typically available to allow assessment for erosions and other ancillary findings. Furthermore, regions of interest can be applied to indeterminate lesions, creating a histogram of effective atomic numbers within the selected voxels (Fig.6). The effective atomic numbers can be compared with reference values and this can act as a problem-solving tool in select cases where conventional assessment is still ambiguous.
Figure 5.
DECT image reconstructions displaying the deposition of MSU crystals. (A) Two-dimensional reformatted image highlighting areas of MSU deposition (green) within the knee. Note both the intra-articular (star) and intratendinous (arrowhead) deposition. (B) Three-dimensional image with MSU deposition highlighted in purple. (C) Three-dimensional image with MSU deposition highlighted in green
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Figure 6.
Application of regions of interest onto DECT images (top row and bottom right) to produce a histogram of effective atomic numbers (bottom left)
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DECT forms part of the ACR/EURLAR classification criteria for gout . A positive DECT scan is defined as the presence of colour-coded MSU deposits at articular or peri-articular sites, after potential causes for false positive results have been accounted for.
Multiple studies have since demonstrated that DECT is both sensitive and specific for the presence of MSU deposits, with pooled sensitivities and specificities in the region of 90% [42–44]. A recent meta-analysis by Gamala et al. found that pooled sensitivity of DECT was 81% when assessing multiple joints for the extent of disease or 83% when DECT is only applied to symptomatic joints. Specificity was high, ranging from 88% to 91%. No major difference was found between patients with gout diagnosed by clinical criteria or by a positive aspirate. There was generally low pooled sensitivity (55%) in patients with recent-onset disease, <6 weeks in duration. Jia et al. demonstrated that the sensitivity of DECT significantly increases with disease duration, reaching 92.86% in those with symptoms for >24 months. In the same study, the volume of MSU crystal deposition was strongly correlated with disease duration and number of flares, which could potentially account for the significantly increased DECT sensitivity in the ≥24 month group.
DECT appears to be more sensitive relative to ultrasound, although some ultrasound features convey slightly higher specificity [31, 47]. This could reflect the ability of DECT to assess for deeper structures that are not readily accessible by ultrasound. However, there is evidence that in recent-onset disease, the sensitivity of DECT is lower than that of ultrasound [46, 48]. This may be due to the limited spatial resolution of DECT, typically 2 mm, which could result in undetected early MSU deposits that may be visible with ultrasound instead. Lee et al. improved overall diagnostic accuracy in early disease by combining DECT and conventional CT images. The authors theorized that articular hyperdensity on conventional CT corresponds to early MSU deposits and was the CT equivalent of the double contour sign.
MSU deposits can be detected by DECT in asymptomatic hyperuricaemic patients, although deposits occur more frequently and in larger volumes in symptomatic patients . Wang et al. also demonstrated that 15% of asymptomatic patients have MSU crystal deposits within the ankles and/or feet. This strongly correlated with age and also with the duration of symptoms, although this was not formally assessed. There is also some evidence that MSU deposits identified by DECT can be predictive of future flares of gout .
In cases of known gout, DECT can illustrate decreases in crystal deposition after the introduction of ULT. In a 2-year prospective study by Uhlig et al. , patients with gout were initiated on ULT with a treat-to-target approach. Successful reduction in semiquantative DECT scoring was demonstrated at 1 and 2 years in all assessed locations, regardless of whether the patient met their treatment target. Dalbeth et al. also demonstrated a volumetric reduction in DECT urate levels in patients receiving ULT, with significantly greater reductions in patients meeting serum urate targets.
Multiple potential artefacts and limitations of DECT exist in the assessment of gout, aside from reduced sensitivity early in the disease. Common artefacts include superfluous identification of MSU in high-density areas such as the nail bed or entheses, apparent submillimetre lesions arising from noise and limited interpretation in the presence of beam hardening artefacts (Fig.7) . Mixed composition deposits, containing calcium as well as MSU, can also impair accurate interpretation . The sensitivity of DECT for non-tophaceous gout is lower than for tophaceous gout, including in joints with aspirate-proven MSU deposits .
Figure 7.
Beam hardening artefact from a joint replacement resulting in significantly degraded (A) CT and (B) DECT images of the knee
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MRI
MRI, by its multiplanar nature and excellent soft tissue spatial resolution, is a key modality in the evaluation of most rheumatological conditions. General inflammatory features such as effusions and synovial hypertrophy, as well as structural changes in the form of erosions and chondral damage, can be identified.
With respect to gout, MRI can provide useful information regarding the presence of tophi, inflammation and joint damage, although its diagnostic role is less well established than DECT or ultrasound . MRI can demonstrate erosions that are not identified on ultrasound or radiographs and reliable quantitative volume assessment of tophi is also possible [58, 59]. Synovial pannus, defined as contrast enhancement and thickening of the synovial lining, can be detected using MRI and was identified in 87.5% of gout patients in a study by Carter et al. during the intercritical period. However, the relationship between the pannus and the disease process is unclear, as no correlation was found between serum urate and the presence or severity of pannus. Furthermore, a small subgroup analysis by McQueen et al. found that MRI had a high specificity (98%) for detecting the presence of tophi when using DECT as a reference standard.
MRI signal characteristics of tophi are heterogeneous, reflecting the variable internal composition, consisting of a central MSU core with surrounding inflammatory, granulation and fibrous tissue (Fig.8). Tophi demonstrate intermediate to low T1 signal intensity with heterogeneous but typically intermediate to low T2 signal intensity, possibly due to the presence of fibrous tissue. Post-contrast enhancement can be homogeneous and intense or heterogeneous and peripheral . Calcification is thought to be a feature of late disease, although this is contentious .
Figure 8.
MRI appearances of tophi. (A) Sagittal and (B) axial images demonstrate a lobulated intermediate signal mass (arrowheads) eroding the anterior aspect of the patella consistent with a tophus. (C) Lobulated intra-articular mass (star) eroding and destroying the second toe distal IP joint consistent with a tophus. (D) Note the well-defined signal change with low-grade oedema on the fluid-sensitive, fat-suppressed image relative to the degree of joint destruction
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Tophi are commonly associated with bony erosions that classically form overhanging edges. Unlike other inflammatory conditions, erosions are associated with little to no bone marrow oedema, although there are limited data in this regard . In a study by McQueen et al. , features of chronic synovitis were found in a large number of patients, but only a small proportion demonstrated any bone marrow oedema and, when present, this was generally graded as mild. Similarly, cartilage damage in gout is associated with the presence of tophi, erosions and synovitis, but unlike RA, is not associated with bone marrow oedema .
Nuclear medicine
Nuclear medicine studies are not routinely used for the assessment of gout and are not currently included in any diagnostic guidelines. A few isolated cases reports have been published in this regard but no large study [66–68]. The presence of inflammatory change and gouty tophi is associated with increased radiotracer uptake; however, this is non-specific finding and reflects increased metabolic and/or osteoblastic rates in these areas. In these case reports, patients only underwent nuclear medicine studies in the workup for suspected malignancy and the diagnosis of gout was confirmed on postoperative specimen analysis.
Conclusion
The diagnosis of gout relies on a combination of clinical, laboratory and imaging findings, of which aspiration of MSU crystals remains the gold standard. Radiographs, CT, DECT, MRI and ultrasound have all been used in the assessment of gout. Of these, ultrasound and DECT hold the most potential in the early diagnosis of gout, with good evidence that both are accurate in assessing disease burden. Both have been incorporated in diagnostic algorithms, with plain radiographs used to assess for chronic established disease. MRI can identify some gout-related changes, such as tophi and erosions, but its exact role is currently unclear and there is limited data in this regard. Nuclear medicine studies are currently not used in the assessment of gout. Aside from establishing the diagnosis of gout, there is evidence that both ultrasound and DECT can identify asymptomatic patients with gout-related changes, predict future gout flares and assess response to treatment. With further evidence, both of these modalities could have expanded roles in future guidelines.
Data availability
No new data were generated or analysed in support of this article.
Authors’ contributions
L.W. and E.R. conceptualized the manuscript and contributed images. L.W. wrote the original draft and subsequent revisions with both L.W. and E.R. reviewing the final manuscript.
Funding
No specific funding was received from any bodies in the public, commercial or not-for-profit sectors to carry out the work described in this article.
Disclosure statement: The authors have declared no conflicts of interest.
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Reflection
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As light reflects from mirrors, we reflect lines and graphs from mirrors in mathematics. Reflections are of great interest in mathematics as they can be used in different areas of geometry to prove many results. In calculus and analysis, there are terms which make use of reflection like even and odd functions, inverse of a function, etc.
Contents
Definition
The reflection of a point, line, or a figure is the mirrored image of it along some line, plane, etc.
For example, in two dimensions, reflecting a line over another line results in a second line. In the figure below, the line (y=3x) is reflected over the line (y=x), resulting in the line (y=\frac{x}{3}.)
Construction
In practical applications, it is sometimes necessary to construct the reflections of points, lines, and even polygons about a line.
One method of doing it is via a compass. The details of this construction are given below.
Reflection of a given point (A) about a given line (l:)
Construction of Image
A proof of this is given below.
If we show that the line joining (A) and (B) is perpendicular to (l) and that these points are equidistant from (l,) then we're done!
So, let's join (A) and (B). Let (AB) intersect (l) at (O).
Note that as (A) and (B) both lie on the circles with centers (O_1) and (O_2), we have
[ O_1A=O_1B, \quad O_2A=O_2B.]
As (O_1O_2=O_1O_2,) we see that
[\Delta O_1O_2A \cong \Delta O_1O_2B.]
Hence (\angle O_1O_2A=\angle O_1O_2B). Also, as (O_2A=O_2B) and (OO_2=OO_2,) we have
[\begin{align}
\Delta OO_2A &\cong \Delta OO_2B \
\Rightarrow OA&=OB \
\Rightarrow \angle AOO_2 &= \angle BOO_2.
\end{align}]
As (\angle AOO_2) and (\angle BOO_2) are angles on a straight line, (\angle AOO_2+\angle BOO_2 = 180^\circ. )
Hence we conclude that (\angle AOO_2=\angle BOO_2=90^\circ). This means that the distance of (A) from (l) is (OA). Similarly the distance of (B) from (l) is (OB). As (OA=OB), both (A) and (B) are equidistant from the line (l).
So, we're done! (_\square)
Some miscellaneous constructions are as follows:
Reflection Across a Line
While reflecting a point about the (x)-axis, the magnitude of its (y)-coordinate remains the same but its sign changes. But the (x)-coordinate remains the same. This is because we are taking it to the quadrant below it. So, we can say that ({\text{R}}_{x}(x,y) = (x,-y)).
What will be the reflection of the point ((3,-5)) in the (x)-axis?
We have
[\begin{align}
{\text{R}}_x (x,y) & = (x,-y)\
{\text{R}}_x (3,-5) & = \big(3,-(-5)\big)\
& =(3,5).\ _\square
\end{align}]
While reflecting a point about the (y)-axis, the magnitude of its (x)-coordinate remains the same but its sign changes. But the (y)-coordinate remains the same. This is because we are taking it to the quadrant beside it. So, we can say that ({\text{R}}_{y}(x,y) = (-x,y)).
What will be the reflection of the point ((2,4)) in the (y)-axis?
We have
[\begin{align}
{\text{R}}_y (x,y) & = (-x,y)\
{\text{R}}_y (2,4) & = (-2,4).\ _\square
\end{align}]
While reflecting a point in the origin, the magnitude of its coordinates remains the same but their signs change. This is because we are taking it to the diagonally opposite quadrant. So, we can say ({\text{R}}_o (x,y) = (-x,-y)).
What will be the reflection of the point ((-13,6)) in the origin?
We have
[\begin{align}
{\text{R}}_o (x,y) & = (-x,-y)\
{\text{R}}_o (-13,6) & = \big(-(-13),-6\big)\
& =(13,-6).\ _\square
\end{align}]
In two dimensions, it is also possible to reflect across any line of the form (y=mx) (()note that the special case (x=0) is the same as reflecting over the (y)-axis(),) and once this is done we can easily extend to (mx+b) by shifting the (y)-coordinates by (b). It is possible to obtain a general formula for the reflected point ((x',y'):)
[(x',y') = \left(2\dfrac{x+(y-b)m}{1+m^2}-x,\ 2\dfrac{x+(y-b)m}{1+m^2}x-y+2b\right).]
The expression above can be obtained using vectors. Again, let (y=mx) or (\lambda \langle 1, m \rangle), and let (\mathbf{p}=\langle x,y \rangle). If the projection of (\mathbf{p}) onto the line is the vector (\mathbf{l}), we can easily find (\mathbf{p'} = 2 \mathbf{l}-\mathbf{p}). Fortunately, there is a simple formula for the projection of a vector onto another, which gives (()here, we can treat the line as a vector by choosing (\lambda>x) and taking that point as (\mathbf{r}))
[\displaystyle \mathbf{l}= \frac{ \mathbf{p} \cdot \mathbf{r} }{ \left|\mathbf{r} \right|^2 } \mathbf{r}.]
This leaves you with a way to calculate ((x',y') = \mathbf{p'} = 2 \frac{ \mathbf{p} \cdot \mathbf{r} }{ \left|\mathbf{r} \right|^2 } \mathbf{r}-\mathbf{p}) without any trig functions (if you use rotation matrices).
Specifically, you can simplify what we have above to ((x',y') = \left(2\frac{x+ym}{1+m^2}-x,\ 2\frac{x+ym}{1+m^2}x-y\right)), which is a nice formula that can be applied in many situations. If we want (mx+b), we apply (y \to y-b) and add (b) to (y') (()these are the coordinate shifts talked about before the proof: we move everything down by (b), apply our formula, then move everything up by (b)) to get
[\displaystyle (x',y') = \left(2\frac{x+(y-b)m}{1+m^2}-x,\ 2\frac{x+(y-b)m}{1+m^2}x-y+2b\right).\ _\square]
Reflection Symmetry
An interesting application of the reflection across a line is that the graph of a function and its inverse function are symmetric about ( x = y ).
As we know, (y=f(x) \implies x=f^{-1}(y)) and this can be interpreted as a change of coordinates (x) and (y), so we can define the graph of (f^{-1}(x)) as the reflection of the graph of (f(x)) across the line (y=x).
Reveal the answer
For the function (f(x)={x}^5+1), find the number of real values of (x) such that (f(x)={f}^{-1}(x)).
The correct answer is: 1
For the function (f(x)={x}^5+1), find the number of real values of (x) such that (f(x)={f}^{-1}(x)).
Problem Solving
You are given two fixed points (A) and (B) on the same side of an arbitrary straight line. Find the minimum perimeter of the triangle formed by the two fixed points and a third point (C) which could be anywhere on the line.
Idea to work on: the third point lies on the intersection of the straight line joining the image of one point and the second point with the given line.
[f (x)=x^{3}]
[f (x)=\cos x]
[f (x)=\tan x]
None of the above
Reveal the answer
About the (y)-axis, for which of the following reflection of (f(x)) is (f(x)?)
The correct answer is: [f (x)=\cos x]
Reveal the answer
Let (ABC) be an equilateral triangle with perimeter (24 \text{ cm}). Let (M) be the midpoint of (AC). Let (L) and (N) be points on (AB) and (BC,) respectively, such that the perimeter of triangle (LMN) is at a minimum.
Find this minimum perimeter.
The correct answer is: 12
[ e^{1 } ]
[ e^{ -1 } ]
[ e^{e } ]
[ e^{ -e } ]
[ e^{ \frac{1}{e} } ]
Reveal the answer
For some constant ( a> 1 ), the two functions (f(x) = a^x) and (g(x) = \log_a x) intersect exactly once.
What is (a?)
The correct answer is: [ e^{ \frac{1}{e} } ]
About the (y)-axis, for which of the following reflection of (f(x)) is (f(x)?)
Let (ABC) be an equilateral triangle with perimeter (24 \text{ cm}). Let (M) be the midpoint of (AC). Let (L) and (N) be points on (AB) and (BC,) respectively, such that the perimeter of triangle (LMN) is at a minimum.
Find this minimum perimeter.
For some constant ( a> 1 ), the two functions (f(x) = a^x) and (g(x) = \log_a x) intersect exactly once.
What is (a?)
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14356 | https://www.geeksforgeeks.org/microeconomics/profit-maximization-meaning-elements-conditions-and-formula/ | Profit Maximization : Meaning, Elements, Conditions and Formula
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23 Jul, 2025
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What is Profit Maximization?
Profit Maximization is the core objective of many businesses that represent the pursuit of strategies to achieve the highest possible net income. This involves identifying optimal production levels, pricing strategies, and cost management practices to ensure revenue exceeds costs, leading to increased profitability. In essence, it's about striking the right balance between income generation and cost management to ensure sustained financial success.
Geeky Takeaways:
Profit maximization is the goal of a business to increase the net income or profit of a business to the highest possible level.
Revenue Maximization, Cost Minimization, Optimal Output Level, and Pricing Strategy are key elements of Profit Maximization.
Profit Maximization is all about generating maximum profit and managing costs while operating at the optimum level of production.
Table of Content
Elements of Profit Maximization
Conditions to attain Profit Maximization
Profit Maximization Formula
Profit Maximization Graph
Reasons for Profit Maximization
Significance of Profit Maximization
Advantages and Disadvantages of Profit Maximization
Conclusion
Frequently Asked Questions (FAQs)
Elements of Profit Maximization
(A) Output and Pricing Optimization:
1. Understanding Demand Elasticity: Businesses should analyze the elasticity of demand for their products or services. Inelastic demand allows for higher prices, while elastic demand may require lower prices to increase overall revenue.
2. Segmentation and Targeting: Identifying specific market segments and targeting them effectively can lead to more precise pricing strategies. Different consumer groups may respond differently to pricing adjustments.
3. Dynamic Pricing Strategies: Implementing dynamic pricing models, such as personalized pricing or time-based pricing, can optimize revenue by adjusting prices based on factors like demand fluctuations, customer behavior, or seasonal trends.
4. Bundling and Cross-Selling: Offering product bundles or implementing cross-selling strategies can encourage customers to purchase more, potentially increasing overall revenue without a proportionate increase in costs.
(B) Cost Control:
1. Technology Integration: Leveraging technology, such as automation and advanced data analytics, can streamline operations, reduce errors, and enhance efficiency, contributing to overall cost reduction.
2. Supply Chain Optimization: Collaborating with suppliers and optimizing the supply chain can lead to cost savings. Negotiating favorable terms, minimizing lead times, and adopting just-in-time inventory practices are examples of supply chain optimization.
3. Energy Efficiency and Sustainability: Investing in energy-efficient technologies not only aligns with sustainability goals but can also result in cost savings over time. Sustainable practices, when integrated into operations, can enhance efficiency and reduce waste.
4. Employee Training and Productivity: Investing in employee training can improve skills and productivity, contributing to cost control. Engaged and skilled employees are often more efficient, reducing the likelihood of errors and rework.
5. Outsourcing and Offshoring: Assessing which functions can be outsourced or offshored to locations with lower labor costs can be a strategic approach to cost reduction. However, careful consideration is needed to balance cost savings with maintaining quality standards.
6. Economies of Scale: Increasing production levels can lead to economies of scale, where the average cost per unit decreases as production volume rises. This can be achieved through efficient production processes and maximizing capacity utilization.
7. Flexible Cost Structure: Maintaining a flexible cost structure allows businesses to adapt to changes in the market. Fixed costs should be kept to a minimum, and variable costs should be structured in a way that allows for adjustments based on demand fluctuations.
8. Benchmarking: Regularly benchmarking costs against industry standards and competitors can highlight areas where a business may be overspending. This can inform cost-cutting initiatives and improve overall efficiency.
Conditions to attain Profit Maximization
Marginal Cost should be equal to Marginal Revenue. (MC = MR)
Marginal Cost should be greater than Marginal Revenue after MC = MR.
Profit Maximization Formula
The profit maximization formula is expressed mathematically as:
MC = MR
Where,
MC (Marginal Cost) represents the additional cost incurred by producing one additional unit of output. It is calculated as the change in total cost divided by the change in quantity of output (MC=ΔTC/ΔQ).
MR (Marginal Revenue) represents the additional revenue gained by selling one additional unit of output. It is calculated as the change in total revenue divided by the change in quantity of output (MR=ΔTR/ΔQ).
Profit Maximization Graph
The profit maximization graph is a visual representation of the relationship between output levels and profits for a given business. A profit maximization graph helps to determine the level of output at which the profit is maximum.
Component of Graph:
1. Marginal Revenue (MR) Curve: Marginal Revenue curve represents the additional revenue earned from the sale of one additional unit of the product.
2. Marginal Cost (MC) Curve: Marginal Cost Curve represents the additional cost incurred by a firm to produce one more unit of a good or service.
3. Break-Even Point: The break-even point occurs where the MC and MR curves intersect, indicating the level of output at which revenue earned is equal to the cost incurred. At this point, the firm is covering all its costs but is not yet making a profit. It is a situation of 'no loss - no profit.'
4. Profit (π) Curve: The profit curve is derived by subtracting cost from revenue at each output level. The point where the profit curve is at its peak indicates the level of output at which profit is maximized.
5. Profit-Maximizing Output: The profit-maximizing output level is found where Marginal Revenue (MR) equals Marginal Cost (MC).
6. Optimal Production Level: The profit maximization point represents the optimal production level for the firm; i.e., the intersection point of the Marginal Cost (MC) curve and the Marginal Revenue (MR) curve. Producing more or fewer units would result in lower profits due to diminishing returns or underutilization of resources.
Explanation:
In the above graph, the X-axis represents the output, and the Y-axis represents the cost and price. A Marginal Revenue Curve is a horizontal line parallel to the X-axis, and a Marginal Cost Curve is U-shaped. Point R and K are two points of intersection where MC equals MR. Since after point R marginal cost remains lower than marginal revenue, the firm will continue to increase its output to earn more. So, point R despite being the point of intersection cannot be considered a Profit Maximization Point. Hence. in the above graph point K is a point of profit maximization because:
1. At point K, Marginal Revenue (MR) = Marginal Cost (MC).
2. Beyond this point marginal cost becomes more than marginal revenue.
So, the firm earns a maximum profit at this level of output, i.e., at OQ1 ( Profit-Maximizing Output).
Reasons for Profit Maximization
The profit maximization point occurs where the Marginal Cost equals the Marginal Revenue. This is because, at this point, the additional cost of producing one more unit is exactly balanced by the additional revenue gained from selling that unit.
Marginal Cost should be greater than Marginal Revenue after MC = MR because if beyond this point MC is lower than MR then, the firm continues to earn profit, hence the point of profit maximization is not attained.
Significance of Profit Maximization
At the profit maximization point, the firm is optimizing its production level to ensure that the cost of production is justified by the revenue generated from selling each additional unit.
The level of output at which MC equals MR is the optimal production level for maximizing profits. Beyond this point, the marginal cost may exceed marginal revenue, leading to a reduction in overall profit.
Advantages and Disadvantages of Profit Maximization
Advantages:
1. Operational Efficiency: Profit maximization often encourages businesses to streamline operations and adopt cost-effective practices, leading to improved operational efficiency.
2. Competitive Edge: Generating higher profits can provide a competitive advantage, allowing businesses to invest in innovation, marketing, and other strategies that enhance their position in the market.
3. Investment Opportunities: Increased profits provide businesses with the financial resources needed for investment in research and development, expansion, and other growth initiatives.
4. Creditworthiness: A consistently profitable business is viewed as more creditworthy by lenders and investors, making it easier to secure loans and attract investment capital.
5. Employee Benefits: Profitable companies are better positioned to provide competitive salaries, benefits, and opportunities for employee growth, contributing to higher employee satisfaction and retention.
Disadvantages:
1. Ethical Concerns: Strict profit maximization, without ethical considerations, may lead to exploitative practices, compromising the well-being of stakeholders, including employees and the environment.
2. Quality Sacrifice: In the pursuit of short-term profits, there may be a risk of sacrificing product or service quality, which can have long-term consequences on customer trust and brand reputation.
3. Limited Innovation: Overemphasis on immediate financial gains might discourage investment in long-term innovation, hindering a company's ability to adapt to evolving market demands.
4. Negative Externalities: Profit maximization strategies may overlook the environmental and social impact of business activities, leading to negative externalities that harm communities and ecosystems.
5. Employee Well-being: Prioritizing profits over employee well-being can result in burnout, low morale, and high turnover, negatively impacting the long-term sustainability of the business.
Conclusion
In conclusion, while profit maximization stands as a fundamental objective for businesses, its pursuit must be approached with strategic foresight and ethical considerations. The delicate balance between short-term gains and long-term sustainability is integral to ensuring the enduring prosperity of a company and the well-being of its stakeholders. By embracing practices that go beyond immediate financial metrics—such as ethical business conduct, environmental responsibility, and a customer-centric approach—businesses can foster lasting success. Moreover, a commitment to corporate social responsibility, adaptability, and continuous improvement is essential in navigating the complexities of a dynamic market. Therefore, profit maximization should not exist in isolation but rather as part of a broader strategy that aligns with ethical principles and envisions the long-term sustainability of the business, contributing to enduring success and positive societal impact.
1. What is the primary goal of profit maximization?
Answer:
The primary goal of profit maximization is to increase the wealth of the owner or the shareholders of the firm by increasing the net profit of the firm.
2. What is the formula for profit maximization?
Answer:
The profit maximization formula involves setting Marginal Cost (MC) equal to Marginal Revenue (MR); i.e., MR = MC.
3. Is profit maximization the same as wealth maximization?
Answer:
No, profit maximization focuses on short-term gains, while wealth maximization considers long-term value creation and factors like sustainability and corporate social responsibility.
4. What are some tips for maximizing profits?
Answer:
Cost efficiency, strategic pricing, and investment in innovation to capture new markets shall be the point of attention to mamize profit.
Also refer to Profit Maximization in Perfect Competition Market and Profit Maximization in Monopoly Market.
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CBSE Class 11 Microeconomics Notes
7 min read
Chapter 1: Introduction
Introduction to Microeconomics
6 min readMicroeconomics and Macroeconomics: Meaning, Scope, and Interdependence
5 min readEconomic Problem & Its Causes
4 min readCentral Problems of an Economy
6 min read
Chapter 2: Consumer's Equilibrium
Theory of Consumer Behaviour
6 min readDifference between Needs and Wants
9 min readUtility Analysis : Total Utility and Marginal Utility
7 min readLaw of Diminishing Marginal Utility (DMU) : Meaning, Assumptions & Example
4 min readConsumer's Equilibrium in case of Single and Two Commodity
9 min readIndifference Curve : Meaning, Assumptions & Properties
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10 min readDifference between Budget Line and Budget Set
5 min readShift in Budget Line
4 min readConsumer€™s Equilibrium by Indifference Curve Analysis
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Chapter 3: Demand
Theory and Determinants of Demand
7 min readIndividual and Market Demand
7 min readDifference between Individual Demand and Market Demand
3 min readWhat is Demand Function and Demand Schedule?
5 min readLaw of Demand
12 min readMovement along Demand Curve and Shift in Demand Curve
7 min readDifference between Expansion in Demand and Increase in Demand
3 min readDifference between Contraction in Demand and Decrease in Demand
3 min readSubstitute Goods and Complementary Goods
6 min readDifference between Substitute Goods and Complementary Goods
2 min readNormal Goods and Inferior Goods
6 min readDifference between Normal Goods and Inferior Goods
3 min readTypes of Demand
3 min readSubstitution and Income Effect
5 min readDifference between Substitution Effect and Income Effect
3 min readDifference between Normal Goods, Inferior Goods, and Giffen Goods
4 min read
Chapter 4: Elasticity of Demand
Price Elasticity of Demand: Meaning, Types, Calculation and Factors Affecting Price Elasticity
7 min readMethods of Measuring Price Elasticity of Demand: Percentage and Geometric Method
6 min readDifference between Elastic and Inelastic Demand
5 min readRelationship between Price Elasticity of Demand and Total Expenditure
3 min read
Chapter 5: Production Function: Returns to a Factor
Production Function: Meaning, Features, and Types
6 min readWhat is TP, AP and MP? Explain with examples.
3 min readLaw of Variable Proportion: Meaning, Assumptions, Phases and Reasons for Variable Proportions
9 min readRelationship between TP, MP, and AP
5 min readLaw of Returns to Scale
3 min readDifference between Returns to Factor and Returns to Scale
2 min read
Chapter 6: Concepts of Cost and Revenue
What is Cost Function?
6 min readDifference between Explicit Cost and Implicit Cost
2 min readTypes of Cost
7 min readWhat is Total Cost ? | Formula, Example and Graph
4 min readWhat is Average Cost ? | Formula, Example and Graph
4 min readWhat is Marginal Cost ? | Formula, Example and Graph
3 min readVariable Cost: Meaning, Formula, Types and Importance
11 min readInterrelation between Costs
6 min readTypes of Cost
7 min readConcepts of Revenue| Total Revenue, Average Revenue and Marginal Revenue
4 min readRelationship between Revenues (AR, MR and TR)
5 min readBreak-even Analysis: Importance, Uses, Components and Calculation
5 min readWhat is Break-even Point and Shut-down Point?
2 min read
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Producer's Equilibrium: Meaning, Assumptions, and Determination
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4 min readDifference Between Expansion of Supply and Increase in Supply
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Chapter 9: Forms of Market
Market : Characteristics & Classification
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6 min readDifference between Perfect Competition and Monopolistic Competition
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6 min readDistinction between the four Forms of Market(Perfect Competition, Monopoly, Monopolistic Competition and Oligopoly)
5 min readLong-Run Equilibrium under Perfect, Monopolistic, and Monopoly Market
4 min readProfit Maximization : Meaning, Elements, Conditions and Formula
9 min readProfit Maximization in Perfect Competition Market
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14357 | https://www.bytelearn.com/articles/combining-like-terms/ | Skip to content
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How To Teach Combining Like Terms
By John Maloney / December 21, 2022 February 7, 2023 / 5 minutes of reading / Math & More
Table Of Contents
Key Concepts of Combining Like Terms
How do you teach combining like terms?
Step 1: Identify the like terms
Step 2: Rearrange the expression so the like terms are next to each other.
Step 3: Simplify the expression.
Examples
Example 1 (Two Constants and One Coefficient):
Example 2 (Two Constants and Two Like Coefficients):
Example 3 (Two Unlike Coefficients):
Why teach combining like terms this way?
Vocabulary for teaching combining like terms
Misconceptions and errors students are likely to have on Combining Like Terms:
Resources:
Frequently Asked Questions for Combining Like Terms
How do you explain like terms?
How do you explain combining like terms?
How do you combine like terms with rational numbers?
Key Concepts of Combining Like Terms
100+ Free Math Worksheets, Practice Tests & Quizzes
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Students are introduced to combining like terms in 6th grade. A big piece of the understanding in that grade is that expressions with different variables cannot be combined and that constant don’t go with anything that has a variable. Combining like terms gets a lot more complicated in 7th grade with integer coefficients and constants and powers more than 1.
Introduce students to combining like terms with a real-world task. Coach students to circle, box, and underline like terms with the sign in front to help them add and subtract with ease.
For more math resources you can click here
How do you teach combining like terms?
Explain to students what like terms are. Like terms are terms whose variables and their exponents (like 2 in x^2) are the same. For example, 5x and -2x are like terms because they are both terms with x. Explain that like terms can be combined and unlike terms cannot be combined. Defining examples and non-examples will help formalize your students’ understanding of math vocabulary.
Step 1: Identify the like terms
Start out circling one set of like terms. Remind students that they should include the + or – sign in front of the number. Then box or underline the remaining sets of like terms.
Step 2: Rearrange the expression so the like terms are next to each other.
Write an equivalent expression with each pair of like terms next to each other.
Step 3: Simplify the expression.
Combine each set of like terms. Provide your students with integer tiles so they practice adding and subtracting positive and negative integers. Encourage students to practice writing expressions with subtraction instead of adding a negative. In this example, have students write -28j – 11 instead of -28 + -11
Examples
Example 1 (Two Constants and One Coefficient):
Step 1:
Step 2:
Step 3:
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Example 2 (Two Constants and Two Like Coefficients):
Step 1:
Step 2:
Step 3:
Example 3 (Two Unlike Coefficients):
Step 1:
Step 2:
Step 3:
Why teach combining like terms this way?
Using shapes and colors can help students visually see the like terms that will be combined. Then students can rearrange the expression as needed to combine the like terms. It also reminds students to take the + or – sign into consideration when marking like terms.
Using Byte’s practice problems and virtual manipulatives will give your students the chance to make mistakes, get feedback, and try again until they master this essential concept!
Vocabulary for teaching combining like terms
Term: Part of an expression separated by + or – operations.
Like Term: Terms that have the same variables, raised to the same power. Constants are like terms.
Unlike Term: Terms that have different variables, or the same variables but raised to different powers.
Coefficient: A number that multiplies a variable. If there is no coefficient in front of a variable, the coefficient is 1.
Constant: A term that does not contain a variable.
Variable: A symbol (usually a letter) that is used to represent a number in an expression or equation.
Misconceptions and errors students are likely to have on Combining Like Terms:
Students might combine terms from left to right and disregard the variables. For example, students could simplify the expression 2 + 3x + 4 as 9xy.
Address this conceptual misunderstanding by assigning each variable with a specific noun. For example, if B represents Beyonce, and D represents Drake, does it make sense that 2B (or 2 Beyonces) + 3D (or 3 Drakes) is the same as 5BD (or 5 Beyonce Drakes?).
Students might combine every term with the same variable. For example, students might think that x + xy is the same as 2xy.
Try assigning a noun for each variable and ask your students if it makes sense to combine two nouns together. If this doesn’t click, try providing your students with integer tiles. It will “click” more easily for students if each set of tiles are color-coded. If this is not possible, provide colored pencils to your students and coach them to write each term in expanded form. For example, 2x + 3y is the same as x + x + y + y + y. Guide students to identify why the terms are unlike versus like.
Students may commit integer mistakes. For example, students may mistake -8x + 3y + 10x as 18x + 3y.
Provide your students with integer chips so they can model canceling out opposites.
If the previous suggestion doesn’t work, coach your students to write each term with a coefficient in expanded form with colored pencils.
Students may not recognize the minus signs as negative terms. For example, in 6x – 5 – 2x + 3, students might combine 6x and 2x and 5 and 3 and ignore the minus signs.
Remind students to circle the + or – in front of a term indicating if the term is positive or negative.
Also read: Solving Equations With Variables On One Side
Resources:
Downloadable worksheet for practicing combining like terms
Practice Problems for assessing combining like terms
Frequently Asked Questions for Combining Like Terms
How do you explain like terms?
Like terms have the same variable to the same power. Constants will always combine with each other.
How do you explain combining like terms?
Terms can only be combined if they have the same variable to the same power. Terms that do not have a variable, or constants, will combine.
How do you combine like terms with rational numbers?
Recognize which terms are positive or negative based on whether there is a plus or minus in front of the term. Combine like terms, and be careful when adding negative terms.
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AP®︎/College Biology
Course: AP®︎/College Biology>Unit 1
Lesson 1: Structure of water and hydrogen bonding
Hydrogen bonding in water
Hydrogen bonds in water
Capillary action and why we see a meniscus
Surface tension
Cohesion and adhesion of water
Water as a solvent
Specific heat, heat of vaporization, and density of water
Importance of water for life
Lesson summary: Water and life
Structure of water and hydrogen bonding
Science>
AP®︎/College Biology>
Chemistry of life>
Structure of water and hydrogen bonding
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Specific heat, heat of vaporization, and density of water
AP.BIO: SYI‑1 (EU), SYI‑1.A (LO), SYI‑1.A.1 (EK), SYI‑1.A.2 (EK), SYI‑1.A.3 (EK)
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Learn about specific heat capacity and heat of vaporization of water, evaporative cooling, and why ice floats.
Introduction
Let’s imagine that it’s a hot day. You’ve just been out in the sun for awhile, and you’re sweating quite a bit as you sit down and grab a glass of cool ice water. You idly notice both the sweat beads on your arms and the chunks of ice floating at the top of your water glass. Thanks to your hard work studying the properties of water, you recognize both the sweat on your arms and the floating ice cubes in your glass as examples of water's amazing capacity for hydrogen bonding.
How does that work? Water molecules are very good at forming hydrogen bonds, weak associations between the partially positive and partially negative ends of the molecules. Hydrogen bonding explains both the effectiveness of evaporative cooling (why sweating cools you off) and the low density of ice (why ice floats).
Here, we’ll take a closer look at the role of hydrogen bonding in temperature changes, freezing, and vaporization of water.
Water: Solid, liquid, and gas
Water has unique chemical characteristics in all three states—solid, liquid, and gas—thanks to the ability of its molecules to hydrogen bond with one another. Since living things, from human beings to bacteria, have a high water content, understanding the unique chemical features of water in its three states is key to biology.
In liquid water, hydrogen bonds are constantly being formed and broken as the water molecules slide past each other. The breaking of these bonds is caused by the energy of motion (kinetic energy) of the water molecules due to the heat contained in the system.
When the heat is raised (for instance, as water is boiled), the higher kinetic energy of the water molecules causes the hydrogen bonds to break completely and allows water molecules to escape into the air as gas. We observe this gas as water vapor or steam.
On the other hand, when the temperature drops and water freezes, water molecules form a crystal structure maintained by hydrogen bonding (as there is too little heat energy left to break the hydrogen bonds). This structure makes ice less dense than liquid water.
Density of ice and water
Water’s lower density in its solid form is due to the way hydrogen bonds are oriented as it freezes. Specifically, in ice, the water molecules are pushed farther apart than they are in liquid water.
That means water expands when it freezes. You may have seen this for yourself if you've ever put a sealed glass container containing a mostly-watery food (soup, soda, etc.) into the freezer, only to have it crack or explode as the liquid water inside froze and expanded.
With most other liquids, solidification—which occurs when the temperature drops and kinetic (motion) energy of molecules is reduced—allows molecules to pack more tightly than in liquid form, giving the solid a greater density than the liquid. Water is an anomaly (that is, a weird standout) in its lower density as a solid.
Image: modified from OpenStax Biology. Modifications of work by Jane Whitney (left), image created using Visual Molecular Dynamics (VMD) software (Humphrey, 1996), and by Carlos Ponte (right).
Because it is less dense, ice floats on the surface of liquid water, as we see for an iceberg or the ice cubes in a glass of iced tea. In lakes and ponds, a layer of ice forms on top of the liquid water, creating an insulating barrier that protects the animals and plant life in the pond below from freezing.
Why is it harmful for living things to freeze? We can understand this by thinking back to the case of a bottle of soda pop cracking in the freezer. When a cell freezes, its watery contents expand and its membrane (just like the soda bottle) is broken into pieces.
Heat capacity of water
It takes a lot of heat to increase the temperature of liquid water because some of the heat must be used to break hydrogen bonds between the molecules. In other words, water has a high specific heat capacity, which is defined as the amount of heat needed to raise the temperature of one gram of a substance by one degree Celsius. The amount of heat needed to raise the temperature of 1 g water by 1 °C is has its own name, the calorie.
Like in nutrition information?
This calorie is similar to that one, but not exactly the same. The “food calorie” you see on the back of a candy bar is is actually a kilocalorie, or a thousand small calories.
Because of its high heat capacity, water can minimize changes in temperature. For instance, the specific heat capacity of water is about five times greater than that of sand. The land cools faster than the sea once the sun goes down, and the slow-cooling water can release heat to nearby land during the night. Water is also used by warm-blooded animals to distribute heat through their bodies: it acts similarly to a car’s cooling system, moving heat from warm places to cool places, helping the body keep an even temperature.
Heat of vaporization of water
Just as it takes a lot of heat to increase the temperature of liquid water, it also takes an unusual amount of heat to vaporize a given amount of water, because hydrogen bonds must be broken in order for the molecules to fly off as gas. That is, water has a high heat of vaporization, the amount of energy needed to change one gram of a liquid substance to a gas at constant temperature.
Water’s heat of vaporization is around 540 cal/g at 100 °C, water's boiling point. Note that some molecules of water – ones that happen to have high kinetic energy – will escape from the surface of the water even at lower temperatures.
As water molecules evaporate, the surface they evaporate from gets cooler, a process called evaporative cooling. This is because the molecules with the highest kinetic energy are lost to evaporation (see the video on evaporative cooling for more info). In humans and other organisms, the evaporation of sweat, which is about 99% water, cools the body to maintain a steady temperature.
Attribution and references
Attribution:
This article is a modified derivative of “Water,” by OpenStax College, Biology (CC BY 3.0). Download the original article for free at
The modified article is licensed under a CC BY-NC-SA 4.0 license.
Additional references:
Evaporation. (2015). In ScienceDaily. Retrieved from
Harvey, A. (2000, July 17). Re: How is heat capacity of water related to hydrogen bonding? [963444050.Ch]. Message posted to
Humphrey, W., Dalke, A., and Schulten, K. (1996). VMD—Visual Molecular Dynamics. Journal of Molecular Graphics, 14, 33-38.
Raven, P. H., Johnson, G. B., Mason, K. A., Losos, J. B., and Singer, S. R. (2014). The nature of molecules and properties of water. In Biology (10th ed., AP ed., pp. 17-30). New York, NY: McGraw-Hill.
Reece, J. B., Urry, L. A., Cain, M. L., Wasserman, S. A., Minorsky, P. V., and Jackson, R. B. (2011). Four emergent properties of water contribute to Earth's suitability for life. In Campbell Biology (10th ed., pp. 45-50). San Francisco, CA: Pearson.
Some energy details related to heating water. (n.d). In HyperPhysics. Retrieved from
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JLD 9 years ago Posted 9 years ago. Direct link to JLD's post “In the last paragraph it ...” more In the last paragraph it says: "In lakes and ponds, a layer of ice forms on top of the liquid water, creating an insulating barrier..."
How does ice provide an insulating barrier?
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Ryan Hoyle 9 years ago Posted 9 years ago. Direct link to Ryan Hoyle's post “Awesome question. Part of...” more Awesome question. Part of the answer is that less dense materials conduct less heat, and thus slow down heat transfer. If you think about using a metal vs wooden spoon in a hot pan of water, it's the metal one that will burn you, because it is more dense and a better conductor of heat. So the transfer of heat from water to air is slowed down by the layer of ice. Another part of the answer is the ice prevents evaporative cooling, the liquid water molecules become physically trapped and so the ones with the highest kinetic energy can't escape, which would reduce the overall average kinetic energy and thus temperature of the water (see Sal's video on evaporative cooling). Because this doesn't happen with the layer of ice in the way, water can stay warmer for longer.
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hannadiop15 2 years ago Posted 2 years ago. Direct link to hannadiop15's post “why do some molecules of ...” more why do some molecules of water escape from the surface of the water at lower temperatures and not others?
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Richard 2 years ago Posted 2 years ago. Direct link to Richard's post “It’s important to underst...” more It’s important to understand what temperature actually is for this question. It is the average kinetic energy of the matter, with the key being average. Not all particles of matter have the same kinetic energy. Some have greater than others, and temperature is simply the mean of those energies. This means that even if some liquid water isn’t at the boiling temperature, there exist some fraction of the water molecules which have sufficient energy to transition to the gas phase. So even though most of the water molecules remain liquid, the most energetic water molecules escape the liquid as water vapor.
Hope that helps.
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Archana 6 years ago Posted 6 years ago. Direct link to Archana's post “At the surface of a liqui...” more At the surface of a liquid, why do some molecules evaporate but others do not?
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Ivana - Science trainee 6 years ago Posted 6 years ago. Direct link to Ivana - Science trainee's post “The layer which is most c...” more The layer which is most closer to the air, interacts with air molecules.
Rest of molecules cannot eadily evaporate because cohesion forces are stronger than cohesion forces.
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onucash 2 years ago Posted 2 years ago. Direct link to onucash's post “If icebergs float, what a...” more If icebergs float, what about the icebergs where most of their structure is underwater?
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Charles LaCour 2 years ago Posted 2 years ago. Direct link to Charles LaCour's post “With an iceberg floating ...” more With an iceberg floating doesn't mean that it is 100% on top of of the water. The amount of the iceberg that is under water is the ratio of the density of the water to the density of the ice. If the ice is 80% the density of the water then the iceberg will be 80% under water.
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Tatewisaacs03 2 years ago Posted 2 years ago. Direct link to Tatewisaacs03's post “do water molecules always...” more do water molecules always want to make the lattice structure found in ice and are just unable to when in liquid form because the energy/heat of each molecule is too great?
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failure 2 years ago Posted 2 years ago. Direct link to failure's post “water molecules always wa...” more water molecules always wanting to make a lattice might be difficult as also external factors affect it. The temperature in the liquid state is high so the molecule's hydrogen bonds are constantly being broken and formed enabling it not to form a lattice, and also the room temperature affects it. all in all, increasing kinetic energy. but lattice structure is the best for it as kinetic energy is very minimal due to low temperature keeping the lattice intact. not hindering/ disturbing the bonds.
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Marwan 9 years ago Posted 9 years ago. Direct link to Marwan's post “what is the difference be...” more what is the difference between heat and temperature?
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Brandon Pon 3 months ago Posted 3 months ago. Direct link to Brandon Pon's post “You can use heat during a...” more You can use heat during a phase change, however the temperature of the substance may not change. This is because the heat is used as energy.
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Kayla Roh a year ago Posted a year ago. Direct link to Kayla Roh's post “Sorry, this is kind of of...” more Sorry, this is kind of off topic, but water expands as it freezes, and thus, as the article states, living beings should avoid freezing because our cells will pop. If this is the case, how do cryonics work? In the case of a human, wouldn't their blood break out of the vessels, and if the blood was drained, wouldn't this cause cell death? Additionally, if the freezing point of the blood were to be heightened by dissolving some chemical in it, wouldn't this eliminate the benefits of cryonics itself, which I think comes from freezing the tissue in the first place? Any help would be appreciated, thank you!
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Charles LaCour a year ago Posted a year ago. Direct link to Charles LaCour's post “With cryonics they try an...” more With cryonics they try and freeze things rapidly which keeps the ice crystals small. Animal cell walls are not too rigid so as long as the ice crystals do not pierce the cell wall they can stay intact with some expansion and by having small ice crystals it will minimize cell ruptures.
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Madison Cawthon 4 years ago Posted 4 years ago. Direct link to Madison Cawthon's post “What exactly does "540 ca...” more What exactly does "540 cal/g at 100 C" mean? I thought each calorie raised the temperature of one gram of water up one degree, so I thought 540 calories would raise one gram 540 degrees; what am I missing?
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David Vegh 2 months ago Posted 2 months ago. Direct link to David Vegh's post “It explains that calories...” more It explains that calories are the amount of heat needed to raise the temperature of 1 g water by 1 °C. So when it mentions that it is "540cal/g at 100 °C", that is the amount of calories it takes to get 1 gram of water to 100 °C, and not to 540 °C. In other words, it takes approx 5,4 cal to heat 1 gram of water by 1 °C.
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JNL06 2 years ago Posted 2 years ago. Direct link to JNL06's post “So ice is less dense than...” more So ice is less dense than water because of the pockets of air formed from the hydrogen bonds. But is the ice itself denser or lighter than its water form? For example, if you were to freeze water and somehow remove all the air from inside the ice (such as freezing water in a vacuum), would the ice actually sink when you place it inside water?
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Charles LaCour 2 years ago Posted 2 years ago. Direct link to Charles LaCour's post “It is not air in between ...” more It is not air in between the water molecules. The average intermolecular distance between water molecules in ice is a constant value of about 0.276 nm (nanometer) between the adjacent oxygen atoms. In liquid water the average is about ~0.253 nm but because of the molecules not being in a rigid lattice like in ice there are some that are further and others closer.
The difference in intermolecular distance of ice and water is 0.023 nm and the size of molecular oxygen, nitrogen, and argon (the three most abundant gasses in air) are 0.299 nm, 0.305 nm, and 0.363 nm. These are way too big to account 0.023 nm increase in distance.
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eltingra000 2 years ago Posted 2 years ago. Direct link to eltingra000's post “How does the bottom of th...” more How does the bottom of the ocean not completely freeze over? If you have a bowl of water for example, the ice forms at the top first of course but eventually the whole thing freezes, so how does a larger body of water not completely freeze?
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BhanuM 2 years ago Posted 2 years ago. Direct link to BhanuM's post “I might be wrong but I th...” more I might be wrong but I think it is because of convection currents. As cold water-- next to the ice --from above sinks down as it is denser due to less kinetic energy, the warmer water beneath, which wasn't exposed to the cold environment yet, rises to the surface. Also note that cold water is denser than ice-- which floats --because it hasn't formed a crystal lattice yet.
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14359 | https://terrytao.wordpress.com/2007/06/05/open-question-the-parity-problem-in-sieve-theory/comment-page-1/ | The parity problem is a notorious problem in sieve theory: this theory was invented in order to count prime patterns of various types (e.g. twin primes), but despite superb success in obtaining upper bounds on the number of such patterns, it has proven to be somewhat disappointing in obtaining lower bounds. [Sieves can also be used to study many other things than primes, of course, but we shall focus only on primes in this post.] Even the task of reproving Euclid’s theorem – that there are infinitely many primes – seems to be extremely difficult to do by sieve theoretic means, unless one of course injects into the theory an estimate at least as strong as Euclid’s theorem (e.g. the prime number theorem). The main obstruction is the parity problem: even assuming such strong hypotheses as the Elliott-Halberstam conjecture (a sort of “super-generalised Riemann Hypothesis” for sieves), sieve theory is largely (but not completely) unable to distinguish numbers with an odd number of prime factors from numbers with an even number of prime factors. This “parity barrier” has been broken for some select patterns of primes by injecting some powerful non-sieve theory methods into the subject, but remains a formidable obstacle in general.
I’ll discuss the parity problem in more detail later in this post, but I want to first discuss how sieves work [drawing in part on some excellent unpublished lecture notes of Iwaniec]; the basic ideas are elementary and conceptually simple, but there are many details and technicalities involved in actually executing these ideas, and which I will try to suppress for sake of exposition.
Let’s consider a basic question in prime number theory, namely how to count the number of primes in a given range, say between N and 2N for some large integer N. (This problem is more or less equivalent to that of counting primes between 1 and N, thanks to dyadic decomposition, but by keeping the magnitude of all numbers comparable to N we can simplify some (very minor) technicalities.) Of course, we know that this particular question can be settled fairly satisfactorily (the answer is ) using known facts about the Riemann zeta function, but let us pretend for now that we do not know about this function. (Once one moves to slightly more complicated additive questions about the primes, such as counting twin primes, the theory of the zeta function and its relatives becomes much less powerful, even assuming such things as the Riemann hypothesis; the problem is that these functions are measuring the multiplicative structure of the primes rather than the additive structure.)
The set of primes does not appear to have enough usable structure in order to perform such counts quickly. However, one can count other sets of numbers between N and 2N with much more ease. For instance, the set of integers between N and 2N can be easily counted with small error:
the error term O(1) in this case is in fact just 1. Similarly, we can count, say, the number of odd numbers between N and 2N,
simply because the set of odd numbers has density and is periodic of period 2. The error term O(1) now depends on the parity of N. More generally, we can count any given residue class in [N,2N] to a reasonable accuracy:
,
where the error term is now more complicated, and depends on what N is doing modulo q. This estimate is quite good as long as q is small compared with N, but once q is very large, the error term O(1) can begin to overwhelm the main term (especially if the main term is going to appear in a delicate summation with lots of cancellation). In general, any summation involving the main term N/q will be relatively easy to manipulate (because it is essentially multiplicative in q, and thus amenable to all the methods of multiplicative number theory, in particular Euler products and zeta functions); it is the error term O(1) which causes all the difficulty.
Once we have figured out how to count these basic sets, we can also count some combinations of these sets, as long as these combinations are simple enough. For instance, suppose we want to count
Well, we know that the total number of integers in [N,2N] is N+O(1). Of this set, we know that are not coprime to 2 (i.e. divisible by 2), and that are not coprime to 3. So we should subtract those two sets from the original set, leaving . But the numbers which are divisible by both 2 and 3 (i.e. divisible by 6) have been subtracted twice, so we have to put them back in; this adds in another , giving a final count of for the quantity (1); this is of course a simple instance of the inclusion-exclusion principle in action. An alternative way to estimate (1) is to use the Chinese remainder theorem to rewrite (1) as
and use our ability to count residue classes modulo 6 to get the same final count of (though the precise bound on the error term will be slightly different). For very small moduli such as 2 and 3, the Chinese remainder theorem is quite efficient, but it is somewhat rigid, and for higher moduli (e.g. for moduli much larger than ) it turns out that the more flexible inclusion-exclusion principle gives much better results (after applying some tricks to optimise the efficiency of that principle).
We can of course continue the example in (1), counting the numbers in [N,2N] which are coprime to 2,3,5,7, etc., which by the sieve of Eratosthenes will eventually give us a count for the primes in [N,2N], but let us pause for a moment to look at the larger picture. We have seen that some sets in [N,2N] are fairly easy to count accurately (e.g. residue classes with small modulus), and others are not (e.g. primes, twin primes). What is the defining characteristic of the former types of sets? One reasonable answer is that the sets that are easy to count are low-complexity, but this is a rather vaguely defined term. I would like to propose instead that sets (or more generally, weight functions – see below) are easy to count (or at least estimate) whenever they are smooth in a certain sense to be made more precise shortly. This terminology comes from harmonic analysis rather than from number theory (though number theory does have the related concept of a smooth number), so I will now digress a little bit to talk about smoothness, as it seems to me that this concept implicitly underlies the basic strategy of sieve theory.
Instead of talking about the problem of (approximately) counting a given set in [N,2N], let us consider instead the analogous problem of (approximately) computing the area of a given region E (e.g. a solid ellipse) in the unit square . As we are taught in high school, one way to do this is to subdivide the square into smaller squares, e.g. squares of length for some n, and count how many of these small squares lie completely or partially in the set E, and multiply by the area of each square; this is of course the prelude to the Riemann integral. It works well as long as the set E is “smooth” in the sense that most of the small squares are either completely inside or completely outside the set E, with few borderline cases; this notion of smoothness can also be viewed as a quantitative version of Lebesgue measurability. Another way of saying this is that if one wants to determine whether a given point (x,y) lies in E, it is usually enough just to compute x and y to the first n significant digits in the decimal expansion.
Now we return to counting sets in [N,2N]. One can also define the notion of a “smooth set” here by again using the most significant digits of the numbers n in the interval [N,2N]; for instance, the set [1.1 N, 1.2 N] would be quite smooth, as one would be fairly confident whether n would lie in this set or not after looking at just the top two or three significant digits. However, with this “Euclidean” or “Archimedean” notion of smoothness, sets such as the primes or the odd numbers are certainly not smooth. However, things look a lot better if we change the metric, or (more informally) if we redefine what “most significant digit” is. For instance, if we view the last digit in the base 10 expansion of a number n (i.e. the value of ) as the most significant one, rather than the first – or more precisely, if we use the 10-adic metric intead of the Euclidean one, thus embedding the integers into rather than into – then the odd numbers become quite smooth (the most significant digit completely determines membership in this set). The primes in [N,2N] are not fully smooth, but they do exhibit some partial smoothness; indeed, if the most significant digit is 0, 2, 4, 5, 6, or 8, this fully determines membership in the set, though if the most significant digit is 1, 3, 7, or 9 then one only has partial information on membership in the set.
Now, the 10-adic metric is not fully satisfactory for characterising the elusive concept of number-theoretic “smoothness”. For instance, the multiples of 3 should be a smooth set, but this is not the case in the 10-adic metric (one really needs all the digits before one can be sure whether a number is a multiple of 3!). Also, we have the problem that the set [N/2,N] itself is now no longer smooth. This can be fixed by working not with just the Euclidean metric or a single n-adic metric, but with the product of all the n-adic metrics and the Euclidean metric at once. Actually, thanks to the Chinese remainder theorem, it is enough to work with the product of the p-adic metrics for primes p and the Euclidean metric, thus embedding the integers in the integer adele ring . For some strange reason, this adele ring is not explicitly used in most treatments of sieve theory, despite its obvious relevance (and despite the amply demonstrated usefulness of this ring in algebraic number theory or in the theory of L-functions, as exhibited for instance by Tate’s thesis). At any rate, we are only using the notion of “smoothness” in a very informal sense, and so we will not need the full formalism of the adeles here. Suffice to say that a set of integers in [N,2N] is “smooth” if membership in that set can be largely determined by its most significant digits in the Euclidean sense, and also in the p-adic senses for small p; roughly speaking, this means that this set is approximately the pullback of some “low complexity” set in the adele ring – a set which can be efficiently fashioned out of a few of basic sets which generate the topology and -algebra of that ring. (Actually, in many applications of sieve theory, we only need to deal with moduli q which are square-free, which means that we can replace the p-adics with the cyclic group , and so it is now just the residues mod p for small p, together with the Euclidean most significant digits, which should control what smooth sets are; thus the adele ring has been replaced by the product .)
Let us now return back to sieve theory, and the task of counting “rough” sets such as the primes in [N,2N]. Since we know how to accurately count “smooth” sets such as with q small, one can try to describe the rough set of primes as some sort of combination of smooth sets. The most direct implementation of this idea is the sieve of Eratosthenes; if one then tries to compute the number of primes using the inclusion-exclusion principle, one obtains the Legendre sieve; we implicitly used this idea previously when counting the quantity (1). However, the number of terms in the inclusion-exclusion formula is very large; if one runs the sieve of Eratosthenes for k steps (i.e. sieving out multiples of the first k primes), there are basically terms in the inclusion-exclusion formula, leading to an error term which in the worst case could be of size . A related issue is that the modulus q in many of the terms in the Legendre sieve become quite large – as large as the product of the first k primes (which turns out to be roughly in size). Since the set one is trying to count is only of size N, we thus see that the Legendre sieve becomes useless after just or so steps of the Eratosthenes sieve, which is well short of what one needs to accurately count primes (which requires that one uses or so steps). More generally, “exact” sieves such as the Legendre sieve are useful for any situation involving only a logarithmically small number of moduli, but are unsuitable for sieving with much larger numbers of moduli.
One can describe the early development of sieve theory as a concerted effort to rectify the drawbacks of the Legendre sieve. The first main idea here is to not try to compute the size of the rough set exactly – as this is too “expensive” in terms of the number of smooth sets required to fully describe the rough set – but instead to just settle for upper or lower bounds on the size of this set, which use fewer smooth sets. There is thus a tradeoff between how well the bounds approximate the original set, and how well one can compute the bounds themselves; by selecting various parameters appropriately one can optimise this tradeoff and obtain a final bound which is non-trivial but not completely exact. For instance, in using the Legendre sieve to try to count primes between N and 2N, one can instead use that sieve to count the much larger set of numbers between N and 2N which are coprime to the first k primes, thus giving an upper bound for the primes between N and 2N. It turns out that the optimal value of k here is roughly or so (after this, the error terms in the Legendre sieve get out of hand), and give an upper bound of for the number of primes between N and 2N – somewhat far from the truth (which is ), but still non-trivial.
In a similar spirit, one can work with various truncated and approximate versions of the inclusion-exclusion formula which involve fewer terms. For instance, to estimate the cardinality of the union of k sets, one can replace the inclusion-exclusion formula
by the obvious upper bound
(also known as the union bound), or by the slightly less obvious lower bound
More generally, if one takes the first n terms on the right-hand side of (2), this will be an upper bound for the left-hand side for odd n and a lower bound for even n. These inequalities, known as the Bonferroni inequalities, are a nice exercise to prove: they are equivalent to the observation that in the binomial identity
for any , the partial sums on the right-hand side alternate in sign between non-negative and non-positive. If one inserts these inequalities into the Legendre sieve and optimises the parameter, one can improve the upper bound for the number of primes in [N,2N] to , which is significantly closer to the truth. Unfortunately, this method does not provide any lower bound other than the trivial bound of 0; either the main term is negative, or the error term swamps the main term. A similar argument was used by Brun to show that the number of twin primes in [N,2N] was (again, the truth is conjectured to be ), which implied his famous theorem that the sum of reciprocals of the twin primes is convergent.
The full inclusion-exclusion expansion is a sum over terms, which one can view as binary strings of 0s and 1s of length k. In the Bonferroni inequalities, one only sums over a smaller collection of strings, namely the Hamming ball of strings which only involve n or fewer 1s. There are other collections of strings one can use which lead to upper or lower bounds; one can imagine revealing such a string one digit at a time and then deciding whether to keep or toss out this string once some threshold rule is reached. There are various ways to select these thresholding rules, leading to the family of combinatorial sieves. One particularly efficient such rule is similar to that given by the Bonferroni inequalities, but instead of using the number of 1s in a string to determine membership in the summation, one uses a weighted number of 1s (giving large primes more weight than small primes, because they tend to increase the modulus too quickly and thus should be removed from the sum sooner than the small primes). This leads to the beta sieve, which for instance gives the correct order of magnitude of for the number of primes in [N,2N] or for the number of twin primes in [N,2N]. This sieve is also powerful enough to give lower bounds, but only if one stops the sieve somewhat early, thus enlarging the set of primes to a set of almost primes (numbers which are coprime to all numbers less than a certain threshold, and thus have a bounded number of prime factors). For instance, this sieve can show that there are an infinite number of twins , each of which has at most nine prime factors (the nine is not optimal, but to get better results requires much more work).
There seems however to be a limit as to what can be accomplished by purely combinatorial sieves. The problem stems from the “binary” viewpoint of such sieves: any given term in the inclusion-exclusion expansion is either included or excluded from the sieve upper or lower bound, and there is no middle ground. This leads to the next main idea in modern sieve theory, which is to work not with the cardinalities of sets in [N,2N], but rather with more flexible notion of sums of weight functions (real-valued functions on [N,2N]). The starting point is the obvious formula
for the cardinality of a set A in [N,2N], where is the indicator function of the set A. Applying this to smooth sets such as , we obtain
;
in particular, specialising to the 0 residue class a = 0 (which is the residue class of importance for counting primes) we have
for any d. Thus if we can obtain a pointwise upper bound on by a divisor sum (which is a number-theoretic analogue of a smooth function), thus
for all n and some real constants (which could be positive or negative), then on summing we obtain the upper bound
One can also hope to obtain lower bounds on |A| by a similar procedure (though in practice, lower bounds for primes have proven to be much more difficult to obtain, due to the parity problem which we will discuss below). These strategies are suited for the task of bounding the number of primes in [N,2N]; if one wants to do something fancier such as counting twin primes n, n+2, one has to either involve more residue classes (e.g. the class a = -2 will play a role in the twin prime problem) or else insert additional weights in the summation (e.g. weighting all summations in n by an additional factor of , where is the von Mangoldt function). To simplify the exposition let us just stick with the plainer problem of counting primes.
These strategies generalise the combinatorial sieve strategy, which is a special case in which the constants are restricted to be +1, 0, or -1. In practice, the sum in (4) is relatively easy to sum by multiplicative number theory techniques (the coefficients , in applications, usually involve the Möbius function (not surprising, since they are encoding some sort of inclusion-exclusion principle) and are often related to the coefficients of a Hasse-Weil zeta function, as they basically count solutions modulo d to some set of algebraic equations), and the main task is to ensure that the error term in (3) does not swamp the main term. To do this, one basically needs the weights to be concentrated on those d which are relatively small compared with N, for instance they might be restricted to some range where the sieve level is some small power of N. Thus for instance, starting with the identity
, (5)
which corresponds to the zeta-function identity
,
where is the von Mangoldt function and is the Möbius function, we obtain the upper bound
where A denotes the primes from N to 2N. This is already enough (together with the elementary asymptotic ) to obtain the weak prime number theorem , but unfortunately this method does not give a nontrivial lower bound for |A|. However, a variant of the method does give a nice asymptotic for almost primes – products of at most two (large) primes [e.g. primes larger than for some fixed ]. Indeed, if one introduces the second von Mangoldt function
which is mostly supported on almost primes (indeed, and for distinct primes p, q, and is mostly zero otherwise), and uses the elementary asymptotic
then one obtains the Selberg symmetry formula
This formula (together with the weak prime number theorem mentioned earlier) easily implies an “ almost prime number theorem”, namely that the number of almost primes less than N is . [This fact is much easier to prove than the prime number theorem itself. In terms of zeta functions, the reason why the prime number theorem is difficult is that the simple pole of at s=1 could conceivably be counteracted by other simple poles on the line . On the other hand, the almost prime number theorem is much easier because the effect of the double pole of at s=1 is not counteracted by the other poles on the line , which are at most simple.]
The almost prime number theorem establishes the prime number theorem “up to a factor of 2”. It is surprisingly difficult to improve upon this factor of 2 by elementary methods, though once one can replace 2 by for some (a fact which is roughly equivalent to the absence of zeroes of on the line ), one can iterate the Selberg symmetry formula (together with the tautological fact that an almost prime is either a prime or the product of two primes) to get the prime number theorem; this is essentially the Erdős-Selberg elementary proof of that theorem.
One can obtain other divisor bounds of the form (3) by various tricks, for instance by modifying the weights in the above formulae (5) and (6). A surprisingly useful upper bound for the primes between N and 2N is obtained by the simple observation that
whenever are arbitrary real numbers with , basically because the square of any real number is non-negative. This leads to the Selberg sieve, which suffices for many applications; for instance, it can prove the Brun-Titchmarsh inequality, which asserts that the number of primes between N and N+M is at most , which is again off by a factor of 2 from the truth when N and M are reasonably comparable. There are also some useful lower bounds for the indicator function of the almost primes of divisor sum type, which can be used for instance to derive Chen’s theorem that there are infinitely many primes p such that p+2 is a almost prime, or the theorem that there are infinitely many almost primes of the form .
In summary, sieve theory methods can provide good upper bounds, lower bounds, and even asymptotics for almost primes, which lead to upper bounds for primes which tend to be off by a constant factor such as 2. Rather frustratingly, though, sieve methods have proven largely unable to count or even lower bound the primes themselves, thus leaving the twin prime conjecture (or the conjecture about infinitely many primes of the form ) still out of reach. The reason for this – the parity problem – was first clarified by Selberg. Roughly speaking, it asserts:
Parity problem. If A is a set whose elements are all products of an odd number of primes (or are all products of an even number of primes), then (without injecting additional ingredients), sieve theory is unable to provide non-trivial lower bounds on the size of A. Also, any upper bounds must be off from the truth by a factor of 2 or more.
Thus we can hope to count almost primes (because they can have either an odd or an even number of factors), or to count numbers which are the product of 6 or 7 primes (which can for instance be done by a sieve of Bombieri), but we cannot hope to use plain sieve theory to just count primes, or just count semiprimes (the product of exactly two primes).
To explain this problem, we introduce the Liouville function (a close relative of the Möbius function), which is equal to +1 when n is the product of an even number of primes and -1 otherwise. Thus the parity problem applies whenever is identically +1 or identically -1 on the set A of interest.
The Liouville function oscillates quite randomly between +1 and -1. Indeed, the prime number theorem turns out to be equivalent to the assertion that is asymptotically of mean zero,
(a fact first observed by Landau), and if the Riemann hypothesis is true then we have a much better estimate
for all .
Assuming the generalised Riemann hypothesis, we have a similar claim for residue classes:
for all .
What this basically means is that the Liouville function is essentially orthogonal to all smooth sets, or all smooth functions. Since sieve theory attempts to estimate everything in terms of smooth sets and functions, it thus cannot eliminate an inherent ambiguity coming from the Liouville function. More concretely, let A be a set where is constant (e.g. is identically -1, which would be the case if A consisted of primes) and suppose we attempt to establish a lower bound for the size of a set A in, say, [N,2N] by setting up a divisor sum lower bound
(6)
where the divisors d are concentrated in for some reasonably small sieve level R. If we sum in n we obtain a lower bound of the form
and we can hope that the main term will be strictly positive and the error term is of lesser order, thus giving a non-trivial lower bound on |A|. Unfortunately, if we multiply both sides of (6) by the non-negative weight and sum in n, we obtain
since we are assuming to equal -1 on A. If we sum this in n, and use the fact that is essentially orthogonal to divisor sums, we obtain
which basically means that the bound (7) cannot improve upon the trivial bound . A similar argument using the weight also shows that any upper bound on |A| obtained via sieve theory has to essentially be at least as large as 2|A|.
Despite this parity problem, there are a few results in which sieve theory, in conjunction with other methods, can be used to count primes. The first of these is the elementary proof of the prime number theorem alluded to earlier, using the multiplicative structure of the primes inside the almost primes. This method unfortunately does not seem to generalise well; for instance, the product of twin primes is not a twin almost prime. Other examples arise if one starts counting certain special two-parameter families of primes; for instance, Friedlander and Iwaniec showed that there are infinitely many primes of the form by a lengthy argument which started with Vaughan’s identity, which is sort of like an exact sieve, but with a (non-smooth) error term which has the form of a bilinear sum, which captures correlation with the Liouville function. The main difficulty is to control this bilinear error term, which after a number of (non-trivial) arithmetic manipulations (in particular, factorising over the Gaussian integers) reduces to understanding some correlations between the Möbius function and the Jacobi symbol, which is then achieved by a variety of number-theoretic tools. The method was then modified by Heath-Brown to also show infinitely many primes of the form . Related results for other cubic forms using similar methods have since been obtained by Heath-Brown and Moroz and by Helfgott (analogous claims for quadratic forms date back to Iwaniec). These methods all seem to require that the form be representable as a norm over some number field and so it does not seem as yet to yield a general procedure to resolve the parity problem.
The parity problem can also be sometimes be overcome when there is an exceptional Siegel zero, which basically means that there is a quadratic character which correlates very strongly with the primes. Morally speaking, this means that the primes can be largely recovered from the almost primes as being those almost primes which are quadratic non-residues modulo the conductor q of , and this additional information seems (in principle, at least) to overcome the parity problem obstacle (related to this is the fact that Siegel zeroes, if they exist, disprove GRH, and so the Liouville function is no longer as uniformly distributed on smooth sets as Selberg’s analysis assumed). For instance, Heath-Brown showed that if a Siegel zero existed, then there are infinitely many prime twins. Of course, assuming GRH then there are no Siegel zeroes, in which case these results would be technically vacuous; however, they do suggest that to break the parity barrier, we may assume without loss of generality that there are no Siegel zeroes.
Another known way to partially get around the parity problem is to combine precise asymptotics on almost primes (or of weight functions concentrated near the almost primes) with a lower bound on the number of primes, and then use combinatorial tools to parlay the lower bound on primes into lower bounds on prime patterns. For instance, suppose you knew could count the set
accurately (where is the set of -almost primes), and also obtain sufficiently good lower bounds on the sets
and more precisely that one obtains
.
(For comparison, the parity problem predicts that one cannot hope to do any better than showing that , so the above inequality is not ruled out by the parity problem obstruction.)
Then, just from the pigeonhole principle, one deduces the existence of such that at least two of n, n+2, n+6 are prime, thus yielding a pair of primes whose gap is at most 6. This naive approach does not quite work directly, but by carefully optimising the argument (for instance, replacing the condition with something more like ), Goldston, Yildirim, and Pintz were able to show unconditionally that prime gaps in could be as small as , and could in fact be as small as 16 infinitely often if one assumes the Elliot-Halberstam conjecture.
In a somewhat similar spirit, my result with Ben Green establishing that the primes contain arbitrarily long progressions proceeds by first using sieve theory methods to show that the almost primes (or more precisely, a suitable weight function concentrated near the almost primes) are very pseudorandomly distributed, in the sense that several self-correlations of can be computed and agree closely with what one would have predicted if the almost primes were distributed randomly (after accounting for some irregularities caused by small moduli). Because of the parity problem, the primes themselves are not known to be as pseudorandomly distributed as the almost primes; however, the prime number theorem does at least tell us that the primes have a positive relative density in the almost primes. The main task is then to show that any set of positive relative density in a sufficiently pseudorandom set contains arithmetic progressions of any specified length; this combinatorial result (a “relative Szemerédi theorem“) plays roughly the same role that the pigeonhole principle did in the work of Goldston-Yildirim-Pintz. (On the other hand, the relative Szemerédi theorem works even for arbitrarily low density, whereas the pigeonhole principle does not; because of this, our sieve theory analysis is far less delicate than that in Goldston-Yildirim-Pintz.)
It is probably premature, with our current understanding, to try to find a systematic way to get around the parity problem in general, but it seems likely that we will be able to find some further ways to get around the parity problem in special cases, and perhaps once we have assembled enough of these special cases, it will become clearer what to do in general.
[Update, June 6: definition of almost prime modified, to specify that all prime factors are large. Reference to profinite integers deleted due to conflict with established notation.]
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87 comments
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5 June, 2007 at 10:25 pm
TK
This blog is the best I’ve seen on the internet for mathematical exposition. Second only to the arxiv as a source of wisdom.
Kudos to the author; I hope he will continue to find time to write these long postings.
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6 June, 2007 at 11:44 am
Rob
How could one go about obtaining a copy of the unpublished lecture notes of Iwaniec referred to above?
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6 June, 2007 at 1:03 pm
Emmanuel Kowalski
This is indeed a wonderful post!
Here is a remark which may also be useful to understand some issues with sieve.
In the “almost prime” setting, there is often a distinction between integers having few prime factors (with multiplicity) and integers having no small prime factors. Most sieves typically produce the second type of integers, more precisely integers such that all primes dividing are larger than for some (fixed) positive constant (often small). So those have at most prime factors.
The reason the distinction is sometimes important lies in the density of the two sets: for instance, there are about integers divisible by at most two primes, but only a constant (depending on ) times integers with no prime factor . In other words, the density of the second type of “almost primes” is comparable to that of the primes themselves, whereas there are rather more of the first type.
This might explain for instance (where the parity problem might suggest a serious difficulty) why Goldston, Graham, Pintz and Yildirim were able to prove that there are infinitely many integers which are products of two primes exactly and at some bounded distance (i.e., what looks like the analogue of bounded distances between primes; but among a set where the mean spacing is not but somewhat smaller).
To reply (partly) to the last comment: Iwaniec and Friedlander are currently finishing
a book on sieve based in large part on those unpublished notes.
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16 December, 2015 at 12:17 pm
Justin
I am very interested in the distribution of integers that are the product of at most two primes. You mention that there are asymptotically integers with no prime factor , where is a constant depending on $c$. Does this directly appear in the literature somewhere? Or is it derived from a more general result? I have been searching high and low for it, but to no avail.
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16 December, 2015 at 12:58 pm
Terence Tao
See . Tenenbaum’s text “Introduction to analytic and probabilistic number theory” should have a thorough treatment of this (probably Montgomery-Vaughan’s “Multiplicative Number Theory” also).
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16 December, 2015 at 11:38 pm
Justin
This is really helpful; thanks for the quick reply!
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6 June, 2007 at 4:02 pm
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[…] Open question: The parity problem in sieve theory The parity problem is a notorious problem in sieve theory: this theory was invented in order to count prime patterns of […] […]
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6 June, 2007 at 5:18 pm
Felipe Voloch
Sometimes one can get lower bounds out of upper bounds in prime number estimates using some Galois theory. Two examples of this (actually closely related) occur in Bombieri’s version of Stepanov’s proof of the Riemann hypothesis for function fields and in Chebotarev’s proof of his density theorem. I’ve never seen these arguments in the context of sieve theory, though.
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6 June, 2007 at 11:11 pm
Emmanuel Kowalski
In sieve, getting lower bounds from upper bounds is usually done with the Buchstab identity, which is again a form of inclusion exclusion and states that
where is the sum of some sequence over integers with no prime factor less than , and is chosen arbitrarily less than , i.e, integers with no prime factor less than are those with no prime factors less than (there are more of them), minus those with smallest prime divisor between and , and no prime factor smaller than this .
Somewhat related is the nice trick of “switching sieve”, found first (I think) in the paper of Iwaniec on primes represented by quadratic forms, and spectacularly used by Chen in his result on : in this second case, roughly, Chen showed first that some weighted counting of primes with is “large”; then he showed that the number of primes where is smaller than this lower bound, so most of the first lower bound had to come from where … The second step is done by sieving the sequence to count how many primes it may represent (there one needs only an upper bound!). It is not obvious a priori that this will work; in Chen’s first version, it depends on getting some numerical inequality right (and this depends of course on the careful choice of weight in the first lower bound).
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7 June, 2007 at 9:05 am
Jordan Ellenberg
You ask why the adelic formulation is rarely used in sieve theory; I think it’s probably just because when you work over Z you
can get away with _not_ using it. It’s when you want to prove things
over arbitrary number fields (or for that matter function fields)
that the adelic point of view becomes really indispensible — so I
think one good outcome of using the adeles systematically is that it
will be clear (if you care) which of these theorems in sieve theory
truly have to do with Z and which are theorems about integers in
general number fields. Of course it is not quite obvious even what the correct questions are over more general fields! (And perhaps here, too, the adelic formulation would be of use.)
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7 June, 2007 at 9:55 am
Emmanuel Kowalski
With respect to adèles, I think even when it would be a good conceptual tool, it is not necessarily a good fit for the foundational part of sieve methods (before applications), which tends to be very “finitistic” (although O() symbols are sometimes used, they can pretty much be replaced by completely explicit bounds if one is careful enough — Iwaniec did this very successfully for the so-called beta-sieve). So adèles could be also be replaced with finite products of completions, and then dissolved in finite quotients…
Recently I’ve been playing with quite general sieve settings, where the integers are replaced with potentially much more complicated-looking things (e.g., discrete groups, algebraic fundamental groups, probability spaces with random walks,…) without seeing infinitary objects like the adèles being obviously necessary (though they can be used to phrase things in a clean way).
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7 June, 2007 at 10:11 am
Terence Tao
Dear Emmanuel and Jordan,
Thanks very much for your comments! I can see that adeles are perhaps not the right tool for really getting into the quantitative and finitary details of error estimates, etc., but they should at least provide a nice way to compute the “main term” that comes out of sieve theory (such as the expressions mentioned in my post). For instance, I have felt that all the standard conjectures (Hardy-Littlewood prime tuples conjecture, Bateman-Horn conjecture, etc.) about the distribution of prime patterns would be very cleanly stated in an adelic setting.
Regarding other number fields, I have a paper establishing infinitely many constellations amongst the Gaussian integer primes (or any dense subset thereof), by modifying the methods of my paper with Ben. For this, I of course had to do some sieving in the Gaussian integers, but this was fairly straightforward to do by hand. I don’t know how to do the same thing for more general integral rings, e.g. , mainly because I am using a rather “low-tech” approach to sieving that relies in particular on unique factorisation (the fundamental theorem of arithmetic). (There are also some mild problems with units, but as long as there are only finitely many of these, this does not seem to be dangerous.) A model problem would be to figure out how to use sieve theory to count constellations of “almost primes” or “almost irreducibles” in such integral rings, though I am not entirely sure exactly what those concepts even mean once one loses unique factorisation. There is also the question of exactly what type of “boxes” one should count the patterns in, though presumably the Archimedean valuations should tell you what to do there.
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7 June, 2007 at 11:22 am
Emmanuel Kowalski
Certainly, as far as the main term is concerned, or for any Euler product more generally, (involving in addition an archimedean term), seeing things adelically is the right thing to do as soon as some generality is desired.
For problems in rings of integers in number fields with no unique factorization, one might look naively for problems involving “irreducibles”, i.e., elements of the ring which generate a prime ideal. Algebraic/analytic number theory proves that there are many of those (a proportion $1/h$ of prime ideals are principal, where $h$ is the class number). If there are infinitely many units, more needs to be done to get nice finite sets in which to sieve, but for imaginary quadratic fields, this makes it possible to write down some nice twin-prime like problems in general.
There might well be more interesting problems remaining to be discovered, on the other hand…
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8 June, 2007 at 10:34 pm
Keith Conrad
This post is about Terry’s remark on what type of “boxes” one should count in.
There certainly is no canonical choice unless the number field has a unique archimedean place (thus limiting you to Q and imagiary quadratics). In other cases, you could pick some region in the product of archimedean completions of the number field — so in R x R if you had in mind Q(sqrt(2)), say, and take uniformly expanded versions of that as some appropriate scaling parameter grows. (If you aren’t careful then you can pick regions that are too thin in some direction.) Even if there’s nothing canonical about such a choice of boxes in which you tabulate statistics on prime values of some expression across the region, numerics shows that the number of prime values you find in the box (if you’re looking at values of a polynomial f(x), generating a principal prime ideal in case h > 1) grows with the scaling parameter in the same asymptotic way as the classical Hardy–Littlewood constant you would attach to the polynomial times the sum of 1/log|N(f(x))| where N is the norm down to Z from your ring of integers and x runs over integers in the box. I am not saying there is any hope of a proof in such generality, but the numerics look okay. (This is just an extrapolation of the usual heuristic probability of f(n) being prime as C_f/log|f(n)| where f is in Z[X] and C_f is a Hardy–Littlewood constant which measures how often f’s values on Z are not divisible by the primes.)
One way to get around the choice of ad hox “boxes” is to forget about any kind of “(un)natural density” for prime values and use the Zariski topology instead. This is done in recent work of Sarnak, Bourgain, Gamburd.
In a somewhat different direction, ordinarily it is said that looking at prime pairs n and n + 1 is silly: one of n and n + 1 is even. But if you work not in Z but in Z[1/2] then the idea makes sense since the problem at 2 goes away. Just as Z is discrete and co-compact in R, Z[1/2] is discrete and co-compact in R x Q_2. If you draw a large region in R x Q_2 which is not thin in some direction then the number of n in the region from Z[1/2] (not just Z) for which n and n+1 are both prime (in Z[1/2], of course) agrees with the sum of 1/log N_2(n(n+1)) over the region times the corresponding Hardy–Littlewood constant with the local factor at 2 dropped out. By N_2(m) for m in Z[1/2] I mean the size of Z[1/2]/(m), just as one measures (nonzero) integers k by |k| = #(Z/k). Here I am not claiming one can actually prove something new, but the numerics look okay. Similarly, for the “triple primes” of the form n, n+2, n+4, which is nonsense in Z because one of the terms is always a multiple of 3, there is meaning to this idea in Z[1/3] and counting triple primes n, n+2, n+4 from Z[1/3] within regions of R x Q_3 returns numerics consistent with analogues of the usual prime-counting heuristics for polynomial values over Z, but here the Hardy–Littlewood constant has no local factor at 3.
Oh, when I say the Hardy–Littlewood constant has its local factor at 2 or 3 taken out depending on n and n + 1 or n, n+2, and n+4 being considered, the effect of 2 and 3 reappears elsewhere: you need to include in the heuristic prime-counting estimate a reciprocal of the residue of a zeta-function at s = 1 with the local Euler factor at 2 or 3 removed, which actually makes something show up from 2 or 3 in the residue itself. (Since
Res_{s=1} zeta(s) = 1, one doesn’t notice zeta-residues in the classical setting over Z.)
Ack, I’ve probably written too much. Good night.
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9 June, 2007 at 1:15 am
Ben Green
This reminds me a little bit of a trick Terry and I use in our work on linear equations in primes which we call the “W-trick”. Suppose you want to find long arithmetic progressions in the odd primes
3,5,7,11,13,17,19….
Then you may as well look for progressions in the sequence
1,2,3,5,6,8,9….
which I got from the previous one by subtracting one from each term and dividing by two. The new sequence doesn’t suffer from the same dislike of the even numbers that the primes do – in fact since asymptotically 50 percent of the primes are 1 mod 4, about half of the elements of this sequence are even and half odd.
Now I’ll just look at the elements of the new sequence which are divisible by 3, and I’ll divide each of them by 3 to get a new sequence
1,2,3,5,6,7,9…..
This sequence is now nicely distributed mod 2 and mod 3 (that is, mod 6). To see why, let’s ask when 6n + r lies in the sequence. This is the case if and only if 6(6n + r) + 1 is prime, that is to say iff 36n + a is prime where a = 6r + 1. But there are, asymptotically, equal numbers of primes in each residue class a mod 36 for which a is coprime to 36.
One can continue in this way mod 5, 7 and so on up to w, obtaining a sequence which is a rescaled version of the primes congruent to 1 mod W, where W = 2 x 3 x 5 x 7… x w (this is why we call it the W-trick).
You can actually play the same game for the primes congruent to a mod W, for any a.
These rescaled primes will have negligible bias mod every prime less than or equal to w. This makes computing the Hardy-Littlewood constant a rather trivial matter – if w is large then it tends to be very close to 1 (actually there are some slight issues with this – the Hardy-Littlewood constant appearing in the Goldbach conjecture can be dominated by the contribution from large primes, say if you’re counting the number of ways to represent N = p + q where N has many large prime factors).
If one can prove that all these Hardy-Littlewood constants are about 1 then one can piece together the information over all progressions a (mod W), for different a, to recover the Hardy-Littlewood constant for the original problem. For example one can use this technique to make the classical prediction of the number of twin primes less than N.
The point about these rescaled primes is that (unlike the primes themselves) they look sort of random. Thus in making predictions about them one can look at what is true for a random set of similar size and just guess that the same ought to be true for the primes.
There is another heuristic which is more accurate than the above, which is the so-called Mobius Randomness Principle. This asserts that the Mobius function is highly orthogonal to everything unless there is some obvious reason why it shouldn’t be (for example, it is not orthogonal to itself or to the von Mangoldt function). This principle can also be used to guess these Hardy-Littlewood constants, and much more – for example the Riemann hypothesis is equivalent to the statement that Mobius is highly orthogonal (with square-root cancellation) to the constant function 1.
Proving even a weak instance of the Mobius randomness principle involves overcoming the parity problem, as Terry explained above. There’s more on all this in Terry’s earlier post (Simons Lecture I) and in some of the comments that follow it.
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9 June, 2007 at 10:00 am
Jonathan Vos Post
Very fine thread! I understood it better, having seen ben Green’s recent lectures to the Caltech Math department.
The same methods seem to work for quasi-Eratosthenesian sieves, such as for Ulam’s “lucky numbers.”
Note the comment at that OEIS link:
COMMENT
An interesting general discussion of the phenomenon of ‘random primes’ (generalizing the lucky numbers) occurs in Hawkins (1958). Heyde (1978) proves that Hawkins’ random primes do not only almost always satisfy the PNT but also the Riemann Hypothesis. – Alf van der Poorten, Jun 27 2002
REFERENCES
M. Gardner, Lucky numbers and 2187, Math. Intellig., 19 (No. 2, 1997), 26-29.
M. Gardner, Gardner’s Workout, Chapter 21 “Lucky Numbers and 2187” pp 149-156 A.K.Peters MA 2002.
V. Gardiner, R. Lazarus, N. Metropolis and S. Ulam, On certain sequences of integers defined by sieves, Math. Mag., 29 (1955), 117-119.
R. K. Guy, Unsolved Problems in Number Theory, C3.
D. Hawkins, The random sieve, Math. Mag. 31 (1958), 1-3.
C. C. Heyde, Ann. Probability, 6 (1978), 850-875.
C. S. Ogilvy, Tomorrow’s Math. 2nd ed., Oxford Univ. Press, 1972, p. 99.
D. Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, 114.
One can examine intersections such as:
A031157 Numbers that are both lucky and prime.
3, 7, 13, 31, 37, 43, 67, 73, 79, 127, 151, 163, 193, 211, 223, 241, 283, 307, 331, 349, 367, 409, 421, 433, 463, 487, 541, 577, 601, 613, 619, 631, 643, 673, 727, 739, 769, 787, 823, 883, 937, 991, 997, 1009, 1021, 1039, 1087, 1093, 1117, 1123, …
Looking mod 10 can be confusing to me, as in:
A129864 Numbers that are both lucky and emirp.
13, 31, 37, 73, 79, 739, 769, 991, 1009, 1021, …
FORMULA
A000959 INTERSECTION A006567. a(n) is n element of A000959 AND a(n) is an element of A000040 AND R(a(n)) = A004086(a(n)) is an element of A000040.
EXAMPLE
a(9) = 1009 because 1009 is a lucky number A000959(154) and 1009 is an emirp because 1009 is prime and R(1009) = 9001 is prime (but is not lucky).
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9 June, 2007 at 12:28 pm
physics_grad_student
Hi Terry,
I was wondering if the techniques you mentioned can be used to solve a problem I’m dealing with but I can’t seem to get it to work after trying for a few days. I would like to prove the following (BTW I’m not a math major)
Suppose we are interested in modular arithmetic mod d. d>1
Let S be the set of numbers coprime to d excluding zero. Let a and b be elements of set S.
Then what is the range of the function x = a+b mod d?
Numerically, for all d less than 2000, the answer is
1) If d is even, x = {0,2,4. . . d-2} (all even numbers less than d)
2) If d is odd, x= {0,1,2,. . .d-1} (all numbers less than d)
I would like to prove that the above result holds for all values of d.
I can prove this for d=odd prime, which can be extended to odd prime powers, which can then be extended to d=product of at most 4 odd prime powers. Any help is greatly appreciated.
BTW is there any connection between this problem and the Goldbach Conjecture?
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9 June, 2007 at 11:31 pm
Noah Snyder
physics_grad_student:
Your conjecture is true, and although it is totally unrelated to the Goldbach Conjecture it is still a cute little excercise in elementary number theory. I’ll just do the d is odd case, though the d is even case is basically identical with a small twist.
By the Chinese remainder theorem it is enough to do the case where d is a power of a prime, say d=p^n. Now, modulo p^n it is easy to compute what fraction of the numbers are relatively prime to p^n, namely (p-1)/p. In particular, if p is odd then more than half of the numbers modulo p^n are relatively prime to p^n.
Now, let’s rephrase the problem slightly. Let U be the set of numbers relatively prime to d=p^n. For some number x, let x-U be the set of elements of the form x-u for u in U. Saying that x is of the form a+b as above is equivalent to saying that U and x-U have a nontrivial intersection. (If y is in U and in x-U then x = y + (x-y) exhibits x as a sum of two elements of U.) If p is odd, since more than half of the numbers are in U and more than half of the numbers are in x-U they must overlap!
The interesting thing about this argument is that it only works if you tackle it one prime at a time. In particular, in Z/105 fewer than half of the integers less than 105 are relatively prime to 105. So a direct pidgeon hole argument won’t work, you need the Chinese remainder theorem.
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10 June, 2007 at 8:11 am
Terence Tao
Actually, there is some connection to the Goldbach conjecture; physics_grad_student’s conjecture is the “local” version. If there was a residue class mod n which was not expressible as a sum of two classes coprime to n, but was expressible as an even number, then Goldbach’s conjecture would easily be seen to be false. In the language of my Simons I lecture, the fact that the above conjecture is true means that there are no local obstructions to solvability of the Goldbach problem. Unfortunately, we don’t currently have a “local to global” principle for this type of problem to then finish the job.
The type of sieve theory that gives near misses to the Goldbach conjecture (e.g. Chen’s theorem that every large even number is the sum of a prime and a P_2) implicitly uses the fact that the above conjecture is true to ensure that the “main term” or “singular series” in the number of solutions is safely bounded away from zero.
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10 June, 2007 at 8:22 am
Jonathan Vos Post
105 doesn’t work as a side-effect of it being the smallest odd sphenic number, i.e. product of 3 distinct odd primes. The next odd sphenic number is 165. Why does the argument break down at these? If we patch the argument, wheat then onstructs that approach to proof?
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10 June, 2007 at 3:34 pm
Jonathan Vos Post
I think these exceptions come from Sylow’s theorems.
A homework problem of Brian Harbourne, U. Nebraska at Lincoln, gives most of the reason.
Problem: Let H and K be subgroups of a group G of order pqr, where p, q and r are distinct primes. If |H| = pq and |K| = qr, prove that the intersection (call it I) of H and K has order q.
Proof: Since |I| divides both |H| and |K|, we see that |I divides both qr and qp. But the only common factors of qr and qp are 1 and q.
So suppose |I| = 1. Let HK denote the subset
{g in G : g = hk for some h in H and some k in K}. Suppose h_1 and h_2 are in H and that k_1 and k_2 are in K.
If h_1 k_1 = h_2 k_2, then (k_1 k_2)^-1 =
(h_1)^-1 h_2, but (k_1 k_2)^-1 is in K and
(h_1)^-1 h_2 is in H, so (k_1 k_2)^-1 =
( h_1)^-1 h_2 is in I, and by assumption must thus be e (since we are assuming |I| = 1).
Thus (k_1 k_2)^-1 = e so k_1 = k_2, and likewise h_1 = h_2. Thus the map f : HxK -> HK defined by f((h, k)) = hk is injective. It is obviously surjective, so it is bijective, hence |HK| = |H||K| = (prq)^2. But HK is a subset of G, and this means |HK| = (prq)^2 is bigger than |G| = pqr, which is impossible. Thus |I| = 1 must be false, so |I| = q.
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8 August, 2007 at 8:01 am
(Emmanuel Kowalski) The large sieve inequalities « What’s new
[…] post may be seen as complementary to the post “The parity problem in sieve theory“. In addition to a survey of another important sieve technique, it might be interesting as a […]
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12 August, 2007 at 8:13 pm
Atle Selberg « What’s new
[…] In working on the zeta function, Selberg developed two powerful tools which are still used routinely in analytic number theory today. The first is the method of mollifiers to smooth out the magnitude oscillations of the zeta function, leaving making the (more interesting) phase oscillation more visible. The second was the method of the Selberg sieve, which is a particularly elegant choice of sieve which allows one to count patterns in almost primes (and hence to upper bound patterns in primes) quite accurately. Variants of the Selberg sieve were a crucial ingredient in the work of Goldston-Yıldırım-Pintz on prime gaps, and the work of Ben Green and myself on arithmetic progressions in primes. (I discuss the Selberg sieve, as well as the Selberg symmetry formula below, in my post on the parity problem.) […]
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6 December, 2007 at 10:11 pm
Milliman Lecture III: Sum-product estimates, expanders, and exponential sums « What’s new
[…] one has to add back in points which are divisible by multiple small factors, and so forth). See my post on sieve theory and the parity problem for further discussion. In order for sieve theory to work well, one needs to be able to accurately […]
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1 January, 2008 at 8:41 pm
Amateur
I’m not a mathematician, so please forgive this intrusion, but I’ve been fascinated by sieve theory
and this seems the right place to ask about an aspect of it that I find very puzzling and would very much appreciate if someone could point out an error in my reasoning.
The phenomenon is already apparent in examples of counting problems given by T.Tao at the beginning of his esssay on parity problem at the top of this thread. The simplest question that exhibits the phenomenon explicitly is a variation on the second example given by T.Tao and asks: How many odd numbers are contained in a set of three cosecutive integers?
The problem is not with determining the answers, but with their plurality. This lack of uniqueness implies lack of functional dependence. Specifically that the cardinality of the presumed set of solutions is not a function of the cardinality of the original set to which the sieve is applied.
Abstractly the sieve can be viewed as an instance of the implicit function problem – actually an explicit formula for its solution, if and when applicable. It seems that in this case the underlying functional relationship neccessary for the existence of solution to the implicit function problem is not in place.
In the generic case of inclusion-exclusion the problem does not appear, because this procedure is applied to sets defined explicitly, or at least, as in the case of the various subsets appearing in the formula, capable of being so defined (explicitly computed), by the use of set operations (union, intersection, relative complement) as functions of the original set and relations defined on it. In this setting there is no problem in coverting the method to handle the corresponding cardinalities if those appearing in the formula can be explicitly determined.
Likewise in number theory, if enough information is supplied to nail the set down explicitly the problem disappears.
If per chance I’m right then the meaning and significance of the “error term” in at least some sieves
might require a re-examination.
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3 January, 2008 at 3:09 pm
Terence Tao
Dear Amateur,
The sieve of Eratosthenes (for instance) does provide an explicit, deterministic formula for the number of primes in any given range. This formula is known as the Legendre identity, see
Unfortunately, the formula contains the greatest integer part function , which is hard to compute with. We can express this function as a combination of the “main term” x and the “error term” . We understand how to sum up the main term very well, but the error terms are what cause all the trouble. Heuristically, the fractional parts of all the terms which arise in the Legendre identity should be uniformly distributed between 0 and 1 – what reason would they have for doing otherwise? – but we unfortunately cannot prove this; similarly for other sieves. So there’s not much we can do with the error terms for other than treat them as being essentially arbitrary numbers between 0 and 1, which is what causes the formulae from sieve theory to become somewhat inexact. But actually, in many cases having a simple but somewhat inexact formula is more useful than an exact but overly complicated formula. For instance, to prove the twin prime conjecture, it is not necessary to count the number of twin primes exactly; having a rough lower bound on this count which goes to infinity would be sufficient, and presumably simpler to establish than an exact count.
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7 January, 2008 at 8:22 pm
AMS lecture: Structure and randomness in the prime numbers « What’s new
[…] well before the mark (sieve levels such as are typical). (The reason for this has to do with the parity problem, which I will not discuss further […]
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19 November, 2008 at 2:41 pm
Marker lecture III: Small gaps between primes « What’s new
[…] But there seems to be a significant obstacle to pushing things all the way to 2. Indeed, the parity problem tells us that for any reasonable definition of “almost prime” which is amenable to […]
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20 November, 2008 at 1:09 pm
Marker lecture IV: sieving for almost primes and expanders « What’s new
[…] fairly typical). The selection of the sieve weights is now a well-developed science (see also my earlier post, and Kowalski’s guest post, on this topic), and Bourgain, Gamburd and Sarnak basically use […]
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1 September, 2009 at 10:47 am
timur
The mathematician Sun Tzu and the military general Sun Tzu seem to be different individuals.
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1 September, 2009 at 1:54 pm
Terence Tao
Hmm, I think you’re right. A shame; the story I had was a good one, but it appears to be apocryphal after all, so I’ve deleted it.
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2 September, 2009 at 12:38 pm
timur
A long time ago in a galaxy far, far away… :))
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24 September, 2009 at 3:19 pm
The prime number theorem in arithmetic progressions, and dueling conspiracies « What’s new
[…] while the genuine primes do not, is a reflection of the parity problem in sieve theory; see this post for further discussion. The symmetry formula is however enough to get “within a factor of […]
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29 December, 2009 at 8:44 am
Craig
I was thinking about your theorem with Ben Green that says that the primes contain arbitrarily long progressions. Here is a strange but simple argument that this must be true:
Suppose that it weren’t true. Then there must exist an N such that there are no N-term arithmetic progressions of primes. Then if we exhibited an arithmetic progression a+bk, k=1,2,3,…,N-1 of all primes, we could know immediately that a is not prime, since if so then a would be part of an N-term progression of primes.
Also note that if this were possible, we could know that a is not prime without having any information about the possible divisors of a, since the fact that each a+bk, k=1,2,3,…,N-1, is prime says nothing about the possible divisors of a. But we need to have some information about the divisors of a in order to know that a is not prime, since the truth of the statement (a is not prime) depends on the divisors of a. This is a contradiction. Hence, your and Ben Green’s theorem must be true.
Comments?
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29 December, 2009 at 8:52 am
jc
I don’t follow this argument. In particular, what you mean by “information about possible divisors” doesn’t make any sense to me. Consider the fact that one can test a number for primality without factoring, for instance.
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29 December, 2009 at 9:10 am
Craig
jc,
Yes, it is true that you can test a number for primality without factoring, but you cannot test a number for primality without having some information about its factors.
For instance, see the Wikipedia entry for the AKS algorithm. When testing whether n is prime, the algorithm checks whether 1<gcd(a,n)<n for a<= r < n. The result of this test gives some information about the factors of n.
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29 December, 2009 at 11:59 am
Terence Tao
Actually, the primality of a+bk does tell us something about the divisors of a. For instance, if a+b and a+2b are both prime (and b is positive), this forces a to be odd, simply because a+2b is also odd.
More generally, an argumentam ad ignorantiam does not rise to the level of a rigorous proof, though it may have some heuristic value: just because we cannot see any obvious way to connect the primality of a+b, a+2b, …, a+(N-1)b to the divisors of a, it does not necessarily follow that no such connection exists (and could possibly be discovered in the future by a sufficiently clever mathematician).
Incidentally, there are plenty of primality tests that do not require any explicit tests of divisibility by potential factors of the number being tested at all; Wilson’s theorem provides a simple (though impractical) such example. Another example is the Lucas-Lehmer test for the primality of a Mersenne number. (Of course, the certificate of primality provided by such tests does imply a posteriori that there is no divisibility by smaller factors, but this does not imply that an explicit trial division is necessary prior to establishing the primality certificate.)
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29 December, 2009 at 12:23 pm
Craig
Terrence,
You are correct about my argument being refuted since a+b, a+2b prime implies that a is odd. But what if I formulate the argument as: a+bk, k=0,1,2,…,N-2, are prime doesn’t say anything about whether a+b(N-1) is prime, since it says nothing about the divisors of a+b(N-1)?
Also, I disagree with you about Wilson’s theorem. Calculating whether (n-1)! mod n is zero or negative one is essentially the same thing as asking whether n is prime, just stated in a different way.
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29 December, 2009 at 12:58 pm
Terence Tao
Essentially the same example applies: if a, a+b, a+2b are prime, and b is positive, then a+3b is odd (because a+b is odd).
In any case, even one could not immediately produce such examples, the basic argument is still an argumentam ad ignorantiam, and could potentially be refuted by an example of this type in the future. For instance, to use Wilson’s theorem as a hypothetical, suppose some clever mathematician found an ingenious formula that connected (a-1)!, (a+b-1)!, (a+2b-1)!, and (a+3b-1)! to each other. Then it is potentially conceivable that one could use Wilson’s theorem and the primality of a+b, a+2b, a+3b to say something about the primality of a. Now, I don’t know of such a formula (and in fact I strongly doubt that one exists); but I cannot rigorously rule out the possibility that something like this (or one of an infinite number of other possibilities, using some other number-theoretic fact than Wilson’s theorem) could actually happen (except by using my theorem with Ben, of course). All I can say is that I do not see any obvious way to pull off anything like this, but to use this ignorance to deduce anything is the logical fallacy of argumentam ad ignorantium.
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29 December, 2009 at 1:09 pm
Craig
OK, that makes sense. Thank you very much for your response, Terence, and sorry for mispelling your name before.
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29 December, 2009 at 9:15 am
Antonios Manoussos
Dear Prof. Tao,
In the text of your very nice article it is written “however, the prime number theorem does at least tell us that the primes have a positive relative density in the almost primes”. I am not an expert in this field but I have a question about that: Since the 2-primes less than a given $n$ are asymptotically $\frac{n}{\log n} \log \log{n}$ it looks like the relative density of the primes in the semi primes is 0. Is this o.k.?
Congratulations for your very interesting blog.
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29 December, 2009 at 12:01 pm
Terence Tao
For sake of this discussion, I define a semiprime to be the product of two large primes – e.g. primes larger than for some fixed threshold . These are the types of almost primes that are generated by sieves (semiprimes with small factors tend to get eliminated quite easily by such sieves). With this definition, the number of semiprimes less than N is comparable to rather than .
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29 December, 2009 at 1:05 pm
Antonios Manoussos
Thank you for the answer! Now it is clear to me.
Antonios
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10 August, 2010 at 3:51 am
Goldbach’s conjecture (to prove or not to prove) | MathFax.com
[…] … ve-theory/ […]
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12 August, 2011 at 8:19 am
Primos gemelos: Un vistazo a algunos resultados. « Lo fascinante de la teoría de números
[…] 5. La conjetura de Elliot-Halberstam es otra conjetura acerca de teoría de números bastante tratada. En palabras de Terence Tao, “Una especie de Hipotesis de Riemann super generalizada“. […]
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28 November, 2011 at 12:11 pm
Craig
Yao’s XOR Lemma shows that computing the parity of several independent weakly unpredictable predicates amplifies the unpredictability to the point where it is almost random. Is there any connection between this and the parity problem in sieve theory?
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1 March, 2012 at 5:25 pm
254B, Notes 7: Sieving and expanders « What’s new
[…] known as the parity problem, which prevents sieve theory from lowering all the way to ; see this blog post for more discussion.) One formulation of this principle was established by Bourgain, Gamburd, and […]
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25 May, 2012 at 7:44 am
Heuristic limitations of the circle method « What’s new
[…] which are much easier) on prime patterns, namely the parity problem. I discuss this problem in this previous blog post, and do not have much more to add here to what I already wrote in that […]
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16 October, 2012 at 8:46 am
Mathematics: What it takes to prove the Goldbach Conjecture? - Quora
[…] know, but finding creative ways around the parity problem in sieve theory is a promising approach: QuoteComment Loading… • Share • Embed • 4m ago Add […]
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15 June, 2013 at 3:38 am
强扭的瓜是甜的 » 素数并不孤独
[…] Open question: The parity problem in sieve theory, Terence Tao, […]
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16 June, 2013 at 12:49 am
oldhat
dear professor Tao,
congratulations for your wonderful site! but it is ‘argumentum ad ignorantiam’, not ‘argumentam ad ignorantium’ (sorry for the nitpickery…).
[Unfortunately I am not able to locate this typo. -T.]
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23 June, 2013 at 1:21 am
素数并不孤独 | fwjmath的相空间
[…] Open question: The parity problem in sieve theory, Terence Tao, […]
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23 June, 2013 at 2:17 am
oldhat
Dear professor Tao
yes, in fact it is not in your post, but it appears in two remarks above, which you wrote on december 29 2009, at 11:59 and at 12:58….forgive the grammatical obsession of an old latinist…and keep up your marvelous work (I can’t begin to tell you how envious I am, as a non mathematician, of the way of doing science you show in your blog)!
[Ah, found it, thanks – T.]
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1 July, 2013 at 10:33 pm
Avi Levy
Regarding the remark immediately following (5):
I am having trouble seeing how (5) corresponds to the zeta-function identity you have posted (in the display following (5)).
Instead, (5) appears more closely related to the “identity” , which I will call (5).
Here’s my reasoning.
First, (5) is just a verbose way of writing (in the notation of recent Polymath8 blog posts)
Then, since corresponds to and corresponds to , (5) is a restatement of (5).
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1 July, 2013 at 10:48 pm
Terence Tao
The first part of (5) (equating with ) does correspond to the identity you state, but the second part of (5) (equating with ) instead corresponds to the identity in the text. Of course both identities are very easy to prove (and easily seen to be equivalent).
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2 July, 2013 at 4:43 pm
Avi Levy
Ah, it seems that the point of (5) is to justify the second equality. I was assuming that (5) was already asserting this statement, which boils down to for all .
Instead, if I understand correctly, you are using the display after (5) in order to justify the second equality of (5)?
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Yes, one line using Liebniz:
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In (4), should be .
[Corrected, thanks – T.]
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Regarding a sieve theoretic proof of Euclid’s theorem: If we sieve the interval by the first primes, then is divisible by all denominators in the Legendre identity; no terms need to be rounded, the error term vanishes, and the number of integers in not divisible by any is given exactly by
Since this holds for any , it follows that there must be infinitely many primes. Shouldn’t this work?
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Yes, this is a valid proof of Euclid’s theorem, albeit one which is very close to Euclid’s original proof. But note that the sifted set that is counting here is not the set of primes in A, so one is not directly establishing the infinitude of primes by the usual sieve-theoretic method of viewing the primes as a sifted set, but instead using the primes as the set of sieving moduli, which is a different type of argument (more multiplicative number theory than sieve theory).
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Thanks for the reply. You are right it is close to Euler’s proof. In fact, an almost replica of his proof appears by sieving the interval and exploiting the -periodicity of to show that .
I agree completely with your point that these proofs lean more or less towards multiplicative number theory. However, as a basis for building up intuition about what structures that develops throughout sieving it seems quite useful to think in these terms (and perhaps even more so pedagogically). From this perspective, proofs of landmark theorems, like Dirichlet’s theorem on arithmetic progressions, your own Green-Tao theorem on arbitrarily long arithmetic progressions of primes, or Zhang’s very recent result on bounded gaps between primes, all boil down to simple variations of the proofs above. Naturally, these proofs hold only for coprimes and not primes. But it is interesting to observe the immense gap between this simplicity and the sieve theoretic proofs.
Another aspect is that the periodicity of introduces certain symmetries and constraints. For example, it is easy to show that for an interval of length , the bounds of and are anti-symmetric, in the sense that
It is merely a playful speculation, but do you think exploring the constraints and symmetries imposed by the periodicity of could be a useful parallel or addition to sieve theory?
Or has this already been done and doomed?
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14360 | https://hal.science/hal-03922884/file/Parity_Permutation_Pattern_Matching__cameraready.pdf | HAL Id: hal-03922884 Submitted on 4 Jan 2023 HAL is a multi-disciplinary open access archive for the deposit and dissemination of sci-entific research documents, whether they are pub-lished or not.
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Parity Permutation Pattern Matching Virginia Ardévol Martínez, Florian Sikora, Stéphane Vialette To cite this version: Virginia Ardévol Martínez, Florian Sikora, Stéphane Vialette. Parity Permutation Pattern Matching.
WALCOM, Mar 2023, Hsinchu, Taiwan. 10.1007/978-3-031-27051-2_32. hal-03922884 Parity Permutation Pattern Matching Virginia Ard´ evol Mart´ ınez1[0000−0002−3703−2335], Florian Sikora1[0000−0003−2670−6258], and St´ ephane Vialette2[0000−0003−2308−6970] 1 Universit´ e Paris-Dauphine, PSL University, CNRS, LAMSADE, 75016 Paris, France {virginia.ardevol-martinez,florian.sikora}@dauphine.fr 2 LIGM, CNRS, Univ Gustave Eiffel, F77454 Marne-la-Vall´ ee, France stephane.vialette@univ-eiffel.fr Abstract. Given two permutations, a pattern σ and a text π, Parity Permutation Pattern Matching asks whether there exists a parity and order preserving embedding of σ into π. While it is known that Permutation Pattern Matching is in FPT, we show that adding the parity constraint to the problem makes it W-hard, even for alternating permutations or for 4321-avoiding patterns. However, it remains in FPT if the text avoids a fixed permutation, thanks to a recent meta-theorem on twin-width. On the other hand, as for the classical version, Parity Permutation Pattern Matching remains polynomial-time solvable when both permutations are separable, or if both are 321-avoiding, but NP-hard if the pattern is 321-avoiding and the text is 4321-avoiding.
Keywords: Permutation Pattern Matching · Fixed Parameter Tractabil-ity · Parameterized hardness · NP-hardness 1 Introduction Permutations are one of the most fundamental objects in discrete mathematics, and in concrete, deciding if a permutation contains another permutation as a pattern is one of the most natural decision problems related to them. More pre-cisely, in the well-known problem Permutation Pattern Matching (PPM), given two permutations σ and π, the task is to determine if σ is a pattern of π, or equivalently, if π contains a subsequence which is order-isomorphic to σ. For example, if π = 31542, it contains σ = 231, as 352 is a subsequence of π with the same relative order as σ, but π does not contain σ = 123, as there are no 3 increasing elements in π. In the latter case, we say that π avoids 123. The notion of avoidance allows to define classes of permutations as sets of permu-tations that avoid certain patterns, for example, 321-avoiding permutations, or (2413,3142)-avoiding permutations, which are known as separable permutations.
PPM was proven to be NP-complete by Bose, Buss, and Lubiw in 1998 .
This motivated the search for exact exponential time algorithms [1,4,13,8]. How-ever, some special cases, such as Longest Increasing Subequence, or the cases where both σ and π are separable or 321-avoiding, are known to be poly-nomial time solvable [10,6,16,2]. In fact, it was shown in that PPM is 2 V. Ard´ evol Mart´ ınez, F. Sikora, S. Vialette always polynomial-time solvable if the pattern avoids any fixed permutation τ ∈{1,12,21,132,231,312,213}, and NP-complete otherwise. This result was then extended in .
Its parameterized complexity was open for a long time, with a series of partial results, but a breakthrough result of Guillemot and Marx showed that it is fixed parameter tractable when parameterized by the size of the pattern σ, using a new width measure structure theory of permutations . They showed that the problem can be solved in time 2O(k2 log k)n, and later on, Fox improved the running time of the algorithm by removing a factor log k from the exponent .
This led to the question of whether a graph-theoretic generalization of their permutation parameter could exist, that was answered positively in , by intro-ducing the notion of twin-width, which has proven huge success recently. They showed that graphs of bounded twin-width define a very natural class with re-spect to computational complexity, as FO model checking becomes linear in them.
Pattern matching for permutations, together with its many variants, has been widely studied in the literature (the best general reference is , see also ).Here we introduce a natural variation of PPM, which we call Par-ity Permutation Pattern Matching, and that incorporates the additional constraint that the elements of σ have to map to elements of π with the same parity, i.e., even (resp. odd) elements of σ have to be mapped to even (resp. odd) elements of π. For one thing, pattern avoidance with additional constraints [3,9], including parity restrictions [20,14], has emerged as a promising research area.
For another, Parity Permutation Pattern Matching aims at providing concrete use cases of the 2-colored extension of PPM introduced in . We show that, surprisingly, it does not fit into the twin-width framework, and this increases the complexity of the problem, as it becomes W-hard parameterized by the length of the pattern.
In fact, the approach used by Guillemot and Marx to prove that PPM is FPT is based on a result that states that given a permutation π, there exists a polynomial time algorithm that either finds an r × r-grid of π or determines that the permutation has bounded width (and returns the merge sequence of the decomposition, which is used to solve the PPM problem in FPT time). This win-win approach works because, if π contains an r × r-grid, it’s not hard to see that it contains every possible pattern σ. However, this cannot be generalized to Parity PPM, as here we have no information on the parity of the elements of the grid, and thus, it is not guaranteed that every pattern maps via a parity respecting embedding into the grid.
Structure of the paper The paper is organized as follows. Section 2 briefly in-troduces the necessary concepts and definitions. In Section 3, we study the pa-rameterized complexity of Parity PPM, showing that it is harder than PPM in general, but that it remains in FPT for some cases, namely when the twin-width of the host permutation is bounded. Finally, in Section 4, we show that concerning the classical P vs NP questions, Parity PPM is similar to PPM. A summary of the complexity of the problems is given in Table 1.
Parity Permutation Pattern Matching 3 PPM Parity PPM General case NP-hard, FPT W-hard Separable permutations P P 321-av σ and 321-av π P P 321-av σ and 4321-av π NP-hard NP-hard 4321-av σ FPT W-hard Alternating π and σ FPT W-hard π is fixed pattern avoiding FPT FPT Table 1. Summary of known results (for PPM) and our results (for Parity PPM).
Due to space constraints, some proofs (marked with (⋆)) are deferred to the full version of this paper.
2 Preliminaries Let [n] = {1,...,n}. A permutation of length n is a bijection f ∶[n] Ð →[n].
Given two permutations σ ∈Sk and π ∈Sn, we say that π (the text, or the host) contains σ (the pattern) if there is an embedding from σ into π, i.e., an injective function f such that for every pair of elements x and y of σ, their images f(x) and f(y) of π are in the same relative order as x and y. Otherwise, we say that π avoids σ. If π contains σ, we write σ ⪯π.
A permutation class is a set C of permutations such that for every permuta-tion π ∈C, every pattern of π is also contained in C. Every permutation class can be defined by the minimal set of permutations that do not lie inside it, and we define this as C = Av(B), where B is the minimal set of avoided permutations.
In this manner, we can define the class Av(4321), which is the set of permu-tations that avoid 4321, Av(321), which is the set of permutations that avoid 321, and Av(2413,3142), i.e., the class of permutations that avoid both 2413 and 3142. As we mentioned in the introduction, the latter is known as the class of separable permutations, and it can also be characterized as the set of permu-tations that have a separating tree. In other words, a permutation is separable if there exists an ordered binary tree T in which the elements of the permuta-tion appear in the leaves and such that the descendants of a tree node form a contiguous subset of these elements.
Furthermore, we define the set of alternating permutations as the set of per-mutations σ ∈Sn such that σ1 > σ2 < σ3 > ....
The problem of determining whether a fixed pattern is contained in a per-mutation has been well studied in the literature, and it is referred to as Per-mutation Pattern Matching. Here, we study a natural variation of PPM, Parity Permutation Pattern Matching, which we define formally below.
Definition 1. Given two permutations, a pattern σ ∈Sk and a text π ∈Sn, the problem Permutation Pattern Matching asks whether π contains σ.
4 V. Ard´ evol Mart´ ınez, F. Sikora, S. Vialette Definition 2. An injective function f from σ to π is a parity respecting embed-ding if for all elements x and y of σ, f(x) and f(y) are in the same relative order as x and y, and for every element x of σ, f(x) has the same parity as x.
We say that an occurrence of a pattern σ in a permutation π respects parity if there is a parity respecting embedding of σ into π. Furthermore, if there is an occurrence of σ in π which respects parity, we say that π parity contains σ, and we write σ ⪯P π. Otherwise, we say that π parity avoids σ.
Definition 3. Parity PPM is the problem of determining whether given a pattern σ and a text π, there exists a parity respecting embedding of σ into π.
As a remark, note that if instead of considering the problem PPM with the constraint that even (resp. odd) elements have to map to even (resp. odd) elements, we require that elements in even (resp. odd) indices (positions) map to elements in even (resp. odd) indices, the problem is equivalent. Indeed, σ parity avoids π if and only if σ−1 parity index avoids π−1.
For example, the parity+order preserving embedding of σ = 2413 into π = 4276315 yields the parity-index+order-preserving embedding of σ−1 = 3142 into π−1 = 6251743 (occurrences are depicted with bold integers).
To see this, assume that there is a parity respecting embedding of σ into π. Denote by Pσ(i) the position in σ of the element with value i and by f the parity respecting injective map between σ and π associated to the embedding.
Since σ−1 = Pσ(1) ... Pσ(k), and f respects parity, if f(i) = j, both i and j have the same parity, and thus, the indices in the inverses will also have the same parity (by definition, odd elements are placed in odd indices in the inverses, and vice versa). Furthermore, since f is an an embedding, for i < j, σi < σj if and only if f(σi) < f(σj). Thus, P(σi) is to the left of P(σj) in both σ−1 and π−1, and by assumption, we also have i < j, so f induces a parity index respecting embedding between the inverses.
In this paper, we focus mainly on the parameterized complexity of the above-mentioned problem. Parameterized complexity allows the classification of NP-hard problems on a finer scale than in the classical setting. Fixed parameter tractable (FPT) algorithms are those with running time O(f(k)⋅poly(n)), where n is the size of the input and f is a computable function that depends only on some well-chosen parameter k. On the other hand, problems for which we believe that there does not exist an algorithm with that running time belong to the W-hierarchy. We refer to for more background on the topic.
3 Parameterized complexity We already saw in the introduction that PPM is in FPT in general, and why the win-win approach of Guillemot and Marx for the parameterized algorithm for PPM doesn’t work for Parity PPM. We show that this intuition is indeed true, proving that the problem is W-hard. In fact, we prove something stronger, which is that Parity PPM is W-hard even when restricted to alternating Parity Permutation Pattern Matching 5 permutations or when the pattern is 4321-avoiding. Note that both results are independent from each other, as alternating permutations and 4321-avoiding permutations are not comparable, but they both imply the W-hardness of the general case.
However, the twin-width framework (on which the parameterized algorithm of Guillemot and Marx is an initial step) will be useful to prove that Parity PPM remains in FPT when the text avoids a fixed pattern.
3.1 Parameterized hardness for alternating permutations Theorem 4. Parity PPM is W-hard parameterized by the length k of the pattern, even for alternating permutations σ and π.
Proof. We reduce from k-Clique in general graphs, which is known to be W-hard parameterized by the size of the clique k . Given as input a graph G and a parameter k, k-Clique asks whether G contains a clique of size k. For our reduction, given a graph G = (V,E), with ∣V ∣= n and ∣E∣= m, and a parameter k, we construct a permutation σ that depends only on the parameter k, and a permutation π, that depends on G, such that there exists a clique of size k in the graph G if and only if there is a parity respecting embedding of σ into π.
Construction We explain the construction of π for a general graph G. The high-level idea is to construct different gadgets to represent the vertices and the edges of the graph, and to somehow link each edge gadget to the corresponding vertex gadgets, that is, we link the gadget associated to edge (u,v) with the gadgets associated to vertices u and v by placing elements of value greater than the minimum element of each vertex gadget and smaller than the maximum element of each vertex gadget between the elements of the edge gadget.
We define the following gadgets (see also Figure 1): – A vertex gadget π[V ], which is a direct sum of n decreasing permutations, all order-isomorphic to 21 and composed of odd elements. It contains 2n elements and starts at element 8m + 3.
– An edge gadget π[E], which is a direct sum of m permutations, all order-isomorphic to 435261 and formed by odd elements. It contains 6m elements and starts at element 3.
– The separator gadget is composed of the four even integers 4(n + 3m) + 4, 8m + 2, 4(n + 3m) + 2 and 2 (and hence is order-isomorphic to 4231). The separator gadget lies between the vertex gadget and the edge gadget.
– Let Even be the 2(n + 3m) −2 even integers between 4 and 4(n + 3m) that do not appear in the vertex gadget, the separator gadget or the edge gadget.
The even garbage gadget is the alternating sequence composed of the even integers of Even. It is constructed recursively from left to right as follows: place (and remove from Even) the maximum of Even, place (and remove from Even) the minimum of Even and recurse. It is placed to the right of the edge gadget.
6 V. Ard´ evol Mart´ ınez, F. Sikora, S. Vialette Fig. 1. Illustration of the construction introduced in the proof of Theorem 4. The permutation in the figure corresponds to a clique of size 3 with vertices v1, v2, v3 and edges (v1, v2), (v1, v3), (v2, v3). The odd elements are represented in black while even elements are colored in red. Furthermore, blue lines delimit the vertex boxes.
– Let Odd be the two odd elements 4(n + 3m) + 3 and 1 that do not appear in the vertex gadget, the separator gadget or the edge gadget. The odd garbage gadget is the decreasing sequence composed of the two odd integers of Odd.
It is constructed as the even garbage gadget and placed directly to its right.
Formally, define, ∀vi ∈V, π[vi] = 8m + 2 + 2∑j<i(deg(vj) + 2) + 2∑i(deg(vi)) + 3 8m + 2 + 2∑j<i(deg(vj) + 2) + 1 (1) Parity Permutation Pattern Matching 7 ∀ek = (i,j) ∈E, π[ek] = 6k + 1 6k −1 8m + 2 + 2∑j′<i(deg(v′ j) + 2) + 2∑(i,j′)/j′<j(1) + 3 6k −3 8m + 2 + 2∑j′<j(deg(v′ j) + 2) + 2∑(i′,j)/i′ vi,2), and the elements of the edge box associated to edge ei as ei,1, ei,2, ei,3 and ei,4, again from left to right. On the other hand, for each edge, we denote the two elements placed in between ei,2 and ei,3, and between ei,3 and ei,4, as hi,1 and hi,2, respectively, where here hi,1 < hi,2 (these are the elements that correspond to the subsequence 56 in σ[ei]).
Finally, the even elements to the right of the edge gadget placed below w2 are referred to as wi,1, wi,2, wi,3 and wi,4, for every edge i ∈{1,...,m}, where wi,1 is the element ei,4 +1, wi,2 is ei,3 +1, wi,3 is ei,2 +1, and wi,4 is ei,1 +1. Note that wi,4 is not defined for the last edge. On the other hand, the even elements to the right of the edge gadget placed above w2 are denoted as xi,t, for every vertex i ∈{1,...,n} and every edge incident to vi, t ∈{1,...,mi} (xi,t = hx,y + 1 for some pair x,y).
Furthermore, we denote by xi,0 and xi,mi+1 the even elements in the extremes such that xi,0 = vi,2 + 1 and xi,mi+1 = vi,1 + 1. Again, note that xn,mn+1 is not defined.
For the elements of π, we follow an analogous notation denoting the elements by v′ i,1, e′ i,1, etc.
8 V. Ard´ evol Mart´ ınez, F. Sikora, S. Vialette Direct Implication Claim 5. (⋆)If there exists a clique of size k in the graph G, then there is a parity respecting embedding of σ into π.
Reverse implication Suppose now that there exists a parity respecting embedding between σ and π and let f be the associated injective mapping. We want to show that we have enough structure in the permutations to infer that there must be a clique of size k in the graph G. In order to do so, we will prove the following sequence of claims that will restrict the map f.
Claim 6. Any parity respecting embedding f from σ to π must map wi to w′ i, for i ∈{1,2,3,4}.
Proof of claim. Since the pattern matching needs to respect parity, f must map the wi’s to even elements of π. Towards a contradiction, assume first that f(wi) ≠w′ j, i,j ∈{1,2,3,4}. That means that f(wi) = w′ i′,j or x′ i′,t, for some indices i′,j or i′,t. But then, the odd elements to the right of wi in σ cannot map to elements to the right of f(wi) in π (as there would be at most 2 odd elements to the right of f(wi) and there are strictly more than 2 odd elements to the right of wi), so f cannot be an embedding of σ into π. Finally, since both the wi’s and the w′ i’s form 4231 subsequences, it is clear that there exists a unique way to embed the wi’s into the w′ i’s, which is mapping each wi to its corresponding w′ i, for every i ∈{1,2,3,4}. Thus, if f(wi) ≠w′ i, f cannot be an embedding.
◁ Claim 7. All the elements to the left (resp. to the right) of the wi’s in σ map to elements to the left (resp. to the right) of w′ i’s in π. Similarly, the elements above (resp. below) w2 in σ map to elements above (resp. below) w′ 2 in π.
Proof of claim. This is a direct corollary of Claim 6.
◁ Claim 8. Any parity respecting embedding f from σ to π must map vertex blocks of σ to vertex blocks of π.
Proof of claim. By Claim 7, since elements to the left of w2 in σ map to elements to the left of w′ 2 in π, we have that f(vi,j) = v′ i′,j′, for i ∈{1,...,k},i′ ∈ {1,...,n} and j,j′ ∈{1,2}. Assume that f(vi,1) = v′ i′,j′ and f(vi,2) = v′ i′′,j′′, with i′ ≠i′′. Since vi,1 is to the left of vi,2, it means that f must map vi,2 to an element placed to the right of f(vi,1) = v′ i′,j′. But vi,1 > vi,2 and every element which is to the right of v′ i′,j′ and which does not belong to the vertex block of v′ i′, is greater than v′ i′,j′. Thus, if i′′ ≠i′, then f would not be an embedding. ◁ Claim 9. Any parity respecting embedding f from σ to π must map edge blocks of σ to edge blocks of π.
Proof of claim. Again, we have that f(ei,j) = e′ i′,j′ for some pair i′,j′, and since the structure of the gadget has the same properties as the vertex gadget, we can use the same argument as in the proof of Claim 8.
◁ Parity Permutation Pattern Matching 9 Claim 10. Any parity respecting embedding f from σ to π must map hi,j to h′ i′,j, where e′ i′ is the edge associated to the edge block where ei,1 maps to.
Proof of claim. By Claim 7, we have that necessarily, f(ei,j) = e′ i′,j, for i ∈ {1,...,l},i′ ∈{1,...,m} and j ∈{1,2,3,4}.
First, since f(ei,2) = e′ i′,2 and f(ei,3) = e′ i′,3, and f is an embedding, the fact that hi,1 is in between ei,2 and ei,3 implies that it must map to an element between e′ i′,2 and e′ i′,3. Similarly, hi,2 must map to an element in between e′ i′,3 and e′ i′,4. Since edge blocks map to edge blocks, there is at most one element that satisfies each of these conditions. And these elements are h′ i′,1 and h′ i′,2, respectively.
◁ Claim 11. All the even elements to the right of the edge gadgets in σ must map to even elements to the right of the edge gadgets in π.
Proof of claim. This follows from Claim 6. Since f(wi) = w′ i for i ∈{1,2,3,4} and f has to respect parity, the rest of the even elements cannot map anywhere else.
◁ Now, suppose that there is a parity respecting embedding f of σ into π and assume, towards a contradiction, that G does not contain a clique of size k. Since there is no clique of size k, it means that we cannot have l = (k 2) edges between the k vertices of G which are in the image of f (that is, the vertices associated to the images of the k vertex boxes of σ).
We know that the k vertex blocks of σ map to k vertex blocks in π and the (k 2) edge blocks of σ map to (k 2) edge blocks of π. Since G does not contain a clique, one of the k vertices corresponding to the k vertex blocks in the image of f will have degree strictly smaller than k −1 when we restrict G to the k selected vertices. Let i′ be the vertex with degree strictly smaller than k −1 and suppose it is the image of vertex block i in σ. Then, there are two possible cases. The first case is that in the image of f, between the values f(vi,1) and f(vi,2), there are less than k odd elements (these elements are necessarily of the form h′ i,j). Since in between vi,1 and vi,2 in σ there are k odd elements of the form hi,j, this would imply that f cannot be a parity respecting embedding. The second possibility is that in between the values f(vi,1) and f(vi,2) there are k odd elements (which again are necessarily of the form h′ i,j) but one of them is not in between f(el,2) and f(el,3), or f(el,3) and f(el,4), for some l ∈{1,...,m}.
This would also contradict the fact that f is a parity respecting isomorphism, as all the hi,j in σ are between some pair el,2,el,3, or el,3,el,4 (with respect to the x-axis). Therefore, if there is a parity respecting embedding of σ into π, it must map the k vertex boxes of σ into k vertex boxes of π associated to k vertices that form a clique in G.
Corollary 12. (⋆)Given a pattern σ ∈Sk and a text π ∈Sn, Parity PPM cannot be solved in time f(k) ⋅no( √ k) for any computable function f, under the Exponential Time Hypothesis (ETH).
10 V. Ard´ evol Mart´ ınez, F. Sikora, S. Vialette Note that reducing from Subgraph Isomorphism instead of k-Clique in the proof of Theorem 4 to get a better lower bound under the ETH is not trivial since there is a notion of order of the pattern in Parity PPM (i.e., two isomorphic subgraphs can result in different permutations depending on the ordering of their vertices).
3.2 Parameterized hardness for 4321-avoiding patterns In this subsection, we complement the previous hardness result by showing that the problem remains hard for patterns belonging to the class of 4321-avoiding permutations. Our proof uses a colored version of PPM defined in , proven W-hard parameterized by k = ∣σ∣in .
Definition 13. 2-colored 2IPP ( 2 Increasing Permutation Pattern) consists on, given a 321-avoiding permutation σ ∈Sk and an arbitrary permu-tation π such that both σ and π are 2-colored permutations, finding a color-preserving embedding of σ into π.
Theorem 14. (⋆) Parity PPM is W-hard parameterized by the length k of the pattern, even if the pattern is 4321-avoiding.
3.3 Parameterized algorithm for fixed pattern avoiding text In the previous subsection, we showed that restricting the pattern does not necessarily reduce the complexity of the problem. However, we now see that restricting the text allows us to use the twin-width meta-theorem to have a positive result. In fact, to see that Parity PPM is FPT if the text avoids a fixed pattern x, it suffices to show that we can describe the problem using first-order (FO) logic, i.e., that we can express it as a formula which uses quantified variables over non-logical objects, and sentences (formulas without free variables) that contain the variables. Indeed, adding unary relations to mark the odd and even values preserves bounded twin-width, and therefore FPT tractability. The result follows from : Lemma 15 ().
FO model checking is FPT on every hereditary proper sub-class of permutation graphs.
This implies that FO model checking is FPT in the class of permutations avoiding a fixed pattern. Here, FO model checking refers to the problem of, given a first-order sentence ϕ of FO and a finite model M of FO (which speci-fies the domain of disclosure of the variables), deciding whether M satisfies ϕ, i.e., whether there exists an assignment of the variables which respects the do-main imposed by M and that satisfies ϕ. Therefore, we can state the following theorem: Theorem 16. (⋆)Parity PPM is in FPT if the text π avoids a fixed permuta-tion.
Parity Permutation Pattern Matching 11 4 Classical complexity Even though Parity PPM is harder than PPM from the parameterized point of view, we will show that this is not the case concerning its classical complexity.
4.1 Hardness A nice quite recent result showed that PPM remains NP-hard, even if the pattern is 321-avoiding and the text is 4321-avoiding . In the following, we show that it remains true for Parity PPM.
Theorem 17. (⋆)Parity PPM is NP-hard, even if σ is a 321-avoiding permu-tation and π is a 4321-avoiding permutation.
4.2 Polynomial-time solvable cases For some specific cases of Permutation Pattern Matching, polynomial time algorithms that solve the problem exactly have been proposed. Here, we show that some of these algorithms can be adapted to solve the problem Parity Permutation Pattern Matching while still running in polynomial time.
Theorem 18. (⋆)Let σ be a permutation in Sk and π be a permutation in Sn.
Parity PPM can be solved in polynomial time in the following cases: 1. If both permutations are separable. In particular, if both permutations are (231, 213)-avoiding, it can be solved in linear time.
2. If both permutations are 321-avoiding.
Acknowledgements Thanks to ´ Edouard Bonnet and Eun Jung Kim for point-ing out the link with the twin-width framework, and to the reviewers for their useful comments.
References 1. Ahal, S., Rabinovich, Y.: On complexity of the subpattern problem. SIAM J Dis-crete Math 22(2), 629–649 (2008) 2. Albert, M.H., Lackner, M., Lackner, M., Vatter, V.: The Complexity of Pattern Matching for 321-Avoiding and Skew-Merged Permutations. Discret. Math. Theor.
Comput. Sci. 18(2) (2016) 3. Alexandersson, P., Fufa, S.A., Getachew, F., Qiu, D.: Pattern-avoidance and Fuss-Catalan numbers. arXiv preprint arXiv:2201.08168 (2022) 4. Berendsohn, B.A., Kozma, L., Marx, D.: Finding and Counting Permutations via CSPs. Algorithmica 83(8), 2552–2577 (2021) 5. Bonnet, ´ E., Kim, E.J., Thomass´ e, S., Watrigant, R.: Twin-width I: tractable FO model checking. J. ACM 69(1), 3:1–3:46 (2022) 6. Bose, P., Buss, J.F., Lubiw, A.: Pattern Matching for Permutations. Inf. Process.
Lett. 65(5), 277–283 (1998) 12 V. Ard´ evol Mart´ ınez, F. Sikora, S. Vialette 7. Bruner, M.L., Lackner, M.: The computational landscape of permutation patterns.
Pure Mathematics and Applications: Special Issue for the Permutation Patterns 2012 Conference 24(2), 83–101 (2013) 8. Bruner, M., Lackner, M.: A Fast Algorithm for Permutation Pattern Matching Based on Alternating Runs. Algorithmica 75(1), 84–117 (2016) 9. Bulteau, L., Fertin, G., Jug´ e, V., Vialette, S.: Permutation Pattern Matching for Doubly Partially Ordered Patterns. In: Proc. of CPM. LIPIcs, vol. 223, pp. 21:1– 21:17 (2022) 10. Cormen, T.H., Leiserson, C.E., Rivest, R.L., Stein, C.: Introduction to algorithms.
MIT press (2022) 11. Cygan, M., Fomin, F.V., Kowalik, L., Lokshtanov, D., Marx, D., Pilipczuk, M., Pilipczuk, M., Saurabh, S.: Parameterized Algorithms. Springer (2015) 12. Fox, J.: Stanley-wilf limits are typically exponential.
arXiv preprint arXiv:1310.8378 (2013) 13. Gawrychowski, P., Rzepecki, M.: Faster Exponential Algorithm for Permutation Pattern Matching. In: 5th SOSA@SODA 2022. pp. 279–284. SIAM (2022) 14. Gil, J.B., Tomasko, J.A.: Restricted Grassmannian permutations. Enumerative Combinatorics and Applications 2(4), Article #S4PP6 (2021) 15. Guillemot, S., Marx, D.: Finding small patterns in permutations in linear time. In: Proc. of SODA. pp. 82–101. SIAM (2014) 16. Guillemot, S., Vialette, S.: Pattern Matching for 321-Avoiding Permutations. In: Proc. of ISAAC. LNCS, vol. 5878, pp. 1064–1073. Springer (2009) 17. Jel´ ınek, V., Kynˇ cl, J.: Hardness of Permutation Pattern Matching. In: Proc. of SODA. pp. 378–396. SIAM (2017) 18. Jel´ ınek, V., Opler, M., Pek´ arek, J.: Griddings of Permutations and Hardness of Pattern Matching. In: Proc. of MFCS. LIPIcs, vol. 202, pp. 65:1–65:22 (2021) 19. Kitaev, S.: Patterns in Permutations and Words. Monographs in Theoretical Com-puter Science. An EATCS Series, Springer (2011) 20. Tanimoto, S.: Combinatorics of the group of parity alternating permutations. Adv.
Appl. Math. 44(3), 225–230 (2010) |
14361 | https://msrads.web.unc.edu/wp-content/uploads/sites/15695/2018/08/Signs-in-Thoracic-Imaging.pdf | Signs in Thoracic Imaging Geoffrey B. Marshall, BSc, MD, Brenda A. Farnquist, BMLSc, MD, John H. MacGregor, MD, FRCP(C), and Paul W. Burrowes, MD, FRCP(C) Abstract: Radiologic signs are recognizable, characteristic patterns used to describe abnormalities visualized on imaging modalities that ultimately aid in the diagnosis and subsequent treatment of disease. This pictorial essay discusses 23 classic roentgenographic signs used in thoracic imaging. Its purpose is to be used as an educational review for residents, whether they are beginning their training or preparing for certification exams, and serve as a refresher and a reference to the practicing radiologist.
Key Words: chest, pulmonary, sign, silhouette, radiograph, computed tomography (J Thorac Imaging 2006;21:76–90) R adiologic signs are recognizable, characteristic pat-terns used to describe abnormalities visualized on imaging modalities that ultimately aid in the diagnosis and subsequent treatment of disease. When the late pulmonary radiologist Benjamin Felson was asked about his propensity for naming signs he responded that ‘‘the name saves time, helps you remember the sign, and advertises ity They say what I mean.’’1 This pictorial essay discusses 23 classic roentgenographic signs used in thoracic imaging and presents them in alphabetic order.
Its purpose is to be used as an educational review for residents, whether they are beginning their training or preparing for certification exams, and serve as a refresher and a reference to the practicing radiologist.
AIR CRESCENT SIGN The air crescent sign appears as a variably sized, peripheral crescentic collection of air surrounding a necrotic central focus of infection on thoracic radiographs (Fig. 1A) and CT (Fig. 1B).2–4 It is often seen in neutropenic patients who have undergone bone marrow or organ transplantation and is most characteristic of infection with invasive pulmonary aspergillosis. The fungus invades the pulmonary vasculature, causing hemorrhage, thrombosis, and infarction. With time, the peripheral necrotic tissue is reabsorbed by leukocytes and air fills the space left peripherally between the devitalized central necrotic tissue and normal lung parenchyma.5 Thus, the presence of the air-crescent sign heralds recovery of granulocytic function.4 Other causes of the air crescent include cavitating neoplasms, bacterial lung abscesses, and infections such as tuberculosis or nocar-diosis.6 BULGING FISSURE SIGN Classically, this sign is associated with consolidation of the right upper lobe (Fig. 2) due to Klebsiella pneumoniae infection.7 Due to the tendency for Klebsiella to produce large volumes of inflammatory exudate, the involved lobe expands and exerts mass effect on the adjacent interlobar fissure.8 The normally straight minor fissure on the lateral view bulges convex posteroinferiorly due to rapid lobar expansion.3,7 Although previously reported in up to 30% of patients with Klebsiella pneumonia,8,9 the finding is identified less commonly today, most likely due to rapid prophylactic implementa-tion of antibiotics.3 Other less common causes of the bulging fissure sign include Hemophilus influenzae, tuber-culosis, pneumococcal pneumonia, large lung abscesses, and lung neoplasms.10 CERVICOTHORACIC SIGN This sign is based on the understanding that if a thoracic mass is in direct contiguity with the soft tissues of the neck, the borders delineating their point of contact will be lost or obscured.11 Anatomically, the thoracic inlet parallels the first ribs, and the posterior aspects of the lung apices extend further superiorly than the anterior portions.12 On PA chest radiographs, a mass clearly delineated on all borders above the level of the clavicles lies posterior to the level of the trachea and completely within the lung.13 When the cephalic border of a mass is obscured at or below the level of the clavicles, it is deemed to be a ‘‘cervicothoracic lesion’’13 involving the anterior mediastinum (Figs. 3A–E). Mediastinal masses posterior to the trachea are well outlined above the level of the clavicles due to the interface with lung in the posterior aspects of the lung apices (Figs. 3F, G).
COMET TAIL SIGN Identified as a curvilinear opacity arising from a rounded, subpleural opacity on CT (Fig. 4A, arrow), the ‘‘comet tail’’ denotes the distorted curving of Copyright r 2006 by Lippincott Williams & Wilkins From the Department of Diagnostic Imaging, Foothills Medical Center, Calgary, Alberta, Canada.
Reprints: Dr Paul Burrowes, Department of Diagnostic Imaging, Foothills Medical Centre, 1403-29 Street NW, Calgary, Alberta, Canada T2N 2T9 (e-mail: paul.burrowes@calgaryhealthregion.ca).
PICTORIAL ESSAY 76 J Thorac Imaging Volume 21, Number 1, March 2006 pulmonary vessels and bronchi toward a region of round atelectasis.14–17 The atelectatic lung typically abuts an area of pleural effusion or thickening, and there are associated signs of volume loss. Round atelectasis is an unusual form of atelectasis most commonly associated with asbestos-related pleural disease (Fig. 4B, arrow), but also seen with other chronic forms of pleural disease, such as tuberculosis, histoplasmosis, pulmonary infarcts, or congestive heart failure.14–17 It is also known as atelec-tatic pseudotumor, folded lung, or Blesovsky syndrome, after the physician who initially postulated an association with asbestos exposure.18 Most often found in the posterior aspect of the lower lobes, it may demonstrate significant contrast enhancement and contain air broncho-grams.14–17 Histopathology demonstrates fibrous thicken-ing of the visceral pleura with extensive pleural folding and invagination.19 The exact cause is still debated. One theory suggests that a pleural effusion is the inciting event causing passive atelectasis and pleural invagination.20 Fibrous adhesions will eventually develop between the visceral and parietal pleural surfaces, so that when the effusion clears, the entrapped lung folds in on itself. An alternate theory suggests that a focal inflammatory pleural fibrotic response secondary to irritants such as asbestos, is solely responsible for causing invagination of the pleura and regional atelectasis of the subja-cent lung.14,19,21 CONTINUOUS DIAPHRAGM SIGN Identifiable on PA chest radiography (Fig. 5, arrows), there is a continuous line of lucency extend-ing across the midline, above the diaphragm.22,23 This is re- presentative of air tracking posterior to the heart in pneumomediastinum and is helpful in differentiating from a pneumothorax.4,22,23 CRAZY PAVING SIGN The ‘‘crazy paving’’ pattern12 is described at HRCT of the lungs (Fig. 6). The pattern is caused by thickening of interlobular septa superimposed on areas of ground glass opacification.24–26 It was first used to describe the pattern of pulmonary alveolar proteinosis, but can be seen in a variety of diseases. Histopathologically, the ground glass attenuation in alveolar proteinosis correlates to lipid-rich, intra-alveolar proteinaceous fluid, with thickening of the interlobular septa due to adjacent inflammatory reaction.13 Infectious causes include Pneu-mocystis carinii pneumonia, organizing pneumonia, usual interstitial pneumonia, non-specific interstitial pneumo-nia, and exogenous lipoid pneumonia. Other entities, such as respiratory bronchiolitis with interstitial lung disease,27 sarcoidosis, adult respiratory distress syndrome, pulmo-nary hemorrhage, and mucinous BAC, also less com-monly demonstrate the crazy paving pattern.12–15 CT HALO SIGN The CT halo sign appears as a zone of ground-glass attenuation around a nodule or mass (Fig. 7) on com-puted tomographic (CT) images.2–4,6,28 In febrile neutro-penic patients, the sign suggests angioinvasive fungal infection, which is associated with a high mortality rate in the immunocompromised host.2–4 The zone of attenuation represents alveolar hemorrhage,2,4,6,28 whereas the nodules represent areas of infarction and necrosis caused by thrombosis of small to medium sized vessels.2–4,6,28,29 Other infectious causes include candidia-sis, cytomegalovirus, herpes simplex virus, and cocci-dioidomycosis.30 The CT halo sign may also be caused by FIGURE 1. Air crescent sign. Frontal radiograph (A) of the chest shows cavitating lesion within the superior aspect of the right lower lobe (arrow). Follow-up enhanced CT of the chest (B) confirming the ‘‘air crescent sign’’ (arrow) in a patient with documented angioinvasive aspergillosis.
J Thorac Imaging Volume 21, Number 1, March 2006 Signs in Thoracic Imaging r 2006 Lippincott Williams & Wilkins 77 non-infectious causes, such as Wegener granulomatosis, metastatic angiosarcoma, Kaposi sarcoma, and brochio-loalveolar carcinoma (BAC).29,30 Due to the lepidic growth pattern of BAC, where the tumor cells spread along the alveolar walls, the typical ground glass halo visualized with the sign results.29 DEEP SULCUS SIGN The presence of radiolucency in a deep costophrenic sulcus on a supine thoracic radiograph (Fig. 8) is characteristic of a pneumothorax in a supine patient.31 Intrapleural air rises to the highest portion of the hemithorax leading to the presence of a lucency in the anteromedial, subpulmonic, and lateral basilar space adjacent to the diaphragm.3,4,31 It is useful in confirming suspected pneumothorax on AP supine radiography in compromised patients, such as those in the intensive care setting.4 DOUBLE DENSITY SIGN On frontal chest radiographs, this sign presents as a curvilinear density projecting over the right retrocardiac FIGURE 3. Cervicothoracic sign. Frontal radiograph of the chest with a coned view (A, B) demonstrates a mass projecting over the right superior mediastinum with indistinct borders along its superior margin. Follow-up enhanced CT of the chest (C, D) reveals a mass extending from the cervical region into the anterior mediastinum representing a multinodular goiter. Conversely, in the case of a posterior mediastinal mass, the supralateral margins project above the level of the clavicles and are clearly defined on the frontal radiograph (E, F) in this patient (different patient from A, B) with a biopsy-proven ganglioneuroma (G, CT image).
FIGURE 2. Bulging fissure sign. PA (A) and lateral (B) radiographs of the chest show dense consolidation of the right middle lobe secondary to Klebsiella pneumoniae.
Although classically the bulging fissure sign is demonstrated with right upper lobe consolidation and bowing of the minor fissure posteroin-feriorly, the consolidated segment in this case is causing upward bowing of the minor fissure (arrows).
Marshall et al J Thorac Imaging Volume 21, Number 1, March 2006 78 r 2006 Lippincott Williams & Wilkins region, signifying left atrial enlargement (Fig. 9).32 The curvilinear line represents the inferolateral margin of the left atrium.33 The double density sign may be observed in patients without cardiac disease; however, there is a semiquantitative measurement to estimate the left atrial diameter and better estimate whether it is a real finding.33 Higgins et al found that, on PA radiographs of adult patients, if the left atrial dimension is defined as the distance from the midpoint of the double density to the inferior wall of the left mainstem bronchus, a distance greater than 7 cm was consistent with a diagnosis of left atrial enlargement, confirmed on echocardiography.33 The measurement was found to be an unreliable sign in the evaluation of pediatric patients with a double density sign on PA radiographs.33 FIGURE 4. Comet tail sign. Rounded atelectasis within the right lower lobe (A) and abutting the posterior pleural surface in a patient with previous asbestos exposure. There is adjacent pleural thickening (B, white arrow), calcified pleural plaques, and signs of volume loss with downward retraction of the major fissure. The swirling of the bronchovascular bundle is thought to resemble a comet’s tail.
FIGURE 5. Continuous diaphragm sign. Frontal radiograph demonstrates a horizontal linear lucency (arrows) overlying the inferior aspect of the heart, typical of pneumomediastinal air tracking posterior to the heart. Air is also seen within the upper mediastinum paralleling the trachea in this patient involved in a motor vehicle accident.
FIGURE 6. Crazy-paving sign. Axial CT of the chest shows thickening of the intralobular and interlobular septa with a superimposed background of ground-glass opacity in a patient with pulmonary alveolar proteinosis.
J Thorac Imaging Volume 21, Number 1, March 2006 Signs in Thoracic Imaging r 2006 Lippincott Williams & Wilkins 79 DOUGHNUT SIGN Although CT is widely accepted as the primary modality for detecting mediastinal lymphadenopathy, chest radiography is generally the initial radiographic investigation performed. On a normal lateral chest radiograph (Fig. 10A), the aortic arch and right and left pulmonary arteries are visualized in an ‘‘inverted horse-shoe’’ configuration.34 In the presence of subcarinal lymphadenopathy (Figs. 10B, C), the inferior portion of the ‘‘horseshoe’’ fills in.34 The lymphadenopathy appears as a mass posterior to the bronchus intermedius and inferior to the tracheal bifurcation, completing the rounded hilar ‘‘doughnut’’ density.34,35 FINGER-IN-GLOVE SIGN Tubular shadows of soft tissue opacity akin to gloved fingers are seen on thoracic radiographs (Figs.
11A, B) and CT (Fig. 11C) and typically originate in the upper lobes in a bronchial distribution.2–4,36,37 The tubular, fingerlike projections represent dilated, mucoid-impacted bronchi surrounded by aerated lung.
When an inciting stenosis or bronchial obstruction occurs, mucous glands will continue to produce fluid, while the secretions are continually taken to the site of narrowing by mucociliary transport.38 As the secretions become inspissated, debris accumulates distal to the point of obstruction, and bronchiectasis ensues. Visualization of the gloved fingers is made possible by collateral air drift through the interalveolar pores of Kohn and canals of Lambert aerating lung distal to the point of mucoid impaction. There are 2 broad etiologic categories: non-obstructive and obstructive.38 Non-obstructive causes, such as allergic bronchopulmonary aspergillosis (ABPA), asthma, or cystic fibrosis, are common considerations.
ABPA is seen most commonly in asthmatic patients and occurs after inhaled Aspergillus organisms are trapped in airway mucus, triggering subsequent type I and type III allergic reactions.38 The acute type I response results in bronchoconstriction, heightened vascular permeability, wall edema, and protracted mucus production, whereas the delayed type III response causes immunopathological damage to the involved bronchi.38 Mucoid impaction in the setting of cystic fibrosis is secondary to mucociliary dysfunction and thick mucous secretions. Benign (bron-chial hamartomas or lipomas) and malignant (broncho-genic carcinoma or carcinoid tumors) neoplasms are considerations in the obstructive category. Congenital obstructive causes, such as bronchial atresia, intralobar sequestration, or bronchogenic cysts, might also be considered.
FLEISCHNER SIGN Felix Fleischner was a German-born radiologist who immigrated to the United States during the second World War.39 The finding that bears his name refers to enlargement of the proximal pulmonary arteries on plain chest radiography (Fig. 12A) or angiography (Fig.
12B) secondary to pulmonary embolism.40,41 This occurs most commonly in the setting of massive pulmonary embolism (defined angiographically as involving 50% or more of the major pulmonary artery branches)42 and has a relatively low sensitivity in diagnosis.43 The enlargement FIGURE 7. CT halo sign. Enhanced CT of the chest in a patient with angioinvasive aspergillosis depicts a left upper lobe lesion with a thin ground glass halo (arrow), presumably relating to hemorrhage.
FIGURE 8. Deep sulcus sign. Anteroposterior supine radio-graph of the chest in a patient with bilateral pneumothoraces.
The costophrenic angles are lucent and extend inferiorly, signifying pleural air that has risen to the dependent portion of the thorax in these regions.
Marshall et al J Thorac Imaging Volume 21, Number 1, March 2006 80 r 2006 Lippincott Williams & Wilkins can be caused by massive embolus enlarging the luminal diameter of the proximal artery in the acute setting but can also be seen in the subacute to chronic setting due to elevated pressures in the pulmonary arterial circulation.44 An important ancillary finding is the abrupt tapering of the occluded pulmonary artery distally, creating the ‘‘knuckle sign.’’44 GOLDEN S SIGN Dr Ross Golden was an American-born radiologist who originally described this sign as a form of right upper lobe collapse associated with a central mass.45 When the right upper lobe bronchus is obstructed by an endobron-chial lesion, there is elevation and medial displacement of the minor fissure with proximal convexity of the fissure due to the mass (Fig. 13). This creates the ‘‘reverse S’’ characteristic of central obstructing bronchogenic carci-nomas.3,4,45 Given the same presentation of a proximal, obstructing endobronchial lesion within the left upper lobe bronchus with associated left upper lobe collapse, the upwardly retracted major fissure will follow an S-shaped contour along its length. Although initially used to describe signs of right upper lobe collapse, the FIGURE 9. Double density sign. Frontal radiograph (A) in a patient with known mitral stenosis shows a retrocardiac curvilinear density paralleling the right heart border, confirmed on the lateral view (B) to be due to left atrial enlarge-ment.
FIGURE 10. Doughnut sign. A normal lateral chest radiograph is shown depicting several of the anatomic landmarks used to localize a subcarinal mass (A: 1, posterior wall of the bronchus intermedius; 2, left upper lobe bronchus; 3, right upper lobe bronchus; 4, approximate level of the carina). Lateral radiograph of the chest (B) shows added density within the infrahilar window representative of subcarinal lymphadenopathy in a patient with non-Hodgkin lymphoma. Follow-up CT of the chest confirmed the finding of subcarinal lymphadenopathy (C).
J Thorac Imaging Volume 21, Number 1, March 2006 Signs in Thoracic Imaging r 2006 Lippincott Williams & Wilkins 81 Golden S sign can be applicable to atelectasis involving any lobe.46 HAMPTON HUMP On plain radiographs (Fig. 14A) and CT (Figs. 14B, C), pulmonary infarcts are typically multifocal, peripheral in location, contiguous with one or more pleural surfaces, and more commonly confined to the lower lungs.3,47,48 The apex of these rounded or triangularly shaped opacities may point toward the hilum.3 The opacities re-solve slowly over a period of several months, akin to ‘‘melting ice cubes,’’ and may leave a residual scar.3 The first documentation of the finding was made by Aubrey Otis Hampton, who was a practicing radiologist in the mid 1920s. He and his co-author Castleman first reported evidence of incomplete pulmonary infarction in the sett-ing of PE in the 1940s.47,48 Autopsy follow-up showed evidence of intra-alveolar hemorrhage without alveolar wall necrosis in the first 2 days of infarction. After 2 days, wall necrosis begins and eventually leads to pulmonary infarction and an organized scar.47,48 Hampton also observed that there were differences in the healing of these incomplete infarcts depending on their premorbid cardiac history.47,48 In patients without heart disease, the incomplete infarcts would generally heal without scarring, whereas patients with congestive failure were more likely to progress to infarction with a persis-ting pulmonary scar.47,48 When pulmonary embolism results in infarction, airspace opacities typically develop within 12 to 24 hours.3,48 FIGURE 11. Finger-in-glove sign. Frontal (A) and lateral (B) chest radiographs show multiple tubular opacities bilaterally (arrows), which were confirmed at CT (C) to be dilated, mucus-impacted bronchi. Bronchoscopy confirmed allergic bronchopulmonary aspergillosis.
FIGURE 12. Fleischner sign.
Frontal radiograph of the chest (A) in a different patient shows enlargement of the right interlobar artery (arrow). The follow-up angiogram (B) confirms the presence of multiple filling defects representing multiple emboli and a distended right interlobar artery (arrow).
Marshall et al J Thorac Imaging Volume 21, Number 1, March 2006 82 r 2006 Lippincott Williams & Wilkins HILUM OVERLAY SIGN In evaluating densities projecting over the hila on PA chest radiographs, it is important to evaluate whether the opacity blends with the normal hilar shadow of the proximal pulmonary arteries and obscures clear visualiza-tion or if the mass and hilar structures are overlapping but distinct from one another. If the hilar vessels are sharply delineated, then it can be assumed that the overlying mass is anterior (Fig. 15) or posterior to the centrally located vascular structures and the intervening aerated lung enables crisp visualization of the pulmonary vessels.12,13 However, when the mass is inseparable from the proximal pulmonary arteries, it is assumed that the structures are immediately adjacent to one another, with the lack of interposing air obscuring the margins of the normal hilar structures. This sign was originally described by Benjamin Felson.12,13 JUXTAPHRENIC PEAK SIGN This sign, first documented by Kattan et al in 1980,49 is an ancillary sign of upper lobe collapse, depicted as a triangular opacity projecting superiorly over the medial half of the diaphragm, at or near its highest point (Fig. 16). It is most commonly related to the presence of an inferior accessory fissure.49,50 Although the mechanism is not certain, one theory suggests it is due to the negative pressure of upper lobe atelectasis causing upward retraction of the visceral pleura and extrapleural fat protruding into the recess of the fissure.49 A recent retrospective analysis of patients who have undergone upper lobectomies suggests that the prevalence of the sign increases in the ensuing weeks after intervention and documents its utility in specifically recognizing the type of surgery.51 LUFTSICHEL SIGN The word Luftsichel is German for ‘‘air crescent.’’ The finding is seen in the setting of left upper lobe collapse. Due to the absence of a minor fissure on the left, as the left upper lobe collapses, the major fissure assumes a vertical position roughly parallel to the anterior chest wall.4 As volume loss progresses, the fissure continues to migrate more anteriorly and medially until the atelectatic lobe is contiguous with the left heart border, effectively obliterating its contour on the frontal radiograph (Figs. 17A–C). With movement of the apical upper lobe segment anteromedially, the superior segment of the left lower lobe hyperinflates and fills the vacated apical space.4,52 Occasionally this segment will insinuate itself between the aortic arch and the collapsed left upper lobe creating a sharp outline, or periaortic lucency, described as the Luftsichel.3,4,52 SCIMITAR SIGN On PA chest radiography, a characteristic broad, curved shadow extends inferiorly toward the diaphragm along the right side of the heart (Figs. 18A, B).53 Its appearance resembles a Turkish sword and signifies a partial anomalous pulmonary venous return, most commonly to the infradiaphragmatic inferior vena cava (Figs. 18C–E). It is part of the congenital hypogenetic lung (scimitar) syndrome.4,53 SIGNET RING SIGN Described on chest CT (Fig. 19, arrows) as a circle of soft-tissue attenuation abutting a lucent ring, this sign depicts the pulmonary artery lying adjacent to a dilated FIGURE 13. Golden S sign. Patient with right upper lobe collapse. Frontal chest radiograph (A) shows an abnormal convexity in the right perihilar region creating a rounded inferior border with the displaced minor fissure and the ‘‘reverse S’’ of right upper lobe collapse. The CT correlates to this finding (B) with retraction of the minor fissure superiorly and collapse of the involved segment medially against the mediastinum due to an adenocarcinoma occluding the right upper lobe bronchus.
J Thorac Imaging Volume 21, Number 1, March 2006 Signs in Thoracic Imaging r 2006 Lippincott Williams & Wilkins 83 bronchus in cross-section.54–57 The sign is found on CT in patients with bronchiectasis, or irreversible, abnormal bronchial dilatation.56 Accompanying findings, such as peribronchial thickening, lack of bronchial tapering, and visualization of bronchi within 1 cm of the pleura, are all contributing findings confirming the diagnosis. In isola-tion, an enlarged bronchoarterial ratio may be a spurious finding, as bronchial diameter varies with parameters such as altitude.58 Measurement of the bronchial luminal diameter is made between the inner walls so as not to overestimate in cases with associated peribronchial thickening. Bronchiectasis occurs due to bronchial wall injury, most often due to necrotizing viral or bacterial bronchitis.4,57,58 It can also occur secondary to bronchial obstruction with subsequent mucus production dilating the involved airways, as mentioned previously with the FIGURE 14. Hampton hump. Frontal radiograph of the chest (A) depicts a peripheral opacity overlying the lateral aspect of the right lower lobe. Follow-up enhanced CT of the chest (B) shows a wedge-shaped peripheral opacity within the right lower lobe caused by infarction due to pulmonary embolus. Mediastinal windows (C) of the same patient show multiple occlusive emboli within the segmental branches of the lower lobes in keeping with pulmonary emboli.
FIGURE 15. Hilum overlay sign. Frontal radiograph of the chest demonstrates a lobulated mass projecting over the left hilum (A). The proximal left pulmonary artery and its branches are clearly visua-lized through the mass, and the lateral view (B) confirms the anterior location of the mass. Enhanced CT of the chest (C, D) further characterizes the mass confirmed on resection to be a thym-oma.
Marshall et al J Thorac Imaging Volume 21, Number 1, March 2006 84 r 2006 Lippincott Williams & Wilkins finger-in-glove sign. Finally, in the setting of pulmonary fibrosis or radiation-induced lung injury, traction on the adjacent bronchi will cause dilatation. Other entities, such as cystic fibrosis and Kartagener syndrome, are also associated.4,57 SILHOUETTE SIGN This classic roentgenographic sign first described by Felson in 1950 states that ‘‘an intrathoracic lesion touching a border of the heart, aorta, or diaphragm will obliterate that border on the roentgenogram. An in-trathoracic lesion not anatomically contiguous with a border of one of these structures will not obliterate that border.’’13,59,60 The underlying premise of the silhouette sign is that visualization of a roentgen shadow depends on a difference in radiographic density between adjacent tissues.13 For example, obscuration of the right heart border is often used to differentiate a right middle lobe process from a lower lobe abnormality (Fig. 20).3,4,60 Similarly, if the left heart border is partially or completely obliterated, the lingula is the region of involvement.
Conversely, the absence of the silhouette sign can help in further localizing a lesion. With lower lobe disease, the right or left heart border on the side of involvement is preserved, whereas the silhouette of the hemidiaphragm is obliterated in cases of lower lobe collapse or consolida-tion (Fig. 21). The basic tenets of the silhouette sign are used to explain various other signs discussed previously, such as the cervicothoracic and hilum overlay sign.
SPLIT PLEURA SIGN Separation and enhancement of the visceral and parietal pleural layers on CT (Fig. 22) is considered strong evidence of empyema.3,61 Normally, individual pleural layers are not discernable as discrete structures.3 Empyemic fluid fills the pleural space, resulting in thickening and enhancement of the pleura with a denotable separation.3,61 TREE-IN-BUD SIGN Best seen on high resolution computed tomography (HRCT) (Fig. 23), this sign appears as small, peripheral, centrilobular soft tissue nodules connected to multi-ple contiguous, linear branching opacities.3,4,62,63 This FIGURE 16. Juxtaphrenic peak sign. Frontal radiograph of the chest shows a ‘‘peaked’’ appearance to the middle portion of the right hemidiaphragm (arrow) secondary to volume loss in the right upper lobe. Upper lobe collapse was secondary to a bronchogenic carcinoma obstructing the right upper lobe bronchus.
FIGURE 17. Luftsichel sign. Left upper lobe collapse secondary to bronchogenic carcinoma causing compensatory hyperinflation of the left lower lobe and a crescent-shaped periaortic lucency (A) representing the expanded superior segment of the left lower lobe. The lateral view confirms the flattened left upper lobe anteriorly (B). Follow-up enhanced CT depicts the collapsed upper lobe against the mediastinum with compensatory hyperinflation of the left lower lobe (C).
J Thorac Imaging Volume 21, Number 1, March 2006 Signs in Thoracic Imaging r 2006 Lippincott Williams & Wilkins 85 radiologic term represents the mucous plugging, bron-chial dilatation, and wall thickening of bronchioli-tis.3,4,62,63 The histopathological correlate demonstrates small airway plugging with mucus, pus, or fluid, with dilated bronchioles, peribronchiolar inflammation, and wall thickening. Anatomically, this sign relates to viscous fluid blocking the intralobular bronchiole of the second-ary pulmonary lobule.63 Ordinarily the intralobular bronchiole is not visualized at HRCT (<1 cm); however, when filled with fluid and inflamed, the bronchiole becomes visible at the resolution of thin-slice CT.63 Although commonly associated with infectious etiologies such as endobronchial spread of Mycobacterium tubercu-losis, recent evidence suggests that the sign is not specific for any one pulmonary disease, but includes other infectious entities such as Pneumocystis jiroveci (Pneumo-cystis carinii) pneumonia and invasive pulmonary asper-gillosis. In addition, various immunodeficiency states, congenital disorders (cystic fibrosis), malignancy (lym-phoma), panbronchiolitis,62 and aspiration are other etiologies in which tree-in-bud may be seen and, thus, clinical correlation is important in elucidating a diag-nosis.3,4,62,63 WESTERMARK SIGN Neil Westermark was a 20th-century German radiologist who first discovered that a certain subset of patients diagnosed with pulmonary embolism (PE) were FIGURE 18. Scimitar sign. Frontal radio-graph of the chest (A) shows a curvi-linear opacity paralleling the right heart border representing an anomalous pul-monary vein. Graphic depiction of the curvilinear anomalous vein and its usual common entry into the inferior vena cava (B). A follow-up unenhanced CT (C, D) and cavagram (E) for central line placement confirmed a supradiaphrag-matic insertion of the anomalous vein into the inferior vena cava just prior to its entry into the right atrium. Used with permission, Canadian Association of Radi-ologists Journal, April 2005.
FIGURE 19. Signet ring sign. Enlarged segmental bronchi (white arrows) adjacent to their paired segmental arteries are seen in this patient with marked cystic bronchiectasis due to Williams-Campbell syndrome.
Marshall et al J Thorac Imaging Volume 21, Number 1, March 2006 86 r 2006 Lippincott Williams & Wilkins FIGURE 20. Silhouette sign, right mid-dle lobe pneumonia. Initial frontal (A) and lateral (B) radiographs in a patient with clinical suspicion of pneumonia demonstrate obliteration of the right heart border.
Follow-up radiographs the next day (C, D) illustrate dense opacification on the lateral view and persisting loss of the right heart border, confirming the presence of a right middle lobe pneumococcal pneumonia.
FIGURE 21. Silhouette sign, left lower lobe collapse. Frontal (A) and lateral (B) radiographs demonstrate features con-sistent with left lower lobe collapse secondary to an infrahilar mass on the left. The left hemidiaphragm is obscured on the frontal view with retrocardiac opacity. The left heart border is pre-served, suggesting lower lobe involve-ment with sparing of the lingula. The lateral view confirms the signs of left lower collapse with upward retraction of the diaphragm and obscuration of the left hemidiaphragm posteriorly due to an underlying pleural effusion. Findings were due to recurrent non-small cell lung carcinoma.
J Thorac Imaging Volume 21, Number 1, March 2006 Signs in Thoracic Imaging r 2006 Lippincott Williams & Wilkins 87 complicated by pulmonary infarction.64 He described the ‘‘anemic’’ or oligemic peripheral regions of lung parench-yma as ‘‘wedge-shaped shadows.’’ Interestingly, he also found that the majority of patients with PE did not have pulmonary infarcts, a finding that has been affirmed in the more recent literature.47 The chest radiograph (Fig. 24A) and CT (Fig. 24B) findings of increased translucency (chest radiograph) or hypoattenuation (CT) correspond-ing to oligemia in the periphery of the lung distal to an occlusive arterial embolus is typical.3,4,64 Visualization typically signifies either occlusion of a larger lobar or segmental artery or widespread small vessel occlusion.4,64 ACKNOWLEDGMENTS The authors would like to thank Mr Ronald Hildebrandt for his assistance in preparing the images for FIGURE 22. Split pleura sign. Enhanced CT of the chest shows a low attenuating pleural fluid collection juxtaposed between the enhancing visceral (white arrow) and parietal (black arrow) pleura, confirming an empyema in this patient.
FIGURE 23. Tree-in-bud sign. Multiple peripheral bud-like opacities (arrow) in a patient with bronchiolitis representing impacted small airway bronchioles.
FIGURE 24. Westermark sign. Frontal radiograph (A) and an enhanced CT of the chest (B) demonstrate lucency within the right upper lobe representing oligemia secondary to pulmo-nary embolus.
Marshall et al J Thorac Imaging Volume 21, Number 1, March 2006 88 r 2006 Lippincott Williams & Wilkins the article. Special thanks to Heather Marshall for designing the schematic diagram of the scimitar vein.
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40. Fleischner FG. Roentgen diagnosis of pulmonary embolism. Heart Bull. 1961;10:104–107.
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Marshall et al J Thorac Imaging Volume 21, Number 1, March 2006 90 r 2006 Lippincott Williams & Wilkins |
14362 | https://flexbooks.ck12.org/cbook/ck-12-interactive-middle-school-math-7-for-ccss/section/4.5/primary/lesson/solving-markup-and-markdown-problems-with-percents-4424719-msm7-ccss/ | Solving Markup and Markdown Problems with Percents | CK-12 Foundation
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4.5
Solving Markup and Markdown Problems with Percents
FlexBooks 2.0>
CK-12 Interactive Middle School Math 7>
Solving Markup and Markdown Problems with Percents
Written by:Lori Jordan
Fact-checked by:The CK-12 Editorial Team
Last Modified: Sep 01, 2025
Lesson Review Asked on Flexi Related Content
Lesson
Retro Video Games
[Figure 1]
In business, an understanding of markups and markdowns is vital to achieving success. In this chapter, you will be assuming the role of a store owner who buys and resells retro video games and video game consoles.Here, you will be exploring some of the ways that you as a business owner c ould use your math skills to increase your profit and sales.
Discussion Questions
What are some ways that can increase the profit you make on an item?
What are some ways that you could bring in more customers to increase sales?
Markups
Many businesses buy their merchandise from wholesalers and resell. Since you don’t make the video games you sell, you need to buy them. To make a profit, you must sell the video games at a higher price than the original price at which you purchased them. The amount which you increase the price to cover the cost is known as markup. For example, if you buy a video game for $20, you must sell it at more than $20 to make a profit.
The amount you increase the price by can be expressed as a percent of the purchase price. I ncreasing the price of an item by 25% is the same as selling it at 100% + 25% = 125% the price at which you purchased it. You can write this using decimals by saying that a markup of 25% is the same as 0.25 + 1 where 1 represents 100% of the purchase price and 0.25 is the 25% markup.In this example, the 25% which the price was increased is known as a markup rate. The selling price would be determined using the equation part = percent⋅whole.
Part:selling price
Percent: 1 + markup rate
Whole:purchase price
Discussion Question
What does the 1 in the percent represent?
Substituting these values into the percent equation gives you the following formula for markups:
selling price = (1 + markup rate)⋅purchase price
Example
You purchase a Nintendo 64 for $45 and plan to resell it for $63. What would the markup rate on this console be?
Use the formula above:
selling price = (1 + markup rate)⋅purchase price
Selling Price: $63
Markup Rate: x
Purchase Price: $45
63=(1+x)⋅45 63 45=1+x 1.4=1+x 0.4=x
The markup rate is 0.4. However, this formula gives us the percent as a decimal.You must convert 0.4 to a percent.
Answer: The markup rate is 40%.
Use the interactive below to e xplore how markup rates affect the sales price of a product.
INTERACTIVE
Markups
Use the sliders to change the Selling Price and the Markup Rate.
See how these affect the Purchase Price.
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Markdown s
A strategy that many businesses use to increase sales is to offer markdowns. A markdown can occur to bring in business, because a product can no longer be sold at its current price, etc. A markdown is an amount by which you decrease the selling price. The amount that you decrease the price by can be expressed as a percent of the selling price, known as the markdown rate.
The selling price would be determined using the equation:part = percent⋅whole.
Percent: 1 - markdown rate
Whole:original price
Part:selling price
Substituting these values into the percent formula gives you the following formula for markdowns:
selling price = (1−markdown rate)⋅original price
Example
You are selling a video game that you initially bought for $30. You decide to reduce the selling price to $21 in hopes to increase the likelihood of having someone buy it from you. What is the markdown rate on the game?
Selling Price: $21
Markdown Rate:x
Original Price: $30
21=(1−x)⋅30 21 30=1−x 0.7=1−x−0.3=−x 0.3=x
The markdown rate is 0.3. However, this formula gives us the percent as a decimal. What is the markdown rate written as a percentage?
Use the interactive below to explore how markdown rates affect the sales price of a product.
INTERACTIVE
Markdowns
Use the sliders to change the Selling Price and the Markdown Rate.
See how these affect the Original Price.
Try It
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Please check your internet connection and try again.
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No
Consecutive Markdowns
Stores will often offer consecutive discounts from multiple promotions.
Example
A thrift store was selling an Atari 2600 for $80 before deciding to have a 40% off promotion. A customer comes in to buy the console and pays with a 10% off any one item coupon. The coupon states that the 10% off coupon applies after any in-store promotions. The customer expects to get 50% off overall and pay $4 0.00. Is the customer correct?
The customer may expect to get 50% off the original price. However, this is not the case. Consecutive markdowns occur one after the other rather than at the same time. To find the price that the customer will pay we must use the markdown formula twice.
selling price = (1−markdown rate)⋅original price
Selling Price: x
Markdown Rate: 40% or 0.4 as a decimal
Original Price: 80
x=(1−0.4)⋅80 x=80(0.6)x=48
Now apply the 10% off coupon.
selling price = (1−markdown rate)⋅original price
Selling Price: x
Markdown Rate: 10% or 0.1 as a decimal
Original Price: 48
x=(1−0.1)⋅48 x=48(0.9)x=43.20
Discussion Question
How did applying the discount after the promotion benefit the business?
When taking a percent of a percent,you can rewrite the formula as follows:
selling price = (1−markdown rate 1)⋅(1−markdown rate 2)⋅original price
Use the interactive below to explore how consecutive markdown rates affect the sales price of a product.
INTERACTIVE
Consecutive Markdowns
Use the sliders to change the Selling Price and the Markdown Rates.
See how these affect the Original Price.
Try It
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Summary
Use the formula: selling price=(1+markup rate)×purchase price to solve problems involving markups.
Use the formula: selling price=(1+markdown rate)×original price to solve problems involving markups.
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Asked by Students
Here are the top questions that students are asking Flexi for this concept:
What is the sale price of a trampoline originally priced at $78.00 after 20% off?
To find the sale price, we need to calculate 20% of the original price and then subtract that from the original price.
First, find 20% of $78.00:
0.20×78.00=15.60
Then, subtract that amount from the original price:
78.00−15.60=62.40
So, the sale price of the trampoline is @$$62.40@$.
How to calculate selling price using markup percentage?
Use the equation
part=(1+markup rate)×whole
In this case, the original price (whole) is $6.43.
The markup rate is 38%, as a decimal 0.38.
part=(1+0.38)×6.43 part=(1.38)×6.43=8.87
So the sale price will be $8.87.
What is the markup price?
The markup price is the amount added to the cost price of a product or service to determine its selling price. It represents the difference between the cost of producing or acquiring an item and the price at which it is sold to customers. The markup price includes the desired profit margin for the seller.
To calculate the markup price, we need the cost price (C)and the markup percentage (M).The formula is:
Markup Price=C+(C×M)
What is the process for calculating percent markup?
To calculate the percent markup on selling price, you need to first determine the cost of the item.
Once you know the cost, subtract it from the selling price to find the profit. Then, divide the profit by the cost and multiply by 100 to get the percentage markup.
Use the equation
markup percent=profit cost×100
A sporting goods store manager was selling a ski set for a certain price. The manager offered the markdowns shown, making the one-day sale price of the ski set $326. Find the original selling price of the ski set.
The problem doesn't provide the markdowns that were applied to the original price of the ski set. Without this information, it is impossible to determine the original selling price.
Overview
Use the formula: selling price=(1+markup rate)×purchase price to solve problems involving markups.
Use the formula: selling price=(1+markdown rate)×original price to solve problems involving markups.
Vocabulary
Amount
Markup
rate
Percent Equation
reduce
percentage
Multiple
Discount
Test Your Knowledge
Question 1
Figure out the sale price if the markup is 20% on $12.00.
a
$10.71
b
$15
c
$14.40
d
$9.60
Check It
Use the equation
part=(1+markup rate)×whole
In this case, the whole is $12.00.
The markup rate is 20%, as a decimal 0.20.
part=(1+0.2)×12 part=(1.2)×12=14.4
So, the new price will be $14.40
FlexCard™
Question 2
Calculate the new price. Original price: $200.00, discount 35%
a
$70
b
$307.69
c
$270
d
$130
Check It
Use the equation
part=(1−markdown rate)×whole
In this case, the whole is $200.
The markdown rate is 35%, as a decimal 0.35.
part=(1−0.35)×200=(0.65)×200=130
So, the price of the item would be $130
FlexCard™
Asked by Students
Ask your question
Here are the top questions that students are asking Flexi for this concept:
What is the sale price of a trampoline originally priced at $78.00 after 20% off?
To find the sale price, we need to calculate 20% of the original price and then subtract that from the original price.
First, find 20% of $78.00:
0.20×78.00=15.60
Then, subtract that amount from the original price:
78.00−15.60=62.40
So, the sale price of the trampoline is @$$62.40@$.
How to calculate selling price using markup percentage?
Use the equation
part=(1+markup rate)×whole
In this case, the original price (whole) is $6.43.
The markup rate is 38%, as a decimal 0.38.
part=(1+0.38)×6.43 part=(1.38)×6.43=8.87
So the sale price will be $8.87.
What is the markup price?
The markup price is the amount added to the cost price of a product or service to determine its selling price. It represents the difference between the cost of producing or acquiring an item and the price at which it is sold to customers. The markup price includes the desired profit margin for the seller.
To calculate the markup price, we need the cost price (C)and the markup percentage (M).The formula is:
Markup Price=C+(C×M)
What is the process for calculating percent markup?
To calculate the percent markup on selling price, you need to first determine the cost of the item.
Once you know the cost, subtract it from the selling price to find the profit. Then, divide the profit by the cost and multiply by 100 to get the percentage markup.
Use the equation
markup percent=profit cost×100
A sporting goods store manager was selling a ski set for a certain price. The manager offered the markdowns shown, making the one-day sale price of the ski set $326. Find the original selling price of the ski set.
The problem doesn't provide the markdowns that were applied to the original price of the ski set. Without this information, it is impossible to determine the original selling price.
Related Content
Prices Involving Discounts
Original Price, Selling Price, Discount, and Percentage Discount - Overview
Discounts - Overview
Let's Make a Deal
A Bridal Bargain
Save Money with Coupons!
Back to Solving Markup and Markdown Problems with Percents
| Image | Reference | Attributions |
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| | [Figure 1] | Credit:Evan-Amos Source:Evan-Amos License:CC0 / Public Domain |
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14363 | https://www.youtube.com/watch?v=JSEPDJfl8m8 | Finding Angle Between 2 Lines (Formula)
Mario's Math Tutoring
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Posted: 18 May 2016
Learn how to find the angle between two lines using the formula we will go over in this video. We also go through 2 example problems in this free math video tutorial by Mario's Math Tutoring.
Timestamps:
00:00 Intro
0:22 Formula for finding the angle between 2 lines
0:39 Example 1
1:48 Example 2 Rewriting the equations in slope intercept form to identify the slope first then using those slopes in the formula.
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Transcript:
Intro hi this is Mario with Mario's math tutoring coming to you with another math video to help you boost your score in your math class improve your understanding and hopefully make learning math a lot less stressful so what we're going to do together in this video is we're going to talk about how to find the angle between two lines okay using the formula okay that we're gonna be discussing in this video so you've got your two lines and you're trying to find that angle between the two lines so Formula for finding the angle between 2 lines here's our formula it's the tangent of theta equals the absolute value of the difference between the two slopes okay of the two lines divided by one plus the product of the two slopes of the two lines so let's take a look at some examples and you'll understand a bit better how this works so in this first Example 1 example our two lines are y equals 2x minus 3 and y equals 5x plus 1 I've graphed the two lines here for us and what we can see is that we're trying to find that angle between the two lines so let's go ahead and apply our formula so you can see that the slopes are two and five so that's where we're going to be using in our formula there so we've got tangent of theta equals 5 minus 2 ok over 1 plus 5 times 2 okay so if we simplify that down we get 3 over okay 10 plus 1 which is 11 okay absolute value of that is still going to be positive 3 11s that's the tangent of theta then what we have to do is just do the tangent inverse okay to find theta to find the angle so it's the tangent inverse of 3 11s let's go to the calculator let's just see what that is so we've got tangent inverse of 3 divided by 11 and we get M 15.3 degrees so that looks about right it's a real small angle here between the two lines so that's going to be 15 point three degrees okay let's look at another example so in this example we know the Example 2 Rewriting the equations in slope intercept form to identify the slope first then using those slopes in the formula. slope of this line is 4 we don't know the slope of the line above here so we're going to rewrite this in slope intercept form so that's going to be negative 6y equals negative 3x plus 2 we're going to divide everything by negative 6 so that's going to give us y equals 1/2 X okay minus 1/3 and you can see the slope of this line is 1/2 the other line it's 4 is going to be 4 so let's go ahead and use our formula so the tangent of theta equals 4 minus 1/2 over 1 plus 4 times 1/2 okay so if we simplify that down what do we get we get this is 8 halves minus 1/2 is 7 halves K over 4 times 1/2 is 2 plus 1 is 3 okay and so then you can see this is actually going to be let's see when you divide by a fraction it's like multiplying by the reciprocal right so this is going to be 7 6 and we're taking the tangent theta equals 7 6 we want to take the inverse tangent to find the angle so theta is going to be tangent inverse of 7 divided by 6 that will give us the angle between the two lines so let's go to the calculator let's see what that comes out to we've got tangent inverse see tangent inverse of 7 sixths and we've got 49 point four degrees so that's our angle so theta equals 49 point four all right so I hope you learn how to work with this formula to find the angle between two lines go ahead and subscribe to the channel check out some of my past videos and I look forward to seeing you in the future ones I'll talk to you soon |
14364 | https://medsys.ai/en/blog/pericarditis-vs-myocarditis-chest-pain-cardiac-markers-inflammation.html | Medsys AI- Pericarditis vs. Myocarditis: Differentiating Inflammatory Origins of Chest Pain and Cardiac Markers
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Pericarditis vs. Myocarditis: Differentiating Inflammatory Origins of Chest Pain and Cardiac Markers
Pericarditis and myocarditis are two inflammatory conditions of the heart that, while sharing some symptoms, have distinct origins and characteristics. Pericardial inflammation affects the pericardium, the membrane surrounding the heart, whereas myocarditis involves the myocardium, the heart muscle itself. Differentiating between these two conditions is crucial for accurate diagnosis and treatment, especially in patients presenting with chest pain and other cardiac symptoms.
Diving Deeper into the Differences
Chest pain is a common symptom in both conditions, but its presentation can vary. In pericarditis, the pain is often sharp and worsens with deep breathing or lying down, while in myocarditis, the pain may be more diffuse and accompanied by symptoms of heart failure, such as dyspnea and fatigue. The use of electrocardiograms (ECG) is essential for identifying specific patterns, such as ST-segment elevation in pericarditis, although these changes can be dynamic and not always present.
Cardiac markers, such as troponin, are useful for assessing myocardial damage. In myocarditis, troponin levels are typically elevated due to myocardial injury, whereas in pericarditis, these levels may be normal or only slightly elevated. The interpretation of troponin levels is crucial to avoid diagnostic errors.
Cardiac magnetic resonance imaging (CMR) is a valuable tool for differentiating between these conditions, as it allows for non-invasive visualization of inflammation and tissue damage. CMR can identify both pericardial and myocardial inflammation, aiding in the distinction between pericarditis and myocarditis, as well as assessing the extent of damage.
Conclusions
Differentiating between pericarditis and myocarditis is essential for appropriate clinical management. Although both conditions may present with chest pain and elevated cardiac markers, differences in clinical presentation, ECG findings, and CMR results are key to accurate diagnosis. Understanding these differences enables physicians to provide more specific treatment and improve patient outcomes.
References
Natural Course of Electrocardiogram Changes and the Value of Multimodality Imaging in Acute Pericarditis
The Main Causes and Mechanisms of Increase in Cardiac Troponin Concentrations Other Than Acute Myocardial Infarction
The prominent role of cardiac magnetic resonance imaging in coronary artery disease
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14365 | https://mathbooks.unl.edu/AppliedCalculus/sec-3-3-absolute-extrema.html | Applied Calculus
Kevin Gonzales, Eric Hopkins, Catherine Zimmitti, Cheryl Kane, Modified to fit Applied Calculus from Coordinated Calculus by Nathan Wakefield et. al., Based upon Active Calculus by Matthew Boelkins
(\newcommand{\dollar}{\$} \DeclareMathOperator{\erf}{erf} \DeclareMathOperator{\arctanh}{arctanh} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} \definecolor{fillinmathshade}{gray}{0.9} \newcommand{\fillinmath}{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}} )
Section 4.3 Absolute Extrema
Motivating Questions
What are the differences between finding relative extreme values and absolute extreme values of a function?
How does the process of finding the absolute maximum or minimum of a function change when we are finding the extrema on a closed interval versus an open interval?
What are the possible points at which extreme values occur? When can we guarantee that a function will have an absolute maximum or minimum?
This section corresponds to 3.6 Absolute Extrema in the workbook.
We have seen that we can use the first derivative of a function to determine where the function is increasing or decreasing, and the second derivative to know where the function is concave up or concave down. This information helps us determine the overall shape and behavior of the graph as well as whether the function has relative extrema. In this section we will explore when a function has an absolute maximum or minimum.
Absolute Maximum and Minimum.
Given a function (f\text{,}) we say that (f(c)) is an absolute maximum of (f) provided that (f(c) \ge f(x)) for all (x) in the domain of (f\text{.}) Similarly, we call (f(c)) an absolute minimum of (f) whenever (f(c) \le f(x)) for all (x) in the domain of (f\text{.})
The difference between a relative maximum and an absolute maximum: there is a relative maximum of (f) at (x = c) if (f(c) \ge f(x)) for all (x) near (c\text{,}) while there is an absolute maximum at (c) if (f(c) \ge f(x)) for all (x) in the domain of (f\text{.})
For instance, in Figure 4.3.1 at the right, we see a function (f) that has an absolute maximum at (x = c) and a relative maximum at (x = a\text{,}) since (f(c)) is greater than or equal to (f(x)) for every value of (x\text{,}) while (f(a)) is only greater than or equal to the value of (f(x)) for (x) near (a\text{.}) Since the function appears to decrease without bound, (f) has no absolute minimum, although there is a relative minimum at (x = b\text{.})
Our emphasis in this section is on finding the absolute extreme values of a function (if they exist), either over its entire domain or on some restricted portion.
Example 4.3.2.
Consider the function (h) given by the graph in Figure 4.1.5 below. Use the graph to answer each of the following questions.
Identify all of the values of (c) for which (h(c)) is a relative maximum of (h\text{.})
Identify all of the values of (c) for which (h(c)) is a relative minimum of (h\text{.})
Does (h) have an absolute maximum on the interval ([-3,3]\text{?}) If so, what is the value of this absolute maximum?
Does (h) have an absolute minimum on the interval ([-3,3]\text{?}) If so, what is its value?
Identify all values of (c) for which (h'(c) = 0\text{.})
Identify all values of (c) for which (h'(c)) does not exist.
True or false: every relative maximum and minimum of (h) occurs at a point where (h'(c)) is either zero or does not exist.
True or false: at every point where (h'(c)) is zero or does not exist, (h) has a relative maximum or minimum.
Hint.
Start by looking for (x)-values where the graph switches from increasing to decreasing. Are there any other points that might work?
Start by looking for (x)-values where the graph switches from decreasing to increasing. Are there any other points that might work?
Is there a highest point on the graph?
Is there a lowest point on the graph?
Remember that horizontal lines have a slope of (0\text{.})
Think back to earlier chapters. What characteristics of (h) or its graph might make it non-differentiable at a certain point?
Compare your answers to (a) and (b) with those to (e) and (f).
Compare your answers to (a) and (b) with those to (e) and (f).
Answer.
(c=-2) and (c=1\text{.})
(c=0\text{.})
(h(1)=2\text{.})
(h(-3)=-3\text{.})
(c=-2.5) and (c=1\text{.})
(c=-2\text{,}) (c=0\text{,}) and (c=1.5\text{.})
True.
False.
Solution.
The graph shows that (h(x)\le-1=h(-2)) for every (x) near (-2) and (h(x)\le2=h(1)) for every (x) near (1\text{.}) We notice that these are points where (h) switches from increasing to decreasing. Since there are no other points on the graph where the (y)-coordinate is larger than everything nearby, it follows that the only relative maxima of (h) on this domain occur at (c=-2) and (c=1\text{.})
The graph shows that (h(x)\ge-2=h(0)) for every (x) near (0\text{,}) a point where (h) switches from decreasing to increasing. Since this is the only point shown where this occurs, we say that the only relative minimum of (h) on this domain occurs at (c=0\text{.})
The highest point shown on the graph is the point ((1,2)\text{.}) Since (h) is continuous on ([-3,3]\text{,}) we don’t have to worry about any points on the interval being higher and not showing up on the displayed grid. Hence we say the absolute maximum of (h) on the interval ([-3,3]) is (2) and occurs at (x=1\text{.})
The lowest point shown on the graph is the point ((-3,-3)\text{.}) Again, since (h) is continuous on ([-3,3]\text{,}) we don’t have to worry about any points on the interval being lower and not showing up in the window that is displayed. Thus we say the absolute minimum of (h) on the interval ([-3,3]) is (-3) and occurs at (x=-3\text{.})
It is worth noting that this was not a point that we found as a local minimum. The difference is that here we are restricting our focus to the interval that is shown, effectively cutting off the graph at its endpoints. Earlier, when we were looking for relative minima, we made the assumption that the graph of (y=h(x)) continues outside of the depicted interval by following the trends displayed and continuing downward on both sides.
5. When looking for points ((c,h(c))) at which (h'(c)=0\text{,}) we are looking for points of horizontal tangency. The graph flattens out at ((-2.5,-2)) and ((1,2)\text{,}) but nowhere else1
Note that at ((1.5,1)) it looks as though the graph is flat to the right of this point. However, the derivative is undefined here because of the cusp, so this is not a point of horizontal tangency
. Thus (h'(c)=0) exactly when (c=-2.5) and (c=1\text{.})
6. Recall that (h'(c)) only exists when all of the following are true:
(h(c)) is defined,
(\lim_{x\to c}h(x)) exists,
(\lim_{x\to c}h(x)=h(c)\text{,}) and
(\lim_{a\to0}\frac{h(c+a)-h(c)}a) exists.
Assuming (h) is continuous at (c) (i.e. the first three points hold), the main consequence of the final point is that a corner or a cusp at ((c,h(c))) makes (h) non-differentiable at (c\text{.})
With all this in mind, we can say that (h'(c)) does not exist at (c=-2\text{,}) (c=0\text{,}) or (c=1.5) because (h) has a cusp at these points.
7. This is true. Every relative extremum of (h) occurs at a point where (h'(c)) is either zero or does not exist.
8. This is false. Some points at which (h'(c)) is zero or undefined are not relative extrema. In particular, (h'(-2.5)) is zero, but the graph is non-decreasing near ((-2.5,-2)\text{.}) Similarly, (h'(1.5)) is undefined, but the graph is non-increasing near ((1.5,1)\text{.}) Neither of these points is a local maximum or minimum.
Subsection 4.3.1 Finding Absolute Extrema
We have seen how to find the relative extrema of a function (f(x)) using the critical points. We will now examine how to find the absolute extrema. We will start by examining cases of a function (f(x)) restricted to a closed domain ([a,b]\text{.}) By restricting the domain to an interval it makes sense that the endpoints of the interval will also be important to consider, as we see in the following example.
Example 4.3.4.
Let (g(x) = \frac{1}{3}x^3 - 2x + 2\text{.})
Find all critical numbers of (g) that lie in the interval (-2 \le x \le 3\text{.})
Use a graphing utility to construct the graph of (g) on the interval (-2 \le x \le 3\text{.})
From the graph, determine the (x)-values at which the absolute minimum and absolute maximum of (g) occur on the interval ([-2,3]\text{.})
How do your answers change if we instead consider the interval (-2 \le x \le 2\text{?})
What if we instead consider the interval (-2 \le x \le 1\text{?})
Hint.
Check that each critical number you find satisfies (-2 \le x \le 3\text{.})
desmos.com2
desmos.com/calculator
is a great choice.
On the graph, look for the lowest and highest possible values of the function.
Ask yourself the same questions as (a)-(c), simply using the new interval.
Ask yourself the same questions as (a)-(c), simply using the new interval.
Answer.
(x = \pm \sqrt{2} \approx \pm 1.414\text{.})
Below is the graph of (y=g(x)) replicated three times, with critical points and endpoints of each interval marked. On the left, we have the interval ([-2,3]) for (c); the middle shows the interval ([-2,2]) for (d); the right displays the interval ([-2,1]) for (e).
On ([-2,3]\text{,}) (g) has an absolute maximum at (x = 3) and an absolute minimum at (x = \sqrt{2}\text{.})
On ([-2,2]\text{,}) (g) has an absolute maximum at (x = -\sqrt{2}) and an absolute minimum at (x = \sqrt{2}\text{.})
On ([-2,3]\text{,}) (g) has an absolute maximum at (x = -\sqrt{2}) and an absolute minimum at (x = 1\text{.})
Solution.
Since (g'(x) = x^2 - 2\text{,}) the critical numbers of (g) are (x = \pm \sqrt{2} \approx \pm 1.414\text{,}) both of which lie in the interval (-2 \le x \le 3\text{.})
Below is the graph of (y=g(x)) replicated three times, with critical points and endpoints of each interval marked. On the left, we have the interval ([-2,3]) for (c); the middle shows the interval ([-2,2]) for (d); the right displays the interval ([-2,1]) for (e).
On ([-2,3]\text{,}) (g) has an absolute maximum at (x = 3) and an absolute minimum at (x = \sqrt{2}\text{.})
On ([-2,2]\text{,}) (g) has an absolute maximum at (x = -\sqrt{2}) and an absolute minimum at (x = \sqrt{2}\text{.})
On ([-2,3]\text{,}) (g) has an absolute maximum at (x = -\sqrt{2}) and an absolute minimum at (x = 1\text{.})
Example 4.3.4 showed how the absolute maximum and absolute minimum of a function on a closed, bounded interval ([a,b]) depend not only on the critical numbers of the function, but also on the values of (a) and (b\text{.}) In fact, the interval we choose has nearly the same influence on extreme values as the function under consideration. For instance, Figure 4.3.5 below depicts again the graph of (y=g(x)) that we looked at in Example 4.3.4.
From left to right, the interval under consideration is changed from ([-2,3]) to ([-2,2]) to ([-2,1]\text{.})
There are two critical numbers on the interval ([-2,3]\text{;}) the absolute minimum is at one critical number and the absolute maximum is at the right endpoint.
On the interval ([-2,2]\text{,}) both critical numbers are again in the interval. One critical value is still the absolute minimum; however, the absolute maximum is now at the other critical number rather than at the right endpoint.
Only one critical number lies on the interval ([-2,1]\text{.}) That critical number gives the absolute maximum; the absolute minimum is at the right endpoint.
These observations demonstrate several important facts that hold more generally.
The Extreme Value Theorem.
If (f) is a continuous function on a closed and bounded interval ([a,b]\text{,}) then (f) has both an absolute minimum and absolute maximum on ([a,b]\text{.})
The Extreme Value Theorem tells us that on any closed and bounded interval ([a,b]\text{,}) a continuous function has to achieve both an absolute minimum and an absolute maximum. The theorem does not tell us where these extreme values occur, only that they must exist. As we saw in Example 4.3.4, the only possible locations for relative extrema are at the endpoints of the interval or at a critical number. It is important to consider only the critical numbers that lie within the interval.
We now have the following approach to finding the absolute maximum and minimum of a continuous function (f) on the interval ([a,b]\text{:})
Find all critical numbers of (f) that lie in the interval.
Evaluate the function (f) at each critical number in the interval and at each endpoint of the interval.
From among these function values, the smallest is the absolute minimum of (f) on the interval, while the largest is the absolute maximum.
Example 4.3.6.
Find the absolute maximum and minimum of each function on the stated interval.
(f(x) =x^4-98x^2+10) on ([-6,15])
(h(x) = xe^{-x}) on ([0,3])
(q(x) = \frac{x^2}{x-2}) on ([3,7])
(h(x) = xe^{-ax}) on (\left[0, \frac{2}{a}\right]\text{;}) assume (a \gt 0)
Answer.
Absolute maximum: (28585\text{;}) Absolute minimum: (-2391\text{.})
Absolute maximum: (e^{-1}\text{;}) Absolute minimum: (0\text{.})
Absolute maximum: (9.8\text{;}) absolute minimum: (8\text{.})
Absolute maximum: ((ae)^{-1}\text{;}) absolute minimum: (0\text{.})
Solution.
Given (f(x)=x^4-98x^2+10\text{,}) it follows that (f'(x) = 4x^3-196x\text{,}) solving (f'(x)= 0=4x(x^2-49)) which gives three critical values: (x=0\text{,}) (x=7) and (x=-7\text{.})
Since we are looking at the interval (-6\leq x\leq 15) we will only test the critical values within this interval, which are (x=0,7\text{.})
Now we plug in these two critical values, along with then endpoints (x=-6) and (x=15) to find the absolute maximum and absolute minimum:
\begin{equation} f(-6)=(-6)^4-98(-6)^2+10=-2222 \end{equation}
\begin{equation} f(0)=(0)^4-98(0)+10=10 \end{equation}
\begin{equation} f(7)=(7)^4-98(7)^2+10=-2391 \end{equation}
\begin{equation} f(15)=(15)^4-98(15)^2+10=28585 \end{equation}
Thus the absolute maximum of (f(x)) is (28585) which occurs at (x=15\text{,}) and the absolute minimum of (f(x)) is (-2391) which occurs at (x=7\text{.})
2. For (h(x) = xe^{-x}\text{,}) we know that (h'(x) = e^{-x}+xe^{-x}(-1) = e^{-x}(1-x)\text{.}) Therefore the only critical number of (h) is (x = 1\text{.}) Next, we compute (h(1)\text{,}) (h(0)\text{,}) and (h(3)\text{.}) Observe that
(\displaystyle h(0) = 0)
(\displaystyle h(1) = e^{-1} \approx 0.36788)
(\displaystyle h(3) = 3e^{-3} \approx 0.14936)
Thus, on ([0,3]\text{,}) the absolute maximum of (h) is (e^{-1}) and the absolute minimum is (0\text{.})
3. With (q(x) = \frac{x^2}{x-2}\text{,}) we have
\begin{equation} q'(x) = \frac{2x(x-2) - x^2}{(x-2)^2} = \frac{x^2-4x}{(x-2)^2} = \frac{x(x-4)}{(x-2)^2}\text{.} \end{equation}
Hence the critical numbers of (q) are (x = 0) and (x = 4\text{.}) Only the latter critical number lies in the interval ([3,7]\text{,}) and thus we evaluate (q) and find
(\displaystyle q(3) = \frac{9}{1} = 9)
(\displaystyle q(4) = \frac{16}{2} = 8)
(\displaystyle q(7) = \frac{49}{5} = 9.8)
We now see that on ([3,7]) the absolute maximum of (q) is (9.8) and the absolute minimum is (8\text{.})
4. We start by differentiating (h\text{,}) finding (h'(x)=e^{-ax}(1-ax)\text{.}) Thus the only critical number of (h) is (\frac1a\text{,}) which is on the interval (\left[0,\frac2a\right]\text{.}) Evaluating (h) at the interval endpoints and the critical number, we find
(\displaystyle h(0)=0)
(\displaystyle h\left(\frac1a\right)=\frac1ae^{-1}\approx\frac{0.368}a\gt0)
(\displaystyle h\left(\frac2a\right)=\frac2ae^{-2}\approx\frac{0.271}a\gt0)
Thus on (\left[0,\frac2a\right]\text{,}) the absolute maximum of (h) is (\frac1ae^{-1}=(ae)^{-1}) and the absolute minimum is (0\text{.})
Up to this point, we have seen how to find the absolute maximum and minimum of a function on a closed interval. But what if we are not restricted to a closed domain ([a,b]\text{?})
In general, a function on an open domain, say ((0,\infty)) or ((-\infty,\infty)) may or may not have an absolute maximum or minimum. Will have to find the critical points and then examine the long term behavior of the function.
Example 4.3.7.
Let (f(x) = 2 + \frac{3}{1+(x+1)^2}\text{.})
Determine all of the critical numbers of (f\text{.})
Construct a first derivative sign chart for (f) and thus determine all intervals on which (f) is increasing or decreasing.
Does (f) have an absolute maximum? If so, why, and what is its value and where is the maximum attained? If not, explain why.
Determine (\lim_{x \to \infty} f(x)) and (\lim_{x \to -\infty} f(x)\text{.})
Explain why (f(x) \gt 2) for every value of (x\text{.})
Does (f) have an absolute minimum? If so, why, and what is its value and where is the minimum attained? If not, explain why.
Hint.
What is true about (f'(c)) when (c) is a critical number of (f\text{?})
Remember that on a given interval of the first derivative sign chart for (f\text{,}) the derivative (f') should have the same sign everywhere, so you only need to test at a single point on the interval.
What does the chart from (b) tell you? Thinking about the algebraic structure of (f(x)) can also support your reasoning.
What happens to (1+(x+1)^2) for large values of (x\text{?}) What happens to its reciprocal?
(1+(x+1)^2\gt0) for every value of (x\text{.})
Think about the long-run behavior of (f) that you found in (d).
Answer.
(-1) is the only critical number of (f\text{.})
(f) is increasing on ((-\infty,-1)) and decreasing on ((-1,\infty)\text{.})
(f) has an absolute maximum of (5) at the point ((-1,5)\text{.})
(\lim_{x\to\pm\infty}f(x)=2\text{.})
(\frac3{1+(x+1)^2}\gt0) for every value of (x\text{.})
(f) has a horizontal asymptote at (y=2) but no absolute minimum.
Solution.
We first notice that the domain of (f) is all real numbers, since the denominator of (f(x)) is always at least (1\text{:})
\begin{align} (x+1)^2\ge\mathstrut\amp0\ 1+(x+1)^2\ge\mathstrut\amp1\text{.} \end{align}
To differentiate (f\text{,}) we will first rewrite (f(x)) as
\begin{equation} f(x)=2+3\left(1+(x+1)^2\right)^{-1} \end{equation}
and then use the chain rule to find
\begin{align} f'(x)=\mathstrut\amp3\left((-1)\left(1+(x+1)^2\right)^{-2}\left(2(x+1)(1)\right)\right)\ =\mathstrut\amp\frac{-6(x+1)}{\left(1+(x+1)^2\right)^2}\text{.} \end{align}
The denominator of (f'(x)) is always at least (1) for the same reason that the denominator of (f(x)) is always at least (1\text{,}) so there are no points (x=c) for which (f'(c)) does not exist. Thus the only critical number(s) of (f) will occur when (f'(c)=0\text{.}) This happens when
\begin{align} -6(x+1)=\mathstrut\amp0\ x+1=\mathstrut\amp0\ x=\mathstrut\amp-1\text{.} \end{align}
Thus the only critical number of (f) is (c=-1\text{,}) with (f'(-1)=0\text{.})
2. As stated in (a), the denominator of (f'(x)) is always at least (1\text{,}) so in particular, it is always positive. On our first derivative sign chart, the only critical number we have is (c=-1\text{,}) which splits the domain of (f) and (f') into the intervals ((-\infty,-1)) and ((-1,\infty)\text{.}) Thinking of the factors of (f'(x)) as (-6\text{,}) ((x+1)\text{,}) and (\frac{1}{\left(1+(x+1)^2\right)^2}\text{,}) the signs of these at (x=-2) are “(--+)”, with the final product positive. Likewise, the signs of these factors at (x=0) are “(-++)”, making the final product negative. Therefore (f'(x)\gt0) on the interval (x\lt-1\text{,}) and (f'(x)\lt0) on the interval (x\gt-1\text{.}) Hence (f) is increasing on ((-\infty,-1)) and decreasing on ((-1,\infty)\text{.})
3. According to the first derivative test, the point ((-1,f(-1))) is a relative maximum of (f\text{.}) Since this is the only critical point of the function and (f) is continuous everywhere, there is nowhere else that (f) can change direction. It follows that this point is actually an absolute maximum.
From an algebraic standpoint, we have already observed that (1+(x+1)^2\ge1) for every value of (x\text{.}) Consequently,
\begin{equation} 0\lt\frac{3}{1+(x+1)^2}\le3 \end{equation}
for every value of (x\text{,}) and this quotient achieves its maximum of (3) when (x=-1\text{.}) Hence
\begin{equation} 2\lt2+\frac{3}{1+(x+1)^2}\le5 \end{equation}
for every value of (x\text{,}) so the maximum value that (f ) can attain is (5 \text{.}) Since (f(-1)=5\text{,}) we conclude that the absolute maximum of (f) is (5) at (x=-1\text{.})
4. The symmetry of (f) about (x=-1) allows us to compute both limits simultaneously. We first note that (\lim_{x\to\pm\infty}\left(1+(x+1)^2\right)=\infty) and (\lim_{x\to\pm\infty}\frac{1}{1+(x+1)^2}=0\text{.}) Thus
\begin{equation} \lim_{x\to\pm\infty}f(x)=\lim_{x\to\pm\infty}\left(2+\frac{3}{1+(x+1)^2}\right)=2+0=2\text{.} \end{equation}
This tells us the graph of (y=f(x)) has a horizontal asymptote of (y=2\text{.})
5. In (c), we stated the inequality
\begin{equation} 2\lt2+\frac{3}{1+(x+1)^2}\le5\text{.} \end{equation}
Since the quotient in the formula for (f(x)) is always strictly greater than zero, adding (2) to the quotient to obtain (f(x)) will always yield a value strictly greater than (2\text{.})
6. Since (f(x)\gt2) for every value of (x\text{,}) an absolute minimum of (f) (if it existed) would necessarily be larger than (2\text{.}) But we showed in (d) that (\lim_{x\to\pm\infty}f(x)=2\text{,}) meaning that (f) gets arbitrarily close to (2) as (x) gets large. So, there is no value larger than (2 ) that is an absolute minimum for (f\text{,}) and thus (f ) has no absolute minimum.
The Extreme Value Theorem on an Open Domain.
Suppose (f(x)) is a continuous function on a open interval, for example ((0,\infty)) or ((\infty,\infty)\text{.}) If (f(x)) has exactly one critical value (x=c) on the interval, then:
If the function has a relative maximum at (x=c\text{,}) then (f(x)) has an absolute maximum at (x=c) and no absolute minimum.
If the function has a relative minimum at (x=c\text{,}) then (f(x)) has an absolute minimum at (x=c) and no absolute maximum.
Example 4.3.8.
Find the absolute maximum and minimum of each function on the stated interval.
(f(x) =4x^2-16x+4) on ((-\infty,\infty))
(\displaystyle h(x) = 4x+\frac{1}{x}) on ((-\infty,0))
(f(x) = 4 - e^{-(x-2)^2}) on ((-\infty, \infty))
(f(x) = b - e^{-(x-a)^2}) on ((-\infty, \infty)\text{;}) assume (a, b \gt 0)
Answer.
Absolute minimum of: (-12) at (x=2\text{,}) no absolute maximum.
Absolute maximum: (-4\text{;}) no absolute minimum.
Absolute minimum: (3\text{;}) no absolute maximum.
Absolute minimum: (b-1\text{;}) no absolute maximum.
Solution.
Given (f(x)=4x^2-16x+4\text{,}) it follows that (f'(x) =8x-16\text{,}) solving (f'(x)= 0) gives a single critical value: (x=2)
Then (f''(x)=8) which is always positive, that is (f(x)) is always concave up and thus by the second derivative test (f(x)) has a relative minimum at (x=2\text{.}) Since this is the only critical value, it follows that (f(x)) has an absolute minimum of (f(2)=4(2)^2-16(2)+4=-12) at (x=2\text{,}) and consequently has no absolute maximum.
2. For (h(x) = 4x+\frac{1}{x}\text{,}) we know that (h'(x) = 4-x^{-2}= 4-\frac{1}{x^2}\text{.}) To solve (h'(x)=4-\frac{1}{x^2}=0) add the (\frac{1}{x^2}) to both sides to get (4=\frac{1}{x^2}\text{,}) then multiply both sides by (x^2) and divide by (4) to get (x^2=\frac{1}{x^2}\text{.}) Solve by taking the square root and we get two critical values (x=\pm \frac{1}{2}\text{.}) Since we are looking at the interval ((-\infty,0)) we only have a single critical value (x=\frac{-1}{2}=-0.5\text{.})
Then (\displaystyle h''(x)=2x^{-3}) and (h''(-.5) \lt 0) (x=-0.5) is a relative maximum. Since there is a single critical value the function (\displaystyle h(x)=4x+\frac{1}{x}) has an absolute maximum of (\displaystyle h(-0.5)=4(-0.5)+\frac{1}{-0.5}=-4) at (x=-0.5) and no absolute minimum.
3. First, since (f(x) = 4 - e^{-(x-2)^2}\text{,}) we see by the chain rule that (f'(x) = -e^{-(x-2)^2}(-2(x-2)) = 2(x-2)e^{-(x-2)^2}\text{.}) Since (e^{-(x-2)^2}) is always positive (and in particular is never zero), it follows that the only critical number of (f) is (x = 2\text{.}) Furthermore, with (f'(x) = 2(x-2)e^{-(x-2)^2}\text{,}) we see that (f'(x) \lt 0) for (x \lt 2\text{,}) and (f'(x) \gt 0) for (x \gt 2\text{.}) Thus (f) is decreasing for (x \lt 2) and increasing for (x \gt 2\text{.}) The first derivative test then tells us that (f) has an absolute minimum at (x = 2) and (f) does not have an absolute maximum.
4. We first find that (f'(x)=2(x-a)e^{-(x-a)^2}) and the only critical number of (f) is (x=a\text{.}) Since (f'(x)\lt0) for (x\lt a) and (f'(x)\gt0) for (x\gt a\text{,}) the first derivative test tells us that (f) has a minimum at (x=a) but has no maximum. Finally, the value of the absolute minimum is (f(a)=b-1\text{.})
Subsection 4.3.2 Summary
To find relative extreme values of a function, we use a first derivative sign chart and classify all of the function’s critical numbers. If instead we are interested in absolute extreme values, we must consider the interval on which we are finding the extreme values (is it a closed interval, an open interval, or the entire domain?).
For a continuous function on a closed, bounded interval, the only possible points at which absolute extreme values occur are the critical numbers and the endpoints. Hence we evaluate the function at each endpoint and at each critical number in the interval and then compare the results to decide which is largest (the absolute maximum) and which is smallest (the absolute minimum).
In the case of finding absolute extrema on an open interval, we first identify all of the critical numbers of the function on that interval, and then use a first or second derivative sign chart to investigate the behavior of the function on that interval.
Exercises 4.3.3 Exercises
1. Finding absolute Extrema.
Consider the function (f(x) = 2 x^3+ 24 x^2 - 120 x+ 6 , \quad - 10 \leq x \leq 3\text{.})
Find the absolute minimum value of this function.
Find the absolute maximum value of this function.
Answer:
2. Finding absolute Extrema.
Consider the function (f(x) = x^4 - 18 x^2+ 1, \quad -2 \leq x \leq 7\text{.})
Find the absolute minimum value of this function.
Answer:
find the absolute maximum value of this function.
Answer:
3. Analyzing Function Behavior.
Let (f(x) = \displaystyle{2+\frac{3}{1+(x+1)^2}}\text{.})
Determine all critical values of (f\text{.}) If there is more than one, enter the values as a comma-separated list.
Critical value(s) =
Construct a first derivative sign chart for (f) and thus determine all intervals on which (f) is increasing or decreasing. If there is more than one, enter the intervals as a comma-separated list. Use interval notation: for example, (-17,20) is the interval (-17 \lt x \lt 20\text{,}) and (-inf, 40) is the interval (x\lt 40\text{.})
Interval(s) where (f) is increasing:
Interval(s) where (f) is decreasing:
Does (f) have a global maximum? If so, enter its value. If not, enter DNE.
Global maximum =
Determine the following limits.
(\displaystyle{\lim_{x\rightarrow \infty}} f(x) =)
(\displaystyle{\lim_{x\rightarrow -\infty}} f(x) =)
Explain why (f(x) > 2) for every value of (x\text{.})
Does (f) have a global minimum? If so, enter its value. If not, enter DNE.
Global minimum =
4. Analyzing Extrema From a Graph.
Identify the marked points as being an absolute maximum or minimum, a relative maximum or minimum, or none of the above. (Select all that apply.)
Relative maximum
Absolute minimum
Absolute maximum
Relative minimum
None of the above
Relative maximum
Absolute maximum
Relative minimum
Absolute minimum
None of the above
Absolute maximum
Absolute minimum
Relative maximum
Relative minimum
None of the above
Relative maximum
Absolute minimum
Relative minimum
Absolute maximum
None of the above
Relative maximum
Relative minimum
Absolute maximum
Absolute minimum
None of the above
Relative minimum
Absolute minimum
Absolute maximum
Relative maximum
None of the above
Absolute minimum
Relative minimum
Absolute maximum
Relative maximum
None of the above
5. Conditions for When absolute Extrema May Occur.
Based on the given information about each function, decide whether the function has an absolute maximum, an absolute minimum, neither, both, or that it is not possible to say without more information. Assume that each function is twice differentiable and defined for all real numbers, unless noted otherwise. In each case, write one sentence to explain your conclusion.
(f) is a function such that (f''(x) \lt 0) for every (x\text{.})
(g) is a function with two critical numbers (a) and (b) (where (a \lt b)), and (g'(x) \lt 0) for (x \lt a\text{,}) (g'(x) \lt 0) for (a \lt x \lt b\text{,}) and (g'(x) \gt 0) for (x \gt b\text{.})
(h) is a function with two critical numbers (a) and (b) (where (a \lt b)), and (h'(x) \lt 0) for (x \lt a\text{,}) (h'(x) \gt 0) for (a \lt x \lt b\text{,}) and (h'(x) \lt 0) for (x \gt b\text{.}) In addition, (\lim_{x \to \infty} h(x) = 0) and (\lim_{x \to -\infty} h(x) = 0\text{.})
(p) is a function differentiable everywhere except at (x = a) and (p''(x) \gt 0) for (x \lt a) and (p''(x) \lt 0) for (x \gt a\text{.})
6. Finding Extrema on Closed and Bounded Intervals.
For each family of functions that depends on one or more parameters, determine the function’s absolute maximum and absolute minimum on the given interval.
(p(x) = x^3 - a^2x) on ([0,a]\text{;}) assume (a \gt 0\text{.})
(r(x) = axe^{-bx}) on (\left[\frac{1}{2b}, \frac{2}{b}\right]\text{;}) assume (a \gt 0, b \gt 1\text{.})
(w(x) = a(1-e^{-bx})) on ([b, 3b]\text{;}) assume (a, b \gt 0\text{.})
7. Conditions for Where absolute Extrema May Occur.
For each of the functions described below (each continuous on ([a,b])), state the location of the function’s absolute maximum and absolute minimum on the interval ([a,b]\text{,}) or say there is not enough information provided to make a conclusion. Assume that any critical numbers mentioned in the problem statement represent all of the critical numbers the function has in ([a,b]\text{.}) In each case, write one sentence to explain your answer.
(f'(x) \le 0) for all (x) in ([a,b]\text{.})
(g) has a critical number at (c) such that (a \lt c\lt b) and (g'(x) \gt 0) for (x \lt c) and (g'(x) \lt 0) for (x \gt c\text{.})
(h(a) = h(b)) and (h''(x) \lt 0) for all (x) in ([a,b]\text{.})
(p(a) \gt 0\text{,}) (p(b) \lt 0\text{,}) and for the critical number (c) such that (a \lt c \lt b\text{,}) (p'(x) \lt 0) for (x \lt c) and (p'(x) \gt 0) for (x \gt c\text{.})
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14366 | https://www.learnamericanenglishonline.com/Yellow%20Level/Y12%20Perfect%20Modals.html | Learn American English Online
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Lesson Twelve:
Perfect Modal Verbs
modal verb + have + past participle
Perfect modals take a modal verb (could, should, must, might/may) and pairs it with a perfect tense phrase (have + past participle), which is how it gets its name. Don’t let the vocabulary confuse you! When we say perfect modals, we are not saying that these are modals that are perfect.
Each of these modal constructions have slightly different meanings, and that’s what you will learn in this lesson. Pay attention to the construction and use of each, as these phrases are very common in English- you will hear them often and will need to use them. Overall, perfect modals allow you to talk about possibilities or regret.
As an added bonus, one way to remember this type of construction? Take a listen to one of the most popular American songs, “It Must Have Been Love” and you will never use it incorrectly!
| | | |
---
| Perfect Modal | Picture | sentence |
| could have _____ couldn’t have _____ past ability | doing the dishesdoing the dishes | The boy could have done the dishes himself, but his father decided to help. |
| should have ____ shouldn’t have ____ You did or didn’t do something that was a good idea. | dizzy girldizzy girl | The girl shouldn’t have spun around so many times. She got dizzy and fell down. spin: turn around many times. spin / spun / spun |
| would have ____ wouldn’t have ____ past condition / past situation | house of cardshouse of cards | This house of cards would have fallen over if the person who built it hadn’t been so careful. |
| may have _____ may not have _____ past possibility | old cameraold camera | My grandfather may have used this camera when he was a young man, but I’m not sure. |
| might have ______ might not have _____ past possibility | woman eatingwoman eating | Her mother might have put mustard on her sandwich. She hopes not. |
| must have ____ must not have ___ past probability. This indicates that something probably happened in the past. | rock bandrock band | They must have practiced a lot because they’re very good musicians. |
could have _____
couldn’t have _____
should have ____
shouldn’t have ____
The girl shouldn’t have spun around so many times. She got dizzy and fell down.
spin: turn around many times.
spin / spun / spun
would have ____
wouldn’t have ____
past condition /
past situation
may have _____
may not have _____
might have ______
might not have _____
must have ____
must not have ___
past probability.
This indicates that something probably happened in the past.
Let’s not forget…
The Future Perfect Tense
Unlike all the other modals above, will is used for the future:
Subject + will + have + past participle
| | |
--- |
| Singular | Plural |
| I will have lived | We will have lived |
| You will have lived | You will have lived |
| He will have lived | |
| She will have lived | They will have lived |
| It will have lived | |
This is a difficult tense to use. It describes an action that will be completed in the future.
| For example: I moved to Minnesota in 1991. The year now is 2008. By 2011, I will have lived in Minnesota for 20 years. |
For example:
I moved to Minnesota in 1991. The year now is 2008.
By 2011, I will have lived in Minnesota for 20 years.
Click here for a quiz
Next: Lesson 13
Improve your English in the Reading Rooms
Listen and write |
14367 | https://pmc.ncbi.nlm.nih.gov/articles/PMC11826717/ | Brazilian guideline for pediatric cycloplegia and mydriasis - PMC
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Brazilian guideline for pediatric cycloplegia and mydriasis
Diretrizes brasileiras para cicloplegia e midríase em crianças
Ian Curi
Ian Curi
1 Department of Ophthalmology, Hospital Federal dos Servidores do Estado do Rio de Janeiro, Rio de Janeiro, RJ, Brazil.
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1, Simone Akiko Nakayama
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2 Department of Ophthalmology and Visual Science, Escola Paulista de Medicina, Universidade Federal de São Paulo, São Paulo, SP, Brazil.
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2, Érika Mota Pereira
Érika Mota Pereira
3 Department of Ophthalmology, Hospital São Geraldo, Universidade Federal de Minas Gerais, Belo Horizonte, MG, Brazil.
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3, Luisa Moreira Hopker
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4 Department of Ophthalmology, Hospital de Olhos do Paraná, Curitiba, PR, Brazil.
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4, Fábio Ejzenbaum
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5 Department of Ophthalmology, Santa Casa de Misericórdia de São Paulo, São Paulo, SP, Brazil.
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5, Ronaldo Boaventura Barcellos
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5, Rosane da Cruz Ferreira
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6 VER Instituto para Prevenção da Cegueira Infantil, Porto Alegre, RS, Brazil.
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2 Department of Ophthalmology and Visual Science, Escola Paulista de Medicina, Universidade Federal de São Paulo, São Paulo, SP, Brazil.
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2, Craig A Mckeown
Craig A Mckeown
7 Pediatric Ophthalmology and Strabismus Service, Bascom Palmer Eye Institute, University of Miami Miller School of Medicine, Miami, FL, USA.
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7, Júlia Dutra Rossetto
Júlia Dutra Rossetto
2 Department of Ophthalmology and Visual Science, Escola Paulista de Medicina, Universidade Federal de São Paulo, São Paulo, SP, Brazil.
8 Pediatric Ophthalmology Department, Instituto de Puericultura e Pediatria Martagão Gesteira, Universidade Federal do Rio de Janeiro, Rio de Janeiro, RJ, Brazil.
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1 Department of Ophthalmology, Hospital Federal dos Servidores do Estado do Rio de Janeiro, Rio de Janeiro, RJ, Brazil.
2 Department of Ophthalmology and Visual Science, Escola Paulista de Medicina, Universidade Federal de São Paulo, São Paulo, SP, Brazil.
3 Department of Ophthalmology, Hospital São Geraldo, Universidade Federal de Minas Gerais, Belo Horizonte, MG, Brazil.
4 Department of Ophthalmology, Hospital de Olhos do Paraná, Curitiba, PR, Brazil.
5 Department of Ophthalmology, Santa Casa de Misericórdia de São Paulo, São Paulo, SP, Brazil.
6 VER Instituto para Prevenção da Cegueira Infantil, Porto Alegre, RS, Brazil.
7 Pediatric Ophthalmology and Strabismus Service, Bascom Palmer Eye Institute, University of Miami Miller School of Medicine, Miami, FL, USA.
8 Pediatric Ophthalmology Department, Instituto de Puericultura e Pediatria Martagão Gesteira, Universidade Federal do Rio de Janeiro, Rio de Janeiro, RJ, Brazil.
Corresponding author:
Julia D. Rossetto. E-mail: julia_rossetto@hotmail.com
Received 2021 Jun 9; Accepted 2021 Sep 20; Issue date 2023 Jul-Aug.
This is an Open Access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
PMC Copyright notice
PMCID: PMC11826717 PMID: 35319660
Abstract
Cycloplegia is crucial for reliable pediatric ophthalmology examinations. This document provides a recommendation for pediatric cycloplegia and mydriasis for Brazilian ophthalmologists. This article was developed based on literature reviews; the clinical experience of Brazilian specialists, as obtained through questionnaires; and the consensus of the Expert Committee of the Brazilian Pediatric Ophthalmology Society. According to the best evidence and formulations available in Brazil, this committee recommends the use of one drop of 1% cyclopentolate plus one drop of 1% tropicamide in children older than 6 months and two drops of 1% tropicamide 0-5 minutes apart for those younger than 6 months. Mydriasis may be increased by a single drop of 2.5% phenylephrine. For retinopathy of prematurity screening, the recommendation is 0.5% or 1% tropicamide, administered two or three times, 5 minutes apart, and 2.5% phenylephrine, used preferably once. In all scenarios, we recommend the use of a prior drop of 0.5% proxymetacaine.
Keywords: Mydriatics; Refraction, ocular; Infant, newborn; Child; diagnostic techniques, ophthalmological
Abstract
A cicloplegia é crucial para um exame oftalmológico pediátrico acurado. Este documento visa a fornecer uma recomendação para cicloplegia e midríase pediátrica para oftalmologistas brasileiros. Foi desenvolvido com base em revisão literária, na experiência clínica de especialistas brasileiros, por meio de questionários, e no consenso do comitê de especialistas da Sociedade Brasileira de Oftalmologia Pediátrica (SBOP). De acordo com as melhores evidências, este comitê recomenda o uso de uma gota de ciclopentolato 1%, mais uma gota de tropicamida 1% em crianças maiores de 6 meses e duas gotas de tropicamida 1% com intervalo de 0-5 minutos para menores de 6 meses. A midríase pode ser potencializada por uma gota de fenilefrina 2,5%. Para o rastreamento da retinopatia da prematuridade, a recomendação é tropicamida 0,5 ou 1%, duas ou três vezes, com 5 minutos de intervalo, e 2,5% de fenilefrina, preferencialmente uma vez. O uso prévio de proxymetacaína 0,5% é sempre recomendado.
Keywords: Midríase, Refração ocular, Recém-nascido, Criança, Técnicas de diagnóstico oftalmológico
INTRODUCTION
To perform an appropriate and accurate pediatric eye examination, adequate cycloplegia is essential, as temporary relaxation of the ciliary muscle enables accurate measurement of the refractive status of children. An ideal cycloplegia would be effective, convenient, and safe. Effectiveness would require maximum ciliary muscle paralysis and adequate mydriasis. Convenience would demand a rapid onset of cycloplegic action and a sufficient duration of effect for the examination to be performed but not so prolonged as to cause patient discomfort. Safety would necessitate the absence of side effects. At present, however, no cycloplegic agent meets all of these qualifications. Therefore, there is a need for a protocol that can provide adequate cycloplegia for the pediatric population.
In Brazil, combination cycloplegic and/or mydriatic drops are not commercially available. The only obtainable medications are 1% tropicamide, 1% cyclopentolate, 0.5% and 1% atropine, and 10% phenylephrine. The absence of combined medication formulations at lower concentrations increases the risk of adverse effects. In addition, no standardized national guideline has been established for pediatric cycloplegia. The objective of the present article is to provide a protocol for cycloplegia and mydriasis in children for use by Brazilian ophthalmologists.
METHODOLOGY
This pediatric cycloplegia/mydriasis guideline was developed based on the medical literature, the clinical experience of Brazilian specialists (as obtained via questionnaires), and the consensus of the Expert Committee of the Brazilian Pediatric Ophthalmology Society (SBOP). PubMed/Medline databases were searched for articles published in peer-reviewed journals written in Portuguese and English using combinations of the following MeSH terms: “cycloplegia,” “cycloplegic,” “mydriasis,” “mydriatic,” “atropine,” “cyclopentolate,” “tropicamide,” “phenylephrine,” “iris color,” “side-effect,” “adverse reaction,” “adverse effect,” “child, preschool,” “infant,” and “pediatric.”
Selected papers were reviewed by the SBOP Expert Committee. Because of the scarcity of well-designed articles on this topic, the documents considered were not restricted to systematic reviews, randomized controlled trials, or observational studies. To classify the level of evidence and the strength of the recommendations, the following method was used: Level I was based on two or more high-quality randomized clinical trials (RCTs). Level II was based on a small number of RCTs; on more than one controlled, but not randomized, study; on more than one RCT of lesser quality; on cohort or case-control studies, preferably from more than one research group or more than one center; or on observations of clear-cut effects in noncontrolled studies. Level III was based on expert opinion, clinical experience, descriptive studies, or cohort or case-control studies of lower quality(1).
A 14-question questionnaire was sent to 337 members of the SBOP and the Brazilian Strabismus Center (CBE), and 136 responses were obtained. Members were asked the following questions about their routines regarding pediatric cycloplegia/mydriasis:
What is your subspecialty?
Do you use a topical anesthetic before cycloplegic eye drops?
At what age do you start using 1% cyclopentolate eye drops?
Describe the adverse effects that you have had with cyclopentolate and their frequency.
Do you perform different cycloplegic routines for patients with and without strabismus?
Do you consider the cycloplegia due to tropicamide equivalent to that obtained with cyclopentolate?
Do you consider the cycloplegia due to cyclopentolate equivalent to that obtained with atropine?
Do you use atropine in any situation for refraction?
Do you use phenylephrine as an adjunct to obtain mydriasis?
How long after the cyclopentolate drop do you perform your exam?
How long after the tropicamide drop do you perform your exam?
Is your routine different for patients with light or dark irises?
Do you use cyclopentolate in children with epilepsy? How long after the last crisis?
Describe your cycloplegic routine.
Another 10-question questionnaire was sent to a sample of specialists in retinopathy of prematurity (ROP) regarding their routine in obtaining mydriasis in neonatal intensive care units (NICUs).
The results of the literature search and the answers to the questionnaire were critically analyzed by this Expert Committee for the preparation of this guideline. Ethics approval was waived because the study did not require human subject participation.
RESULTS/RECOMMENDATIONS
Anesthetic eye drops: are they recommended?
The use of anesthetic drops is intended to reduce the discomfort caused by cycloplegic/mydriatic drops, to improve the child’s experience during the ophthalmological visit, and to enhance the absorption of the drops(2,3). The responses to our questionnaire revealed that 56.6% of the specialists instill anesthetic eye drops before performing cycloplegia. Two double-masked studies with small samples of adults showed a significant reduction in total discomfort scores when both tropicamide (pain scale 5.15 placebo × 2.5 proxymetacaine) and cyclopentolate (pain scale 4.29 placebo × 1.16 proxymetacaine) were instilled after using 0.5% proxymetacaine (proparacaine) compared with instillation after a placebo(4,5).
Other benefits associated with the use of topical anesthetics include a decreased time to reach the maximum cycloplegic effect(3) and maintenance of the peak of cycloplegia and mydriasis for a longer time.(6) With regard to which anesthetic eye drops are preferable, a randomized, double-masked protocol with 23 adults compared proxymetacaine and tetracaine topical anesthetics. The study showed that the mean pain score was significantly lower with the use of proxymetacaine, albeit with no difference in the cost of medications(7). A survey conducted by the American Association for Pediatric Ophthalmology and Strabismus among its members regarding the choice of the best anesthetic agent for ROP assessments indicated that 63% prefer the use of proxymetacaine, 25% prefer tetracaine, and 3% prefer oxybuprocaine(8). The time of onset and the duration of action of an anesthetic are also relevant for its clinical use. Proxymetacaine has an onset of anesthetic effect of 30 seconds after instillation, and the effect lasts for 15-25 minutes(9).
Recommendations. Based on these findings, this Expert Committee recommends the use of one drop of 0.5% proxymetacaine 30 seconds to 1 minute before the instillation of the first mydriatic/cycloplegic eye drop (Level of Recommendation II).
Atropine, cyclopentolate, or tropicamide. Which is the best choice?
Anticholinergic agents, such as atropine, cyclopentolate, and tropicamide, inhibit the muscarinic actions of acetylcholine in the ciliary muscle (cycloplegia) and the iris sphincter (mydriasis). The human eye has five subtypes of muscarinic receptors (M1 to M5). The ciliary muscle and the iris sphincter each have a complex arrangement composed of all five subtypes, but the M3 type predominates (60-75%)(10,11).
Atropine is a nonselective muscarinic antagonist. By contrast, tropicamide and cyclopentolate probably have different affinities for each receptor subtype, which results in a synergistic effect when used together. The maximum cycloplegic effect occurs within 20-45 minutes with tropicamide, within 30-60 minutes with cyclopentolate, and within 60-180 minutes with atropine. Complete recovery occurs in 6 hours, 24 hours, and 7-12 days, respectively(11).
However, despite the similarity between the effects of tropicamide and cyclopentolate, the window of opportunity for performing the examination is much narrower for tropicamide than for cyclopentolate. One study showed that 2 hours after administration of 1% cyclopentolate, the subjects had regained only 25% of their initial amplitude of accommodation, whereas subjects who had received 1% tropicamide had regained about 70% of their baseline accommodation(12). Thus, although some studies have suggested that cycloplegia is achieved as effectively with tropicamide as with cyclopentolate(13), tropicamide is not recommended for use on its own when refraction is a priority, because its effect is extremely transient.
When used in combination, tropicamide and cyclopentolate allow for a faster cycloplegic onset, a more ample window of peak activity, and a reduced duration of the drug effect in comparison with their use alone. In addition to its greater convenience, the combined formulation is highly effective, as it produces cycloplegia as successfully as cyclopentolate alone and at no additional cost(14-16).
In some countries, 1% atropine remains the gold standard for performing cycloplegia. In Brazil, 41.9% of the specialists recognize that the cycloplegia obtained with 1% cyclopentolate is inferior to that obtained with 1% atropine. Despite this, only 9.6% of pediatric ophthalmologists still use 1% atropine and only in some situations for refraction, as they believe that cyclopentolate is sufficient in most cases.
Atropine is known to have a greater cycloplegic effect than that of other eye drops.(17,18) However, for most routine examinations, this difference might not be clinically significant, and atropine has inconveniently prolonged effects. Thus, cyclopentolate and tropicamide, because of their shorter duration of action, have become more popular options. However, it is important to remember that the use of atropine also leads to a significantly lower mean residual accommodation than is achieved with cyclopentolate and tropicamide combined or with cyclopentolate alone(15). Furthermore, about one-fifth of patients may present a difference of +1.00 diopter or more between the cyclopentolate and atropine refractions in at least one eye(17), and this difference can be meaningful in some clinical scenarios. Conversely, in most cases, the mean spherical equivalent difference is <0.50D(19). Thus, the use of atropine can be reserved for particular circumstances, such as for a suspected residual accommodative component after cyclopentolate and tropicamide cycloplegia in esotropic patients or for accommodation spasms.
There is currently no consensus regarding atropine dosages. A comparison between atropinization administered as two drops (5 minutes apart) in the office (measured 90 minutes afterward) versus 3 days (3 times daily) revealed, on average, values 0.5 diopters higher in the latter group(20). Another study showed no significant difference in cycloplegic refraction between an eight-drop versus a four-drop regimen(21).
In Brazil, Bicas et al conducted a study on the dosage of 1% atropine for cycloplegic refraction. In this study, patients received two drops of 1% atropine, 5 minutes apart, on day zero and used one drop of 1% atropine in each eye, three times a day for the following ten days. Refraction was assessed on days 0, 3, 5, 7, and 10. The study found no significant differences in the refraction measurements. Such findings suggest that no or only low augmentation occurs with cumulative atropine doses(22,23).
Nevertheless, in our questionnaire, 87.50% of the participants responded that tropicamide had an inferior cycloplegic effect as compared with cyclopentolate. This result is in accordance with studies in the literature that showed the superiority of cyclopentolate in obtaining a greater accommodation blockage(24,25).
The American Academy of Ophthalmology (AAO) suggests the use of 1% cyclopentolate only in children older than 6 months and recommends 0.2% cyclopentolate plus 1% phenylephrine in infants younger than 6 months. The AAO states that the required dosage can be higher in heavily pigmented irises and that 1% tropicamide can be used as an adjunct(26). In this context, one important point to highlight is that 78% of Brazilian specialists do not change their protocol according to iris color. One prospective randomized trial compared one, two, and three drops of 1% cyclopentolate and found no statistically significant differences between the treatment groups. The authors concluded that a single drop of 1% cyclopentolate suffices for cycloplegic refraction in children(27). The responses of the Brazilian specialists regarding the age at which they use 1% cyclopentolate indicated that 11.76% use it from birth, 5.88% use it from 3 months onward, 22.06% from 6 months, and 3.68% from 9 months, whereas 44.85% use it only after 12 months.
Recommendations. In accordance with the evidence presented, this Expert Committee recommends the use of one drop of cyclopentolate 1% plus one drop of 1% tropicamide to obtain an optimal cycloplegic effect in children older than 6 months (Level of Recommendation II). The use of two drops of 1% tropicamide 0-5 minutes apart is advocated for those younger than 6 months (Level of Recommendation II; Tables 1 and 2). In pediatric cases, 1% atropine can be used as an alternative for cycloplegia in patients with accommodation spasms or with a suspected residual accommodative component not revealed by cyclopentolate plus tropicamide (Level of Recommendation III). The use of 1% atropine twice a day for 3 days seems to be a reasonable dosage to fulfill that goal (Level of Recommendation III).
Table 1.
SBOP-recommended protocols for infants younger than 6 months
| Time | Eye drop |
:---: |
| 0 min | Proxymetacaine (0.5%) |
| 30 sec-1 min | Tropicamide (1%) |
| 1-6 min | Tropicamide (1%) |
| 30-40 min | Examination |
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Notes: (i) A single drop of 2.5% phenylephrine should be used to maximize mydriasis when needed (compounded in specialized pharmacies). (ii) The use of a third drop of 1% tropicamide is acceptable if needed.
Table 2.
SBOP-recommended protocols for children older than 6 months
| Time | Eye drop |
:---: |
| 0 min | Proxymetacaine (0.5%) |
| 30 sec-1 min | Cyclopentolate (1%) |
| 1-6 min | Tropicamide (1%) |
| 30-40 min | Examination |
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Notes: (i) A single drop of 2.5% phenylephrine (compounded in specialized pharmacies) should be used to maximize mydriasis when needed (ii) The use of a third drop of tropicamide (1%) is acceptable, if needed. (iii) In some specific clinical settings, atropine (1% twice a day for 3 days) can be used according to the decision of the doctor.
Should we use phenylephrine as an adjuvant to maximize mydriasis?
Phenylephrine is a sympathomimetic agent that acts on alpha-1 adrenergic receptors and has little or no effect on beta-adrenergic receptors. Its mydriatic action occurs due to contraction of the iris dilator muscle, with constriction of the conjunctival arterioles (conjunctival whitening) and activation of the Müller muscle (enlargement of the eyelid fissure) as secondary ocular effects. Its maximum effect is reached after 45-60 minutes of instillation, and the reversal occurs in 6 hours(28). Phenylephrine has no cycloplegic effect; hence, its usefulness is restricted to increasing mydriasis for the evaluation of the extreme retinal periphery (ROP or retinoblastoma) or in cases of poor dilation (uveitis).
Studies have compared different concentrations and combinations of phenylephrine for effective mydriasis and a good safety profile in newborns(29,30). Most guidelines recommend the use of a drop of 2.5% phenylephrine for ROP examination(31-33). However, a systematic review suggests one drop of 1% phenylephrine and 0.2% cyclopentolate as the lowest effective combination regimen(34). Conversely, an RCT that compared three drops of 0.5% tropicamide versus two drops of 0.5% tropicamide plus one drop of 5% phenylephrine reported finding a 1.9 times greater pupil surface area with the latter combination(35).
A prospective randomized study in neonates showed that the use of two drops of 2.5% phenylephrine (at an interval of 5 minutes) is sufficient to significantly increase heart rate and blood pressure. However, when only one drop of 2.5% phenylephrine was used in combination with 0.5% tropicamide and 0.5% cyclopentolate, the mydriasis achieved was sufficient (average of 6.4 mm), and this regimen caused no significant increase in heart rate or systolic pressure in relation to the control group. The treatment group showed only a slight increase in diastolic pressure; this was statistically significant but apparently insignificant from a clinical point of view(36).
Particular attention should be paid to extremely premature infants, extremely low-weight infants, and patients with respiratory distress, as these patients are more susceptible to gastrointestinal side effects. In such patients, vasoconstriction of the blood supply and anticholinergic effect can reduce peristalsis, causing slow gastric emptying, emesis, abdominal distension, and even necrotizing enterocolitis. In older infants and children, the use of one drop of 2.5% phenylephrine, although rarely necessary, presents an adequate safety profile. However, the use of a 10% concentration seems to promote a dangerous increase in the number and severity of side effects, with reports of cardiorespiratory arrest(37,38).
The responses to our survey revealed that 45% of Brazilian pediatric ophthalmologists use phenylephrine as an adjunct to obtain mydriasis under certain circumstances. In Brazil, 2.5% phenylephrine is not commercially available and must be compounded in specialized pharmacies according to current legislation.
Recommendations. Based on this evidence, this Expert Committee recommends the use of one drop of adjuvant 2.5% phenylephrine, in addition to the regular protocol, whenever the evaluation of the extreme retinal periphery is necessary or in the presence of poor mydriasis (Level of Recommendation II).
Premature infant examinations in NICUs
Fifty-nine ROP specialists from different parts of Brazil answered a questionnaire about their dilation routine for premature infants in the NICU. Of these specialists, 35% were pediatric ophthalmologists, 61.4% were retina specialists, and 3.5% were general ophthalmologists. With regard to the use of anesthetic drops, 57.6% declared that they do not use them, whereas 42.4% do use them (27.1% use proxymetacaine, 3.4% use tetracaine, and 11.9% use whichever is available). For cyclopentolate, 94.9% do not use it, whereas 1.7% use 0.2% cyclopentolate, 1.7% use 0.5% cyclopentolate, and 1.7% use 1% cyclopentolate.
Tropicamide is used by most (98%) Brazilian ROP specialists; 59.3% prefer the 0.5% formulation, whereas 37.3% use 1% tropicamide. Phenylephrine is used by 83.1% of those consulted; most (74.6%) advocate the use of the 2.5% concentration, while 5.1% prefer the 1% formulation.
With regard to the number of instillations, 6.8% use each type of drop only once, 44% use it twice, 44% use it three times, and 3.4% use it four times.
Indirect ophthalmoscopy is performed within 20 minutes or less after instillation by 8.4% of the specialists, whereas 27.1% wait for 30 minutes, 42.4% wait for 40 minutes, and 22% wait for 60 minutes.
The questionnaire responses also revealed that, for the drops with concentrations not commercially available, 41.1% of the participants order these from a compounding pharmacy, 39.3% acquire them from their hospital pharmacy, and 16.6% obtain them on their own.
Recommendations. For the evaluation of ROP in premature infants with optimal mydriasis, this Expert Committee recommends the use of one drop of 0.5% proxymetacaine, followed by a single drop of 2.5% phenylephrine and two or three drops of 0.5% or 1.0% tropicamide 5 minutes apart (Table 3). The examination should be performed at least 30-40 minutes after the first drop (Level of Recommendation III).
Table 3.
Mydriasis for premature infants in the neonatal intensive care unit (ROP screening)
| Time | Eye drop |
:---: |
| 0 min | Proxymetacaine (0.5%) |
| 30 sec-1 min | Phenylephrine (2.5%) |
| 6 min | Tropicamide (0.5% or 1%) |
| 11 min | Tropicamide (0.5% or 1%) |
| 30-40 min | Examination |
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Note: (i) The use of a third drop of tropicamide (0.5%-1%) is acceptable if needed.
Ideal interval between eye drop instillation and examination
The usual recommendation is to wait 5 minutes between the first and second eye drops to avoid washing out the first drop. This assumption was proven experimentally with albino rabbits, but it should be carefully applied in clinical practice(39).
One randomized study compared two regimens, one with tropicamide plus phenylephrine drops instilled 10 minutes apart and one with concurrent application, and found no statistically significant difference in the pupillary diameter(40). Another clinical study compared the relative pupil surface before and after the administration of one drop of 10% phenylephrine and one drop of 0.5% tropicamide, either immediately or with a 5-minute time interval. The protocol with instillations 5 minutes apart yielded only a 5.6% gain in pupil surface(41).
The timing of the peak action of each medication must be considered when determining the optimum interval between drop instillation and examination under mydriasis. The great majority of Brazilian pediatric ophthalmologists wait at least 30 minutes between cyclopentolate instillation and the examination. Overall, 38.24% wait 30 minutes and 47.79% wait 40 minutes. In the case of tropicamide, 31.62% wait 20 minutes and 44.85% wait 30 minutes.
Recommendations. Considering the lack of strong evidence regarding the ideal interval time, this Expert Committee accepts an interval between cyclopentolate and tropicamide drops of 0-5 minutes and an interlude for the examination of 30-40 minutes after the first drop (Level of Recommendation III).
Side effects and risks of medications
Any medication can cause adverse effects. Therefore, its use should be endorsed when the benefits exceed the risks. However, in addition to the rarity of the occurrence of adverse effects, their severity must be taken into account.
Among SBOP/CBE members, 75% reported mild side effects with the use of 1% cyclopentolate. A large number mention facial flushing, drowsiness, and agitation as sporadic, whereas hallucination is rare. Seven of 136 reported at least one episode of convulsion. When asked about the use of cyclopentolate eye drops in children with a history of epilepsy, 41.18% of doctors contraindicate their use but 58.82% retain their use, with 25.74% using them in any context, 10.29% if seizures have been controlled for 30 days, 3.68% if the control has lasted 60 days, 5.88% if the control has lasted 90 days, and 11.03% if the control has lasted 180 days.
A multicenter survey of German speakers was performed to estimate the likelihood of severe complications (i.e., had to be monitored for several hours) and very severe complications (i.e., caused patients to be admitted to a hospital). A total of 1.7 million cumulative cycloplegias over 1112 years of cumulative cycloplegic experience were analyzed. The estimated rates were 1.1:100,000 and 2.7:100.000 for severe psychiatric side effects and 0.5:100,000 and 8.7:100.000 for severe physical side effects using cyclopentolate and atropine, respectively. Therefore, during 30 years of cycloplegic experience with an average of 34 cycloplegias per week, only two severe to very severe complications could be expected with the sole use of cyclopentolate and only 10 with the sole use of atropine. Severe mental complications included intoxication, hallucination, agitation, and depression. Severe physical complications were seizures, asthma, fever, circulatory impairment, and tachycardia. No deaths or permanent damage were reported. However, because of the high diagnostic value of cycloplegic refraction in children, these frequent and short-lasting side effects are viewed as medically acceptable(42).
A Japanese study investigated the incidence rate and side effects of topical 0.25%, 0.5%, or 1% atropine and 1% cyclopentolate for cycloplegia in children aged 15 years or younger. Among 811 patients who received atropine, 8.8% had side effects, most frequently (53.6%) occurring following the initiation of the instillation on the first day. The symptoms included flushing (40.8%), fever (30.0%), and both (15.5%). The risk of adverse effects was higher with the 1% concentration and in the group younger than 1 year. However, no serious reactions were described. Of a total of 2238 patients who received 1% cyclopentolate (one or two instillations), 1.2% had side effects, including drowsiness (37.0%), red eye (14.8%), flushing (11.1%), and redness (11.1%). Hyperactivity, irritable mood, skin sores, and conjunctivitis were also reported. No serious reactions were reported. The reactions were slightly more likely in children younger than 1 year and in patients with systemic diseases, such as Down syndrome(43).
In a Dutch cohort of 3-14-year-old children, 504 patients were administered 1% cyclopentolate + 1% tropicamide (C + T) and 408 had 1% cyclopentolate twice (C + C). Adverse reactions were reported for C + C in 10.3% of the patients and for C + T in 4.8%. Repeated instillation of 1% cyclopentolate, younger age, and low body mass index were associated with higher incidences of side effects, but no serious side effects were reported(44).
Aside from using the proper dosage, studies also recommend the application of pressure to the nasolacrimal sac when adding cycloplegic drops to reduce systemic side effects(45).
Recommendations. This Expert Committee does not recommend the use of 1% atropine or 1% cyclopentolate in children younger than 6 months. Although rare, these patients are more vulnerable to severe side effects. For those greater than 6 months of age, the use of 1% atropine or 1% cyclopentolate is safe, except in patients with Down syndrome, neurological problems, history of seizures, or closed-angle glaucoma (Level of Recommendation III).
DISCUSSION
No protocols currently exist in Brazil for cycloplegia and mydriasis for pediatric eye examination. The purpose of this article is to provide a guideline based on the best scientific evidence, expert consensus, and Brazilian pharmacologic availability.
Cycloplegic refraction is the gold standard for clinical assessment of refractive error in young children(46). There is a statistically significant difference between cycloplegic and noncycloplegic refraction in children, with significantly more hyperopia (less myopia) in the cycloplegic group. This difference is also higher in younger children and in children with greater hyperopia(47). In the pediatric group, up to 18% of all eyes with noncycloplegic myopia become emmetropic after cycloplegia, and 15.7% become hyperopic under cycloplegia(48). Refraction without cycloplegia or with inadequate cycloplegia may alter outcomes in an underplus or overminus direction or can result in an unbalanced prescription between the eyes. This can lead to undesirable consequences, such as amblyopia.
Internationally, formal cycloplegic recommendations can be found only in broad protocols, such as ROP guidelines and strabismus management protocols. For example, the Royal College of Ophthalmologists advocates the use of 0.5% proxymetacaine, followed by cyclopentolate (0.5% in children under 6 months and 1% in children older than 6 months) and examination after 30 minutes in pediatric strabismus assessment(49). The College also proposes a mydriatic combination of 2.5% phenylephrine and 0.5% cyclopentolate, instilled as one drop each, in two to three doses, each 5 minutes apart, 1 hour before ROP screening(33).
In Brazil, no combination mydriatic formulations have concentrations suitable for the pediatric public. The available drugs are 1% tropicamide, 1% cyclopentolate, 0.5% and 1% atropine, and 10% phenylephrine. Considering that several of these formulations have the potential for serious adverse effects, especially in infants younger than 6 months, the SBOP recommends the use of only tropicamide in this age group. In children older than 6 months, for efficient cycloplegia, the SBOP recommends the use of one drop of 1% cyclopentolate and one drop of 1% tropicamide. The first instillation should be preceded by one drop of 0.5% proximetacaine, and the examinations should ideally be performed between 30 and 40 minutes of application of the first drop. A single drop of 2.5% phenylephrine should be considered when the retinal periphery assessment is crucial or when dilation is poor. However, the latter medication must be compounded in specialized pharmacies, according to current legislation.
Guidelines are not intended to provide step-by-step medical care or to replace clinical judgment. On the contrary, their intention is to support standards of practice(50). This guideline written by the SBOP should therefore be considered in this context. Adhering to its recommendations will not necessarily produce successful results in all cases. This guideline is also not intended to define or serve as a legal standard for medical care; therefore, it should not be used as a legal resource, as its general nature cannot provide individualized guidance for all patients in all circumstances(50,51).
Footnotes
Funding: This study received no specific financial support.
a
Sociedade Brasileira de Oftalmologia Pediátrica (SBOP).
b
Conselho Brasileiro de Oftalmologia (CBO).
c
Centro Brasileiro de Estrabismo (CBE).
d
Sociedade Brasileira de Pediatria (SBP).
Disclosure of potential conflicts of interest: None of the authors have any potential conflicts of interest to disclose.
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14368 | https://byjus.com/english/syntax/ | What is syntax? This term has significant meanings in English grammar and linguistics, and computer science and programming languages. In this article, you will be introduced to the concept of syntax in linguistics, the meaning and definition of the term, the importance of syntax, the rules to be followed and the types of syntactic patterns. Furthermore, there are a number of examples to help you clearly understand how syntax is applied.
Table of Contents
What Is Syntax? – Meaning and Definition
The Role of Syntax in the English Language
How to Apply Syntax in Sentences? – Rules
Some Exceptions You Can Make
Types of Syntactic Patterns with Examples
Check Your Understanding of Syntax
Frequently Asked Questions on Syntax in Linguistics
What Is Syntax? – Meaning and Definition
‘Syntax’ is the term used to refer to the arrangement of words, phrases and clauses in a sentence to make them sound and look meaningful. The Oxford Learner’s Dictionary defines syntax as “the way that words and phrases are put together to form sentences in a language”. According to The Concise Oxford Dictionary of Literary Terms, syntax is “the way in which words and clauses are ordered and connected so as to form sentences”; it also refers to “the set of grammatical rules governing such word order”.
The Role of Syntax in the English Language
Now that you know what syntax means, try and understand how it works in the English language. Gathering all the words and phrases alone would not do you any good. You should know how to arrange them in an order that conveys the message exactly in the way you want it. Parts of speech, like adverbs, can be used at the beginning, middle and/or end of sentences. The meaning of the sentence can change depending on where the word is placed. In addition to this, there are words which have the same spellings but different meanings and words that have multiple functions. For these reasons and many more, it is important that you understand the concept of syntax and practise how to write syntactically correct sentences.
How to Apply Syntax in Sentences? – Rules
To make it simpler and easier for you to understand, here is a list of rules you have to bear in mind when forming sentences. Go through them and analyse your sentences to see if they have proper syntax.
Rule 1:
A sentence should have a subject and a verb. Other than imperative sentences, all other sentences must, for sure, have at least one subject and one verb. Imperative sentences do not require a subject.
For example:
It / was raining. (Declarative sentence)
Subject / Verb
Switch off / your mobile phones. (Imperative sentence)
Verb / Object
Rule 2:
The subject should come first, followed by the verb. Only in the case of interrogative sentences do subjects follow the verb. In other words, interrogative sentences begin with a verb.
For example:
The meeting / starts / in half an hour. (Declarative sentence)
Subject / Verb / Adjunct
Does / the meeting / start / in half an hour? (Interrogative sentence)
Helping verb / Subject / Main verb / Adjunct
Rule 3:
Describing words like adjectives and adverbs should be positioned before the words they describe.
For example:
It is very cold today. (Adverb – describes the intensity of the cold weather)
I wore a blue dress for my sister’s wedding. (Adjective – describes the colour of the dress)
Rule 4:
Objects follow the subject and verb in a sentence. If there is a direct object and an indirect object, the indirect object comes after the verb, and the direct object follows the indirect object.
For example:
Bianca / gave / me / a book.
Subject / Verb / Indirect object / Direct object
Nandhu / bought / us / dinner.
Subject / Verb / Indirect object / Direct object
Rule 5:
Clauses, irrespective of whether they are dependent clauses or independent clauses, must have a subject and a verb. Phrases, on the other hand, do not require a subject.
For example:
Phrase / Clause
On the eve of Christmas, / we left for Madrid.
The clause ‘we left for Madrid’ has a subject (we), verb (left) and adjunct (for Madrid).
Dependent clause / Independent clause
Though Lakshmi was tired, / she did not take a day off from work.
The dependent clause has a subject, ‘Lakshmi’, and a verb, ‘was’. The independent clause also has a subject, ‘she’, and a verb, ‘did not take’.
Some Exceptions You Can Make
Though it is almost always necessary to follow the rules of syntax, there are some forms of writing where you can make a few exceptions. Poetry, for instance, uses rhyming words to create a rhythm and possesses a metre. So, when writing a poem, you have the liberty to alter the syntax of sentences to attain the impact you intend to make in the minds of your readers. Likewise, in drama, dialogues need not always strictly be in a set syntactic pattern. Playwrights are allowed to make some alterations in order to enhance the scene and bring an effect. This is often referred to as one of the aspects of ‘creative freedom’.
As a postscript to what you have learnt, keep this in mind as well. When you alter sentences to suit the genre or your writing style, analyse if it delivers the message you are trying to convey. Do it only if your writing demands it.
Types of Syntactic Patterns with Examples
In the English language, there are a few basic syntactic patterns based on which longer sentences are built. Let us look at the basic sentence patterns in English.
Syntactic Pattern 1:
The first pattern is the basic one – sentences with just a subject and a verb.
For example:
It / worked.
Subject / Verb
Diya / was crying.
Subject / Verb
Syntactic Pattern 2:
The second syntactic pattern has another component added to the first one – an object.
For example:
It / is / a puppy.
Subject / Verb / Object
Finola / liked / the dress.
Subject / Verb / Object
Syntactic Pattern 3:
Sentences can have two objects, as discussed earlier, out of which one would be indirect and the other direct. In such a case, the indirect object is preceded by the verb and followed by the direct object.
For example:
Neena / gave / her father / a smartwatch.
Subject / Verb / Indirect object / Direct object
Suhail / showed / me / his collection of stamps.
Subject / Verb / Indirect object / Direct object
Syntactic Pattern 4:
Sentences having a subject, verb, and complement follow the fourth pattern. Subject complements are placed immediately after the verb in the sentence.
For example:
Manisha / is / an actor.
Subject / Verb / Subject complement
They / are / artists.
Subject / Verb / Subject complement
Syntactic Pattern 5:
Object complements are placed after the object in the sentence.
For example:
Randall / found / the book / boring.
Subject / Verb / Object / Object complement
They / made / their friend / happy.
Subject / Verb / Object / Object complement
Syntactic Pattern 6:
Adjuncts can come after the object, before the subject, and before or after the verb in a sentence.
For example:
She / walked / quickly.
Subject / Verb / Adjunct
Yesterday, / we / won.
Adjunct/ Subject / Verb
Syntactic Pattern 7:
An adjunct and a complement can be used in a sentence. Such a sentence can be in either of the following patterns:
i) Adjunct followed by subject, verb and complement.
ii) Subject, verb, complement and then the adjunct.
iii) Subject, verb, adjunct and then the complement.
For example:
Immediately, / she / got / moody.
Adjunct / Subject / Verb / Complement
She / was elected / president of the Rotary Club / this week.
Subject / Verb / Complement / Adjunct
They / were / completely / happy.
Subject / Verb / Adjunct / Complement
Syntactic Pattern 8:
If multiple adjuncts are used in a sentence, the adjuncts can be placed, one at the beginning and the other at the end of the sentence. Another way to do it would be to position it one after the other.
For example:
Last week, / Anu, Teena and Tessa / were / in London.
Adjunct/ Subject / Verb / Adjunct
Anu, Teena and Tessa / travelled / to London / by train.
Subject / Verb / Adjunct / Adjunct
Syntactic Pattern 9:
If a sentence in the subject, verb and object pattern takes an adjunct, it can be placed after the object or before the subject.
For example:
Mia / bought / a smartphone / yesterday.
Subject / Verb / Object / Adjunct
Last week,/ the restaurant / served / kunafa.
Adjunct / Subject / Verb / Object
Check Your Understanding of Syntax
Rewrite the following jumbled sentences using the syntactic structure and also punctuate them appropriately.
served/Tharani/a/soup/me/of/soup
sad/was/Riya
ring/me/a/buy
where/the/hospital/is/nearest/general
Portugal/Prashanth/today/reached
this/is/street/your/stays/friend/the/in
twenty-five/there/in/students/my/are/class
a/is/cricketer/Jaysal
partially/bottle/covered/was/the
hockey/plays/well/Karthik
Here are the answers; check if you have rearranged the words correctly to form meaningful sentences.
Tharini served me a bowl of soup.
Riya was sad.
Buy me a ring.
Where is the nearest general hospital?
Prashanth reached Portugal today.
Is this the street your friend stays in?
There are twenty-five students in my class.
Jaysal is a cricketer.
The bottle was partially covered.
Karthik plays hockey well.
Frequently Asked Questions on Syntax in Linguistics
Q1
What is syntax?
‘Syntax’ is the term used to refer to the arrangement of words, phrases and clauses in a sentence to make them sound and look meaningful.
Q2
What is the definition of syntax?
The Oxford Learner’s Dictionary defines syntax as “the way that words and phrases are put together to form sentences in a language”. According to The Concise Oxford Dictionary of Literary Terms, syntax is “the way in which words and clauses are ordered and connected so as to form sentences”; it also refers to “the set of grammatical rules governing such word order”.
Q3
How does syntax differ from word order?
Syntax refers to the arrangement of words in a sentence. Word order is part of syntax; different arrangements of words help in the formation of different types of sentences.
Q4
Give 5 examples of syntactically correct sentences.
It is sunny today.
The birds flew at once.
When I met you, I thought you were an introvert.
Keep the door open.
When is the next train to Nagpur? |
14369 | https://en.wikibooks.org/wiki/Jet_Propulsion/Propulsive_efficiency | Jet Propulsion/Propulsive efficiency - Wikibooks, open books for an open world
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<Jet Propulsion
The propulsive efficiency is defined as:
η p=P r o p u l s i v e W o r k D e l i v e r e d T o V e h i c l e M e c h a n i c a l W o r k G e n e r a t e d B y E n g i n e{\displaystyle \eta _{p}={\frac {PropulsiveWorkDeliveredToVehicle}{MechanicalWorkGeneratedByEngine}}}
The mechanical work generated by the engine is crudely split between the work difference between the input and exit mass flows from the engine as well as to the work imparted to the vehicle. For a jet engine in the air frame of reference the input mass flow has zero kinetic energy. Thus for a jet engine with a constant relative exit velocity (a close approximation of a turbojet) the propulsive efficiency increases with aircraft velocity.
Mathematically the propulsive efficiency:
η p=2 u 0 u e+u 0{\displaystyle \eta {p}={\frac {2u{0}}{u_{e}+u_{0}}}}
where
u 0={\displaystyle u_{0}=} velocity of the vehicle
u e={\displaystyle u_{e}=} velocity of the exhaust
Thus reducing the exhaust velocity will give a higher propulsive efficiency. The thrust of the engine is given by
F=m˙(u e−u 0){\displaystyle F={\dot {m}}(u_{e}-u_{0})}
where m˙{\displaystyle {\dot {m}}} is the mass flow through the engine. Thus to for a given thrust the exhaust velocity can be decreased if the mass flow is increased. This is usually done by adding a fan or a propeller to the turbojet, resulting in a turbofan or turboprop engine. The reduction in exhaust velocity also reduces the exhaust noise, but has the disadvantage that the thrust reduces more markedly with speed than for a turbojet.
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Orgo lab Report 6
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Organic Chemistry Laboratory I (CHEM 233)
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Lab 6- Preparation of Alkyl Halides by substitution Reac tion: NaI and Agno3 Tests for
Alkyl Halides
Murtuza Khambhati
Lab Day: March 13, 2020
Lab partners: Srilasya Reddy
Introduction:
The SN1 reaction is a substituti on reaction in organic chemistry. "SN" stands for
"nucleophilic substituti on", and the "1" says that the rate-determining step is unimolecular. Thus,
the rate equation is often shown as having first-order depende nce on electrophile and zero-order
dependence on nucleophile. This relationship holds for situations where the amount of
nucleophile is much greater than that of the interm ediate. Instead, the rate equation may be more
accurately described using steady-state kinetics. The reaction involve s a carbocation intermediate
and is commonly seen in reactions of secondary or terti ary alkyl halides under strongly basic
conditions or, under strongly acidic conditions, with secondary or tert iary alcohols (Wade, 2005)
The SN2 reaction is a type of reaction mecha nism that is common in organic chemistry.
In this mechanism, one bond is broken, and one bond is formed synchronously, i.e., in one step.
SN2 is a kind of nucleophilic substitut ion reaction mechanism. Since two reacting species are
involved in the slow (rate-determi ning) step, this leads to the term substitution nucleophilic (bi-
molecular) or SN2; the other major kind is SN1. Many other more special ized mechanisms
describe substitution reactions. (John, 1992)
Methods and Background
The goal of this lab is to prepare a terti ary alkyl halide by SN1 reaction and primary alkyl
halide by SN2 reaction. We used simple distill ation technique to isolate liquid product. The
objective of this lab was to distinguish between primary, secondary, te rtiary alkyl halides using
iodine and nitrate tests. For SN1 reacti on we wash the organic layer with saturated sodium
bicarbonate to neutralize HCl. For SN2 reacti on, Reflux apparatus was set up which can be
defined as state of boiling liquid whose vapors are conti nually returned by condensation;
typically used to hold the reactions at the boiling point of the solvent for exte nded periods of
time without evaporation.
The sodium iodide test and the silver nitrate test, whic h help distinguish between primary
and tertiary alkyl halides, also operate by nucleophili c substitution mechanism. Silver is a strong
enough Lewis acid to cause the dissociati on of a C-Cl bond during the silver nitrate test.
Similarly, HCl protonate s an alcohol to form an oxonium ion, which causes dissociation of C-O
bond during a substitution reaction. Both reactions then involve the format ion of carbocation
intermediate; therefore, both reactions are first order substitut ions sine the rate determining step
involves only one molecule. Since each reaction involve s carbocation intermediates, the rates of
the reaction will increase with the stabil ity and thus increasing substitution.
Between the molecule s of the solvent and the carbon centers are what helps to start the
reactions. Following this, the second step is that the le aving group will receive the pair of
bonding bond electrons while the organic fragments will become a carbocat ion making it a
positive charge. The carbocation can go through a rearrangement to form a more stable
carbocation or lose a proton resulting in an alkene by elim ination. This can occur or as well
combine with the nucleophile to give a substit ution product. This reaction is called solvolysis
only if the nucleophile is the solvent.
Figure 1- SN1 Mechanism
SN2 consists of a direct attack on the substrate by the nucleophil e. This is usually
backside attack. The carbon leaving group is broken down as a formati on of carbon and therefore
creating a nucleophile bond. (Figure 2). This reacti on is called bimolecular, therefore the rate of
reaction depending on the concertation of the nucleophil e and substrate making it second order.
There is difference between SN1 and SN2 in terms of reacti vity. SN2 occurs
simultaneously sure to their nucleophi lic attack from back and detachment of leaving group at
the same time. While in SN1 reacti on, first the Leaving group detaches, and then nucleophilic
attack occur from either side.
Figure 2 – SN2 Mechanism
Procedure:
Synthesis of 1- bromobutane
We started the procedure with measuring of 11.2 grams of sodium bromide and tra nsfer it to a
100ml round bottom flask. We also add 10 ML of water and 10 mL of 1 butanol and 1 boil ing
stone to the flask. By mixing all the contents with a gentl e swirl, we placed the flask in an ice
water bath. We then slowly added 10 mL of concentrated sulfur acid to the cold rea ction mixture
while at the same continuing to mix.
A stopwatch was kept, keeping the time measured to 45 minut es. During this period, the reflux
apparatus would be assembled to start on the SN1 react ion. Section is named “Synthesis of 2 -
Chloro-2 Methyl Butane
After the 45 minutes. The water hoses from the Hampel column are rem oved causing the water
to drain from the outside of the column. Afterwards, the apparat us for simple distillation would
be set up using a 25 mL receiving flask. We collected distil l product only until the temperature
reached at 115 degree Celsius. Collec ting product material above 115 degrees Celsius is not
permitted since it wouldn’t be accurate data. The recei ving flask containing the distilled product
was always kept in the ice bath. Next, the disti llate was transferred to a separatory funnel and
washed with 10 mL of water. Afterward, the 2 layers were separa ted and put in 2 different
beakers. The organic layer was washed one by one with 4 different solutions: 8mL of
concentrated sulfuric acid, two x 5mL of 2M NaOH, 10 mL of water, and 0 mL of saturate d
sodium chloride (brine). Each of these solutions was washed one by one to compl ete the
extraction. Only when sulfuric acid was added, the organic laye r was the top layer. Within the
rest of the solutions, the bottom layer was always the organic layer be ing the denser one.
Finally, the 1-bromobutane was transferred to a clea n Erlenmeyer flask and was mixed with
Na2SO4 until the clumps of solid disappear. After drying, the dried 1-Bromobut ane was
weighted in a flask. That same flask was weighted prior to putt ing the dried 1-Bromobutane. The
percent yield was obtained, and silver nitrat e test and sodium iodide test were carried out to
obtain results.
Synthesis of 2-chloro-2-Methylbutane
We started the preparation of 2-chloror-2-methylbuta ne by adding 10ml of 2 methyl 2 butanol
and 25 ml of concentrated 12M hydrochloric acid in the separatory funnel. The conte nts inside
the separatory funnel was mixed for about a minute gentl y without the stopper on the funnel.
After that stopper was opened continuously after shaking to release pressure. Sepa rated layers
helped to determine the organic layer from the aqueous layer. The organic laye r was washed with
10 mL portions of saturated aqueous sodium chloride and cold saturat ed aqueous sodium
bicarbonate. Afterwards, again the separatory funnel was inverted and opening the stop coc ker
slowly to release gas pressure. This creat ed 2 layers, in which only the organic layer was
separated and washed with 10ml of water and then with aqueous sodium chl oride. The aqueous
layer was transferred to the 2-chloro 2-mehtl ybutane chloride to an Erlenmeyer flask. The 2-
chloro 2-methlynbutane in the Erlenm eyer flask was swirled for about 10 minutes with Na2SO4
until the product became dry.
Finally, the product was transferred to a round bottom flask and was disti lled with simple
distillation technique. The fraction was colle cted with a boiling point of about 75 degrees
Celsius. The product was weighed, and percent yield was cal culated. Silver nitrate test and
sodium iodide test were performed to obtain results.
Figure 3- Simple Distillati on Setup
Sodium Iodide Test:
3 test tubes were prepared by adding 1mL of the sodium iodide reagents in acetone to ea ch. In
one test tube, 1mL of the sodium iodide reagent is added in acetone. To one test tube, we added
1-2 drops of primary alkyl which is 1-bromobutane. The tube was shaken and then l et to stand
for about 3 minutes. If positive, it would be indica ted as precipitate formation. If no precipitate,
then negative. This procedure was repeate d with secondary (2-chlorobutane) and tertiary (2-
chloro2-methlybutane) alkyl halides. 1- bromobutane as well as 2-chloro-2-me htylbutane did
form white precipitates.
Silver Nitrate Test
3 test tubes were prepared by adding 1mL of the silver nitrate reagent in ethanol to eac h. To one
of the test tubes, we added 1-2 drops of primary alkyl which is 1-bromobut ane. The tube was
shaken and then let to stand for about 3 minutes. If posit ive, it would be indicated as precipitate
formation. If no precipitate, then negative. This procedure was repeat ed with secondary (2-
chlorobutane) and tertiary (2-chloro2-methlybutane) alkyl halides. 1- bromobutane forms white
precipitate which is positive while 2-chloro-2-met hylbutane, tertiary alkyl halide results also in
positive with yellow precipitat e formation.
[Sodium iodide test for 2-chloro-2-methyl butane (left) (Result: positive, formation of
precipitates), Silver Nitrate test for 2-chloro-2-methyl butane (Right)(Result: Positive, white
cloudy precipitates)
[Silver nitrate test for 1- bromobutane (left) (Result: positive, yellow precipitate formation,
Sodium iodide test for 1- bromobutane (right) (result: posit ive, white precipitate formation)]
Data Acquisition:
Observation Table 1:
NaI- Sodium Iodide AgNO3- Silver Nitrate Test
1 Bromobutane
primary alkyl halide
White cloudy precipitate
formation
Positive
Fade pale yellow color
formation without
precipitate
Negative
2-Chlorobutane
secondary alkyl halide
Clear liquid but no
precipitate formation
No reaction
Minimal cloudiness in clear
liquid with no precipitate
formation
No reaction
2-Chlorobutane 2-methylbutane
tertiary alkyl halide
white liquid precipitate
formed but didn’t stayed
for minute
Negative
White cloudy precipitate
formation
Positive
Observation Table 2:
Calculations:
Data Acquisition Table 3:
1-Butanol 1 bromobutane 2-methyl-2-
butanol
2-chloro-
2methylbutane
Recorded
Weight or
volume
10 mL 3.8 grams 10 mL 3.15 grams
Density 0.81 g/mol 1.27 g/mol
0.805 g/mol
0.86 g/ml
Molar Mass
(g/mol)
74.121 g/mol 137.02 g/mol 88.15 g/mol 106.53 g/mol
Data Calculations:
Theoretical Yield Calculat ions:
Weight SN2 Reaction-
1 bromobutane
SN1 Reaction-
2 chloro2-methlybutanol
Flask alone 66.70 g 63.87 g
With Solution 70.50 g 67.02 g
Subtracted: (grams
collected)
3.8 grams 3.15 grams
Recorded weight or volume x density of compound x 1 mol of compound = mol of compound
1ml molar mass
mol of compound x 1 mol of other compound x molar mass of other compound = grams of
1 mol of compound 1 mol of other compound theoretical yield
Percent Yield= Actual Yield / Theoreti cal Yield
Theoretical Yield for 1- bromobutane:
10mL 1 butanol x 0.81 g x 1 mol 1-butanol = 0.109 mol 1 butanol
1mL 74.121 g/mol
0.109 mol 1 butanol x 1 mol 1 bromobutane x 137.02 g bromobutane = 14.95 g 1bromobutane
1 mol 1-butanol 1 mol 1-bromobutane
Actual Yield =3.8 grams
Theoretical Yield = 14.95 grams
Percent Yield = 3.8 grams / 14.95 grams = 25.41%
Theoretical Yield for 2-chloro-2methlybutane
10mL 2 methly-2butanol x 0.805g x 1 mol 2methlybutane = 0.091 mol -2methyl2-butane
1mL 88.15 g
0.091 mol 2-methly-2butane x 1 mol 2 chloro2methylbutane x 106.53 g 2chloro-2methlybutane
1mol 2-methly-2-butanol 1 mol 2-chloro-2methlybuta ne
=9.69g of 2-chloro-2methylbutane
Actual Yield = 3.15 grams
Theoretical Yield = 9.69 grams
Percent Yield = 3.15 grams/9.69grams = 32.50%
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Preview text
Lab 6- Preparation of Alkyl Halides by substitution Reaction: NaI and Agno3 Tests forAlkyl Halides
Murtuza Khambhati
Lab Day: March 13, 2020
Lab partners: Srilasya Reddy
Introduction:
The SN1 reaction is a substitution reaction in organic chemistry. "SN" stands for "nucleophilic substitution", and the "1" says that the rate-determining step is unimolecular. Thus, the rate equation is often shown as having first-order dependence on electrophile and zero-order dependence on nucleophile. This relationship holds for situations where the amount of nucleophile is much greater than that of the intermediate. Instead, the rate equation may be more accurately described using steady-state kinetics. The reaction involves a carbocation intermediate and is commonly seen in reactions of secondary or tertiary alkyl halides under strongly basic conditions or, under strongly acidic conditions, with secondary or tertiary alcohols[ CITATION LGW05 \l 1033 ]
The SN2 reaction is a type of reaction mechanism that is common in organic chemistry. In this mechanism, one bond is broken, and one bond is formed synchronously, i., in one step. SN2 is a kind of nucleophilic substitution reaction mechanism. Since two reacting species are involved in the slow (rate-determining) step, this leads to the term substitution nucleophilic (bi- molecular) or SN2; the other major kind is SN1. Many other more specialized mechanisms describe substitution reactions.[ CITATION McM92 \l 1033 ]
Methods and Background
The goal of this lab is to prepare a tertiary alkyl halide by SN1 reaction and primary alkyl halide by SN2 reaction. We used simple distillation technique to isolate liquid product. The objective of this lab was to distinguish between primary, secondary, tertiary alkyl halides using iodine and nitrate tests. For SN1 reaction we wash the organic layer with saturated sodium bicarbonate to neutralize HCl. For SN2 reaction, Reflux apparatus was set up which can be defined as state of boiling liquid whose vapors are continually returned by condensation; typically used to hold the reactions at the boiling point of the solvent for extended periods of time without evaporation.
The sodium iodide test and the silver nitrate test, which help distinguish between primary and tertiary alkyl halides, also operate by nucleophilic substitution mechanism. Silver is a strong enough Lewis acid to cause the dissociation of a C-Cl bond during the silver nitrate test. Similarly, HCl protonates an alcohol to form an oxonium ion, which causes dissociation of C-O bond during a substitution reaction. Both reactions then involve the formation of carbocation intermediate; therefore, both reactions are first order substitutions sine the rate determining step
involves only one molecule. Since each reaction involves carbocation intermediates, the rates of the reaction will increase with the stability and thus increasing substitution.
Between the molecules of the solvent and the carbon centers are what helps to start the reactions. Following this, the second step is that the leaving group will receive the pair of bonding bond electrons while the organic fragments will become a carbocation making it a positive charge. The carbocation can go through a rearrangement to form a more stable carbocation or lose a proton resulting in an alkene by elimination. This can occur or as well combine with the nucleophile to give a substitution product. This reaction is called solvolysis only if the nucleophile is the solvent.
Figure 1- SN1 Mechanism SN2 consists of a direct attack on the substrate by the nucleophile. This is usually backside attack. The carbon leaving group is broken down as a formation of carbon and therefore creating a nucleophile bond. (Figure 2). This reaction is called bimolecular, therefore the rate of reaction depending on the concertation of the nucleophile and substrate making it second order.
There is difference between SN1 and SN2 in terms of reactivity. SN2 occurs simultaneously sure to their nucleophilic attack from back and detachment of leaving group at the same time. While in SN1 reaction, first the Leaving group detaches, and then nucleophilic attack occur from either side.
Figure 2 – SN2 Mechanism
chloro 2-methlynbutane in the Erlenmeyer flask was swirled for about 10 minutes with Na2SO until the product became dry.
Finally, the product was transferred to a round bottom flask and was distilled with simple distillation technique. The fraction was collected with a boiling point of about 75 degrees Celsius. The product was weighed, and percent yield was calculated. Silver nitrate test and sodium iodide test were performed to obtain results.
Figure 3- Simple Distillation Setup
Sodium Iodide Test:
3 test tubes were prepared by adding 1mL of the sodium iodide reagents in acetone to each. In one test tube, 1mL of the sodium iodide reagent is added in acetone. To one test tube, we added 1-2 drops of primary alkyl which is 1-bromobutane. The tube was shaken and then let to stand for about 3 minutes. If positive, it would be indicated as precipitate formation. If no precipitate, then negative. This procedure was repeated with secondary (2-chlorobutane) and tertiary (2- chloro2-methlybutane) alkyl halides. 1- bromobutane as well as 2-chloro-2-mehtylbutane did form white precipitates.
Silver Nitrate Test
3 test tubes were prepared by adding 1mL of the silver nitrate reagent in ethanol to each. To one of the test tubes, we added 1-2 drops of primary alkyl which is 1-bromobutane. The tube was shaken and then let to stand for about 3 minutes. If positive, it would be indicated as precipitate
formation. If no precipitate, then negative. This procedure was repeated with secondary (2- chlorobutane) and tertiary (2-chloro2-methlybutane) alkyl halides. 1- bromobutane forms white precipitate which is positive while 2-chloro-2-methylbutane, tertiary alkyl halide results also in positive with yellow precipitate formation.
[Sodium iodide test for 2-chloro-2-methyl butane (left) (Result: positive, formation of precipitates), Silver Nitrate test for 2-chloro-2-methylbutane (Right)(Result: Positive, white cloudy precipitates)
[Silver nitrate test for 1- bromobutane (left) (Result: positive, yellow precipitate formation, Sodium iodide test for 1- bromobutane (right) (result: positive, white precipitate formation)]
Data Calculations:
Theoretical Yield Calculations:
Recorded weight or volume x density of compound x 1 mol of compound = mol of compound
1ml molar mass
mol of compound x 1 mol of other compound x molar mass of other compound = grams of
1 mol of compound 1 mol of other compound theoretical yield
Percent Yield= Actual Yield / Theoretical Yield
Theoretical Yield for 1- bromobutane:
10mL 1 butanol x 0 g x 1 mol 1-butanol = 0 mol 1 butanol
1mL 74 g/mol
0 mol 1 butanol x 1 mol 1 bromobutane x 137 g bromobutane = 14 g 1bromobutane
1 mol 1-butanol 1 mol 1-bromobutane
Actual Yield =3 grams
Theoretical Yield = 14 grams
Percent Yield = 3 grams / 14 grams = 25%
Theoretical Yield for 2-chloro-2methlybutane
10mL 2 methly-2butanol x 0 x 1 mol 2methlybutane = 0 mol -2methyl2-butane
1mL 88 g
0 mol 2-methly-2butane x 1 mol 2 chloro2methylbutane x 106 g 2chloro-2methlybutane
1mol 2-methly-2-butanol 1 mol 2-chloro-2methlybutane
=9 of 2-chloro-2methylbutane
Actual Yield = 3 grams
Theoretical Yield = 9 grams
Percent Yield = 3 grams/9 = 32%
Data Collection Table 4:
Product Name 2-Chloro- Methyl-Butane
2-ChloroButane 1-Bromobutane
% Yield 32% 25%
Substitution Tertiary n/a provided Primary
Boiling Point 85-86 degrees Celsius
70 degrees Celsius
101 degrees Celsius AgNO3 Test Positive Negative Negative NaI Test Negative Negative Positive
Conclusion:
The purpose of this lab was to prepare a tertiary alkyl halide by SN1 reaction and primary alkyl halide by SN2 reaction. The goals of this lab were to use those products to calculate the percent yield. A positive silver nitrate test indicates for SN1 reaction while a positive sodium iodide test indicates for SN2 reaction.
The silver nitrate test for alkyl chlorides can be inferred that AgCl did precipitate in ethanol following the SN1 mechanism. This is due to 2-Chlorobutane 2-methylbutane showing that the reactivity for alkyl halides is strongest for tertiary. Ag+ is strong enough to remove Cl, making tertiary the most reactivity, and primary alkyl halides the least reactive.
Iodide test for alkyl chlorides, showed positive test when observing NaCl precipitate in acetone following the SN2 mechanism. Based on our mechanisms this is true due to Na+ not being a strong Acid to remove Cl. This leads to iodide having to attack the electrophilic carbon at the same time that Cl is leaving. Primary alkyl halides would be the fastest.
The percent yield for 2-chloro-2-butane was 32 %. While 1-Bromobutane was obtained as 25%. (Final Data Table 4). The percent yields were poor meaning that they had a low percent making the possibility of an error in the data. This meant that after doing the experiment not much too little of the sample was recovered from the reaction. This could have happened when distillation might have inappropriately collected since it had reached a high temperature before it was even collected or some of the sample was lost within the flask when drying and extracting.
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14371 | https://www.reddit.com/r/mathematics/comments/1fb9dcz/is_there_a_mathematical_symbol_that_means_compare/?tl=de | Gibt es ein mathematisches Symbol, das „vergleichen“ bedeutet, ohne anzugeben, ob ein Term größer, kleiner oder gleich einem anderen ist? : r/mathematics
Skip to main contentGibt es ein mathematisches Symbol, das „vergleichen“ bedeutet, ohne anzugeben, ob ein Term größer, kleiner oder gleich einem anderen ist? : r/mathematics
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Gibt es ein mathematisches Symbol, das „vergleichen“ bedeutet, ohne anzugeben, ob ein Term größer, kleiner oder gleich einem anderen ist?
Ich studiere Chemie und als ich mir Notizen machte, brauchte ich ein solches Symbol, da die beiden Größen je nach Situation unterschiedliche Werte haben. Ich wollte ein Symbol, das einfach die Notwendigkeit zum Ausdruck bringt, die beiden Begriffe zu vergleichen. Ich habe mir dieses Symbol spontan ausgedacht, frage mich aber, ob es tatsächlich etwas Ähnliches gibt.
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14372 | https://fiveable.me/key-terms/cell-biology/proton-motive-force | printables
🦠cell biology review
key term - Proton Motive Force
Citation:
MLA
Definition
Proton motive force refers to the energy stored as a result of a proton gradient across a membrane, which is crucial for ATP synthesis in both mitochondria and chloroplasts. This electrochemical gradient is created by the active transport of protons (H+ ions) out of the mitochondrial or thylakoid lumen, resulting in a higher concentration of protons outside than inside, generating potential energy that drives ATP synthase to produce ATP during cellular respiration and photosynthesis.
5 Must Know Facts For Your Next Test
The proton motive force is generated during both oxidative phosphorylation in mitochondria and photophosphorylation in chloroplasts, playing a central role in energy conversion.
The inner mitochondrial membrane and thylakoid membrane are key sites where proton gradients are established, highlighting the importance of membrane structure in bioenergetics.
The strength of the proton motive force is influenced by both the concentration gradient and the electrical potential difference across the membrane, known as membrane potential.
Proton motive force not only drives ATP synthesis but also powers other processes like active transport of metabolites across membranes.
Inhibitors such as oligomycin can disrupt ATP production by blocking ATP synthase, illustrating how vital proton motive force is for cellular energy metabolism.
Review Questions
How does the generation of proton motive force contribute to ATP production in mitochondria?
The generation of proton motive force involves pumping protons across the inner mitochondrial membrane during electron transport. This creates a high concentration of protons in the intermembrane space compared to the mitochondrial matrix. When protons flow back into the matrix through ATP synthase, their movement drives the conversion of ADP and inorganic phosphate into ATP, illustrating the direct link between proton motive force and ATP production.
Compare and contrast the mechanisms by which proton motive force is generated in mitochondria and chloroplasts.
In mitochondria, proton motive force is generated through oxidative phosphorylation where electrons from NADH and FADH2 pass through the electron transport chain, leading to proton pumping into the intermembrane space. In chloroplasts, during photosynthesis, light energy drives electrons through a similar chain in thylakoid membranes, creating a proton gradient in the thylakoid lumen. Both processes harness an electrochemical gradient to synthesize ATP but differ in their energy sources—chemical versus light.
Evaluate the significance of proton motive force in cellular metabolism and how its disruption could impact cellular functions.
Proton motive force is critical for ATP synthesis, which is essential for various cellular processes including muscle contraction, biosynthesis, and active transport. If proton motive force is disrupted, such as by uncouplers that dissipate the gradient or inhibitors blocking ATP synthase, it can lead to reduced ATP levels. This decline in energy availability can compromise vital cellular functions and overall metabolism, potentially resulting in cell death if energy demands are not met.
Related terms
ATP Synthase: An enzyme that catalyzes the formation of ATP from ADP and inorganic phosphate using the energy provided by the flow of protons back into the mitochondrial matrix or thylakoid lumen.
Electron Transport Chain: A series of protein complexes and other molecules that transfer electrons from electron donors to electron acceptors, creating the proton gradient necessary for generating proton motive force.
Chemiosmosis: The process through which ATP is produced as protons move down their concentration gradient through ATP synthase, utilizing the energy of the proton motive force.
"Proton Motive Force" also found in:
Subjects (2)
Biological Chemistry I
General Biology I |
14373 | https://www.nice.org.uk/guidance/qs44/chapter/quality-statement-7-treatment-of-eczema-herpeticum | Quality statement 7: Treatment of eczema herpeticum | Atopic eczema in under 12s | Quality standards | NICE
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Eczema
Atopic eczema in under 12s
Quality standard
Reference number: QS44
Published: 05 September 2013
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Quality statements
Quality statement 1: Assessment at diagnosis
Quality statement 2: Stepped approach to management
Quality statement 3: Psychological wellbeing and quality of life
Quality statement 4: Provision of emollients
Quality statement 5: Referral for specialist dermatological advice
Quality statement 6: Specialist allergy investigation
Quality statement 7: Treatment of eczema herpeticum
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Quality statement
Rationale
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What the quality statement means for different audiences
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Definitions of terms used in this quality statement
Quality statement 7: Treatment of eczema herpeticum
Quality statement
Children with atopic eczema who have suspected eczema herpeticum receive immediate treatment with systemic aciclovir and are referred for same-day specialist dermatological advice.
Rationale
Eczema herpeticum (widespread herpes simplex virus) is a serious under-recognised condition and, if not diagnosed promptly, the child's condition may deteriorate rapidly. Eczema herpeticum can be fatal or can lead to blindness if not treated, and should therefore be an indication for urgent referral.
Quality measures
The following measures can be used to assess the quality of care or service provision specified in the statement. They are examples of how the statement can be measured, and can be adapted and used flexibly.
Structure
Evidence of local arrangements to ensure that children with atopic eczema who have suspected eczema herpeticum receive immediate treatment with systemic aciclovir and are referred for same-day specialist dermatological advice.
Data source: No routinely collected national data for this measure has been identified. Data can be collected from information recorded locally by healthcare professionals and provider organisations, for example from referral pathways.
Process
The proportion of children with atopic eczema who have suspected eczema herpeticum who receive immediate treatment with systemic aciclovir and are referred for same-day specialist dermatological advice.
Numerator – the number of children in the denominator who receive immediate treatment with systemic aciclovir and are referred for same-day specialist dermatological advice.
Denominator – the number of children with atopic eczema who have suspected eczema herpeticum.
Data source: No routinely collected national data for this measure has been identified. Data can be collected from information recorded locally by healthcare professionals and provider organisations, for example from patient records.
What the quality statement means for different audiences
Service providers ensure that local arrangements are in place for children with atopic eczema who have suspected eczema herpeticum to receive immediate treatment with systemic aciclovir and to be referred for same-day specialist dermatological advice.
Healthcare practitioners ensure that children with atopic eczema who have suspected eczema herpeticum receive immediate treatment with systemic aciclovir and are referred for same-day specialist dermatological advice.
Commissioners ensure that they commission services with local arrangements to give children with atopic eczema who have suspected eczema herpeticum immediate treatment with systemic aciclovir and to refer them for same-day specialist dermatological advice.
Children with atopic eczema who have suspected eczema herpeticum (a rare but serious infection caused by the same virus that causes cold sores) receive immediate treatment with an antiviral drug (called systemic aciclovir), which can be given as medicine or an injection, and are referred immediately for same-day specialist advice.
Source guidance
Atopic eczema in under 12s: diagnosis and management. NICE guideline CG57 (2007, updated 2023), recommendations 1.5.1.40 to 1.5.1.42 and 1.7.1.1
Definitions of terms used in this quality statement
Suspected eczema herpeticum
Eczema herpeticum is a widespread herpes simplex virus. Signs of eczema herpeticum are:
areas of rapidly worsening, painful eczema
clustered blisters consistent with early-stage cold sores
punched-out erosions (circular, depressed, ulcerated lesions) usually 1 mm to 3 mm that are uniform in appearance (these may coalesce to form larger areas of erosion with crusting)
possible fever, lethargy or distress.
[Adapted from NICE's guideline on atopic eczema in under 12s, recommendation 1.5.1.42]
Specialist dermatological advice
The referral should be to a specialist dermatological unit dealing with paediatric patients, for example, a clinician with experience and qualifications in paediatric dermatology. This could include a specialist nurse or a GP with a specialist interest if they are working within a dermatological unit and trained in paediatric dermatology. If eczema herpeticum involves the skin around the eyes, the child should be referred for same-day ophthalmological and dermatological advice. [Adapted from NICE's guideline on atopic eczema in under 12s, recommendations 1.5.1.40, 1.5.1.41, 1.7.1.1 and expert opinion]
Systemic aciclovir
Oral or intravenous aciclovir can be given depending on the clinical situation. Aciclovir is likely to be given orally in primary care and intravenously in secondary care. NICE's full guideline on atopic eczema in under 12s recommends that if a child with atopic eczema has a lesion on the skin suspected to be herpes simplex virus, treatment with oral aciclovir should be started even if the infection is localised. [Adapted from NICE's guideline on atopic eczema in under 12s, recommendations 1.5.1.39 to 1.5.1.41 and expert opinion]
Next page Update informationPrevious page Quality statement 6: Specialist allergy investigation
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14374 | https://innovationspace.ansys.com/courses/wp-content/uploads/sites/5/2020/05/Lesson-3-Engineering-Strain-and-True-Strain.pdf | Engineering Strain and True Strain Mechanical Strain in Deformation Analysis – Lesson 3 • DECEMBER 2019 Strain Measurement – Engineering Strain Engineering Strain Engineering strain is defined as the ratio between the change in length and the original length.
Assumption: for each step during deformation, the change of deformation is uniform, resulting in step-incremental deformation.
Strain Measurement – Engineering Strain (cont.) step-incremental deformation Engineering Strain True strain is the change in length with respect to the instant length.
Strain Measurement – True Strain True Strain True strain: deformation is continuous instead of stepwise.
Strain Measurement – True Strain (cont.) Deformation is continuous!
Infinitesimal True Strain Stepwise vs. Continuous Let’s put the two strain definitions on the same page and discuss the relationship between them.
Strain Measurement True Strain Relationship between true and engineering strain: Engineering Strain Strain Measurement (cont.) No Deformation Let’s check whether these two strain measurements meet the requirements of a strain definition: True Strain Engineering Strain No deformation Small deformation Zero Strain Same strain value Strain Measurement at Small Deformation Small Deformation No deformation Small deformation Zero Strain Same strain value True Strain Engineering Strain Strain Measurement at Large Deformation Large Deformation However, when the deformation is comparatively large, the difference between the two strain measurements will show up.
When the change in length is 30% of the original length, the engineering strain is 0.3, but the true strain is calculated to be 0.26. That’s more than a 10% difference.
True Strain Engineering Strain Strain Measurement Let’s plot the two strain measurements with length change as the horizontal axis and strain value as the vertical axis.
Engineering strain is a straight line, while true strain is nonlinear. They deviate from each other with increasing deformation.
True Strain Engineering Strain |
14375 | https://www.youtube.com/watch?v=fUszhfJynd4 | intersecting chords of circles (KristaKingMath)
Krista King
273000 subscribers
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Description
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Posted: 22 Dec 2014
► My Geometry course:
In this video we'll learn about what happens when two chords intersect each other inside a circle. Remember that a chord is just a line segment that has its endpoints on the circle. When two chords intersect each other, each chord is divided into two segments. If we multiply the lengths of each segment together, the products will always be equal.
● ● ● GET EXTRA HELP ● ● ●
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● ● ● CONNECT WITH KRISTA ● ● ●
Hi, I’m Krista! I make math courses to keep you from banging your head against the wall. ;)
Math class was always so frustrating for me. I’d go to a class, spend hours on homework, and three days later have an “Ah-ha!” moment about how the problems worked that could have slashed my homework time in half. I’d think, “WHY didn’t my teacher just tell me this in the first place?!”
So I started tutoring to keep other people out of the same aggravating, time-sucking cycle. Since then, I’ve recorded tons of videos and written out cheat-sheet style notes and formula sheets to help every math student—from basic middle school classes to advanced college calculus—figure out what’s going on, understand the important concepts, and pass their classes, once and for all. Interested in getting help? Learn more here:
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45 comments
Transcript:
in this video we're talking about the interesting thing that happens when two chords intersect one another inside a circle so if we look at this first figure here we have a circle and we have two chords in the circle remember that a chord is basically just a line segment that's inside of the circle that has its end points on the perimeter of the circles so for example this line segment here line segment AB is a chord because it's just a line segment inside the circle with each end point and point a and end point B on the edge of the circle so we have chord ab and we have chord CD and they intersect each other at this point here in the middle so when we have two intersecting chords here's the interesting thing that happens you can take the two measurements of each segment of the chords so for example chord AB this portion of it up until the intersection Point starting at a and up to the intersection point is 10 units from the intersection point to B we've got eight units so the total length of the chord is 18 but if we multiply 10 by 8 we can actually set that equal to the product of the lengths of the segments of the other cord so in other words here let's look at the two segments for chord AB well that's 10 and 8 so if we say 10 mtip 8 we can set that equal to the lengths of these other segments for chord CD which are X and 16 so we'll do 16 X and then we can solve for this unknown value of x so we get 80 is equal to 16x if we divide both sides by 16 we get X is equal to 5 and we can go ahead and say that the length of this segment here from C to the intersection point is five so that's the unique relationship that we have between intersecting chords within a circle we can do this here with another example we have chord QR and chord St we know these lengths 9 6 4 and X so if we look at the chord QR we have lengths 9 and four so we'll say 9 4 and then set that equal to the lengths from the two segments of the other chord 6 and X so 6 X that's going to give us 36 is = to 6X and dividing both sides by 6 we get x equal to 6 and we can even do this when we have multiple unknown lengths we just take the two lengths from chords EF so x and x - 2 so we'll get x x - 2 multiply those together set that equal to the product of the lengths from the other chord so three and 8 so I get 3 8 when I multiply out here I'm going to get x^2 minus 2x is equal to 24 If I subtract 24 from both sides I'm going to get x^2 - 2x - 24 is equal to zero now I can go ahead and Factor the leftand side so I'm going to get x - 6 x + 4 when I factor and then if I set each of these factors individually equal to zero I can say x - 6 = 0 or x = 6 and x + 4al 0 or X is equal to -4 so I have two possible solutions for X I need to test them in in my circle here to figure out which one is the correct solution and I can see right away that xal 6 is going to be the correct solution because I can't possibly have a negative value this distance here from e to the intersection point along chord EF this distance is X well I can't have a distance of-4 that's impossible you can't identify a negative distance so xal -4 can't possibly be a real solution to this problem that means that xal 6 is going to be the unknown value X and if we want to go ahead and plug in here we can say that this length is going to be 6 when we plug 6 into x - 2 we're going to get 6 - 2 or 4 so we can say that this length is equal to four and then if we just want to double check ourselves we should be able to say 6 4 or 24 is equal to 3 8 or 24 so in fact it does check out as a value for x that satisfies this formula of intersecting chords |
14376 | https://www.greenemath.com/College_Algebra/58/Completing-the-SquarePracticeTest.html | GreeneMath.com Master Math & Ace Your Exam!
Completing the Square Practice Test
About Completing the Square:
We previously learned how to solve quadratic equations by factoring. In many cases, we must utilize a different method. When this occurs, we can turn to a method known as completing the square. This method creates a perfect square trinomial on one side and sets it equal to a constant on the other. We can then solve using the square root property.
Test Objectives
Demonstrate the ability to use the square root property
Demonstrate the ability to solve a quadratic equation by completing the square
Demonstrate the ability to solve a quadratic equation with a complex solution
Completing the Square Practice Test:
1:
Instructions: solve each equation.
$$a)\hspace{.2em}x^2 - 4x - 32=0$$
$$b)\hspace{.2em}x^2 - 4x - 60=0$$
Watch the Step by Step Video Solution | View the Written Solution
2:
Instructions: solve each equation.
$$a)\hspace{.2em}x^2 - 10x - 36=0$$
$$b)\hspace{.2em}3x^2 + 6x - 70=-10$$
Watch the Step by Step Video Solution | View the Written Solution
3:
Instructions: solve each equation.
$$a)\hspace{.2em}x^2 - 4x + 48=-5$$
$$b)\hspace{.2em}2x^2 + 4x + 4=10$$
Watch the Step by Step Video Solution | View the Written Solution
4:
Instructions: solve each equation.
$$a)\hspace{.2em}4x^2 + 73=6 - 2x$$
$$b)\hspace{.2em}8x^2 - 10x + 5=-8x$$
Watch the Step by Step Video Solution | View the Written Solution
5:
Instructions: solve each equation.
$$a)\hspace{.2em}69 - 17x=4x - 2x^2$$
$$b)\hspace{.2em}10x^2 - 4x=142$$
Watch the Step by Step Video Solution | View the Written Solution
Written Solutions:
1:
Solutions:
$$a)\hspace{.2em}x=-4, 8$$
$$b)\hspace{.2em}x=-6, 10$$
Watch the Step by Step Video Solution
2:
Solutions:
$$a)\hspace{.2em}x=5 \pm \sqrt{61}$$
$$b)\hspace{.2em}x=-1 \pm \sqrt{21}$$
Watch the Step by Step Video Solution
3:
Solutions:
$$a)\hspace{.2em}x=2 \pm 7i$$
$$b)\hspace{.2em}x=-3, 1$$
Watch the Step by Step Video Solution
4:
Solutions:
$$a)\hspace{.2em}x=\frac{-1 \pm i\sqrt{267}}{4}$$
$$b)\hspace{.2em}x=\frac{1 \pm i\sqrt{39}}{8}$$
Watch the Step by Step Video Solution
5:
Solutions:
$$a)\hspace{.2em}x=\frac{21 \pm i\sqrt{111}}{4}$$
$$b)\hspace{.2em}x=\frac{1 \pm 2\sqrt{89}}{5}$$
Watch the Step by Step Video Solution |
14377 | https://blog.csdn.net/tMb8Z9Vdm66wH68VX1/article/details/109140168 | 理解贝叶斯定理-CSDN博客
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两道例题详解贝叶斯定理
理解贝叶斯定理
最新推荐文章于 2025-08-05 16:35:27 发布
转载 于 2020-10-17 17:00:00 发布·6.5k 阅读
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来源：大数据DT
本文约3400字，建议阅读5分钟。
本文首先讲解条件概率及贝叶斯定理，之后有两道例题，看看你都能答对吗？
首先来看一些概率的重要准则，这些准则大多数都和我们的直觉相符。
概率在0～1之间变化，即0≤P（A）≤1，其中P（A）代表事件A发生的概率。
如果事件确定发生，则其发生概率为1。即，当事件A确定发生时，P（A）=1。
如果事件确定不会发生，则其发生概率为0。即，当事件A确定不发生时，P（A）=0。
如果事件A和事件B不能同时发生，则称它们是互斥的。当事件A和事件B互斥时，任一事件（事件A或事件B）发生的概率就是每个事件发生概率之和，即：P（A或B）=P（A）+P（B）
概率计算规则中还有一个重要的概念是条件概率（conditional probability）。条件概率是指当某一事件（B）发生时，另一事件（A）发生的概率，即P（A|B）。当事件A和事件B互斥时，则有：P（A|B）=0，因为如果事件B发生，事件A就不可能发生（同样地，P（B|A）=0）。当事件A和事件B不互斥时，条件概率的表达公式为：
需要注意的是，一般来说P（A|B）不一定与P（B|A）相等，因为后者的表达公式中分母不同：
但是，我们可以根据条件概率P（A|B）推算出条件概率P（B|A），反之亦然。这样，通过一个简单的代数运算后，我们就得出了决策理论界最著名的定理之一——贝叶斯定理，有时也称为贝叶斯规则。18世纪，英格兰坦布里奇韦尔斯（Tunbridge Wells）城的牧师托马斯·贝叶斯首先提出此定律：
上面公式中仅有“～A”符号前面未出现过，“～A”表示“非A”，即事件A不发生。因此，P（～A）是指事件A不发生的概率。
所有这些概率规则都很重要，而对于判断和决策来说，贝叶斯定理尤为重要。贝叶斯定理是我们应该如何基于新证据来更新我们的特定信念的准则。将公式中的A替换为“焦点假设”（标记为H），将B替换为“一组收集的与假设相关的数据”（标记为D），就会得到下面的公式：
公式中，P（H）是指在收集数据之前焦点假设为真的概率估计，P（～H）是指在收集数据之前的备择假设（即～H）为真的概率估计。
公式中还包括其他一些条件概率：P（H|D）表示焦点假设在观察到特定数据之后为真的概率（有时称为后验概率）, P（D|H）是在焦点假设为真的情况下观察到特定数据的概率，P（D|～H）是在备择假设为真的情况下观察到特定数据的概率。
需要强调的是，P（D|H）和P（D|～H）并不是互补的（它们加起来不等于1.0）。因为数据有时会看起来既支持焦点假设也支持备择假设，或者两者都不支持。
在一些实际案例中，人们常常很难遵守贝叶斯法则。但需要强调的是，当人们在贝叶斯推理上犯了错误时，并不是指他们计算错误，或得到的数据不准确，而是指他们在概率估计过程中犯了严重的性质上的错误。
简而言之，这些案例考察人们是否具有贝叶斯思维的意识，即他们是否对呈现的重要信息和正确的思考方向有足够的敏锐度。标准贝叶斯统计必然包括计算在内，但要避免概率相关的思维错误，人们只需要学会如何基于概率逻辑正确思考即可。
下面两个案例将详细介绍一些常见概率推理错误。
01 出租车问题
第一个被称为出租车问题，学术界对这个问题的研究已经超过30年。
某个夜晚，一辆出租车肇事后逃逸。该城市共有两家出租车公司，一家公司的出租车均为绿色（“绿色”公司），拥有出租车数量为全市出租车总数的85%；另一家公司的出租车均为蓝色（“蓝色”公司），拥有出租车数量为全市出租车总数的15%。一名目击者称肇事出租车是“蓝色”公司的。法院对目击者的证词进行了测试，发现目击者在出事当时那种情况下正确识别两种颜色的概率是80%。那么肇事出租车是蓝色的概率是多少（用百分数表示，范围从0%到100%）？
被试被告知不必精确计算答案，只需要给出一个大致的估计值。考察的关键点不在于答案的精确度，而在于人们的估计是否在一个大致正确的范围内。很遗憾，许多人的答案并不在这个范围内。
在出租车问题上，贝叶斯定理提供了一个最佳方法，即将给定的以下两条信息结合起来分析：
15%的出租车是蓝色。
目击者认为该出租车是蓝色的（识别准确率为80%）。
大多数人并不能自然地将两条信息综合考虑。事实上，很多人在知道了肇事出租车为蓝色的概率只有0.41后感到很震惊，因为他们没有意识到尽管目击者声称肇事车辆是蓝色的，但是肇事出租车仍更可能是绿色的（0.59），而非蓝色的（0.41）。原因是出租车是绿色的先验概率（85%）高于目击者识别出租车为蓝色的可信度（80%）。
如果不使用贝叶斯计算公式，我们来看一下0.41的概率是如何得到的：
在100起此类事故中，15辆出租车是蓝色的，而目击者能够正确辨认其中的80%（12辆）；同样在这100起事故中，有85辆出租车是绿色的，而目击者会将其中的20%（17辆）辨认为蓝色。因此，将会有29（12+17）辆出租车被辨认为蓝色，而事实上只有12辆是蓝色的，所以肇事出租车是蓝色的概率为41%。
根据贝叶斯法则，计算公式如下：
只有不到一半的被试给出的答案介于0.20～0.70，而大部分答案接近0.80。简而言之，他们的答案依据的是目击者正确辨认的概率，而没有根据较低的先验概率（0.15）对目击者的判断概率打折扣。
大多数人高估了肇事出租车是蓝色的可能性，因为他们将注意力集中在目击者的辨认上，而忽视了蓝色出租车的基础概率。在需要结合抽象概率信息进行判断时，人们往往会倾向于高估具体的鲜活的个案信息。
02 医疗风险评估
第二个例子与出租车问题的逻辑相同，但是更贴近日常生活，涉及医疗风险评估的问题，同样被许多研究所关注：
假设XYZ病毒能够引起严重的疾病，该病发病率为千分之一。假设有一种化验方法，可以精准地检测到该病毒。也就是说，如果一个人携带XYZ病毒，一定可以被检测出来。但是该项化验的假阳性率为5%，即健康人接受该项化验，会有5%的可能性被误诊为病毒携带者。假设从人群中随机选择一人进行检测，化验结果为阳性（阳性意味着受检者可能是XYZ病毒携带者）。那么，在不考虑具体症状、病史等情况下，此人携带XYZ病毒的概率是多少？（用百分数表示，范围从0到100%。）
最常见的答案是95%，而正确答案是约为2%！人们极大地高估了阳性结果代表个体为XYZ病毒携带者的概率，这与出租车问题一样，人们倾向于重视具体信息，而忽视基础概率信息。
尽管使用贝叶斯法则能够计算出正确答案，但是简单的数学推理也能帮助我们厘清基础概率对预估结果产生的巨大影响。我们已知的信息是：每1000人中只有1人是真正的XYZ病毒携带者。如果另外999位未携带病毒者全部接受化验，由于化验的假阳性率为5%，那么将有约50人的检测结果呈假阳性（0.05乘以999），因此有51人检测结果呈阳性，而实际上只有1人（约2%）为真的病毒携带者。
总之，由于XYZ病毒的基础感染率非常低，绝大多数人并未感染，再加上较高的化验假阳性率，因此可以推断大部分检查结果为阳性的人并非病毒携带者。
根据贝叶斯法则，计算如下：
小结
在以上两个例子中，很多人高估了个案证据，而低估了统计证据。对大多数人来说，个案证据（目击者辨认、实验室化验结果）看起来更“具体”、更“触手可及”、更“活灵活现”。相比来说，概率证据看起来好像太过抽象了，因而无法使用！
这种想法当然是错误的，因为个案证据本身也是概率性的。目击者只能在一定程度上做出正确辨认，临床检验也有一定程度的误诊率。人们若想做出正确决策，就必须同时考虑情境中涉及的两种概率：个案证据正确的概率和先验概率。
在所有的基础概率问题中，若想做出正确决策，人们必须将个案证据的判断概率与先验概率结合起来。正确的做法是，要么用贝叶斯公式进行计算，要么基于贝叶斯法则将基础概率和个案信息结合起来进行推理。
对于贝叶斯推理的讨论，我们并不是期望人们使用贝叶斯公式进行具体计算，而只是希望人们学会从定性的角度进行“贝叶斯式思考”，或者形成“贝叶斯直觉”。
比如，仅仅意识到基础概率的重要性，就足以使人们洞察到隐藏在XYZ病毒问题中的关键点：假阳性率较高的检查方法应用于基础发病率很低的疾病时，多数检查结果呈阳性的个体其实可能并未患病。
简而言之，这部分题目测量的是人们对概率的自然判断是否或近似遵循了贝叶斯定理。很明显，人们的概率判断往往采用“大致估计”的策略。我们提倡在“大致估计”时，应遵循贝叶斯法则。遵循贝叶斯法则无须知道贝叶斯公式，也不需要进行任何有意识的计算。
关于作者：
基思·斯坦诺维奇(Keith E.Stanovich)，加拿大多伦多大学人类发展与应用心理学终身荣誉教授，曾任加拿大应用认知科学首席科学家。他的研究领域是推理、决策和阅读的心理学机制，至今已发表200多篇科学论文，出版8部著作。丹尼尔·卡尼曼在其诺贝尔奖致辞中也多次引用斯坦诺维奇的研究成果。
本文摘编自《理商：如何评估理性思维》，经出版方授权发布。
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Why is Boolean a lattice?
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$\begingroup$
I've had minimal exposure to lattice theory but I must answer this question due to a project I'm working in. If anyone could answer this question in the simplest explanation possible with examples included then I'd highly appreciate it!
I looked up the definition of lattice: a poset in which every two elements have a unique sup and inf. Can someone give me an example of a poset?
What is the binary relation in a Boolean?
Sorry for asking these basic lattice theory questions. For someone who hasn't taken boolean algebra class, this stuff is pretty advanced and confusing! I don't really need rigorous proof!
order-theory
boolean-algebra
lattice-orders
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edited Jun 9, 2016 at 4:41
Arman Malekzadeh
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asked Jun 9, 2016 at 4:14
RainroadRainroad
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$\begingroup$ What is your definition of "Boolean algebra"? $\endgroup$
Eric Wofsey
– Eric Wofsey
2016-06-09 04:17:41 +00:00
Commented Jun 9, 2016 at 4:17
$\begingroup$ If you want to ask about Boolean algebra, the Wikipedia article mentions which part of the definition corresponds to the definition of a lattice. To be more specific $\sup{a,b} = a\vee b$ and $\inf{a,b} = a\wedge b$. $\endgroup$
Martin Sleziak
– Martin Sleziak
2016-06-09 06:39:04 +00:00
Commented Jun 9, 2016 at 6:39
$\begingroup$ Could you give me an example of a poset? Also the part you are referring to, is it in the definition section? $\endgroup$
Rainroad
– Rainroad
2016-06-09 19:07:25 +00:00
Commented Jun 9, 2016 at 19:07
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1 Answer 1
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$\begingroup$
I'm a bit confused by your question. Let's start from the beginning.
A poset is a partially ordered set, that is, a set $S$ and a binary relation $\leq \subseteq S \times S$ such that $\leq$ is reflexive, antisymmetric, and transitive.
Examples of posets:
$(\mathbb{N},\leq)$, where $\leq$ is the standard "less or equal" order.
$(\mathcal{P}(S),\subseteq)$, where $\mathcal{P}(S)$ is the set of all subsets of some set $S$ and $\subseteq$ is the standard inclusion relation.
$(\mathbb{N},\mid)$, where $\mid$ is the "divides" relation.
A poset doesn't have to be a lattice, but can be, assuming that infima and suprema exist for every pair of elements in $S$. Infimum and supremum of $a,b$ is standardly denoted as $a \land b$ and $a \lor b$.
A lattice $(S,\leq)$ is called a Boolean lattice if:
there exist elements $0,1 \in S$ such that $0 \leq a$ and $a \leq 1$ for every $a \in S$.
for every $a \in S$, there exists $a' \in S$ such that $a \land a' = 0$ and $a \lor a' = 1$.
$S$ is distributive, ie. $a \lor (b \land c) = (a \lor b) \land (a \lor c)$ for every $a,b,c \in S$.
$S$ being distributive implies that $a'$ is in fact unique for every $a \in S$.
This set of requirements coincidentally (or perhaps not coincidentally) exactly corresponds to the standard definitions of boolean logic, so we use Boolean lattices to "model" it.
To answer your question, the binary operation can be any partial order, assuming that it satisfies all the conditions for Boolean lattices.
To give you a good example, $(\mathcal{P}(S),\subseteq)$ is a boolean lattice for every finite set $S$.
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answered Jun 10, 2016 at 10:21
AtheusAtheus
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$\begingroup$ Moreover, $(\mathcal{P}(S),\subseteq)$ is a boolean lattice for every set $S$, finite or not. $\endgroup$
amrsa
– amrsa
2016-06-28 11:39:48 +00:00
Commented Jun 28, 2016 at 11:39
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14379 | https://blog.csdn.net/weixin_32199769/article/details/117982582 | 增长率用计算机怎么算,增长率计算公式（excel公式来计算平均增长率的方法）...-CSDN博客
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增长率用计算机怎么算,增长率计算公式（excel公式来计算平均增长率的方法）...
最新推荐文章于 2022-03-29 15:58:58 发布
Jack Weavi于 2021-06-16 18:38:31 发布
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文章标签：增长率用计算机怎么算
本文介绍了如何使用Excel公式计算平均增长率，包括计算两年和多年平均增长率的公式，如SQRT函数和POWER函数的运用。通过理解平均增长率的数学公式x=(c/a)^(1/n)-1，可以在Excel中方便地进行批量数据处理，避免手工计算的错误，提高工作效率。
增长率计算公式(excel 公式来计算平均增长率的方法)作为一名财务人员经常在财务的数据中分析计算平均增长率，如果说我们要是通过手工来计算的话可以使用带开放的计算器计算出平均增长率，但是如果要有很多数据那我们该怎么计算平均增长值呢，手工计算容易出错，今天我们就告诉你通过excel公式来计算平均增长率的方法。
我们在使用excel公式计算平均增长率之前需要了解什么是excel平均增长率计算公式，平均增长率是指我们单位从第一年到第N年的每一年的产值、利润、营业额等的平均增长率，计算平均增长率的公式是：a(1+x)^n=c，其中a是基期数额，n为年限，c是期末数额，x为平均增长率。
这么说如果我们需要计算X的话，数学公式里：x=(c/a)^(1/n)-1,意思就是说我们用期末数额除以基期数额开年限次方减1，而开年限次方就是乘年限倒数次方。
所以说在excel中我们计算平均增长率的方法就有两种了，就是以下两种：
一、我们使用excel函数计算平均增长率也是有两种方法：
1、用excel计算两年的平均增长率，因为是excel计算两年的增长率所以我们只用开平方就行了，公式这样写：=SQRT(c/a)-1，SQRT是EXCEL的开方函数，这个只是在excel中计算两年平均增长率使用的公式
2、在excel中计算多年平均增长率的使用公式是：=POWER(10,log10(c/a)/n)-1。POWER函数是返回给定数字的乘幂，POWER(10,log10(c/a)/n)等同于10^log10(c/a)/n，也就是10的log10(c/a)/n次方。 Log10是返回以10为底任意数的对数，把这个公式写入EXCEL计算平均增长率的单元格里，就可以计算任意年限的平均增长率了。
二、第二种方法就是把前面我们所说的excel平均增长率公式直接写进去，x=(c/a)^(1/n)-1直接写入需要计算平均增长率的EXCEL单元格中，因为excel可以使用数学公式，所以我们在excel表格中直接写数据公式来计算平均增长值就可以了。
以上就是我们今天所说的通过excel公式来计算平均增长率的方法，其中我们利用了excel函数和excel公式讲解了今天的计算平均增长率的方法，其实excel能帮我们做的不只只是这些 数据分析 运算，还有很多更强大的功能需要我们去发掘。
相关资源：( 完整版 ) Excel 常用函数 公式 大全 ( 实用 ) . docx_ excel 函数 公式 大全 . . .
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资源浏览查阅136次。excel 常用函数 公式 及技巧搜集 ( 常用的 ) 【身份证信息提取】从身份证号码中提取出生年月日 =TEXT ( MID ( A1,7,6+( LEN ( A1 )=18 )2 ),"#-00-00")+0 =TEXT ( MID ( A1,7,6+( LEN ( A1 )=18 ),更多下载资源、学习资料请访问CSDN下载频道
( 转 ) 超棒的 EXCEL 使用技巧_在g2单元格中输入 公式“=e2f2", 复制 公式 到 . . .
9-22
Excel 中如果运 算 对象是空白单元格,Excel 将此空值当作零值 )。 解决 方法:修改单元格引用,或者在用作除数的单元格中输入不为零的值。 原因二:输入的 公式 中包含明显的除数零,例如:=5/0。 解决 方法:将零改为非零值。 4 .#NAME? 在 公式 中使用了 Excel 不能识别的文本时将产生错误值#NAME?。 原因一:删除了 公式 中使用 . . .
增长率 用 计算机 怎么 算,操作 方法：Excel 使用 公式 来 计算 增长率 教程
weixin_35772420的博客
06-16 8438
有关使用 公式 计算 增长率 的 Excel 教程。 Excel 经常需要使用 公式 来 计算 增长率。如何使用 公式 来 计算 增长率？以下是有关使用 公式 计算 增长率 的 excel 教程。希望阅读后能为您带来启发！ Excel 使用 公式 来 计算 增长率 教程：计算 增长率 步骤1：在单元格D2中输入 公式=( C2-B2 )/ C2，然后双击单元格右下角的填充手柄，或者按住鼠标左键，将填充的框拖到指定的单元格区域，如下图所示：计算 增长率 步骤2：为了 . . .
年 平均 增长速度 excel 计算公式
jidawanghao的专栏
03-29 2125
= ( POWER ( ROUND ( D6/C6,3 ) ROUND ( E6/D6,3 ) ROUND ( F6/E6,3 ) ROUND ( G6/F6,3 ) ROUND ( H6/G6,3 ) ROUND ( I6/H6,3 ) ROUND ( J6/I6,3 ) ROUND ( K6/J6,3 ) ROUND ( L6/K6,3 ) ROUND ( M6/L6,3 ),1/10 )-1 )100
【Excel】身份证号最后一位“X”怎么 计算 身份证最后一位的 _计算 函数-CS . . .
9-26
看了这么多 计算 步骤是不是感觉有点麻烦,其实一个 Excel 公式 就可以搞定! =VLOOKUP ( MOD ( SUM ( MID ( A1,{1;2;3;4;5;6;7;8;9;10;11;12;13;14;15;16;17},1 ){7;9;10;5;8;4;2;1;6;3;7;9;10;5;8;4;2}),11 ),{0,1;1,0;2,“X”;3,9;4,8;5,7;6,6;7,5;8,4;9,3;10,2},2,. . .
如何使用 Excel 对一元一次或多次方程 求 解,单变量 求 解工具的 使用
9-20
需要使用到 Excel 中 数据-模拟分析-单变量 求 解 以下为详细教程: 在L15单元格中输入 公式:=2L16 ( 或者使用鼠标单击L16单元格 ) +1,然后Enter ( 回车 ) 点击数据-模拟分析 -单变量 求 解 目标单元格 为 公式 ( L15 ) 所在单元格 目标值为所给定的Y值 ( 9 ) 可变单元格 为X ( L16 ) 所在单元格 . . .
平均 增长率 不用 计算机,官方数据：平均 增长率 计算公式 如何使用 excel 计算 平均 增长率 . . .
weixin_35857552的博客
06-27 1万+
在财务部门，我们每天都要处理数据，因此在处理数据时需要学习各种处理数据的 方法。最简单，最常用的 方法 是 计算 平均 增长率。如今，编辑器很简单介绍 平均 增长率 的 计算 方法。平均 增长率 计算公式-使用 excel 计算 平均 增长率 平均 增长率 是指从第一年到第N年，我们单位的产值，利润，营业额等的 平均 增长率。用于 计算 平均 增长率 的 公式 为：a ( 1 + x )^ n = c，其中a是基期金额，n是年数，c是结束金额，x是 平均 增长率 . . .
一个 计算 年 平均 递 增长率 的 EXCEL 公式!那些经常做经营分析什么的同志或许有用哦!------( 1 ) . pdf
07-22
其中，"计算 年 平均 递 增长率 的 EXCEL 公式"就是我们需要掌握的一个非常实用的技能。 在 EXCEL 中，年 平均 递 增长率 可以通过复合 增长率 的 公式 来 计算。复合 增长率 的 计算公式 是：[ ( \frac{FV}{PV})^{1/n} - 1 ]。其中，FV . . .
经验:EXCEL 生成y=kx+b函数图,有x,y原始数据,求 平均 系数k
9-26
经验:EXCEL 生成y=kx+b函数图,有x,y原始数据,求 平均 系数k 1 . 数据录入 2 . 生成图表 3 . 加入趋势线 4 . 得出 公式 1 . 数据录入 将需要生成的数据录入 excel 2 . 生成图表 框选数据,点击《插入》——《X,Y散点图 ( 带平滑线与数据标记 )》,就会生成图表 3 . 加入趋势线 . . .
Excel 应用技巧之三——常用技巧_通配符 excel 设置格式
9-3
本文分享 Excel 工作中的关键技巧,包括 公式 求 值步骤、如何快速改变行列顺序、删除空行、序列更新、插入空行等。深入探讨了技巧如日期显示、下拉菜单设置、带单位 求 和及工作表安全设置。同时涵盖了通配符、切片器、格式刷和外部数据导入等内容,以及实用插件的介绍。 摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往 . . .
如何用 计算 器或 Excel 计算 年均 增长率 ( 1 ) . pdf
07-28
在 计算 年均 增长率 时，无论是使用 计算 器还是 Excel 电子表格软件，核心都是利用复利增长的原理来估 算。具体到 方法 上，年均 增长率（CAGR，Compound Annual Growth Rate）是评估投资一段时间内 平均 年增长速度的指标。计算 . . .
如何用 计算 器或 Excel 计算 年均 增长率 ( 20220207164501 ) ( 1 ) . pdf
07-25
然后，使用 Excel 计算 年均 增长率，可以通过内置的复合年 增长率 ( CAGR ) 函数来实现，Excel 中并没有直接的CAGR函数，但可以使用以下 公式 进行 计算： = ( 最终值 / 初始值 )^( 1 / 年数 ) - 1 在 Excel 中，如果将初始值放在A1 . . .
Excel 求 解多元一次方程组 ( 线代 ) _ excel 解方程
9-26
假如在 Excel 的A2:N5区域中以下图 方法 输入了一个四元一次方程组。 在P2:S5区域用 公式 得到其系数矩阵,T2:T5的返回值为常数项向量。如P2单元格中的 公式 为: = OFFSET ($B$1,ROW ( A1 ),COLUMN ( A1 )3-3 )IF ( OFFSET ($A$1,ROW ( A1 ),COLUMN ( A1 )3-3 ) = "-",-1,1 ) . . .
复合 增长率 计算公式 . pdf
10-07
复合 增长率 计算公式 CAGR 复合 增长率 计算公式 CAGR是财务分析中一个重要的概念，它表示投资的复合 增长率。该 公式 的 计算 方式为： CAGR = ( Ending Value / Beginning Value )^( 1/# of years ) - 1 其中，Ending Value为 . . .
Excel 模板比较行业 增长率 趋势 . zip
06-13
总的来说，"Excel 模板比较行业 增长率 趋势"提供了一个结构化的 方法 来比较不同行业的增长动态，通过有效的数据分析，帮助企业做出更明智的战略决策，无论是投资、市场进入还是产品开发。熟练掌握这种分析 方法，对于 . . .
增长率 用 计算机 怎么 算,excel 如何利用 公式 来 计算 平均 增长率 的 方法
weixin_28717431的博客
06-16 5868
作为一名财务人员经常在财务的数据中分析 计算 平均 增长率，如果说我们要是通过手工来 计算 的话可以使用带开放的 计算 器 计算 出 平均 增长率，但是如果要有很多数据那我们该怎么 计算 平均 增长值呢，手工 计算 容易出错，今天我们就告诉你通过 excel 公式 来 计算 平均 增长率 的 方法。我们在使用 excel 公式 计算 平均 增长率 之前需要了解什么是 excel 平均 增长率 计算公式，平均 增长率 是指我们单位从第一年到第N年的每一年的产值、利润、营 . . .
五年 平均 增速用 计算机 怎么 算,行测重要知识点：年均 增长率 的 计算 热门推荐
weixin_27945229的博客
07-08 1万+
行测题目中 求 解年均 增长率 时会涉及到多次方的运 算，在实际操作中如不借助 计算 器很难精确运 算。而考试是明确规定不能携带 计算 器的，因而考生遇到这类 计算 问题就比较头疼。今天中公教育跟大家一起来看看年均 增长率 如何来进行估 算。一、年均 增长率 公式 二、年均 增长率 的估 算 1、二项式展开当q<5%时，结合 可得 ( 1 ) 2、实际 增长率 的 平均 数3、特征数字在使用 公式 ( 1 ) 和 公式 ( 2 ) 计算 年均 增长率 时，计算 结果是比实际结 . . .
用 计算机 算 出增长比例 公式,复合 增长率 excel 公式（年均复合 增长率 计算 器）
weixin_29009501的博客
06-27 8660
老兵直接拿历年 增长率 的和除以年限作为复合 增长率 了。其实，我的本意就是想 求 个均值，和前面的几个项目一样都是均值，但是潜意识地写成了复合 增长率，毕竟复合 增长率 更加常用且更加合理。虽然是个小错误，但是有必要拿出来说明一下，因为，前期很多股友加入我的公众号 ( 老兵学堂 ) 后都索要了这个财务数据分析的 EXCEL 模板，我给的是贵州茅台的模板。为了不给股友带来影响，特说明一下。顺便吆喝一下，加入公众号，私信留下邮箱 . . .
我国使用 计算机 增长率 表格,excel 表格如何 计算 数据的 增长率-怎么用 Excel 计算 年均复合 增长率？. . .
weixin_39536630的博客
06-25 3003
已知六年数据，如何用 EXCEL 求 年 平均 增长率 1、以2010版 excel，如下图中需要年 平均 增长要在E1单中体现出来，先选中E1单元格；2、在E1单元格中输入 计算公式“=POWER ( B6/B1,1/( A6-A1 ) )-1”；3、按下回车键后可以得出年 平均 增长率 的 计算 结果了；4、最后选中E1单元格，然后点击“数字”栏里面的百分比“%”，就可以得出年 平均 增长率 是“1%”。计算机 一级用 excel 表格 计算 增长 . . .
计算机 一级总产量怎么 算,excel 表格数据增长百分比-Excel 表格如何 计算 产量增减百分比？. . .
weixin_34998630的博客
07-17 3666
Excel 表格如何 计算 产量增减百分比？1、在 excel 的单中输入需要计长比例的数据。2、然后在C1单元格内输入 计算：=( B1-A1 )/A1。3、点击回车，即可将 计算公式 生成结果，此时 计算 结果显示的不是比例。4、右键点击C1单元格，在弹出的选项中点击“设置单元格格式”。5、然后在打开的对话框中选择“百分比”，并将小数位数设置为“2”，点击确定。6、即可将C1单元格的数据更改为百分比显示。计算机 一级用 . . .
springboot基于安卓企业网络主机IP地址管理系统1-c8z929m7-qc031【附万字论文+PPT+包部署+录制讲解视频】. zip
09-28
标题SpringBoot与安卓融合的企业网络主机IP管理系统研究AI更换标题第1章引言介绍企业网络主机IP地址管理的研究背景、意义、现状，论文的 方法 及创新点。1 . 1研究背景与意义阐述企业网络中IP地址管理的重要性和当前存在的问题。1 . 2国内外研究现状分析国内外在IP地址管理系统方面的研究进展和技术应用。1 . 3研究 方法 以及创新点介绍采用SpringBoot与安卓融合技术的研究 方法 及其创新之处。第2章相关理论总结和评述与IP地址管理系统相关的现有理论和技术。2 . 1网络通信基础介绍TCPIP协议、IP地址分类及子网划分等基础知识。2 . 2SpringBoot框架理论阐述SpringBoot框架的核心概念、特点及其在Web开发中的应用。2 . 3安卓开发理论介绍安卓系统的架构、开发环境及主要组件。第3章系统设计详细描述基于SpringBoot与安卓的企业网络主机IP管理系统的设计方案。3 . 1系统架构设计介绍系统的总体架构、模块划分及模块间的交互方式。3 . 2数据库设计设计数据库结构，包括IP地址信息表、设备信息表等。3 . 3安卓客户端设计阐述安卓客户端的功能设计、界面布局及交互逻辑。第4章系统实现详细描述系统的实现过程，包括开发环境搭建、代码实现及调试等。4 . 1SpringBoot服务端实现介绍服务端的功能实现，包括IP地址分配、回收、查询等。4 . 2安卓客户端实现阐述客户端的具体实现，包括界面开发、网络通信及数据处理。4 . 3系统集成与测试介绍系统的集成过程及测试 方法，包括单元测试、集成测试等。第5章研究结果呈现系统实现后的实验分析结果，包括性能评估和功能验证。5 . 1系统性能评估通过实验数据评估系统的响应时间、吞吐量等性能指标。5 . 2系统功能验证通过实际使用场景验证系统的各项功能是否满足需 求。5 . 3对比 方法 分析对比传统IP地址管理方式，分析新系统的优势和不足。第6章结论与展望总结系统研究成果，
基于robei框架构建的轻量级分布式微服务架构项目_采用SpringCloudAlibabaNacosDubboRocketMQRedisMySQL等技术栈_实现高可用弹性伸缩的服 . zip
09-28
基于robei框架构建的轻量级分布式微服务架构项目_采用SpringCloudAlibabaNacosDubboRocketMQRedisMySQL等技术栈_实现高可用弹性伸缩的服
旧型区域图转为js图七 最新发布
09-28
Webji-741 . html
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14380 | https://www.proprep.com/questions/what-is-the-compound-name-for-if7-and-what-does-its-structure-tell-us-about-the-exceptions-to-the-oc | What is the compound name for (IF_7), and what does its structure tell us about the exceptions to the octet rule?
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The compound I F 7 is named iodine heptafluoride. It is a compound consisting of one iodine atom covalently bonded to seven fluorine atoms. The name "heptafluoride" indicates that there are seven fluorine atoms (from the Greek 'hepta' meaning seven) bonded to the iodine.
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Naming inorganic compounds 1
Naming inorganic compounds 2
Exercise 2 - Name the following inorganic compounds: a. HI (aq) b. NiCr₂O₇ c.Al(ClO₄)₃ d. HNO₂(aq)
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Summary Show Transcript
00:00 In this video and in the next one,
00:02 we'll learn about the names and formulas of inorganic compounds.
00:07 Let's begin with a binary compounds of metals and nonmetals.
00:12 A binary compound involves 2 elements, that's the binary.
00:17 Binary compounds of a metal and non-metal are ionic compounds,
00:22 and they consist of a metal cation and a non-metal anion.
00:27 We write the name of the metal first and then the name of
00:31 the non-metal with the suffix ide.
00:35 Instead of chlorine, we'll have chloride.
00:38 We should note throughout and always check that the compound is electrically neutral.
00:45 Many metal ions have only one possibility,
00:49 that means there's only one oxidation state.
00:52 These include lithium plus, sodium plus,
00:56 potassium plus, rubidium plus,
00:58 cesium plus, magnesium 2 plus,
01:01 calcium 2 plus,
01:02 strontium 2 plus,
01:03 barium 2 plus,
01:05 aluminum 3 plus,
01:07 zinc 2 plus,
01:08 silver plus, and gold plus.
01:12 However, there are other metals that have more than one possibility.
01:17 That means there's more than one oxidation state.
01:20 For example, chromium can be 2 plus or 3 plus,
01:23 iron can be 2 plus or 3 plus,
01:26 cobalt can be 2 plus or 3 plus,
01:28 copper can be plus or 2 plus,
01:31 mercury can have 2 mercury atoms together,
01:36 Hg_2 and that's 2 plus or just one mercury atom Hg 2 plus.
01:43 Tin can be 2 plus or 4 plus,
01:47 and lead can be 2 plus or 4 plus.
01:51 Now there are far fewer nonmetal ions.
01:54 We have H minus,
01:55 that's called hydride, F minus,
01:58 fluoride, Cl minus,
02:00 chloride, Br minus, bromide,
02:03 I minus, iodide,
02:06 O 2 minus,
02:07 that's oxide, S^2 minus,
02:10 sulfide, N^3 minus nitrite.
02:14 Let's take some examples.
02:16 Here's one that you're very familiar with.
02:18 NaCl is sodium chloride, MgO,
02:23 magnesium oxide, Al_2O_3 aluminum oxide.
02:29 We can check that this is the correct formula.
02:33 Al is 3 plus,
02:35 that's 2 times 3 plus and oxygen is 2 minus,
02:39 that's 3 times 2 minus,
02:42 so we have plus 6 plus minus 6,
02:46 which gives us 0.
02:47 So 6 plus minus 6 giving us 0.
02:51 Now, when we have the possibility of more than one oxidation state,
02:56 then we have to distinguish between the different oxidation states in the name.
03:01 For example, FeO, we write iron II.
03:05 We use the II oxide.
03:08 That means iron has a plus 2 oxidation state.
03:13 If it's Fe_2O_3,
03:16 that's iron III oxide,
03:18 that means iron has a plus 3 oxidation state.
03:22 Remember for ions, the oxidation state and the ionic charge are the same,
03:29 so it's Fe_2 plus an oxidation state plus 2.
03:35 Where we have Cu_2O,
03:37 that's copper I oxide,
03:39 that means copper has the oxidation state of plus 1,
03:43 and CuO, that's copper II oxide,
03:48 that means copper has the oxidation state of plus 2.
03:52 Now we go on to the binary compounds of 2 nonmetals.
03:56 These are molecular compounds consisting of 2 nonmetals before we add ionic compounds.
04:04 We write the name of the element with the positive oxidation state
04:08 first and then the name of the negative oxidation state,
04:12 and we again have the suffix ide.
04:16 Now, very often an element has several oxidation states,
04:21 and then we use prefixes.
04:24 Here's a list of them.
04:27 Greek if it come from Greek.
04:28 Mono, di,
04:30 tri, tetra, penta,
04:31 hexa, hepta, octa, nona, and deca.
04:35 I will have some examples,
04:36 for example, carbon monoxide.
04:39 Here the carbon has the oxidation state of plus 2.
04:42 We don't write usually,
04:44 mono carbon monoxide, we just leave it as carbon.
04:48 Here is carbon dioxide,
04:51 the 2 oxygens,
04:52 and the oxidation state of carbon is plus 4.
04:55 If we're given a formula,
04:57 we can always check what the oxidation state is,
05:00 because oxygen is minus 2 and carbon therefore must be plus 4.
05:06 Now nitrogen has a lot of possible oxidation states and here are a few examples.
05:11 Dinitrogen oxide, that's N_2O.
05:15 The oxidation state of N must be plus 1.
05:19 NO_2 nitrogen dioxide.
05:23 The oxidation state of N must be plus 4.
05:27 N_2O_5 dinitrogen pentoxide.
05:31 The oxidation state here of N is plus 5.
05:35 We can check if that's correct.
05:37 There are 2 nitrogens,
05:39 so it's 2 times 5 and 5 oxygens,
05:43 so that's 5 times minus 2.
05:46 That gives us 10 plus,
05:48 minus 10 total of 0.
05:50 Now we get to phosphorus.
05:52 We can have phosphorus trichloride, PCl_3,
05:56 where the oxidation state of the phosphorus is plus 3 and PCl_5,
06:01 that's phosphorous pentachloride, where the oxidation states of phosphorus is plus 5.
06:08 Now we go into binary acids.
06:11 Binary acid is a compound consisting of hydrogen and a non-metal in an aqueous solution.
06:19 We write the prefix hydro,
06:22 followed by the name of the non-metal with the suffix ic.
06:27 Here's some examples.
06:29 HF is hydrofluoric acid,
06:33 HBr is hydrobromic acid, HCl,
06:36 with which you're all familiar,
06:38 is hydrochloric acid,
06:40 HI is hydroiodic acid,
06:43 H_2S is hydrosulfuric acid.
06:46 You have to remember all of these are in aqueous media,
06:50 they're all in water.
06:52 In this video, we talked about the names of inorganic compounds,
06:57 and we'll continue this in the next video.
This video provides an overview of the names and formulas of inorganic compounds. It begins with binary compounds of metals and nonmetals, which consist of a metal cation and a non-metal anion. The metal is written first, followed by the non-metal with the suffix “ide”. Many metal ions have only one oxidation state, while others have more than one. Nonmetal ions include hydride, fluoride, chloride, bromide, iodide, oxide, sulfide, and nitrite. Examples of binary compounds include sodium chloride, magnesium oxide, and aluminum oxide. Binary compounds of two nonmetals are molecular compounds, and the element with the positive oxidation state is written first. Prefixes are used to distinguish between different oxidation states. Examples of binary acids include hydrofluoric acid, hydrobromic acid, hydrochloric acid, hydroiodic acid, and hydrosulfuric acid.
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14381 | https://physics.solver360.com/solvers/fluid-mechanics/viscosity-calculator | Viscosity Calculator - Dynamic & Kinematic Viscosity | PhysicsSolver360°
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Viscosity Calculator
Calculate dynamic and kinematic viscosity of fluids using shear stress and velocity gradient relationships
Parameters
shearStress:
▲▼
Paⓘ
velocityGradient:
▲▼
s⁻¹ⓘ
density:
▲▼
kg/m³ⓘ
temperature:
▲▼
°Cⓘ
Show Trail
Calculated Values
Dynamic Viscosity:
0.10;P a⋅s 0.10;Pa·s 0.10;P a⋅s
Kinematic Viscosity:
0.00;m 2/s 0.00;m²/s 0.00;m 2/s
Reynolds Number:
100.00;100.00;100.00;
Viscosity Index:
2.00;2.00;2.00;
Examples
Example 1: Water at 20°C
Water flowing between parallel plates with shear stress.
Dynamic Viscosity: 0.10 0.10 0.10
Kinematic Viscosity: 0.00 0.00 0.00
Reynolds Number: 100.00 100.00 100.00
Example 2: Oil Flow
Engine oil with higher viscosity.
Dynamic Viscosity: 5.00 5.00 5.00
Kinematic Viscosity: 0.01 0.01 0.01
Reynolds Number: 1.70 1.70 1.70
Example 3: Air Flow
Air with low viscosity at room temperature.
Dynamic Viscosity: 0.00 0.00 0.00
Kinematic Viscosity: 0.00 0.00 0.00
Reynolds Number: 122.50 122.50 122.50
Viscosity
Viscosity is a measure of a fluid's resistance to flow and deformation. It describes the internal friction between fluid layers moving at different velocities. Viscosity is crucial in understanding fluid behavior, from simple pipe flow to complex industrial processes.
Dynamic viscosity (μ) is defined as the ratio of shear stress (τ) to velocity gradient (du/dy): μ = τ/(du/dy). It has units of Pa·s (Pascal-seconds) or N·s/m². Dynamic viscosity measures the fluid's resistance to shear deformation.
Kinematic viscosity (ν) is the ratio of dynamic viscosity to fluid density: ν = μ/ρ. It has units of m²/s. Kinematic viscosity is often more useful in fluid dynamics as it combines the effects of viscosity and density.
Newtonian fluids have constant viscosity regardless of shear rate, while non-Newtonian fluids have viscosity that changes with shear rate. Most common fluids like water, air, and oils are approximately Newtonian at moderate conditions.
Viscosity typically decreases with increasing temperature for liquids and increases with temperature for gases. This temperature dependence is crucial for many engineering applications and is often described by empirical relationships.
Key Concepts
Dynamic Viscosity: μ = τ/(du/dy) [Pa·s]
Kinematic Viscosity: ν = μ/ρ [m²/s]
Shear Stress: τ = F/A [Pa]
Velocity Gradient: du/dy [s⁻¹]
Newtonian Fluid: Constant viscosity
Temperature Dependence: Viscosity changes with temperature
Real-World Applications
Pipe Flow: Pressure drop calculations
Lubrication: Oil viscosity selection
Coating: Paint and ink flow
Food Processing: Rheology control
Blood Flow: Cardiovascular analysis
Physics Equations
Dynamic Viscosity:
μ=τ d u d y\mu = \frac{\tau}{\frac{du}{dy}}μ=d y d uτ
Kinematic Viscosity:
ν=μ ρ\nu = \frac{\mu}{\rho}ν=ρ μ
Shear Stress:
τ=μ d u d y\tau = \mu \frac{du}{dy}τ=μ d y d u
Reynolds Number:
R e=ρ v L μ Re = \frac{\rho v L}{\mu}R e=μ ρ vL
Temperature Correction (Water):
μ=μ 0 e−0.024(T−T 0)\mu = \mu_0 e^{-0.024(T-T_0)}μ=μ 0e−0.024(T−T 0)
Step-by-Step Solution
See how the main results are calculated.
Hide Calculations Copy Solution
1
Step 1: Identify Parameters
First, we identify the parameters needed for viscosity calculation:
2
Step 2: Calculate Dynamic Viscosity
Using the definition of dynamic viscosity:
3
Step 3: Calculate Kinematic Viscosity
Kinematic viscosity is dynamic viscosity divided by density:
4
Step 4: Calculate Reynolds Number (Example)
Reynolds number indicates flow regime (example: v=1 m/s, L=0.01 m):
5
Step 5: Interpret Results
Analyze the calculated values:
Frequently Asked Questions (FAQ)
What is the difference between dynamic and kinematic viscosity?
Dynamic viscosity (μ) measures resistance to shear deformation and has units Pa·s. Kinematic viscosity (ν) is μ/ρ and has units m²/s. Kinematic viscosity is often more useful in fluid dynamics.
How does temperature affect viscosity?
For liquids, viscosity typically decreases with increasing temperature. For gases, viscosity increases with temperature. This is due to changes in molecular interactions and thermal energy.
What is a Newtonian fluid?
A Newtonian fluid has constant viscosity regardless of shear rate. Most common fluids like water, air, and oils are approximately Newtonian under normal conditions.
What is the Reynolds number?
The Reynolds number (Re = ρvL/μ) indicates the relative importance of inertial to viscous forces. It determines whether flow is laminar (Re < 2300) or turbulent (Re > 4000).
How is viscosity measured?
Viscosity is measured using viscometers that apply known shear stress and measure resulting velocity gradient, or by measuring flow through capillary tubes under controlled conditions.
Practice MCQs
Dynamic viscosity is defined as: A) τ/(du/dy) B) μ/ρ C) ρvL/μ D) F/A
Kinematic viscosity has units of: A) Pa·s B) m²/s C) N·s/m² D) kg/m³
For a Newtonian fluid: A) Viscosity changes with shear rate B) Viscosity is constant C) Viscosity is zero D) Viscosity is infinite
As temperature increases, liquid viscosity typically: A) Increases B) Decreases C) Remains constant D) Becomes negative
The Reynolds number indicates: A) Pressure ratio B) Temperature ratio C) Inertial to viscous force ratio D) Density ratio
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14382 | https://en.wikipedia.org/wiki/Sphere_packing | Sphere packing - Wikipedia
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(Top)
1 Classification and terminology
2 Regular packingToggle Regular packing subsection
2.1 Dense packing
2.2 Other common lattice packings
2.3 Jammed packings with a low density
3 Irregular packing
4 Hypersphere packing
5 Unequal sphere packing
6 Hyperbolic space
7 Touching pairs, triplets, and quadruples
8 Other spaces
9 See also
10 References
11 Bibliography
12 External links
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Sphere packing
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From Wikipedia, the free encyclopedia
Geometrical structure
Sphere packing finds practical application in the stacking of cannonballs.
In geometry, a sphere packing is an arrangement of non-overlapping spheres within a containing space. The spheres considered are usually all of identical size, and the space is usually three-dimensional Euclidean space. However, sphere packing problems can be generalised to consider unequal spheres, spaces of other dimensions (where the problem becomes circle packing in two dimensions, or hypersphere packing in higher dimensions) or to non-Euclidean spaces such as hyperbolic space.
A typical sphere packing problem is to find an arrangement in which the spheres fill as much of the space as possible. The proportion of space filled by the spheres is called the packing density of the arrangement. As the local density of a packing in an infinite space can vary depending on the volume over which it is measured, the problem is usually to maximise the average or asymptotic density, measured over a large enough volume.
For equal spheres in three dimensions, the densest packing uses approximately 74% of the volume. A random packing of equal spheres generally has a density around 63.5%.
Classification and terminology
[edit]
A lattice arrangement (commonly called a regular arrangement) is one in which the centers of the spheres form a very symmetric pattern which needs only n vectors to be uniquely defined (in n-dimensional Euclidean space). Lattice arrangements are periodic. Arrangements in which the spheres do not form a lattice (often referred to as irregular) can still be periodic, but also aperiodic (properly speaking non-periodic) or random. Because of their high degree of symmetry, lattice packings are easier to classify than non-lattice ones. Periodic lattices always have well-defined densities.
Regular packing
[edit]
Regular arrangement of equal spheres in a plane changing to an irregular arrangement of unequal spheres (bubbles).
HCP lattice (left) and the FCC lattice (right) are the two most common highest density arrangements.
Two ways to stack three planes made of spheres
Dense packing
[edit]
Main article: Close-packing of equal spheres
In three-dimensional Euclidean space, the densest packing of equal spheres is achieved by a family of structures called close-packed structures. One method for generating such a structure is as follows. Consider a plane with a compact arrangement of spheres on it. Call it A. For any three neighbouring spheres, a fourth sphere can be placed on top in the hollow between the three bottom spheres. If we do this for half of the holes in a second plane above the first, we create a new compact layer. There are two possible choices for doing this, call them B and C. Suppose that we chose B. Then one half of the hollows of B lies above the centers of the balls in A and one half lies above the hollows of A which were not used for B. Thus the balls of a third layer can be placed either directly above the balls of the first one, yielding a layer of type A, or above the holes of the first layer which were not occupied by the second layer, yielding a layer of type C. Combining layers of types A, B, and C produces various close-packed structures.
Two simple arrangements within the close-packed family correspond to regular lattices. One is called cubic close packing (or face-centred cubic, "FCC")—where the layers are alternated in the ABCABC... sequence. The other is called hexagonal close packing ("HCP"), where the layers are alternated in the ABAB... sequence.[dubious – discuss] But many layer stacking sequences are possible (ABAC, ABCBA, ABCBAC, etc.), and still generate a close-packed structure. In all of these arrangements each sphere touches 12 neighboring spheres, and the average density is
π 3 2 ≈ 0.74048. {\displaystyle {\frac {\pi }{3{\sqrt {2}}}}\approx 0.74048.}
In 1611, Johannes Kepler conjectured that this is the maximum possible density amongst both regular and irregular arrangements—this became known as the Kepler conjecture. Carl Friedrich Gauss proved in 1831 that these packings have the highest density amongst all possible lattice packings. In 1998, Thomas Callister Hales, following the approach suggested by László Fejes Tóth in 1953, announced a proof of the Kepler conjecture. Hales' proof is a proof by exhaustion involving checking of many individual cases using complex computer calculations. Referees said that they were "99% certain" of the correctness of Hales' proof. On 10 August 2014, Hales announced the completion of a formal proof using automated proof checking, removing any doubt.
Other common lattice packings
[edit]
Some other lattice packings are often found in physical systems. These include the cubic lattice with a density of π 6 ≈ 0.5236 {\displaystyle {\frac {\pi }{6}}\approx 0.5236} , the hexagonal lattice with a density of π 3 3 ≈ 0.6046 {\displaystyle {\frac {\pi }{3{\sqrt {3}}}}\approx 0.6046} and the tetrahedral lattice with a density of π 3 16 ≈ 0.3401 {\displaystyle {\frac {\pi {\sqrt {3}}}{16}}\approx 0.3401} .
Jammed packings with a low density
[edit]
Packings where all spheres are constrained by their neighbours to stay in one location are called rigid or jammed. The strictly jammed (mechanically stable even as a finite system) regular sphere packing with the lowest known density is a diluted ("tunneled") fcc crystal with a density of only π√2/9 ≈ 0.49365. The loosest known regular jammed packing has a density of approximately 0.0555.
Irregular packing
[edit]
Main article: Random close pack
If we attempt to build a densely packed collection of spheres, we will be tempted to always place the next sphere in a hollow between three packed spheres. If five spheres are assembled in this way, they will be consistent with one of the regularly packed arrangements described above. However, the sixth sphere placed in this way will render the structure inconsistent with any regular arrangement. This results in the possibility of a random close packing of spheres which is stable against compression. Vibration of a random loose packing can result in the arrangement of spherical particles into regular packings, a process known as granular crystallisation. Such processes depend on the geometry of the container holding the spherical grains.
When spheres are randomly added to a container and then compressed, they will generally form what is known as an "irregular" or "jammed" packing configuration when they can be compressed no more. This irregular packing will generally have a density of about 64%. Recent research predicts analytically that it cannot exceed a density limit of 63.4% This situation is unlike the case of one or two dimensions, where compressing a collection of 1-dimensional or 2-dimensional spheres (that is, line segments or circles) will yield a regular packing.
Hypersphere packing
[edit]
The sphere packing problem is the three-dimensional version of a class of ball-packing problems in arbitrary dimensions. In two dimensions, the equivalent problem is packing circles on a plane. In one dimension it is packing line segments into a linear universe.
In dimensions higher than three, the densest lattice packings of hyperspheres are known for 8 and 24 dimensions. Very little is known about irregular hypersphere packings; it is possible that in some dimensions the densest packing may be irregular. Some support for this conjecture comes from the fact that in certain dimensions (e.g. 10) the densest known irregular packing is denser than the densest known regular packing.
In 2016, Maryna Viazovska announced a proof that the E8 lattice provides the optimal packing (regardless of regularity) in eight-dimensional space, and soon afterwards she and a group of collaborators announced a similar proof that the Leech lattice is optimal in 24 dimensions. This result built on and improved previous methods which showed that these two lattices are very close to optimal. The new proofs involve using the Laplace transform of a carefully chosen modular function to construct a radially symmetric function f such that f and its Fourier transform f̂ both equal 1 at the origin, and both vanish at all other points of the optimal lattice, with f negative outside the central sphere of the packing and f̂ positive. Then, the Poisson summation formula for f is used to compare the density of the optimal lattice with that of any other packing. Before the proof had been formally refereed and published, mathematician Peter Sarnak called the proof "stunningly simple" and wrote that "You just start reading the paper and you know this is correct."
Another line of research in high dimensions is trying to find asymptotic bounds for the density of the densest packings. It is known that for large n, the densest lattice in dimension n has density θ ( n ) {\displaystyle \theta (n)} between cn ⋅ 2−n (for some constant c) and 2−(0.599+o(1))n. Conjectural bounds lie in between. In a 2023 preprint, Marcelo Campos, Matthew Jenssen, Marcus Michelen and Julian Sahasrabudhe announced an improvement to the lower bound of the maximal density to θ ( n ) ≥ ( 1 − o ( 1 ) ) n ln n 2 n + 1 {\displaystyle \theta (n)\geq (1-o(1)){\frac {n\ln n}{2^{n+1}}}} , among their techniques they make use of the Rödl nibble.
Unequal sphere packing
[edit]
See also: Unequal circle packing
A dense packing of spheres with a radius ratio of 0.64799 and a density of 0.74786
Many problems in the chemical and physical sciences can be related to packing problems where more than one size of sphere is available. Here there is a choice between separating the spheres into regions of close-packed equal spheres, or combining the multiple sizes of spheres into a compound or interstitial packing. When many sizes of spheres (or a distribution) are available, the problem quickly becomes intractable, but some studies of binary hard spheres (two sizes) are available.
When the second sphere is much smaller than the first, it is possible to arrange the large spheres in a close-packed arrangement, and then arrange the small spheres within the octahedral and tetrahedral gaps. The density of this interstitial packing depends sensitively on the radius ratio, but in the limit of extreme size ratios, the smaller spheres can fill the gaps with the same density as the larger spheres filled space. Even if the large spheres are not in a close-packed arrangement, it is always possible to insert some smaller spheres of up to 0.29099 of the radius of the larger sphere.
When the smaller sphere has a radius greater than 0.41421 of the radius of the larger sphere, it is no longer possible to fit into even the octahedral holes of the close-packed structure. Thus, beyond this point, either the host structure must expand to accommodate the interstitials (which compromises the overall density), or rearrange into a more complex crystalline compound structure. Structures are known which exceed the close packing density for radius ratios up to 0.659786.
Upper bounds for the density that can be obtained in such binary packings have also been obtained.
In many chemical situations such as ionic crystals, the stoichiometry is constrained by the charges of the constituent ions. This additional constraint on the packing, together with the need to minimize the Coulomb energy of interacting charges leads to a diversity of optimal packing arrangements.
The upper bound for the density of a strictly jammed sphere packing with any set of radii is 1 – an example of such a packing of spheres is the Apollonian sphere packing. The lower bound for such a sphere packing is 0 – an example is the Dionysian sphere packing.
Hyperbolic space
[edit]
Although the concept of circles and spheres can be extended to hyperbolic space, finding the densest packing becomes much more difficult. In a hyperbolic space there is no limit to the number of spheres that can surround another sphere (for example, Ford circles can be thought of as an arrangement of identical hyperbolic circles in which each circle is surrounded by an infinite number of other circles). The concept of average density also becomes much more difficult to define accurately. The densest packings in any hyperbolic space are almost always irregular.
Despite this difficulty, K. Böröczky gives a universal upper bound for the density of sphere packings of hyperbolic n-space where n ≥ 2. In three dimensions the Böröczky bound is approximately 85.327613%, and is realized by the horosphere packing of the order-6 tetrahedral honeycomb with Schläfli symbol {3,3,6}. In addition to this configuration at least three other horosphere packings are known to exist in hyperbolic 3-space that realize the density upper bound.
Touching pairs, triplets, and quadruples
[edit]
The contact graph of an arbitrary finite packing of unit balls is the graph whose vertices correspond to the packing elements and whose two vertices are connected by an edge if the corresponding two packing elements touch each other. The cardinality of the edge set of the contact graph gives the number of touching pairs, the number of 3-cycles in the contact graph gives the number of touching triplets, and the number of tetrahedrons in the contact graph gives the number of touching quadruples (in general for a contact graph associated with a sphere packing in n dimensions that the cardinality of the set of n-simplices in the contact graph gives the number of touching (n + 1)-tuples in the sphere packing). In the case of 3-dimensional Euclidean space, non-trivial upper bounds on the number of touching pairs, triplets, and quadruples were proved by Karoly Bezdek and Samuel Reid at the University of Calgary.
The problem of finding the arrangement of n identical spheres that maximizes the number of contact points between the spheres is known as the "sticky-sphere problem". The maximum is known for n ≤ 11, and only conjectural values are known for larger n.
Other spaces
[edit]
Sphere packing on the corners of a hypercube (with Hamming balls, spheres defined by Hamming distance) corresponds to designing error-correcting codes: if the spheres have radius t, then their centers are codewords of a (2_t_ + 1)-error-correcting code. Lattice packings correspond to linear codes. There are other, subtler relationships between Euclidean sphere packing and error-correcting codes. For example, the binary Golay code is closely related to the 24-dimensional Leech lattice.
For further details on these connections, see the book Sphere Packings, Lattices and Groups by Conway and Sloane.
See also
[edit]
Close-packing of equal spheres
Apollonian sphere packing
Finite sphere packing
Hermite constant
Inscribed sphere
Kissing number
Sphere-packing bound
Random close pack
Cylinder sphere packing
Sphere packing in a sphere
References
[edit]
^ Wu, Yugong; Fan, Zhigang; Lu, Yuzhu (1 May 2003). "Bulk and interior packing densities of random close packing of hard spheres". Journal of Materials Science. 38 (9): 2019–2025. doi:10.1023/A:1023597707363. ISSN 1573-4803. S2CID 137583828.
^ a b Dai, Weijing; Reimann, Joerg; Hanaor, Dorian; Ferrero, Claudio; Gan, Yixiang (13 March 2019). "Modes of wall induced granular crystallisation in vibrational packing". Granular Matter. 21 (2): 26. arXiv:1805.07865. doi:10.1007/s10035-019-0876-8. ISSN 1434-7636. S2CID 254106945.
^ Gauß, C. F. (1831). "Besprechung des Buchs von L. A. Seeber: Untersuchungen über die Eigenschaften der positiven ternären quadratischen Formen usw" [Discussion of L. A. Seeber's book: Studies on the characteristics of positive ternary quadratic forms etc]. Göttingsche Gelehrte Anzeigen.
^ "Long-term storage for Google Code Project Hosting". Google Code Archive.
^ "Wolfram Math World, Sphere packing".
^ Torquato, S.; Stillinger, F. H. (2007). "Toward the jamming threshold of sphere packings: Tunneled crystals". Journal of Applied Physics. 102 (9): 093511–093511–8. arXiv:0707.4263. Bibcode:2007JAP...102i3511T. doi:10.1063/1.2802184. S2CID 5704550.
^ "Wolfram Math World, Sphere packing".
^ Chaikin, Paul (June 2007). "Random thoughts". Physics Today. 60 (6). American Institute of Physics: 8. Bibcode:2007PhT....60f...8C. doi:10.1063/1.2754580. ISSN 0031-9228.
^ Song, C.; Wang, P.; Makse, H. A. (29 May 2008). "A phase diagram for jammed matter". Nature. 453 (7195): 629–632. arXiv:0808.2196. Bibcode:2008Natur.453..629S. doi:10.1038/nature06981. PMID 18509438. S2CID 4420652.
^ Griffith, J.S. (1962). "Packing of equal 0-spheres". Nature. 196 (4856): 764–765. Bibcode:1962Natur.196..764G. doi:10.1038/196764a0. S2CID 4262056.
^ Weisstein, Eric W. "Hypersphere Packing". MathWorld.
^ Sloane, N. J. A. (1998). "The Sphere-Packing Problem". Documenta Mathematica. 3: 387–396. arXiv:math/0207256. Bibcode:2002math......7256S.
^ Viazovska, Maryna (1 January 2017). "The sphere packing problem in dimension 8". Annals of Mathematics. 185 (3): 991–1015. arXiv:1603.04246. doi:10.4007/annals.2017.185.3.7. ISSN 0003-486X. S2CID 119286185.
^ Cohn, Henry; Kumar, Abhinav; Miller, Stephen; Radchenko, Danylo; Viazovska, Maryna (1 January 2017). "The sphere packing problem in dimension 24". Annals of Mathematics. 185 (3): 1017–1033. arXiv:1603.06518. doi:10.4007/annals.2017.185.3.8. ISSN 0003-486X. S2CID 119281758.
^ Cohn, Henry; Kumar, Abhinav (2009), "Optimality and uniqueness of the Leech lattice among lattices", Annals of Mathematics, 170 (3): 1003–1050, arXiv:math.MG/0403263, doi:10.4007/annals.2009.170.1003, ISSN 1939-8980, MR 2600869, S2CID 10696627, Zbl 1213.11144 Cohn, Henry; Kumar, Abhinav (2004), "The densest lattice in twenty-four dimensions", Electronic Research Announcements of the American Mathematical Society, 10 (7): 58–67, arXiv:math.MG/0408174, Bibcode:2004math......8174C, doi:10.1090/S1079-6762-04-00130-1, ISSN 1079-6762, MR 2075897, S2CID 15874595
^ Miller, Stephen D. (4 April 2016), The solution to the sphere packing problem in 24 dimensions via modular forms, Institute for Advanced Study, archived from the original on 21 December 2021. Video of an hour-long talk by one of Viazovska's co-authors explaining the new proofs.
^ Klarreich, Erica (30 March 2016), "Sphere Packing Solved in Higher Dimensions", Quanta Magazine
^ Cohn, Henry (2017), "A conceptual breakthrough in sphere packing" (PDF), Notices of the American Mathematical Society, 64 (2): 102–115, arXiv:1611.01685, doi:10.1090/noti1474, ISSN 0002-9920, MR 3587715, S2CID 16124591
^ Torquato, S.; Stillinger, F. H. (2006), "New conjectural lower bounds on the optimal density of sphere packings", Experimental Mathematics, 15 (3): 307–331, arXiv:math/0508381, doi:10.1080/10586458.2006.10128964, MR 2264469, S2CID 9921359
^ Campos, Marcelo; Jenssen, Matthew; Michelen, Marcus; Sahasrabudhe, Julian (2023). "A new lower bound for sphere packing". arXiv:2312.10026 [math.MG].
^ Houston-Edwards, Kelsey (30 April 2024). "To Pack Spheres Tightly, Mathematicians Throw Them at Random". Quanta Magazine. Retrieved 30 April 2024.
^ a b O'Toole, P. I.; Hudson, T. S. (2011). "New High-Density Packings of Similarly Sized Binary Spheres". The Journal of Physical Chemistry C. 115 (39): 19037. doi:10.1021/jp206115p.
^ Hudson, D. R. (1949). "Density and Packing in an Aggregate of Mixed Spheres". Journal of Applied Physics. 20 (2): 154–162. Bibcode:1949JAP....20..154H. doi:10.1063/1.1698327.
^ Zong, C. (2002). "From deep holes to free planes". Bulletin of the American Mathematical Society. 39 (4): 533–555. doi:10.1090/S0273-0979-02-00950-3.
^ Marshall, G. W.; Hudson, T. S. (2010). "Dense binary sphere packings". Contributions to Algebra and Geometry. 51 (2): 337–344.
^ de Laat, David; de Oliveira Filho, Fernando Mário; Vallentin, Frank (12 June 2012). "Upper bounds for packings of spheres of several radii". Forum of Mathematics, Sigma. 2. arXiv:1206.2608. doi:10.1017/fms.2014.24. S2CID 11082628.
^ Dennis, Robert; Corwin, Eric (2 September 2021). "Dionysian Hard Sphere Packings Are Mechanically Stable at Vanishingly Low Densities". Physical Review. 128 (1): 018002. arXiv:2006.11415. doi:10.1103/PhysRevLett.128.018002.
^ Bowen, L.; Radin, C. (2002). "Densest Packing of Equal Spheres in Hyperbolic Space". Discrete and Computational Geometry. 29: 23–39. doi:10.1007/s00454-002-2791-7.
^ Böröczky, K. (1978). "Packing of spheres in spaces of constant curvature". Acta Mathematica Academiae Scientiarum Hungaricae. 32 (3–4): 243–261. doi:10.1007/BF01902361. S2CID 122561092.
^ Böröczky, K.; Florian, A. (1964). "Über die dichteste Kugelpackung im hyperbolischen Raum". Acta Mathematica Academiae Scientiarum Hungaricae. 15 (1–2): 237–245. doi:10.1007/BF01897041. S2CID 122081239.
^ Kozma, R. T.; Szirmai, J. (2012). "Optimally dense packings for fully asymptotic Coxeter tilings by horoballs of different types". Monatshefte für Mathematik. 168: 27–47. arXiv:1007.0722. doi:10.1007/s00605-012-0393-x. S2CID 119713174.
^ Bezdek, Karoly; Reid, Samuel (2013). "Contact Graphs of Sphere Packings Revisited". Journal of Geometry. 104 (1): 57–83. arXiv:1210.5756. doi:10.1007/s00022-013-0156-4. S2CID 14428585.
^ "The Science of Sticky Spheres". American Scientist. 6 February 2017. Retrieved 14 July 2020.
^ Conway, John H.; Sloane, Neil J. A. (1998). Sphere Packings, Lattices and Groups (3rd ed.). Springer Science & Business Media. ISBN 0-387-98585-9.
Bibliography
[edit]
Aste, T.; Weaire, D. (2000). The Pursuit of Perfect Packing. London: Institute of Physics Publishing. ISBN 0-7503-0648-3.
Conway, J. H.; Sloane, N. J. H. (1998). Sphere Packings, Lattices and Groups (3rd ed.). Springer. ISBN 0-387-98585-9.
Sloane, N. J. A. (1984). "The Packing of Spheres". Scientific American. 250: 116–125. Bibcode:1984SciAm.250e.116G. doi:10.1038/scientificamerican0584-116.
External links
[edit]
Dana Mackenzie (May 2002) "A fine mess" (New Scientist)
A non-technical overview of packing in hyperbolic space.
Weisstein, Eric W. "Circle Packing". MathWorld.
"Kugelpackungen (Sphere Packing)" (T. E. Dorozinski)
"3D Sphere Packing Applet" Archived 26 April 2009 at the Wayback Machine Sphere Packing java applet
"Densest Packing of spheres into a sphere" java applet
"Database of sphere packings" (Erik Agrell)
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14383 | https://pmc.ncbi.nlm.nih.gov/articles/PMC7686821/ | The diagnostic accuracy of digital, infrared and mercury-in-glass thermometers in measuring body temperature: a systematic review and network meta-analysis - PMC
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. 2020 Nov 25;16(4):1071–1083. doi: 10.1007/s11739-020-02556-0
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The diagnostic accuracy of digital, infrared and mercury-in-glass thermometers in measuring body temperature: a systematic review and network meta-analysis
Valentina Pecoraro
Valentina Pecoraro
1 Department of Laboratory Medicine and Pathology, Ospedale Civile Sant’Agostino Estense, AUSL Modena, Modena, Italy
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1, Davide Petri
Davide Petri
2 Department of Clinical and Experimental Medicine, University of Pisa, Via Roma, 10, 56126 Pisa, Italy
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2, Giorgio Costantino
Giorgio Costantino
3 IRCCS Fondazione Ca’ Granda, Ospedale Maggiore Policlinico, UOC Pronto Soccorso e Medicina D’Urgenza, Università Degli Studi di Milano, Milan, Italy
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3, Alessandro Squizzato
Alessandro Squizzato
4 Department of Medicine and Surgery, University of Insubria, Como, Italy
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4, Lorenzo Moja
Lorenzo Moja
5 Department of Biomedical Sciences for Health, University of Milan, Milan, Italy
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5, Gianni Virgili
Gianni Virgili
6 Department of Neurosciences, Psychology, Drug Research and Child Health (NEUROFARBA), AOU Careggi, Florence, Italy
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6, Ersilia Lucenteforte
Ersilia Lucenteforte
2 Department of Clinical and Experimental Medicine, University of Pisa, Via Roma, 10, 56126 Pisa, Italy
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1 Department of Laboratory Medicine and Pathology, Ospedale Civile Sant’Agostino Estense, AUSL Modena, Modena, Italy
2 Department of Clinical and Experimental Medicine, University of Pisa, Via Roma, 10, 56126 Pisa, Italy
3 IRCCS Fondazione Ca’ Granda, Ospedale Maggiore Policlinico, UOC Pronto Soccorso e Medicina D’Urgenza, Università Degli Studi di Milano, Milan, Italy
4 Department of Medicine and Surgery, University of Insubria, Como, Italy
5 Department of Biomedical Sciences for Health, University of Milan, Milan, Italy
6 Department of Neurosciences, Psychology, Drug Research and Child Health (NEUROFARBA), AOU Careggi, Florence, Italy
✉
Corresponding author.
Received 2020 Aug 15; Accepted 2020 Oct 28; Issue date 2021.
© The Author(s) 2020
Open AccessThis article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit
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PMCID: PMC7686821 PMID: 33237494
Abstract
Not much is known about how accurate and reproducible different thermometers are at diagnosing patients with suspected fever. The study aims at evaluating which peripheral thermometers are more accurate and reproducible. We searched Medline, Embase, Scopus, WOS, CENTRAL, and Cinahl to perform: (1) diagnostic accuracy meta-analysis (MA) using rectal mercury-in-glass or digital thermometry as reference, and bivariate models for pooling; (2) network MA to estimate differences in mean temperature between devices; (3) Bland–Altman method to estimate 95% coefficient of reproducibility. PROSPERO registration: CRD42020174996. We included 46 studies enrolling more than 12,000 patients. Using 38°C (100.4℉) as cut-off temperature, temporal infrared thermometry had a sensitivity of 0.76 (95% confidence interval, 0.65,0.84; low certainty) and specificity of 0.96 (0.92,0.98; moderate certainty); tympanic infrared thermometry had a sensitivity of 0.77 (0.60,0.88; low certainty) and specificity of 0.98 (0.95,0.99; moderate certainty). For all the other index devices, it was not possible to pool the estimates. Compared to the rectal mercury-in-glass thermometer, mean temperature differences were not statistically different from zero for temporal or tympanic infrared thermometry; the median coefficient of reproducibility ranged between 0.53°C [0.95℉] for infrared temporal and 1.2°C [2.16℉] for axillary digital thermometry. Several peripheral thermometers proved specific, but not sensitive for diagnosing fever with rectal thermometry as a reference standard, meaning that finding a temperature below 38°C does not rule out fever. Fixed differences between temperatures together with random error means facing differences between measurements in the order of 2°C [4.5℉]. This study informs practitioners of the limitations associated with different thermometers; peripheral ones are specific but not sensitive.
Electronic supplementary material
The online version of this article (10.1007/s11739-020-02556-0) contains supplementary material, which is available to authorized users.
Keywords: Body temperature, Diagnostic tests, Fever, Systematic review, Thermometers
Introduction
Body temperature is a vital parameter. Fever (or pyrexia) is the temporary pathological state that involves an alteration of the hypothalamic thermoregulation system and a consequent elevation of body temperature above the value considered normal. Many diseases begin with increased body temperature, determining a febrile state. Although there is no single agreed threshold for diagnosing fever, a value above the interval between 37.7°C and 38.3°C is usually considered a febrile response . Measurement of body temperature depends on the selection of the anatomical area, with marked differences between the body core temperature and the surface temperature . Another important source of variability is that body temperature changes during the course of the day and depends on a person’s activity.
Fever originates from infections (e.g. viral, bacterial) and from non-infectious conditions (e.g. inflammation, malignancies, autoimmune disease, drug adverse events), and in some cases, its aetiology is of unknown origin. Fever is also a common symptom of COVID-19, typically appearing 2–14 days after exposure. Therefore, clinical electronic thermometers are an important screening and diagnostic tool to assist in the identification of those individuals who may be infected with COVID-19 .
Determination of body temperature is a key clinical action in the management of patients: the presence of fever affects the decision of clinicians, patients, and caregivers, impacting diagnosis, investigations, and therapies (e.g. antibiotic administration). So, accurate measurement of temperature is essential, and thermometers should accurately measure body temperature oscillations.
The US Food and Drug Administration acknowledges the fact that non-contact temperature assessment devices are not effective if used as the only means of detecting a COVID-19 infection. This failure is not only related to the absence of fever in some affected patients, but also because devices fail to identify elevated temperature, or misread normal temperature as elevated. Moreover, failure to follow the manufacturer’s instructions for use, such as for set-up, operation, and training, is also reported as a limitation of non-contact thermometer use .
There are several types of medical thermometers. Mercury-in-glass thermometers were the standard reference method for decades [5–7], until the late 2000s when they were banned from the market due to the environmental toxicity of mercury . Alternative thermometers have come into use, such as digital tympanic or axillary, infrared skin scan, temporal artery thermometers, and non-contact infrared thermometers.
Despite the central role of thermometers in clinical practice, our knowledge of the relative performance of alternative thermometers, including differences in measured temperature, is limited. Consequently, it is necessary to understand the characteristics and diagnostic accuracy of different thermometers, appreciating their limitations as tools that guide patient management. This is particularly important given the triage role of fever measurement in several clinical settings, particularly in emergency care settings, with the aim of sending potential COVID-19 patients to appropriate care pathways.
We systematically reviewed studies comparing the accuracy of digital, infrared and mercury-in-glass thermometers, estimating body temperature on different anatomical sites, both in adults and children.
Methods
We performed a systematic review and meta-analysis according to the recommendations indicated in the Cochrane Handbook for Diagnostic Test Accuracy Reviews . Moreover, we used NMA methods to compare multiple diagnostic tests and body sites in one simultaneous analysis. For this purpose, we extracted between-test differences and used them as a continuous variable to fit standard NMA techniques.
The reporting was in accordance with the Preferred Reporting Items for Systematic reviews and Meta-Analyses of Diagnostic Test Accuracy Studies (PRISMA-DTA) criteria .
This systematic review has been registered on PROSPERO 2020 (CRD42020174996).
Search strategy
We performed a systematic search up to March 2020 on six electronic databases: Medline, Embase, Web of Science (WOS), Scopus, The Cochrane Central Register of Controlled Trials (CENTRAL), and Cinhal, to identify all possible eligible studies. These databases were searched using the following search keywords: “sensitivity”, “specificity”, “body temperature”, “thermometer”. The search strategy was first developed for Medline and then adapted to all other databases. Finally, we checked the reference list of all selected studies.
Patients
We included adult and child patients screened for fever in emergency and hospital in-patient departments.
Index and reference standard thermometer categories
The thermometer type was classified as mercury-in-glass, infrared or other digital devices . The body sites considered were grouped as peripheral (i.e. tympanic, temporal artery, axillary, and oral) or central (i.e. rectal, pulmonary artery, urinary bladder, and oesophageal sites) [12, 13]. For diagnostic accuracy analyses, we assumed that mercury-in-glass or digital thermometry at the rectal site was the reference standard. Because of the limited number of studies, we conducted separate analyses for body site and thermometer type. For network meta-analyses of mean differences, we considered rectal mercury-in-glass as the reference category.
Outcomes
The primary outcome was the diagnostic accuracy of digital, infrared, and mercury-in-glass thermometers defined as the number of true positives (TP), false positives (FP), false negatives (FN), and true negatives (TN) reported in each study. When these data were not available, they were calculated from sensitivity and specificity data. We also evaluated the mean difference in temperature determined using different types of thermometers (fixed bias) and reported as 1.96 times the standard deviation (SD) of these differences (random error), which are the two components of the 95% coefficient of reproducibility .
Study selection
We included studies which respected the following eligibility criteria: (i) randomized clinical trial, observational cohort or cross-sectional study; (ii) enrolled adults or children accessing an emergency department (ED), enrolled adult or child patients hospitalized in hospital or in neonatal departments; (iii) studies that considered rectal or axillary temperature as the reference standard, measured with mercury-in-glass or digital thermometers; (iv) body temperature measured by clinicians or nurses; (v) studies that provided sensitivity and specificity data and temperature measured with each thermometer used; (vi) published in English, Italian, Spanish, or French. We excluded surgical patients, studies where body temperature was measured by mothers or using only one type of thermometer. After removing duplicates, two independent authors screened titles and abstracts and identified all potentially eligible studies. The full text of selected citations was then reviewed according to the inclusion criteria.
Data extraction
One author used a standardized data extraction form to collect relevant publication details regarding study methods and results, and the second author checked the data. The authors collected data about: (i) study characteristics (i.e. authors, year of publication, title, reference, study design, eligibility criteria and setting); (ii) patient characteristics (i.e. age, number of enrolled patients, and site of measurement of body temperature); (iii) detailed information about the index test (i.e. any other type of body thermometer) and reference standard (i.e. mercury-in-glass or digital thermometer measuring rectal temperature or temperature in other body sites). Other details collected were the type of thermometer, the cut-off used and the method of measuring body temperature; (iv) diagnostic study data (i.e. sensitivity, specificity, TP, TN, FP, FN); (v) mean and standard deviation (SD) of the body temperature measured.
Quality assessment
The methodological quality of each selected study was assessed according to the Quality Assessment of Diagnostic Accuracy Studies (QUADAS-2) checklist which considers four domains (patient selection, index test, reference standard, and flow and timing), each rated in terms of risk of bias and applicability to the research question. The risk of bias was judged as “low”, “high”, or “unclear”. Each domain included different signalling questions guiding the risk of bias assessment. If all signalling questions received a favorable answer, then the risk of bias was judged as “low”. Concerning applicability, the authors recorded the information on why the study may not have matched the review question. Concerns regarding applicability were rated as “low”, “high”, or “unclear”. At any review stage, disagreements were resolved by discussion or by the involvement of a third investigator.
Data analysis
For each study, we constructed two-by-two tables and pooled TP, FP, TN, and FN to create separate forest plots to examine the accuracy of different devices to diagnose fever. We used mixed models to fit bivariate meta-analyses, which model sensitivity and specificity while accounting for their correlation . For this purpose, we pooled data at a 38°C threshold and adopted rectal temperature detected using mercury-in-glass or digital thermometry as the reference standard. We performed a meta-analysis if data were available from at least five studies.
As reported above, for our secondary objective we used NMA methods to use direct and indirect evidence and compare the mean difference of each device using the rectal mercury thermometer as the reference technique. We generated standard errors (SEs) from SDs of the differences or from p-values as appropriate; then we used available between-test correlation coefficients, or their median, to compute adjusted SEs that could not be obtained by conversion of published SDs .
We considered the 95% coefficient of reproducibility as a measure of reliability between two tests (i.e. different thermometers) with measurements obtained on the same person . The coefficient of reproducibility is defined as the mean difference (MD) ± 1.96 SD of differences (SDD). In our study, the mean difference is the fixed bias and was estimated using NMA techniques. Once the fixed bias is taken into account, 1.96 × SDDs inform on the random error measurement component. However, meta-analytic methods to estimate pooled SDDs have not yet been developed to the best of our knowledge. Therefore, we presented 1.96 × SDDs for each direct comparison and reported on their variation and the median value for each comparison.
The software STATA 15.2 (StataCorp, 2011; Stata Statistical Software: Release 15. College Station, TX) was used for all analyses. In particular, the ‘network’ suite of commands was used to fit NMAs .
Evidence profile
We evaluated the evidence using the GRADE approach and produced a’Summary of findings’ table for studies that assessed the accuracy of tympanic infrared and temporal artery thermometers to diagnose fever. Studies were initially considered of high quality but were downgraded according to their risk of bias, the directness of evidence (generalizability), consistency, and precision of results across all trials that measured a given specific outcome. Directness refers to the extent to which trial participants, interventions, and outcome measures considered in the included trials are relevant to the review question. Consistency concerns the degree of homogeneity (direction and magnitude) of results across the different studies. Precision describes the grade of uncertainty around the effect estimate, in other words, the width of estimated CIs .
We used the STATA metandi package to fit bivariate models, the STATA network routine to perform NMA and the STATA metan function to obtain pairwise meta-analyses .
The funder of the study had no role in study design, data collection, data analysis, data interpretation, or writing of the report. The corresponding author had full access to all the data in the study and had final responsibility for the decision to submit for publication.
Results
Studies identification and selection
The literature search on Medline, Embase, Web of Science (WOS), Scopus, CENTRAL, and Cinhal, after the exclusion of duplicates and irrelevant records, identified 1279 references. Of these, 1201 were excluded because they did not meet the inclusion criteria. Seventy-eight studies were considered eligible for inclusion and their full texts were evaluated for details. Of these, 32 were excluded because (i) diagnostic accuracy data were not reported (n = 16); (ii) comparison between different types of thermometers was not performed (n = 5); (iii) they were narrative reviews (n = 3); (iv) considered interventions different from those provided as inclusion criteria (n = 4); (v) considered other body temperature sites as a reference standard (i.e. bladder temperature) (n = 2); were a letter (n = 1); were a questionnaire (n = 1). Finally, a total of 46 studies [23–68] were included in this systematic review (Fig.1).
Fig. 1.
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PRISMA flow diagram
Study characteristics
We included 46 studies (12,602 patients), of whom 32 studies (8321 patients, 66%) enrolled only children, 11 enrolled only adults (1856 patients, 15%) and three studies (2425 patients, 19%) enrolled both adults and children. Nineteen studies (4391 patients) enrolled patients admitted to ED. We included 30 cohort studies, 12 cross-sectional studies, and four randomized controlled trials. Table 1 reports details of the studies included. The number of participants ranged from 15 to 2000. The selected studies were published between 1991 and 2019. We included 43 studies in quantitative analyses, three studies [25, 27, 37] did not provide data allowing the DTA analysis and NMA analysis. Six studies [24, 41, 45, 47, 55, 66] reporting results on measurements were excluded from DTA analyses because the unit of analysis was patients, but was included in the NMA because the unit of analysis was means. For one of them , however, we calculated a 2 × 2 table using reported estimates on measurements and prevalence of fever in these patients.
Table 1.
Characteristics of the individual included studies
| Author | Study design | Setting | Population | No. patients enrolled | Reference standard (body temperature, device) | Index test (body temperature, device) |
:--- :--- :---
| Allegaert 2014 | RCT | ED | Children | 294 | Rectal, digital | Tympanic, infrared Forehead, infrared Temporal artery, infrared |
| Apa 2013 | Cohort | Pediatric unit | Children | 50 | Axillary, digital | Tympanic, infrared Forehead, infrared |
| Balla 2019 | Cross-sectional | Neurological and infection wards | Adult | 15 | Rectal, digital | Axillary, thermistor |
| Batra 2013 | Cohort | Pediatric ED | Children | 100 | Rectal, mercury-in-glass | Axillary, digital Tympanic, infrared Temporal artery, infrared |
| Berksoy 2018 | Cohort | Pediatric ED | Children | 319 | Axillary, digital | Forehead, infrared |
| Brennan 1995 | Cohort | ED | Children | 370 | Rectal, digital | Tympanic, infrared |
| Brosinski 2018 | Cohort | ED | Adult and children | 251 | Rectal, digital | Temporal artery, infrared |
| Chiappini 2011 | Cross sectional | Pediatric ED | Children | 252 | Axillary, mercury-in-glass | Forehead, infrared |
| Dakappa 2016 | RCT | Hospital | Adult | 55 | Axillary, mercury-in-glass | Tympanic, infrared |
| Devrim 2007 | Cross sectional | Pediatric hospital | Children | 102 | Axillary, mercury-in-glass | Tympanic, infrared |
| Duru 2012 | Cohort | Pediatric unit and ED | Children | 300 | Rectal, mercury-in-glass | Tympanic, infrared |
| Edelu 2011 | Cohort | Hospital | Children | 800 | Rectal, mercury-in-glass | Tympanic, infrared |
| Forrest 2017 | Cohort | ED | Children | 85 | Rectal, digital | Axillary, digital Temporal artery, infrared |
| Gasim 2013 | Cross sectional | ED | Adult and children | 174 | Axillary, mercury-in-glass | Tympanic, infrared |
| Goswami 2017 | Cohort | Pediatric unit | Children | 210 | Rectal, digital | Axillary, digital Temporal artery, infrared |
| Greenes 2001 | Cross-sectional | ED | Children | 304 | Rectal, digital | Tympanic, infrared Temporal artery, infrared |
| Hamilton 2013 | Cohort | ED | Children | 205 | Rectal or oral, digital | Tympanic, infrared Temporal artery, infrared |
| Hay 2004 | Cohort | Primary care | Children | 94 | Axillary, mercury-in-glass | Tympanic, infrared |
| Hebbar 2005 | Cohort | Pediatric unit | Children | 44 | Rectal, digital | Axillary, digital Temporal artery, infrared |
| Isler 2014 | Cohort | Pediatric ED | Children | 218 | Temporal artery, infrared | Axillary, mercury-in-glass Axillary, digital |
| Jean Mary 2002 | Cohort | Hospital | Children | 198 | Rectal, digital | Axillary, infrared Tympanic, infrared |
| Jensen 2000 | RCT | Surgical unit | Adult | 200 | Rectal, mercury-in-glass | Tympanic, infrared Oral, digital Axillary, digital Rectal, digital |
| Kara 2009 | RCT | Pediatric hospital | Children | 61 | Axillary, mercury-in-glass | Axillary, digital |
| Kocoglu 2002 | Cohort | Hospital | Children | 110 | Rectal, mercury-in-glass | Axillary, mercury-in-glass Tympanic, infrared |
| Leon 2005 | Cross sectional | Intensive care unit | Adult | 50 | Axillary, mercury-in-glass | Tympanic, infrared |
| Mogensen 2018 | Cross-sectional | Pediatric department | Children | 995 | Rectal, digital | Tympanic, infrared Temporal artery, infrared |
| Mogensen 2018b | Cohort | ED | Adult | 599 | Rectal, digital | Tympanic, infrared |
| Morley 1998 | Cohort | Hospital | Children | 1090 | Axillary, mercury-in-glass | Forehead, infrared |
| Muma 1991 | Cross sectional | Pediatric ED | Children | 224 | Rectal, digital | Axillary, digital Tympanic, infrared |
| Odinaka 2014 | Cohort | Pediatric unit | Children | 156 | Rectal, mercury-in-glass | Forehead, infrared |
| Oncel 2013 | Cohort | Maternity unit | Children | 120 | Rectal, mercury-in-glass | Axillary, digital Forehead, infrared |
| Paes 2010 | Cohort | Pediatric unit | Children | 100 | Rectal, digital | Tympanic, infrared Temporal, infrared |
| Petersen 1997 | Cohort | Neurosurgical unit | Adult | 65 | Rectal, mercury-in-glass | Tympanic, infrared |
| Rabbani 2010 | Cohort | Hospital | Adult and children | 2000 | Oral, mercury-in-glass | Tympanic, infrared |
| Rajee 2006 | Cohort | ED | Adult | 200 | Oral, mercury-in-glass | Tympanic, infrared |
| Schreiber 2013 | Cross-sectional | Pediatric ED | Children | 284 | Axillary, mercury | Axillary, digital Axillary, galinstan |
| Schuh 2004 | Cohort | ED | Children | 327 | Rectal | Forehead |
| Sehgal 2002 | Cohort | ED | Children | 60 | Rectal, digital | Tympanic, infrared |
| Singler 2013 | Cohort | ED | Adult | 427 | Rectal, digital | Tympanic, infrared Temporal artery, infrared |
| Smitz 2009 | Cohort | Geriatric unit | Adult | 100 | Rectal, digital | Tympanic, infrared |
| Smitz 2000 | Cohort | Geriatric unit | Adult | 45 | Rectal, mercury-in-glass | Tympanic, infrared |
| Teller 2014 | Cross-sectional | Private pediatric practice | Children | 254 | Rectal, digital | Tympanic, infrared Forehead, infrared |
| Teran 2012 | Cross-sectional | Inpatient unit and ED | Children | 434 | Rectal, mercury-in-glass | Forehead, infrared Temporal artery, infrared |
| Van Staaij 2003 | Cohort | Pediatric unit | Children | 41 | Rectal, digital | Tympanic, infrared |
| Wilshaw 1999 | Cohort | Clinic | Children | 120 | Rectal, mercury | Tympanic, infrared Axillary, digital |
| Yaron 1995 | RCT | ED | Adult | 100 | Rectal, digital | Tympanic, infrared |
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Risk of bias assessment
The results of the methodological quality of the included studies are shown in Appendix 1. The majority of the studies were judged low risk of bias for patient selection and flow and timing. Twenty-six studies (56%) enrolled consecutive or a random sample of patients. Patient enrolment was unclear in 17 studies. The index test domain was judged as unclear in four studies and at high risk in seven studies. Assessors deemed blinding was adequate only in seven studies, and five studies were not blinded regarding the results of the index test and reference standard, but this aspect seemed not to influence the applicability of the study results. In all studies except one, all patients received the same reference standard. Concerns regarding applicability were low for most of the evaluated studies.
Diagnostic accuracy estimates
Twenty-eight studies [23, 26, 28–30, 33–35, 38–40, 43, 47–52, 54, 56–59, 62–65, 68] provided data which permitted the extraction of sensitivity and specificity in 10,207 participants, of whom 2729 (27%) had fever according to the reference standard used. The reference standard was a mercury-in-glass or digital thermometer at the rectal site in 19 studies, mercury-in-glass at the axillary or oral sites (seven studies), the digital thermometer at oral/rectal site (one study), or rectal sites with no information on the device (one study). Fifteen studies out of 19 used 38°C as the cut-off value of temperature for reference devices, four studies used lower values, two studies higher values.
In order to make our results transferrable, we included only studies using a cut-off of 38°C and a reference standard verification at the rectal site, whether using a digital or a mercury-in-glass thermometer.
In 9 studies (2533 participants, 885 with fever) using temporal artery infrared thermometry at a threshold of 38°C, sensitivity varied between 0.41 and 0.91, while a high specificity (from 0.85 to 1.00) was achieved (Fig.2). The meta-analytic estimates were 0.76 (95% CI 0.65,0.84) for sensitivity and 0.96 (0.92,0.98) for specificity (Table 2). This means that adopting a 38°C index test threshold, there are very few false positives, even at the relatively high prevalence of fever at 30%, but there are several false negatives, so the test is useful to rule in the disease when positive. The certainty of the evidence, after downgrading by one level for risk of bias, was low for patients with fever due to imprecision of sensitivity estimates, and moderate for patients without fever (Table 2).
Fig. 2.
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Forest plots of the accuracy of infrared thermometer at the temporal and the tympanic sites (index devices) versus mercury-in-glass or digital thermometer at the rectal site (reference standard device) among studies using 38°C as cut-off values of temperature for index and reference standard devices. TP true positive, FP false positive, FN false negative, TN true negative, CI confidence interval
Table 2.
Summary of findings tables
| DTA (Mercury-in-glass or digital rectal at 38°C as cut-off as reference device) | NMA |
:--- |
| Device | | No. of studies (no. of patients) | Estimate (95% CI) | Effect per 1000 patients tested | Certainty of evidence | No. of direct studies (no. of patients) | Mean difference (95% CI) | Coefficient of reproducibility, median (range) |
| Infrared temporal | TP | 9 (885) | Sensitivity: 0.76 (0.65, 0.84) | 228 (180–252) | ⨁⨁◯◯ Low | 1 (434) vs. Mercury-in-glass rectal | −0.09 (−0.42,–0.24) | 0.53 (0.53,0.53) |
| FN | 72 (48–120) |
| TN | 9 (1648) | Specificity: 0.96 (0.92,0.98) | 672 (644–686) | ⨁⨁⨁◯ Moderate | 8 (2206) vs. Digital rectal | −0.09 (−0.33, –0.16) | 1.2 (0.81,1.5) |
| FP | 28 (14–56) |
| Infrared tympanic | TP | 9 (1,279) | Sensitivity: 0.77 (0.60,0.88) | 231 (180–264) | ⨁⨁◯◯ Low | 6 (840) vs. Mercury-in-glass rectal | −0.22 (−0.49,–0.04) | 0.73 (0.56,1.9) |
| FN | 69 (36–120) |
| TN | 9 (2583) | Specificity: 0.98 (0.95,0.99) | 686 (665–693) | ⨁⨁⨁◯ Moderate | 12 (3520) vs. Digital rectal | −0.22 (−0.43,−0.01) | 1.1 (0.24,1.5) |
| FP | 14 (7–35) |
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Similarly, in nine studies (3862 participants, 1279 with fever) using tympanic infrared thermometry at a threshold of 38°C, high specificity was achieved (from 0.92 to 1.00); however, sensitivity varied between 0.49 and 0.98 in eight studies and was 0.23 in the study with perfect specificity (Fig.2). The meta-analytic estimates were 0.77 (0.60,0.88) for sensitivity and 0.98 (0.95,0.99) for specificity (Table 2). After downgrading by one level for risk of bias, the certainty of the evidence was low for patients with fever due to imprecision of sensitivity estimates, and moderate for patients without fever.
No statistically significant difference was found between sensitivity and specificity estimates with infrared tympanic vs. temporal artery thermometry, which is unsurprising given the high heterogeneity in sensitivity.
There were three or fewer studies on other devices/sites and cut-offs (Appendix 2), thus meta-analyses were not possible.
Mean differences between thermometers using network meta-analyses
Thirty-six studies [24, 28–33, 35, 36, 38, 40–56, 58, 60–63, 65–68] provided data that permitted the extraction of temperature means in 9,878 participants.
Twenty-one studies included in this analysis were incompletely reported regarding the correlation between measurements on the same person. Specifically, four studies reported the SD of the differences between each pair of device-site, 10 studies reported p-values of paired tests, and seven studies reported correlation coefficients, with one study reporting two parameters and the other studies reporting no data to extract the within-subject correlation. We adopted the strategy reported in the Methods to overcome this issue.
Appendix 3 presents a network map. There was no sign of overall (p = 0.9795) or loop-specific between-study heterogeneity, possibly also due to the precision of within-study estimates (small SDs) as compared to the between-study SD (tau) which was 0.397°C, meaning that the additional uncertainty due to heterogeneity was almost + −0.8°C in any 95% predictive interval.
Appendix 4 shows all studies in direct meta-analyses (boxes with horizontal bars) together with the NMA estimate (diamonds). The direct comparison between axillary mercury and infrared tympanic thermometry was very heterogeneous in 6 studies, with extreme values of mean differences ranging from less than −1°C to + 1°C. On the contrary, infrared tympanic thermometry, compared to rectal mercury-in-glass (five studies) or rectal digital thermometry (eight studies), showed consistent differences, suggesting less variation in their results.
Figure3 presents all pairwise mixed estimates with 95% CIs. Assuming rectal mercury-in-glass thermometry as the reference, axillary digital thermometry was significantly lower by −0.67°C (−0.98, −0.37), as was also axillary mercury-in-glass thermometry (−0.55°C [−0.87, −0.23]); a similar difference was obtained for oral digital, and axillary galinstan thermometry, but with greater imprecision crossing significance (−0.56°C [−1.21, 0.08] and −0.52°C [−1.25, 0.21], respectively). All other differences were also in the direction of a lower temperature with respect to rectal thermometry by −0.22°C to 0.00°C, but none was statistically significant. All other pairwise differences among devices were small in most cases but imprecisely estimated.
Fig. 3.
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Forest plot of mean differences among thermometers at different anatomical sites from network meta-analysis
Description of random error for each direct comparison
Figure4 presents the 95% coefficient of reproducibility (95% CR). The mean 95% CR value of 73 direct comparisons between devices was 1.06°C, with 19 comparisons below 0.82°C, 17 between 0.82 and 1.08°C, 19 between 1.08 and 1.24°C, and 18 exceeding 1.24°C. The median 95%CR vs. rectal mercury-in-glass thermometry was 1.16°C for axillary digital thermometry (three studies), 0.79°C for digital oral thermometry (one study), 0.70°C for digital rectal thermometry (one study), 0.73°C for tympanic infrared thermometry (six studies), 1.08°C for infrared forehead thermometry (three studies) and 0.53°C for infrared temporal thermometry (one study).
Fig. 4.
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Forest plot of 95% coefficient of reproducibility (95% CR) of thermometers at different anatomical sites from the meta-analysis
Sensitivity analyses
We restricted the NMA to 24 studies conducted on children and found a similar pattern of differences, although they were less precise due to the reduced size. There were too few studies to estimate accuracy in a specific setting, such as ED.
Discussion
This systematic review summarizes published data from 46 studies evaluating different types of thermometers to measure body temperature. The gold standard to measure core temperature is the rectal temperature as it better reflects a true central temperature. However, it has several drawbacks including impracticability, discomfort, and, although rare, possible complications, such as perforation or transmission of microorganisms.
Our meta-analysis showed that alternative peripheral thermometers were not always accurate at estimating central core temperature, with a tendency to underestimate it up to one degree Celsius. Another challenge is the pervasive presence of a random error that afflicts all thermometers and that can be estimated to add an extra degree of error. The interplay between the fixed and random error originating by the use of different thermometers might generate, in the worst case, clinically relevant differences in the order of two degrees Celsius. The uncertainty associated with thermometers and the resulting implications for decision-making led researchers to use a relatively high fever threshold of 38°C for both the index and the reference test. With this value, the specificity of peripheral thermometers is high and adequate to confirm fever when detected, but the sensitivity is much lower, making it difficult to exclude fever for temperatures below 38°C.
Our network meta-analysis showed that axillary temperature, determined with both mercury and digital thermometers, was significantly lower by −0.65°C and −0.67°C, respectively, than body temperature measured with rectal mercury-in-glass thermometry, while infrared (tympanic, temporal artery, forehead) devices were slightly better estimators of body temperature, showing smaller, non-significant differences. It is to be noted that the mean difference of the rectal temperatures measured with mercury-in-glass or digital thermometers is nil with mild variability and this makes the choice of these two devices as mixed reference standard reasonable. When the aim is to diagnose a febrile state, both in children and adults, the accuracy estimates of both infrared tympanic and temporal thermometry are the best in our review, and they are supported by the largest body of evidence. Previous reviews, despite the variability in the methodology used and in the included studies, also concluded that tympanic and temporal artery thermometers are more accurate, achieving high specificity but insufficient sensitivity when assessed against rectal thermometry [2, 69, 70]. Some of these reviews also conducted meta-analyses of the mean difference between peripheral and rectal thermometry and found that the mean difference was about 0.2°C . Niven et al. calculated 95% coefficients of reproducibility as twice the SDs but did not explain how SDs were pooled across studies .
Rectal temperature is just a proxy of the real (and latent) body temperature. For instance, if the reference device tended to overestimate the real temperature, the “real” sensitivities of the index devices could be possibly higher than the ones illustrated in the paper, because some of the reported false negatives would in fact be true negatives.
According to the GRADE evaluation, the overall certainty of our estimates was moderate, due to some limitations in the design of several studies, or considerable heterogeneity across studies. Heterogeneity could be due to several reasons: measuring temperature in different body sites; concomitant inclusion of children and adults; a threshold effect caused by the use of different offsets by manufacturers to obtain adjusted temperatures according to thermometer technology; and intra- and inter-operator variability of measurements. The risk of bias assessment showed that the study populations were in general selected with convenience samples of participants. Blinding was almost nonexistent, but we considered this as non-fundamental since most technologies give a digital result that has to be recorded without interpretation. The timing between the index and reference methods was usually reported. Finally, there was great heterogeneity among included studies which reduced the quality of evidence.
We suggest that in future studies temperatures should be measured independently at specific sites in a consecutive series of eligible individuals. All thermometers should be previously calibrated. Details on placement time, patient stabilization, and mode of use of thermometers should be provided. Temperature readings should be carried out concurrently or sequentially and the time between measurements clearly documented. However, body temperature should be evaluated in relation to individual variability, since it varies with respect to age, gender, site of measurement, type of thermometer and presence of disease.
As shown above, we found a cut-off of 38°C was highly specific, but not sensitive enough to detect fever with an equivalent rectal cut-off so that thermometry could be used to exclude or rule out fever. If the aim of body temperature measurement is to triage subjects with high sensitivity (confirm or rule in fever, SpIN approach), future research should use an external body site cut-off of about 37°C to confirm a rectal temperature exceeding 38°C. We highlight that the balance of sensitivity and specificity should not be assumed to be stable when the cut-off is changed on the basis of our data, since ROC curves are often asymmetric and the overall accuracy (e.g. DOR) at high specificity may not match the value found at high sensitivity.
Fever is one of the most common patient complaints and signs in emergency departments and is often caused by infection. Other sources include pulmonary embolism, intracranial hemorrhage, medication, or malignancy. Determining a fever represents a fundamental step of health status assessment, with a bearing on medical decisions; for instance, fever can contribute to the empirical assessment of bacterial infections, leading to the prescription of antibiotics. The presence of fever might lead to quarantine in patients suspected of Covid-19 infection or admission to the hospital. Temperature measurement is imperfect and requires awareness and appreciation of its limits. Health professionals should consider that large errors are found when measuring temperature. Therefore, they should complement temperature with additional clinical elements (e.g. medical history, heart rate, and palpitations). Health professionals should adopt quality assurance procedures for fever diagnosis in order to limit variation in clinical practice, enhancing the education on thermometer use and measurement interpretation, similarly to what has been done with the promotion of hand hygiene practice. A simple approach to decrease random error would be to increase the number of measurements, an action that should be considered when the temperature has strong decision-making implications.
When a temperature cut-off of 38°C is used to define fever, several peripheral thermometers proved to be specific, but not sensitive when rectal thermometry is used as a reference standard, meaning that finding a temperature below 38°C does not rule out fever. Among all devices, infrared tympanic and temporal thermometers were better estimators of central temperature and achieved consistent performances across studies. Most thermometers are afflicted with substantial random error. The under-appreciation of the uncertainty in measuring temperature while practicing medicine might have serious consequences: the limited accuracy and reproducibility of thermometers may translate into weak decision-making, a huge waste of resources, and suboptimal patient and population health outcomes.
Electronic supplementary material
Below is the link to the electronic supplementary material.
Supplementary file1 (DOCX 799 kb) (798.5KB, docx)
Authors’ contribution
VP,LM, GV and EL conceived and designed the systematic review. VP, DP,GV and EL performed the systematic review. EL and GV analysed the data. VP, DP,LM,GV and EL wrote the paper. All authors were involved in the critical revision of the intellectual content of the manuscript.
Funding
Open access funding provided by Università di Pisa within the CRUI-CARE Agreement. This study has been funded by “Fondazione CR Firenze” (#24383).
Compliance with ethical standards
Conflict of interest
The authors declared no potential conflict of interest.
Statement of human and animal rights
This article does not contain any studies with human participants or animals performed by any of the authors.
Footnotes
Publisher's Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
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14384 | https://lessonplanet.com/teachers/ck-12-angle-bisectors-in-triangles | Ck 12: Angle Bisectors in Triangles Interactive for 9th - 10th Grade | Lesson Planet
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Grade 9th - 10th
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Interactive
Ck 12: Angle Bisectors in Triangles
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[Free Registration/Login Required] Students will use interactive triangles to explore how a line bisects an angle. Click learn more to watch a video and click challenge me to answer a practice question.
3 Views 0 Downloads CCSS:Adaptable
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types of lines, angle bisector
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Knovation Readability Score: 3 (1 low difficulty, 5 high difficulty)
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HSG-C.A.2
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14385 | https://www-backup.salemstate.edu/alkane-alkene-alkyne | Alkane Alkene Alkyne - Salem State Vault
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Home/Plus/Alkane Alkene Alkyne
Alkane Alkene Alkyne
Ashley
October 20, 2024 8 min read
The world of organic chemistry is vast and complex, with numerous compounds and reactions that form the foundation of life and industry. Among these, hydrocarbons are a fundamental class of compounds, consisting entirely of hydrogen and carbon atoms. Hydrocarbons can be broadly classified into several categories, including alkanes, alkenes, and alkynes, each with its unique properties, structures, and applications. Understanding these compounds is crucial for chemists, engineers, and researchers as they are the building blocks of more complex molecules and materials.
Table of Contents
Introduction to Alkanes, Alkenes, and Alkynes
Properties and Reactions of Alkanes
Properties and Reactions of Alkenes
Properties and Reactions of Alkynes
Applications and Implications
Introduction to Alkanes, Alkenes, and Alkynes
Alkanes, alkenes, and alkynes are types of unsaturated and saturated hydrocarbons. The primary distinction among them lies in the type of bonds between the carbon atoms. Alkanes are saturated hydrocarbons, meaning they only contain single bonds between carbon atoms, following the general formula CnH2n+2. Alkenes, on the other hand, are unsaturated hydrocarbons with at least one carbon-to-carbon double bond, adhering to the formula CnH2n. Alkynes, another category of unsaturated hydrocarbons, contain at least one carbon-to-carbon triple bond, with the general formula CnH2n-2.
Key Points
Alkanes are saturated hydrocarbons with single bonds between carbon atoms.
Alkenes are unsaturated hydrocarbons characterized by at least one double bond between carbon atoms.
Alkynes are unsaturated hydrocarbons with at least one triple bond between carbon atoms.
The general formulas for alkanes, alkenes, and alkynes are CnH2n+2, CnH2n, and CnH2n-2, respectively.
Understanding the structure and properties of these hydrocarbons is essential for their applications in chemistry and industry.
Properties and Reactions of Alkanes
Alkanes are generally non-polar, making them insoluble in water but soluble in non-polar solvents. They are also relatively unreactive due to their saturated nature, which makes them less prone to chemical reactions compared to alkenes and alkynes. However, they can undergo substitution reactions, where a hydrogen atom is replaced by another functional group, and combustion reactions, where they react with oxygen to produce carbon dioxide and water. The energy density of alkanes, especially in the form of fossil fuels like methane (natural gas), propane, and butane, makes them crucial for energy production.
Properties and Reactions of Alkenes
Alkenes, due to the presence of a double bond, exhibit a higher reactivity compared to alkanes. This double bond allows them to participate in addition reactions, where another molecule adds across the double bond, resulting in a single bond. A common example is the addition of hydrogen (hydrogenation) or bromine across the double bond. Alkenes also undergo polymerization reactions, where multiple alkene molecules combine to form a large polymer chain, which is a critical process in the production of plastics like polyethylene and polypropylene.
Properties and Reactions of Alkynes
Alkynes, with their triple bond, are even more reactive than alkenes. The triple bond can participate in addition reactions similar to alkenes but with more vigor due to the higher reactivity. Alkynes can undergo hydrogenation to first form an alkene and then an alkane, demonstrating the stepwise addition of hydrogen atoms across the triple bond. They can also react with water in the presence of acid to form ketones, showcasing their versatility in organic synthesis.
| Type of Hydrocarbon | General Formula | Example |
---
| Alkane | CnH2n+2 | Methane (CH4) |
| Alkene | CnH2n | Ethene (C2H4) |
| Alkyne | CnH2n-2 | Ethyne (C2H2) |
💡 The distinction between alkanes, alkenes, and alkynes not only lies in their structures but also in their applications and reactivities. Understanding these differences is pivotal for advancing in fields like organic chemistry, materials science, and energy production.
Applications and Implications
The applications of alkanes, alkenes, and alkynes are diverse and critical to various industries. Alkanes serve as primary energy sources and are used in the production of other chemicals. Alkenes are the backbone of the plastics industry, with polyethylene and polypropylene being two of the most widely used plastics globally. Alkynes, while less common in everyday applications, play a significant role in organic synthesis and the production of pharmaceuticals and other complex molecules.
Given the central role of these hydrocarbons in energy and manufacturing, their sustainability and environmental impact are under scrutiny. The combustion of alkanes contributes significantly to greenhouse gas emissions, and the production of plastics from alkenes has led to substantial plastic waste and pollution issues. Efforts to find more sustainable sources of energy and to develop biodegradable alternatives to traditional plastics are underway, highlighting the evolving landscape of hydrocarbon usage and application.
What is the primary difference between alkanes, alkenes, and alkynes?
+
The primary difference lies in the type of carbon-carbon bonds: alkanes have single bonds, alkenes have at least one double bond, and alkynes have at least one triple bond.
What are some common applications of alkanes, alkenes, and alkynes?
+
Alkanes are primarily used as energy sources, alkenes are used in the production of plastics, and alkynes are used in organic synthesis and the production of pharmaceuticals.
Why are alkenes more reactive than alkanes?
+
Alkenes are more reactive due to the presence of a double bond between carbon atoms, which makes them more susceptible to addition reactions compared to the single-bonded alkanes.
In conclusion, the study of alkanes, alkenes, and alkynes offers insights into the fundamental principles of organic chemistry and their applications in various industries. As research and technology advance, the role of these hydrocarbons in sustainable development and environmental conservation will continue to evolve, highlighting the need for a deep understanding of their properties, reactions, and implications.
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14386 | https://programs.iowadnr.gov/lakemanagement/fishiowa/fishdetails/ROB | Fish Iowa - Fish Species - Rock Bass
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Fish Species Profile DNR Homepage
Species Information
Characteristics
A stout, heavy-bodied sunfish with a large mouth, which extends beyond mid-eye when the mouth is closed. The spiny dorsal fin and soft dorsal fin are broadly connected, but without a notch. The dorsal fin is much longer and more noticeable than the anal fin. Six anal fin spines and 12 dorsal fin spines distinguish this fish from all other sunfish. The body is olive with brassy reflections and dark mottlings along the sides. The breast and belly are whitish, and the lower side has spots that form horizontal lines. There is brown mottling and faint banding on the anal, dorsal and tail fins. The pectoral fins are rounded, set low and are amber-colored.
Foods
aquatic insects, minnows and other small fish
Expert Tip
True to their name, the Rock Bass can be caught in streams by fishing close to the rocks near the current.
Details
The Rock Bass is mainly a sedentary and inactive fish, spending much of its activity hiding in the shadows of underwater structures.
Spawning matches that of Smallmouth Bass. Spawning starts when the water temperature is 65 to 75 degrees, usually in May and June. Males build nests over gravel and sand bottoms where the females lay several thousand eggs. Females have an average of 5,000 eggs, but one or several fish may deposit part or all of their eggs in a single nest. After hatching, the young fish are found only in quiet-waters protected from waves and strong current. Rock bass grow 1 1/2- to 2-inches the first year and reach 5- to 7-inches long after three years. Rock Bass have lived 13 years in nature and weighed 1 1/2 pounds, but most seldom exceed 10-inches.
Recent stream sampling information is available fromIowa DNR's biological monitoring and assessment program.
Species Distribution
Distribution Map
Limited to the clearer, rocky habitats of the interior streams in northeast Iowa. Rarely found in the upper parts of the Des Moines and Mississippi rivers. Occasionally found in the large, natural lakes.
See our most recent distribution data for this species on the Iowa DNR's Bionet application.
Record Fish
State Record(s)
Jim Driscoll
June 01, 1973: 1.50 lbs. - 10.50 in.
Pool 11, Mississippi River
View more state records in Master Angler
Master Angler Catches
RYAN STAUDT
August 29, 2025: 8.00 in.
Lizard Creek
BRENT BAERENWALD
August 29, 2025: 8.50 in.
Boone River
Bradford Lee
August 29, 2025: 1.00 lbs. - 8.50 in.
White Fox Creek
Michael Gyles
August 19, 2025: 0.75 lbs. - 8.00 in.
Lake Ahquabi
View more Master Angler entries
Populations and Stocking
Fish Surveys
Tip: Click Species Length by Site, then use the dropdown to filter by fish species of interest.
Where this Fish Is Found
Pool 16, Mississippi River
Pool 17, Mississippi River
Turkey River (above Clermont)
Volga River
Upper Iowa River (above Decorah)
Maquoketa River (above Monticello)
Turkey River (below Clermont)
Cedar River (above Nashua)
Upper Iowa River (below Decorah)
Wapsipinicon River (Troy Mills to Oxford Junction)
Lidtke Impoundment
Vernon Springs Impoundment
Elkader Impoundment
Lake Oelwein
Otranto Impoundment
Waucoma Impoundment
Bear Creek
Bohemian Creek
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Version: 2.4.1.21023 |
14387 | https://www.youtube.com/watch?v=excg9IC-yGA | How do you write asinx+bcosx as ksin(x+α)?
Diane R Koenig
10200 subscribers
65 likes
Description
3737 views
Posted: 22 Jun 2020
When graphing or solving trig equations, it is sometimes helpful to write two-term expressions involving sine and cosine of an angle as an expression with just the sine function. This video describes the process and gives an example of rewriting and then graphing the function y=-sinx+√3 cosx
from y=ksin(x+α).
10 comments
Transcript:
Intro there are some situations where we have an addition or a subtraction of a sine and cosine function acting on the same angle and it's harder to work with in that form if we could get it where it was just maybe an amplitude change and a phase shift on one of those trig functions and still mean the same thing it would be much easier to work with whether we're solving equations or graphing functions or whichever those cases might be in this video we're going to talk about how to take an expression that is in the form of a numerical multiplication of sine of an angle and then plus a numerical multiplication of cosine of that same angle that the sine was acting on and actually take that and write that as an equivalent equate expression that is just a multiple of sine of that angle with a phase shift now in doing this there are specifics K about how you get each of these components when making that transition the first one is K this new coefficient in front of your sine function is equal to the square root of a squared plus B squared so you get K by taking the square root of the coefficient that's in front of the sine at by angle quantity squared plus the coefficient that's in front of the cosine of the angle that's where so our K is the square root of a squared plus B squared now alpha how do Alpha we find that angle in the phase shift well it has to be the angle that meets both the conditions that if I were to take the sine of that angle I would get the ratio of the coefficient in front of the cosine over that K value the square root of a squared plus B squared also if I took cosine of that angle it would be the same as the ratio of the coefficient from in front of the sine of the original expression over the square root of a squared plus B squared so sometimes in different books they might have this a little bit rearranged you don't want to get it tied too much to a and B as letters in the numerator of these ratios you just want to think for the sine of the angle for the phase shift you're going to take the coefficient from in front of the cosine term over the K and then for the cosine of the angle you're going to take the coefficient in front of the sign over the term so you're you're making this kind of mixture back to the roots where it comes from when we were looking at our identity for actually the sine of the sum of two angles and kind of working backwards on it but in this Example we're not going to like develop it we're just going to show this example so here it says to graph y equal negative 1 sine X plus the square root of 3 cosine X by writing it in the form so I'm going to write this in the form K sine X plus alpha and then graph this y equal K times the sine of X plus alpha so let's start by just looking at this expression and bringing it through so I have my Y is equal to well that's negative 1 times sine X plus the square root of 3 times the cosine of X so here my a is negative 1 and my B is the square root of 3 so K is going to be the square root of negative 1 in parenthesis squared plus the square root of 3 in parenthesis squared now remember you have to follow your order of operations and you have to make sure that you are careful about realizing that it's negative 1 quantity squared so negative 1 quantity squared is 1 plus the square root of 3 squared is 3 and 1 plus 3 under there is 4 so K is the square root of 4 which gives me a K value of 2 so that gives me back a I Finding the Alpha already have this coefficient then so next up I need to find the Alpha well the Alpha has to be the angle such that the sine of that angle alpha is the coefficient in front of the cosine over the K so it's going to be the square root of three over two and it's also true that the cosine of that alpha is equal to the coefficient from in front of the sine over two so negative half so I want to think okay as far as the reference angles go what reference angle will have a cosine ratio of 1/2 and a sine ratio of the square root of three over two well that's a nice angle that's PI over 3 but that's the reference angle now what quadrant does an angle have to be in so that the cosine ratio is negative and the sine ratio is positive well from the origin like with the unit circle you would have to go left and up so it would be in the second quadrant so the over three angle that's in the second quadrant is my alpha being 2 PI over 3 so now I'm ready to write my equation it's y is equal to K so to in our problem times the sine of X plus my angle is 2 PI over 3 so that made it much easier to graph because we can think back to when we were graphing trig functions and we have our amplitude changes than our phase shifts than all of that so our basic function here is the sine function so remember we're gonna like kind of build off of doing the work with our basic function so our basic basic function of sine X well sine of 0 is 0 sine of PI over 2 is 1 sine of pi is 0 sine of 3 PI over 2 is negative 1 and sine of 2 pi is 0 so here's our sine of X now this also has an amplitude of the absolute value of 2 which is 2 so every one of those white outputs we're going to multiply by 2 so 2 times 0 is 0 2 times 1 is 2 2 times 0 is 0 2 times negative 1 is negative 2 and 2 times 0 is 0 so here is our y equal to sine X and then the last thing we need to take care of is our phase shift now remember our phase shift we take what the trig function is acting on and set it equal to 0 so if X plus 2 PI over 3 is equal to 0 subtract your 2 PI over 3 from both sides X is negative 2 PI over 3 that's the one that would be comparable to being your like zero angle and where it got shifted to so this y equal to sine of X plus 2 PI over 3 has our amplitude of basic function or basic graph of sine amplitude of 2 and a phase shift of 2 PI over 3 to the left ok so now each of these angles we're going to take the dot that's at the value right now we're going move it 2 PI over 3 to the left so if I'm at an angle of 0 and I go left 2 PI over 3 I'm at an angle of negative 2 PI over 3 and if I'm at an angle of PI over 2 and I go left 2 PI over 3 well think about it with fractions so if I'm at an angle of PI over 2 and I go left 2 PI over 3 I need to get common denominators to put this together better so comment and I'm near 6 I'll multiply top and bottom of PI root 2 by 3 that's 3 PI over 6 minus top and bottom of the next fraction by 2 so that's 4 PI over 6 and 3 PI over 6 minus 4 PI over 6 is a negative PI over 6 so this dot think has a y-coordinate of positive 2 is going to move to negative PI over 6 angle so negative PI over 6 comma 2 and then 4 PI subtract 2 PI over 3 you're going to be at PI over 3 to where it has its 0 and then when I have 3 PI over 2 and I subtract the 2 PI over 3 so 3 PI over 2 subtract the 2 PI over 3 again get your common denominators so 9 PI over 6 minus 4 PI over 6 gives me a 5 PI over 6 that's where I'm going to have the negative two and then finally when I take 2 pi and subtract 2 PI over 3 well 2 pi is 6 PI over 3 minus 2 PI over 3 gives me a 4 PI over 3 so I'm going to move that to 4 PI over 3 and that's where it's 0 again so we have then our graph of our final function that is here so I was able to use all of the information that I'd already gained when I was doing the graphing with my ships and my amplitude change in all of that by transitioning this to term with sine and cosine acting on the same angle to amplitude change and phase shift of a sine function and again it's not just going to be something that's useful in the graphing problems it will also show up again when we're looking at solving trig equations and maybe we have a trig equation that's involved in a form such as that |
14388 | https://www.semanticscholar.org/paper/Le-Cam%27s-inequality-and-poisson-approximations-Steele/80abb7ce8a0a41c8b1d14b1173333c063b274577 | Le Cam's inequality and poisson approximations | Semantic Scholar
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DOI:10.2307/2325124
Corpus ID: 122052595
Le Cam's inequality and poisson approximations
@article{Steele1994LeCI,
title={Le Cam's inequality and poisson approximations},
author={J. Michael Steele},
journal={American Mathematical Monthly},
year={1994},
volume={101},
pages={48-54},
url={
}
J. Steele
Published 1994
Mathematics
American Mathematical Monthly
where A = P1 + P2 + .. + Pn Naturally, this inequality contains the classical Poisson limit law (Just set pi = A/n and note that the right side simplifies to 2A2/n), but it also achieves a great deal more. In particular, Le Cam's inequality identifies the sum of the squares of the pi as a quantity governing the quality of the Poisson approximation. Le Cam's inequality also seems to be one of those facts that repeatedly calls to be proved-and improved. Almost before the ink was dry on Le Cam's…Expand
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14389 | https://lib.zu.edu.pk/ebookdata/Engineering/Energy%20System/Fluid%20Mechanics_%20Fundamentals%20and%20Applications%20by%20Yunus%20A.%20Cengel%20Dr.,%20John%20M.%20Cimbala%20.pdf | F L U I D M E C H A N I C S FUNDAMENTALS AND APPLICATIONS FOURTH EDITION cen96534_fm_i_xxvi.indd 1 01/12/16 10:56 am This page intentionally left blank F L U I D M E C H A N I C S FUNDAMENTALS AND APPLICATIONS FOURTH EDITION YUNUS A. ÇENGEL Department of Mechanical Engineering University of Nevada, Reno JOHN M. CIMBALA Department of Mechanical and Nuclear Engineering The Pennsylvania State University cen96534_fm_i_xxvi.indd 3 01/12/16 10:56 am FLUID MECHANICS: FUNDAMENTALS AND APPLICATIONS, FOURTH EDITION Published by McGraw-Hill Education, 2 Penn Plaza, New York, NY 10121. Copyright © 2018 by McGraw-Hill Education. All rights reserved. Printed in the United States of America. Previous editions © 2014, 2010, and 2006. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of McGraw-Hill Education, including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning.
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Library of Congress Cataloging-in-Publication Data Names: Çengel, Yunus A., author. | Cimbala, John M., author.
Title: Fluid mechanics : fundamentals and applications / Yunus A. Çengel (Department of Mechanical Engineering, University of Nevada, Reno), John M. Cimbala (Department of Mechanical and Nuclear Engineering, The Pennsylvania State University).
Description: Fourth edition. | New York, NY : McGraw-Hill Education, | Includes bibliographical references and index.
Identifiers: LCCN 2016050135| ISBN 9781259696534 (alk. paper) | ISBN 1259696537 (alk. paper) Subjects: LCSH: Fluid dynamics.
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mheducation.com/highered cen96534_fm_i_xxvi.indd 4 01/12/16 10:56 am S N L v cen96534_fm_i_xxvi.indd 5 01/12/16 10:56 am A b o u t t h e A u t h o r s Yunus A. Çengel is Professor Emeritus of Mechanical Engineering at the University of Nevada, Reno. He received his B.S. in mechanical engineering from Istanbul Technical University and his M.S. and Ph.D. in mechanical engineering from North Carolina State University. His research areas are renewable energy, desalination, exergy analysis, heat transfer enhancement, radiation heat transfer, and energy conservation. He served as the director of the Industrial Assessment Center (IAC) at the University of Nevada, Reno, from 1996 to 2000. He has led teams of engineering students to numerous manufacturing facilities in Northern Nevada and California to do industrial assessments, and has prepared energy conservation, waste minimization, and productivity enhancement reports for them. Dr. Çengel is the coauthor of the widely adopted textbook Thermodynamics: An Engi neering Approach, 8th edition (2015), published by McGraw-Hill Education. He is also the coauthor of the textbook Heat and Mass Transfer: Fundamentals & Applications, 5th Edition (2015), and the coauthor of the textbook Fundamentals of Thermal-Fluid Sciences, 5th edition (2017), both published by McGraw-Hill Education. Some of his textbooks have been translated to Chinese, Japanese, Korean, Spanish, Turkish, Italian, and Greek. Dr. Çengel is the recipient of several outstanding teacher awards, and he has received the ASEE Meriam/Wiley Distinguished Author Award for excellence in authorship in 1992 and again in 2000.
Dr. Çengel is a registered Professional Engineer in the State of Nevada, and is a member of the American Society of Mechanical Engineers (ASME) and the Ameri can Society for Engineering Education (ASEE). John M. Cimbala is Professor of Mechanical Engineering at The Pennsyl vania State University, University Park. He received his B.S. in Aerospace Engi neering from Penn State and his M.S. in Aeronautics from the California Institute of Technology (CalTech). He received his Ph.D. in Aeronautics from CalTech in 1984 under the supervision of Professor Anatol Roshko, to whom he will be forever grateful. His research areas include experimental and computational fluid mechan ics and heat transfer, turbulence, turbulence modeling, turbomachinery, indoor air quality, and air pollution control. Professor Cimbala completed sabbatical leaves at NASA Langley Research Center (1993–94), where he advanced his knowledge of computational fluid dynamics (CFD), and at Weir American Hydo (2010–11), where he performed CFD analyses to assist in the design of hydroturbines.
Dr. Cimbala is the coauthor of three other textbooks: Indoor Air Quality Engi neering: Environmental Health and Control of Indoor Pollutants (2003), pub lished by Marcel-Dekker, Inc.; Essentials of Fluid Mechanics: Fundamentals and Applications (2008); and Fundamentals of Thermal-Fluid Sciences, 5th edition (2017), both published by McGraw-Hill Education. He has also contributed to parts of other books, and is the author or coauthor of dozens of journal and conference papers. He has also recently ventured into writing novels. More information can be found at www.mne.psu.edu/cimbala.
Professor Cimbala is the recipient of several outstanding teaching awards and views his book writing as an extension of his love of teaching. He is a member of the American Society of Mechanical Engineers (ASME), the American Society for Engineering Education (ASEE), and the American Physical Society (APS).
cen96534_fm_i_xxvi.indd 6 01/12/16 10:56 am B r i e f C o n t e n t s c h a p t e r o n e INTRODUCTION AND BASIC CONCEPTS 1 c h a p t e r t w o PROPERTIES OF FLUIDS 37 c h a p t e r t h r e e PRESSURE AND FLUID STATICS 77 c h a p t e r f o u r FLUID KINEMATICS 137 c h a p t e r f i v e BERNOULLI AND ENERGY EQUATIONS 189 c h a p t e r s i x MOMENTUM ANALYSIS OF FLOW SYSTEMS 249 c h a p t e r s e v e n DIMENSIONAL ANALYSIS AND MODELING 297 c h a p t e r e i g h t INTERNAL FLOW 351 c h a p t e r n i n e DIFFERENTIAL ANALYSIS OF FLUID FLOW 443 c h a p t e r t e n APPROXIMATE SOLUTIONS OF THE NAVIER–STOKES EQUATION 519 c h a p t e r e l e v e n EXTERNAL FLOW: DRAG AND LIFT 611 c h a p t e r t w e l v e COMPRESSIBLE FLOW 667 c h a p t e r t h i r t e e n OPEN-CHANNEL FLOW 733 c h a p t e r f o u r t e e n TURBOMACHINERY 793 c h a p t e r f i f t e e n INTRODUCTION TO COMPUTATIONAL FLUID DYNAMICS 885 cen96534_fm_i_xxvi.indd 7 01/12/16 10:56 am C o n t e n t s Preface xv c h a p t e r o n e INTRODUCTION AND BASIC CONCEPTS 1 1–1 Introduction 2 What Is a Fluid? 2 Application Areas of Fluid Mechanics 4 1–2 A Brief History of Fluid Mechanics 6 1–3 The No-Slip Condition 8 1–4 Classification of Fluid Flows 9 Viscous versus Inviscid Regions of Flow 10 Internal versus External Flow 10 Compressible versus Incompressible Flow 10 Laminar versus Turbulent Flow 11 Natural (or Unforced) versus Forced Flow 11 Steady versus Unsteady Flow 12 One-, Two-, and Three-Dimensional Flows 13 Uniform versus Nonuniform Flow 14 1–5 System and Control Volume 15 1–6 Importance of Dimensions and Units 16 Some SI and English Units 17 Dimensional Homogeneity 19 Unity Conversion Ratios 21 1–7 Modeling in Engineering 22 1–8 Problem-Solving Technique 24 Step 1: Problem Statement 24 Step 2: Schematic 24 Step 3: Assumptions and Approximations 24 Step 4: Physical Laws 24 Step 5: Properties 25 Step 6: Calculations 25 Step 7: Reasoning, Verification, and Discussion 25 1–9 Engineering Software Packages 26 Equation Solvers 27 CFD Software 28 1–10 Accuracy, Precision, and Significant Digits 28 Application Spotlight: What Nuclear Blasts and Raindrops Have in Common 32 Summary 33 References and Suggested Reading 33 Problems 33 c h a p t e r t w o PROPERTIES OF FLUIDS 37 2–1 Introduction 38 Continuum 38 2–2 Density and Specific Gravity 39 Density of Ideal Gases 40 2–3 Vapor Pressure and Cavitation 41 2–4 Energy and Specific Heats 43 2–5 Compressibility and Speed of Sound 45 Coefficient of Compressibility 45 Coefficient of Volume Expansion 46 Speed of Sound and Mach Number 49 2–6 Viscosity 51 2–7 Surface Tension and Capillary Effect 56 Capillary Effect 59 Summary 62 Application Spotlight: Cavitation 63 References and Suggested Reading 64 Problems 64 c h a p t e r t h r e e PRESSURE AND FLUID STATICS 77 3–1 Pressure 78 Pressure at a Point 79 Variation of Pressure with Depth 80 3–2 Pressure Measurement Devices 84 The Barometer 84 The Manometer 87 Other Pressure Measurement Devices 90 3–3 Introduction to Fluid Statics 91 3–4 Hydrostatic Forces on Submerged Plane Surfaces 92 Special Case: Submerged Rectangular Plate 95 3–5 Hydrostatic Forces on Submerged Curved Surfaces 97 3–6 Buoyancy and Stability 100 Stability of Immersed and Floating Bodies 104 cen96534_fm_i_xxvi.indd 8 01/12/16 10:56 am CONTENTS ix 3–7 Fluids in Rigid-Body Motion 106 Special Case 1: Fluids at Rest 108 Special Case 2: Free Fall of a Fluid Body 108 Acceleration on a Straight Path 108 Rotation in a Cylindrical Container 110 Summary 114 References and Suggested Reading 115 Problems 115 c h a p t e r f o u r FLUID KINEMATICS 137 4–1 Lagrangian and Eulerian Descriptions 138 Acceleration Field 140 Material Derivative 143 4–2 Flow Patterns and Flow Visualization 145 Streamlines and Streamtubes 145 Pathlines 146 Streaklines 148 Timelines 150 Refractive Flow Visualization Techniques 151 Surface Flow Visualization Techniques 152 4–3 Plots of Fluid Flow Data 152 Profile Plots 153 Vector Plots 153 Contour Plots 154 4–4 Other Kinematic Descriptions 155 Types of Motion or Deformation of Fluid Elements 155 4–5 Vorticity and Rotationality 160 Comparison of Two Circular Flows 163 4–6 The Reynolds Transport Theorem 164 Alternate Derivation of the Reynolds Transport Theorem 169 Relationship between Material Derivative and RTT 172 Summary 172 Application Spotlight: Fluidic Actuators 173 Application Spotlight: Smelling Food; the Human Airway 174 References and Suggested Reading 175 Problems 175 c h a p t e r f i v e BERNOULLI AND ENERGY EQUATIONS 189 5–1 Introduction 190 Conservation of Mass 190 The Linear Momentum Equation 190 Conservation of Energy 190 5–2 Conservation of Mass 191 Mass and Volume Flow Rates 191 Conservation of Mass Principle 193 Moving or Deforming Control Volumes 195 Mass Balance for Steady-Flow Processes 195 Special Case: Incompressible Flow 196 5–3 Mechanical Energy and Efficiency 198 5–4 The Bernoulli Equation 203 Acceleration of a Fluid Particle 204 Derivation of the Bernoulli Equation 204 Force Balance across Streamlines 206 Unsteady, Compressible Flow 207 Static, Dynamic, and Stagnation Pressures 207 Limitations on the Use of the Bernoulli Equation 208 Hydraulic Grade Line (HGL) and Energy Grade Line (EGL) 210 Applications of the Bernoulli Equation 212 5–5 General Energy Equation 219 Energy Transfer by Heat, Q 220 Energy Transfer by Work, W 220 5–6 Energy Analysis of Steady Flows 223 Special Case: Incompressible Flow with No Mechanical Work Devices and Negligible Friction 226 Kinetic Energy Correction Factor, 𝛼 226 Summary 233 References and Suggested Reading 234 Problems 235 c h a p t e r s i x MOMENTUM ANALYSIS OF FLOW SYSTEMS 249 6–1 Newton’s Laws 250 6–2 Choosing a Control Volume 251 6–3 Forces Acting on a Control Volume 252 6–4 The Linear Momentum Equation 255 Special Cases 257 Momentum-Flux Correction Factor, β 257 Steady Flow 259 Flow with No External Forces 260 6–5 Review of Rotational Motion and Angular Momentum 269 6–6 The Angular Momentum Equation 272 Special Cases 274 Flow with No External Moments 275 Radial-Flow Devices 275 cen96534_fm_i_xxvi.indd 9 01/12/16 10:56 am x FLUID MECHANICS Application Spotlight: Manta Ray Swimming 280 Summary 282 References and Suggested Reading 282 Problems 283 c h a p t e r s e v e n DIMENSIONAL ANALYSIS AND MODELING 297 7–1 Dimensions and Units 298 7–2 Dimensional Homogeneity 299 Nondimensionalization of Equations 300 7–3 Dimensional Analysis and Similarity 305 7–4 The Method of Repeating Variables and the Buckingham Pi Theorem 309 Historical Spotlight: Persons Honored by Nondimensional Parameters 317 7–5 Experimental Testing, Modeling, and Incomplete Similarity 325 Setup of an Experiment and Correlation of Experimental Data 325 Incomplete Similarity 326 Wind Tunnel Testing 326 Flows with Free Surfaces 329 Application Spotlight: How a Fly Flies 332 Summary 333 References and Suggested Reading 333 Problems 333 c h a p t e r e i g h t INTERNAL FLOW 351 8–1 Introduction 352 8–2 Laminar and Turbulent Flows 353 Reynolds Number 354 8–3 The Entrance Region 355 Entry Lengths 356 8–4 Laminar Flow in Pipes 357 Pressure Drop and Head Loss 359 Effect of Gravity on Velocity and Flow Rate in Laminar Flow 361 Laminar Flow in Noncircular Pipes 362 8–5 Turbulent Flow in Pipes 365 Turbulent Shear Stress 366 Turbulent Velocity Profile 368 The Moody Chart and Its Associated Equations 370 Types of Fluid Flow Problems 372 8–6 Minor Losses 379 8–7 Piping Networks and Pump Selection 386 Series and Parallel Pipes 386 Piping Systems with Pumps and Turbines 388 8–8 Flow Rate and Velocity Measurement 396 Pitot and Pitot-Static Probes 396 Obstruction Flowmeters: Orifice, Venturi, and Nozzle Meters 398 Positive Displacement Flowmeters 401 Turbine Flowmeters 402 Variable-Area Flowmeters (Rotameters) 403 Ultrasonic Flowmeters 404 Electromagnetic Flowmeters 406 Vortex Flowmeters 407 Thermal (Hot-Wire and Hot-Film) Anemometers 408 Laser Doppler Velocimetry 410 Particle Image Velocimetry 411 Introduction to Biofluid Mechanics 414 Application Spotlight: PIV Applied to Cardiac Flow 420 Application Spotlight: Multicolor Particle Shadow Velocimetry/Accelerometry 421 Summary 423 References and Suggested Reading 424 Problems 425 c h a p t e r n i n e DIFFERENTIAL ANALYSIS OF FLUID FLOW 443 9–1 Introduction 444 9–2 Conservation of Mass—The Continuity Equation 444 Derivation Using the Divergence Theorem 445 Derivation Using an Infinitesimal Control Volume 446 Alternative Form of the Continuity Equation 449 Continuity Equation in Cylindrical Coordinates 450 Special Cases of the Continuity Equation 450 9–3 The Stream Function 456 The Stream Function in Cartesian Coordinates 456 The Stream Function in Cylindrical Coordinates 463 The Compressible Stream Function 464 cen96534_fm_i_xxvi.indd 10 01/12/16 10:56 am CONTENTS xi 9–4 The Differential Linear Momentum Equation— Cauchy’s Equation 465 Derivation Using the Divergence Theorem 465 Derivation Using an Infinitesimal Control Volume 466 Alternative Form of Cauchy’s Equation 469 Derivation Using Newton’s Second Law 469 9–5 The Navier–Stokes Equation 470 Introduction 470 Newtonian versus Non-Newtonian Fluids 471 Derivation of the Navier–Stokes Equation for Incompressible, Isothermal Flow 472 Continuity and Navier–Stokes Equations in Cartesian Coordinates 474 Continuity and Navier–Stokes Equations in Cylindrical Coordinates 475 9–6 Differential Analysis of Fluid Flow Problems 476 Calculation of the Pressure Field for a Known Velocity Field 476 Exact Solutions of the Continuity and Navier–Stokes Equations 481 Differential Analysis of Biofluid Mechanics Flows 499 Summary 502 References and Suggested Reading 502 Application Spotlight: The No-Slip Boundary Condition 503 Problems 504 c h a p t e r t e n APPROXIMATE SOLUTIONS OF THE NAVIER–STOKES EQUATION 519 10–1 Introduction 520 10–2 Nondimensionalized Equations of Motion 521 10–3 The Creeping Flow Approximation 524 Drag on a Sphere in Creeping Flow 527 10–4 Approximation for Inviscid Regions of Flow 529 Derivation of the Bernoulli Equation in Inviscid Regions of Flow 530 10–5 The Irrotational Flow Approximation 533 Continuity Equation 533 Momentum Equation 535 Derivation of the Bernoulli Equation in Irrotational Regions of Flow 535 Two-Dimensional Irrotational Regions of Flow 538 Superposition in Irrotational Regions of Flow 542 Elementary Planar Irrotational Flows 542 Irrotational Flows Formed by Superposition 549 10–6 The Boundary Layer Approximation 558 The Boundary Layer Equations 563 The Boundary Layer Procedure 568 Displacement Thickness 572 Momentum Thickness 575 Turbulent Flat Plate Boundary Layer 576 Boundary Layers with Pressure Gradients 582 The Momentum Integral Technique for Boundary Layers 587 Summary 595 References and Suggested Reading 596 Application Spotlight: Droplet Formation 597 Problems 598 c h a p t e r e l e v e n EXTERNAL FLOW: DRAG AND LIFT 611 11–1 Introduction 612 11–2 Drag and Lift 614 11–3 Friction and Pressure Drag 618 Reducing Drag by Streamlining 619 Flow Separation 620 11–4 Drag Coefficients of Common Geometries 621 Biological Systems and Drag 622 Drag Coefficients of Vehicles 625 Superposition 627 11–5 Parallel Flow over Flat Plates 629 Friction Coefficient 631 11–6 Flow over Cylinders and Spheres 633 Effect of Surface Roughness 636 11–7 Lift 638 Finite-Span Wings and Induced Drag 642 Lift Generated by Spinning 643 Flying in Nature! 647 Summary 650 Application Spotlight: Drag Reduction 652 References and Suggested Reading 653 Problems 653 c h a p t e r t w e l v e COMPRESSIBLE FLOW 667 12–1 Stagnation Properties 668 12–2 One-Dimensional Isentropic Flow 671 Variation of Fluid Velocity with Flow Area 673 Property Relations for Isentropic Flow of Ideal Gases 675 cen96534_fm_i_xxvi.indd 11 01/12/16 10:56 am xii FLUID MECHANICS 12–3 Isentropic Flow through Nozzles 677 Converging Nozzles 678 Converging–Diverging Nozzles 682 12–4 Shock Waves and Expansion Waves 685 Normal Shocks 686 Oblique Shocks 691 Prandtl–Meyer Expansion Waves 696 12–5 Duct Flow with Heat Transfer and Negligible Friction (Rayleigh Flow) 701 Property Relations for Rayleigh Flow 706 Choked Rayleigh Flow 708 12–6 Adiabatic Duct Flow with Friction (Fanno Flow) 710 Property Relations for Fanno Flow 713 Choked Fanno Flow 716 Application Spotlight: Shock-Wave/ Boundary-Layer Interactions 720 Summary 721 References and Suggested Reading 722 Problems 722 c h a p t e r t h i r t e e n OPEN-CHANNEL FLOW 733 13–1 Classification of Open-Channel Flows 734 Uniform and Varied Flows 734 Laminar and Turbulent Flows in Channels 735 13–2 Froude Number and Wave Speed 737 Speed of Surface Waves 739 13–3 Specific Energy 741 13–4 Conservation of Mass and Energy Equations 744 13–5 Uniform Flow in Channels 745 Critical Uniform Flow 747 Superposition Method for Nonuniform Perimeters 748 13–6 Best Hydraulic Cross Sections 751 Rectangular Channels 753 Trapezoidal Channels 753 13–7 Gradually Varied Flow 755 Liquid Surface Profiles in Open Channels, y(x) 757 Some Representative Surface Profiles 760 Numerical Solution of Surface Profile 762 13–8 Rapidly Varied Flow and the Hydraulic Jump 765 13–9 Flow Control and Measurement 769 Underflow Gates 770 Overflow Gates 772 Application Spotlight: Bridge Scour 779 Summary 780 References and Suggested Reading 781 Problems 781 c h a p t e r f o u r t e e n TURBOMACHINERY 793 14–1 Classifications and Terminology 794 14–2 Pumps 796 Pump Performance Curves and Matching a Pump to a Piping System 797 Pump Cavitation and Net Positive Suction Head 803 Pumps in Series and Parallel 806 Positive-Displacement Pumps 809 Dynamic Pumps 812 Centrifugal Pumps 812 Axial Pumps 822 14–3 Pump Scaling Laws 830 Dimensional Analysis 830 Pump Specific Speed 833 Affinity Laws 835 14–4 Turbines 839 Positive-Displacement Turbines 840 Dynamic Turbines 840 Impulse Turbines 841 Reaction Turbines 843 Gas and Steam Turbines 853 Wind Turbines 853 14–5 Turbine Scaling Laws 861 Dimensionless Turbine Parameters 861 Turbine Specific Speed 864 Application Spotlight: Rotary Fuel Atomizers 867 Summary 868 References and Suggested Reading 869 Problems 869 c h a p t e r f i f t e e n INTRODUCTION TO COMPUTATIONAL FLUID DYNAMICS 885 15–1 Introduction and Fundamentals 886 Motivation 886 Equations of Motion 886 Solution Procedure 887 Additional Equations of Motion 889 cen96534_fm_i_xxvi.indd 12 01/12/16 10:56 am CONTENTS xiii Grid Generation and Grid Independence 889 Boundary Conditions 894 Practice Makes Perfect 899 15–2 Laminar CFD Calculations 899 Pipe Flow Entrance Region at Re = 500 899 Flow around a Circular Cylinder at Re = 150 903 15–3 Turbulent CFD Calculations 908 Flow around a Circular Cylinder at Re = 10,000 911 Flow around a Circular Cylinder at Re = 107 913 Design of the Stator for a Vane-Axial Flow Fan 913 15–4 CFD with Heat Transfer 921 Temperature Rise through a Cross-Flow Heat Exchanger 921 Cooling of an Array of Integrated Circuit Chips 923 15–5 Compressible Flow CFD Calculations 928 Compressible Flow through a Converging–Diverging Nozzle 929 Oblique Shocks over a Wedge 933 CFD Methods for Two-Phase Flows 934 15–6 Open-Channel Flow CFD Calculations 936 Flow over a Bump on the Bottom of a Channel 936 Flow through a Sluice Gate (Hydraulic Jump) 937 Summary 938 Application Spotlight: A Virtual Stomach 939 References and Suggested Reading 940 Problems 940 a p p e n d i x 1 PROPERTY TABLES AND CHARTS (SI UNITS) 947 TABLE A–1 Molar Mass, Gas Constant, and Ideal-Gas Specific Heats of Some Substances 948 TABLE A–2 Boiling and Freezing Point Properties 949 TABLE A–3 Properties of Saturated Water 950 TABLE A–4 Properties of Saturated Refrigerant-134a 951 TABLE A–5 Properties of Saturated Ammonia 952 TABLE A–6 Properties of Saturated Propane 953 TABLE A–7 Properties of Liquids 954 TABLE A–8 Properties of Liquid Metals 955 TABLE A–9 Properties of Air at 1 atm Pressure 956 TABLE A–10 Properties of Gases at 1 atm Pressure 957 TABLE A–11 Properties of the Atmosphere at High Altitude 959 FIGURE A–12 The Moody Chart for the Friction Factor for Fully Developed Flow in Circular Pipes 960 TABLE A–13 One-Dimensional Isentropic Compressible Flow Functions for an Ideal Gas with k = 1.4 961 TABLE A–14 One-Dimensional Normal Shock Functions for an Ideal Gas with k = 1.4 962 TABLE A–15 Rayleigh Flow Functions for an Ideal Gas with k = 1.4 963 TABLE A–16 Fanno Flow Functions for an Ideal Gas with k = 1.4 964 a p p e n d i x 2 PROPERTY TABLES AND CHARTS (ENGLISH UNITS) 965 TABLE A–1E Molar Mass, Gas Constant, and Ideal-Gas Specific Heats of Some Substances 966 TABLE A–2E Boiling and Freezing Point Properties 967 TABLE A–3E Properties of Saturated Water 968 TABLE A–4E Properties of Saturated Refrigerant-134a 969 TABLE A–5E Properties of Saturated Ammonia 970 TABLE A–6E Properties of Saturated Propane 971 TABLE A–7E Properties of Liquids 972 TABLE A–8E Properties of Liquid Metals 973 TABLE A–9E Properties of Air at 1 atm Pressure 974 TABLE A–10E Properties of Gases at 1 atm Pressure 975 TABLE A–11E Properties of the Atmosphere at High Altitude 977 Glossary 979 Index 993 Conversion Factors 1019 Nomenclature 1021 cen96534_fm_i_xxvi.indd 13 06/12/16 11:31 am This page intentionally left blank P r e f a c e B A C K GR OU ND Fluid mechanics is an exciting and fascinating subject with unlimited practi cal applications ranging from microscopic biological systems to automobiles, airplanes, and spacecraft propulsion. Fluid mechanics has also historically been one of the most challenging subjects for undergraduate students because proper analysis of fluid mechanics problems requires not only knowledge of the concepts but also physical intuition and experience. Our hope is that this book, through its careful explanations of concepts and its use of numer ous practical examples, sketches, figures, and photographs, bridges the gap between knowledge and the proper application of that knowledge. Fluid mechanics is a mature subject; the basic equations and approxima tions are well established and can be found in any introductory textbook. Our book is distinguished from other introductory books because we present the subject in a progressive order from simple to more difficult, building each chapter upon foundations laid down in earlier chapters. We provide more dia grams and photographs than other books because fluid mechanics is, by its nature, a highly visual subject. Only by illustrating the concepts discussed, can students fully appreciate the mathematical significance of the material. O B J ECT IVES This book has been written for the first fluid mechanics course for under graduate engineering students. There is sufficient material for a two-course sequence, if desired. We assume that readers will have an adequate back ground in calculus, physics, engineering mechanics, and thermodynamics. The objectives of this text are ⬤ ⬤To present the basic principles and equations of fluid mechanics.
⬤ ⬤To show numerous and diverse real-world engineering examples to give the student the intuition necessary for correct application of fluid mechanics principles in engineering applications.
⬤ ⬤To develop an intuitive understanding of fluid mechanics by emphasiz ing the physics, and reinforcing that understanding through illustrative figures and photographs.
The book contains enough material to allow considerable flexibility in teach ing the course. Aeronautics and aerospace engineers might emphasize poten tial flow, drag and lift, compressible flow, turbomachinery, and CFD, while mechanical or civil engineering instructors might choose to emphasize pipe flows and open-channel flows, respectively. N E W T O T H E FOURTH EDITIO N All the popular features of the previous editions have been retained while new ones have been added. The main body of the text remains largely unchanged. A noticeable change is the addition of a number of exciting new pictures throughout the book.
cen96534_fm_i_xxvi.indd 15 01/12/16 10:56 am xvi FLUID MECHANICS Four new subsections have been added: “Uniform versus Nonuniform Flow” and “Equation Solvers” to Chap. 1, “Flying in Nature” by guest author Azar Eslam Panah of Penn State Berks to Chap. 11, and “CFD Methods for Two-Phase Flows” by guest author Alex Rattner of Penn State to Chap. 15. In Chap. 8, we now highlight the explicit Churchill equation as an alternative to the implicit Colebrook equation. Two new Application Spotlights, have been added: “Smelling Food; the Human Airway” by Rui Ni of Penn State, to Chap. 4, and “Multicolor Par ticle Shadow Velocimetry/Accelerometry” by Michael McPhail and Michael Krane of Penn State to Chap. 8. A large number of the end-of-chapter problems in the text have been mod ified and many problems were replaced by new ones. Also, several of the solved example problems have been replaced.
P HILO SO PHY AND G O AL The Fourth Edition of Fluid Mechanics: Fundamentals and Applications has the same goals and philosophy as the other texts by lead author Yunus Çengel.
⬤ ⬤Communicates directly with tomorrow’s engineers in a simple yet precise manner ⬤ ⬤Leads students toward a clear understanding and firm grasp of the basic principles of fluid mechanics ⬤ ⬤Encourages creative thinking and development of a deeper understand ing and intuitive feel for fluid mechanics ⬤ ⬤Is read by students with interest and enthusiasm rather than merely as a guide to solve homework problems The best way to learn is by practice. Special effort is made throughout the book to reinforce the material that was presented earlier (in each chapter as well as in material from previous chapters). Many of the illustrated example problems and end-of-chapter problems are comprehensive and encourage students to review and revisit concepts and intuitions gained previously. Throughout the book, we show examples generated by computational fluid dynamics (CFD). We also provide an introductory chapter on the subject. Our goal is not to teach the details about numerical algorithms associated with CFD—this is more properly presented in a separate course. Rather, our intent is to introduce undergraduate students to the capabilities and limitations of CFD as an engineering tool. We use CFD solutions in much the same way as experimental results are used from wind tunnel tests (i.e., to reinforce understanding of the physics of fluid flows and to provide quality flow visual izations that help explain fluid behavior). With dozens of CFD end-of-chapter problems posted on the website, instructors have ample opportunity to intro duce the basics of CFD throughout the course.
CO NTENT AND O RG ANIZATIO N This book is organized into 15 chapters beginning with fundamental concepts of fluids, fluid properties, and fluid flows and ending with an introduction to computational fluid dynamics.
⬤ ⬤Chapter 1 provides a basic introduction to fluids, classifications of fluid flow, control volume versus system formulations, dimensions, units, significant digits, and problem-solving techniques.
cen96534_fm_i_xxvi.indd 16 01/12/16 10:56 am xvii PREFACE ⬤ ⬤Chapter 2 is devoted to fluid properties such as density, vapor pressure, specific heats, speed of sound, viscosity, and surface tension.
⬤ ⬤Chapter 3 deals with fluid statics and pressure, including manometers and barometers, hydrostatic forces on submerged surfaces, buoyancy and stability, and fluids in rigid-body motion.
⬤ ⬤Chapter 4 covers topics related to fluid kinematics, such as the differ ences between Lagrangian and Eulerian descriptions of fluid flows, flow patterns, flow visualization, vorticity and rotationality, and the Reynolds transport theorem.
⬤ ⬤Chapter 5 introduces the fundamental conservation laws of mass, momentum, and energy, with emphasis on the proper use of the mass, Bernoulli, and energy equations and the engineering applications of these equations.
⬤ ⬤Chapter 6 applies the Reynolds transport theorem to linear momentum and angular momentum and emphasizes practical engineering applica tions of finite control volume momentum analysis.
⬤ ⬤Chapter 7 reinforces the concept of dimensional homogeneity and intro duces the Buckingham Pi theorem of dimensional analysis, dynamic similarity, and the method of repeating variables—material that is use ful throughout the rest of the book and in many disciplines in science and engineering.
⬤ ⬤Chapter 8 is devoted to flow in pipes and ducts. We discuss the dif ferences between laminar and turbulent flow, friction losses in pipes and ducts, and minor losses in piping networks. We also explain how to properly select a pump or fan to match a piping network. Finally, we discuss various experimental devices that are used to measure flow rate and velocity, and provide a brief introduction to biofluid mechanics.
⬤ ⬤Chapter 9 deals with differential analysis of fluid flow and includes der ivation and application of the continuity equation, the Cauchy equation, and the Navier–Stokes equation. We also introduce the stream function and describe its usefulness in analysis of fluid flows, and we provide a brief introduction to biofluids. Finally, we point out some of the unique aspects of differential analysis related to biofluid mechanics.
⬤ ⬤Chapter 10 discusses several approximations of the Navier–Stokes equa tion and provides example solutions for each approximation, including creeping flow, inviscid flow, irrotational (potential) flow, and boundary layers.
⬤ ⬤Chapter 11 covers forces on living and non-living bodies (drag and lift), explaining the distinction between friction and pressure drag, and providing drag coefficients for many common geometries. This chapter emphasizes the practical application of wind tunnel mea surements coupled with dynamic similarity and dimensional analysis concepts introduced earlier in Chap. 7.
⬤ ⬤Chapter 12 extends fluid flow analysis to compressible flow, where the behavior of gases is greatly affected by the Mach number. In this chapter, the concepts of expansion waves, normal and oblique shock waves, and choked flow are introduced.
⬤ ⬤Chapter 13 deals with open-channel flow and some of the unique fea tures associated with the flow of liquids with a free surface, such as surface waves and hydraulic jumps.
cen96534_fm_i_xxvi.indd 17 01/12/16 10:56 am xviii FLUID MECHANICS ⬤ ⬤Chapter 14 examines turbomachinery in more detail, including pumps, fans, and turbines. An emphasis is placed on how pumps and turbines work, rather than on their detailed design. We also discuss overall pump and turbine design, based on dynamic similarity laws and simplified velocity vector analyses. ⬤ ⬤Chapter 15 describes the fundamental concepts of computational fluid dyamics (CFD) and shows students how to use commercial CFD codes as tools to solve complex fluid mechanics problems. We emphasize the application of CFD rather than the algorithms used in CFD codes. Each chapter contains a wealth of end-of-chapter homework problems. Most of the problems that require calculation use the SI system of units; how ever, about 20 percent use English units. A comprehensive set of appendices is provided, giving the thermodynamic and fluid properties of several mate rials, in addition to air and water, along with some useful plots and tables. Many of the end-of-chapter problems require the use of material properties from the appendices to enhance the realism of the problems.
L EARNING TO O LS EMPHASIS ON PHYSICS A distinctive feature of this book is its emphasis on the physical aspects of the subject matter in addition to mathematical representations and manipulations. The authors believe that the emphasis in undergraduate education should remain on developing a sense of underlying physical mechanisms and a mastery of solving practical problems that an engineer is likely to face in the real world. Developing an intuitive understanding should also make the course a more motivating and worthwhile experi ence for the students.
EFFECTIVE USE OF ASSOCIATION An observant mind should have no difficulty understanding engineering sciences. After all, the principles of engineering sciences are based on our everyday experiences and experimental observations. Therefore, a physi cal, intuitive approach is used throughout this text. Frequently, parallels are drawn between the subject matter and students’ everyday experiences so that they can relate the subject matter to what they already know.
SELF-INSTRUCTING The material in the text is introduced at a level that an average student can follow comfortably. It speaks to students, not over students. In fact, it is self-instructive. Noting that the principles of science are based on experimental observations, most of the derivations in this text are largely based on physical arguments, and thus they are easy to follow and understand.
EXTENSIVE USE OF ARTWORK AND PHOTOGRAPHS Figures are important learning tools that help the students “get the picture,” and the text makes effective use of graphics. It contains more figures, photo graphs, and illustrations than any other book in this category. Figures attract attention and stimulate curiosity and interest. Most of the figures in this text are intended to serve as a means of emphasizing some key concepts that would otherwise go unnoticed; some serve as page summaries.
cen96534_fm_i_xxvi.indd 18 01/12/16 10:56 am xix PREFACE CONSISTENT COLOR SCHEME FOR FIGURES The figures have a consistent color scheme applied for all arrows.
⬤ ⬤Blue: ( ) motion related, like velocity vectors ⬤ ⬤Green: ( ) force and pressure related, and torque ⬤ ⬤Black: ( ) distance related arrows and dimensions ⬤ ⬤Red: ( ) energy related, like heat and work ⬤ ⬤Purple: ( ) acceleration and gravity vectors, vorticity, and miscellaneous NUMEROUS WORKED-OUT EXAMPLES All chapters contain numerous worked-out examples that both clarify the material and illustrate the use of basic principles in a context that helps develop the student’s intuition. An intuitive and systematic approach is used in the solution of all example problems. The solution methodology starts with a statement of the problem, and all objectives are identified. The assumptions and approximations are then stated together with their justifications. Any properties needed to solve the problem are listed separately. Numerical values are used together with numbers to emphasize that without units, numbers are meaningless. The significance of each example’s result is discussed following the solution. This methodical approach is also followed and provided in the solutions to the end-of-chapter problems, available to instructors.
A WEALTH OF REALISTIC END-OF-CHAPTER PROBLEMS The end-of-chapter problems are grouped under specific topics to make problem selection easier for both instructors and students. Within each group of problems are Concept Questions, indicated by “C,” to check the students’ level of understanding of basic concepts. Problems under Funda mentals of Engineering (FE) Exam Problems are designed to help students prepare for the Fundamentals of Engineering exam, as they prepare for their Professional Engineering license. The problems under Review Problems are more comprehensive in nature and are not directly tied to any specific section of a chapter—in some cases they require review of material learned in previous chapters. Problems designated as Design and Essay are intended to encourage students to make engineering judgments, to conduct independent exploration of topics of interest, and to communicate their findings in a professional manner. Problems designated by an “E” are in English units, and SI users can ignore them. Problems with the icon are comprehensive in nature and are intended to be solved with a computer, using appropriate software. Several economics- and safety-related problems are incorporated throughout to enhance cost and safety awareness among engineering students. Answers to selected problems are listed imme diately following the problem for convenience to students.
USE OF COMMON NOTATION The use of different notation for the same quantities in different engineering courses has long been a source of discontent and confusion. A student taking both fluid mechanics and heat transfer, for example, has to use the notation Q for volume flow rate in one course, and for heat transfer in the other. The need to unify notation in engineering education has often been raised, even in some reports of conferences sponsored by the National Science Foundation through cen96534_fm_i_xxvi.indd 19 01/12/16 10:56 am xx FLUID MECHANICS Foundation Coalitions, but little effort has been made to date in this regard. For example, refer to the final report of the Mini-Conference on Energy Stem Innovations, May 28 and 29, 2003, University of Wisconsin. In this text we made a conscious effort to minimize this conflict by adopting the familiar thermodynamic notation V ˙ for volume flow rate, thus reserving the notation Q for heat transfer. Also, we consistently use an overdot to denote time rate. We think that both students and instructors will appreciate this effort to promote a common notation.
A CHOICE OF SI ALONE OR SI/ENGLISH UNITS In recognition of the fact that English units are still widely used in some industries, both SI and English units are used in this text, with an emphasis on SI. The material in this text can be covered using combined SI/English units or SI units alone, depending on the preference of the instructor. The property tables and charts in the appendices are presented in both units, except the ones that involve dimensionless quantities. Problems, tables, and charts in English units are designated by “E” after the number for easy recognition, and they can be ignored easily by the SI users.
COMBINED COVERAGE OF BERNOULLI AND ENERGY EQUATIONS The Bernoulli equation is one of the most frequently used equations in fluid mechanics, but it is also one of the most misused. Therefore, it is important to emphasize the limitations on the use of this idealized equation and to show how to properly account for imperfections and irreversible losses. In Chap. 5, we do this by introducing the energy equation right after the Bernoulli equation and demonstrating how the solutions of many practical engineering problems differ from those obtained using the Bernoulli equa tion. This helps students develop a realistic view of the Bernoulli equation.
A SEPARATE CHAPTER ON CFD Commercial Computational Fluid Dynamics (CFD) codes are widely used in engineering practice in the design and analysis of flow systems, and it has become exceedingly important for engineers to have a solid understanding of the fundamental aspects, capabilities, and limitations of CFD. Recognizing that most undergraduate engineering curriculums do not have room for a full course on CFD, a separate chapter is included here to make up for this defi ciency and to equip students with an adequate background on the strengths and weaknesses of CFD.
APPLICATION SPOTLIGHTS Throughout the book are highlighted examples called Application Spotlights where a real-world application of fluid mechanics is shown. A unique fea ture of these special examples is that they are written by guest authors. The Application Spotlights are designed to show students how fluid mechanics has diverse applications in a wide variety of fields. They also include eye-catching photographs from the guest authors’ research.
cen96534_fm_i_xxvi.indd 20 01/12/16 10:56 am xxi PREFACE GLOSSARY OF FLUID MECHANICS TERMS Throughout the chapters, when an important key term or concept is introduced and defined, it appears in black boldface type. Fundamental fluid mechanics terms and concepts appear in red boldface type, and these fundamental terms also appear in a comprehensive end-of-book glossary developed by Professor Emeritus James Brasseur of The Pennsylvania State University. This unique glossary is an excellent learning and review tool for students as they move forward in their study of fluid mechanics. CONVERSION FACTORS Frequently used conversion factors, physical constants, and properties of air and water at 20°C and atmospheric pressure are listed at the very end of the book for easy reference.
NOMENCLATURE A list of the major symbols, subscripts, and superscripts used in the text is provided near the end of the book for easy reference.
A C KNOWL EDGM EN TS The authors would like to acknowledge with appreciation the numerous and valuable comments, suggestions, constructive criticisms, and praise from the following evaluators and reviewers: Bass Abushakra Milwaukee School of Engineering John G. Cherng University of Michigan—Dearborn Peter Fox Arizona State University Sathya Gangadbaran Embry Riddle Aeronautical University Jonathan Istok Oregon State University Tim Lee McGill University Nagy Nosseir San Diego State University Robert Spall Utah State University We also thank those who were acknowledged in the first, second, and third editions of this book, but are too numerous to mention again here. The authors are particularly grateful to Mehmet Kanoğlu of University of Gaziantep for his valuable contributions, particularly his modifications of end-of-chapter problems, his editing and updating of the solutions man ual, and his critical review of the entire manuscript. We also thank Tahsin Engin of Sakarya University and Suat Canbazoğlu of Inonu University for contributing several end-of-chapter problems, and Mohsen Hassan Vand for reviewing the book and pointing out a number of errors. Finally, special thanks must go to our families, especially our wives, Zehra Çengel and Suzanne Cimbala, for their continued patience, understanding, and support throughout the preparation of this book, which involved many long hours when they had to handle family concerns on their own because their husbands’ faces were glued to a computer screen.
Yunus A. Çengel John M. Cimbala cen96534_fm_i_xxvi.indd 21 01/12/16 10:56 am Required=Results ® McGraw-Hill Connect® Learn Without Limits Connect is a teaching and learning platform that is proven to deliver better results for students and instructors. Connect empowers students by continually adapting to deliver precisely what they need, when they need it, and how they need it, so your class time is more engaging and effective.
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www.mheducation.com cen96534_fm_i_xxvi.indd 23 01/12/16 10:56 am Online Resources for Instructors Online Resources available at www.mhhe.com/cengel Your home page for teaching fluid mechanics, the Fluid Mechanics: Fundamentals and Applications text-specific website is password protected and offers resources for instructors. ■ ■Electronic Solutions Manual—provides PDF files with detailed typed solutions to all text homework problems.
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cen96534_fm_i_xxvi.indd 24 01/12/16 10:56 am F L U I D M E C H A N I C S FUNDAMENTALS AND APPLICATIONS FOURTH EDITION cen96534_fm_i_xxvi.indd 25 01/12/16 10:56 am This page intentionally left blank 1 CHAPTER 1 I NT R O D UC T I O N AN D B AS I C C O N C E P TS I n this introductory chapter, we present the basic concepts commonly used in the analysis of fluid flow. We start this chapter with a discussion of the phases of matter and the numerous ways of classification of fluid flow, such as viscous versus inviscid regions of flow, internal versus exter nal flow, compressible versus incompressible flow, laminar versus turbulent flow, natural versus forced flow, and steady versus unsteady flow. We also discuss the no-slip condition at solid–fluid interfaces and present a brief his tory of the development of fluid mechanics. After presenting the concepts of system and control volume, we review the unit systems that will be used. We then discuss how mathematical mod els for engineering problems are prepared and how to interpret the results obtained from the analysis of such models. This is followed by a presenta tion of an intuitive systematic problem-solving technique that can be used as a model in solving engineering problems. Finally, we discuss accuracy, pre cision, and significant digits in engineering measurements and calculations. OBJECTIVES When you finish reading this chapter, you should be able to ■ ■ Understand the basic concepts of fluid mechanics ■ ■ Recognize the various types of fluid flow problems encountered in practice ■ ■ Model engineering problems and solve them in a systematic manner ■ ■ Have a working knowledge of accuracy, precision, and significant digits, and recognize the importance of dimensional homogeneity in engineering calculations Schlieren image showing the thermal plume produced by Professor Cimbala as he welcomes you to the fascinating world of fluid mechanics.
Courtesy of Michael J. Hargather and John Cimbala.
cen96537_ch01_001-036.indd 1 29/12/16 10:06 am 2 Introduction and basic concepts 1–1 ■ INTRODUCTION Mechanics is the oldest physical science that deals with both stationary and moving bodies under the influence of forces. The branch of mechanics that deals with bodies at rest is called statics, while the branch that deals with bodies in motion under the action of forces is called dynamics. The subcat egory fluid mechanics is defined as the science that deals with the behavior of fluids at rest (fluid statics) or in motion (fluid dynamics), and the interac tion of fluids with solids or other fluids at the boundaries. Fluid mechanics is also referred to as fluid dynamics by considering fluids at rest as a spe cial case of motion with zero velocity (Fig. 1–1). Fluid mechanics itself is also divided into several categories. The study of the motion of fluids that can be approximated as incompressible (such as liquids, especially water, and gases at low speeds) is usually referred to as hydrodynamics. A subcategory of hydrodynamics is hydraulics, which deals with liquid flows in pipes and open channels. Gas dynamics deals with the flow of fluids that undergo significant density changes, such as the flow of gases through nozzles at high speeds. The category aerodynamics deals with the flow of gases (especially air) over bodies such as aircraft, rockets, and automobiles at high or low speeds. Some other specialized categories such as meteorology, oceanography, and hydrology deal with naturally occurring flows.
What Is a Fluid?
You will recall from physics that a substance exists in three primary phases: solid, liquid, and gas. (At very high temperatures, it also exists as plasma.) A substance in the liquid or gas phase is referred to as a fluid. Distinction between a solid and a fluid is made on the basis of the substance’s abil ity to resist an applied shear (or tangential) stress that tends to change its shape. A solid can resist an applied shear stress by deforming, whereas a fluid deforms continuously under the influence of a shear stress, no matter how small. In solids, stress is proportional to strain, but in fluids, stress is proportional to strain rate. When a constant shear force is applied, a solid eventually stops deforming at some fixed strain angle, whereas a fluid never stops deforming and approaches a constant rate of strain. Consider a rectangular rubber block tightly placed between two plates. As the upper plate is pulled with a force F while the lower plate is held fixed, the rubber block deforms, as shown in Fig. 1–2. The angle of deformation α (called the shear strain or angular displacement) increases in proportion to the applied force F. Assuming there is no slip between the rubber and the plates, the upper surface of the rubber is displaced by an amount equal to the displacement of the upper plate while the lower surface remains station ary. In equilibrium, the net force acting on the upper plate in the horizontal direction must be zero, and thus a force equal and opposite to F must be acting on the plate. This opposing force that develops at the plate–rubber interface due to friction is expressed as F = τA, where τ is the shear stress and A is the contact area between the upper plate and the rubber. When the force is removed, the rubber returns to its original position. This phenome non would also be observed with other solids such as a steel block provided that the applied force does not exceed the elastic range. If this experiment were repeated with a fluid (with two large parallel plates placed in a large Contact area, A Shear stress τ = F/A Shear strain, α Force, F α Deformed rubber FIGURE 1–2 Deformation of a rubber block placed between two parallel plates under the influence of a shear force. The shear stress shown is that on the rubber—an equal but opposite shear stress acts on the upper plate.
FIGURE 1–1 Fluid mechanics deals with liquids and gases in motion or at rest. © Goodshoot/Fotosearch RF cen96537_ch01_001-036.indd 2 29/12/16 10:06 am 3 CHAPTER 1 body of water, for example), the fluid layer in contact with the upper plate would move with the plate continuously at the velocity of the plate no mat ter how small the force F. The fluid velocity would decrease with depth because of friction between fluid layers, reaching zero at the lower plate. You will recall from statics that stress is defined as force per unit area and is determined by dividing the force by the area upon which it acts. The normal component of a force acting on a surface per unit area is called the normal stress, and the tangential component of a force acting on a surface per unit area is called shear stress (Fig. 1–3). In a fluid at rest, the normal stress is called pressure. A fluid at rest is at a state of zero shear stress. When the walls are removed or a liquid container is tilted, a shear develops as the liquid moves to re-establish a horizontal free surface. In a liquid, groups of molecules can move relative to each other, but the vol ume remains relatively constant because of the strong cohesive forces between the molecules. As a result, a liquid takes the shape of the container it is in, and it forms a free surface in a larger container in a gravitational field. A gas, on the other hand, expands until it encounters the walls of the container and fills the entire available space. This is because the gas molecules are widely spaced, and the cohesive forces between them are very small. Unlike liquids, a gas in an open container cannot form a free surface (Fig. 1–4). Although solids and fluids are easily distinguished in most cases, this distinc tion is not so clear in some borderline cases. For example, asphalt appears and behaves as a solid since it resists shear stress for short periods of time. When these forces are exerted over extended periods of time, however, the asphalt deforms slowly, behaving as a fluid. Some plastics, lead, and slurry mixtures exhibit similar behavior. Such borderline cases are beyond the scope of this text. The fluids we deal with in this text will be clearly recognizable as fluids. Intermolecular bonds are strongest in solids and weakest in gases. One reason is that molecules in solids are closely packed together, whereas in gases they are separated by relatively large distances (Fig. 1–5). The mole cules in a solid are arranged in a pattern that is repeated throughout. Because of the small distances between molecules in a solid, the attractive forces of molecules on each other are large and keep the molecules at fixed positions. The molecular spacing in the liquid phase is not much different from that of Free surface Liquid Gas FIGURE 1–4 Unlike a liquid, a gas does not form a free surface, and it expands to fill the entire available space.
(a) (b) (c) FIGURE 1–5 The arrangement of atoms in different phases: (a) molecules are at relatively fixed positions in a solid, (b) groups of molecules move about each other in the liquid phase, and (c) individual molecules move about at random in the gas phase.
FIGURE 1–3 The normal stress and shear stress at the surface of a fluid element. For fluids at rest, the shear stress is zero and pressure is the only normal stress.
dFn dFt dF Normal to surface Tangent to surface Force acting on area dA dA Normal stress: 𝜎= dFn dA Shear stress: 𝜏= dFt dA cen96537_ch01_001-036.indd 3 29/12/16 10:07 am 4 Introduction and basic concepts the solid phase, except the molecules are no longer at fixed positions relative to each other and they can rotate and translate freely. In a liquid, the inter molecular forces are weaker relative to solids, but still strong compared with gases. The distances between molecules generally increase slightly as a solid turns liquid, with water being a notable exception. In the gas phase, the molecules are far apart from each other, and molecu lar ordering is nonexistent. Gas molecules move about at random, continu ally colliding with each other and the walls of the container in which they are confined. Particularly at low densities, the intermolecular forces are very small, and collisions are the only mode of interaction between the mole cules. Molecules in the gas phase are at a considerably higher energy level than they are in the liquid or solid phase. Therefore, the gas must release a large amount of its energy before it can condense or freeze. Gas and vapor are often used as synonymous words. The vapor phase of a substance is customarily called a gas when it is above the critical tempera ture. Vapor usually implies that the current phase is not far from a state of condensation. Any practical fluid system consists of a large number of molecules, and the properties of the system naturally depend on the behavior of these molecules. For example, the pressure of a gas in a container is the result of momentum transfer between the molecules and the walls of the container. However, one does not need to know the behavior of the gas molecules to determine the pres sure in the container. It is sufficient to attach a pressure gage to the container (Fig. 1–6). This macroscopic or classical approach does not require a knowl edge of the behavior of individual molecules and provides a direct and easy way to analyze engineering problems. The more elaborate microscopic or sta tistical approach, based on the average behavior of large groups of individual molecules, is rather involved and is used in this text only in a supporting role.
Application Areas of Fluid Mechanics It is important to develop a good understanding of the basic principles of fluid mechanics, since fluid mechanics is widely used both in everyday activities and in the design of modern engineering systems from vacuum cleaners to supersonic aircraft. For example, fluid mechanics plays a vital role in the human body. The heart is constantly pumping blood to all parts of the human body through the arteries and veins, and the lungs are the sites of airflow in alternating directions. All artificial hearts, breathing machines, and dialysis systems are designed using fluid dynamics (Fig. 1–7). An ordinary house is, in some respects, an exhibition hall filled with appli cations of fluid mechanics. The piping systems for water, natural gas, and sewage for an individual house and the entire city are designed primarily on the basis of fluid mechanics. The same is also true for the piping and ducting network of heating and air-conditioning systems. A refrigerator involves tubes through which the refrigerant flows, a compressor that pressurizes the refrig erant, and two heat exchangers where the refrigerant absorbs and rejects heat. Fluid mechanics plays a major role in the design of all these components. Even the operation of ordinary faucets is based on fluid mechanics. We can also see numerous applications of fluid mechanics in an automo bile. All components associated with the transportation of the fuel from the fuel tank to the cylinders—the fuel line, fuel pump, and fuel injectors or Pressure gage FIGURE 1–6 On a microscopic scale, pressure is determined by the interaction of individual gas molecules. However, we can measure the pressure on a macroscopic scale with a pressure gage.
FIGURE 1–7 Fluid dynamics is used extensively in the design of artificial hearts. Shown here is the Penn State Electric Total Artificial Heart.
Courtesy of the Biomedical Photography Lab, Penn State Biomedical Engineering Institute. Used by permission.
cen96537_ch01_001-036.indd 4 29/12/16 10:07 am 5 CHAPTER 1 carburetors—as well as the mixing of the fuel and the air in the cylinders and the purging of combustion gases in exhaust pipes—are analyzed using fluid mechanics. Fluid mechanics is also used in the design of the heating and air-conditioning system, the hydraulic brakes, the power steering, the automatic transmission, the lubrication systems, the cooling system of the engine block including the radiator and the water pump, and even the tires. The sleek streamlined shape of recent model cars is the result of efforts to minimize drag by using extensive analysis of flow over surfaces. On a broader scale, fluid mechanics plays a major part in the design and analysis of aircraft, boats, submarines, rockets, jet engines, wind turbines, biomedical devices, cooling systems for electronic components, and trans portation systems for moving water, crude oil, and natural gas. It is also considered in the design of buildings, bridges, and even billboards to make sure that the structures can withstand wind loading. Numerous natural phe nomena such as the rain cycle, weather patterns, the rise of ground water to the tops of trees, winds, ocean waves, and currents in large water bodies are also governed by the principles of fluid mechanics (Fig. 1–8). FIGURE 1–8 Some application areas of fluid mechanics.
Cars © Ingram Publishing RF Power plants U.S. Nuclear Regulatory Commission (NRC) Human body © Jose Luis Pelaez Inc/Blend Images LLC RF Piping and plumbing systems Photo by John M. Cimbala Wind turbines © Mlenny Photography/Getty Images RF Industrial applications © 123RF Aircraft and spacecraft © Purestock/SuperStock/RF Natural flows and weather © Jochen Schlenker/Getty Images RF Boats © Doug Menuez/Getty Images RF cen96537_ch01_001-036.indd 5 29/12/16 10:07 am 6 Introduction and basic concepts 1–2 ■ A BRIEF HISTORY OF FLUID MECHANICS1 One of the first engineering problems humankind faced as cities were devel oped was the supply of water for domestic use and irrigation of crops. Our urban lifestyles can be retained only with abundant water, and it is clear from archeology that every successful civilization of prehistory invested in the construction and maintenance of water systems. The Roman aqueducts, some of which are still in use, are the best known examples. However, per haps the most impressive engineering from a technical viewpoint was done at the Hellenistic city of Pergamon in present-day Turkey. There, from 283 to 133 bc, they built a series of pressurized lead and clay pipelines (Fig. 1–9), up to 45 km long that operated at pressures exceeding 1.7 MPa (180 m of head). Unfortunately, the names of almost all these early builders are lost to history. The earliest recognized contribution to fluid mechanics theory was made by the Greek mathematician Archimedes (285–212 bc). He formulated and applied the buoyancy principle in history’s first nondestructive test to deter mine the gold content of the crown of King Hiero II. The Romans built great aqueducts and educated many conquered people on the benefits of clean water, but overall had a poor understanding of fluids theory. (Perhaps they shouldn’t have killed Archimedes when they sacked Syracuse.) During the Middle Ages, the application of fluid machinery slowly but steadily expanded. Elegant piston pumps were developed for dewatering mines, and the watermill and windmill were perfected to grind grain, forge metal, and for other tasks. For the first time in recorded human history, sig nificant work was being done without the power of a muscle supplied by a person or animal, and these inventions are generally credited with enabling the later industrial revolution. Again the creators of most of the progress are unknown, but the devices themselves were well documented by several technical writers such as Georgius Agricola (Fig. 1–10). The Renaissance brought continued development of fluid systems and machines, but more importantly, the scientific method was perfected and adopted throughout Europe. Simon Stevin (1548–1617), Galileo Galilei (1564–1642), Edme Mariotte (1620–1684), and Evangelista Torricelli (1608–1647) were among the first to apply the method to fluids as they investigated hydrostatic pressure distributions and vacuums. That work was integrated and refined by the brilliant mathematician and philosopher, Blaise Pascal (1623–1662). The Italian monk, Benedetto Castelli (1577– 1644) was the first person to publish a statement of the continuity principle for fluids. Besides formulating his equations of motion for solids, Sir Isaac Newton (1643–1727) applied his laws to fluids and explored fluid inertia and resistance, free jets, and viscosity. That effort was built upon by Daniel Bernoulli (1700–1782), a Swiss, and his associate Leonard Euler (1707– 1783). Together, their work defined the energy and momentum equations. Bernoulli’s 1738 classic treatise Hydrodynamica may be considered the first fluid mechanics text. Finally, Jean d’Alembert (1717–1789) developed the idea of velocity and acceleration components, a differential expression of 1 This section is contributed by Professor Glenn Brown of Oklahoma State University.
FIGURE 1–9 Segment of Pergamon pipeline. Each clay pipe section was 13 to 18 cm in diameter. Courtesy of Gunther Garbrecht. Used by permission.
FIGURE 1–10 A mine hoist powered by a reversible water wheel. © Universal History Archive/Getty Images cen96537_ch01_001-036.indd 6 29/12/16 10:07 am 7 CHAPTER 1 continuity, and his “paradox” of zero resistance to steady uniform motion over a body. The development of fluid mechanics theory through the end of the eigh teenth century had little impact on engineering since fluid properties and parameters were poorly quantified, and most theories were abstractions that could not be quantified for design purposes. That was to change with the development of the French school of engineering led by Riche de Prony (1755–1839). Prony (still known for his brake to measure shaft power) and his associates in Paris at the École Polytechnique and the École des Ponts et Chaussées were the first to integrate calculus and scientific theory into the engineering curriculum, which became the model for the rest of the world. (So now you know whom to blame for your painful freshman year.) Antonie Chezy (1718–1798), Louis Navier (1785–1836), Gaspard Coriolis (1792–1843), Henry Darcy (1803–1858), and many other contributors to fluid engineering and theory were students and/or instructors at the schools. By the mid nineteenth century, fundamental advances were coming on several fronts. The physician Jean Poiseuille (1799–1869) had accurately measured flow in capillary tubes for multiple fluids, while in Germany Gotthilf Hagen (1797–1884) had differentiated between laminar and turbu lent flow in pipes. In England, Lord Osborne Reynolds (1842–1912) con tinued that work (Fig. 1–11) and developed the dimensionless number that bears his name. Similarly, in parallel to the early work of Navier, George Stokes (1819–1903) completed the general equation of fluid motion (with friction) that takes their names. William Froude (1810–1879) almost single-handedly developed the procedures and proved the value of physical model testing. American expertise had become equal to the Europeans as demon strated by James Francis’ (1815–1892) and Lester Pelton’s (1829–1908) pio neering work in turbines and Clemens Herschel’s (1842–1930) invention of the Venturi meter. In addition to Reynolds and Stokes, many notable contributions were made to fluid theory in the late nineteenth century by Irish and English scientists, including William Thomson, Lord Kelvin (1824–1907), William Strutt, Lord Rayleigh (1842–1919), and Sir Horace Lamb (1849–1934). These individu als investigated a large number of problems, including dimensional analysis, irrotational flow, vortex motion, cavitation, and waves. In a broader sense, FIGURE 1–11 Osborne Reynolds’ original apparatus for demonstrating the onset of turbu lence in pipes, being operated by John Lienhard at the University of Manchester in 1975. Courtesy of John Lienhard, University of Houston. Used by Permission.
cen96537_ch01_001-036.indd 7 29/12/16 10:07 am 8 Introduction and basic concepts their work also explored the links between fluid mechanics, thermodynam ics, and heat transfer. The dawn of the twentieth century brought two monumental developments. First, in 1903, the self-taught Wright brothers (Wilbur, 1867–1912; Orville, 1871–1948) invented the airplane through application of theory and deter mined experimentation. Their primitive invention was complete and contained all the major aspects of modern aircraft (Fig. 1–12). The Navier–Stokes equa tions were of little use up to this time because they were too difficult to solve. In a pioneering paper in 1904, the German Ludwig Prandtl (1875–1953) showed that fluid flows can be divided into a layer near the walls, the bound ary layer, where the friction effects are significant, and an outer layer where such effects are negligible and the simplified Euler and Bernoulli equations are applicable. His students, Theodor von Kármán (1881–1963), Paul Blasius (1883–1970), Johann Nikuradse (1894–1979), and others, built on that theory in both hydraulic and aerodynamic applications. (During World War II, both sides benefited from the theory as Prandtl remained in Germany while his best student, the Hungarian-born von Kármán, worked in America.) The mid twentieth century could be considered a golden age of fluid mechanics applications. Existing theories were adequate for the tasks at hand, and fluid properties and parameters were well defined. These sup ported a huge expansion of the aeronautical, chemical, industrial, and water resources sectors; each of which pushed fluid mechanics in new directions. Fluid mechanics research and work in the late twentieth century were dominated by the development of the digital computer in America. The ability to solve large complex problems, such as global climate mod eling or the optimization of a turbine blade, has provided a benefit to our society that the eighteenth-century developers of fluid mechanics could never have imagined (Fig. 1–13). The principles presented in the following pages have been applied to flows ranging from a moment at the micro scopic scale to 50 years of simulation for an entire river basin. It is truly mind-boggling. Where will fluid mechanics go in the twenty-first century and beyond? Frankly, even a limited extrapolation beyond the present would be sheer folly. However, if history tells us anything, it is that engineers will be applying what they know to benefit society, researching what they don’t know, and having a great time in the process.
1–3 ■ THE NO-SLIP CONDITION Fluid flow is often confined by solid surfaces, and it is important to under stand how the presence of solid surfaces affects fluid flow. We know that water in a river cannot flow through large rocks, and must go around them. That is, the water velocity normal to the rock surface must be zero, and water approaching the surface normally comes to a complete stop at the sur face. What is not as obvious is that water approaching the rock at any angle also comes to a complete stop at the rock surface, and thus the tangential velocity of water at the surface is also zero. Consider the flow of a fluid in a stationary pipe or over a solid surface that is nonporous (i.e., impermeable to the fluid). All experimental observa tions indicate that a fluid in motion comes to a complete stop at the surface FIGURE 1–12 The Wright brothers take flight at Kitty Hawk.
Courtesy Library of Congress Prints & Photographs Division [LC-DIG-ppprs-00626].
FIGURE 1–13 Old and new wind turbine technologies north of Woodward, OK. The modern turbines have up to 8 MW capacities.
Photo courtesy of the Oklahoma Wind Power Initiative. Used by permission.
cen96537_ch01_001-036.indd 8 29/12/16 10:07 am 9 CHAPTER 1 and assumes a zero velocity relative to the surface. That is, a fluid in direct contact with a solid “sticks” to the surface, and there is no slip. This is known as the no-slip condition. The fluid property responsible for the no-slip condition and the development of the boundary layer is viscosity and is discussed in Chap. 2. The photograph in Fig. 1–14 clearly shows the evolution of a velocity gra dient as a result of the fluid sticking to the surface of a blunt nose. The layer that sticks to the surface slows the adjacent fluid layer because of viscous forces between the fluid layers, which slows the next layer, and so on. A consequence of the no-slip condition is that all velocity profiles must have zero values with respect to the surface at the points of contact between a fluid and a solid surface (Fig. 1–15). Therefore, the no-slip condition is responsible for the development of the velocity profile. The flow region adjacent to the wall in which the viscous effects (and thus the velocity gra dients) are significant is called the boundary layer. Another consequence of the no-slip condition is the surface drag, or skin friction drag, which is the force a fluid exerts on a surface in the flow direction. When a fluid is forced to flow over a curved surface, such as the back side of a cylinder, the boundary layer may no longer remain attached to the sur face and separates from the surface—a process called flow separation (Fig. 1–16). We emphasize that the no-slip condition applies everywhere along the surface, even downstream of the separation point. Flow separation is discussed in greater detail in Chap. 9. A phenomenon similar to the no-slip condition occurs in heat transfer. When two bodies at different temperatures are brought into contact, heat transfer occurs such that both bodies assume the same temperature at the points of contact. Therefore, a fluid and a solid surface have the same tem perature at the points of contact. This is known as no-temperature-jump condition.
1–4 ■ CLASSIFICATION OF FLUID FLOWS Earlier we defined fluid mechanics as the science that deals with the behav ior of fluids at rest or in motion, and the interaction of fluids with solids or other fluids at the boundaries. There is a wide variety of fluid flow prob lems encountered in practice, and it is usually convenient to classify them on the basis of some common characteristics to make it feasible to study them in groups. There are many ways to classify fluid flow problems, and here we present some general categories.
FIGURE 1–14 The development of a velocity profile due to the no-slip condition as a fluid flows over a blunt nose.
“Hunter Rouse: Laminar and Turbulence Flow Film.” Copyright IIHR-Hydroscience & Engineer ing. The University of Iowa. Used by permission.
Relative velocities of fluid layers Uniform approach velocity, V Zero velocity at the surface Plate FIGURE 1–15 A fluid flowing over a stationary surface comes to a complete stop at the surface because of the no-slip condition.
Separation point FIGURE 1–16 Flow separation during flow over a curved surface.
From Head, Malcolm R. 1982 in Flow Visualization II, W. Merzkirch. Ed., 399–403, Washington: Hemisphere.
cen96537_ch01_001-036.indd 9 29/12/16 10:07 am 10 Introduction and basic concepts Viscous versus Inviscid Regions of Flow When two fluid layers move relative to each other, a friction force devel ops between them and the slower layer tries to slow down the faster layer. This internal resistance to flow is quantified by the fluid property viscosity, which is a measure of internal stickiness of the fluid. Viscosity is caused by cohesive forces between the molecules in liquids and by molecular colli sions in gases. There is no fluid with zero viscosity, and thus all fluid flows involve viscous effects to some degree. Flows in which the frictional effects are significant are called viscous flows. However, in many flows of practi cal interest, there are regions (typically regions not close to solid surfaces) where viscous forces are negligibly small compared to inertial or pressure forces. Neglecting the viscous terms in such inviscid flow regions greatly simplifies the analysis without much loss in accuracy. The development of viscous and inviscid regions of flow as a result of inserting a flat plate parallel into a fluid stream of uniform velocity is shown in Fig. 1–17. The fluid sticks to the plate on both sides because of the no-slip condition, and the thin boundary layer in which the viscous effects are signifi cant near the plate surface is the viscous flow region. The region of flow on both sides away from the plate and largely unaffected by the presence of the plate is the inviscid flow region. Internal versus External Flow A fluid flow is classified as being internal or external, depending on whether the fluid flows in a confined space or over a surface. The flow of an unbounded fluid over a surface such as a plate, a wire, or a pipe is external flow. The flow in a pipe or duct is internal flow if the fluid is bounded by solid surfaces. Water flow in a pipe, for example, is internal flow, and airflow over a ball or over an exposed pipe during a windy day is external flow (Fig. 1–18). The flow of liquids in a duct is called open-channel flow if the duct is only partially filled with the liquid and there is a free surface. The flows of water in rivers and irrigation ditches are examples of such flows. Internal flows are dominated by the influence of viscosity throughout the flow field. In external flows the viscous effects are limited to boundary layers near solid surfaces and to wake regions downstream of bodies. Compressible versus Incompressible Flow A flow is classified as being compressible or incompressible, depending on the level of variation of density during flow. Incompressibility is an approximation, in which the flow is said to be incompressible if the density remains nearly constant throughout. Therefore, the volume of every portion of fluid remains unchanged over the course of its motion when the flow is approximated as incompressible. The densities of liquids are essentially constant, and thus the flow of liq uids is typically incompressible. Therefore, liquids are usually referred to as incompressible substances. A pressure of 210 atm, for example, causes the density of liquid water at 1 atm to change by just 1 percent. Gases, on the other hand, are highly compressible. A pressure change of just 0.01 atm, for example, causes a change of 1 percent in the density of atmospheric air.
FIGURE 1–18 External flow over a tennis ball, and the turbulent wake region behind.
Courtesy of NASA and Cislunar Aerospace, Inc.
Inviscid flow region Viscous flow region Inviscid flow region FIGURE 1–17 The flow of an originally uniform fluid stream over a flat plate, and the regions of viscous flow (next to the plate on both sides) and inviscid flow (away from the plate).
Fundamentals of Boundry Layers, National Committee for Fluid Mechanics Films, © Education Development Center.
cen96537_ch01_001-036.indd 10 29/12/16 10:07 am 11 CHAPTER 1 When analyzing rockets, spacecraft, and other systems that involve high-speed gas flows (Fig. 1–19), the flow speed is often expressed in terms of the dimensionless Mach number defined as Ma = V c = Speed of flow Speed of sound where c is the speed of sound whose value is 346 m/s in air at room tempera ture at sea level. A flow is called sonic when Ma = 1, subsonic when Ma < 1, supersonic when Ma > 1, and hypersonic when Ma ≫ 1. Dimensionless parameters are discussed in detail in Chap. 7. Compressible flow is discussed in detail in Chap. 12. Liquid flows are incompressible to a high level of accuracy, but the level of variation of density in gas flows and the consequent level of approxi mation made when modeling gas flows as incompressible depends on the Mach number. Gas flows can often be approximated as incompressible if the density changes are under about 5 percent, which is usually the case when Ma < 0.3. Therefore, the compressibility effects of air at room tempera ture can be neglected at speeds under about 100 m/s. Compressibility effects should never be neglected for supersonic flows, however, since compress ible flow phenomena like shock waves occur (Fig. 1–19). Small density changes of liquids corresponding to large pressure changes can still have important consequences. The irritating “water hammer” in a water pipe, for example, is caused by the vibrations of the pipe generated by the reflection of pressure waves following the sudden closing of the valves.
Laminar versus Turbulent Flow Some flows are smooth and orderly while others are rather chaotic. The highly ordered fluid motion characterized by smooth layers of fluid is called laminar. The word laminar comes from the movement of adjacent fluid particles together in “laminae.” The flow of high-viscosity fluids such as oils at low velocities is typically laminar. The highly disordered fluid motion that typically occurs at high velocities and is characterized by velocity fluc tuations is called turbulent (Fig. 1–20). The flow of low-viscosity fluids such as air at high velocities is typically turbulent. A flow that alternates between being laminar and turbulent is called transitional. The experiments conducted by Osborne Reynolds in the 1880s resulted in the establishment of the dimensionless Reynolds number, Re, as the key parameter for the determination of the flow regime in pipes (Chap. 8). Natural (or Unforced) versus Forced Flow A fluid flow is said to be natural or forced, depending on how the fluid motion is initiated. In forced flow, a fluid is forced to flow over a surface or in a pipe by external means such as a pump or a fan. In natural flows, fluid motion is due to natural means such as the buoyancy effect, which manifests itself as the rise of warmer (and thus lighter) fluid and the fall of cooler (and thus denser) fluid (Fig. 1–21). In solar hot-water systems, for example, the thermosiphoning effect is commonly used to replace pumps by placing the water tank sufficiently above the solar collectors.
Laminar Transitional Turbulent FIGURE 1–20 Laminar, transitional, and turbulent flows over a flat plate.
Courtesy of ONERA. Photo by Werlé.
FIGURE 1–19 Schlieren image of the spherical shock wave produced by a bursting ballon at the Penn State Gas Dynamics Lab. Several secondary shocks are seen in the air surrounding the ballon.
© G.S. Settles, Gas Dynamics Lab, Penn State University. Used with permission.
cen96537_ch01_001-036.indd 11 29/12/16 10:07 am 12 Introduction and basic concepts Steady versus Unsteady Flow The terms steady and uniform are used frequently in engineering, and thus it is important to have a clear understanding of their meanings. The term steady implies no change of properties, velocity, temperature, etc., at a point with time. The opposite of steady is unsteady. The term uniform implies no change with location over a specified region. These meanings are consistent with their everyday use (steady girlfriend, uniform distribution, etc.). The terms unsteady and transient are often used interchangeably, but these terms are not synonyms. In fluid mechanics, unsteady is the most general term that applies to any flow that is not steady, but transient is typically used for developing flows. When a rocket engine is fired up, for example, there are tran sient effects (the pressure builds up inside the rocket engine, the flow accelerates, etc.) until the engine settles down and operates steadily. The term periodic refers to the kind of unsteady flow in which the flow oscillates about a steady mean. Many devices such as turbines, compressors, boilers, condensers, and heat exchangers operate for long periods of time under the same conditions, and they are classified as steady-flow devices. (Note that the flow field near the rotating blades of a turbomachine is of course unsteady, but we consider the overall flow field rather than the details at some localities when we classify devices.) During steady flow, the fluid properties can change from point to point within a device, but at any fixed point they remain constant. Therefore, the volume, the mass, and the total energy content of a steady-flow device or flow section remain constant in steady operation. A simple analogy is shown in Fig. 1–22. Steady-flow conditions can be closely approximated by devices that are intended for continuous operation such as turbines, pumps, boilers, condens ers, and heat exchangers of power plants or refrigeration systems. Some cyclic devices, such as reciprocating engines or compressors, do not sat isfy the steady-flow conditions since the flow at the inlets and the exits is pulsating and not steady. However, the fluid properties vary with time in a FIGURE 1–21 In this schlieren image of a girl in a swimming suit, the rise of lighter, warmer air adjacent to her body indicates that humans and warm-blooded animals are surrounded by thermal plumes of rising warm air.
© G.S. Settles, Gas Dynamics Lab, Penn State University. Used with permission.
FIGURE 1–22 Comparison of (a) instantaneous snapshot of an unsteady flow, and (b) long exposure picture of the same flow.
Photos by Eric G. Paterson. Used by permission.
(a) (b) cen96537_ch01_001-036.indd 12 29/12/16 10:07 am 13 CHAPTER 1 periodic manner, and the flow through these devices can still be analyzed as a steady-flow process by using time-averaged values for the properties. Some fascinating visualizations of fluid flow are provided in the book An Album of Fluid Motion by Milton Van Dyke (1982). A nice illustration of an unsteady-flow field is shown in Fig. 1–23, taken from Van Dyke’s book. Figure 1–23a is an instantaneous snapshot from a high-speed motion picture; it reveals large, alternating, swirling, turbulent eddies that are shed into the periodi cally oscillating wake from the blunt base of the object. The unsteady wake pro duces waves that move upstream alternately over the top and bottom surfaces of the airfoil in an unsteady fashion. Figure 1–23b shows the same flow field, but the film is exposed for a longer time so that the image is time averaged over 12 cycles. The resulting time-averaged flow field appears “steady” since the details of the unsteady oscillations have been lost in the long exposure. One of the most important jobs of an engineer is to determine whether it is sufficient to study only the time-averaged “steady” flow features of a problem, or whether a more detailed study of the unsteady features is required. If the engi neer were interested only in the overall properties of the flow field (such as the time-averaged drag coefficient, the mean velocity, and pressure fields), a time-averaged description like that of Fig. 1–23b, time-averaged experimental mea surements, or an analytical or numerical calculation of the time-averaged flow field would be sufficient. However, if the engineer were interested in details about the unsteady-flow field, such as flow-induced vibrations, unsteady pres sure fluctuations, or the sound waves emitted from the turbulent eddies or the shock waves, a time-averaged description of the flow field would be insufficient. Most of the analytical and computational examples provided in this text book deal with steady or time-averaged flows, although we occasionally point out some relevant unsteady-flow features as well when appropriate.
One-, Two-, and Three-Dimensional Flows A flow field is best characterized by its velocity distribution, and thus a flow is said to be one-, two-, or three-dimensional if the flow velocity varies in one, two, or three primary dimensions, respectively. A typical fluid flow involves a three-dimensional geometry, and the velocity may vary in all three dimensions, rendering the flow three-dimensional [V ›(x, y, z) in rectangular or V ›(r, θ, z) in cylindrical coordinates]. However, the variation of velocity in certain directions can be small relative to the variation in other directions and can be ignored with negligible error. In such cases, the flow can be modeled conveniently as being one- or two-dimensional, which is easier to analyze. Consider steady flow of a fluid entering from a large tank into a circular pipe. The fluid velocity everywhere on the pipe surface is zero because of the no-slip condition, and the flow is two-dimensional in the entrance region of the pipe since the velocity changes in both the r- and z-directions, but not in the θ-direction. The velocity profile develops fully and remains unchanged after some distance from the inlet (about 10 pipe diameters in turbulent flow, and typically farther than that in laminar pipe flow, as in Fig. 1–24), and the flow in this region is said to be fully developed. The fully developed flow in a circular pipe is one-dimensional since the velocity varies in the radial r-direction but not in the angular θ- or axial z-directions, as shown in Fig. 1–24. That is, the velocity profile is the same at any axial z-location, and it is symmetric about the axis of the pipe.
(a) (b) FIGURE 1–23 Oscillating wake of a blunt-based airfoil at Mach number 0.6. Photo (a) is an instantaneous image, while photo (b) is a long-exposure (time-averaged) image.
(a) Dyment, A., Flodrops, J. P. & Gryson, P. 1982 in Flow Visualization II, W. Merzkirch, ed., 331–336. Washington: Hemisphere. Used by permission of Arthur Dyment.
(b) Dyment, A. & Gryson, P. 1978 in Inst. Mèc. Fluides Lille, No. 78-5. Used by permission of Arthur Dyment.
cen96537_ch01_001-036.indd 13 29/12/16 10:07 am 14 Introduction and basic concepts Note that the dimensionality of the flow also depends on the choice of coordinate system and its orientation. The pipe flow discussed, for example, is one-dimensional in cylindrical coordinates, but two-dimensional in Cartesian coordinates—illustrating the importance of choosing the most appropriate coordinate system. Also note that even in this simple flow, the velocity cannot be uniform across the cross section of the pipe because of the no-slip condi tion. However, at a well-rounded entrance to the pipe, the velocity profile may be approximated as being nearly uniform across the pipe, since the velocity is nearly constant at all radii except very close to the pipe wall. A flow may be approximated as two-dimensional when the aspect ratio is large and the flow does not change appreciably along the longer dimension. For example, the flow of air over a car antenna can be considered two-dimensional except near its ends since the antenna’s length is much greater than its diam eter, and the airflow hitting the antenna is fairly uniform (Fig. 1–25).
FIGURE 1–25 Flow over a car antenna is approximately two-dimensional except near the top and bottom of the antenna.
Axis of symmetry r z θ FIGURE 1–26 Axisymmetric flow over a bullet.
z r Developing velocity profile, V(r, z) Fully developed velocity profile, V(r) FIGURE 1–24 The development of the velocity profile in a circular pipe. V = V(r, z) and thus the flow is two-dimensional in the entrance region, and becomes one-dimensional downstream when the velocity profile fully develops and remains unchanged in the flow direction, V = V(r).
EXAMPLE 1–1 Axisymmetric Flow over a Bullet Consider a bullet piercing through calm air during a short time interval in which the bullet’s speed is nearly constant. Determine if the time-averaged airflow over the bullet during its flight is one-, two-, or three-dimensional (Fig. 1–26).
SOLUTION It is to be determined whether airflow over a bullet is one-, two-, or three-dimensional.
Assumptions There are no significant winds and the bullet is not spinning.
Analysis The bullet possesses an axis of symmetry and is therefore an axisym metric body. The airflow upstream of the bullet is parallel to this axis, and we expect the time-averaged airflow to be rotationally symmetric about the axis—such flows are said to be axisymmetric. The velocity in this case varies with axial dis tance z and radial distance r, but not with angle θ. Therefore, the time-averaged airflow over the bullet is two-dimensional.
Discussion While the time-averaged airflow is axisymmetric, the instantaneous airflow is not, as illustrated in Fig. 1–23. In Cartesian coordinates, the flow would be three-dimensional. Finally, many bullets also spin.
Uniform versus Nonuniform Flow Uniform flow implies that all fluid properties, such as velocity, pressure, tem perature, etc., do not vary with position. A wind tunnel test section, for exam ple, is designed such that the air flow is as uniform as possible. Even then, however, the flow does not remain uniform as we approach the wind tun nel walls, due to the no-slip condition and the presence of a boundary layer, cen96537_ch01_001-036.indd 14 29/12/16 10:07 am 15 CHAPTER 1 as mentioned previously. The flow just downstream of a well-rounded pipe entrance (Fig. 1–24) is nearly uniform, again except for a very thin bound ary layer near the wall. In engineering practice, it is common to approximate the flow in ducts and pipes and at inlets and outlets as uniform, even when it is not, for simplicity in calculations. For example, the fully developed pipe flow velocity profile of Fig. 1–24 is certainly not uniform, but for calculation purposes we sometimes approximate it as the uniform profile at the far left of the pipe, which has the same average velocity. Although this makes the calculations easier, it also introduces some errors that require correction fac tors; these are discussed in Chaps. 5 and 6 for kinetic energy and momentum, respectively.
1–5 ■ SYSTEM AND CONTROL VOLUME A system is defined as a quantity of matter or a region in space chosen for study. The mass or region outside the system is called the surroundings. The real or imaginary surface that separates the system from its surround ings is called the boundary (Fig. 1–27). The boundary of a system can be fixed or movable. Note that the boundary is the contact surface shared by both the system and the surroundings. Mathematically speaking, the bound ary has zero thickness, and thus it can neither contain any mass nor occupy any volume in space. Systems may be considered to be closed or open, depending on whether a fixed mass or a volume in space is chosen for study. A closed system (also known as a control mass or simply a system when the context makes it clear) consists of a fixed amount of mass, and no mass can cross its boundary. But energy, in the form of heat or work, can cross the boundary, and the volume of a closed system does not have to be fixed. If, as a special case, even energy is not allowed to cross the boundary, that system is called an isolated system. Consider the piston–cylinder device shown in Fig. 1–28. Let us say that we would like to find out what happens to the enclosed gas when it is heated. Since we are focusing our attention on the gas, it is our system. The inner surfaces of the piston and the cylinder form the boundary, and since no mass is crossing this boundary, it is a closed system. Notice that energy may cross the boundary, and part of the boundary (the inner surface of the piston, in this case) may move. Everything outside the gas, including the piston and the cylinder, is the surroundings. An open system, or a control volume, as it is often called, is a selected region in space. It usually encloses a device that involves mass flow such as a compressor, turbine, or nozzle. Flow through these devices is best stud ied by selecting the region within the device as the control volume. Both mass and energy can cross the boundary (the control surface) of a control volume. A large number of engineering problems involve mass flow in and out of an open system and, therefore, are modeled as control volumes. A water heater, a car radiator, a turbine, and a compressor all involve mass flow and should be analyzed as control volumes (open systems) instead of as control masses (closed systems). In general, any arbitrary region in space can be selected as a control volume. There are no concrete rules for the SURROUNDINGS BOUNDARY SYSTEM FIGURE 1–27 System, surroundings, and boundary.
GAS 2 kg 1.5 m3 GAS 2 kg 1 m3 Moving boundary Fixed boundary FIGURE 1–28 A closed system with a moving boundary.
cen96537_ch01_001-036.indd 15 29/12/16 10:07 am 16 Introduction and basic concepts selection of control volumes, but a wise choice certainly makes the analysis much easier. If we were to analyze the flow of air through a nozzle, for example, a good choice for the control volume would be the region within the nozzle, or perhaps surrounding the entire nozzle. A control volume can be fixed in size and shape, as in the case of a noz zle, or it may involve a moving boundary, as shown in Fig. 1–29. Most con trol volumes, however, have fixed boundaries and thus do not involve any moving boundaries. A control volume may also involve heat and work inter actions just as a closed system, in addition to mass interaction.
1–6 ■ IMPORTANCE OF DIMENSIONS AND UNITS Any physical quantity can be characterized by dimensions. The magnitudes assigned to the dimensions are called units. Some basic dimensions such as mass m, length L, time t, and temperature T are selected as primary or fundamental dimensions, while others such as velocity V, energy E, and volume V are expressed in terms of the primary dimensions and are called secondary dimensions, or derived dimensions. A number of unit systems have been developed over the years. Despite strong efforts in the scientific and engineering community to unify the world with a single unit system, two sets of units are still in common use today: the English system, which is also known as the United States Cus tomary System (USCS), and the metric SI (from Le Système International d’ Unités), which is also known as the International System. The SI is a simple and logical system based on a decimal relationship between the vari ous units, and it is being used for scientific and engineering work in most of the industrialized nations, including England. The English system, however, has no apparent systematic numerical base, and various units in this system are related to each other rather arbitrarily (12 in = 1 ft, 1 mile = 5280 ft, 4 qt = 1 gal, etc.), which makes it confusing and difficult to learn. The United States is the only industrialized country that has not yet fully con verted to the metric system. The systematic efforts to develop a universally acceptable system of units dates back to 1790 when the French National Assembly charged the French Academy of Sciences to come up with such a unit system. An early version of the metric system was soon developed in France, but it did not find universal acceptance until 1875 when The Metric Convention Treaty was prepared and signed by 17 nations, including the United States. In this international treaty, meter and gram were established as the metric units for length and mass, respectively, and a General Conference of Weights and Measures (CGPM) was established that was to meet every six years. In 1960, the CGPM produced the SI, which was based on six fundamental quantities, and their units were adopted in 1954 at the Tenth General Con ference of Weights and Measures: meter (m) for length, kilogram (kg) for mass, second (s) for time, ampere (A) for electric current, degree Kelvin (°K) for temperature, and candela (cd) for luminous intensity (amount of light). In 1971, the CGPM added a seventh fundamental quantity and unit: mole (mol) for the amount of matter.
FIGURE 1–29 A control volume may involve fixed, moving, real, and imaginary boundaries.
CV Moving boundary Fixed boundary Real boundary (b) A control volume (CV) with fixed and moving boundaries as well as real and imaginary boundaries (a) A control volume (CV) with real and imaginary boundaries Imaginary boundary CV (a nozzle) cen96537_ch01_001-036.indd 16 29/12/16 10:07 am 17 CHAPTER 1 Based on the notational scheme introduced in 1967, the degree sym bol was officially dropped from the absolute temperature unit, and all unit names were to be written without capitalization even if they were derived from proper names (Table 1–1). However, the abbreviation of a unit was to be capitalized if the unit was derived from a proper name. For example, the SI unit of force, which is named after Sir Isaac Newton (1647–1723), is newton (not Newton), and it is abbreviated as N. Also, the full name of a unit may be pluralized, but its abbreviation cannot. For example, the length of an object can be 5 m or 5 meters, not 5 ms or 5 meter. Finally, no period is to be used in unit abbreviations unless they appear at the end of a sentence. For example, the proper abbreviation of meter is m (not m.). The recent move toward the metric system in the United States seems to have started in 1968 when Congress, in response to what was happening in the rest of the world, passed a Metric Study Act. Congress continued to promote a voluntary switch to the metric system by passing the Metric Conversion Act in 1975. A trade bill passed by Congress in 1988 set a September 1992 deadline for all federal agencies to convert to the metric system. However, the deadlines were relaxed later with no clear plans for the future. As pointed out, the SI is based on a decimal relationship between units. The prefixes used to express the multiples of the various units are listed in Table 1–2. They are standard for all units, and the student is encouraged to memorize some of them because of their widespread use (Fig. 1–30).
Some SI and English Units In SI, the units of mass, length, and time are the kilogram (kg), meter (m), and second (s), respectively. The respective units in the English system are the pound-mass (lbm), foot (ft), and second (s). The pound symbol lb is actually the abbreviation of libra, which was the ancient Roman unit of weight. The English retained this symbol even after the end of the Roman occupation of Britain in 410. The mass and length units in the two systems are related to each other by 1 lbm = 0.45359 kg 1 ft = 0.3048 m In the English system, force is often considered to be one of the primary dimensions and is assigned a nonderived unit. This is a source of confu sion and error that necessitates the use of a dimensional constant (gc) in many formulas. To avoid this nuisance, we consider force to be a secondary dimension whose unit is derived from Newton’s second law, i.e., Force = (Mass) (Acceleration) or F = ma (1–1) In SI, the force unit is the newton (N), and it is defined as the force required to accelerate a mass of 1 kg at a rate of 1 m/s2. In the English system, the force unit is the pound-force (lbf) and is defined as the force required to TABLE 1–1 The seven fundamental (or primary) dimensions and their units in SI Dimension Unit Length Mass Time Temperature Electric current Amount of light Amount of matter meter (m) kilogram (kg) second (s) kelvin (K) ampere (A) candela (cd) mole (mol) TABLE 1–2 Standard prefixes in SI units Multiple Prefix 1024 1021 1018 1015 1012 109 106 103 102 101 10−1 10−2 10−3 10−6 10−9 10−12 10−15 10−18 10−21 10−24 yotta, Y zetta, Z exa, E peta, P tera, T giga, G mega, M kilo, k hecto, h deka, da deci, d centi, c milli, m micro, μ nano, n pico, p femto, f atto, a zepto, z yocto, y 1 kg 200 mL (0.2 L) (103 g) 1 MΩ (106 Ω) FIGURE 1–30 The SI unit prefixes are used in all branches of engineering.
cen96537_ch01_001-036.indd 17 29/12/16 10:07 am 18 Introduction and basic concepts accelerate a mass of 1 slug (32.174 lbm) at a rate of 1 ft/s2 (Fig. 1–31). That is, 1 N = 1 kg·m/s2 1 lbf = 32.174 lbm·ft/s2 A force of 1 N is roughly equivalent to the weight of a small apple (m = 102 g), whereas a force of 1 lbf is roughly equivalent to the weight of four medium apples (mtotal = 454 g), as shown in Fig. 1–32. Another force unit in common use in many European countries is the kilogram-force (kgf), which is the weight of 1 kg mass at sea level (1 kgf = 9.807 N). The term weight is often incorrectly used to express mass, particularly by the “weight watchers.” Unlike mass, weight W is a force. It is the gravi tational force applied to a body, and its magnitude is determined from an equation based on Newton’s second law, W = mg (N) (1–2) where m is the mass of the body, and g is the local gravitational accel eration (g is 9.807 m/s2 or 32.174 ft/s2 at sea level and 45° latitude). An ordinary bathroom scale measures the gravitational force acting on a body. The weight per unit volume of a substance is called the specific weight γ and is determined from γ = ρg, where ρ is density. The mass of a body remains the same regardless of its location in the uni verse. Its weight, however, changes with a change in gravitational accelera tion. A body weighs less on top of a mountain since g decreases (by a small amount) with altitude. On the surface of the moon, an astronaut weighs about one-sixth of what she or he normally weighs on earth (Fig. 1–33). At sea level a mass of 1 kg weighs 9.807 N, as illustrated in Fig. 1–34. A mass of 1 lbm, however, weighs 1 lbf, which misleads people to believe that pound-mass and pound-force can be used interchangeably as pound (lb), which is a major source of error in the English system. It should be noted that the gravity force acting on a mass is due to the attraction between the masses, and thus it is proportional to the mag nitudes of the masses and inversely proportional to the square of the dis tance between them. Therefore, the gravitational acceleration g at a loca tion depends on latitude, the distance to the earth, and to a lesser extent, the positions of the moon and the sun. The value of g varies with location from 9.8295 m/s2 at 4500 m below sea level to 7.3218 m/s2 at 100,000 m above sea level. However, at altitudes up to 30,000 m, the variation of g from the sea-level value of 9.807 m/s2 is less than 1 percent. Therefore, for most practical purposes, the gravitational acceleration can be assumed to be constant at 9.807 m/s2, often rounded to 9.81 m/s2. It is interesting to note that the value of g increases with distance below sea level, reaches a maxi mum at about 4500 m below sea level, and then starts decreasing. (What do you think the value of g is at the center of the earth?) The primary cause of confusion between mass and weight is that mass is usually measured indirectly by measuring the gravity force it exerts. This approach also assumes that the forces exerted by other effects such as air buoyancy and fluid motion are negligible. This is like measuring the dis tance to a star by measuring its red shift, or measuring the altitude of an airplane by measuring barometric pressure. Both of these are also indirect measurements. The correct direct way of measuring mass is to compare it m = 1 kg m = 32.174 lbm a = 1 m/s2 a = 1 ft/s2 F = 1 lbf F = 1 N FIGURE 1–31 The definition of the force units.
1 kgf 10 apples m ≈ 1 kg 4 apples m ≈ 1 lbm 1 lbf 1 apple m ≈ 102 g 1 N FIGURE 1–32 The relative magnitudes of the force units newton (N), kilogram-force (kgf), and pound-force (lbf).
FIGURE 1–33 A body weighing 150 lbf on earth will weigh only 25 lbf on the moon.
cen96537_ch01_001-036.indd 18 29/12/16 10:07 am 19 CHAPTER 1 to a known mass. This is cumbersome, however, and it is mostly used for calibration and measuring precious metals. Work, which is a form of energy, can simply be defined as force times distance; therefore, it has the unit “newton-meter (N.m),” which is called a joule (J). That is, 1 J = 1 N·m (1–3) A more common unit for energy in SI is the kilojoule (1 kJ = 103 J). In the English system, the energy unit is the Btu (British thermal unit), which is defined as the energy required to raise the temperature of 1 lbm of water at 68°F by 1°F. In the metric system, the amount of energy needed to raise the temperature of 1 g of water at 14.5°C by 1°C is defined as 1 calorie (cal), and 1 cal = 4.1868 J. The magnitudes of the kilojoule and Btu are very nearly the same (1 Btu = 1.0551 kJ). Here is a good way to get a feel for these units: If you light a typical match and let it burn itself out, it yields approximately one Btu (or one kJ) of energy (Fig. 1–35). The unit for time rate of energy is joule per second (J/s), which is called a watt (W). In the case of work, the time rate of energy is called power. A commonly used unit of power is horsepower (hp), which is equivalent to 745.7 W. Electrical energy typically is expressed in the unit kilowatt-hour (kWh), which is equivalent to 3600 kJ. An electric appliance with a rated power of 1 kW consumes 1 kWh of electricity when running continu ously for one hour. When dealing with electric power generation, the units kW and kWh are often confused. Note that kW or kJ/s is a unit of power, whereas kWh is a unit of energy. Therefore, statements like “the new wind turbine will generate 50 kW of electricity per year” are meaningless and incorrect. A correct statement should be something like “the new wind tur bine with a rated power of 50 kW will generate 120,000 kWh of electricity per year.” Dimensional Homogeneity We all know that you cannot add apples and oranges. But we somehow manage to do it (by mistake, of course). In engineering, all equations must be dimensionally homogeneous. That is, every term in an equation must have the same dimensions. If, at some stage of an analysis, we find our selves in a position to add two quantities that have different dimensions or units, it is a clear indication that we have made an error at an earlier stage. So checking dimensions (or units) can serve as a valuable tool to spot errors.
g = 9.807 m/s2 W = 9.807 kg·m/s2 = 9.807 N = 1 kgf W = 32.174 lbm·ft/s2 = 1 lbf g = 32.174 ft/s2 kg lbm FIGURE 1–34 The weight of a unit mass at sea level.
FIGURE 1–35 A typical match yields about one Btu (or one kJ) of energy if completely burned.
Photo by John M. Cimbala.
FIGURE 1–36 A wind turbine, as discussed in Example 1–2.
Photo by Andrew Cimbala.
EXAMPLE 1–2 Electric Power Generation by a Wind Turbine A school is paying $0.09/kWh for electric power. To reduce its power bill, the school installs a wind turbine (Fig. 1–36) with a rated power of 30 kW. If the turbine operates 2200 hours per year at the rated power, determine the amount of electric power generated by the wind turbine and the money saved by the school per year.
SOLUTION A wind turbine is installed to generate electricity. The amount of electric energy generated and the money saved per year are to be determined.
cen96537_ch01_001-036.indd 19 29/12/16 10:07 am 20 Introduction and basic concepts We all know from experience that units can give terrible headaches if they are not used carefully in solving a problem. However, with some attention and skill, units can be used to our advantage. They can be used to check formulas; sometimes they can even be used to derive formulas, as explained in the following example.
Analysis The wind turbine generates electric energy at a rate of 30 kW or 30 kJ/s. Then the total amount of electric energy generated per year becomes Total energy = (Energy per unit time)(Time interval) = (30 kW)(2200 h) = 66,000 kWh The money saved per year is the monetary value of this energy determined as Money saved = (Total energy)(Unit cost of energy) = (66,000 kWh)($0.09/kWh) = $5940 Discussion The annual electric energy production also could be determined in kJ by unit manipulations as Total energy = (30 kW)(2200 h)( 3600 s 1 h )( 1 kJ/s 1 kW ) = 2.38 × 108 kJ which is equivalent to 66,000 kWh (1 kWh = 3600 kJ).
EXAMPLE 1–3 Obtaining Formulas from Unit Consideration The drag force exerted on a car by air depends on a dimensionless drag coefficient, the density of air, the car velocity, and the frontal area of the car (Fig. 1–37). That is, FD = FD (Cdrag, Afront, ρ, V ). Based on unit considerations alone, obtain a relation for the drag force.
SOLUTION A relation for the air drag exerted on a car is to be obtained in terms of the drag coefficient, the air density, the car velocity, and the frontal area of the car.
Analysis The drag force depends on a dimensionless drag coefficient, the air density, the car velocity, and the frontal area. Also, the unit of force F is newton N, which is equivalent to kg·m/s2. Therefore, the independent quantities should be arranged such that we end up with the unit kg·m/s2 for the drag force. Putting the given information into perspective, we have FD[kg·m/s2] = Cdrag[–], Afront[m2], ρ[kg/m3], and V[m/s] It is obvious that the only way to end up with the unit “kg·m/s2” for drag force is to multiply density with the square of the velocity and the frontal area, with the drag coefficient serving as the constant of proportionality. Therefore, the desired relation is FD = CdragρAfrontV 2 ⤡ kg · m/s2 = [kg/m3] [m2] [m2/s2] FIGURE 1–37 Schematic for Example 1–3.
Air V cen96537_ch01_001-036.indd 20 29/12/16 10:07 am 21 CHAPTER 1 You should keep in mind that a formula that is not dimensionally homo geneous is definitely wrong (Fig. 1 –38), but a dimensionally homogeneous formula is not necessarily right.
Unity Conversion Ratios Just as all nonprimary dimensions can be formed by suitable combinations of primary dimensions, all nonprimary units (secondary units) can be formed by combinations of primary units. Force units, for example, can be expressed as N = kg m s2 and lbf = 32.174 lbm ft s2 They can also be expressed more conveniently as unity conversion ratios as N kg·m/s2 = 1 and lbf 32.174 lbm·ft/s2 = 1 Unity conversion ratios are identically equal to 1 and are unitless, and thus such ratios (or their inverses) can be inserted conveniently into any calculation to properly convert units (Fig. 1–39). You are encouraged to always use unity conversion ratios such as those given here when converting units. Some text books insert the archaic gravitational constant gc defined as gc = 32.174 lbm·ft/ lbf·s2 = kg·m/N·s2 = 1 into equations in order to force units to match. This practice leads to unnecessary confusion and is strongly discouraged by the present authors. We recommend that you instead use unity conversion ratios.
Discussion Note that the drag coefficient is dimensionless, so we cannot be sure whether it goes in the numerator or denominator, or has some exponent, etc. Com mon sense dictates, however, that the drag force should be linearly proportional to the drag coefficient.
FIGURE 1–38 Always check the units in your calculations.
0.3048 m 1 ft 1 min 60 s 1 lbm 0.45359 kg 32.174 lbm·ft/s2 1 lbf 1 kg·m/s2 1 N 1 kPa 1000 N/m2 1 kJ 1000 N·m 1 W 1 J/s FIGURE 1–39 Every unity conversion ratio (as well as its inverse) is exactly equal to one. Shown here are a few commonly used unity conversion ratios, each within its own set of parentheses.
EXAMPLE 1–4 The Weight of One Pound-Mass Using unity conversion ratios, show that 1.00 lbm weighs 1.00 lbf on earth (Fig. 1–40).
SOLUTION A mass of 1.00 lbm is subjected to standard earth gravity. Its weight in lbf is to be determined.
Assumptions Standard sea-level conditions are assumed.
Properties The gravitational constant is g = 32.174 ft/s2.
Analysis We apply Newton’s second law to calculate the weight (force) that cor responds to the known mass and acceleration. The weight of any object is equal to its mass times the local value of gravitational acceleration. Thus, W = mg = (1.00 lbm)(32.174 ft/s2)( 1 lbf 32.174 lbm·ft/s2) = 1.00 lbf Discussion The quantity in large parentheses in this equation is a unity conver sion ratio. Mass is the same regardless of its location. However, on some other planet with a different value of gravitational acceleration, the weight of 1 lbm would differ from that calculated here.
lbm FIGURE 1–40 A mass of 1 lbm weighs 1 lbf on earth.
cen96537_ch01_001-036.indd 21 29/12/16 10:07 am 22 Introduction and basic concepts When you buy a box of breakfast cereal, the printing may say “Net weight: One pound (454 grams).” (See Fig. 1–41.) Technically, this means that the cereal inside the box weighs 1.00 lbf on earth and has a mass of 453.6 g (0.4536 kg). Using Newton’s second law, the actual weight of the cereal on earth is W = mg = (453.6 g)(9.81 m/s2)( 1 N 1 kg·m/s2) ( 1 kg 1000 g) = 4.49 N 1–7 ■ MODELING IN ENGINEERING An engineering device or process can be studied either experimentally (test ing and taking measurements) or analytically (by analysis or calculations). The experimental approach has the advantage that we deal with the actual physical system, and the desired quantity is determined by measurement, within the limits of experimental error. However, this approach is expen sive, time-consuming, and often impractical. Besides, the system we are studying may not even exist. For example, the entire heating and plumbing systems of a building must usually be sized before the building is actu ally built on the basis of the specifications given. The analytical approach (including the numerical approach) has the advantage that it is fast and inexpensive, but the results obtained are subject to the accuracy of the assumptions, approximations, and idealizations made in the analysis. In engineering studies, often a good compromise is reached by reduc ing the choices to just a few by analysis, and then verifying the findings experimentally. The descriptions of most scientific problems involve equations that relate the changes in some key variables to each other. Usually the smaller the increment chosen in the changing variables, the more general and accurate the description. In the limiting case of infinitesimal or differential changes in variables, we obtain differential equations that provide precise math ematical formulations for the physical principles and laws by represent ing the rates of change as derivatives. Therefore, differential equations are used to investigate a wide variety of problems in sciences and engineering (Fig. 1–42). However, many problems encountered in practice can be solved without resorting to differential equations and the complications associated with them. The study of physical phenomena involves two important steps. In the first step, all the variables that affect the phenomena are identified, reason able assumptions and approximations are made, and the interdependence of these variables is studied. The relevant physical laws and principles are invoked, and the problem is formulated mathematically. The equation itself is very instructive as it shows the degree of dependence of some variables on others, and the relative importance of various terms. In the second step, the problem is solved using an appropriate approach, and the results are interpreted. Many processes that seem to occur in nature randomly and without any order are, in fact, being governed by some visible or not-so-visible physi cal laws. Whether we notice them or not, these laws are there, governing Identify important variables Make reasonable assumptions and approximations Apply relevant physical laws Physical problem A differential equation Apply applicable solution technique Apply boundary and initial conditions Solution of the problem FIGURE 1–42 Mathematical modeling of physical problems.
Net weight: One pound (454 grams) Weight?
I thought gram was a unit of mass!
FIGURE 1–41 A quirk in the metric system of units.
cen96537_ch01_001-036.indd 22 29/12/16 10:07 am 23 CHAPTER 1 consistently and predictably over what seem to be ordinary events. Most of these laws are well defined and well understood by scientists. This makes it possible to predict the course of an event before it actually occurs or to study various aspects of an event mathematically without actually running expensive and time-consuming experiments. This is where the power of analysis lies. Very accurate results to meaningful practical problems can be obtained with relatively little effort by using a suitable and realistic mathe matical model. The preparation of such models requires an adequate knowl edge of the natural phenomena involved and the relevant laws, as well as sound judgment. An unrealistic model will obviously give inaccurate and thus unacceptable results. An analyst working on an engineering problem often finds himself or her self in a position to make a choice between a very accurate but complex model, and a simple but not-so-accurate model. The right choice depends on the situation at hand. The right choice is usually the simplest model that yields satisfactory results (Fig. 1–43). Also, it is important to consider the actual operating conditions when selecting equipment. Preparing very accurate but complex models is usually not so difficult. But such models are not much use to an analyst if they are very difficult and time-consuming to solve. At the minimum, the model should reflect the essential features of the physical problem it represents. There are many sig nificant real-world problems that can be analyzed with a simple model. But it should always be kept in mind that the results obtained from an analysis are at best as accurate as the assumptions made in simplifying the problem. Therefore, the solution obtained should not be applied to situations for which the original assumptions do not hold. A solution that is not quite consistent with the observed nature of the problem indicates that the mathematical model used is too crude. In that FIGURE 1–43 Simplified models are often used in fluid mechanics to obtain approximate solutions to difficult engineering problems. Here, the helicopter’s rotor is modeled by a disk, across which is imposed a sudden change in pressure. The helicopter’s body is modeled by a simple ellipsoid. This simplified model yields the essential features of the overall air flow field in the vicinity of the ground.
(a) Photo by John M. Cimbala.
Ground Rotor disk Simplified body (a) Actual engineering problem (b) Minimum essential model of the engineering problem cen96537_ch01_001-036.indd 23 29/12/16 10:07 am 24 Introduction and basic concepts case, a more realistic model should be prepared by eliminating one or more of the questionable assumptions. This will result in a more complex problem that, of course, is more difficult to solve. Thus any solution to a problem should be interpreted within the context of its formulation.
1–8 ■ PROBLEM-SOLVING TECHNIQUE The first step in learning any science is to grasp the fundamentals and to gain a sound knowledge of it. The next step is to master the fundamentals by test ing this knowledge. This is done by solving significant real-world problems. Solving such problems, especially complicated ones, requires a systematic approach. By using a step-by-step approach, an engineer can reduce the solution of a complicated problem into the solution of a series of simple problems (Fig. 1–44). When you are solving a problem, we recommend that you use the following steps zealously as applicable. This will help you avoid some of the common pitfalls associated with problem solving.
Step 1: Problem Statement In your own words, briefly state the problem, the key information given, and the quantities to be found. This is to make sure that you understand the problem and the objectives before you attempt to solve the problem.
Step 2: Schematic Draw a realistic sketch of the physical system involved, and list the relevant information on the figure. The sketch does not have to be something elabo rate, but it should resemble the actual system and show the key features. Indicate any energy and mass interactions with the surroundings. Listing the given information on the sketch helps one to see the entire problem at once. Also, check for properties that remain constant during a process (such as temperature during an isothermal process), and indicate them on the sketch.
Step 3: Assumptions and Approximations State any appropriate assumptions and approximations made to simplify the problem to make it possible to obtain a solution. Justify the ques tionable assumptions. Assume reasonable values for missing quantities that are necessary. For example, in the absence of specific data for atmo spheric pressure, it can be taken to be 1 atm. However, it should be noted in the analysis that the atmospheric pressure decreases with increasing elevation. For example, it drops to 0.83 atm in Denver (elevation 1610 m) (Fig. 1–45).
Step 4: Physical Laws Apply all the relevant basic physical laws and principles (such as the con servation of mass), and reduce them to their simplest form by utilizing the assumptions made. However, the region to which a physical law is applied must be clearly identified first. For example, the increase in speed of water SOLUTION HARD WAY EASY WAY PROBLEM FIGURE 1–44 A step-by-step approach can greatly simplify problem solving.
Given: Air temperature in Denver To be found: Density of air Missing information: Atmospheric pressure Assumption #1: Take P = 1 atm (Inappropriate. Ignores effect of altitude. Will cause more than 15% error.) Assumption #2: Take P = 0.83 atm (Appropriate. Ignores only minor effects such as weather.) FIGURE 1–45 The assumptions made while solving an engineering problem must be reasonable and justifiable.
cen96537_ch01_001-036.indd 24 29/12/16 10:07 am 25 CHAPTER 1 flowing through a nozzle is analyzed by applying conservation of mass between the inlet and outlet of the nozzle.
Step 5: Properties Determine the unknown properties at known states necessary to solve the problem from property relations or tables. List the properties separately, and indicate their source, if applicable.
Step 6: Calculations Substitute the known quantities into the simplified relations and perform the calculations to determine the unknowns. Pay particular attention to the units and unit cancellations, and remember that a dimensional quantity without a unit is meaningless. Also, don’t give a false implication of high precision by copying all the digits from the screen of the calculator—round the final results to an appropriate number of significant digits (Section 1–10).
Step 7: Reasoning, Verification, and Discussion Check to make sure that the results obtained are reasonable and intuitive, and verify the validity of the questionable assumptions. Repeat the calcula tions that resulted in unreasonable values. For example, under the same test conditions the aerodynamic drag acting on a car should not increase after streamlining the shape of the car (Fig. 1–46). Also, point out the significance of the results, and discuss their implica tions. State the conclusions that can be drawn from the results, and any rec ommendations that can be made from them. Emphasize the limitations under which the results are applicable, and caution against any possible misunder standings and using the results in situations where the underlying assumptions do not apply. For example, if you determined that using a larger-diameter pipe in a proposed pipeline will cost an additional $5000 in materials, but it will reduce the annual pumping costs by $3000, indicate that the larger-diameter pipeline will pay for its cost differential from the electricity it saves in less than two years. However, also state that only additional material costs associated with the larger-diameter pipeline are considered in the analysis. Keep in mind that the solutions you present to your instructors, and any engineering analysis presented to others, is a form of communication. Therefore neatness, organization, completeness, and visual appearance are of utmost importance for maximum effectiveness (Fig. 1–47). Besides, neat ness also serves as a great checking tool since it is very easy to spot errors and inconsistencies in neat work. Carelessness and skipping steps to save time often end up costing more time and unnecessary anxiety. The approach described here is used in the solved example problems with out explicitly stating each step, as well as in the Solutions Manual of this text. For some problems, some of the steps may not be applicable or neces sary. For example, often it is not practical to list the properties separately. However, we cannot overemphasize the importance of a logical and orderly approach to problem solving. Most difficulties encountered while solving a problem are not due to a lack of knowledge; rather, they are due to a lack of organization. You are strongly encouraged to follow these steps in problem solving until you develop your own approach that works best for you.
Before streamlining V V After streamlining Unreasonable!
Before streamlining Before streamlining V V After streamlining After streamlining Unreasonable!
Unreasonable!
FD FD FIGURE 1–46 The results obtained from an engineering analysis must be checked for reasonableness.
FIGURE 1–47 Neatness and organization are highly valued by employers.
cen96537_ch01_001-036.indd 25 29/12/16 10:07 am 26 Introduction and basic concepts 1–9 ■ ENGINEERING SOFTWARE PACKAGES You may be wondering why we are about to undertake an in-depth study of the fundamentals of another engineering science. After all, almost all such problems we are likely to encounter in practice can be solved using one of several sophisticated software packages readily available in the market today. These software packages not only give the desired numerical results, but also supply the outputs in colorful graphical form for impressive presen tations. It is unthinkable to practice engineering today without using some of these packages. This tremendous computing power available to us at the touch of a button is both a blessing and a curse. It certainly enables engi neers to solve problems easily and quickly, but it also opens the door for abuses and misinformation. In the hands of poorly educated people, these software packages are as dangerous as sophisticated powerful weapons in the hands of poorly trained soldiers. Thinking that a person who can use the engineering software packages without proper training in the fundamentals can practice engineering is like thinking that a person who can use a wrench can work as a car mechanic. If it were true that the engineering students do not need all these fundamental courses they are taking because practically everything can be done by com puters quickly and easily, then it would also be true that the employers would no longer need high-salaried engineers since any person who knows how to use a word-processing program can also learn how to use those software packages. However, the statistics show that the need for engineers is on the rise, not on the decline, despite the availability of these powerful packages. We should always remember that all the computing power and the engi neering software packages available today are just tools, and tools have meaning only in the hands of masters. Having the best word-processing pro gram does not make a person a good writer, but it certainly makes the job of a good writer much easier and makes the writer more productive (Fig. 1–48). Hand calculators did not eliminate the need to teach our children how to add or subtract, and sophisticated medical software packages did not take the place of medical school training. Neither will engineering software pack ages replace the traditional engineering education. They will simply cause a shift in emphasis in the courses from mathematics to physics. That is, more time will be spent in the classroom discussing the physical aspects of the problems in greater detail, and less time on the mechanics of solution procedures. All these marvelous and powerful tools available today put an extra bur den on today’s engineers. They must still have a thorough understanding of the fundamentals, develop a “feel” of the physical phenomena, be able to put the data into proper perspective, and make sound engineering judg ments, just like their predecessors. However, they must do it much better, and much faster, using more realistic models because of the powerful tools available today. The engineers in the past had to rely on hand calculations, slide rules, and later hand calculators and computers. Today they rely on software packages. The easy access to such power and the possibility of a simple misunderstanding or misinterpretation causing great damage make it more important today than ever to have solid training in the fundamentals of engineering. In this text we make an extra effort to put the emphasis on FIGURE 1–48 An excellent word-processing program does not make a person a good writer; it simply makes a good writer a more efficient writer.
© Caia Images/Glow Images RF cen96537_ch01_001-036.indd 26 29/12/16 10:07 am 27 CHAPTER 1 developing an intuitive and physical understanding of natural phenomena instead of on the mathematical details of solution procedures.
Equation Solvers You are probably familiar with the equation solving capabilities of spread sheets such as Microsoft Excel. Despite its simplicity, Excel is commonly used in solving systems of equations in engineering as well as finance. It enables the user to conduct parametric studies, plot the results, and ask “what if ” questions. It can also solve simultaneous equations if properly set up. There are also numerous sophisticated equation solvers commonly used in engineering practice such as the Engineering Equation Solver (EES) which is a program that easily solves systems of linear or nonlinear alge braic or differential equations numerically. It has a large library of built-in thermodynamic property functions as well as mathematical functions, and allows the user to supply additional property data. Unlike some software packages, equation solvers do not solve engineer ing problems; they only solve the equations supplied by the user. Therefore, the user must understand the problem and formulate it by applying any rel evant physical laws and relations. Equation solvers save the user considerable time and effort by simply solving the resulting mathematical equations. This makes it possible to attempt significant engineering problems not suitable for hand calculations and to conduct parametric studies quickly and conveniently.
FIGURE 1–49 EES screen images for Example 1–5.
EXAMPLE 1–5 Solving a System of Equations Numerically The difference of two numbers is 4, and the sum of the squares of these two numbers is equal to the sum of the numbers plus 20. Determine these two numbers.
SOLUTION Relations are given for the difference and the sum of the squares of two numbers. The two numbers are to be determined.
Analysis We first solve the problem using EES. We start the EES program by double-clicking on its icon, open a new file, and type the following on the blank screen that appears: x – y = 4 x ˆ 2 + y ˆ 2 = x + y + 20 which is an exact mathematical expression of the problem statement with x and y denoting the unknown numbers. The solution to this system of equations (one linear and one nonlinear) with two unknowns is obtained by a single click on the “ calculator” icon on the taskbar. It gives (Fig. 1–49) x = 5 and y = 1 We now solve the same problem using Excel. Start Excel. File/Options/Add-Ins/Solver Add-In/OK, where the underline means to click on that option and the slash separates each sequential option. Choose a cell for x and a cell for y and enter initial guesses there (we chose cells C25 and D25 and guessed 0.5 and 0.5). We must rewrite the two equations so that no variables are on cen96537_ch01_001-036.indd 27 29/12/16 10:07 am 28 Introduction and basic concepts CFD Software Computational fluid dynamics (CFD) is used extensively in engineering and research, and we discuss CFD in detail in Chap. 15. We also show exam ple solutions from CFD throughout the textbook since CFD graphics are great for illustrating flow streamlines, velocity, and pressure distributions, etc. beyond what we are able to visualize in the laboratory (Fig. 1–51). However, because there are several different commercial CFD packages available for users, and student access to these codes is highly dependent on departmental licenses, we do not provide end-of-chapter CFD problems that are tied to any particular CFD package. Instead, we provide some gen eral CFD problems in Chap. 15, and we also maintain a website (see link at www.mhhe.com/cengel) containing CFD problems that can be solved with a number of different CFD programs. Students are encouraged to work through some of these problems to become familiar with CFD.
1–10 ■ ACCURACY, PRECISION, AND SIGNIFICANT DIGITS In engineering calculations, the supplied information is not known to more than a certain number of significant digits, usually three digits. Conse quently, the results obtained cannot possibly be precise to more significant digits. Reporting results in more significant digits falsely implies greater precision than exists, and it should be avoided. Regardless of the system of units employed, engineers must be aware of three principles that govern the proper use of numbers: accuracy, precision, and significant digits. For engineering measurements, they are defined as follows: • Accuracy error (inaccuracy) is the value of one reading minus the true value. In general, accuracy of a set of measurements refers to the closeness of the average reading to the true value. Accuracy is generally associated with repeatable, fixed errors. the right-hand side (RHS): x − y = 4 and x2 + y2 − x − y = 20. Choose a cell for the RHS of each equation and enter the formula there (we chose cells D20 and D21; see the equations in Fig. 1–50a). Data/Solver. Set the cell for the RHS of the first equation (D20) as the “Objective” with a value of 4, set the cells for x and y (C25:D25) as those subject to constraints, and set the con straint such that the cell for the RHS of the second equation (D21) must equal 20. Solve/OK. The solution iterates to the correct final values of x = 5 and y = 1, respectively (Fig. 1–50b). Note: For better convergence, the precision, number of allowed iterations, etc. can be changed in Data/Solver/Options.
Discussion Note that all we did is formulate the problem as we would on paper; EES or Excel took care of all the mathematical details of the solution. Also note that equations can be linear or nonlinear, and they can be entered in any order with unknowns on either side. Friendly equation solvers such as EES allow the user to concentrate on the physics of the problem without worrying about the mathematical complexities associated with the solution of the resulting system of equations.
(a) (b) FIGURE 1–50 Excel screen images for Example 1–5. (a) Equations, with initial guesses highlighted. (b) Final results after using Excel’s Solver, with converged values highlighted.
FIGURE 1–51 The unsteady vortex rope formed in the draft tube of a model Francis turbine operating at a discharge coefficient of 0.34. Rope simulated using the commercial CFD software, ANSYS-FLUENT. Shown are isocontours of swirling strength. © Girish Kumar Rajan. Used by permission.
cen96537_ch01_001-036.indd 28 29/12/16 10:07 am 29 CHAPTER 1 • Precision error is the value of one reading minus the average of readings. In general, precision of a set of measurements refers to the fineness of the resolution and the repeatability of the instrument. Precision is generally associated with unrepeatable, random errors.
• Significant digits are digits that are relevant and meaningful. A measurement or calculation can be very precise without being very accurate, and vice versa. For example, suppose the true value of wind speed is 25.00 m/s. Two anemometers A and B take five wind speed readings each: Anemometer A: 25.50, 25.69, 25.52, 25.58, and 25.61 m/s. Average of all readings = 25.58 m/s.
Anemometer B: 26.3, 24.5, 23.9, 26.8, and 23.6 m/s. Average of all readings = 25.02 m/s.
Clearly, anemometer A is more precise, since none of the readings differs by more than 0.11 m/s from the average. However, the average is 25.58 m/s, 0.58 m/s greater than the true wind speed; this indicates significant bias error, also called constant error or systematic error. On the other hand, anemometer B is not very precise, since its readings swing wildly from the average; but its overall average is much closer to the true value. Hence, anemometer B is more accurate than anemometer A, at least for this set of readings, even though it is less precise. The difference between accuracy and precision can be illustrated effectively by analogy to shooting arrows at a target, as sketched in Fig. 1–52. Shooter A is very precise, but not very accurate, while shooter B has better overall accuracy, but less precision. Many engineers do not pay proper attention to the number of significant digits in their calculations. The least significant numeral in a number implies the precision of the measurement or calculation. For example, a result written as 1.23 (three significant digits) implies that the result is precise to within one digit in the second decimal place; i.e., the number is somewhere between 1.22 and 1.24. Expressing this number with any more digits would be misleading. The number of significant digits is most easily evaluated when the number is written in exponential notation; the number of signifi cant digits can then simply be counted, including zeroes. Alternatively, the least significant digit can be underlined to indicate the author’s intent. Some examples are shown in Table 1–3. When performing calculations or manipulations of several parameters, the final result is generally only as precise as the least precise parameter in the problem. For example, suppose A and B are multiplied to obtain C. If A = 2.3601 (five significant digits), and B = 0.34 (two significant digits), then C = 0.80 (only two digits are significant in the final result). Note that most students are tempted to write C = 0.802434, with six significant digits, since that is what is displayed on a calculator after multiplying these two numbers. Let’s analyze this simple example carefully. Suppose the exact value of B is 0.33501, which is read by the instrument as 0.34. Also suppose A is exactly 2.3601, as measured by a more accurate and precise instrument. In this case, C = A × B = 0.79066 to five significant digits. Note that our first answer, C = 0.80 is off by one digit in the second decimal place. Likewise, if B is 0.34499, and is read by the instrument as 0.34, the product of A and B would be 0.81421 to five significant digits. Our original answer of 0.80 A B + + + + + + + + + + + + + ++ FIGURE 1–52 Illustration of accuracy versus precision. Shooter A is more precise, but less accurate, while shooter B is more accurate, but less precise.
TABLE 1–3 Significant digits Number Exponential Notation Number of Significant Digits 12.3 123,000 0.00123 40,300 40,300 0.005600 0.0056 0.006 1.23 × 101 1.23 × 105 1.23 × 10−3 4.03 × 104 4.0300 × 104 5.600 × 10−3 5.6 × 10−3 6. × 10−3 3 3 3 3 5 4 2 1 cen96537_ch01_001-036.indd 29 29/12/16 10:07 am 30 Introduction and basic concepts is again off by one digit in the second decimal place. The main point here is that 0.80 (to two significant digits) is the best one can expect from this multiplication since, to begin with, one of the values had only two signifi cant digits. Another way of looking at this is to say that beyond the first two digits in the answer, the rest of the digits are meaningless or not signifi cant. For example, if one reports what the calculator displays, 2.3601 times 0.34 equals 0.802434, the last four digits are meaningless. As shown, the final result may lie between 0.79 and 0.81—any digits beyond the two sig nificant digits are not only meaningless, but misleading, since they imply to the reader more precision than is really there. As another example, consider a 3.75-L container filled with gasoline whose density is 0.845 kg/L, and determine its mass. Probably the first thought that comes to your mind is to multiply the volume and density to obtain 3.16875 kg for the mass, which falsely implies that the mass so determined is precise to six significant digits. In reality, however, the mass cannot be more precise than three significant digits since both the volume and the density are precise to three significant digits only. Therefore, the result should be rounded to three significant digits, and the mass should be reported to be 3.17 kg instead of what the calculator displays (Fig. 1–53). The result 3.16875 kg would be correct only if the volume and density were given to be 3.75000 L and 0.845000 kg/L, respectively. The value 3.75 L implies that we are fairly confident that the volume is precise within ±0.01 L, and it cannot be 3.74 or 3.76 L. However, the volume can be 3.746, 3.750, 3.753, etc., since they all round to 3.75 L. You should also be aware that sometimes we knowingly introduce small errors in order to avoid the trouble of searching for more accurate data. For example, when dealing with liquid water, we often use the value of 1000 kg/m3 for density, which is the density value of pure water at 0°C. Using this value at 75°C will result in an error of 2.5 percent since the density at this tem perature is 975 kg/m3. The minerals and impurities in the water will intro duce additional error. This being the case, you should have no reservation in rounding the final results to a reasonable number of significant digits. Besides, having a few percent uncertainty in the results of engineering anal ysis is usually the norm, not the exception. When writing intermediate results in a computation, it is advisable to keep several “extra” digits to avoid round-off errors; however, the final result should be written with the number of significant digits taken into consideration. You must also keep in mind that a certain number of sig nificant digits of precision in the result does not necessarily imply the same number of digits of overall accuracy. Bias error in one of the readings may, for example, significantly reduce the overall accuracy of the result, perhaps even rendering the last significant digit meaningless, and reducing the over all number of reliable digits by one. Experimentally determined values are subject to measurement errors, and such errors are reflected in the results obtained. For example, if the density of a substance has an uncertainty of 2 percent, then the mass determined using this density value will also have an uncertainty of 2 percent. Finally, when the number of significant digits is unknown, the accepted engineering standard is three significant digits. Therefore, if the length of a pipe is given to be 40 m, we will assume it to be 40.0 m in order to justify using three significant digits in the final results.
Given: Also, 3.75 × 0.845 = 3.16875 Volume: Density: Find: Mass: m = V = 3.16875 kg Rounding to 3 significant digits: m = 3.17 kg (3 significant digits) V = 3.75 L = 0.845 kg/L FIGURE 1–53 A result with more significant digits than that of given data falsely implies more precision.
cen96537_ch01_001-036.indd 30 29/12/16 10:07 am 31 CHAPTER 1 EXAMPLE 1–6 Significant Digits and Volume Flow Rate Jennifer is conducting an experiment that uses cooling water from a garden hose. In order to calculate the volume flow rate of water through the hose, she times how long it takes to fill a container (Fig. 1–54). The volume of water collected is V = 1.1 gal in time period Δt = 45.62 s, as measured with a stopwatch. Calculate the volume flow rate of water through the hose in units of cubic meters per minute.
SOLUTION Volume flow rate is to be determined from measurements of volume and time period.
Assumptions 1 Jennifer recorded her measurements properly, such that the volume measurement is precise to two significant digits while the time period is precise to four significant digits. 2 No water is lost due to splashing out of the container.
Analysis Volume flow rate V . is volume displaced per unit time and is expressed as Volume flow rate: V · = ΔV Δt Substituting the measured values, the volume flow rate is determined to be V · = 1.1 gal 45.62 s ( 3.7854 × 10−3 m3 1 gal ) ( 60 s 1 min) = 5.5 × 10−3 m3/min Discussion The final result is listed to two significant digits since we cannot be confident of any more precision than that. If this were an intermediate step in subsequent calculations, a few extra digits would be carried along to avoid accumulated round-off error. In such a case, the volume flow rate would be written as V . = 5.4765 × 10−3 m3/min. Based on the given information, we cannot say anything about the accuracy of our result, since we have no infor mation about systematic errors in either the volume measurement or the time measurement. Also keep in mind that good precision does not guarantee good accuracy. For example, if the stopwatch had not been properly calibrated, its accuracy could be quite poor, yet the readout would still be displayed to four significant digits of precision. In common practice, precision is often associated with resolution, which is a measure of how finely the instrument can report the measurement. For example, a digital voltmeter with five digits on its display is said to be more precise than a digital voltmeter with only three digits. However, the number of displayed digits has nothing to do with the overall accuracy of the measurement. An instrument can be very precise without being very accurate when there are significant bias errors. Likewise, an instrument with very few displayed digits can be more accurate than one with many digits (Fig. 1–55).
FIGURE 1–54 Photo for Example 1–6 for the measurement of volume flow rate.
Photo by John M. Cimbala.
Exact time span = 45.623451 . . . s (a) TIMEXAM 46.
s (b) TIMEXAM 43.
s (c) TIMEXAM 44.189 s (d) TIMEXAM 45.624 s FIGURE 1–55 An instrument with many digits of resolution (stopwatch c) may be less accurate than an instrument with few digits of resolution (stopwatch a). What can you say about stopwatches b and d?
cen96537_ch01_001-036.indd 31 29/12/16 10:07 am 32 Introduction and basic concepts Guest Author: Lorenz Sigurdson, Vortex Fluid Dynamics Lab, University of Alberta Why do the two images in Fig. 1–56 look alike? Figure 1–56b shows an above-ground nuclear test performed by the U.S. Department of Energy in 1957. A nuclear blast created a fireball on the order of 100 m in diameter. Expansion is so quick that a compressible flow feature occurs: an expanding spherical shock wave. The image shown in Fig. 1–56a is an everyday innocuous event: an inverted image of a dye-stained water drop after it has fallen into a pool of water, looking from below the pool surface. It could have fallen from your spoon into a cup of coffee, or been a secondary splash after a raindrop hit a lake. Why is there such a strong similarity between these two vastly different events? The application of fundamental principles of fluid mechanics learned in this book will help you understand much of the answer, although one can go much deeper. The water has higher density (Chap. 2) than air, so the drop has experienced negative buoyancy (Chap. 3) as it has fallen through the air before impact. The fireball of hot gas is less dense than the cool air surrounding it, so it has posi tive buoyancy and rises. The shock wave (Chap. 12) reflecting from the ground also imparts a positive upward force to the fireball. The primary structure at the top of each image is called a vortex ring. This ring is a mini-tornado of concentrated vorticity (Chap. 4) with the ends of the tornado looping around to close on itself. The laws of kinematics (Chap. 4) tell us that this vortex ring will carry the fluid in a direction toward the top of the page. This is expected in both cases from the forces applied and the law of conservation of momentum applied through a control volume analysis (Chap. 5). One could also analyze this problem with differential analysis (Chaps. 9 and 10) or with computational fluid dynamics (Chap. 15). But why does the shape of the tracer material look so similar? This occurs if there is approximate geometric and kinematic simi larity (Chap. 7), and if the flow visualization (Chap. 4) technique is similar. The passive tracers of heat and dust for the bomb, and fluorescent dye for the drop, were introduced in a similar manner as noted in the figure caption. Further knowledge of kinematics and vortex dynamics can help explain the similarity of the vortex structure in the images to much greater detail, as dis cussed by Sigurdson (1997) and Peck and Sigurdson (1994). Look at the lobes dangling beneath the primary vortex ring, the striations in the “stalk,” and the ring at the base of each structure. There is also topological similarity of this structure to other vortex structures occurring in turbulence. Comparison of the drop and bomb has given us a better understanding of how turbulent structures are created and evolve. What other secrets of fluid mechanics are left to be revealed in explaining the similarity between these two flows?
References Peck, B., and Sigurdson, L.W., “The Three-Dimensional Vortex Structure of an Impacting Water Drop,” Phys. Fluids, 6(2) (Part 1), p. 564, 1994.
Peck, B., Sigurdson, L.W., Faulkner, B., and Buttar, I., “An Apparatus to Study Drop-Formed Vortex Rings,” Meas. Sci. Tech., 6, p. 1538, 1995.
Sigurdson, L.W., “Flow Visualization in Turbulent Large-Scale Structure Research,” Chapter 6 in Atlas of Visualization, Vol. III, Flow Visualization Society of Japan, eds., CRC Press, pp. 99–113, 1997.
APPLICATION SPOTLIGHT ■ What Nuclear Blasts and Raindrops Have in Common FIGURE 1–56 Comparison of the vortex structure created by: (a) a water drop after impacting a pool of water (inverted, from Peck and Sigurdson, 1994), and (b) an above-ground nuclear test in Nevada in 1957 (U.S. Department of Energy). The 2.6 mm drop was dyed with fluorescent tracer and illuminated by a strobe flash 50 ms after it had fallen 35 mm and impacted the clear pool. The drop was approximately spherical at the time of impact with the clear pool of water. Interruption of a laser beam by the falling drop was used to trigger a timer that controlled the time of the strobe flash after im pact of the drop. Details of the careful experimental procedure necessary to create the drop photograph are given by Peck and Sigurdson (1994) and Peck et al. (1995). The tracers added to the flow in the bomb case were primarily heat and dust. The heat is from the original fireball which for this particular test (the “Priscilla” event of Operation Plumbob) was large enough to reach the ground from where the bomb was initially suspended. Therefore, the tracer’s initial geometric condition was a sphere intersecting the ground.
(a) From Peck B., and Sigurdson, L.W., Phys. Fluids, 6(2)(Part 1), 564, 1994. Used with Permission.
(b) © Galerie Bilderwelt/Getty Images (a) (b) cen96537_ch01_001-036.indd 32 29/12/16 10:07 am 33 CHAPTER 1 SUMMARY In this chapter some basic concepts of fluid mechanics are introduced and discussed. A substance in the liquid or gas phase is referred to as a fluid. Fluid mechanics is the science that deals with the behavior of fluids at rest or in motion and the interaction of fluids with solids or other fluids at the boundaries. The flow of an unbounded fluid over a surface is external flow, and the flow in a pipe or duct is internal flow if the fluid is completely bounded by solid surfaces. A fluid flow is classified as being compressible or incompress ible, depending on the density variation of the fluid dur ing flow. The densities of liquids are essentially constant, and thus the flow of liquids is typically incompressible. The term steady implies no change with time. The opposite of steady is unsteady. The term uniform implies no change with location over a specified region. A flow is said to be one-dimensional when the properties or variables change in one dimension only. A fluid in direct contact with a solid surface sticks to the surface and there is no slip. This is known as the no-slip condition, which leads to the forma tion of boundary layers along solid surfaces. In this book we concentrate on steady incompressible viscous flows— both internal and external. A system of fixed mass is called a closed system, and a system that involves mass transfer across its boundaries is called an open system or control volume. A large number of engineering problems involve mass flow in and out of a system and are therefore modeled as control volumes. In engineering calculations, it is important to pay particular attention to the units of the quantities to avoid errors caused by inconsistent units, and to follow a systematic approach. It is also important to recognize that the information given is not known to more than a certain number of significant digits, and the results obtained cannot possibly be accurate to more significant digits. The information given on dimensions and units; problem-solving technique; and accuracy, preci sion, and significant digits will be used throughout the entire text.
REFERENCES AND SUGGESTED READING 1. American Society for Testing and Materials. Standards for Metric Practice. ASTM E 380-79, January 1980.
2. G. M. Homsy, H. Aref, K. S. Breuer, S. Hochgreb, J. R. Koseff, B. R. Munson, K. G. Powell, C. R. Robertson, and S. T. Thoroddsen. Multi-Media Fluid Mechanics (CD). Cambridge: Cambridge University Press, 2000.
3. M. Van Dyke. An Album of Fluid Motion. Stanford, CA: The Parabolic Press, 1982.
PROBLEMS Introduction, Classification, and System 1–1C What is a fluid? How does it differ from a solid? How does a gas differ from a liquid? 1–2C Define internal, external, and open-channel flows.
1–3C Define incompressible flow and incompressible fluid. Must the flow of a compressible fluid necessarily be treated as compressible?
1–4C Consider the flow of air over the wings of an aircraft. Is this flow internal or external? How about the flow of gases through a jet engine?
1–5C What is forced flow? How does it differ from natural flow? Is flow caused by winds forced or natural flow?
1–6C How is the Mach number of a flow defined? What does a Mach number of 2 indicate?
1–7C When an airplane is flying at a constant speed relative to the ground, is it correct to say that the Mach number of this airplane is also constant?
1–8C Consider the flow of air at a Mach number of 0.12. Should this flow be approximated as being incompressible?
1–9C What is the no-slip condition? What causes it?
1–10C What is a boundary layer? What causes a boundary layer to develop?
1–11C What is a steady-flow process?
1–12C Define stress, normal stress, shear stress, and pressure.
1–13C What are system, surroundings, and boundary?
Problems designated by a “C” are concept questions, and stu dents are encouraged to answer them all. Problems designated by an “E” are in English units, and SI users can ignore them. Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software.
cen96537_ch01_001-036.indd 33 29/12/16 10:07 am 34 INTRODUCTION AND BASIC CONCEPTS 1–14C When analyzing the acceleration of gases as they flow through a nozzle, what would you choose as your sys tem? What type of system is this?
1–15C When is a system a closed system, and when is it a control volume?
1–16C You are trying to understand how a reciprocating air compressor (a piston-cylinder device) works. What system would you use? What type of system is this?
Mass, Force, and Units 1–17C What is the difference between pound-mass and pound-force?
1–18C In a news article, it is stated that a recently devel oped geared turbofan engine produces 15,000 pounds of thrust to propel the aircraft forward. Is “pound” mentioned here lbm or lbf? Explain.
1–19C Explain why the light-year has the dimension of length.
1–20C What is the net force acting on a car cruising at a constant velocity of 70 km/h (a) on a level road and (b) on an uphill road?
1–21 A man goes to a traditional market to buy a steak for dinner. He finds a 12-oz steak (1 lbm = 16 oz) for $3.15. He then goes to the adjacent international market and finds a 320-g steak of identical quality for $3.30. Which steak is the better buy?
1–22 What is the weight, in N, of an object with a mass of 150 kg at a location where g = 9.6 m/s2?
1–23 What is the weight of a 1-kg substance in N, kN, kg·m/s2, kgf, lbm·ft/s2, and lbf?
1–24 Determine the mass and the weight of the air con tained in a room whose dimensions are 3 m × 5 m × 7 m. Assume the density of the air is 1.16 kg/m3. Answers: 122 kg, 1195 N 1–25 A 3-kW resistance heater in a water heater runs for 2 hours to raise the water temperature to the desired level. Determine the amount of electric energy used in both kWh and kJ.
1–26E A 195-lbm astronaut took his bathroom scale (a spring scale) and a beam scale (compares masses) to the moon where the local gravity is g = 5.48 ft/s2. Determine how much he will weigh (a) on the spring scale and (b) on the beam scale. Answers: (a) 33.2 lbf, (b) 195 lbf 1–27 The acceleration of high-speed aircraft is sometimes expressed in g’s (in multiples of the standard acceleration of gravity). Determine the net force, in N, that a 90-kg man would experience in an aircraft whose acceleration is 6 g’s.
1–28 A 10-kg rock is thrown upward with a force of 280 N at a location where the local gravitational acceleration is 9.79 m/s2. Determine the acceleration of the rock, in m/s2.
1–29 Solve Prob. 1–30 using appropriate software. Print out the entire solution, including the numerical results with proper units.
1–30 The value of the gravitational acceleration g decreases with elevation from 9.807 m/s2 at sea level to 9.767 m/s2 at an altitude of 13,000 m, where large passenger planes cruise. Determine the percent reduction in the weight of an airplane cruising at 13,000 m relative to its weight at sea level.
1–31 At 45° latitude, the gravitational acceleration as a function of elevation z above sea level is given by g = a − bz, where a = 9.807 m/s2 and b = 3.32 × 10−6 s−2. Determine the height above sea level where the weight of an object will decrease by 1 percent. Answer: 29,500 m 1–32 The gravitational constant g is 9.807 m/s2 at sea level, but it decreases as you go up in elevation. A useful equation for this decrease in g is g = a – bz, where z is the elevation above sea level, a = 9.807 m/s2, and b = 3.32 × 10–6 1/s2. An astronaut “weighs” 80.0 kg at sea level. [Technically this means that his/her mass is 80.0 kg.] Calculate this person’s weight in N while floating around in the International Space Station (z = 354 km). If the Space Station were to suddenly stop in its orbit, what gravitational acceleration would the astronaut feel immediately after the satellite stopped moving? In light of your answer, explain why astronauts on the Space Station feel “weightless.” 1–33 On average, an adult person breathes in about 7.0 liters of air per minute. Assuming atmospheric pressure and 20°C air temperature, estimate the mass of air in kilograms that a person breathes in per day.
1–34 While solving a problem, a person ends up with the equation E = 16 kJ + 7 kJ/kg at some stage. Here E is the total energy and has the unit of kilojoules. Determine how to correct the error and discuss what may have caused it.
1–35 An airplane flies horizontally at 70 m/s. Its propeller delivers 1500 N of thrust (forward force) to overcome aerodynamic drag (backward force). Using dimensional reasoning and unity converstion ratios, calculate the use ful power delivered by the propeller in units of kW and horsepower.
1–36 If the airplane of Prob. 1–35 weighs 1700 lbf, estimate the lift force produced by the airplane’s wings (in lbf and newtons) when flying at 70.0 m/s.
1–37E The boom of a fire truck raises a fireman (and his equipment—total weight 280 lbf) 40 ft into the air to fight a building fire. (a) Showing all your work and using unity conversion ratios, calculate the work done by the boom on the fireman in units of Btu. (b) If the useful power supplied by the boom to lift the fireman is 3.50 hp, estimate how long it takes to lift the fireman.
1–38 A 6-kg plastic tank that has a volume of 0.18 m3 is filled with liquid water. Assuming the density of water is 1000 kg/m3, determine the weight of the combined system.
cen96537_ch01_001-036.indd 34 29/12/16 10:07 am 35 CHAPTER 1 1–39 Water at 15°C from a garden hose fills a 1.5 L con tainer in 2.85 s. Using unity conversion ratios and showing all your work, calculate the volume flow rate in liters per minute (Lpm) and the mass flow rate in kg/s.
1–40 A forklift raises a 90.5 kg crate 1.80 m. (a) Showing all your work and using unity conversion ratios, calculate the work done by the forklift on the crane, in units of kJ. (b) If it takes 12.3 seconds to lift the crate, calculate the useful power supplied to the crate in kilowatts.
1–41 The gas tank of a car is filled with a nozzle that dis charges gasoline at a constant flow rate. Based on unit con siderations of quantities, obtain a relation for the filling time in terms of the volume V of the tank (in L) and the discharge rate of gasoline (V ˙, in L/s).
1–42 A pool of volume V (in m3) is to be filled with water using a hose of diameter D (in m). If the average discharge velocity is V (in m/s) and the filling time is t (in s), obtain a relation for the volume of the pool based on unit consider ations of quantities involved.
1–43 Based on unit considerations alone, show that the power needed to accelerate a car of mass m (in kg) from rest to velocity V (in m/s) in time interval t (in s) is proportional to mass and the square of the velocity of the car and inversely proportional to the time interval.
Modeling and Solving Engineering Problems 1–44C What is the importance of modeling in engineering? How are the mathematical models for engineering processes prepared?
1–45C What is the difference between the analytical and experimental approach to engineering problems? Discuss the advantages and disadvantages of each approach.
1–46C When modeling an engineering process, how is the right choice made between a simple but crude and a complex but accurate model? Is the complex model necessarily a bet ter choice since it is more accurate?
1–47C What is the difference between precision and accu racy? Can a measurement be very precise but inaccurate? Explain.
1–48C How do the differential equations in the study of a physical problem arise?
1–49C What is the value of the engineering software packages in (a) engineering education and (b) engineering practice?
1–50 Solve this system of three equations with three unknowns using appropriate software: 2x −y + z = 9 3x2 + 2y = z + 2 xy + 2z = 14 1–51 Solve this system of two equations with two unknowns using appropriate software: x3 −y2 = 10.5 3xy + y = 4.6 1–52 Determine a positive real root of this equation using appropriate software: 3.5x3 −10x0.5 −3x = −4 1–53 Solve this system of three equations with three unknowns using appropriate software: x2y −z = 1.5 x −3y0.5 + xz = −2 x + y −z = 4.2 Review Problems 1–54E A student buys a 5000 Btu window air conditioner for his apartment bedroom. He monitors it for one hour on a hot day and determines that it operates approximately 60 per cent of the time (duty cycle = 60 percent) to keep the room at nearly constant temperature. (a) Showing all your work and using unity conversion ratios, calculate the rate of heat transfer into the bedroom through the walls, windows, etc. in units of Btu/h and in units of kW. (b) If the energy efficiency ratio (EER) of the air conditioner is 9.0 and electricity costs 7.5 cents per kilowatt-hr, calculate how much it costs (in cents) for him to run the air conditioner for one hour.
1–55 The weight of bodies may change somewhat from one location to another as a result of the variation of the gravita tional acceleration g with elevation. Accounting for this varia tion using the relation in Prob. 1–31, determine the weight of an 65-kg person at sea level (z = 0), in Denver (z = 1610 m), and on the top of Mount Everest (z = 8848 m).
1–56E The reactive force developed by a jet engine to push an airplane forward is called thrust, and the thrust developed by the engine of a Boeing 777 is about 85,000 lbf. Express this thrust in N and kgf.
1–57 For liquids, the dynamic viscosity μ, which is a measure of resistance against flow is approximated as μ = a10b/(T−c), where T is the absolute temperature, and a, b and c are experi mental constants. Using the data listed in Table A–7 for metha nol at 20ºC, 40ºC and 60ºC, determine the constant a, b and c.
1–58 An important design consideration in two-phase pipe flow of solid-liquid mixtures is the terminal settling velocity below, which the flow becomes unstable and eventually the pipe becomes clogged. On the basis of extended transportation tests, the terminal settling velocity of a solid particle in the rest water given by VL = FL√2gD(S −1), where FL is an experi mental coefficient, g the gravitational acceleration, D the pipe diameter, and S the specific gravity of solid particle. What is the dimension of FL? Is this equation dimensionally homogeneous?
cen96537_ch01_001-036.indd 35 29/12/16 10:07 am 36 INTRODUCTION AND BASIC CONCEPTS 1–59 Consider the flow of air through a wind turbine whose blades sweep an area of diameter D (in m). The average air velocity through the swept area is V (in m/s). On the bases of the units of the quantities involved, show that the mass flow rate of air (in kg/s) through the swept area is proportional to air density, the wind velocity, and the square of the diameter of the swept area.
1–60 A tank is filled with oil whose density is ρ = 850 kg/m3. If the volume of the tank is V = 2 m3, determine the amount of mass m in the tank.
FIGURE P1–60 Oil V = 2 m3 m = ?
ρ = 850 kg/m3 Fundamentals of Engineering (FE) Exam Problems 1–61 If mass, heat, and work are not allowed to cross the boundaries of a system, the system is called (a) Isolated (b) Isothermal (c) Adiabatic (d) Control mass (e) Control volume 1–62 The speed of an aircraft is given to be 260 m/s in air. If the speed of sound at that location is 330 m/s, the flight of aircraft is (a) Sonic (b) Subsonic (c) Supersonic (d) Hypersonic 1–63 One J/kg is equal to (a) 1 kPa ⋅ m3 (b) 1 kN ⋅ m/kg (c) 0.001 kJ (d) 1 N ⋅ m (e) 1 m2/s2 1–64 Which is a unit for power?
(a) Btu (b) kWh (c) kcal (d) hph (e) kW 1–65 The speed of an aircraft is given to be 950 km/h. If the speed of sound at that location is 315 m/s, the Mach number is (a) 0.63 (b) 0.84 (c) 1.0 (d) 1.07 (e) 1.20 1–66 The weight of a l0-kg mass at sea level is (a) 9.81 N (b) 32.2 kgf (c) 98.1 N (d) 10 N (e) l00 N 1–67 The weight of a 1-lbm mass is (a) 1 lbm⋅ft/s2 (b) 9.81 lbf (c) 9.81 N (d) 32.2 lbf (e) 1 lbf 1–68 A hydroelectric power plant operates at its rated power of 12 MW. If the plant has produced 26 million kWh of electricity in a specified year, the number of hours the plant has operated that year is (a) 2167 h (b) 2508 h (c) 3086 h (d) 3710 h (e) 8760 h Design and Essay Problems 1–69 Write an essay on the various mass- and volume-measurement devices used throughout history. Also, explain the development of the modern units for mass and volume.
1–70 Search the Internet to find out how to properly add or subtract numbers while taking into consideration the number of significant digits. Write a summary of the proper technique, then use the technique to solve the following cases: (a) 1.006 + 23.47, (b) 703,200 − 80.4, and (c) 4.6903 − 14.58. Be careful to express your final answer to the appropriate number of significant digits.
1–71 Another unit is kgf, which is a force unit used mostly in Europe, and is defined as kp (kilopond). Explain the difference between kilopond (kp = kgf) and kilopound (103 lbf) from force units and write the unity conversion factor between them. The density of water at 4°C is = 1000 kg/m3. Express this density value in units of kp ⋅ m−4 ⋅ s2. 1–72 Discuss why pressure tests of pressurized tanks such as steam boilers, pipes, and tanks including gases such as nitrogen, air, oxygen, etc. with high pressure are carried out hydrostatically by using liquids such as water and hydraulic oil.
cen96537_ch01_001-036.indd 36 29/12/16 10:07 am 2 CHAPTER 37 P RO P E RT I E S O F FLUI D S I n this chapter, we discuss properties that are encountered in the analy sis of fluid flow. First we discuss intensive and extensive properties and define density and specific gravity. This is followed by a discussion of the properties vapor pressure, energy and its various forms, the specific heats of ideal gases and incompressible substances, the coefficient of com pressibility, and the speed of sound. Then we discuss the property viscos ity, which plays a dominant role in most aspects of fluid flow. Finally, we present the property surface tension and determine the capillary rise from static equilibrium conditions. The property pressure is discussed in Chap. 3 together with fluid statics. OBJECTIVES When you finish reading this chapter, you should be able to ■ ■ Have a working knowledge of the basic properties of fluids and understand the continuum approximation ■ ■ Have a working knowledge of viscosity and the consequences of the frictional effects it causes in fluid flow ■ ■ Calculate the capillary rise (or drop) in tubes due to the surface tension effect A drop forms when liquid is forced out of a small tube. The shape of the drop is determined by a balance of pressure, gravity, and surface tension forces.
© Corbis RF cen96537_ch02_037-076.indd 37 29/12/16 5:59 pm 38 PROPERTIES OF FLUIDS 2–1 ■ INTRODUCTION Any characteristic of a system is called a property. Some familiar proper ties are pressure P, temperature T, volume V, and mass m. The list can be extended to include less familiar ones such as viscosity, thermal conductiv ity, modulus of elasticity, thermal expansion coefficient, electric resistivity, and even velocity and elevation. Properties are considered to be either intensive or extensive. Intensive properties are those that are independent of the mass of the system, such as temperature, pressure, and density. Extensive properties are those whose values depend on the size—or extent—of the system. Total mass, total volume V, and total momentum are some examples of extensive properties. An easy way to determine whether a property is intensive or extensive is to divide the system into two equal parts with an imaginary partition, as shown in Fig. 2–1. Each part will have the same value of intensive properties as the original system, but half the value of the extensive properties. Generally, uppercase letters are used to denote extensive properties (with mass m being a major exception), and lowercase letters are used for intensive properties (with pressure P and temperature T being the obvious exceptions). Extensive properties per unit mass are called specific properties. Some examples of specific properties are specific volume (v = V/m) and specific total energy (e = E/m). The state of a system is described by its properties. But we know from experience that we do not need to specify all the properties in order to fix a state. Once the values of a sufficient number of properties are specified, the rest of the properties assume certain values. That is, specifying a certain number of properties is sufficient to fix a state. The number of properties required to fix the state of a system is given by the state postulate: The state of a simple compressible system is completely specified by two inde pendent, intensive properties. Two properties are independent if one property can be varied while the other one is held constant. Not all properties are independent, and some are defined in terms of others, as explained in Section 2–2.
Continuum A fluid is composed of molecules which may be widely spaced apart, espe cially in the gas phase. Yet it is convenient to disregard the atomic nature of the fluid and view it as continuous, homogeneous matter with no holes, that is, a continuum. The continuum idealization allows us to treat proper ties as point functions and to assume that the properties vary continually in space with no jump discontinuities. This idealization is valid as long as the size of the system we deal with is large relative to the space between the mol ecules (Fig. 2–2). This is the case in practically all problems, except some specialized ones. The continuum idealization is implicit in many statements we make, such as “the density of water in a glass is the same at any point.” To have a sense of the distances involved at the molecular level, consider a container filled with oxygen at atmospheric conditions. The diameter of an oxygen molecule is about 3 × 10−10 m and its mass is 5.3 × 10−26 kg. Also, the mean free path 𝜆 of oxygen at 1 atm pressure and 20°C is 6.3 × 10−8 m. That is, an oxygen molecule travels, on average, a distance of 6.3 × 10−8 m (about 200 times its diameter) before it collides with another molecule.
m V T P ρ 1 2 m 1 2 V T P ρ 1 2 m 1 2 V T P ρ Extensive properties Intensive properties FIGURE 2–1 Criterion to differentiate intensive and extensive properties.
FIGURE 2–2 The length scale associated with most flows, such as seagulls in flight, is orders of magnitude larger than the mean free path of the air molecules. Therefore, here, and for all fluid flows considered in this book, the continuum idealization is appropriate.
© PhotoLink/Getty Images RF cen96537_ch02_037-076.indd 38 29/12/16 5:59 pm 39 CHAPTER 2 Also, there are about 3 × 1016 molecules of oxygen in the tiny volume of 1 mm3 at 1 atm pressure and 20°C (Fig. 2–3). The continuum model is applicable as long as the characteristic length of the system (such as its diameter) is much larger than the mean free path of the molecules. At very low pressure, e.g., at very high elevations, the mean free path may become large (for example, it is about 0.1 m for atmospheric air at an elevation of 100 km). For such cases the rarefied gas flow theory should be used, and the impact of individual molecules should be considered. In this text we limit our consideration to substances that can be modeled as a continuum. Quantitatively, a dimensionless number called the Knudsen number Kn = 𝜆/L is defined, where 𝜆 is the mean free path of the fluid molecules and L is some characteristic length scale of the fluid flow. If Kn is very small (typically less than about 0.01), the fluid medium can be approximated as a continuum medium.
2–2 ■ DENSITY AND SPECIFIC GRAVITY Density is defined as mass per unit volume (Fig. 2–4). That is, Density: ρ = m V (kg/m3) (2–1) The reciprocal of density is the specific volume v, which is defined as volume per unit mass. That is, v = V/m = 1/𝜌. For a differential volume element of mass δm and volume δV, density can be expressed as 𝜌 = δm/δV. The density of a substance, in general, depends on temperature and pressure. The density of most gases is proportional to pressure and inversely proportional to temperature. Liquids and solids, on the other hand, are essentially incompressible substances, and the variation of their density with pressure is usually negligible. At 20°C, for example, the density of water changes from 998 kg/m3 at 1 atm to 1003 kg/m3 at 100 atm, a change of just 0.5 percent. The density of liquids and solids depends more strongly on temperature than it does on pressure. At 1 atm, for example, the density of water changes from 998 kg/m3 at 20°C to 975 kg/m3 at 75°C, a change of 2.3 percent, which can still be neglected in many engineering analyses. Sometimes the density of a substance is given relative to the density of a well-known substance. Then it is called specific gravity, or relative density, and is defined as the ratio of the density of a substance to the density of some standard substance at a specified temperature (usually water at 4°C, for which 𝜌H2O = 1000 kg/m3). That is, Specific gravity: SG = ρ ρH2O (2–2) Note that the specific gravity of a substance is a dimensionless quantity. However, in SI units, the numerical value of the specific gravity of a sub stance is exactly equal to its density in g/cm3 or kg/L (or 0.001 times the density in kg/m3) since the density of water at 4°C is 1 g/cm3 = 1 kg/L = 1000 kg/m3. The specific gravity of mercury at 20°C, for example, is 13.6. Therefore, its density at 20°C is 13.6 g/cm3 = 13.6 kg/L = 13,600 kg/m3. The specific gravities of some substances at 20°C are given in Table 2–1. Note that substances with specific gravities less than 1 are lighter than water, and thus they would float on water (if immiscible).
VOID 1 atm, 20°C O2 3 × 1016 molecules/mm3 FIGURE 2–3 Despite the relatively large gaps between molecules, a gas can usually be treated as a continuum because of the very large number of molecules even in an extremely small volume.
V = 12 m3 m = 3 kg ρ = 0.25 kg/m3 v = = 4 m3/kg 1 ρ FIGURE 2–4 Density is mass per unit volume; specific volume is volume per unit mass.
TABLE 2–1 The specific gravity of some substances at 20°C and 1 atm unless stated otherwise Substance SG Water Blood (at 37°C) Seawater Gasoline Ethyl alcohol Mercury Balsa wood Dense oak wood Gold Bones Ice (at 0°C) Air 1.0 1.06 1.025 0.68 0.790 13.6 0.17 0.93 19.3 1.7–2.0 0.916 0.001204 cen96537_ch02_037-076.indd 39 29/12/16 5:59 pm 40 PROPERTIES OF FLUIDS The weight of a unit volume of a substance is called specific weight or weight density and is expressed as Specific weight: 𝛾s = ρg (N/m3) (2–3) where g is the gravitational acceleration. Recall from Chap. 1 that the densities of liquids are essentially constant, and thus they can often be approximated as being incompressible substances during most processes without sacrificing much in accuracy.
Density of Ideal Gases Property tables provide very accurate and precise information about the properties, but sometimes it is convenient to have some simple relations among the properties that are sufficiently general and reasonably accurate. Any equation that relates the pressure, temperature, and density (or specific volume) of a substance is called an equation of state. The simplest and best-known equation of state for substances in the gas phase is the ideal-gas equation of state, expressed as Pv = RT or P = ρRT (2–4) where P is the absolute pressure, v is the specific volume, T is the thermo dynamic (absolute) temperature, 𝜌 is the density, and R is the gas constant. The gas constant R is different for each gas and is determined from R = Ru /M, where Ru is the universal gas constant whose value is Ru = 8.314 kJ/ kmol·K = 1.986 Btu/lbmol·R, and M is the molar mass (also called molecu lar weight) of the gas. The values of R and M for several substances are given in Table A–1. The thermodynamic temperature scale in the SI is the Kelvin scale, and the temperature unit on this scale is the kelvin, designated by K. In the Eng lish system, it is the Rankine scale, and the temperature unit on this scale is the rankine, R. Various temperature scales are related to each other by T(K) = T(°C) + 273.15 = T(R)/1.8 (2–5) T(R) = T(°F) + 459.67 = 1.8 T(K) (2–6) It is common practice to round the constants 273.15 and 459.67 to 273 and 460, respectively, but we do not encourage this practice. Equation 2–4, the ideal-gas equation of state, is also called simply the ideal-gas relation, and a gas that obeys this relation is called an ideal gas. For an ideal gas of volume V, mass m, and number of moles N = m/M, the ideal-gas equation of state can also be written as PV = mRT or PV = NRuT. For a fixed mass m, writing the ideal-gas relation twice and simplifying, the properties of an ideal gas at two different states are related to each other by P1V1/T1 = P2V2/T2. An ideal gas is a hypothetical substance that obeys the relation Pv = RT. It has been experimentally observed that the ideal-gas relation closely approxi mates the P-v-T behavior of real gases at low densities. At low pressures and high temperatures, the density of a gas decreases and the gas behaves like an ideal gas (Fig. 2 –5). In the range of practical interest, many familiar FIGURE 2–5 Air behaves as an ideal gas, even at very high speeds. In this schlieren image, a bullet traveling at about the speed of sound bursts through both sides of a balloon, forming two expanding shock waves. The turbulent wake of the bullet is also visible.
© G.S. Settles, Gas Dynamics Lab, Penn State University. Used with permission.
cen96537_ch02_037-076.indd 40 29/12/16 5:59 pm 41 CHAPTER 2 gases such as air, nitrogen, oxygen, hydrogen, helium, argon, neon, and car bon dioxide and even heavier gases such as krypton can be treated as ideal gases with negligible error (often less than 1 percent). Dense gases such as water vapor in steam power plants and refrigerant vapor in refrigerators, air conditioners, and heat pumps, however, should not be treated as ideal gases since they usually exist at a state near saturation.
EXAMPLE 2–1 Density, Specific Gravity, and Mass of Air in a Room Determine the density, specific gravity, and mass of the air in a room whose dimensions are 4 m × 5 m × 6 m at 100 kPa and 25°C (Fig. 2–6).
SOLUTION The density, specific gravity, and mass of the air in a room are to be determined.
Assumptions At specified conditions, air can be treated as an ideal gas.
Properties The gas constant of air is R = 0.287 kPa⋅m3/kg⋅K.
Analysis The density of the air is determined from the ideal-gas relation P = 𝜌RT to be ρ = P RT = 100 kPa (0.287 kPa·m3/kg·K)(25 + 273.15) K = 1.17 kg/m3 Then the specific gravity of the air becomes SG = ρ ρH2O = 1.17 kg/m3 1000 kg/m3 = 0.00117 Finally, the volume and the mass of the air in the room are V = (4 m)(5 m)(6 m) = 120 m3 m = ρV = (1.17 kg/m3)(120 m3) = 140 kg Discussion Note that we converted the temperature to (absolute) unit K from (relative) unit °C before using it in the ideal-gas relation.
2–3 ■ VAPOR PRESSURE AND CAVITATION It is well-established that temperature and pressure are dependent proper ties for pure substances during phase-change processes, and there is one-to-one correspondence between temperature and pressure. At a given pres sure, the temperature at which a pure substance changes phase is called the saturation temperature Tsat. Likewise, at a given temperature, the pressure at which a pure substance changes phase is called the saturation pressure Psat. At an absolute pressure of 1 standard atmosphere (1 atm or 101.325 kPa), for example, the saturation temperature of water is 100°C. Conversely, at a temperature of 100°C, the saturation pressure of water is 1 atm. The vapor pressure Pv of a pure substance is defined as the pressure exerted by its vapor in phase equilibrium with its liquid at a given tempera ture (Fig. 2–7). Pv is a property of the pure substance, and turns out to be identical to the saturation pressure Psat of the liquid (Pv = Psat). We must be 6 m 4 m 5 m Air P = T = 100 kPa 25°C FIGURE 2–6 Schematic for Example 2–1.
Water molecules—vapor phase Water molecules—liquid phase FIGURE 2–7 The vapor pressure (saturation pressure) of a pure substance (e.g., water) is the pressure exerted by its vapor molecules when the system is in phase equilibrium with its liquid molecules at a given temperature.
cen96537_ch02_037-076.indd 41 29/12/16 6:00 pm 42 PROPERTIES OF FLUIDS careful not to confuse vapor pressure with partial pressure. Partial pressure is defined as the pressure of a gas or vapor in a mixture with other gases. For example, atmospheric air is a mixture of dry air and water vapor, and atmospheric pressure is the sum of the partial pressure of dry air and the par tial pressure of water vapor. The partial pressure of water vapor constitutes a small fraction (usually under 3 percent) of the atmo spheric pressure since air is mostly nitrogen and oxygen. The partial pressure of a vapor must be less than or equal to the vapor pressure if there is no liquid present. However, when both vapor and liquid are present and the system is in phase equilib rium, the partial pressure of the vapor must equal the vapor pressure, and the system is said to be saturated. The rate of evaporation from open water bodies such as lakes is controlled by the difference between the vapor pres sure and the partial pressure. For example, the vapor pressure of water at 20°C is 2.34 kPa. Therefore, a bucket of water at 20°C left in a room with dry air at 1 atm will continue evaporating until one of two things happens: the water evaporates away (there is not enough water to establish phase equi librium in the room), or the evaporation stops when the partial pressure of the water vapor in the room rises to 2.34 kPa at which point phase equilib rium is established. For phase-change processes between the liquid and vapor phases of a pure substance, the saturation pressure and the vapor pressure are equivalent since the vapor is pure. Note that the pressure value would be the same whether it is measured in the vapor or liquid phase (provided that it is measured at a loca tion close to the liquid–vapor interface to avoid any hydrostatic effects). Vapor pressure increases with temperature. Thus, a substance at higher pressure boils at higher temperature. For example, water boils at 134°C in a pressure cooker operating at 3 atm absolute pressure, but it boils at 93°C in an ordinary pan at a 2000-m elevation, where the atmospheric pressure is 0.8 atm. The saturation (or vapor) pressures are given in Appendices 1 and 2 for various substances. An abridged table for water is given in Table 2–2 for easy reference. The reason for our interest in vapor pressure is the possibility of the liquid pressure in liquid-flow systems dropping below the vapor pressure at some locations, and the resulting unplanned vaporization. For example, water at 10°C may vaporize and form bubbles at locations (such as the tip regions of impellers or suction sides of pumps) where the pressure drops below 1.23 kPa. The vapor bubbles (called cavitation bubbles since they form “cavities” in the liquid) collapse as they are swept away from the low-pressure regions, generating highly destructive, extremely high-pressure waves. This phenom enon, which is a common cause for drop in performance and even the erosion of impeller blades, is called cavitation, and it is an important consideration in the design of hydraulic turbines and pumps. Cavitation must be avoided (or at least minimized) in most flow systems since it reduces performance, generates annoying vibrations and noise, and causes damage to equipment. We note that some flow systems use cavita tion to their advantage, e.g., high-speed “supercavitating” torpedoes. The pressure spikes resulting from the large number of bubbles collapsing near a solid surface over a long period of time may cause erosion, surface pit ting, fatigue failure, and the eventual destruction of the components or machinery (Fig. 2–8). The presence of cavitation in a flow system can be sensed by its characteristic tumbling sound.
TABLE 2–2 Saturation (or vapor) pressure of water at various temperatures Temperature T, °C Saturation Pressure Psat, kPa −10 −5 0 5 10 15 20 25 30 40 50 100 150 200 250 300 0.260 0.403 0.611 0.872 1.23 1.71 2.34 3.17 4.25 7.38 12.35 101.3 (1 atm) 475.8 1554 3973 8581 FIGURE 2–8 Cavitation damage on a 16-mm by 23-mm aluminum sample tested at 60 m/s for 2.5 hours. The sample was located at the cavity collapse region downstream of a cavity generator specifically designed to produce high damage potential.
Photograph by David Stinebring, ARL/ Pennsylvania State University. Used by permission.
cen96537_ch02_037-076.indd 42 29/12/16 6:00 pm 43 CHAPTER 2 2–4 ■ ENERGY AND SPECIFIC HEATS Energy can exist in numerous forms such as thermal, mechanical, kinetic, potential, electrical, magnetic, chemical, and nuclear (Fig. 2–9) and their sum constitutes the total energy E (or e on a unit mass basis) of a system. The forms of energy related to the molecular structure of a system and the degree of the molecular activity are referred to as the microscopic energy. The sum of all microscopic forms of energy is called the internal energy of a system, and is denoted by U (or u on a unit mass basis). The macroscopic energy of a system is related to motion and the influence of some external effects such as gravity, magnetism, electricity, and surface tension. The energy that a system possesses as a result of its motion is called kinetic energy. When all parts of a system move with the same velocity, the kinetic energy per unit mass is expressed as ke = V 2/2 where V denotes the velocity of the system relative to some fixed reference frame. The energy that a system possesses as a result of its elevation in a gravitational field is called potential energy and is expressed on a per-unit mass basis as pe = gz where g is the gravitational acceleration and z is the elevation of the center of gravity of the system relative to some arbitrarily selected reference plane. In daily life, we frequently refer to the sensible and latent forms of internal energy as heat, and we talk about the heat content of bodies. In engineering, however, those forms of energy are usually referred to as thermal energy to prevent any confusion with heat transfer. The international unit of energy is the joule (J) or kilojoule (1 kJ = 1000 J). A joule is 1 N times 1 m. In the English system, the unit of energy is the British thermal unit (Btu), which is defined as the energy needed to raise the temperature of 1 lbm of water at 68°F by 1°F. The magnitudes of kJ and Btu are almost identical (1 Btu = 1.0551 kJ). Another well-known unit of energy is the calorie (1 cal = 4.1868 J), which is defined as the energy needed to raise the temperature of 1 g of water at 14.5°C by 1°C.
EXAMPLE 2–2 Danger of Cavitation in a Propeller The analysis of a propeller that operates in water at 20°C shows that the pressure at the tips of the propeller drops to 2 kPa at high speeds. Determine if there is a danger of cavitation for this propeller. SOLUTION The minimum pressure in a propeller is given. It is to be determined if there is a danger of cavitation.
Properties The vapor pressure of water at 20°C is 2.34 kPa (Table 2–2).
Analysis To avoid cavitation, the pressure everywhere in the flow should remain above the vapor (or saturation) pressure at the given temperature, which is Pv = Psat@20°C = 2.34 kPa The pressure at the tip of the propeller is 2 kPa, which is less than the vapor pressure. Therefore, there is a danger of cavitation for this propeller.
Discussion Note that the vapor pressure increases with increasing temperature, and thus there is a greater danger of cavitation at higher fluid temperatures.
(a) FIGURE 2–9 At least six different forms of energy are encountered in bringing power from a nuclear plant to your home, nuclear, thermal, mechanical, kinetic, magnetic, and electrical.
(a) © Creatas/PunchStock RF (b) Comstock Images/Jupiterimages RF (b) cen96537_ch02_037-076.indd 43 29/12/16 6:00 pm 44 PROPERTIES OF FLUIDS In the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and Pv. For convenience, this combination is called enthalpy h. That is, Enthalpy: h = u + Pv = u + P ρ (2–7) where P/𝜌 is the flow energy, also called the flow work, which is the energy per unit mass needed to move the fluid and maintain flow. In the energy analysis of flowing fluids, it is convenient to treat the flow energy as part of the energy of the fluid and to represent the microscopic energy of a fluid stream by enthalpy h (Fig. 2–10). Note that enthalpy is a quantity per unit mass, and thus it is a specific property. In the absence of such effects as magnetic, electric, and surface tension, a system is called a simple compressible system. The total energy of a simple compressible system consists of three parts: internal, kinetic, and potential energies. On a unit-mass basis, it is expressed as e = u + ke + pe. The fluid entering or leaving a control volume possesses an additional form of energy—the flow energy P/𝜌. Then the total energy of a flowing fluid on a unit-mass basis becomes eflowing = P/ρ + e = h + ke + pe = h + V 2 2 + gz (kJ/kg) (2–8) where h = P/𝜌 + u is the enthalpy, V is the magnitude of velocity, and z is the elevation of the system relative to some external reference point. By using the enthalpy instead of the internal energy to represent the energy of a flowing fluid, we do not need to be concerned about the flow work. The energy associated with pushing the fluid is automatically taken care of by enthalpy. In fact, this is the main reason for defining the property enthalpy. The differential and finite changes in the internal energy and enthalpy of an ideal gas can be expressed in terms of the specific heats as du = cv dT and dh = cp dT (2–9) where cv and cp are the constant-volume and constant-pressure specific heats of the ideal gas. Using specific heat values at the average temperature, the finite changes in internal energy and enthalpy can be expressed approximately as Δu ≅cv,avg ΔT and Δh ≅cp,avg ΔT (2–10) For incompressible substances, the constant-volume and constant-pressure specific heats are identical. Therefore, cp ≅ cv ≅ c for liquids, and the change in the internal energy of liquids can be expressed as Δu ≅ cavg ΔT. Noting that 𝜌 = constant for incompressible substances, the differentiation of enthalpy h = u + P/𝜌 gives dh = du + dP/𝜌. Integrating, the enthalpy change becomes Δh = Δu + ΔP/ρ ≅cavg ΔT + ΔP/ρ (2–11) Therefore, Δh ≅ Δu ≅ cavg ΔT for constant-pressure processes, and Δh = ΔP/𝜌 for constant-temperature processes in liquids.
FIGURE 2–10 The internal energy u represents the microscopic energy of a nonflowing fluid per unit mass, whereas enthalpy h represents the microscopic energy of a flowing fluid per unit mass.
Energy = h Flowing fluid Energy = u Stationary fluid cen96537_ch02_037-076.indd 44 29/12/16 6:00 pm 45 CHAPTER 2 2–5 ■ COMPRESSIBILITY AND SPEED OF SOUND Coefficient of Compressibility We know from experience that the volume (or density) of a fluid changes with a change in its temperature or pressure. Fluids usually expand as they are heated or depressurized and contract as they are cooled or pressurized. But the amount of volume change is different for different fluids, and we need to define properties that relate volume changes to the changes in pres sure and temperature. Two such properties are the bulk modulus of elasticity κ and the coefficient of volume expansion 𝛽. It is a common observation that a fluid contracts when more pressure is applied on it and expands when the pressure acting on it is reduced (Fig. 2–11). That is, fluids act like elastic solids with respect to pressure. Therefore, in an analogous manner to Young’s modulus of elasticity for solids, it is appropriate to define a coefficient of compressibility κ (also called the bulk modulus of compressibility or bulk modulus of elasticity) for fluids as 𝜅= −v( ∂P ∂v) T = ρ( ∂P ∂ρ) T (Pa) (2–12) It can also be expressed approximately in terms of finite changes as 𝜅≅−ΔP Δv/v ≅ΔP Δρ/ρ (T = constant) (2–13) Noting that Δv/v or Δ𝜌/𝜌 is dimensionless, κ must have the dimension of pressure (Pa or psi). Also, the coefficient of compressibility represents the change in pressure corresponding to a fractional change in volume or density of the fluid while the temperature remains constant. Then it follows that the coefficient of compressibility of a truly incompressible substance (v = constant) is infinity. A large value of κ indicates that a large change in pressure is needed to cause a small fractional change in volume, and thus a fluid with a large κ is essentially incompressible. This is typical for liquids, and explains why liquids are usually considered to be incompressible. For example, the pres sure of water at normal atmospheric conditions must be raised to 210 atm to compress it 1 percent, corresponding to a coefficient of compressibility value of κ = 21,000 atm. Small density changes in liquids can still cause interesting phenom ena in piping systems such as the water hammer—characterized by a sound that resembles the sound produced when a pipe is “hammered.” This occurs when a liquid in a piping network encounters an abrupt flow restriction (such as a closing valve) and is locally compressed. The acous tic waves that are produced strike the pipe surfaces, bends, and valves as they propagate and reflect along the pipe, causing the pipe to vibrate and produce the familiar sound. In addition to the irritating sound, water hammering can be quite destructive, leading to leaks or even structural damage. The effect can be suppressed with a water hammer arrestor P2 > P1 P1 FIGURE 2–11 Fluids, like solids, compress when the applied pressure is increased from P1 to P2.
cen96537_ch02_037-076.indd 45 29/12/16 6:00 pm 46 PROPERTIES OF FLUIDS (Fig. 2–12), which is a volumetric chamber containing either a bellows or piston to absorb the shock. For large pipes that carry liquids (e.g., hydro power penstocks), a vertical tube called a surge tower is often used. A surge tower has a free air surface at the top and is virtually maintenance free. For other pipelines, a closed hydraulic accumulator tank is used, in which a gas such as air or nitrogen is compressed in the accumulator tank to soften the shock. Note that volume and pressure are inversely proportional (volume decreases as pressure is increased and thus ∂P/∂v is a negative quantity), and the negative sign in the definition (Eq. 2–12) ensures that κ is a posi tive quantity. Also, differentiating 𝜌 = 1/v gives d𝜌 = −dv/v2, which can be rearranged as dρ ρ = −dv v (2–14) That is, the fractional changes in the specific volume and the density of a fluid are equal in magnitude but opposite in sign. For an ideal gas, P = 𝜌RT and (∂P/∂𝜌)T = RT = P/𝜌, and thus 𝜅ideal gas = P (Pa) (2–15) Therefore, the coefficient of compressibility of an ideal gas is equal to its absolute pressure, and the coefficient of compressibility of the gas increases with increasing pressure. Substituting κ = P into the definition of the coef ficient of compressibility and rearranging gives Ideal gas: Δρ ρ = ΔP P (T = constant) (2–16) Therefore, the percent increase of density of an ideal gas during isothermal compression is equal to the percent increase in pressure. For air at 1 atm pressure, κ = P = 1 atm and a decrease of 1 percent in volume (ΔV/V = −0.01) corresponds to an increase of ΔP = 0.01 atm in pressure. But for air at 1000 atm, κ = 1000 atm and a decrease of 1 percent in volume corresponds to an increase of ΔP = 10 atm in pressure. Therefore, a small fractional change in the volume of a gas can cause a large change in pressure at very high pressures. The inverse of the coefficient of compressibility is called the isothermal compressibility α and is expressed as 𝛼= 1 𝜅= −1 v ( ∂v ∂P) T = 1 ρ ( ∂ρ ∂P) T (1/Pa) (2–17) The isothermal compressibility of a fluid represents the fractional change in volume or density corresponding to a unit change in pressure.
Coefficient of Volume Expansion The density of a fluid, in general, depends more strongly on temperature than it does on pressure, and the variation of density with temperature is respon sible for numerous natural phenomena such as winds, currents in oceans, rise of plumes in chimneys, the operation of hot-air balloons, heat transfer by nat ural convection, and even the rise of hot air and thus the phrase “heat rises” FIGURE 2–12 Water hammer arrestors: (a) A large surge tower built to protect the pipeline against water hammer damage.
Photo by Arris S. Tijsseling, visitor of the University of Adelaide, Australia. Used by permission (b) Much smaller arrestors used for supplying water to a household washing machine.
Photo provided courtesy of Oatey Company (a) (b) cen96537_ch02_037-076.indd 46 29/12/16 6:00 pm 47 CHAPTER 2 (Fig. 2–13). To quantify these effects, we need a property that represents the variation of the density of a fluid with temperature at constant pressure. The property that provides that information is the coefficient of volume expansion (or volume expansivity) 𝛽, defined as (Fig. 2–14) 𝛽 = 1 v ( ∂v ∂T) P = −1 ρ ( ∂ρ ∂T) P (1/K) (2–18) It can also be expressed approximately in terms of finite changes as 𝛽≈Δv/v ΔT = −Δρ/ρ ΔT (at constant P) (2–19) A large value of 𝛽 for a fluid means a large change in density with tempera ture, and the product 𝛽 ΔT represents the fraction of volume change of a fluid that corresponds to a temperature change of ΔT at constant pressure. It can be shown that the volume expansion coefficient of an ideal gas (P = 𝜌RT) at a temperature T is equivalent to the inverse of the tempe rature: 𝛽ideal gas = 1 T (1/K) (2–20) where T is the absolute temperature. In the study of natural convection currents, the condition of the main fluid body that surrounds the finite hot or cold regions is indicated by the sub script “infinity” to serve as a reminder that this is the value at a distance where the presence of the hot or cold region is not felt. In such cases, the volume expansion coefficient can be expressed approximately as 𝛽≈ −(ρ∞−ρ)/ρ T∞−T or ρ∞−ρ = ρ𝛽(T −T∞) (2–21) where 𝜌∞ is the density and T∞ is the temperature of the quiescent fluid away from the confined hot or cold fluid pocket. We will see in Chap. 3 that natural convection currents are initiated by the buoyancy force, which is proportional to the density difference, which is in turn proportional to the temperature difference at constant pressure. Therefore, the larger the temperature difference between the hot or cold fluid pocket and the surrounding main fluid body, the larger the buoyancy force and thus the stronger the natural convection currents. A related phe nomenon sometimes occurs when an aircraft flies near the speed of sound. The sudden drop in temperature produces condensation of water vapor on a visible vapor cloud (Fig. 2–15). The combined effects of pressure and temperature changes on the volume change of a fluid can be determined by taking the specific volume to be a function of T and P. Differentiating v = v(T, P) and using the definitions of the compression and expansion coefficients α and 𝛽 give dv = ( ∂v ∂T) P dT + ( ∂v ∂P) T dP = (𝛽 dT −𝛼 dP)v (2–22) Then the fractional change in volume (or density) due to changes in pres sure and temperature can be expressed approximately as Δv v = −Δρ ρ ≅ 𝛽 ΔT −𝛼 ΔP (2–23) FIGURE 2–13 Natural convection over a woman’s hand. © G.S. Settles, Gas Dynamics Lab, Penn State University. Used with permission 20°C 100 kPa 1 kg 21°C 100 kPa 1 kg 20°C 100 kPa 1 kg 21°C 100 kPa 1 kg ΔT (a) A substance with a large β (b) A substance with a small β 휕v 휕T P ΔT 휕v 휕T P FIGURE 2–14 The coefficient of volume expansion is a measure of the change in volume of a substance with temperature at constant pressure.
cen96537_ch02_037-076.indd 47 29/12/16 6:00 pm 48 PROPERTIES OF FLUIDS FIGURE 2–15 Vapor cloud around an F/A-18F Super Hornet as it flies near the speed of sound.
U.S. Navy photo by Photographer's Mate 3rd Class Jonathan Chandler.
EXAMPLE 2–3 Variation of Density with Temperature and Pressure Consider water initially at 20°C and 1 atm. Determine the final density of the water (a) if it is heated to 50°C at a constant pressure of 1 atm, and (b) if it is compressed to 100-atm pressure at a constant temperature of 20°C. Take the isothermal com pressibility of water to be α = 4.80 × 10−5 atm−1.
SOLUTION Water at a given temperature and pressure is considered. The densities of water after it is heated and after it is compressed are to be determined.
Assumptions 1 The coefficient of volume expansion and the isothermal compressibility of water are constant in the given temperature range. 2 An approximate analysis is performed by replacing differential changes in quantities by finite changes.
Properties The density of water at 20°C and 1 atm pressure is 𝜌1 = 998.0 kg/m3. The coefficient of volume expansion at the average temperature of (20 + 50)/2 = 35°C is 𝛽 = 0.337 × 10−3 K−1. The isothermal compressibility of water is given to be α = 4.80 × 10−5 atm−1.
Analysis When differential quantities are replaced by differences and the properties α and 𝛽 are assumed to be constant, the change in density in terms of the changes in pressure and temperature is expressed approximately as (Eq. 2–23) Δρ = 𝛼ρ ΔP − 𝛽ρ ΔT (a) The change in density due to the change of temperature from 20°C to 50°C at constant pressure is Δρ = −𝛽ρ ΔT = −(0.337 × 10−3 K−1)(998 kg/m3)(50 −20) K = −10.0 kg/m3 Noting that Δ𝜌 = 𝜌2 − 𝜌1, the density of water at 50°C and 1 atm is ρ2 = ρ1 + Δρ = 998.0 + (−10.0) = 988.0 kg/m3 which is almost identical to the listed value of 988.1 kg/m3 at 50°C in Table A–3. This is mostly due to 𝛽 varying with temperature almost linearly, as shown in Fig. 2–16.
(b) The change in density due to a change of pressure from 1 atm to 100 atm at constant temperature is Δρ = 𝛼ρ ΔP = (4.80 × 10−5 atm−1)(998 kg/m3)(100 −1) atm = 4.7 kg/m3 Then the density of water at 100 atm and 20°C becomes ρ2 = ρ1 + Δρ = 998.0 + 4.7 = 1002.7 kg/m3 Discussion Note that the density of water decreases while being heated and increases while being compressed, as expected. This problem can be solved more accurately using differential analysis when functional forms of properties are available.
FIGURE 2–16 The variation of the coefficient of volume expansion 𝛽 of water with temperature in the range of 20°C to 50°C. Data were generated and plotted using EES.
0.00050 0.00020 50 45 40 35 30 25 T, °C 20 0.00025 0.00030 0.00035 0.00040 0.00045 β, 1/K cen96537_ch02_037-076.indd 48 29/12/16 6:00 pm 49 CHAPTER 2 Speed of Sound and Mach Number An important parameter in the study of compressible flow is the speed of sound (or the sonic speed), defined as the speed at which an infinitesi mally small pressure wave travels through a medium. The pressure wave may be caused by a small disturbance, which creates a slight change in local pressure. To obtain a relation for the speed of sound in a medium, consider a duct that is filled with a fluid at rest, as shown in Fig. 2–17. A piston fitted in the duct is now moved to the right with a constant incremental velocity dV, creating a sonic wave. The wave front moves to the right through the fluid at the speed of sound c and separates the moving fluid adjacent to the piston from the fluid still at rest. The fluid to the left of the wave front experiences an incremental change in its thermodynamic properties, while the fluid on the right of the wave front maintains its original thermodynamic properties, as shown in Fig. 2–17. To simplify the analysis, consider a control volume that encloses the wave front and moves with it, as shown in Fig. 2–18. To an observer traveling with the wave front, the fluid to the right appears to be moving toward the wave front with a speed of c and the fluid to the left to be moving away from the wave front with a speed of c − dV. Of course, the observer sees the control volume that encloses the wave front (and herself or himself ) as stationary, and the observer is witnessing a steady-flow process. The mass balance for this single-stream, steady-flow process is expressed as m · right = m · left or ρAc = (ρ + dρ)A(c −dV) By canceling the cross-sectional (or flow) area A and neglecting the higher-order terms, this equation reduces to c dρ −ρ dV = 0 No heat or work crosses the boundaries of the control volume during this steady-flow process, and the potential energy change can be neglected. Then the steady-flow energy balance ein = eout becomes h + c2 2 = h + dh + (c −dV)2 2 which yields dh −c dV = 0 where we have neglected the second-order term (dV)2. The amplitude of the ordinary sonic wave is very small and does not cause any appreciable change in the pressure and temperature of the fluid. Therefore, the propagation of a sonic wave is not only adiabatic but also very nearly isentropic. Then the thermody namic relation T ds = dh − dP/𝜌 (see Çengel and Boles, 2015) reduces to T ds = dh −dP ρ x dV + dρ ρ ρ Moving wave front Piston Stationary fluid P + dP h + dh P h dV V x 0 P + dP P P c FIGURE 2–17 Propagation of a small pressure wave along a duct. + ρ ρ ρ d Control volume traveling with the wave front P + dP h + dh P h c – dV c FIGURE 2–18 Control volume moving with the small pressure wave along a duct. 0 → cen96537_ch02_037-076.indd 49 29/12/16 6:00 pm 50 PROPERTIES OF FLUIDS or dh = dP ρ Combining the above equations yields the desired expression for the speed of sound as c2 = dP dρ at s = constant or c2 = ( ∂P ∂ρ) s (2–24) It is left as an exercise for the reader to show, by using thermodynamic property relations, that Eq. 2–24 can also be written as c2 = k( ∂P ∂ρ) T (2–25) where k = cp /cv is the specific heat ratio of the fluid. Note that the speed of sound in a fluid is a function of the thermodynamic properties of that fluid Fig. 2–19. When the fluid is an ideal gas (P = 𝜌RT), the differentiation in Eq. 2–25 can be performed to yield c2 = k( ∂P ∂ρ) T = k[ ∂(ρRT ) ∂ρ ] T = kRT or c = √kRT (2–26) Noting that the gas constant R has a fixed value for a specified ideal gas and the specific heat ratio k of an ideal gas is, at most, a function of tempera ture, we see that the speed of sound in a specified ideal gas is a function of temperature alone (Fig. 2–20). A second important parameter in the analysis of compressible fluid flow is the Mach number Ma, named after the Austrian physicist Ernst Mach (1838–1916). It is the ratio of the actual speed of the fluid (or an object in still fluid) to the speed of sound in the same fluid at the same state: Ma = V c (2–27) Mach number can also be defined as the ratio of inertial forces to elastic forces. If Ma is less than about 1/3, the flow may be approximated as incom pressible since the effects of compressibility become significant only when the Mach number exceeds this value. Note that the Mach number depends on the speed of sound, which depends on the state of the fluid. Therefore, the Mach number of an aircraft cruising at constant velocity in still air may be different at different locations (Fig. 2–21). Fluid flow regimes are often described in terms of the flow Mach number. The flow is called sonic when Ma = 1, subsonic when Ma < 1, supersonic when Ma > 1, hypersonic when Ma > > 1, and transonic when Ma ≅ 1.
FIGURE 2–19 The speed of sound in air increases with temperature. At typical outside temperatures, c is about 340 m/s. In round numbers, therefore, the sound of thunder from a lightning strike travels about 1 km in 3 seconds. If you see the lightning and then hear the thunder less than 3 seconds later, you know that the lightning is close, and it is time to go indoors!
© Bear Dancer Studios/Mark Dierker Air Helium 347 m/s 634 m/s 200 K 300 K 1000 K 284 m/s 1861 m/s 1019 m/s 832 m/s FIGURE 2–20 The speed of sound changes with temperature and varies with the fluid.
cen96537_ch02_037-076.indd 50 29/12/16 6:00 pm 51 CHAPTER 2 2–6 ■ VISCOSITY When two solid bodies in contact move relative to each other, a friction force develops at the contact surface in the direction opposite to motion. To move a table on the floor, for example, we have to apply a force to the table in the horizontal direction large enough to overcome the friction force. The magnitude of the force needed to move the table depends on the friction coefficient between the table legs and the floor. The situation is similar when a fluid moves relative to a solid or when two fluids move relative to each other. We move with relative ease in air, but not so in water. Moving in oil would be even more difficult, as can be observed by the slower downward motion of a glass ball dropped in a tube filled with oil. It appears that there is a property that represents the internal resistance of a fluid to motion or the “fluidity,” and that property is the viscosity. The force a flowing fluid exerts on a body in the flow direction is called the drag force, and the magnitude of this force depends, in part, on viscosity (Fig. 2–23). To obtain a relation for viscosity, consider a fluid layer between two very large parallel plates (or equivalently, two parallel plates immersed in a large body of a fluid) separated by a distance 𝓁 (Fig. 2–24). Now a constant par allel force F is applied to the upper plate while the lower plate is held fixed. After the initial transients, it is observed that the upper plate moves continu ously under the influence of this force at a constant speed V. The fluid in contact with the upper plate sticks to the plate surface and moves with it at the same speed, and the shear stress 𝜏 acting on this fluid layer is 𝜏= F A (2–28) EXAMPLE 2–4 Mach Number of Air Entering a Diffuser Air enters a diffuser shown in Fig. 2–22 with a speed of 200 m/s. Determine (a) the speed of sound and (b) the Mach number at the diffuser inlet when the air tempera ture is 30°C.
SOLUTION Air enters a diffuser at high speed. The speed of sound and the Mach number are to be determined at the diffuser inlet.
Assumption Air at the specified conditions behaves as an ideal gas.
Properties The gas constant of air is R = 0.287 kJ/kg·K, and its specific heat ratio at 30°C is 1.4.
Analysis We note that the speed of sound in a gas varies with temperature, which is given to be 30°C.
(a) The speed of sound in air at 30°C is determined from Eq. 2–26 to be c = √kRT = √(1.4)(0.287 kJ/kg·K)(303 K)( 1000 m2/s2 1 kJ/kg ) = 349 m/s (b) Then the Mach number becomes Ma = V c = 200 m/s 349 m/s = 0.573 Discussion The flow at the diffuser inlet is subsonic since Ma < 1.
V = 320 m/s Air 220 K Ma = 1.08 V = 320 m/s Air 300 K Ma = 0.92 FIGURE 2–21 The Mach number can be different at different temperatures even if the flight speed is the same.
© Purestock/SuperStock RF Diffuser V = 200 m/s T = 30°C Air FIGURE 2–22 Schematic for Example 12–4.
Water Air Drag force Drag force Drag force V V Water FIGURE 2–23 A fluid moving relative to a body exerts a drag force on the body, partly because of friction caused by viscosity.
Top: © Photodisc/Getty Images RF Bottom: © Digital Vision/Getty Images RF cen96537_ch02_037-076.indd 51 29/12/16 6:00 pm 52 PROPERTIES OF FLUIDS where A is the contact area between the plate and the fluid. Note that the fluid layer deforms continuously under the influence of shear stress. The fluid in contact with the lower plate assumes the velocity of that plate, which is zero (because of the no-slip condition—see Section 1–2). In steady laminar flow, the fluid velocity between the plates varies linearly between 0 and V, and thus the velocity profile and the velocity gradient are u( y) = y ℓV and du dy = V ℓ (2–29) where y is the vertical distance from the lower plate. During a differential time interval dt, the sides of fluid particles along a vertical line MN rotate through a differential angle d𝛽 while the upper plate moves a differential distance da = V dt. The angular displacement or defor mation (or shear strain) can be expressed as d𝛽 ≈ tan d𝛽= da ℓ= V dt ℓ = du dy dt (2–30) Rearranging, the rate of deformation under the influence of shear stress 𝜏 becomes d𝛽 dt = du dy (2–31) Thus we conclude that the rate of deformation of a fluid element is equiva lent to the velocity gradient du/dy. Further, it can be verified experimentally that for most fluids the rate of deformation (and thus the velocity gradient) is directly proportional to the shear stress 𝜏, 𝜏 ∝ d𝛽 dt or 𝜏 ∝ du dy (2–32) Fluids for which the rate of deformation is linearly proportional to the shear stress are called Newtonian fluids after Sir Isaac Newton, who expressed it first in 1687. Most common fluids such as water, air, gasoline, and oils are Newtonian fluids. Blood and liquid plastics are examples of non-Newtonian fluids. In one-dimensional shear flow of Newtonian fluids, shear stress can be expressed by the linear relationship Shear stress: 𝜏 = 𝜇 du dy (N/m2) (2–33) where the constant of proportionality 𝜇 is called the coefficient of viscosity or the dynamic (or absolute) viscosity of the fluid, whose unit is kg/m·s, or equivalently, N·s/m2 (or Pa⋅s where Pa is the pressure unit pascal). A common viscosity unit is poise, which is equivalent to 0.1 Pa⋅s (or cen tipoise, which is one-hundredth of a poise). The viscosity of water at 20°C is 1.002 centipoise, and thus the unit centipoise serves as a useful reference. A plot of shear stress versus the rate of deformation (velocity gradient) for a Newtonian fluid is a straight line whose slope is the viscosity of the fluid, as shown in Fig. 2–25. Note that viscosity is independent of the rate of defor mation for Newtonian fluids. Since the rate of deformation is proportional to the strain rate, Fig. 2–25 reveals that viscosity is actually a coefficient in a stress–strain relationship.
V Velocity V u(y) = u = 0 u = V y ℓ ℓ N da M N″ Velocity profile Force F x y dβ Area A FIGURE 2–24 The behavior of a fluid in laminar flow between two parallel plates when the upper plate moves with a constant velocity.
Rate of deformation, du/dy Shear stress, τ Oil Water Air Viscosity = Slope μ = = a a b b du/dy τ FIGURE 2–25 The rate of deformation (velocity gradient) of a Newtonian fluid is proportional to shear stress, and the constant of proportionality is the viscosity.
cen96537_ch02_037-076.indd 52 29/12/16 6:00 pm 53 CHAPTER 2 The shear force acting on a Newtonian fluid layer (or, by Newton’s third law, the force acting on the plate) is Shear force: F = 𝜏A = 𝜇A du dy (N) (2–34) where again A is the contact area between the plate and the fluid. Then the force F required to move the upper plate in Fig. 2–24 at a constant speed of V while the lower plate remains stationary is F = 𝜇A V ℓ (N) (2–35) This relation can alternately be used to calculate 𝜇 when the force F is measured. Therefore, the experimental setup just described can be used to measure the viscosity of fluids. Note that under identical conditions, the force F would be very different for different fluids. For non-Newtonian fluids, the relationship between shear stress and rate of deformation is not linear, as shown in Fig. 2–26. The slope of the curve on the 𝜏 versus du/dy chart is referred to as the apparent viscosity of the fluid. Fluids for which the apparent viscosity increases with the rate of deformation (such as solutions with suspended starch or sand) are referred to as dilatant or shear thickening fluids, and those that exhibit the oppo site behavior (the fluid becoming less viscous as it is sheared harder, such as some paints, polymer solutions, and fluids with suspended particles) are referred to as pseudoplastic or shear thinning fluids. Some materials such as toothpaste can resist a finite shear stress and thus behave as a solid, but deform continuously when the shear stress exceeds the yield stress and behave as a fluid. Such materials are referred to as Bingham plastics after Eugene C. Bingham (1878–1945), who did pioneering work on fluid viscos ity for the U.S. National Bureau of Standards in the early twentieth century. In fluid mechanics and heat transfer, the ratio of dynamic viscosity to density appears frequently. For convenience, this ratio is given the name kinematic viscosity ν and is expressed as ν = 𝜇/𝜌. Two common units of kinematic viscosity are m2/s and stoke (1 stoke = 1 cm2/s = 0.0001 m2/s). In general, the viscosity of a fluid depends on both temperature and pres sure, although the dependence on pressure is rather weak. For liquids, both the dynamic and kinematic viscosities are practically independent of pres sure, and any small variation with pressure is usually disregarded, except at extremely high pressures. For gases, this is also the case for dynamic vis cosity (at low to moderate pressures), but not for kinematic viscosity since the density of a gas is proportional to its pressure (Fig. 2–27). The viscosity of a fluid is a measure of its resistance to the rate of defor mation. Viscosity is due to the internal frictional force that develops between different layers of fluids as they are forced to move relative to each other. The viscosity of a fluid is directly related to the pumping power needed to transport a fluid in a pipe or to move a body (such as a car in air or a submarine in the sea) through a fluid. Viscosity is caused by the cohe sive forces between the molecules in liquids and by the molecular collisions in gases, and it varies greatly with temperature. The viscosity of liquids decreases with temperature, whereas the viscosity of gases increases with Rate of deformation, du/dy Shear stress, τ Bingham plastic Pseudoplastic Newtonian Dilatant FIGURE 2–26 Variation of shear stress with the rate of deformation for Newtonian and non-Newtonian fluids (the slope of a curve at a point is the apparent viscosity of the fluid at that point).
Air at 20°C and 1 atm: μ = 1.83 × 10–5 kg/m⋅s ν = 1.52 × 10–5 m2/s Air at 20°C and 4 atm: μ = 1.83 × 10–5 kg/m⋅s ν = 0.380 × 10–5 m2/s FIGURE 2–27 Dynamic viscosity, in general, does not depend on pressure, but kinematic viscosity does.
cen96537_ch02_037-076.indd 53 29/12/16 6:00 pm 54 PROPERTIES OF FLUIDS temperature (Fig. 2–28). This is because in a liquid the molecules possess more energy at higher temperatures, and they can oppose the large cohesive intermolecular forces more strongly. As a result, the energized liquid mol ecules can move more freely. In a gas, on the other hand, the intermolecular forces are negligible, and the gas molecules at high temperatures move randomly at higher velocities. This results in more molecular collisions per unit volume per unit time and therefore in greater resistance to flow. The kinetic theory of gases predicts the viscosity of gases to be proportional to the square root of temperature. That is, 𝜇gas ∝√T. This prediction is confirmed by practical observations, but deviations for different gases need to be accounted for by incorporat ing some correction factors. The viscosity of gases is expressed as a func tion of temperature by the Sutherland correlation (from The U.S. Standard Atmosphere) as Gases: 𝜇= aT 1/2 1 + b/T (2–36) where T is absolute temperature and a and b are experimentally determined constants. Note that measuring viscosity at two different temperatures is sufficient to determine these constants. For air at atmospheric conditions, the values of these constants are a = 1.458 × 10−6 kg/(m⋅s⋅K1/2) and b = 110.4 K. The viscosity of gases is independent of pressure at low to moderate pressures (from a few percent of 1 atm to several atm). But vis cosity increases at high pressures due to the increase in density. For liquids, the viscosity is approximated as Liquids: 𝜇= a10 b/(T−c) (2–37) where again T is absolute temperature and a, b, and c are experimentally determined constants. For water, using the values a = 2.414 × 10−5 N⋅s/m2, b = 247.8 K, and c = 140 K results in less than 2.5 percent error in viscosity in the temperature range of 0°C to 370°C (Touloukian et al., 1975). The viscosities of some fluids at room temperature are listed in Table 2–3. They are plotted against temperature in Fig. 2–29. Note that the viscosities of different fluids differ by several orders of magnitude. Also note that it is more difficult to move an object in a higher-viscosity fluid such as engine oil than it is in a lower-viscosity fluid such as water. Liquids, in general, are much more viscous than gases. Consider a fluid layer of thickness ℓ within a small gap between two con centric cylinders, such as the thin layer of oil in a journal bearing. The gap between the cylinders can be modeled as two parallel flat plates separated by the fluid. Noting that torque is T = FR (force times the moment arm, which is the radius R of the inner cylinder in this case), the tangential velocity is V = 𝜔R (angular velocity times the radius), and taking the wetted surface area of the inner cylinder to be A = 2πRL by disregarding the shear stress acting on the two ends of the inner cylinder, torque can be expressed as T = FR = 𝜇 2𝜋R3𝜔L ℓ = 𝜇 4𝜋2R3n . L ℓ (2–38) where L is the length of the cylinder and n . is the number of revolutions per unit time, which is usually expressed in rpm (revolutions per minute). Note Liquids Gases Temperature Viscosity FIGURE 2–28 The viscosity of liquids decreases and the viscosity of gases increases with temperature.
TABLE 2–3 Dynamic viscosity of some fluids at 1 atm and 20°C (unless otherwise stated) Fluid Dynamic Viscosity 𝜇, kg/m⋅s Glycerin: −20°C 0°C 20°C 40°C Engine oil: SAE 10W SAE 10W30 SAE 30 SAE 50 Mercury Ethyl alcohol Water: 0°C 20°C 100°C (liquid) 100°C (vapor) Blood, 37°C Gasoline Ammonia Air Hydrogen, 0°C 134.0 10.5 1.52 0.31 0.10 0.17 0.29 0.86 0.0015 0.0012 0.0018 0.0010 0.00028 0.000012 0.00040 0.00029 0.00015 0.000018 0.0000088 cen96537_ch02_037-076.indd 54 29/12/16 6:00 pm 55 CHAPTER 2 that the angular distance traveled during one rotation is 2π rad, and thus the relation between the angular velocity in rad/min and the rpm is 𝜔 = 2πn .. Equation 2–38 can be used to calculate the viscosity of a fluid by measuring torque at a specified angular velocity. Therefore, two concentric cylinders can be used as a viscometer, a device that measures viscosity.
0.5 0.4 0.3 0.2 SAE 10 oil Crude oil (SG = 0.86) Kerosene Aniline Mercury Castor oil Glycerin SAE 30 oil 0.1 0.06 0.04 0.03 0.02 0.01 6 4 3 2 1 × 10–3 6 4 3 2 1 × 10–4 4 3 2 1 × 10–5 5 –20 0 20 40 Temperature, °C 60 80 100 120 6 Carbon tetrachloride Ethyl alcohol Water Gasoline (SG = 0.68) Helium Air Carbon Dioxide Hydrogen Benzene Absolute viscosity μ, N⋅s/m2 FIGURE 2–29 The variation of dynamic (absolute) viscosity of common fluids with temperature at 1 atm (1 N⋅s/m2 = 1 kg/m⋅s = 0.020886 lbf⋅s/ft2).
Data from EES and F. M. White, Fluid Mechanics 7e. Copyright © 2011 The McGraw-Hill Companies, Inc. R Shaft Stationary cylinder Fluid ℓ n = 300 rpm .
FIGURE 2–30 Schematic for Example 2–5 (not to scale).
EXAMPLE 2–5 Determining the Viscosity of a Fluid The viscosity of a fluid is to be measured by a viscometer constructed of two 40-cm-long concentric cylinders (Fig. 2–30). The outer diameter of the inner cylinder is 12 cm, and the gap between the two cylinders is 0.15 cm. The inner cylinder is rotated at 300 rpm, and the torque is measured to be 1.8 N⋅m. Determine the viscosity of the fluid.
SOLUTION The torque and the rpm of a double cylinder viscometer are given. The viscosity of the fluid is to be determined.
Assumptions 1 The inner cylinder is completely submerged in the fluid. 2 The viscous effects on the two ends of the inner cylinder are negligible.
cen96537_ch02_037-076.indd 55 29/12/16 6:00 pm 56 PROPERTIES OF FLUIDS 2–7 ■ SURFACE TENSION AND CAPILLARY EFFECT It is often observed that a drop of blood forms a hump on a horizontal glass; a drop of mercury forms a near-perfect sphere and can be rolled just like a steel ball over a smooth surface; water droplets from rain or dew hang from branches or leaves of trees; a liquid fuel injected into an engine forms a mist of spherical droplets; water dripping from a leaky faucet falls as nearly spherical droplets; a soap bubble released into the air forms a nearly spherical shape; and water beads up into small drops on flower pet als (Fig. 2–31a). In these and other observances, liquid droplets behave like small balloons filled with the liquid, and the surface of the liquid acts like a stretched elas tic membrane under tension. The pulling force that causes this tension acts parallel to the surface and is due to the attractive forces between the mol ecules of the liquid. The magnitude of this force per unit length is called surface tension or coefficient of surface tension 𝜎s and is usually expressed in the unit N/m (or lbf/ft in English units). This effect is also called surface energy (per unit area) and is expressed in the equivalent unit of N⋅m/m2 or J/m2. In this case, 𝜎s represents the stretching work that needs to be done to increase the surface area of the liquid by a unit amount. To visualize how surface tension arises, we present a microscopic view in Fig. 2–32 by considering two liquid molecules, one at the surface and one deep within the liquid body. The attractive forces applied on the inte rior molecule by the surrounding molecules balance each other because of symmetry. But the attractive forces acting on the surface molecule are not symmetric, and the attractive forces applied by the gas molecules above are usually very small. Therefore, there is a net attractive force act ing on the molecule at the surface of the liquid, which tends to pull the molecules on the surface toward the interior of the liquid. This force is balanced by the repulsive forces from the molecules below the surface that are trying to be compressed. The result is that the liquid minimizes its surface area. This is the reason for the tendency of liquid droplets to attain a spherical shape, which has the minimum surface area for a given volume.
Analysis The velocity profile is linear only when the curvature effects are neg ligible, and the profile can be approximated as being linear in this case since ℓ/R = 0.025 < < 1. Solving Eq. 2–38 for viscosity and substituting the given values, the viscosity of the fluid is determined to be 𝜇= Tℓ 4𝜋2R3n . L = (1.8 N·m)(0.0015 m) 4𝜋2(0.06 m)3(300 1 min) ( 1 min 60 s )(0.4 m) = 0.158 N·s/m2 Discussion Viscosity is a strong function of temperature, and a viscosity value without a corresponding temperature is of little usefulness. Therefore, the tem perature of the fluid should have also been measured during this experiment, and reported with this calculation.
FIGURE 2–31 Some consequences of surface tension: (a) drops of water beading up on a leaf, (b) a water strider sitting on top of the surface of water, and (c) a color schlieren image of the water strider revealing how the water surface dips down where its feet contact the water (it looks like two insects but the second one is just a shadow).
(a) © Don Paulson Photography/Purestock/ SuperStock RF (b) NPS Photo by Rosalie LaRue (c) © G.S. Settles, Gas Dynamics Lab, Penn State University. Used with permission.
(a) (b) (c) A molecule on the surface A molecule inside the liquid FIGURE 2–32 Attractive forces acting on a liquid molecule at the surface and deep inside the liquid.
cen96537_ch02_037-076.indd 56 29/12/16 6:00 pm 57 CHAPTER 2 You also may have observed, with amusement, that some insects can land on water or even walk on water (Fig. 2–31b) and that small steel needles can float on water. These phenomena are made possible by surface tension which balances the weights of these objects. To understand the surface tension effect better, consider a liquid film (such as the film of a soap bubble) suspended on a U-shaped wire frame with a movable side (Fig. 2–33). Normally, the liquid film tends to pull the movable wire inward in order to minimize its surface area. A force F needs to be applied on the movable wire in the opposite direction to balance this pull ing effect. Both sides of the thin film are surfaces exposed to air, and thus the length along which the surface tension acts in this case is 2b. Then a force balance on the movable wire gives F = 2b𝜎s, and thus the surface tension can be expressed as 𝜎s = F 2b (2–39) Note that for b = 0.5 m, the measured force F (in N) is simply the surface tension in N/m. An apparatus of this kind with sufficient precision can be used to measure the surface tension of various liquids. In the U-shaped wire frame apparatus, the movable wire is pulled to stretch the film and increase its surface area. When the movable wire is pulled a distance Δx, the surface area increases by ΔA = 2b Δx, and the work W done during this stretching process is W = Force × Distance = F Δx = 2b𝜎s Δx = 𝜎s ΔA where we have assumed that the force remains constant over the small distance. This result can also be interpreted as the surface energy of the film is increased by an amount 𝜎s ΔA during this stretching process, which is consistent with the alternative interpretation of 𝜎s as surface energy per unit area. This is similar to a rubber band having more potential (elastic) energy after it is stretched further. In the case of liquid film, the work is used to move liquid molecules from the interior parts to the surface against the attraction forces of other molecules. Therefore, surface tension also can be defined as the work done per unit increase in the surface area of the liquid. The surface tension varies greatly from substance to substance, and with temperature for a given substance, as shown in Table 2–4. At 20°C, for example, the surface tension is 0.073 N/m for water and 0.440 N/m for mercury surrounded by atmospheric air. The surface tension of mercury is large enough that mercury droplets form nearly spherical balls that can be rolled like a solid ball on a smooth surface. The surface tension of a liq uid, in general, decreases with temperature and becomes zero at the critical point (and thus there is no distinct liquid–vapor interface at temperatures above the critical point). The effect of pressure on surface tension is usually negligible. The surface tension of a substance can be changed considerably by impurities. Therefore, certain chemicals, called surfactants, can be added to a liquid to decrease its surface tension. For example, soaps and deter gents lower the surface tension of water and enable it to penetrate the small openings between fibers for more effective washing. But this also Δx F F Movable wire Rigid wire frame Liquid film Wire Surface of film b x σs σs FIGURE 2–33 Stretching a liquid film with a U-shaped wire, and the forces acting on the movable wire of length b.
TABLE 2–4 Surface tension of some fluids in air at 1 atm and 20°C (unless otherwise stated) Fluid Surface Tension 𝜎s, N/m †Water: 0°C 20°C 100°C 300°C Glycerin SAE 30 oil Mercury Ethyl alcohol Blood, 37°C Gasoline Ammonia Soap solution Kerosene 0.076 0.073 0.059 0.014 0.063 0.035 0.440 0.023 0.058 0.022 0.021 0.025 0.028 Multiply by 0.06852 to convert to lbf/ft.
† See Appendices for more precise data for water.
cen96537_ch02_037-076.indd 57 29/12/16 6:00 pm 58 PROPERTIES OF FLUIDS means that devices whose operation depends on surface tension (such as heat pipes) can be destroyed by the presence of impurities due to poor workmanship. We speak of surface tension for liquids only at liquid–liquid or liquid– gas interfaces. Therefore, it is imperative that the adjacent liquid or gas be specified when specifying surface tension. Surface tension determines the size of the liquid droplets that form, and so a droplet that keeps growing by the addition of more mass breaks down when the surface tension can no longer hold it together. This is like a balloon that bursts while being inflated when the pressure inside rises above the strength of the balloon material. A curved interface indicates a pressure difference (or “pressure jump”) across the interface with pressure being higher on the concave side. Consider, for example, a droplet of liquid in air, an air (or other gas) bubble in water, or a soap bubble in air. The excess pressure ΔP above atmospheric pressure can be determined by considering a free-body diagram of half the droplet or bubble (Fig. 2–34). Noting that surface tension acts along the cir cumference and the pressure acts on the area, horizontal force balances for the droplet or air bubble and the soap bubble give Droplet or air bubble: (2𝜋R)𝜎s = (𝜋R2)ΔPdroplet → ΔPdroplet = Pi −Po = 2𝜎s R (2–40) Soap bubble: 2(2𝜋R)𝜎s = (𝜋R2)ΔPbubble → ΔPbubble = Pi −Po = 4𝜎s R (2–41) where Pi and Po are the pressures inside and outside the droplet or bubble, respectively. When the droplet or bubble is in the atmosphere, Po is simply atmospheric pressure. The extra factor of 2 in the force balance for the soap bubble is due to the existence of a soap film with two surfaces (inner and outer surfaces) and thus two circumferences in the cross section. The excess pressure in a droplet of liquid in a gas (or a bubble of gas in a liquid) can also be determined by considering a differential increase in the radius of the droplet due to the addition of a differential amount of mass and interpreting the surface tension as the increase in the surface energy per unit area. Then the increase in the surface energy of the droplet during this differential expansion process becomes 𝛿Wsurface = 𝜎s dA = 𝜎s d(4𝜋R 2) = 8𝜋R𝜎s dR The expansion work done during this differential process is determined by multiplying the force by distance to obtain 𝛿Wexpansion = Force × Distance = F dR = (ΔPA) dR = 4𝜋R2 ΔP dR Equating the two expressions above gives ΔPdroplet = 2𝜎s/R, which is the same relation obtained before and given in Eq. 2–40. Note that the excess pressure in a droplet or bubble is inversely proportional to the radius.
(a) Half of a droplet or air bubble (2πR)σs (πR2)ΔPdroplet (b) Half of a soap bubble 2(2πR)σs (πR2)ΔPbubble FIGURE 2–34 The free-body diagram of half of a droplet or air bubble and half of a soap bubble.
cen96537_ch02_037-076.indd 58 29/12/16 6:00 pm 59 CHAPTER 2 Capillary Effect Another interesting consequence of surface tension is the capillary effect, which is the rise or fall of a liquid in a small-diameter tube inserted into the liquid. Such narrow tubes or confined flow channels are called capillaries. The rise of kerosene through a cotton wick inserted into the reservoir of a kerosene lamp is due to this effect. The capillary effect is also partially responsible for the rise of water to the top of tall trees. The curved free sur face of a liquid in a capillary tube is called the meniscus. It is commonly observed that water in a glass container curves up slightly at the edges where it touches the glass surface; but the opposite occurs for mercury: it curves down at the edges (Fig. 2–35). This effect is usually expressed by saying that water wets the glass (by sticking to it) while mer cury does not. The strength of the capillary effect is quantified by the con tact (or wetting) angle ϕ, defined as the angle that the tangent to the liq uid surface makes with the solid surface at the point of contact. The surface tension force acts along this tangent line toward the solid surface. A liquid is said to wet the surface when ϕ < 90° and not to wet the surface when ϕ > 90°. In atmospheric air, the contact angle of water (and most other organic liquids) with glass is nearly zero, ϕ ≈ 0° (Fig. 2–36). Therefore, the surface tension force acts upward on water in a glass tube along the circum ference, tending to pull the water up. As a result, water rises in the tube until the weight of the liquid in the tube above the liquid level of the reservoir bal ances the surface tension force. The contact angle is 130° for mercury–glass and 26° for kerosene–glass in air. Note that the contact angle, in general, is different in different environments (such as another gas or liquid in place of air). The phenomenon of the capillary effect can be explained microscopically by considering cohesive forces (the forces between like molecules, such as water and water) and adhesive forces (the forces between unlike molecules, such as water and glass). The liquid molecules at the solid–liquid interface are subjected to both cohesive forces by other liquid molecules and adhesive forces by the molecules of the solid. The relative magnitudes of these forces determine whether a liquid wets a solid surface or not. Obviously, the water molecules are more strongly attracted to the glass molecules than they are to other water molecules, and thus water tends to rise along the glass surface. The opposite occurs for mercury, which causes the liquid surface near the glass wall to be suppressed (Fig. 2–37). The magnitude of the capillary rise in a circular tube can be determined from a force balance on the cylindrical liquid column of height h in the tube (Fig. 2–38). The bottom of the liquid column is at the same level as the free surface of the reservoir, and thus the pressure there must be atmospheric pressure. This balances the atmospheric pressure acting at the top surface of the liquid column, and thus these two effects cancel each other. The weight of the liquid column is approximately W = mg = ρVg = ρg(𝜋R2h) Equating the vertical component of the surface tension force to the weight gives W = Fsurface → ρg(𝜋R2h) = 2𝜋R𝜎s cos 𝜙 (a) Wetting fluid Water (b) Nonwetting fluid Mercury ϕ ϕ FIGURE 2–35 The contact angle for wetting and nonwetting fluids.
FIGURE 2–36 The meniscus of colored water in a 4-mm-inner-diameter glass tube. Note that the edge of the meniscus meets the wall of the capillary tube at a very small contact angle.
Photo by Gabrielle Trembley, Pennsylvania State University. Used by permission.
Meniscus Water Mercury h > 0 h < 0 Meniscus FIGURE 2–37 The capillary rise of water and the capillary fall of mercury in a small-diameter glass tube.
cen96537_ch02_037-076.indd 59 29/12/16 6:00 pm 60 PROPERTIES OF FLUIDS Solving for h gives the capillary rise to be Capillary rise: h = 2𝜎s ρgR cos 𝜙 (R = constant) (2–42) This relation is also valid for nonwetting liquids (such as mercury in glass) and gives the capillary drop. In this case ϕ > 90° and thus cos ϕ < 0, which makes h negative. Therefore, a negative value of capillary rise corresponds to a capillary drop (Fig. 2–37). Note that the capillary rise is inversely proportional to the radius of the tube. Therefore, the thinner the tube is, the greater the rise (or fall) of the liquid in the tube. In practice, the capillary effect for water is usually negligible in tubes whose diameter is greater than 1 cm. When pressure measurements are made using manometers and barometers, it is impor tant to use sufficiently large tubes to minimize the capillary effect. The capillary rise is also inversely proportional to the density of the liquid, as expected. Therefore, in general, lighter liquids experience greater capil lary rises. Finally, it should be kept in mind that Eq. 2–42 is derived for constant-diameter tubes and should not be used for tubes of variable cross section.
h W 2R Liquid 2πRσs ϕ FIGURE 2–38 The forces acting on a liquid column that has risen in a tube due to the capillary effect.
h W Water Air 2πRσs ϕ FIGURE 2–39 Schematic for Example 2–6.
EXAMPLE 2–6 The Capillary Rise of Water in a Tube A 0.6-mm-diameter glass tube is inserted into water at 20°C in a cup. Determine the capillary rise of water in the tube (Fig. 2–39).
SOLUTION The rise of water in a slender tube as a result of the capillary effect is to be determined.
Assumptions 1 There are no impurities in the water and no contamination on the surfaces of the glass tube. 2 The experiment is conducted in atmospheric air.
Properties The surface tension of water at 20°C is 0.073 N/m (Table 2–4). The contact angle of water with glass is approximately 0° (from preceding text). We take the density of liquid water to be 1000 kg/m3.
Analysis The capillary rise is determined directly from Eq. 2–42 by substituting the given values, yielding h = 2𝜎s ρgR cos 𝜙= 2(0.073 N/m) (1000 kg/m3)(9.81 m/s2)(0.3 × 10−3 m) (cos 0°)( 1 kg·m/s2 1 N ) = 0.050 m = 5.0 cm Therefore, water rises in the tube 5 cm above the liquid level in the cup.
Discussion Note that if the tube diameter were 1 cm, the capillary rise would be 0.3 mm, which is hardly noticeable to the eye. Actually, the capillary rise in a large-diameter tube occurs only at the rim. The center does not rise at all. Therefore, the capillary effect can be ignored for large-diameter tubes.
cen96537_ch02_037-076.indd 60 29/12/16 6:00 pm 61 CHAPTER 2 Water h Water to turbine Air 1 2 FIGURE 2–40 Schematic for Example 2–7.
EXAMPLE 2–7 Using Capillary Rise to Generate Power in a Hydraulic Turbine Reconsider Example 2–6. Realizing that water rises by 5 cm under the influence of surface tension without requiring any energy input from an external source, a person conceives the idea that power can be generated by drilling a hole in the tube just below the water level and feeding the water spilling out of the tube into a turbine (Fig. 2–40). The person takes this idea even further by suggesting that a series of tube banks can be used for this purpose and cascading can be incorporated to achieve practically feasible flow rates and elevation differences. Determine if this idea has any merit. SOLUTION Water that rises in tubes under the influence of the capillary effect is to be used to generate power by feeding it into a turbine. The validity of this sug gestion is to be evaluated.
Analysis The proposed system may appear like a stroke of genius, since the commonly used hydroelectric power plants generate electric power by simply capturing the potential energy of elevated water, and the capillary rise provides the mechanism to raise the water to any desired height without requiring any energy input. When viewed from a thermodynamic point of view, the proposed system immediately can be labeled as a perpetual motion machine (PMM) since it con tinuously generates electric power without requiring any energy input. That is, the proposed system creates energy, which is a clear violation of the first law of thermodynamics or the conservation of energy principle, and it does not war rant any further consideration. But the fundamental principle of conservation of energy did not stop many from dreaming about being the first to prove nature wrong, and to come up with a trick to permanently solve the world’s energy problems. Therefore, the impossibility of the proposed system should be demon strated. As you may recall from your physics courses (also to be discussed in the next chapter), the pressure in a static fluid varies in the vertical direction only and increases with increasing depth linearly. Then the pressure difference across the 5-cm-high water column in the tube becomes ΔPwater column in tube = P2 −P1 = ρwater gh = (1000 kg/m2)(9.81 m/s2)(0.05 m)( 1 kN 1000 kg·m/s2) = 0.49 kN/m2 ( ≈0.005 atm) That is, the pressure at the top of the water column in the tube is 0.005 atm less than the pressure at the bottom. Noting that the pressure at the bottom of the water column is atmospheric pressure (since it is at the same horizontal line as the water surface in the cup) the pressure anywhere in the tube is below atmospheric pressure with the difference reaching 0.005 atm at the top. Therefore, if a hole were drilled at some elevation in the tube, the top of the meniscus would fall until its elevation was the same as that of the hole.
Discussion The water column in the tube is motionless, and thus, there cannot be any unbalanced force acting on it (zero net force). The force due to the pressure dif ference across the meniscus between the atmospheric air and the water at the top of water column is balanced by the surface tension. If this surface-tension force were to disappear, the water in the tube would drop down under the influence of atmo spheric pressure to the level of the free surface in the tube.
cen96537_ch02_037-076.indd 61 29/12/16 6:00 pm 62 PROPERTIES OF FLUIDS SUMMARY In this chapter, various properties commonly used in fluid mechanics are discussed. The mass-dependent properties of a system are called extensive properties and the others, intensive properties. Density is mass per unit volume, and specific volume is volume per unit mass. The specific grav ity is defined as the ratio of the density of a substance to the density of water at 4°C, SG = ρ ρH2O The ideal-gas equation of state is expressed as P = ρRT where P is the absolute pressure, T is the thermodynamic tem perature, 𝜌 is the density, and R is the gas constant. At a given temperature, the pressure at which a pure sub stance changes phase is called the saturation pressure. For phase-change processes between the liquid and vapor phases of a pure substance, the saturation pressure is commonly called the vapor pressure Pv. Vapor bubbles that form in the low-pressure regions in a liquid (a phenomenon called cavitation) collapse as they are swept away from the low-pressure regions, generating highly destructive, extremely high-pressure shock waves. Energy can exist in numerous forms, and their sum con stitutes the total energy E (or e on a unit-mass basis) of a system. The sum of all microscopic forms of energy is called the internal energy U of a system. The energy that a system possesses as a result of its motion relative to some reference frame is called kinetic energy expressed per unit mass as ke = V 2/2, and the energy that a system possesses as a result of its elevation in a gravitational field is called potential energy expressed per unit mass as pe = gz. The compressibility effects in a fluid are represented by the coefficient of compressibility κ (also called the bulk mod ulus of elasticity) defined as 𝜅= −v( ∂P ∂v) T = ρ( ∂P ∂ρ) T ≅ −ΔP Δv/v The property that represents the variation of the density of a fluid with temperature at constant pressure is the volume expansion coefficient (or volume expansivity) 𝛽, defined as 𝛽= 1 v ( ∂v ∂T) P = −1 ρ( ∂ρ ∂T) P ≅ −Δρ/ρ ΔT The velocity at which an infinitesimally small pressure wave travels through a medium is the speed of sound. For an ideal gas it is expressed as c = √( ∂P ∂ρ) s = √kRT The Mach number is the ratio of the actual speed of the fluid to the speed of sound at the same state: Ma = V c The flow is called sonic when Ma = 1, subsonic when Ma < 1, supersonic when Ma > 1, hypersonic when Ma > > 1, and transonic when Ma ≅ 1. The viscosity of a fluid is a measure of its resistance to deformation. The tangential force per unit area is called shear stress and is expressed for simple shear flow between plates (one-dimensional flow) as 𝜏= 𝜇 du dy where 𝜇 is the coefficient of viscosity or the dynamic (or absolute) viscosity of the fluid, u is the velocity component in the flow direction, and y is the direction normal to the flow direction. Fluids that obey this linear relationship are called Newtonian fluids. The ratio of dynamic viscosity to density is called the kinematic viscosity ν. The pulling effect on the liquid molecules at an interface caused by the attractive forces of molecules per unit length is called surface tension 𝜎s. The excess pressure ΔP inside a spherical droplet or soap bubble, respectively, is given by ΔPdroplet = Pi −Po = 2𝜎s R and ΔPsoap bubble = Pi −Po = 4𝜎s R where Pi and Po are the pressures inside and outside the droplet or soap bubble. The rise or fall of a liquid in a small-diameter tube inserted into the liquid due to surface tension is called the capillary effect. The capillary rise or drop is given by h = 2𝜎s ρgR cos 𝜙 where ϕ is the contact angle. The capillary rise is inversely proportional to the radius of the tube; for water, it is negli gible for tubes whose diameter is larger than about 1 cm. Density and viscosity are two of the most fundamental properties of fluids, and they are used extensively in the chapters that follow. In Chap. 3, the effect of density on the variation of pressure in a fluid is considered, and the hydrostatic forces acting on surfaces are determined. In Chap. 8, the pressure drop caused by viscous effects dur ing flow is calculated and used in the determination of the pumping power requirements. Viscosity is also used as a key property in the formulation and solutions of the equations of fluid motion in Chaps. 9 and 10.
cen96537_ch02_037-076.indd 62 29/12/16 6:00 pm 63 CHAPTER 2 Guest Authors: G. C. Lauchle and M. L. Billet, Penn State University Cavitation is the rupture of a liquid, or of a fluid–solid interface, caused by a reduction of the local static pressure produced by the dynamic action of the fluid in the interior and/or boundaries of a liquid system. The rupture is the forma tion of a visible bubble. Liquids, such as water, contain many microscopic voids that act as cavitation nuclei. Cavitation occurs when these nuclei grow to a sig nificant, visible size. Although boiling is also the formation of voids in a liquid, we usually separate this phenomenon from cavitation because it is caused by an increase in temperature, rather than by a reduction in pressure. Cavitation can be used in beneficial ways, such as in ultrasonic cleaners, etchers, and cutters. But more often than not, cavitation is to be avoided in fluid flow applications because it spoils hydrodynamic performance, it causes extremely loud noise and high vibration levels, and it damages (erodes) the surfaces that support it. When cavitation bubbles enter regions of high pressure and collapse, the underwater shock waves sometimes create minute amounts of light. This phenomenon is called sonoluminescence. Body cavitation is illustrated in Fig. 2–41. The body is a model of the under water bulbulous bow region of a surface ship. It is shaped this way because located within it is a sound navigation and ranging (sonar) system that is spher ical in shape. This part of the surface ship is thus called a sonar dome. As ship speeds get faster and faster some of these domes start to cavitate and the noise created by the cavitation renders the sonar system useless. Naval architects and fluid dynamicists attempt to design these domes so that they will not cavi tate. Model-scale testing allows the engineer to see first hand whether a given design provides improved cavitation performance. Because such tests are con ducted in water tunnels, the conditions of the test water should have sufficient nuclei to model those conditions in which the prototype operates. This assures that the effect of liquid tension (nuclei distribution) is minimized. Important variables are the gas content level (nuclei distribution) of the water, the tem perature, and the hydrostatic pressure at which the body operates. Cavitation first appears—as either the speed V is increased, or as the submergence depth h is decreased—at the minimum pressure point Cpmin of the body. Thus, good hydrodynamic design requires 2(P∞ − Pv)/𝜌V2 > Cpmin, where 𝜌 is density, P∞ = 𝜌gh is the reference to static pressure, Cp is the pressure coefficient (Chap. 7), and Pv is the vapor pressure of water.
References Lauchle, G. C., Billet, M. L., and Deutsch, S., “High-Reynolds Number Liquid Flow Measurements,” in Lecture Notes in Engineering, Vol. 46, Frontiers in Experimental Fluid Mechanics, Springer-Verlag, Berlin, edited by M. Gad-el-Hak, Chap. 3, pp. 95–158, 1989.
Ross, D., Mechanics of Underwater Noise, Peninsula Publ., Los Altos, CA, 1987.
Barber, B. P., Hiller, R. A., Löfstedt, R., Putterman, S. J., and Weninger, K. R., “Defining the Unknowns of Sonoluminescence,” Physics Reports, Vol. 281, pp. 65–143, 1997.
APPLICATION SPOTLIGHT ■ Cavitation (a) (b) FIGURE 2–41 (a) Vaporous cavitation occurs in water that has very little entrained gas, such as that found very deep in a body of water. Cavitation bubbles are formed when the speed of the body—in this case the bulbulous bow region of a surface ship sonar dome— increases to the point where the local static pressure falls below the vapor pressure of the water. The cavitation bubbles are filled essentially with water vapor. This type of cavitation is very violent and noisy. (b) On the other hand, in shallow water, there is much more entrained gas in the water to act as cavitation nuclei. That’s because of the proximity of the dome to the atmosphere at the free surface. The cavitation bubbles first appear at a slower speed, and hence at a higher local static pressure. They are predom inantly filled with the gases that are entrained in the water, so this is known as gaseous cavitation.
Reprinted by permission of G.C. Lauchle and M.L. Billet, Pennsylvania State University.
cen96537_ch02_037-076.indd 63 29/12/16 6:00 pm 64 PROPERTIES OF FLUIDS PROBLEMS Density and Specific Gravity 2–1C What is the difference between intensive and exten sive properties?
2–2C For a substance, what is the difference between mass and molar mass? How are these two related?
2–3C What is specific gravity? How is it related to density?
2–4C The specific weight of a system is defined as the weight per unit volume (note that this definition violates the normal specific property-naming convention). Is the specific weight an extensive or intensive property?
2–5C Under what conditions is the ideal-gas assumption suitable for real gases?
2–6C What is the difference between R and Ru? How are these two related?
2–7 A 75-L container is filled with 1 kg of air at a tempera ture of 27°C. What is the pressure in the container?
2–8E A mass of 1-lbm of argon is maintained at 200 psia and 100°F in a tank. What is the volume of the tank?
2–9E What is the specific volume of oxygen at 40 psia and 80°F?
2–10 A fluid that occupies a volume of 24 L weighs 225 N at a location where the gravitational acceleration is 9.80 m/s2. Determine the mass of this fluid and its density.
2–11E The air in an automobile tire with a volume of 2.60 ft3 is at 70°F and 22 psig. Determine the amount of air that must be added to raise the pressure to the recommended value of 30 psig. Assume the atmospheric pressure to be 14.6 psia and the tem perature and the volume to remain constant. Answer: 0.106 lbm 2–12 The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25°C, the pressure gage reads 210 kPa. If the volume of the tire is 0.025 m3, determine the pressure rise in the tire when the air temperature in the tire rises to 50°C. Also, determine the amount of air that must be bled off to restore pressure to its original value at this temperature. Assume the atmospheric pressure to be 100 kPa.
FIGURE P2–12 © Stockbyte/Getty Images RF REFERENCES AND SUGGESTED READING 1. J. D. Anderson. Modern Compressible Flow with Histori cal Perspective, 3rd ed. New York: McGraw-Hill, 2003.
2. E. C. Bingham. “An Investigation of the Laws of Plastic Flow,” U.S. Bureau of Standards Bulletin, 13, pp. 309–353, 1916.
3. Y. A. Çengel and M. A. Boles. Thermodynamics: An Engineering Approach, 8th ed. New York: McGraw-Hill Education, 2015.
4. D. C. Giancoli. Physics, 6th ed. Upper Saddle River, NJ: Pearson, 2004.
5. Y. S. Touloukian, S. C. Saxena, and P. Hestermans. Ther mophysical Properties of Matter, The TPRC Data Series, Vol. 11, Viscosity. New York: Plenum, 1975.
6. L. Trefethen. “Surface Tension in Fluid Mechanics.” In Illustrated Experiments in Fluid Mechanics. Cambridge, MA: MIT Press, 1972.
7. The U.S. Standard Atmosphere. Washington, DC: U.S. Government Printing Office, 1976.
8. M. Van Dyke. An Album of Fluid Motion. Stanford, CA: Parabolic Press, 1982.
9. C. L. Yaws, X. Lin, and L. Bu. “Calculate Viscosities for 355 Compounds. An Equation Can Be Used to Calculate Liquid Viscosity as a Function of Temperature,” Chemi cal Engineering, 101, no. 4, pp. 1110–1128, April 1994.
10. C. L. Yaws. Handbook of Viscosity. 3 Vols. Houston, TX: Gulf Publishing, 1994.
Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and SI users can ignore them. Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software.
cen96537_ch02_037-076.indd 64 29/12/16 6:00 pm 65 CHAPTER 2 2–13 A spherical balloon with a diameter of 9 m is filled with helium at 20°C and 200 kPa. Determine the mole num ber and the mass of the helium in the balloon. Answers: 31.3 kmol, 125 kg 2–14 Reconsider Prob. 2–13. Using appropriate soft ware, investigate the effect of the balloon diam eter on the mass of helium contained in the balloon for the pressures of (a) 100 kPa and (b) 200 kPa. Let the diameter vary from 5 m to 15 m. Plot the mass of helium against the diameter for both cases.
2–15 A cylindrical tank of methanol has a mass of 60 kg and a volume of 75 L. Determine the methanol’s weight, den sity, and specific gravity. Take the gravitational acceleration to be 9.81 m/s2. Also, estimate how much force is needed to accelerate this tank linearly at 0.25 m/s2.
2–16 The combustion in a gasoline engine may be approxi mated by a constant volume heat addition process, and the con tents of the combustion chamber both before and after com bustion as air. The conditions are 1.80 MPa and 450°C before the combustion and 1500°C after it. Determine the pressure at the end of the combustion process. Answer: 4414 kPa Combustion chamber 1.80 MPa 450°C FIGURE P2–16 2–17 The density of atmospheric air varies with ele vation, decreasing with increasing altitude. (a) Using the data given in the table, obtain a relation for the variation of density with elevation, and calculate the density at an elevation of 7000 m. (b) Calculate the mass of the atmosphere using the correlation you obtained. Assume the earth to be a perfect sphere with a radius of 6377 km, and take the thickness of the atmosphere to be 25 km. r, km 𝜌, kg/m3 6377 1.225 6378 1.112 6379 1.007 6380 0.9093 6381 0.8194 6382 0.7364 6383 0.6601 6385 0.5258 6387 0.4135 6392 0.1948 6397 0.08891 6402 0.04008 2–18 The density of saturated liquid refrigerant–134a for –20°C ≤ T ≤ 100°C is given in Table A–4. Using this value develop an expression in the form 𝜌 = aT 2 + bT + c for the density of refrigerant–134a as a function of absolute tempera ture, and determine relative error for each data set.
2–19 Consider Table 2–1 in the textbook, which lists the specific gravities of various substances. (a) Explain the differ ence between specific gravity and specific weight. Which one (if any) is dimensionless? (b) Calculate the specific weight of all the substances in Table 2–1. For the case of bones, give answers for both the low and high values given in the table. Note: Excel is recommended for this kind of problem in which there is much repetition of calculations, but you are welcome to do the calculations by hand or with any other software. If using software like Excel, do not worry about the number of significant digits, since this is not something easy to modify in Excel. (c) As discussed in the text, there is another related property called specific volume. Calculate the specific volume of a liquid with SG = 0.592. Vapor Pressure and Cavitation 2–20C What is vapor pressure? How is it related to satura tion pressure?
2–21C Does water boil at higher temperatures at higher pressures? Explain.
2–22C If the pressure of a substance is increased during a boiling process, will the temperature also incre ase or will it remain constant? Why?
2–23C What is cavitation? What causes it?
2–24E The pressure on the suction side of pumps is typi cally low, and the surfaces on that side of the pump are sus ceptible to cavitation, especially at high fluid temperatures. If the minimum pressure on the suction side of a water pump is 0.70 psia absolute, determine the maximum water tempera ture to avoid the danger of cavitation.
2–25 A pump is used to transport water to a higher reser voir. If the water temperature is 20°C, determine the lowest pressure that can exist in the pump without cavitation.
2–26 In a piping system, the water temperature remains under 30°C. Determine the minimum pressure allowed in the system to avoid cavitation.
Energy and Specific Heats 2–27C What is total energy? Identify the different forms of energy that constitute the total energy.
2–28C List the forms of energy that contribute to the inter nal energy of a system.
2–29C How are heat, internal energy, and thermal energy related to each other?
2–30C What is flow energy? Do fluids at rest possess any flow energy?
cen96537_ch02_037-076.indd 65 29/12/16 6:00 pm 66 PROPERTIES OF FLUIDS 2–31C How do the energies of a flowing fluid and a fluid at rest compare? Name the specific forms of energy associ ated with each case.
2–32C Using average specific heats, explain how internal energy changes of ideal gases and incompressible substances can be determined.
2–33C Using average specific heats, explain how enthalpy changes of ideal gases and incompressible substances can be determined.
2–34E Ignoring any losses, estimate how much energy (in units of Btu) is required to raise the temperature of water in a 75-gallon hot-water tank from 60°F to 110°F.
2–35 Saturated water vapor at 150°C (enthalpy h = 2745.9 kJ/kg) flows in a pipe at 35 m/s at an elevation of z = 25 m. Determine the total energy of vapor in J/kg relative to the ground level.
Compressibility 2–36C What does the coefficient of volume expansion of a fluid represent? How does it differ from the coefficient of compressibility?
2–37C What does the coefficient of compressibility of a fluid represent? How does it differ from isothermal com pressibility?
2–38C Can the coefficient of compressibility of a fluid be negative? How about the coefficient of volume expansion?
2–39 Use the coefficient of volume expansion to estimate the density of water as it is heated from 60°F to 130°F at 1 atm. Compare your result with the actual density (from the appendices).
2–40 The volume of an ideal gas is to be reduced by half by compressing it isothermally. Determine the required change in pressure.
2–41 Water at 1 atm pressure is compressed to 400 atm pressure isothermally. Determine the increase in the density of water. Take the isothermal compressibility of water to be 4.80 × 10−5 atm−1.
2–42 It is observed that the density of an ideal gas increases by 10 percent when compressed isothermally from 10 atm to 11 atm. Determine the percent increase in density of the gas if it is compressed isothermally from 100 atm to 101 atm.
2–43 Saturated refrigerant-134a liquid at 10°C is cooled to 0°C at constant pressure. Using coefficient of volume expan sion data, determine the change in the density of the refrigerant.
2–44 A water tank is completely filled with liquid water at 20°C. The tank material is such that it can withstand ten sion caused by a volume expansion of 0.8 percent. Determine the maximum temperature rise allowed without jeopardizing safety. For simplicity, assume 𝛽 = constant = 𝛽 at 40°C.
2–45 Repeat Prob. 2–44 for a volume expansion of 0.4 per cent for water.
2–46 The density of seawater at a free surface where the pressure is 98 kPa is approximately 1030 kg/m3. Taking the bulk modulus of elasticity of seawater to be 2.34 × 109 N/m2 and expressing variation of pressure with depth z as dP = 𝜌g dz determine the density and pressure at a depth of 2500 m. Disregard the effect of temperature.
2–47E Taking the coefficient of compressibility of water to be 7 × 105 psia, determine the pressure increase required to reduce the volume of water by (a) 1 percent and (b) 2 percent.
2–48 A frictionless piston-cylinder device contains 10 kg of water at 20°C at atmospheric pressure. An external force F is then applied on the piston until the pressure inside the cylinder increases to 100 atm. Assuming the coefficient of compressibility of water remains unchanged during the com pression; estimate the energy needed to compress the water isothermally. Answer: 29.4 J FIGURE P2–48 Water Pressure gauge F 2–49 Reconsider Prob. 2–48. Assuming a linear pressure increase during the compression, estimate the energy needed to compress the water isothermally.
2–50 When modeling fluid flows with small changes in temperature and pressure, the Boussinesq approximation is often used in which the fluid density is assumed to vary lin early with changes in temperature. The Boussinesq approxi mation is 𝜌 = 𝜌0[1 − 𝛽(T − T0)], where 𝛽 is assumed to be constant over the given temperature range; 𝛽 is evaluated at reference temperature T0, taken as some average or mid-value temperature in the flow; and 𝜌0 is a reference density, also evaluated at T0. The Boussinesq approximation is used to model a flow of air at nearly constant pressure, P = 95.0 kPa, but the temperature varies between 20°C and 60°C. Using the mid-way point (40°C) as the reference temperature, calculate cen96537_ch02_037-076.indd 66 29/12/16 6:00 pm 67 CHAPTER 2 the density at the two temperature extremes using the Bouss inesq approximation, and compare with the actual density at these two temperatures obtained from the ideal gas law. In particular, for both temperatures calculate the percentage error caused by the Boussinesq approximation.
2–51 Using the definition of the coefficient of volume expansion and the expression 𝛽ideal gas = 1/T, show that the percent increase in the specific volume of an ideal gas during isobaric expansion is equal to the percent increase in absolute temperature.
Speed of Sound 2–52C What is sound? How is it generated? How does it travel? Can sound waves travel in a vacuum?
2–53C In which medium does a sound wave travel faster: in cool air or in warm air?
2–54C In which medium will sound travel fastest for a given temperature: air, helium, or argon?
2–55C In which medium does a sound wave travel faster: in air at 20°C and 1 atm or in air at 20°C and 5 atm?
2–56C Does the Mach number of a gas flowing at a con stant velocity remain constant? Explain.
2–57C Is it realistic to approximate that the propagation of sound waves is an isentropic process? Explain.
2–58C Is the sonic velocity in a specified medium a fixed quantity, or does it change as the properties of the medium change? Explain.
2–59 The isentropic process for an ideal gas is expressed as Pv k = constant. Using this process equation and the defini tion of the speed of sound (Eq. 2–24), obtain the expression for the speed of sound for an ideal gas (Eq. 2–26).
2–60 Carbon dioxide enters an adiabatic nozzle at 1200 K with a velocity of 50 m/s and leaves at 400 K. Assuming constant specific heats at room temperature, determine the Mach number (a) at the inlet and (b) at the exit of the nozzle. Assess the accuracy of the constant specific heat approximation. Answers: (a) 0.0925, (b) 3.73 2–61 Nitrogen enters a steady-flow heat exchanger at 150 kPa, 10°C, and 100 m/s, and it receives heat in the amount of 120 kJ/kg as it flows through it. Nitrogen leaves the heat exchanger at 100 kPa with a velocity of 200 m/s. Determine the Mach number of the nitrogen at the inlet and the exit of the heat exchanger.
2–62 Assuming ideal gas behavior, determine the speed of sound in refrigerant-134a at 0.8 MPa and 70°C.
2–63 Determine the speed of sound in air at (a) 300 K and (b) 800 K. Also determine the Mach number of an aircraft moving in air at a velocity of 330 m/s for both cases.
2–64E Steam flows through a device with a pressure of 120 psia, a temperature of 700°F, and a velocity of 900 ft/s. Determine the Mach number of the steam at this state by assuming ideal-gas behavior with k = 1.3. Answer: 0.441 2–65E Reconsider Prob. 2–64E. Using appropriate software, compare the Mach number of steam flow over the temperature range 350 to 700°F. Plot the Mach number as a function of temperature.
2–66 Air expands isentropically from 2.2 MPa and 77°C to 0.4 MPa. Calculate the ratio of the initial to the final speed of sound. Answer: 1.28 2–67 Repeat Prob. 2–66 for helium gas.
2–68 The Airbus A-340 passenger plane has a maximum takeoff weight of about 260,000 kg, a length of 64 m, a wing span of 60 m, a maximum cruising speed of 945 km/h, a seating capacity of 271 passengers, a maximum cruising alti tude of 14,000 m, and a maximum range of 12,000 km. The air temperature at the crusing altitude is about −60°C. Deter mine the Mach number of this plane for the stated limiting conditions.
Viscosity 2–69C What is a Newtonian fluid? Is water a Newtonian fluid?
2–70C What is viscosity? What is the cause of it in liquids and in gases? Do liquids or gases have higher dynamic vis cosities?
2–71C How does the kinematic viscosity of (a) liquids and (b) gases vary with temperature?
2–72C Consider two identical small glass balls dropped into two identical containers, one filled with water and the other with oil. Which ball will reach the bottom of the con tainer first? Why?
2–73E The viscosity of a fluid is to be measured by a vis cometer constructed of two 5-ft-long concentric cylinders. The inner diameter of the outer cylinder is 6 in, and the gap between the two cylinders is 0.035 in. The outer cylinder is rotated at 250 rpm, and the torque is measured to be 1.2 lbf⋅ft. Determine the viscosity of the fluid. Answer: 0.000272 lbf⋅s/ft2 2–74 The dynamic viscosity of carbon dioxide at 50°C and 200°C are 1.612 × 10–5 Pa·s and 2.276 × 10–5 Pa·s, respec tively. Determine the constants a and b of Sutherland correla tion for carbon dioxide at atmospheric pressure. Then predict the viscosity of carbon dioxide at 100°C and compare your result against the value given in Table A–10.
2–75 Consider the flow of a fluid with viscosity 𝜇 through a circular pipe. The velocity profile in the pipe is given as u(r) = umax(1 − rn/Rn), where umax is the maximum flow velocity, which occurs at the centerline; r is the radial distance from the centerline; and u(r) is the flow velocity at any position r. cen96537_ch02_037-076.indd 67 29/12/16 6:00 pm 68 PROPERTIES OF FLUIDS Develop a relation for the drag force exerted on the pipe wall by the fluid in the flow direction per unit length of the pipe.
r R umax u(r) = umax(1 – rn/Rn) 0 FIGURE P2–75 2–76 The viscosity of a fluid is to be measured by a vis cometer constructed of two 75-cm-long concentric cylinders. The outer diameter of the inner cylinder is 15 cm, and the gap between the two cylinders is 1 mm. The inner cylin der is rotated at 300 rpm, and the torque is measured to be 0.8 N·m. Determine the viscosity of the fluid.
1 mm Fluid 300 rpm Stationary cylinder FIGURE P2–76 2–77 A thin 30-cm × 30-cm flat plate is pulled at 3 m/s hori zontally through a 3.6-mm-thick oil layer sandwiched between two plates, one stationary and the other moving at a constant velocity of 0.3 m/s, as shown in Fig. P2–77. The dynamic vis cosity of the oil is 0.027 Pa⋅s. Assuming the velocity in each oil layer to vary linearly, (a) plot the velocity profile and find the location where the oil velocity is zero and (b) determine the force that needs to be applied on the plate to maintain this motion.
F Fixed wall Moving wall V = 3 m/s h1 = 1 mm h2 = 2.6 mm Vw = 0.3 m/s FIGURE P2–77 2–78 A rotating viscometer consists of two concentric cylinders—an inner cylinder of radius Ri rotating at angular velocity (rotation rate) 𝜔i, and a stationary outer cylinder of inside radius Ro. In the tiny gap between the two cylinders is the fluid of viscosity 𝜇. The length of the cylinders (into the page in Fig. P2–78) is L. L is large such that end effects are negligible (we can treat this as a two-dimensional prob lem). Torque (T) is required to rotate the inner cylinder at con stant speed. (a) Showing all of your work and algebra, gener ate an approximate expression for T as a function of the other variables. (b) Explain why your solution is only an approxi mation. In particular, do you expect the velocity profile in the gap to remain linear as the gap becomes larger and larger (i.e., if the outer radius Ro were to increase, all else staying the same)?
Liquid: ρ, μ Rotating inner cylinder Stationary outer cylinder Ro Ri ωi FIGURE P2–78 2–79 The clutch system shown in Fig. P2–79 is used to trans mit torque through a 2-mm-thick oil film with 𝜇 = 0.38 N⋅s/m2 between two identical 30-cm-diameter disks. When the driv ing shaft rotates at a speed of 1200 rpm, the driven shaft is observed to rotate at 1125 rpm. Assuming a linear velocity profile for the oil film, determine the transmitted torque.
30 cm Driving shaft Driven shaft SAE 30W oil 2 mm FIGURE P2–79 cen96537_ch02_037-076.indd 68 29/12/16 6:00 pm 69 CHAPTER 2 2–80 Reconsider Prob. 2–79. Using appropriate soft ware, investigate the effect of oil film thickness on the torque transmitted. Let the film thickness vary from 0.1 mm to 10 mm. Plot your results, and state your conclusions. 2–81 A 50-cm × 30-cm × 20-cm block weighing 150 N is to be moved at a constant velocity of 1.10 m/s on an inclined surface with a friction coefficient of 0.27. (a) Determine the force F that needs to be applied in the horizontal direction. (b) If a 0.40-mm-thick oil film with a dynamic viscosity of 0.012 Pa⋅s is applied between the block and inclined surface, determine the percent reduction in the required force.
150 N F 30 cm 50 cm 20º V = 1.10 m/s FIGURE P2–81 2–82 For flow over a plate, the variation of velocity with vertical distance y from the plate is given as u(y) = ay − by2 where a and b are constants. Obtain a relation for the wall shear stress in terms of a, b, and 𝜇.
2–83 In regions far from the entrance, fluid flow through a circular pipe is one dimensional, and the velocity profile for laminar flow is given by u(r) = umax(1 − r 2/R2), where R is the radius of the pipe, r is the radial distance from the center of the pipe, and umax is the maximum flow velocity, which occurs at the center. Obtain (a) a relation for the drag force applied by the fluid on a section of the pipe of length L and (b) the value of the drag force for water flow at 20°C with R = 0.08 m, L = 30 m, umax = 3 m/s, and 𝜇 = 0.0010 kg/m⋅s.
r R umax umax 1 – r2 R2 o FIGURE P2–83 2–84 Repeat Prob. 2–83 for umax = 6 m/s. Answer: (b) 2.26 N 2–85 A frustum-shaped body is rotating at a constant angu lar speed of 200 rad/s in a container filled with SAE 10W oil at 20°C (𝜇 = 0.100 Pa⋅s), as shown in Fig. P2–85. If the thickness of the oil film on all sides is 1.2 mm, determine the power required to maintain this motion. Also determine the reduction in the required power input when the oil tem perature rises to 80°C (𝜇 = 0.0078 Pa⋅s).
D = 12 cm L = 12 cm d = 4 cm Case SAE 10W oil r ω z FIGURE P2–85 2–86 A rotating viscometer consists of two concentric cylinders—a stationary inner cyliner of radius Ri and an outer cylinder of inside radius Ro rotating at angular veloc ity (rotation rate) 𝜔o. In the tiny gap between the two cylin ders is the fluid whose viscosity (𝜇) is to be measured. The length of the cylinders (into the page in Fig. P2–86) is L. L is large such that end effects are negligible (we can treat this as a two-dimensional problem). Torque (T) is required to rotate the inner cylinder at constant speed. Showing all your work and algebra, generate an approximate expression of T as a function of the other variables.
Liquid: ρ, μ Stationary inner cylinder Rotating outer cylinder Ro Ri ωo FIGURE P2–86 2–87 A thin plate moves between two parallel, horizontal, stationary flat surfaces at a constant velocity of 5 m/s. The two stationary surfaces are spaced 4 cm apart, and the medium between them is filled with oil whose viscosity is 0.9 N⋅s/m2. The part of the plate immersed in oil at any given time is 2-m cen96537_ch02_037-076.indd 69 29/12/16 6:00 pm 70 PROPERTIES OF FLUIDS long and 0.5-m wide. If the plate moves through the mid-plane between the surfaces, determine the force required to maintain this motion. What would your response be if the plate was 1 cm from the bottom surface (h2) and 3 cm from the top surface (h1)?
F Stationary surface Stationary surface h1 h2 V = 5 m/s FIGURE P2–87 2–88 Reconsider Prob. 2–87. If the viscosity of the oil above the moving plate is 4 times that of the oil below the plate, determine the distance of the plate from the bottom sur face (h2) that will minimize the force needed to pull the plate between the two oils at constant velocity.
2–89 A cylinder of mass m slides down from rest in a verti cal tube whose inner surface is covered by a viscous oil of film thickness h. If the diameter and height of the cylinder are D and L, respectively, derive an expression for the veloc ity of the cylinder as a function of time, t. Discuss what will happen as t → ∞. Can this device serve as a viscometer?
FIGURE P2–89 Oil film, h Cylinder L D Surface Tension and Capillary Effect 2–90C What is surface tension? What is its cause? Why is the surface tension also called surface energy?
2–91C What is the capillary effect? What is its cause? How is it affected by the contact angle?
2–92C A small-diameter tube is inserted into a liquid whose contact angle is 110°. Will the level of liquid in the tube be higher or lower than the level of the rest of the liquid? Explain.
2–93C Consider a soap bubble. Is the pressure inside the bubble higher or lower than the pressure outside?
2–94C Is the capillary rise greater in small- or large-diameter tubes?
2–95 Determine the gage pressure inside a soap bubble of diameter (a) 0.2 cm and (b) 5 cm at 20°C.
2–96E A 2.4-in-diameter soap bubble is to be enlarged by blowing air into it. Taking the surface tension of soap solu tion to be 0.0027 lbf/ft, determine the work input required to inflate the bubble to a diameter of 2.7 in.
2–97 A 1.6-mm-diameter tube is inserted into an unknown liquid whose density is 960 kg/m3, and it is observed that the liquid rises 5 mm in the tube, making a contact angle of 15°. Determine the surface tension of the liquid.
2–98 Consider a 0.15-mm diameter air bubble in a liquid. Determine the pressure difference between the inside and outside of the air bubble if the surface tension at the air-liquid interface is (a) 0.080 N/m and (b) 0.12 N/m.
2–99 The surface tension of a liquid is to be measured using a liquid film suspended on a U-shaped wire frame with an 8-cm-long movable side. If the force needed to move the wire is 0.030 N, determine the surface tension of this liquid in air.
2–100 A capillary tube of 1.2 mm diameter is immersed vertically in water exposed to the atmosphere. Determine how high water will rise in the tube. Take the contact angle at the inner wall of the tube to be 6° and the surface tension to be 1.00 N/m. Answer: 0.338 m 2–101E A 0.018-in-diameter glass tube is inserted into mercury, which makes a contact angle of 140° with glass. Determine the capillary drop of mercury in the tube at 68°F. Answer: 0.874 in 2–102 A capillary tube is immersed vertically in a water container. Knowing that water starts to evaporate when the pressure drops below 2 kPa, determine the maximum capil lary rise and tube diameter for this maximum-rise case. Take the contact angle at the inner wall of the tube to be 6° and the surface tension to be 1.00 N/m.
2–103 Contrary to what you might expect, a solid steel ball can float on water due to the surface tension effect. Deter mine the maximum diameter of a steel ball that would float on water at 10°C. What would your answer be for an alumi num ball? Take the densities of steel and aluminum balls to be 7800 kg/m3 and 2700 kg/m3, respectively.
2–104 Nutrients dissolved in water are carried to upper parts of plants by tiny tubes partly because of the capillary effect. Determine how high the water solution will rise in a tree in a 0.0026-mm-diameter tube as a result of the capil lary effect. Treat the solution as water at 20°C with a contact angle of 15°. Answer: 11.1 m cen96537_ch02_037-076.indd 70 29/12/16 6:00 pm 71 CHAPTER 2 0.0026 mm Water solution FIGURE P2–104 Review Problems 2–105 Consider a 55-cm-long journal bearing that is lubri cated with oil whose viscosity is 0.1 kg/m⋅s at 20°C at the beginning of operation and 0.008 kg/m⋅s at the anticipated steady operating temperature of 80°C. The diameter of the shaft is 8 cm, and the average gap between the shaft and the journal is 0.08 cm. Determine the torque needed to overcome the bearing friction initially and during steady operation when the shaft is rotated at 1500 rpm.
2–106 The diameter of one arm of a U-tube is 5 mm while the other arm is large. If the U-tube contains some water, and both surfaces are exposed to atmospheric pressure, determine the difference between the water levels in the two arms.
2–107E A rigid tank contains 40 lbm of air at 20 psia and 70°F. More air is added to the tank until the pressure and temperature rise to 35 psia and 90°F, respectively. Determine the amount of air added to the tank. Answer: 27.4 lbm 2–108 A 10-m3 tank contains nitrogen at 25°C and 800 kPa. Some nitrogen is allowed to escape until the pressure in the tank drops to 600 kPa. If the temperature at this point is 20°C, determine the amount of nitrogen that has escaped. Answer: 21.5 kg 2–109 The absolute pressure of an automobile tire is mea sured to be 320 kPa before a trip and 335 kPa after the trip. Assuming the volume of the tire remains constant at 0.022 m3, determine the percent increase in the absolute temperature of the air in the tire.
2–110E The analysis of a propeller that operates in water at 60°F shows that the pressure at the tips of the propeller drops to 0.1 psia at high speeds. Determine if there is a danger of cavitation for this propeller.
2–111 A closed tank is partially filled with water at 70°C. If the air above the water is completely evacuated, determine the absolute pressure in the evacuated space. Assume the temperature to remain constant.
2–112 The specific gravities of solids and carrier fluids of a slurry are usually known, but the specific gravity of the slurry depends on the concentration of the solid particles. Show that the specific gravity of a water-based slurry can be expressed in terms of the specific gravity of the solid SGs and the mass concentration of the suspended solid particles Cs, mass as SGm = 1 1 + Cs, mass(1/SGs −1) 2–113 A rigid tank contains an ideal gas at 300 kPa and 600 K. Half of the gas is withdrawn from the tank and the gas is at 100 kPa at the end of the process. Determine (a) the final temperature of the gas and (b) the final pressure if no mass were withdrawn from the tank and the same final tem perature were reached at the end of the process.
2–114 The composition of a liquid with suspended solid particles is generally characterized by the fraction of solid particles either by weight or mass, Cs, mass = ms/mm or by vol ume, Cs, vol = Vs /Vm where m is mass and V is volume. The subscripts s and m indicate solid and mixture, respectively. Develop an expression for the specific gravity of a water-based suspension in terms of Cs, mass and Cs, vol.
2–115 The variation of the dynamic viscosity of water with absolute temperature is given as T, K 𝜇, Pa⋅s 273.15 1.787 × 10−3 278.15 1.519 × 10−3 283.15 1.307 × 10−3 293.15 1.002 × 10−3 303.15 7.975 × 10−4 313.15 6.529 × 10−4 333.15 4.665 × 10−4 353.15 3.547 × 10−4 373.15 2.828 × 10−4 Using these tabulated data, develop a relation for viscosity in the form of 𝜇 = 𝜇(T ) = A + BT + CT 2 + DT 3 + ET 4. Using the relation developed, predict the dynamic viscosity of water at 50°C at which the reported value is 5.468 × 10−4 Pa⋅s. Compare your result with the results of Andrade’s equa tion, which is given in the form of 𝜇 = D⋅e B/T, where D and B are constants whose values are to be determined using the viscosity data given.
2–116 A newly produced pipe with diameter of 3 m and length 15 m is to be tested at 10 MPa using water at 15°C. After sealing both ends, the pipe is first filled with water and then the pressure is increased by pumping additional water into the test pipe until the test pressure is reached. Assuming no deformation in the pipe, determine how much additional water needs to be pumped into the pipe. Take the coefficient of compressibility to be 2.10 × 109 Pa. Answer: 505 kg cen96537_ch02_037-076.indd 71 29/12/16 6:00 pm 72 PROPERTIES OF FLUIDS 2–117 Prove that the coefficient of volume expansion for an ideal gas is 𝛽ideal gas = 1/T.
2–118 Although liquids, in general, are hard to compress, the compressibility effect (variation in the density) may become unavoidable at the great depths in the oceans due to enormous pressure increase. At a certain depth the pressure is reported to be 100 MPa and the average coefficient of com pressibility is about 2350 MPa.
(a) Taking the liquid density at the free surface to be 𝜌0 = 1030 kg/m3, obtain an analytical relation between density and pressure, and determine the density at the specified pressure. Answer: 1074 kg/m3 (b) Use Eq. 2–13 to estimate the density for the specified pressure and compare your result with that of part (a).
2–119E Air expands isentropically from 200 psia and 240°F to 60 psia. Calculate the ratio of the initial to final speed of sound. Answer: 1.19 2–120 A shaft with a diameter of D = 80 mm and a length of L = 400 mm, shown in Fig. P2–120 is pulled with a constant velocity of U = 5 m/s through a bearing with vari able diameter. The clearance between shaft and bearing, which varies from h1 = 1.2 mm to h2 = 0.4 mm, is filled with a Newtonian lubricant whose dynamic viscosity is 0.10 Pa⋅s. Determine the force required to maintain the axial movement of the shaft. Answer: 69 N FIGURE P2–120 y h1 h2 x L D U Bearing Shaft Viscous oil, μ 2–121 Reconsider Prob. 2–120. The shaft now rotates with a constant angular speed of n . = 1450 rpm in a bearing with variable diameter. The clearance between shaft and bearing, which varies from h1 = 1.2 mm to h2 = 0.4 mm, is filled with a Newtonian lubricant whose dynamic viscosity is 0.1 Pa⋅s. Determine the torque required to maintain the motion.
2–122 Derive a relation for the capillary rise of a liquid between two large parallel plates a distance t apart inserted into the liquid vertically. Take the contact angle to be f.
2–123 A 10-cm-diameter cylindrical shaft rotates inside a 50-cm-long 10.3-cm diameter bearing. The space between the shaft and the bearing is completely filled with oil whose vis cosity at anticipated operating temperature is 0.300 N⋅s/m2. Determine the power required to overcome friction when the shaft rotates at a speed of (a) 600 rpm and (b) 1200 rpm.
2–124 A large plate is pulled at a constant speed of U = 4 m/s over a fixed plate on 5-mm-thick engine oil film at 20°C. Assuming a half-parabolic velocity profile in the oil film, while the flow is developing as sketched, determine the shear stress on the upper plate and its direction. Repeat for the linear profile (dashed line) that develops after a long time.
FIGURE P2–124 U = 4 m/s h = 5 mm Engine oil y 2–125 Some rocks or bricks contain small air pockets in them and have a spongy structure. Assuming the air spaces form col umns of an average diameter of 0.006 mm, determine how high water can rise in such a material. Take the surface tension of the air–water interface in that material to be 0.085 N/m.
2–126 A fluid between two very long parallel plates is heated in a way that its viscosity decreases linearly from 0.90 Pa⋅s at the lower plate to 0.50 Pa⋅s at the upper plate. The spac ing between the two plates is 0.4 mm. The upper plate moves steadily at a velocity of 10 m/s, in a direction parallel to both plates. The pressure is constant everywhere, the fluid is New tonian, and assumed incompressible. Neglect gravitational effects. (a) Obtain the fluid velocity u as a function of y, u(y), where y is the vertical axis perpendicular to the plates. Plot the velocity profile across the gap between the plates. (b) Calculate the value of the shear stress. Show the direction of the shear stress on the moving plate and on the top surface of the fluid element adjacent to the moving plate.
FIGURE P2–126 U = 10 m/s y x 0 h = 0.4 mm 2–127 The rotating parts of a hydroelectric power plant hav ing power capacity W . have a rotational synchronous speed n . . The weight of the rotating parts (the hydroturbine and its electric generator) is supported in a thrust bearing having annular form between D and d diameters as sketched. The thrust bearing is operated with a very thin oil film of cen96537_ch02_037-076.indd 72 29/12/16 6:00 pm 73 CHAPTER 2 thickness e and dynamic viscosity 𝜇. It is assumed that the oil is a Newtonian fluid and the velocity is approximated as linear in the bearing. Calculate the ratio of lost power in the thrust bearing to the produced power in the hydraulic power plant. Use W . = 48.6 MW, 𝜇 = 0.035 Pa⋅s, n . = 500 rpm, e = 0.25 mm, D = 3.2 m, and d = 2.4 m.
FIGURE P2–127 r e e Oil Oil O d ω = const.
D F(load) 2–128 The viscosity of some fluids changes when a strong electric field is applied on them. This phenomenon is known as the electrorheological (ER) effect, and fluids that exhibit such behavior are known as ER fluids. The Bingham plastic model for shear stress, which is expressed as 𝜏 = 𝜏y + 𝜇(du/dy) is widely used to describe ER fluid behavior because of its simplicity. One of the most promising applications of ER flu ids is the ER clutch. A typical multidisk ER clutch consists of several equally spaced steel disks of inner radius R1 and outer radius R2, N of them attached to the input shaft. The gap h between the parallel disks is filled with a viscous fluid. (a) Find a relationship for the torque generated by the clutch when the output shaft is stationary and (b) calculate the torque for an ER clutch with N = 11 for R1 = 50 mm, R2 = 200 mm, and n . = 2400 rpm if the fluid is SAE 10 with 𝜇 = 0.1 Pa·s, 𝜏y = 2.5 kPa, and h = 1.2 mm. Answer: (b) 2060 N·m FIGURE P2–128 Shell Input shaft Plates mounted on input shaft Variable magnetic field Output shaft R2 R1 Plates mounted on shell h = 1.2 mm 2–129 The viscosity of some fluids, called magnetorheolog ical (MR) fluids, changes when a magnetic field is applied. Such fluids involve micron-sized magnetizable particles suspended in an appropriate carrier liquid, and are suitable for use in controllable hydraulic clutches. See Fig. P2–128. The MR fluids can have much higher viscosities than the ER fluids, and they often exhibit shear-thinning behavior in which the viscosity of the fluid decreases as the applied shear force increases. This behavior is also known as pseudoplastic behavior, and can be successfully represented by Herschel– Bulkley constitutive model expressed as 𝜏 = 𝜏y + K(du /dy)m. Here 𝜏 is the shear stress applied, 𝜏y is the yield stress, K is the consistency index, and m is the power index. For a Herschel–Bulkley fluid with 𝜏y = 900 Pa, K = 58 Pa·sm, and m = 0.82, (a) find a relationship for the torque transmitted by an MR clutch for N plates attached to the input shaft when the input shaft is rotating at an angular speed of 𝜔 while the output shaft is stationary and (b) calculate the torque trans mitted by such a clutch with N = 11 plates for R1 = 50 mm, R2 = 200 mm, n . = 3000 rpm, and h = 1.5 mm.
2–130 Some non-Newtonian fluids behave as a Bingham plastic for which shear stress can be expressed as 𝜏 = 𝜏y + 𝜇(du/dr). For laminar flow of a Bingham plastic in a hori zontal pipe of radius R, the velocity profile is given as u(r) = (ΔP/4𝜇L)(r2 − R2) + (𝜏y /𝜇)(r − R), where ΔP/L is the con stant pressure drop along the pipe per unit length, 𝜇 is the dynamic viscosity, r is the radial distance from the center line, and 𝜏y is the yield stress of Bingham plastic. Determine (a) the shear stress at the pipe wall and (b) the drag force act ing on a pipe section of length L.
2–131 In some damping systems, a circular disk immersed in oil is used as a damper, as shown in Fig. P2–131. Show that the damping torque is proportional to angular speed in accordance with the relation Tdamping = C𝜔 where C = 0.5𝜋𝜇(1/a + 1/b)R4. Assume linear velocity profiles on both sides of the disk and neglect the tip effects.
Disk Damping oil a b R FIGURE P2–131 2–132 Oil of viscosity 𝜇 = 0.0357 Pa⋅s and density 𝜌 = 0.796 kg/m3 is sandwiched in the small gap between two very large parallel flat plates. A third flat plate of surface area A = 20.0 cm × 20.0 cm (on one side) is dragged through the oil at steady velocity V = 1.00 m/s to the right as sketched. cen96537_ch02_037-076.indd 73 29/12/16 6:01 pm 74 PROPERTIES OF FLUIDS The top plate is stationary, but the bottom plate is moving at velocity V = 0.300 m/s to the left as sketched. The heights are h1 = 1.00 mm and h2 = 1.65 mm. The force required to pull the plate through the oil is F. (a) Sketch the velocity pro files and calculate the distance yA where the velocity is zero. Hint: Since the gaps are small and the oil is very viscous, the velocity profiles are linear in both gaps. Use the no-slip con ditions at the walls to determine the velocity profile in each gap. (b) Calculate force F in newtons (N) required to keep the middle plate moving at constant speed.
FIGURE P2–132 F V = 1.00 m/s Vw = 0.300 m/s yA h1 h2 Area = A Oil μ, ρ μ, ρ Oil Area = A Fundamentals of Engineering (FE) Exam Problems 2–133 The specific gravity of a fluid is specified to be 0.82. The specific volume of this fluid is (a) 0.00100 m3/kg (b) 0.00122 m3/kg (c) 0.0082 m3/kg (d ) 82 m3/kg (e) 820 m3/kg 2–134 The specific gravity of mercury is 13.6. The specific weight of mercury is (a) 1.36 kN/m3 (b) 9.81 kN/m3 (c) 106 kN/m3 (d ) 133 kN/m3 (e) 13,600 kN/m3 2–135 A 0.08-m3 rigid tank contains air at 3 bar and 127°C. The mass of the air in the tank is (a) 0.209 kg (b) 0.659 kg (c) 0.8 kg (d) 0.002 kg (e) 0.066 kg 2–136 The pressure of water is increased from 100 kPa to 700 kPa by a pump. The density of the water is 1 kg/L. If the water temperature does not change during this process, the change of specific enthalpy of the water is (a) 400 kJ/kg (b) 0.4 kJ/kg (c) 600 kJ/kg (d) 800 kJ/kg (e) 0.6 kJ/kg 2–137 An ideal gas flows in a pipe at 37°C. The density of the gas is 1.9 kg/m3 and its molar mass is 44 kg/kmol. The pressure of the gas is (a) 13 kPa (b) 79 kPa (c) 111 kPa (d ) 490 kPa (e) 4900 kPa 2–138 Liquid water vaporizes into water vapor as it flows in the piping of a boiler. If the temperature of water in the pipe is 180°C, the vapor pressure of the water in the pipe is (a) 1002 kPa (b) 180 kPa (c) 101.3 kPa (d ) 18 kPa (e) 100 kPa 2–139 In a water distribution system, the pressure of water can be as low as 1.4 psia. The maximum temperature of water allowed in the piping to avoid cavitation is (a) 50°F (b) 77°F (c) 100°F (d ) 113°F (e) 140°F 2–140 The pressure of water is increased from 100 kPa to 900 kPa by a pump. The temperature of water also increases by 0.15°C. The density of water is 1 kg/L and its specific heat is cp = 4.18 kJ/kg⋅°C. The enthalpy change of the water during this process is (a) 900 kJ/kg (b) 1.43 kJ/kg (c) 4.18 kJ/kg (d ) 0.63 kJ/kg (e) 0.80 kJ/kg 2–141 An ideal gas is compressed isothermally from 100 kPa to 170 kPa. The percent increase in the density of this gas during this process is (a) 70% (b) 35% (c) 17% (d) 59% (e) 170% 2–142 The variation of the density of a fluid with tempera ture at constant pressure is represented by (a) Bulk modulus of elasticity (b) Coefficient of compress ibility (c) Isothermal compressibility (d) Coefficient of volume expansion (e) None of these 2–143 Water is heated from 2°C to 78°C at a constant pres sure of 100 kPa. The initial density of water is 1000 kg/m3 and the volume expansion coefficient of water is 𝛽 = 0.377 × 10–3 K–1. The final density of the water is (a) 28.7 kg/m3 (b) 539 kg/m3 (c) 997 kg/m3 (d) 984 kg/m3 (e) 971 kg/m3 2–144 The viscosity of liquids and the vis cosity of gases with temperature.
(a) Increases, increases (b) Increases, decreases (c) Decreases, increases (d) Decreases, decreases (e) Decreases, remains the same 2–145 The pressure of water at atmospheric pressure must be raised to 210 atm to compress it by 1 percent. Then, the coefficient of compressibility value of water is (a) 209 atm (b) 20,900 atm (c) 21 atm (d ) 0.21 atm (e) 210,000 atm 2–146 The density of a fluid decreases by 3 percent at con stant pressure when its temperature increases by 10°C. The coefficient of volume expansion of this fluid is (a) 0.03 K−1 (b) 0.003 K−1 (c) 0.1 K−1 (d ) 0.5 K−1 (e) 3 K−1 2–147 The speed of a spacecraft is given to be 1250 km/h in atmospheric air at −40°C. The Mach number of this flow is (a) 35 .9 (b) 0.85 (c) 1.0 (d ) 1.13 (e) 2.74 2–148 The dynamic viscosity of air at 20°C and 200 kPa is 1.83 × 10−5 kg/m⋅s. The kinematic viscosity of air at this state is (a) 0.525 × 10−5 m2/s (b) 0.77 × 10−5 m2/s (c) 1.47 × 10−5 m2/s (d ) 1.83 × 10−5 m2/s (e) 0.380 × 10−5 m2/s 2–149 A viscometer constructed of two 30-cm-long con centric cylinders is used to measure the viscosity of a fluid. The outer diameter of the inner cylinder is 9 cm, and the gap between the two cylinders is 0.18 cm. The inner cylinder is rotated at 250 rpm, and the torque is measured to be 1.4 N⋅m. The viscosity of the fluid is (a) 0.0084 N⋅s/m2 (b) 0.017 N⋅s/m2 (c) 0.062 N⋅s/m2 (d ) 0.0049 N⋅s/m2 (e) 0.56 N⋅s/m2 cen96537_ch02_037-076.indd 74 29/12/16 6:01 pm 75 CHAPTER 2 2–150 A 0.6-mm-diameter glass tube is inserted into water at 20°C in a cup. The surface tension of water at 20°C is 𝜎s = 0.073 N/m. The contact angle can be taken as zero degrees. The capillary rise of water in the tube is (a) 2.6 cm (b) 7.1 cm (c) 5.0 cm (d ) 9.7 cm (e) 12.0 cm 2–151 A liquid film suspended on a U-shaped wire frame with a 6-cm-long movable side is used to measure the surface tension of a liquid. If the force needed to move the wire is 0.028 N, the surface tension of this liquid exposed to air is (a) 0.00762 N/m (b) 0.096 N/m (c) 0.168 N/m (d) 0.233 N/m (e) 0.466 N/m 2–152 It is observed that water at 20°C rises up to 20 m height in a tree due to capillary effect. The surface tension of water at 20°C is 𝜎s = 0.073 N/m and the contact angle is 20°. The maximum diameter of the tube in which water rises is (a) 0.035 mm (b) 0.016 mm (c) 0.02 mm (d) 0.002 mm (e) 0.0014 mm Design and Essay Problems 2–153 Design an experiment to measure the viscosity of liquids using a vertical funnel with a cylindrical reservoir of height h and a narrow flow section of diameter D and length L. Making appropriate assumptions, obtain a relation for viscos ity in terms of easily measurable quantities such as density and volume flow rate.
2–154 Write general relationship between shear stress 𝜏 and rate of deformation du/dy for non-Newtonian fluids. Also, write a report on how to measure the viscosity of non-Newtonian fluids.
2–155 Write an essay on the rise of the fluid to the top of trees by capillary and other effects.
2–156 Write an essay on the oils used in car engines in different seasons and their viscosities.
2–157 Consider the flow of water through a clear tube. It is sometimes possible to observe cavitation in the throat created by pinching off the tube to a very small diameter as sketched. We assume incompressible flow with neg ligible gravitational effects and negligible irreversibili ties. You will learn later (Chap. 5) that as the duct cross-sectional area decreases, the velocity increases and the pressure decreases according to V1A1 = V2 A2 and P1 + ρ V 2 1 2 = P2 + ρ V 2 2 2 respectively, where V1 and V2 are the average velocities through cross-sectional areas A1 and A2. Thus, both the maximum velocity and minimum pressure occur at the throat. (a) If the water is at 20°C, the inlet pressure is 20.803 kPa, and the throat diameter is one-twentieth of the inlet diameter, estimate the minimum average inlet velocity at which cavitation is likely to occur in the throat. (b) Repeat at a water temperature of 50°C. Explain why the required inlet velocity is higher or lower than that of part (a).
Throat P2 P1 V2 V1 Inlet FIGURE P2–157 2–158 Even though steel is about 7 to 8 times denser than water, a steel paper clip or razor blade can be made to float on water! Explain and discuss. Predict what would happen if you mix some soap with the water.
FIGURE P2–158 Photo by John M. Cimbala.
cen96537_ch02_037-076.indd 75 29/12/16 6:01 pm cen96537_ch02_037-076.indd 76 29/12/16 6:01 pm This page intentionally left blank 3 CHAPTER 77 P RE S S UR E A N D F LUI D STAT I CS T his chapter deals with forces applied by fluids at rest or in rigid-body motion. The fluid property responsible for those forces is pressure, which is a normal force exerted by a fluid per unit area. We start this chapter with a detailed discussion of pressure, including absolute and gage pressures, the pressure at a point, the variation of pressure with depth in a gravitational field, the barometer, the manometer, and other pressure mea surement devices. This is followed by a discussion of the hydrostatic forces applied on submerged bodies with plane or curved surfaces. We then con sider the buoyant force applied by fluids on submerged or floating bodies, and discuss the stability of such bodies. Finally, we apply Newton’s second law of motion to a body of fluid in motion that acts as a rigid body and ana lyze the variation of pressure in fluids that undergo linear acceleration and in rotating containers. This chapter makes extensive use of force balances for bodies in static equilibrium, and it would be helpful if the relevant topics from statics are first reviewed.
OBJECTIVES When you finish reading this chapter, you should be able to ■ ■ Determine the variation of pressure in a fluid at rest ■ ■ Calculate pressure using various kinds of manometers ■ ■ Calculate the forces and moments exerted by a fluid at rest on plane or curved submerged surfaces ■ ■ Analyze the stability of floating and submerged bodies ■ ■ Analyze the rigid-body motion of fluids in containers during linear acceleration or rotation John Ninomiya flying a cluster of 72 helium-filled balloons over Temecula, California in April of 2003. The helium balloons displace approxi mately 230 m3 of air, providing the necessary buoyant force. Don’t try this at home!
Photo by Susan Dawson. Used by permission.
cen96537_ch03_077-136.indd 77 14/01/17 2:22 pm 78 pressure and fluid statics 3–1 ■ PRESSURE Pressure is defined as a normal force exerted by a fluid per unit area. We speak of pressure only when we deal with a gas or a liquid. The counterpart of pressure in solids is normal stress. Since pressure is defined as force per unit area, it has units of newtons per square meter (N/m2), which engineers call a pascal (Pa). That is, 1 Pa = 1 N/m2 The pressure unit pascal is too small for most pressures encountered in practice. Therefore, its multiples kilopascal (1 kPa = 103 Pa) and megapascal (1 MPa = 106 Pa) are commonly used. Three other pressure units commonly used in practice, especially in Europe, are bar, standard atmosphere, and kilogram-force per square centimeter: 1 bar = 105 Pa = 0.1 MPa = 100 kPa 1 atm = 101,325 Pa = 101.325 kPa = 1.01325 bars 1 kgf/cm2 = 9.807 N/cm2 = 9.807 × 104 N/m2 = 9.807 × 104 Pa = 0.9807 bar = 0.9679 atm Note the pressure units bar, atm, and kgf/cm2 are almost equivalent to each other. In the English system, the pressure unit is pound-force per square inch (lbf/in2, or psi), and 1 atm = 14.696 psi. The pressure units kgf/cm2 and lbf/in2 are also denoted by kg/cm2 and lb/in2, respectively, and they are com monly used in tire gages. It can be shown that 1 kgf/cm2 = 14.223 psi.
Pressure is also used on solid surfaces as synonymous to normal stress, which is the force acting perpendicular to the surface per unit area. For example, a 150-pound person with a total foot imprint area of 50 in2 exerts a pressure of 150 lbf/50 in2 = 3.0 psi on the floor (Fig. 3–1). If the person stands on one foot, the pressure doubles. If the person gains excessive weight, he or she is likely to encounter foot discomfort because of the increased pressure on the foot (the size of the bottom of the foot does not change with weight gain). This also explains how a person can walk on fresh snow without sinking by wearing large snowshoes, and how a person cuts with little effort when using a sharp knife.
The actual pressure at a given position is called the absolute pressure, and it is measured relative to absolute vacuum (i.e., absolute zero pressure). Most pressure-measuring devices, however, are calibrated to read zero in the atmosphere (Fig. 3–2), and so they indicate the difference between the abso lute pressure and the local atmospheric pressure. This difference is called the gage pressure. Pgage can be positive or negative, but pressures below atmospheric pressure are sometimes called vacuum pressures and are mea sured by vacuum gages that indicate the difference between the atmospheric pressure and the absolute pressure. Absolute, gage, and vacuum pressures are related to each other by Pgage = Pabs −Patm (3–1) Pvac = Patm −Pabs (3–2) This is illustrated in Fig. 3–3.
150 pounds Afeet = 50 in2 P = 3 psi P = 6 psi 300 pounds W Afeet 150 lbf 50 in2 P = σn = = 3 psi = FIGURE 3–1 The normal stress (or “pressure”) on the feet of a chubby person is much greater than on the feet of a slim person.
FIGURE 3–2 Some basic pressure gages.
© Ashcroft Inc. Used by permission.
cen96537_ch03_077-136.indd 78 14/01/17 2:22 pm 79 CHAPTER 3 Like other pressure gages, the gage used to measure the air pressure in an automobile tire reads the gage pressure. Therefore, the common reading of 32.0 psi (2.25 kgf/cm2) indicates a pressure of 32.0 psi above the atmo spheric pressure. At a location where the atmospheric pressure is 14.3 psi, for example, the absolute pressure in the tire is 32.0 + 14.3 = 46.3 psi.
In thermodynamic relations and tables, absolute pressure is almost always used. Throughout this text, the pressure P will denote absolute pressure unless specified otherwise. Often the letters “a” (for absolute pressure) and “g” (for gage pressure) are added to pressure units (such as psia and psig) to clarify what is meant.
Absolute vacuum Absolute vacuum Pabs Pvac Patm Patm Patm Pgage Pabs Pabs = 0 FIGURE 3–3 Absolute, gage, and vacuum pressures.
P P P P P FIGURE 3–4 Pressure is a scalar quantity, not a vector; the pressure at a point in a fluid is the same in all directions.
EXAMPLE 3–1 Absolute Pressure of a Vacuum Chamber A vacuum gage connected to a chamber reads 5.8 psi at a location where the atmospheric pressure is 14.5 psi. Determine the absolute pressure in the chamber.
SOLUTION The gage pressure of a vacuum chamber is given. The absolute pressure in the chamber is to be determined.
Analysis The absolute pressure is easily determined from Eq. 3–2 to be Pabs = Patm −Pvac = 14.5 −5.8 = 8.7 psi Discussion Note that the local value of the atmospheric pressure is used when determining the absolute pressure.
Pressure at a Point Pressure is the compressive force per unit area, and it gives the impression of being a vector. However, pressure at any point in a fluid is the same in all directions (Fig. 3–4). That is, it has magnitude but not a specific direction, and thus it is a scalar quantity. This can be demonstrated by considering a small wedge-shaped fluid element of unit length (Δy = 1 into the page) in equilibrium, as shown in Fig. 3–5. The mean pressures at the three surfaces are P1, P2, and P3, and the force acting on a surface is the product of mean cen96537_ch03_077-136.indd 79 14/01/17 2:22 pm 80 pressure and fluid statics pressure and the surface area. From Newton’s second law, a force balance in the x- and z-directions gives ∑Fx = max = 0: P1 ΔyΔz −P3 Δyl sin 𝜃 = 0 (3–3a) ∑Fz = maz = 0: P2 ΔyΔx −P3 Δyl cos 𝜃 −1 2 ρg Δx Δy Δz = 0 (3–3b) where 𝜌 is the density and W = mg = 𝜌g Δx Δy Δz/2 is the weight of the fluid element. Noting that the wedge is a right triangle, we have Δx = l cos 𝜃 and Δz = l sin 𝜃. Substituting these geometric relations and dividing Eq. 3–3a by Δy Δz and Eq. 3–3b by Δx Δy gives P1 −P3 = 0 (3–4a) P2 −P3 −1 2 ρg Δz = 0 (3–4b) The last term in Eq. 3–4b drops out as Δz → 0 and the wedge becomes infinitesimal, and thus the fluid element shrinks to a point. Then combining the results of these two relations gives P1 = P2 = P3 = P (3–5) regardless of the angle 𝜃. We can repeat the analysis for an element in the yz-plane and obtain a similar result. Thus we conclude that the pressure at a point in a fluid has the same magnitude in all directions. This result is applicable to fluids in motion as well as fluids at rest since pressure is a scalar, not a vector.
Variation of Pressure with Depth It will come as no surprise to you that pressure in a fluid at rest does not change in the horizontal direction. This can be shown easily by considering a thin horizontal layer of fluid and doing a force balance in any horizontal direction. However, this is not the case in the vertical direction in a gravity field. Pressure in a fluid increases with depth because more fluid rests on deeper layers, and the effect of this “extra weight” on a deeper layer is bal anced by an increase in pressure (Fig. 3–6).
To obtain a relation for the variation of pressure with depth, consider a rectangular fluid element of height Δz, length Δx, and unit depth (Δy = 1 into the page) in equilibrium, as shown in Fig. 3–7. Assuming the density of the fluid 𝜌 to be constant, a force balance in the vertical z-direction gives ∑Fz = maz = 0: P1 Δx Δy −P2 Δx Δy −ρg Δx Δy Δz = 0 where W = mg = 𝜌g Δx Δy Δz is the weight of the fluid element and Δz = z2 − z1. Dividing by Δx Δy and rearranging gives ΔP = P2 −P1 = −ρg Δz = −𝛾s Δz (3–6) where 𝛾s = 𝜌g is the specific weight of the fluid. Thus, we conclude that the pressure difference between two points in a constant density fluid is propor tional to the vertical distance Δz between the points and the density 𝜌 of the fluid. Noting the negative sign, pressure in a static fluid increases linearly with depth. This is what a diver experiences when diving deeper in a lake.
z x P1 ΔyΔz P2 Δx Δy P3Δyl l g Δz Δx (Δy = 1) θ θ FIGURE 3–5 Forces acting on a wedge-shaped fluid element in equilibrium.
P gage FIGURE 3–6 The pressure of a fluid at rest increases with depth (as a result of added weight).
cen96537_ch03_077-136.indd 80 14/01/17 2:22 pm 81 CHAPTER 3 An easier equation to remember and apply between any two points in the same fluid under hydrostatic conditions is Pbelow = Pabove + ρg|Δz| = Pabove + 𝛾s|Δz| (3–7) where “below” refers to the point at lower elevation (deeper in the fluid) and “above” refers to the point at higher elevation. If you use this equation consistently, you should avoid sign errors.
For a given fluid, the vertical distance Δz is sometimes used as a measure of pressure, and it is called the pressure head.
We also conclude from Eq. 3–6 that for small to moderate distances, the variation of pressure with height is negligible for gases because of their low density. The pressure in a tank containing a gas, for example, can be con sidered to be uniform since the weight of the gas is too small to make a significant difference. Also, the pressure in a room filled with air can be approximated as a constant (Fig. 3–8).
If we take the “above” point to be at the free surface of a liquid open to the atmosphere (Fig. 3–9), where the pressure is the atmospheric pressure Patm, then from Eq. 3–7 the pressure at a depth h below the free surface becomes P = Patm + ρgh or Pgage = ρgh (3–8) Liquids are essentially incompressible substances, and thus the variation of density with depth is negligible. This is also the case for gases when the elevation change is not very large. The variation of density of liquids or gases with temperature can be significant, however, and may need to be considered when high accuracy is desired. Also, at great depths such as those encountered in oceans, the change in the density of a liquid can be significant because of the compression by the tremendous amount of liquid weight above.
The gravitational acceleration g varies from 9.807 m/s2 at sea level to 9.764 m/s2 at an elevation of 14,000 m where large passenger planes cruise. This is a change of just 0.4 percent in this extreme case. Therefore, g can be approximated as a constant with negligible error.
For fluids whose density changes significantly with elevation, a relation for the variation of pressure with elevation can be obtained by dividing Eq. 3–6 by Δz, and taking the limit as Δz → 0. This yields dP dz = −ρg (3–9) Note that dP is negative when dz is positive since pressure decreases in an upward direction. When the variation of density with elevation is known, the pressure difference between any two points 1 and 2 can be determined by integration to be ΔP = P2 −P1 = −∫ 2 1 ρg dz (3–10) For constant density and constant gravitational acceleration, this relation reduces to Eq. 3–6, as expected. Pressure in a fluid at rest is independent of the shape or cross section of the container. It changes with the vertical distance, but remains constant P1 W P2 0 z z2 z1 x Δx Δz g FIGURE 3–7 Free-body diagram of a rectangular fluid element in equilibrium.
Ptop = 1 atm Air (5-m-high room) Pbottom = 1.006 atm FIGURE 3–8 In a room filled with a gas, the variation of pressure with height is negligible.
cen96537_ch03_077-136.indd 81 14/01/17 2:22 pm 82 pressure and fluid statics in other directions. Therefore, the pressure is the same at all points on a horizontal plane in a given fluid. The Dutch mathematician Simon Stevin (1548–1620) published in 1586 the principle illustrated in Fig. 3–10. Note that the pressures at points A, B, C, D, E, F, and G are the same since they are at the same depth, and they are interconnected by the same static fluid. However, the pressures at points H and I are not the same since these two points cannot be interconnected by the same fluid (i.e., we cannot draw a curve from point I to point H while remaining in the same fluid at all times), although they are at the same depth. (Can you tell at which point the pressure is higher?) Also notice that the pressure force exerted by the fluid is always normal to the surface at the specified points. A consequence of the pressure in a fluid remaining constant in the hori zontal direction is that the pressure applied to a confined fluid increases the pressure throughout by the same amount. This is called Pascal’s law, after Blaise Pascal (1623–1662). Pascal also knew that the force applied by a fluid is proportional to the surface area. He realized that two hydrau lic cylinders of different areas could be connected, and the larger could be used to exert a proportionally greater force than that applied to the smaller. “Pascal’s machine” has been the source of many inventions that are a part of our daily lives such as hydraulic brakes and lifts. This is what enables us to lift a car easily by one arm, as shown in Fig. 3–11. Noting that P1 = P2 since both pistons are at the same level (the effect of small height differ ences is negligible, especially at high pressures), the ratio of output force to input force is determined to be P1 = P2 → F1 A1 = F2 A2 → F2 F1 = A2 A1 (3–11) The area ratio A2/A1 is called the ideal mechanical advantage of the hydraulic lift. Using a hydraulic car jack with a piston area ratio of A2/A1 = 100, for Pabove = Patm Pbelow = Patm + ρgh h FIGURE 3–9 Pressure in a liquid at rest increases linearly with vertical distance from the free surface.
FIGURE 3–10 Under hydrostatic conditions, the pressure is the same at all points on a horizontal plane in a given fluid regardless of geometry, provided that the points are interconnected by the same fluid.
h A B C D E Water Mercury F G I H Patm PA = PB = PC = PD = PE = PF = PG = Patm + ρgh PH ≠ PI cen96537_ch03_077-136.indd 82 14/01/17 2:22 pm 83 CHAPTER 3 example, a person can lift a 1000-kg car by applying a force of just 10 kgf (= 90.8 N). Related devices called pressure intensifiers are used to increase pressure through a combination of pistons and cylinders in tandem. Wide applica tions of Pascal’s law are also seen in hydraulic testing of pressurized tanks, calibration of pressure gages, pressing of oils such as olive, hazelnut, and sunflower oils, compression of wood stocks, etc.
EXAMPLE 3–2 Operation of a Hydraulic Jack Consider a hydraulic jack being used in a car repair shop, as in Fig. 3–12. The pistons have an area of A1 = 0.8 cm2 and A2 = 0.04 m2. Hydraulic oil with a specific gravity of 0.870 is pumped in as the small piston on the left side is pushed up and down, slowly raising the larger piston on the right side. A car that weighs 13,000 N is to be jacked up.
(a) At the beginning, when both pistons are at the same elevation (h = 0), calculate the force F1 in newtons required to hold the weight of the car. (b) Repeat the calculation after the car has been lifted two meters (h = 2 m).
Compare and discuss.
SOLUTION We are to determine the force required to lift a car with a hydraulic jack at two different elevations.
Assumptions 1 The oil is incompressible. 2 The system is at rest during the analysis (hydrostatics). 3 Air density is negligible compared to oil density.
Analysis (a) When h = 0, the pressure at the bottom of each piston must be the same. Thus, P1 = F1 A1 = P2 = F2 A2 →F1 = F2 A1 A2 = (13,000 N) 0.8 cm2 0.0400 m2 ( 1 m 100 cm) 2 = 26.0 N At the beginning, when h = 0, the required force is thus F1 = 26.0 N.
(b) When h ≠ 0, the hydrostatic pressure due to the elevation difference must be taken into account, namely, P1 = F1 A1 = P2 + ρgh = F2 A2 + ρgh F1 = F2 A1 A2 + ρghA1 = (13,000 N) 0.00008 m2 0.04 m2 + (870 kg/m3)(9.807 m/s2)(2.00 m)(0.00008 m2)( 1 N 1 kg·m/s2) = 27.4 N Thus, after the car has been raised 2 meters, the required force is 27.4 N.
Comparing the two results, it takes more force to keep the car elevated than it does to hold it at h = 0. This makes sense physically because the elevation difference generates a higher pressure (and thus a higher required force) at the lower piston due to hydrostatics.
Discussion When h = 0, the specific gravity (or density) of the hydraulic fluid does not enter the calculation—the problem simplifies to setting the two pressures equal. However, when h ≠ 0, there is a hydrostatic head and therefore the density of the fluid enters the calculation. The air pressure on the right side of the jack is actually slightly lower than that on the left side, but we have neglected this effect.
F1 = P1A1 1 2 A1 P1 F2 = P2A2 A2 P2 FIGURE 3–11 Lifting of a large weight by a small force by the application of Pascal’s law. A common example is a hydraulic jack.
(Top) © Stockbyte/Getty Images RF FIGURE 3–12 Schematic for Example 3–2.
Hydraulic oil SG = 0.870 h A1 F1 A2 F2 cen96537_ch03_077-136.indd 83 14/01/17 2:22 pm 84 pressure and fluid statics 3–2 ■ PRESSURE MEASUREMENT DEVICES The Barometer Atmospheric pressure is measured by a device called a barometer; thus, the atmospheric pressure is often referred to as the barometric pressure. The Italian Evangelista Torricelli (1608–1647) was the first to conclu sively prove that the atmospheric pressure can be measured by inverting a mercury-filled tube into a mercury container that is open to the atmosphere, as shown in Fig. 3–13. The pressure at point B is equal to the atmospheric pressure, and the pressure at point C can be taken to be zero since there is only mercury vapor above point C and the pressure is very low relative to Patm and can be neglected to an excellent approximation. Writing a force balance in the vertical direction gives Patm = ρgh (3–12) where 𝜌 is the density of mercury, g is the local gravitational acceleration, and h is the height of the mercury column above the free surface. Note that the length and the cross-sectional area of the tube have no effect on the height of the fluid column of a barometer (Fig. 3–14). A frequently used pressure unit is the standard atmosphere, which is defined as the pressure produced by a column of mercury 760 mm in height at 0°C (𝜌Hg = 13,595 kg/m3) under standard gravitational acceleration (g = 9.807 m/s2). If water instead of mercury were used to measure the standard atmospheric pressure, a water column of about 10.3 m would be needed. Pressure is sometimes expressed (especially by weather forecasters) in terms of the height of the mercury column. The standard atmospheric pressure, for example, is 760 mmHg (29.92 inHg) at 0°C. The unit mmHg is also called the torr in honor of Torricelli. Therefore, 1 atm = 760 torr and 1 torr = 133.3 Pa. Atmospheric pressure Patm changes from 101.325 kPa at sea level to 89.88, 79.50, 54.05, 26.5, and 5.53 kPa at altitudes of 1000, 2000, 5000, 10,000, and 20,000 meters, respectively. The typical atmospheric pressure in Denver (elevation = 1610 m), for example, is 83.4 kPa. Remember that the atmospheric pressure at a location is simply the weight of the air above that location per unit surface area. Therefore, it changes not only with elevation but also with weather conditions. The decline of atmospheric pressure with elevation has far-reaching rami fications in daily life. For example, cooking takes longer at high altitudes since water boils at a lower temperature at lower atmospheric pressures. Nose bleeding is a common experience at high altitudes since the difference between the blood pressure and the atmospheric pressure is larger in this case, and the delicate walls of veins in the nose are often unable to with stand this extra stress. For a given temperature, the density of air is lower at high altitudes, and thus a given volume contains less air and less oxygen. So it is no surprise that we tire more easily and experience breathing problems at high altitudes. To compensate for this effect, people living at higher altitudes develop more efficient lungs. Similarly, a 2.0-L car engine will act like a 1.7-L car engine at 1500 m altitude (unless it is turbocharged) because of the 15 percent drop A2 A1 A3 FIGURE 3–14 The length or the cross-sectional area of the tube has no effect on the height of the fluid column of a barometer, provided that the tube diameter is large enough to avoid surface tension (capillary) effects.
h A h B P ≈ 0 Vacuum Mercury C Patm W = ρghA FIGURE 3–13 The basic barometer.
cen96537_ch03_077-136.indd 84 14/01/17 2:22 pm 85 CHAPTER 3 in pressure and thus 15 percent drop in the density of air (Fig. 3–15). A fan or compressor will displace 15 percent less air at that altitude for the same volume displacement rate. Therefore, larger cooling fans may need to be selected for operation at high altitudes to ensure the specified mass flow rate. The lower pressure and thus lower density also affects lift and drag: airplanes need a longer runway at high altitudes to develop the required lift, and they climb to very high altitudes for cruising in order to reduce drag and thus achieve better fuel efficiency.
Engine Lungs FIGURE 3–15 At high altitudes, a car engine generates less power and a person gets less oxygen because of the lower density of air.
FIGURE 3–16 Schematic for Example 3–4.
1.2 m Patm IV bottle EXAMPLE 3–3 Measuring Atmospheric Pressure with a Barometer Determine the atmospheric pressure at a location where the barometric reading is 740 mmHg and the gravitational acceleration is g = 9.805 m/s2. Assume the temperature of mercury to be 10°C, at which its density is 13,570 kg/m3.
SOLUTION The barometric reading at a location in height of mercury column is given. The atmospheric pressure is to be determined.
Assumptions The temperature of mercury is assumed to be 10°C.
Properties The density of mercury is given to be 13,570 kg/m3.
Analysis From Eq. 3–12, the atmospheric pressure is determined to be Patm = ρgh = (13,570 kg/m3)(9.805 m/s2)(0.740 m)( 1 N 1 kg·m/s2) ( 1 kPa 1000 N/m2) = 98.5 kPa Discussion Note that density changes with temperature, and thus this effect should be considered in calculations.
EXAMPLE 3–4 Gravity Driven Flow from an IV Bottle Intravenous infusions usually are driven by gravity by hanging the fluid bottle at sufficient height to counteract the blood pressure in the vein and to force the fluid into the body (Fig. 3–16). The higher the bottle is raised, the higher the flow rate of the fluid will be. (a) If it is observed that the fluid and the blood pressures balance each other when the bottle is 1.2 m above the arm level, determine the gage pres sure of the blood. (b) If the gage pressure of the fluid at the arm level needs to be 20 kPa for sufficient flow rate, determine how high the bottle must be placed. Take the density of the fluid to be 1020 kg/m3.
SOLUTION It is given that an IV fluid and the blood pressures balance each other when the bottle is at a certain height. The gage pressure of the blood and elevation of the bottle required to maintain flow at the desired rate are to be deter mined.
cen96537_ch03_077-136.indd 85 14/01/17 2:22 pm 86 pressure and fluid statics EXAMPLE 3–5 Hydrostatic Pressure in a Solar Pond with Variable Density Solar ponds are small artificial lakes of a few meters deep that are used to store solar energy. The rise of heated (and thus less dense) water to the surface is pre vented by adding salt at the pond bottom. In a typical salt gradient solar pond, the density of water increases in the gradient zone, as shown in Fig. 3–17, and the den sity can be expressed as ρ = ρ0√1 + tan2( 𝜋 4 s H) where 𝜌0 is the density on the water surface, s is the vertical distance measured downward from the top of the gradient zone (s = −z), and H is the thickness of the gradient zone. For H = 4 m, 𝜌0 = 1040 kg/m3, and a thickness of 0.8 m for the surface zone, calculate the gage pressure at the bottom of the gradient zone.
SOLUTION The variation of density of saline water in the gradient zone of a solar pond with depth is given. The gage pressure at the bottom of the gradient zone is to be determined.
Assumptions The density in the surface zone of the pond is constant.
Assumptions 1 The IV fluid is incompressible. 2 The IV bottle is open to the atmosphere.
Properties The density of the IV fluid is given to be 𝜌 = 1020 kg/m3.
Analysis (a) Noting that the IV fluid and the blood pressures balance each other when the bottle is 1.2 m above the arm level, the gage pressure of the blood in the arm is simply equal to the gage pressure of the IV fluid at a depth of 1.2 m, Pgage, arm = Pabs −Patm = ρgharm−bottle = (1020 kg/m3)(9.81 m/s2)(1.20 m)( 1 kN 1000 kg·m/s2) ( 1 kPa 1 kN/m2) = 12.0 kPa (b) To provide a gage pressure of 20 kPa at the arm level, the height of the surface of the IV fluid in the bottle from the arm level is again determined from Pgage, arm = ρgharm−bottle to be harm−botttle = Pgage, arm ρg = 20 kPa (1020 kg/m3)(9.81 m/s2) ( 1000 kg·m/s2 1 kN ) ( 1 kN/m2 1 kPa ) = 2.00 m Discussion Note that the height of the reservoir can be used to control flow rates in gravity-driven flows. When there is flow, the pressure drop in the tube due to frictional effects also should be considered. For a specified flow rate, this requires raising the bottle a little higher to overcome the pressure drop.
Increasing salinity and density Surface zone Sun H = 4 m s Gradient zone Storage zone 1 2 ρ0 = 1040 kg/m3 FIGURE 3–17 Schematic for Example 3–5.
cen96537_ch03_077-136.indd 86 14/01/17 2:22 pm 87 CHAPTER 3 Properties The density of brine on the surface is given to be 1040 kg/m3.
Analysis We label the top and the bottom of the gradient zone as 1 and 2, respectively. Noting that the density of the surface zone is constant, the gage pressure at the bottom of the surface zone (which is the top of the gradient zone) is P1 = ρgh1 = (1040 kg/m3)(9.81 m/s2)(0.8 m)( 1 kN 1000 kg·m/s2) = 8.16 kPa since 1 kN/m2 = 1 kPa. Since s = −z, the differential change in hydrostatic pressure across a vertical distance of ds is given by dP = ρg ds Integrating from the top of the gradient zone (point 1 where s = 0) to any location s in the gradient zone (no subscript) gives P −P1 = ∫ s 0 ρg ds → P = P1 + ∫ s 0 ρ0√1 + tan2( 𝜋 4 s H)g ds Performing the integration gives the variation of gage pressure in the gradient zone to be P = P1 + ρ0g 4H 𝜋 sinh−1(tan 𝜋 4 s H) Then the pressure at the bottom of the gradient zone (s = H = 4 m) becomes P2 = 8.16 kPa + (1040 kg/m3)(9.81 m/s2) 4(4 m) 𝜋 sinh−1(tan 𝜋 4 4 4) ( 1 kN 1000 kg·m/s2) = 54.0 kPa (gage) Discussion The variation of gage pressure in the gradient zone with depth is plotted in Fig. 3–18. The dashed line indicates the hydrostatic pressure for the case of constant density at 1040 kg/m3 and is given for reference. Note that the variation of pressure with depth is not linear when density varies with depth. That is why integration was required.
The Manometer We notice from Eq. 3–6 that an elevation change of −Δz in a fluid at rest corresponds to ΔP/𝜌g, which suggests that a fluid column can be used to measure pressure differences. A device based on this principle is called a manometer, and it is commonly used to measure small and moderate pres sure differences. A manometer consists of a glass or plastic U-tube contain ing one or more fluids such as mercury, water, alcohol, or oil (Fig. 3–19). To keep the size of the manometer to a manageable level, heavy fluids such as mercury are used if large pressure differences are anticipated. Consider the manometer shown in Fig. 3–20 that is used to measure the pressure in the tank. Since the gravitational effects of gases are negligible, the pressure anywhere in the tank and at position 1 has the same value. Fur thermore, since pressure in a fluid does not vary in the horizontal direction within a fluid, the pressure at point 2 is the same as the pressure at point 1, P2 = P1.
4 3 Constant density Variable density 2 3.5 2.5 1.5 1 0.5 0 0 10 20 30 P, kPa s, m 40 50 60 FIGURE 3–18 The variation of gage pressure with depth in the gradient zone of the solar pond.
FIGURE 3–19 A simple U-tube manometer, with high pressure applied to the right side.
Photo by John M. Cimbala.
cen96537_ch03_077-136.indd 87 14/01/17 2:22 pm 88 pressure and fluid statics The differential fluid column of height h is in static equilibrium, and it is open to the atmosphere. Then the pressure at point 2 is determined directly from Eq. 3–7 to be P2 = Patm + ρgh (3–13) where 𝜌 is the density of the manometer fluid in the tube. Note that the cross-sectional area of the tube has no effect on the differential height h, and thus the pressure exerted by the fluid. However, the diameter of the tube should be large enough (more than several millimeters) to ensure that the surface tension effect and thus the capillary rise is negligible.
P SG = 0.85 = ?
h = 55 cm Patm = 96 kPa FIGURE 3–21 Schematic for Example 3–6.
Gas h 1 2 FIGURE 3–20 The basic manometer.
EXAMPLE 3–6 Measuring Pressure with a Manometer A manometer is used to measure the pressure of a gas in a tank. The fluid used has a specific gravity of 0.85, and the manometer column height is 55 cm, as shown in Fig. 3–21. If the local atmospheric pressure is 96 kPa, determine the absolute pressure within the tank.
SOLUTION The reading of a manometer attached to a tank and the atmospheric pressure are given. The absolute pressure in the tank is to be determined.
Assumptions The density of the gas in the tank is much lower than the density of the manometer fluid.
Properties The specific gravity of the manometer fluid is given to be 0.85. We take the standard density of water to be 1000 kg/m3.
Analysis The density of the fluid is obtained by multiplying its specific gravity by the density of water, ρ = SG (ρH2O) = (0.85)(1000 kg/m3) = 850 kg/m3 Then from Eq. 3–13, P = Patm + ρgh = 96 kPa + (850 kg/m3)(9.81 m/s2)(0.55 m)( 1 N 1 kg·m/s2) ( 1 kPa 1000 N/m2) = 100.6 kPa Discussion Note that the gage pressure in the tank is 4.6 kPa.
Some manometers use a slanted or inclined tube in order to increase the resolution (precision) when reading the fluid height. Such devices are called inclined manometers. Many engineering problems and some manometers involve multiple immiscible fluids of different densities stacked on top of each other. Such systems can be analyzed easily by remembering that (1) the pressure change across a fluid column of height h is ΔP = 𝜌gh, (2) pressure increases downward in a given fluid and decreases upward (i.e., Pbottom > Ptop), and (3) two points at the same elevation in a continuous fluid at rest are at the same pressure.
cen96537_ch03_077-136.indd 88 14/01/17 2:22 pm 89 CHAPTER 3 The last principle, which is a result of Pascal’s law, allows us to “jump” from one fluid column to the next in manometers without worrying about pressure change as long as we stay in the same continuous fluid and the fluid is at rest. Then the pressure at any point can be determined by start ing with a point of known pressure and adding or subtracting 𝜌gh terms as we advance toward the point of interest. For example, the pressure at the bottom of the tank in Fig. 3–22 can be determined by starting at the free surface where the pressure is Patm, moving downward until we reach point 1 at the bottom, and setting the result equal to P1. It gives Patm + ρ1gh1 + ρ2gh2 + ρ3gh3 = P1 In the special case of all fluids having the same density, this relation reduces to Patm + 𝜌g(h1 + h2 + h3) = P1. Manometers are particularly well-suited to measure pressure drops across a horizontal flow section between two specified points due to the presence of a device such as a valve or heat exchanger or any resistance to flow. This is done by connecting the two legs of the manometer to these two points, as shown in Fig. 3–23. The working fluid can be either a gas or a liquid whose density is 𝜌1. The density of the manometer fluid is 𝜌2, and the differential fluid height is h. The two fluids must be immiscible, and 𝜌2 must be greater than 𝜌1. A relation for the pressure difference P1 − P2 can be obtained by starting at point 1 with P1, moving along the tube by adding or subtracting the 𝜌gh terms until we reach point 2, and setting the result equal to P2: P1 + ρ1g(a + h) −ρ2gh −ρ1ga = P2 (3–14) Note that we jumped from point A horizontally to point B and ignored the part underneath since the pressure at both points is the same. Simplifying, P1 −P2 = (ρ2 −ρ1)gh (3–15) Note that the distance a must be included in the analysis even though it has no effect on the result. Also, when the fluid flowing in the pipe is a gas, then 𝜌1 ≪ 𝜌2 and the relation in Eq. 3–15 simplifies to P1 − P2 ≅ 𝜌2gh.
Patm 1 h3 h2 h1 Fluid 2 Fluid 1 Fluid 3 FIGURE 3–22 In stacked-up fluid layers at rest, the pressure change across each fluid layer of density 𝜌 and height h is 𝜌gh.
EXAMPLE 3–7 Measuring Pressure with a Multifluid Manometer The water in a tank is pressurized by air, and the pressure is measured by a mul tifluid manometer as shown in Fig. 3–24. The tank is located on a mountain at an altitude of 1400 m where the atmospheric pressure is 85.6 kPa. Determine the air pressure in the tank if h1 = 0.1 m, h2 = 0.2 m, and h3 = 0.35 m. Take the densi ties of water, oil, and mercury to be 1000 kg/m3, 850 kg/m3, and 13,600 kg/m3, respectively.
SOLUTION The pressure in a pressurized water tank is measured by a multi fluid manometer. The air pressure in the tank is to be determined.
Assumption The air pressure in the tank is uniform (i.e., its variation with eleva tion is negligible due to its low density), and thus we can determine the pressure at the air–water interface.
a h ρ1 A B Fluid flow A flow section or flow device 1 2 ρ2 FIGURE 3–23 Measuring the pressure drop across a flow section or a flow device by a differential manometer.
cen96537_ch03_077-136.indd 89 14/01/17 2:22 pm 90 pressure and fluid statics Properties The densities of water, oil, and mercury are given to be 1000 kg/m3, 850 kg/m3, and 13,600 kg/m3, respectively.
Analysis Starting with the pressure at point 1 at the air–water interface, moving along the tube by adding or subtracting the 𝜌gh terms until we reach point 2, and setting the result equal to Patm since the tube is open to the atmosphere gives P1 + ρwatergh1 + ρoilgh2 −ρmercurygh3 = P2 = Patm Solving for P1 and substituting, P1 = Patm −ρwatergh1 −ρoilgh2 + ρmercurygh3 = Patm + g(ρmercuryh3 −ρwaterh1 −ρoilh2) = 85.6 kPa + (9.81 m/s2)(13,600 kg/m3)(0.35 m) −(1000 kg/m3)(0.1 m) −(850 kg/m3)(0.2 m) ( 1 kPa 1000 N/m2) = 130 kPa Discussion Note that jumping horizontally from one tube to the next and real izing that pressure remains the same in the same fluid simplifies the analysis con siderably. Also note that mercury is a toxic fluid, and mercury manometers and thermometers are being replaced by ones with safer fluids because of the risk of exposure to mercury vapor during an accident.
Other Pressure Measurement Devices Another type of commonly used mechanical pressure measurement device is the Bourdon tube, named after the French engineer and inventor Eugene Bourdon (1808–1884), which consists of a bent, coiled, or twisted hollow metal tube whose end is closed and connected to a dial indicator needle (Fig. 3–25). When the tube is open to the atmosphere, the tube is undeflected, and the needle on the dial at this state is calibrated to read zero (gage pressure). When the fluid inside the tube is pressurized, the tube stretches and moves the needle in proportion to the applied pressure. Electronics have made their way into every aspect of life, including pres sure measurement devices. Modern pressure sensors, called pressure trans ducers, use various techniques to convert the pressure effect to an electri cal effect such as a change in voltage, resistance, or capacitance. Pressure transducers are smaller and faster, and they can be more sensitive, reliable, and precise than their mechanical counterparts. They can measure pressures from less than a millionth of 1 atm to several thousands of atm. A wide variety of pressure transducers is available to measure gage, abso lute, and differential pressures in a wide range of applications. Gage pres sure transducers use the atmospheric pressure as a reference by venting the back side of the pressure-sensing diaphragm to the atmosphere, and they give a zero signal output at atmospheric pressure regardless of altitude. Absolute pressure transducers are calibrated to have a zero signal output at full vacuum. Differential pressure transducers measure the pressure difference between two locations directly instead of using two pressure transducers and taking their difference.
h1 h2 h3 Oil Mercury Water Air 1 2 FIGURE 3–24 Schematic for Example 3–7; drawing not to scale.
C-type Spiral Twisted tube Tube cross section Helical FIGURE 3–25 Various types of Bourdon tubes used to measure pressure. They work on the same principle as party noise-makers (bottom photo) due to the flat tube cross section.
(Bottom) Photo by John M. Cimbala.
cen96537_ch03_077-136.indd 90 14/01/17 2:22 pm 91 CHAPTER 3 Strain-gage pressure transducers work by having a diaphragm deflect between two chambers open to the pressure inputs. As the diaphragm stretches in response to a change in pressure difference across it, the strain gage stretches and a Wheatstone bridge circuit amplifies the output. A capacitance transducer works similarly, but capacitance change is measured instead of resistance change as the diaphragm stretches. Piezoelectric transducers, also called solid-state pressure transducers, work on the principle that an electric potential is generated in a crystalline substance when it is subjected to mechanical pressure. This phenomenon, first discovered by brothers Pierre and Jacques Curie in 1880, is called the piezoelectric (or press-electric) effect. Piezoelectric pressure transducers have a much faster frequency response compared to diaphragm units and are very suitable for high-pressure applications, but they are generally not as sensitive as diaphragm-type transducers, especially at low pressures. Another type of mechanical pressure gage called a deadweight tester is used primarily for calibration and can measure extremely high pres sures (Fig. 3–26). As its name implies, a deadweight tester measures pres sure directly through application of a weight that provides a force per unit area—the fundamental definition of pressure. It is constructed with an inter nal chamber filled with a fluid (usually oil), along with a tight-fitting piston, cylinder, and plunger. Weights are applied to the top of the piston, which exerts a force on the oil in the chamber. The total force F acting on the oil at the piston–oil interface is the sum of the weight of the piston plus the applied weights. Since the piston cross-sectional area Ae is known, the pres sure is calculated as P = F/Ae. The only significant source of error is that due to static friction along the interface between the piston and cylinder, but even this error is usually negligibly small. The reference pressure port is connected to either an unknown pressure that is to be measured or to a pres sure sensor that is to be calibrated.
3–3 ■ INTRODUCTION TO FLUID STATICS Fluid statics deals with problems associated with fluids at rest. The fluid can be either gaseous or liquid. Fluid statics is generally referred to as hydrostatics when the fluid is a liquid and as aerostatics when the fluid is a gas. In fluid statics, there is no relative motion between adjacent fluid layers, and thus there are no shear (tangential) stresses in the fluid trying to deform it. The only stress we deal with in fluid statics is the normal stress, which is the pressure, and the variation of pressure is due only to the weight of the fluid. Therefore, the topic of fluid statics has significance only in gravity fields, and the force relations developed naturally involve the gravi tational acceleration g. The force exerted on a surface by a fluid at rest is normal to the surface at the point of contact since there is no relative motion between the fluid and the solid surface, and thus there are no shear forces acting parallel to the surface. Fluid statics is used to determine the forces acting on floating or sub merged bodies and the forces developed by devices like hydraulic presses and car jacks. The design of many engineering systems such as water dams and liquid storage tanks requires the determination of the forces acting on their surfaces using fluid statics. The complete description of the resultant FIGURE 3–26 A deadweight tester is able to measure extremely high pressures (up to 10,000 psi in some applications). Ae F Oil reservoir Adjustable plunger Crank Oil Internal chamber Reference pressure port Piston Weights cen96537_ch03_077-136.indd 91 14/01/17 2:22 pm 92 pressure and fluid statics hydrostatic force acting on a submerged surface requires the determination of the magnitude, the direction, and the line of action of the force. In the following two sections, we consider the forces acting on both plane and curved surfaces of submerged bodies due to pressure.
3–4 ■ HYDROSTATIC FORCES ON SUBMERGED PLANE SURFACES A plate (such as a gate valve in a dam, the wall of a liquid storage tank, or the hull of a ship at rest) is subjected to fluid pressure distributed over its surface when exposed to a liquid (Fig. 3–27). On a plane surface, the hydro static forces form a system of parallel forces, and we often need to deter mine the magnitude of the force and its point of application, which is called the center of pressure. In most cases, the other side of the plate is open to the atmosphere (such as the dry side of a gate), and thus atmospheric pres sure acts on both sides of the plate, yielding a zero resultant. In such cases, it is convenient to subtract atmospheric pressure and work with the gage pres sure only (Fig. 3–28). For example, Pgage = 𝜌gh at the bottom of the lake. Consider the top surface of a flat plate of arbitrary shape completely sub merged in a liquid, as shown in Fig. 3–29 together with its normal view. The plane of this surface (normal to the page) intersects the horizontal free surface at angle 𝜃, and we take the line of intersection to be the x-axis (out of the page). The absolute pressure above the liquid is P0, which is the local atmospheric pressure Patm if the liquid is open to the atmosphere (but P0 may be different than Patm if the space above the liquid is evacuated or pres surized). Then the absolute pressure at any point on the plate is P = P0 + ρgh = P0 + ρgy sin 𝜃 (3–16) where h is the vertical distance of the point from the free surface and y is the distance of the point from the x-axis (from point O in Fig. 3–29). The resultant hydrostatic force FR acting on the surface is determined by h h Patm Patm + ρgh (a) Patm considered (b) Patm subtracted ρgh = FIGURE 3–28 When analyzing hydrostatic forces on submerged surfaces, the atmospheric pressure can be subtracted for simplicity when it acts on both sides of the structure.
dA C CP Centroid Center of pressure Plane surface of area A h = y sin 휃 P = P0 + ρgy sin 휃 O PC = Pavg FR = PC A Pressure prism Pressure distribution y y z yC A dA Plane surface P = P0 + ρgh FR = ∫ P dA 휃 FIGURE 3–29 Hydrostatic force on an inclined plane surface completely submerged in a liquid.
FIGURE 3–27 Hoover Dam.
© Comstock Images/Jupiterimages RF cen96537_ch03_077-136.indd 92 14/01/17 2:22 pm 93 CHAPTER 3 integrating the force P dA acting on a differential area dA over the entire surface area, FR = ∫A P dA = ∫A (P0 + ρgy sin 𝜃 ) dA = P0A + ρg sin 𝜃 ∫A y dA (3–17) But the first moment of area ∫A y dA is related to the y-coordinate of the centroid (or center) of the surface by yC = 1 A ∫A y dA (3–18) Substituting, FR = (P0 + ρgyC sin 𝜃 )A = (P0 + ρghC)A = PC A = Pavg A (3–19) where PC = P0 + 𝜌ghC is the pressure at the centroid of the surface, which is equivalent to the average pressure Pavg on the surface, and hC = yC sin 𝜃 is the vertical distance of the centroid from the free surface of the liquid (Fig. 3–30). Thus we conclude that: The magnitude of the resultant force acting on a plane surface of a completely submerged plate in a homogeneous (constant density) fluid is equal to the product of the pressure PC at the centroid of the surface and the area A of the surface (Fig. 3–31). The pressure P0 is usually atmospheric pressure, which can be ignored in most force calculations since it acts on both sides of the plate. When this is not the case, a practical way of accounting for the contribution of P0 to the resultant force is simply to add an equivalent depth hequiv = P0/𝜌g to hC; that is, to assume the presence of an additional liquid layer of thick ness hequiv on top of the liquid with absolute vacuum above. Next we need to determine the line of action of the resultant force FR. Two parallel force systems are equivalent if they have the same magnitude and the same moment about any point. The line of action of the resultant hydrostatic force, in general, does not pass through the centroid of the sur face—it lies underneath where the pressure is higher. The point of inter section of the line of action of the resultant force and the surface is the center of pressure. The vertical location of the line of action is determined by equating the moment of the resultant force to the moment of the distrib uted pressure force about the x-axis: yPFR = ∫A yP dA = ∫A y(P0 + ρgy sin 𝜃 ) dA = P0 ∫A y dA + ρg sin 𝜃 ∫A y2 dA or yPFR = P0 yC A + ρg sin 𝜃 Ixx, O (3–20) where yP is the distance of the center of pressure from the x-axis (point O in Fig. 3–31) and Ixx, O = ∫A y 2 dA is the second moment of area (also called the area moment of inertia) about the x-axis. The second moments of area are widely available for common shapes in engineering handbooks, but they are usually given about the axes passing through the centroid of the area. Fortunately, the second moments of area about two parallel axes are Free surface hC Patm = PC = Patm + ρghC Pavg Centroid of surface FIGURE 3–30 The pressure at the centroid of a plane surface is equivalent to the average pressure on the surface.
Center of pressure Centroid of area Line of action 0 FR = PC A yC yP z 휃 FIGURE 3–31 The resultant force acting on a plane surface is equal to the product of the pressure at the centroid of the surface and the surface area, and its line of action passes through the center of pressure.
cen96537_ch03_077-136.indd 93 14/01/17 2:22 pm 94 pressure and fluid statics related to each other by the parallel axis theorem, which in this case is expressed as Ixx, O = Ixx, C + y 2 C A (3–21) where Ixx, C is the second moment of area about the x-axis passing through the centroid of the area and yC (the y-coordinate of the centroid) is the distance between the two parallel axes. Substituting the FR relation from Eq. 3–19 and the Ixx, O relation from Eq. 3–21 into Eq. 3–20 and solving for yP yields yP = yC + Ixx, C [yC + P0 /(ρg sin 𝜃 )]A (3–22a) For P0 = 0, which is usually the case when the atmospheric pressure is ignored, it simplifies to yP = yC + Ixx, C yCA (3–22b) Knowing yP, the vertical distance of the center of pressure from the free surface is determined from hP = yP sin 𝜃.
The Ixx, C values for some common areas are given in Fig. 3–32. For areas that possess symmetry about the y-axis, the center of pressure lies on the y-axis directly below the centroid. The location of the center of pressure in such cases is simply the point on the surface of the vertical plane of sym metry at a distance hP from the free surface.
b/2 b C C C b/2 A = ab, Ixx, C = ab3/12 a/2 a/2 y x (a) Rectangle R R R A = πR2, Ixx, C = πR4/4 b a y x (b) Circle A = πab, Ixx, C = πab3/4 y x (c) Ellipse 2b/3 b/3 C C C A = ab/2, Ixx, C = ab3/36 a/2 a/2 y x (d) Triangle R A = πR2/2, Ixx, C = 0.109757R4 a y x (e) Semicircle A = πab/2, Ixx, C = 0.109757ab3 y x (f ) Semiellipse b 4b 3π 4R 3π FIGURE 3–32 The centroid and the centroidal moments of inertia for some common geometries.
cen96537_ch03_077-136.indd 94 14/01/17 2:22 pm 95 CHAPTER 3 Pressure acts normal to the surface, and the hydrostatic forces acting on a flat plate of any shape form a volume whose base is the plate area and whose length is the linearly varying pressure, as shown in Fig. 3–33. This virtual pressure prism has an interesting physical interpretation: its volume is equal to the magnitude of the resultant hydrostatic force acting on the plate since FR = ʃ P dA, and the line of action of this force passes through the centroid of this homogeneous prism. The projection of the centroid on the plate is the pressure center. Therefore, with the concept of pressure prism, the problem of describing the resultant hydrostatic force on a plane surface reduces to find ing the volume and the two coordinates of the centroid of this pressure prism.
Special Case: Submerged Rectangular Plate Consider a completely submerged rectangular flat plate of height b and width a tilted at an angle 𝜃 from the horizontal and whose top edge is hori zontal and is at a distance s from the free surface along the plane of the plate, as shown in Fig. 3–34a. The resultant hydrostatic force on the upper surface is equal to the average pressure, which is the pressure at the mid point of the surface, times the surface area A. That is, Tilted rectangular plate: FR = PC A = [P0 + ρg(s + b/2) sin 𝜃 ]ab (3–23) The force acts at a vertical distance of hP = yP sin 𝜃 from the free surface directly beneath the centroid of the plate where, from Eq. 3–22a, yP = s + b 2 + ab3/12 [s + b/2 + P0 /(ρg sin 𝜃 )]ab = s + b 2 + b2 12[s + b/2 + P0 /(ρg sin 𝜃 )] (3–24) When the upper edge of the plate is at the free surface and thus s = 0, Eq. 3–23 reduces to Tilted rectangular plate (s = 0): FR = [P0 + ρg(b sin 𝜃 )/2]ab (3–25) Surface a b P Pressure prism FIGURE 3–33 The hydrostatic forces acting on a plane surface form a pressure prism whose base (left face) is the surface and whose length is the pressure.
O s yp b O s yp b F R = [P0 + ρg(s + b/2) sin 휃 ]ab F R = [P0 + ρg(s + b/2)]ab F R = (P0 + ρgh)ab (b) Vertical plate (c) Horizontal plate (a) Tilted plate a h P0 P0 P0 휃 FIGURE 3–34 Hydrostatic force acting on the top surface of a submerged rectangular plate for tilted, vertical, and horizontal cases.
cen96537_ch03_077-136.indd 95 14/01/17 2:22 pm 96 pressure and fluid statics For a completely submerged vertical plate (𝜃 = 90°) whose top edge is hori zontal, the hydrostatic force can be obtained by setting sin 𝜃 = 1 (Fig. 3–34b) Vertical rectangular plate: FR = [P0 + ρg(s + b/2)]ab (3–26) Vertical rectangular plate (s = 0): FR = (P0 + ρgb/2)ab (3–27) When the effect of P0 is ignored since it acts on both sides of the plate, the hydrostatic force on a vertical rectangular surface of height b whose top edge is horizontal and at the free surface is FR = 𝜌gab2/2 acting at a dis tance of 2b/3 from the free surface directly beneath the centroid of the plate. The pressure distribution on a submerged horizontal surface is uniform, and its magnitude is P = P0 + 𝜌gh, where h is the distance of the surface from the free surface. Therefore, the hydrostatic force acting on a horizontal rectangular surface is Horizontal rectangular plate: FR = (P0 + ρgh)ab (3–28) and it acts through the midpoint of the plate (Fig. 3–33c).
1.2 m 8 m Lake 1 m FIGURE 3–35 Schematic for Example 3–8.
EXAMPLE 3–8 Hydrostatic Force Acting on the Door of a Submerged Car A heavy car plunges into a lake during an accident and lands at the bottom of the lake on its wheels (Fig. 3–35). The door is 1.2 m high and 1 m wide, and the top edge of the door is 8 m below the free surface of the water. Determine the hydro static force on the door and the location of the pressure center, and discuss if the driver can open the door.
SOLUTION A car is submerged in water. The hydrostatic force on the door is to be determined, and the likelihood of the driver opening the door is to be assessed.
Assumptions 1 The bottom surface of the lake is horizontal. 2 The passenger cabin is well-sealed so that no water leaks inside. 3 The door can be approximated as a vertical rectangular plate. 4 The pressure in the passenger cabin remains at atmospheric value since there is no water leaking in, and thus no compression of the air inside. Therefore, atmospheric pressure cancels out in the calculations since it acts on both sides of the door. 5 The weight of the car is larger than the buoyant force acting on it.
Properties We take the density of lake water to be 1000 kg/m3 throughout.
Analysis The average (gage) pressure on the door is the pressure value at the centroid (midpoint) of the door and is determined to be Pavg = PC = ρghC = ρg(s + b/2) = (1000 kg/m3)(9.81 m/s2)(8 + 1.2/2 m)( 1 kN 1000 kg·m/s2) = 84.4 kN/m2 Then the resultant hydrostatic force on the door becomes FR = PavgA = (84.4 kN/m2) (1 m × 1.2 m) = 101.3 kN cen96537_ch03_077-136.indd 96 14/01/17 2:22 pm 97 CHAPTER 3 3–5 ■ HYDROSTATIC FORCES ON SUBMERGED CURVED SURFACES In many practical applications, submerged surfaces are not flat (Fig. 3–36). For a submerged curved surface, the determination of the resultant hydro static force is more involved since it typically requires integration of the pressure forces that change direction along the curved surface. The concept of the pressure prism in this case is not much help either because of the complicated shapes involved. The easiest way to determine the resultant hydrostatic force FR acting on a two-dimensional curved surface is to determine the horizontal and vertical components FH and FV separately. This is done by considering the free-body diagram of the liquid block enclosed by the curved surface and the two plane surfaces (one horizontal and one vertical) passing through the two ends of the curved surface, as shown in Fig. 3–37. Note that the vertical surface of the liquid block considered is simply the projection of the curved surface on a vertical plane, and the horizontal surface is the projection of the curved surface on a horizontal plane. The resultant force acting on the curved solid surface is then equal and opposite to the force acting on the curved liquid surface (Newton’s third law). The force acting on the imaginary horizontal or vertical plane surface and its line of action can be determined as discussed in Section 3–4. The weight of the enclosed liquid block of volume V is simply W = 𝜌gV, and it acts downward through the centroid of this volume. Noting that the fluid block is in static equilibrium, the force balances in the horizontal and vertical direc tions give Horizontal force component on curved surface: FH = Fx (3–29) Vertical force component on curved surface: FV = Fy ± W (3–30) FIGURE 3–36 In many structures of practical application, the submerged surfaces are not flat, but curved as here at Glen Canyon Dam in Utah and Arizona.
© Corbis RF The pressure center is directly under the midpoint of the door, and its dis tance from the surface of the lake is determined from Eq. 3–24 by setting P0 = 0, yielding yP = s + b 2 + b2 12(s + b/2) = 8 + 1.2 2 + 1.22 12(8 + 1.2/2) = 8.61 m Discussion A strong person can lift 100 kg, which is a weight of 981 N or about 1 kN. Also, the person can apply the force at a point farthest from the hinges (1 m farther) for maximum effect and generate a moment of 1 kN·m. The resultant hydrostatic force acts under the midpoint of the door, and thus a dis tance of 0.5 m from the hinges. This generates a moment of 50.6 kN·m, which is about 50 times the moment the driver can possibly generate. Therefore, it is impossible for the driver to open the door of the car. The driver’s best bet is to let some water in (by rolling the window down a little, for example) and to keep his or her head close to the ceiling. The driver should be able to open the door shortly before the car is filled with water since at that point the pressures on both sides of the door are nearly the same and opening the door in water is almost as easy as opening it in air.
cen96537_ch03_077-136.indd 97 14/01/17 2:23 pm 98 pressure and fluid statics where the summation Fy ± W is a vector addition (i.e., add magnitudes if both act in the same direction and subtract if they act in opposite directions). Thus, we conclude that 1. The horizontal component of the hydrostatic force acting on a curved surface is equal (in both magnitude and the line of action) to the hydrostatic force acting on the vertical projection of the curved surface. 2. The vertical component of the hydrostatic force acting on a curved surface is equal to the hydrostatic force acting on the horizontal projection of the curved surface, plus (minus, if acting in the opposite direction) the weight of the fluid block. The magnitude of the resultant hydrostatic force acting on the curved sur face is FR = √F2 H + F2 V, and the tangent of the angle it makes with the hori zontal is tan 𝛼 = FV /FH. The exact location of the line of action of the resul tant force (e.g., its distance from one of the end points of the curved surface) can be determined by taking a moment about an appropriate point. These discussions are valid for all curved surfaces regardless of whether they are above or below the liquid. Note that in the case of a curved surface above a liquid, the weight of the liquid is subtracted from the vertical component of the hydrostatic force since they act in opposite directions (Fig. 3–38). When the curved surface is a circular arc (full circle or any part of it), the resultant hydrostatic force acting on the surface always passes through the center of the circle. This is because the pressure forces are normal to the sur face, and all lines normal to the surface of a circle pass through the center of the circle. Thus, the pressure forces form a concurrent force system at the cen ter, which can be reduced to a single equivalent force at that point (Fig. 3–39). Finally, the hydrostatic force acting on a plane or curved surface submerged in a multilayered fluid of different densities can be determined by consid ering different parts of surfaces in different fluids as different surfaces, find ing the force on each part, and then adding them using vector addition. For a plane surface, it can be expressed as (Fig. 3–40) Plane surface in a multilayered fluid: FR = ∑FR, i = ∑PC, i Ai (3–31) Horizontal projection of the curved surface Curved surface Liquid Liquid block Vertical projection of the curved surface Free-body diagram of the enclosed liquid block a A B A B C C b FH FR Fx Fy W FV FIGURE 3–37 Determination of the hydrostatic force acting on a submerged curved surface.
Curved surface W Fx Fy FIGURE 3–38 When a curved surface is above the liquid, the weight of the liquid and the vertical component of the hydrostatic force act in the opposite directions.
cen96537_ch03_077-136.indd 98 14/01/17 2:23 pm 99 CHAPTER 3 where PC, i = P0 + 𝜌ighC, i is the pressure at the centroid of the portion of the surface in fluid i and Ai is the area of the plate in that fluid. The line of action of this equivalent force can be determined from the requirement that the moment of the equivalent force about any point is equal to the sum of the moments of the individual forces about the same point.
O FR Resultant force Circular surface Pressure forces FIGURE 3–39 The hydrostatic force acting on a circular surface always passes through the center of the circle since the pressure forces are normal to the surface and they all pass through the center.
FR1 b1 FR2 b2 Oil Water FIGURE 3–40 The hydrostatic force on a surface submerged in a multilayered fluid can be determined by considering parts of the surface in different fluids as different surfaces.
EXAMPLE 3–9 A Gravity-Controlled Cylindrical Gate A long solid cylinder of radius 0.8 m hinged at point A is used as an automatic gate, as shown in Fig. 3–41. When the water level reaches 5 m, the gate opens by turning about the hinge at point A. Determine (a) the hydrostatic force acting on the cylin der and its line of action when the gate opens and (b) the weight of the cylinder per m length of the cylinder.
SOLUTION The height of a water reservoir is controlled by a cylindrical gate hinged to the reservoir. The hydrostatic force on the cylinder and the weight of the cylinder per m length are to be determined.
Assumptions 1 Friction at the hinge is negligible. 2 Atmospheric pressure acts on both sides of the gate, and thus it cancels out.
Properties We take the density of water to be 1000 kg/m3 throughout.
Analysis (a) We consider the free-body diagram of the liquid block enclosed by the circular surface of the cylinder and its vertical and horizontal projections. The hydrostatic forces acting on the vertical and horizontal plane surfaces as well as the weight of the liquid block are determined as Horizontal force on vertical surface: FH = Fx = Pavg A = ρghCA= ρg(s + R/2)A = (1000 kg/m3)(9.81 m/s2)(4.2 + 0.8/2 m)(0.8 m × 1 m)( 1 kN 1000 kg·m/s2) = 36.1 kN Vertical force on horizontal surface (upward): Fy = Pavg A = ρghC A = ρghbottom A = (1000 kg/m3)(9.81 m/s2)(5 m)(0.8 m × 1 m)( 1 kN 1000 kg·m/s2) = 39.2 kN Weight (downward) of fluid block for one m width into the page: W = mg = ρgV = ρg(R2 −𝜋R2/4)(1 m) = (1000 kg/m3)(9.81 m/s2)(0.8 m)2(1 −𝜋/4)(1 m)( 1 kN 1000 kg·m/s2) = 1.3 kN Therefore, the net upward vertical force is FV = Fy −W = 39.2 −1.3 = 37.9 kN Then the magnitude and direction of the hydrostatic force acting on the cylindrical surface become FR = √F 2 H + F 2 V = √36.12 + 37.92 = 52.3 kN tan 𝜃 = FV/FH = 37.9/36.1 = 1.05 → 𝜃 = 46.4° cen96537_ch03_077-136.indd 99 14/01/17 2:23 pm 100 pressure and fluid statics 3–6 ■ BUOYANCY AND STABILITY It is a common experience that an object feels lighter and weighs less in a liq uid than it does in air. This can be demonstrated easily by weighing a heavy object in water by a waterproof spring scale. Also, objects made of wood or other light materials float on water. These and other observations suggest that a fluid exerts an upward force on a body immersed in it. This force that tends to lift the body is called the buoyant force and is denoted by FB. The buoyant force is caused by the increase of pressure with depth in a fluid. Consider, for example, a flat plate of thickness h submerged in a liq uid of density 𝜌f parallel to the free surface, as shown in Fig. 3–42. The area of the top (and also bottom) surface of the plate is A, and its distance to the free surface is s. The gage pressures at the top and bottom surfaces of the plate are 𝜌fgs and 𝜌fg(s + h), respectively. Then the hydrostatic force Ftop = 𝜌fgsA acts downward on the top surface, and the larger force Fbottom = 𝜌fg(s + h)A acts upward on the bottom surface of the plate. The difference between these two forces is a net upward force, which is the buoyant force, FB = Fbottom −Ftop = ρf g(s + h)A −ρf gsA = ρf ghA = ρf gV (3–32) where V = hA is the volume of the plate. But the relation 𝜌fgV is simply the weight of the liquid whose volume is equal to the volume of the plate. Thus, we conclude that the buoyant force acting on the plate is equal to the weight of the liquid displaced by the plate. For a fluid with constant density, the buoyant force is independent of the distance of the body from the free surface. It is also independent of the density of the solid body. The relation in Eq. 3–32 is developed for a simple geometry, but it is valid for any body regardless of its shape. This can be shown mathemati cally by a force balance, or simply by this argument: Consider an arbitrarily shaped solid body submerged in a fluid at rest and compare it to a body of fluid of the same shape indicated by dashed lines at the same vertical location (Fig. 3–43). The buoyant forces acting on these two bodies are the same since the pressure distributions, which depend only on elevation, are the same at the boundaries of both. The imaginary fluid body is in static Wcyl Fx FR A W R = 0.8 m 0.8 m 5 m s = 4.2 m Fy FR FH FV 휃 FIGURE 3–41 Schematic for Example 3–9 and the free-body diagram of the liquid underneath the cylinder.
ρfgsA s h ρfg(s + h)A A FIGURE 3–42 A flat plate of uniform thickness h submerged in a liquid parallel to the free surface.
Therefore, the magnitude of the hydrostatic force acting on the cylinder is 52.3 kN per m length of the cylinder, and its line of action passes through the center of the cylinder making an angle 46.4° with the horizontal.
(b) When the water level is 5 m high, the gate is about to open and thus the reaction force at the bottom of the cylinder is zero. Then the forces other than those at the hinge acting on the cylinder are its weight, acting through the center, and the hydro static force exerted by water. Taking a moment about point A at the location of the hinge and equating it to zero gives FRR sin 𝜃 −Wcyl R = 0 → Wcyl = FR sin 𝜃 = (52.3 kN) sin 46.4° = 37.9 kN Discussion The weight of the cylinder per m length is determined to be 37.9 kN. It can be shown that this corresponds to a mass of 3863 kg per m length and to a density of 1921 kg/m3 for the material of the cylinder.
cen96537_ch03_077-136.indd 100 14/01/17 2:23 pm 101 CHAPTER 3 equilibrium, and thus the net force and net moment acting on it are zero. Therefore, the upward buoyant force must be equal to the weight of the imaginary fluid body whose volume is equal to the volume of the solid body. Further, the weight and the buoyant force must have the same line of action to have a zero moment. This is known as Archimedes’ principle, after the Greek mathematician Archimedes (287–212 bc), and is expressed as The buoyant force acting on a body of uniform density immersed in a fluid is equal to the weight of the fluid displaced by the body, and it acts upward through the centroid of the displaced volume. For floating bodies, the weight of the entire body must be equal to the buoyant force, which is the weight of the fluid whose volume is equal to the volume of the submerged portion of the floating body. That is, FB = W → ρfgVsub = ρavg, bodygVtotal → Vsub Vtotal = ρavg, body ρf (3–33) Therefore, the submerged volume fraction of a floating body is equal to the ratio of the average density of the body to the density of the fluid. Note that when the density ratio is equal to or greater than one, the floating body becomes completely submerged. It follows from these discussions that a body immersed in a fluid (1) remains at rest at any location in the fluid where its average density is equal to the density of the fluid, (2) sinks to the bottom when its average density is greater than the density of the fluid, and (3) rises to the surface of the fluid and floats when the average density of the body is less than the density of the fluid (Fig. 3–44). For a body floating on the surface of a liquid, the total weight of the body must obviously be less than that of the liquid it displaces. It turns out that a portion of the body volume is submerged (volume Vsubmerged), while the remaining portion is positioned above the surface of the liquid. Since the system is stationary, the two vertical forces W and FB must still balance, W = FB = 𝜌fgVsubmerged → Vsubmerged = W/𝜌f g For a body of known weight W, we see that as the liquid density 𝜌f increases, a smaller percentage of the body volume is submerged since 𝜌f is in the denominator (see Fig. 3–45). The buoyant force is proportional to the density of the fluid, and thus we might think that the buoyant force exerted by gases such as air is negligible. This is certainly the case in general, but there are significant exceptions. For example, the volume of a person is about 0.1 m3, and taking the density of air to be 1.2 kg/m3, the buoyant force exerted by air on the person is FB = ρf gV = (1.2 kg/m3)(9.81 m/s2)(0.1 m3) ≅1.2 N The weight of an 80-kg person is 80 × 9.81 = 788 N. Therefore, ignoring the buoyancy in this case results in an error in weight of just 0.15 percent, which is negligible. But the buoyancy effects in gases dominate some important natural phenomena such as the rise of warm air in a cooler envi ronment and thus the onset of natural convection currents, the rise of hot-air or helium balloons, and air movements in the atmosphere. A helium bal loon, for example, rises as a result of the buoyancy effect until it reaches an altitude where the density of air (which decreases with altitude) equals Fluid Fluid Solid Ws FB FB W C C FIGURE 3–43 The buoyant forces acting on a solid body submerged in a fluid and on a fluid body of the same shape at the same depth are identical. The buoyant force FB acts upward through the centroid C of the displaced volume and is equal in magnitude to the weight W of the displaced fluid, but is opposite in direction. For a solid of uniform density, its weight Ws also acts through the centroid, but its magnitude is not necessarily equal to that of the fluid it displaces. (Here Ws > W and thus Ws > FB; this solid body would sink.) Fluid Sinking body Suspended body (neutrally buoyant) Floating body ρ < ρf ρ = ρf ρ > ρf ρf FIGURE 3–44 A solid body dropped into a fluid will sink, float, or remain at rest at any point in the fluid, depending on its average density relative to the density of the fluid.
cen96537_ch03_077-136.indd 101 14/01/17 2:23 pm 102 pressure and fluid statics EXAMPLE 3–10 Measuring Specific Gravity by a Hydrometer If you have a seawater aquarium, you have probably used a small cylindrical glass tube with a lead-weight at its bottom to measure the salinity of the water by simply watching how deep the tube sinks. Such a device that floats in a vertical position and is used to measure the specific gravity of a liquid is called a hydrometer (Fig. 3–47). The top part of the hydrometer extends above the liquid surface, and the divisions on it allow one to read the specific gravity directly. The hydrometer is calibrated such that in pure water it reads exactly 1.0 at the air–water interface. (a) Obtain a relation for the specific gravity of a liquid as a function of distance Δz from the mark corresponding to pure water and (b) determine the mass of lead that must be poured into a 1-cm-diameter, 20-cm-long hydrometer if it is to float half way (the 10-cm mark) in pure water.
SOLUTION The specific gravity of a liquid is to be measured by a hydrom eter. A relation between specific gravity and the vertical distance from the reference level is to be obtained, and the amount of lead that needs to be added into the tube for a certain hydrometer is to be determined.
Assumptions 1 The weight of the glass tube is negligible relative to the weight of the lead added. 2 The curvature of the tube bottom is disregarded.
Properties We take the density of pure water to be 1000 kg/m3.
Analysis (a) Noting that the hydrometer is in static equilibrium, the buoy ant force FB exerted by the liquid must always be equal to the weight W of the hydrometer. In pure water (subscript w), we let the vertical distance between the bottom of the hydrometer and the free surface of water be z0. Setting FB, w = W in this case gives Whydro = FB, w = ρwgVsub = ρwgAz0 (1) where A is the cross-sectional area of the tube, and 𝜌w is the density of pure water. In a fluid lighter than water (𝜌f < 𝜌w), the hydrometer will sink deeper, and the liquid level will be a distance of Δz above z0. Again setting FB = W gives Whydro = FB, f = ρf gVsub = ρf gA(z0 + Δz) (2) This relation is also valid for fluids heavier than water by taking Δz to be a negative quantity. Setting Eqs. (1) and (2) here equal to each other since the weight of the hydrometer is constant and rearranging gives ρwgAz0 = ρf gA(z0 + Δz) → SGf = ρf ρw = z0 z0 + Δz which is the relation between the specific gravity of the fluid and Δz. Note that z0 is constant for a given hydrometer and Δz is negative for fluids heavier than pure water.
(b) Disregarding the weight of the glass tube, the amount of lead that needs to be added to the tube is determined from the requirement that the weight of the lead FIGURE 3–45 The density of the water in the Dead Sea is about 24% higher than that of pure water. Therefore, people float much more easily (with more of their bodies above the water) in the Dead Sea than in fresh water or in normal seawater.
Photo by Andy Cimbala. Used with permission.
the density of helium in the balloon—assuming the balloon does not burst by then, and ignoring the weight of the balloon’s skin. Hot air balloons (Fig. 3–46) work by similar principles. Archimedes’ principle is also used in geology by considering the conti nents to be floating on a sea of magma.
FIGURE 3–46 The altitude of a hot air balloon is controlled by the temperature difference between the air inside and outside the balloon, since warm air is less dense than cold air. When the balloon is neither rising nor falling, the upward buoyant force exactly balances the downward weight.
© PhotoLink/Getty Images RF cen96537_ch03_077-136.indd 102 14/01/17 2:23 pm 103 CHAPTER 3 Hydrometer Lead 1.0 W FB z0 Δz FIGURE 3–47 Schematic for Example 3–10.
FIGURE 3–48 Schematic for Example 3–11.
W h 25 cm Sea Cubic ice block FB be equal to the buoyant force. When the hydrometer is floating with half of it sub merged in water, the buoyant force acting on it is FB = ρwgVsub Equating FB to the weight of lead gives W = mg = ρwgVsub Solving for m and substituting, the mass of lead is determined to be m = ρwVsub = ρw(𝜋R2hsub) = (1000 kg/m3)[𝜋(0.005 m)2(0.1 m)] = 0.00785 kg Discussion Note that if the hydrometer were required to sink only 5 cm in water, the required mass of lead would be one-half of this amount. Also, the assumption that the weight of the glass tube is negligible is questionable since the mass of lead is only 7.85 g.
EXAMPLE 3–11 Height of Ice Block Below the Water Surface Consider a large cubic ice block floating in seawater. The specific gravities of ice and seawater are 0.92 and 1.025, respectively. If a 25-cm-high portion of the ice block extends above the surface of the water, determine the height of the ice block below the surface.
SOLUTION The height of the portion of a cubic ice block that extends above the water surface is measured. The height of the ice block below the surface is to be determined.
Assumptions 1 The buoyancy force in air is negligible. 2 The top surface of the ice block is parallel to the surface of the sea. Properties The specific gravities of ice and seawater are given to be 0.92 and 1.025, respectively, and thus the corresponding densities are 920 kg/m3 and 1025 kg/m3.
Analysis The weight of a body floating in a fluid is equal to the buoyant force acting on it (a consequence of vertical force balance from static equilibrium), as shown in Fig. 3–48. Therefore, W = FB → ρbodygVtotal = ρfluidgVsubmerged Vsubmerged Vtotal = ρbody ρfluid The cross-sectional area of a cube is constant, and thus the “volume ratio” can be replaced by “height ratio.” Then, hsubmerged htotal = ρbody ρfluid → h h + 0.25 = ρice ρwater → h h + 0.25 m = 920 kg/m3 1025 kg/m3 where h is the height of the ice block below the surface. Solving for h gives h = (920 kg/m3)(0.25 m) (1025 − 920) kg/m3 = 2.19 m cen96537_ch03_077-136.indd 103 14/01/17 2:23 pm 104 pressure and fluid statics Discussion Note that 0.92/1.025 = 0.898, so approximately 90% of the volume of an ice block remains under water. For symmetrical ice blocks this also represents the fraction of height that remains under water. This also applies to icebergs; the vast majority of an iceberg is submerged.
Stability of Immersed and Floating Bodies An important application of the buoyancy concept is the assessment of the stability of immersed and floating bodies with no external attachments. This topic is of great importance in the design of ships and submarines (Fig. 3–49). Here we provide some general qualitative discussions on verti cal and rotational stability. We use the classic “ball on the floor” analogy to explain the fundamental concepts of stability and instability. Shown in Fig. 3–50 are three balls at rest on the floor. Case (a) is stable since any small disturbance (someone moves the ball to the right or left) generates a restoring force (due to gravity) that returns it to its initial position. Case (b) is neutrally stable because if some one moves the ball to the right or left, it would stay put at its new location. It has no tendency to move back to its original location, nor does it continue to move away. Case (c) is a situation in which the ball may be at rest at the moment, but any disturbance, even an infinitesimal one, causes the ball to roll off the hill—it does not return to its original position; rather it diverges from it. This situation is unstable. What about a case where the ball is on an inclined floor? It is not appropriate to discuss stability for this case since the ball is not in a state of equilibrium. In other words, it cannot be at rest and would roll down the hill even without any disturbance. For an immersed or floating body in static equilibrium, the weight and the buoyant force acting on the body balance each other, and such bodies are inherently stable in the vertical direction. If an immersed neutrally buoyant body is raised or lowered to a different depth in an incompressible fluid, the body will remain in equilibrium at that location. If a floating body is raised or lowered somewhat by a vertical force, the body will return to its original position as soon as the external effect is removed. Therefore, a floating body possesses vertical stability, while an immersed neutrally buoyant body is neu trally stable since it does not return to its original position after a disturbance. The rotational stability of an immersed body depends on the relative loca tions of the center of gravity G of the body and the center of buoyancy B, which is the centroid of the displaced volume. An immersed body is stable if the body is bottom-heavy and thus point G is directly below point B (Fig. 3–51a). A rotational disturbance of the body in such cases produces a restoring moment to return the body to its original stable position. Thus, a stable design for a submarine calls for the engines and the cabins for the crew to be located at the lower half in order to shift the weight to the bot tom as much as possible. Hot-air or helium balloons (which can be viewed as being immersed in air) are also stable since the heavy cage that carries the load is at the bottom. An immersed body whose center of gravity G is directly above point B is unstable, and any disturbance will cause this body to turn upside down (Fig 3–51c). A body for which G and B coincide is FIGURE 3–49 For floating bodies such as ships, stability is an important consideration for safety.
© Corbis RF (c) Unstable (a) Stable (b) Neutrally stable FIGURE 3–50 Stability is easily understood by analyzing a ball on the floor.
cen96537_ch03_077-136.indd 104 14/01/17 2:23 pm 105 CHAPTER 3 neutrally stable (Fig 3–51b). This is the case for bodies whose density is constant throughout. For such bodies, there is no tendency to overturn or right themselves. What about a case where the center of gravity is not vertically aligned with the center of buoyancy, as in Fig. 3–52? It is not appropriate to discuss sta bility for this case since the body is not in a state of equilibrium. In other words, it cannot be at rest and would rotate toward its stable state even with out any disturbance. The restoring moment in the case shown in Fig. 3–52 is counterclockwise and causes the body to rotate counterclockwise so as to align point G vertically with point B. Note that there may be some oscilla tion, but eventually the body settles down at its stable equilibrium state [case (a) of Fig. 3–51]. The initial stability of the body of Fig. 3–52 is analogous to that of the ball on an inclined floor. Can you predict what would happen if the weight in the body of Fig. 3–52 were on the opposite side of the body? The rotational stability criteria are similar for floating bodies. Again, if the floating body is bottom-heavy and thus the center of gravity G is directly below the center of buoyancy B, the body is always stable. But unlike immersed bodies, a floating body may still be stable when G is directly above B (Fig. 3–53). This is because the centroid of the displaced volume shifts to the side to a point Bʹ during a rotational disturbance while the center of grav ity G of the body remains unchanged. If point Bʹ is sufficiently far, these two forces create a restoring moment and return the body to the original position. A measure of stability for floating bodies is the metacentric height GM, which is the distance between the center of gravity G and the metacenter M—the intersection point of the lines of action of the buoyant force through the body before and after rotation. The metacenter may be considered to be a fixed point for most hull shapes for small rolling angles up to about 20°. Typical values of metacentric height are 0.3–0.7 m for cruise ships, 0.9–1.5 m for sailboats, 0.6–0.9 m for cargo ships, and 0.75–1.3 m for war ships. A floating body is stable if point M is above point G, and thus GM is positive, and unstable if point M is below point G, and thus GM is negative. In the (a) Stable (b) Neutrally stable (c) Unstable FB B G W FB B W G FB B W G Fluid Weight Weight FIGURE 3–51 An immersed neutrally buoyant body is (a) stable if the center of gravity G is directly below the center of buoyancy B of the body, (b) neutrally stable if G and B are coincident, and (c) unstable if G is directly above B.
Restoring moment Weight FB B W G FIGURE 3–52 When the center of gravity G of an immersed neutrally buoyant body is not vertically aligned with the center of buoyancy B of the body, it is not in an equilibrium state and would rotate to its stable state, even without any disturbance.
Metacenter Restoring moment (a) Stable (b) Stable (c) Unstable W B G G B′ M FB Overturning moment G B′′ M FIGURE 3–53 A floating body is stable if the body is (a) bottom-heavy and thus the center of gravity G is below the centroid B of the body, or (b) if the metacenter M is above point G. However, the body is (c) unstable if point M is below point G.
cen96537_ch03_077-136.indd 105 14/01/17 2:23 pm 106 pressure and fluid statics latter case, the weight and the buoyant force acting on the tilted body gener ate an overturning moment instead of a restoring moment, causing the body to capsize. The length of the metacentric height GM above G is a measure of the stability: the larger it is, the more stable is the floating body. As already discussed, a boat can tilt to some maximum angle without capsizing, but beyond that angle it overturns (and sinks). We make a final analogy between the stability of floating objects and the stability of a ball rolling along the floor. Namely, imagine the ball in a trough between two hills (Fig. 3–54). The ball returns to its stable equilibrium position after being perturbed—up to a limit. If the perturbation amplitude is too great, the ball rolls down the opposite side of the hill and does not return to its equilibrium position. This situation is described as stable up to some limit ing level of disturbance, but unstable beyond.
3–7 ■ FLUIDS IN RIGID-BODY MOTION We showed in Section 3–1 that pressure at a given point has the same magnitude in all directions, and thus it is a scalar function. In this section we obtain relations for the variation of pressure in fluids moving like a solid body with or without acceleration in the absence of any shear stresses (i.e., no motion between fluid layers relative to each other). Many fluids such as milk and gasoline are transported in tankers. In an accelerating tanker, the fluid rushes to the back, and some initial splashing occurs. But then a new free surface (usually nonhorizontal) is formed, each fluid particle assumes the same acceleration, and the entire fluid moves like a rigid body. No shear stresses exist within the fluid body since there is no deformation and thus no change in shape. Rigid-body motion of a fluid also occurs when the fluid is contained in a tank that rotates about an axis. Consider a differential rectangular fluid element of side lengths dx, dy, and dz in the x-, y-, and z-directions, respectively, with the z-axis being upward in the vertical direction (Fig. 3–55). Noting that the differential fluid element behaves like a rigid body, Newton’s second law of motion for this element can be expressed as 𝛿F ›= 𝛿m·a › (3–34) where δm = 𝜌 dV = 𝜌 dx dy dz is the mass of the fluid element, a › is the acceleration, and 𝛿F › is the net force acting on the element. The forces acting on the fluid element consist of body forces such as grav ity that act throughout the entire body of the element and are proportional to the volume of the body (and also electrical and magnetic forces, which will not be considered in this text), and surface forces such as the pressure forces that act on the surface of the element and are proportional to the surface area (shear stresses are also surface forces, but they do not apply in this case since the relative positions of fluid elements remain unchanged). The sur face forces appear as the fluid element is isolated from its surroundings for analysis, and the effect of the detached body is replaced by a force at that location. Note that pressure represents the compressive force applied on the fluid element by the surrounding fluid and is always normal to the surface and inward toward the surface.
FIGURE 3–54 A ball in a trough between two hills is stable for small disturbances, but unstable for large disturbances.
dx dz dy P(x, y, z) y x z ρg dx dy dz P + dx dy dz 2 ( g ) дP дz P – dx dy dz 2 ( ) дP дz FIGURE 3–55 The surface and body forces acting on a differential fluid element in the vertical direction.
cen96537_ch03_077-136.indd 106 14/01/17 2:23 pm 107 CHAPTER 3 Taking the pressure at the center of the element to be P, the pressures at the top and bottom surfaces of the element can be expressed as P + (∂P/∂z) dz/2 and P − (∂P/∂z) dz/2, respectively, using a truncated Taylor series expansion (Fig. 3–56). Noting that the pressure force acting on a surface is equal to the average pressure multiplied by the surface area, the net surface force acting on the element in the z-direction is the difference between the pressure forces acting on the bottom and top faces, 𝛿FS, z = (P −∂P ∂z dz 2 ) dx dy −(P + ∂P ∂z dz 2 ) dx dy = −∂P ∂z dx dy dz (3–35) Similarly, the net surface forces in the x- and y-directions are 𝛿FS, x = −∂P ∂x dx dy dz and 𝛿FS, y = −∂P ∂y dx dy dz (3–36) Then the surface force (which is simply the pressure force) acting on the entire element can be expressed in vector form as 𝛿F › S = 𝛿FS, x i ›+ 𝛿FS, y j ›+ 𝛿FS, z k › = −( ∂P ∂x i ›+ ∂P ∂y j ›+ ∂P ∂z k › ) dx dy dz = −∇ › P dx dy dz (3–37) where i ›, j ›, and k › are the unit vectors in the x-, y-, and z-directions, respec tively, and ∇ › P = ∂P ∂x i ›+ ∂P ∂y j ›+ ∂P ∂z k › (3–38) is the pressure gradient. Note that ∇ › or “del” is a vector operator that is used to express the gradients of a scalar function compactly in vector form. Also, the gradient of a scalar function is expressed in a given direction and thus it is a vector quantity. The only body force acting on the fluid element is the weight of the element acting in the negative z-direction, and it is expressed as δFB, z = −gδm = −𝜌g dx dy dz or in vector form as 𝛿F › B, z = −g𝛿mk ›= −ρg dx dy dzk › (3–39) Then the total force acting on the element becomes 𝛿F ›= 𝛿F › S + 𝛿F › B = −(∇ › P + ρgk › ) dx dy dz (3–40) Substituting into Newton’s second law of motion δF → = δm ⋅ a › = 𝜌 dx dy dz ⋅ a › and canceling dx dy dz, the general equation of motion for a fluid that acts as a rigid body (no shear stresses) is determined to be Rigid-body motion of fluids: ∇ › P + ρgk ›= −ρa › (3–41) Resolving the vectors into their components, this relation can be expressed more explicitly as ∂P ∂x i ›+ ∂P ∂y j ›+ ∂P ∂z k ›+ ρgk ›= −ρ(ax i ›+ ay j ›+ azk › ) (3–42) x a x f f(x) = f(a) + f′(a)Δx Δx3 + ...
2!
Δx + + Δx2 f ′′(a) 3!
f ′′′(a) FIGURE 3–56 Taylor series expansion of f from point a to some nearby point x. As x gets small, it is common to truncate the series to first order, keeping only the first two terms on the right side.
cen96537_ch03_077-136.indd 107 14/01/17 2:23 pm 108 pressure and fluid statics or, in scalar form in the three orthogonal directions as Accelerating fluids: ∂P ∂x = −ρax, ∂P ∂y = −ρay, and ∂P ∂z = −ρ(g + az) (3–43) where ax, ay, and az are accelerations in the x-, y-, and z-directions, respectively.
Special Case 1: Fluids at Rest For fluids at rest or moving on a straight path at constant velocity, all com ponents of acceleration are zero, and the relations in Eqs. 3–43 reduce to Fluids at rest: ∂P ∂x = 0, ∂P ∂y = 0, and dP dz = −ρg (3–44) which confirm that, in fluids at rest, the pressure remains constant in any horizontal direction (P is independent of x and y) and varies only in the vertical direction as a result of gravity [and thus P = P(z)]. These relations are applicable for both compressible and incompressible fluids (Fig. 3–57).
Special Case 2: Free Fall of a Fluid Body A freely falling body accelerates under the influence of gravity. When the air resistance is negligible, the acceleration of the body equals the gravi tational acceleration, and acceleration in any horizontal direction is zero. Therefore, ax = ay = 0 and az = −g. Then the equations of motion for accel erating fluids (Eqs. 3–43) reduce to Free-falling fluids: ∂P ∂x = ∂P ∂y = ∂P ∂z = 0 → P = constant (3–45) Therefore, in a frame of reference moving with the fluid, it behaves like it is in an environment with zero gravity. (This is the situation in an orbit ing spacecraft, by the way. Gravity is not zero up there, despite what many people think!) Also, the gage pressure in a drop of liquid in free fall is zero throughout. (Actually, the gage pressure is slightly above zero due to surface tension, which holds the drop intact.) When the direction of motion is reversed and the fluid is forced to accel erate vertically with az = +g by placing the fluid container in an elevator or a space vehicle propelled upward by a rocket engine, the pressure gradient in the z-direction is ∂P/∂z = −2𝜌g. Therefore, the pressure difference across a fluid layer now doubles relative to the stationary fluid case (Fig. 3–58).
Acceleration on a Straight Path Consider a container partially filled with a liquid. The container is moving on a straight path with a constant acceleration. We take the projection of the path of motion on the horizontal plane to be the x-axis, and the projection on the vertical plane to be the z-axis, as shown in Fig. 3–59. The x- and z-components of acceleration are ax and az. There is no movement in the y-direction, and thus the acceleration in that direction is zero, ay = 0. Then the equations of motion for accelerating fluids (Eqs. 3–43) reduce to ∂P ∂x = −ρax, ∂P ∂y = 0, and ∂P ∂z = −ρ(g + az) (3–46) FIGURE 3–57 A glass of water at rest is a special case of a fluid in rigid-body motion. If the glass of water were moving at constant velocity in any direction, the hydrostatic equations would still apply.
© Imagestate Media (John Foxx)/Imagestate RF az = –g z z az = g P2 = P1 P1 P1 P2 = P1 + 2ρgh (a) Free fall of a liquid (b) Upward acceleration of a liquid with az = +g Liquid, ρ Liquid, ρ h h FIGURE 3–58 The effect of acceleration on the pressure of a liquid during free fall and upward acceleration.
cen96537_ch03_077-136.indd 108 14/01/17 2:23 pm 109 CHAPTER 3 Therefore, pressure is independent of y. Then the total differential of P = P(x, z), which is (∂P/∂x)dx + (∂P/∂z) dz, becomes dP = −ρax dx −ρ(g + az) dz (3–47) For 𝜌 = constant, the pressure difference between two points 1 and 2 in the fluid is determined by integration to be P2 −P1 = −ρax(x2 −x1) −ρ(g + az)(z2 −z1) (3–48) Taking point 1 to be the origin (x = 0, z = 0) where the pressure is P0 and point 2 to be any point in the fluid (no subscript), the pressure at that point is expressed as Pressure at a point: P = P0 −ρaxx −ρ(g + az)z (3–49) The vertical rise (or drop) of the free surface at point 2 relative to point 1 is determined by choosing both 1 and 2 on the free surface (so that P1 = P2), and solving Eq. 3–48 for z2 − z1 (Fig. 3–60), Vertical rise of surface: Δzs = zs2 −zs1 = − ax g + az (x2 −x1) (3–50) where zs is the z-coordinate of the liquid’s free surface. The equation for surfaces of constant pressure, called isobars, is obtained from Eq. 3–47 by setting dP = 0 and replacing z by zisobar, which is the z-coordinate (the verti cal distance) of the surface as a function of x. It gives Surfaces of constant pressure: dzisobar dx = − ax g + az = constant (3–51) Thus we conclude that the isobars (including the free surface) in an incom pressible fluid with constant acceleration in linear motion are parallel sur faces whose slope in the xz-plane is Slope of isobars: Slope = dzisobar dx = − ax g + az = −tan 𝜃 (3–52) Obviously, the free surface of such a fluid is a plane surface, and it is inclined unless ax = 0 (the acceleration is in the vertical direction only). Also, conservation of mass, together with the assumption of incompressibility (𝜌 = constant), requires that the volume of the fluid remain constant before and during acceleration. Therefore, the rise of fluid level on one side must be balanced by a drop of fluid level on the other side. This is true regardless of the shape of the container, provided that the liquid is continuous through out the container (Fig. 3–61).
Δzmax z x b ho az g g – a ax Liquid Free surface a 휃 FIGURE 3–59 Rigid-body motion of a liquid in a linearly accelerating tank. The system behaves like a fluid at rest except that g ›−a › replaces g › in the hydrostatic equations.
x1 – x2 az a g ax Constant pressure lines Free surface 2 1 z x Δzs = zs2 – zs1 휃 FIGURE 3–60 Lines of constant pressure (which are the projections of the surfaces of constant pressure on the xz-plane) in a linearly accelerating liquid. Also shown is the vertical rise.
EXAMPLE 3–12 Overflow from a Water Tank During Acceleration An 80-cm-high fish tank of cross section 2 m × 0.6 m that is partially filled with water is to be transported on the back of a truck (Fig. 3–62). The truck accelerates from 0 to 90 km/h in 10 s. If it is desired that no water spills during cen96537_ch03_077-136.indd 109 14/01/17 2:23 pm 110 pressure and fluid statics acceleration, determine the allowable initial water height in the tank. Would you recommend the tank to be aligned with the long or short side parallel to the direction of motion?
SOLUTION A fish tank is to be transported on a truck. The allowable water height to avoid spill of water during acceleration and the proper orientation are to be determined.
Assumptions 1 The road is horizontal during acceleration so that acceleration has no vertical component (az = 0). 2 Effects of splashing, braking, shifting gears, driving over bumps, climbing hills, etc., are assumed to be secondary and are not considered. 3 The acceleration remains constant.
Analysis We take the x-axis to be the direction of motion, the z-axis to be the upward vertical direction, and the origin to be the lower left corner of the tank. Noting that the truck goes from 0 to 90 km/h in 10 s, the acceleration of the truck is ax = ΔV Δt = (90 −0) km/h 10 s ( 1 m/s 3.6 km/h) = 2.5 m/s2 The tangent of the angle the free surface makes with the horizontal is tan 𝜃 = ax g + az = 2.5 9.81 + 0 = 0.255 (and thus 𝜃 = 14.3°) The maximum vertical rise of the free surface occurs at the back of the tank, and the vertical midplane experiences no rise or drop during acceleration since it is a plane of symmetry. Then the vertical rise at the back of the tank relative to the mid plane for the two possible orientations becomes Case 1: The long side is parallel to the direction of motion: Δzs1 = (b1/2) tan 𝜃 = [(2 m)/2] × 0.255 = 0.255 m = 25.5 cm Case 2: The short side is parallel to the direction of motion: Δzs2 = (b2/2) tan 𝜃 = [(0.6 m)/2] × 0.255 = 0.076 m = 7.6 cm Therefore, assuming tipping is not a problem, the tank should definitely be ori ented such that its short side is parallel to the direction of motion. Emptying the tank such that its free surface level drops just 7.6 cm in this case will be adequate to avoid spilling during acceleration.
Discussion Note that the orientation of the tank is important in controlling the ver tical rise. Also, the analysis is valid for any fluid with constant density, not just water, since we used no information that pertains to water in the solution.
FIGURE 3–61 When a continuous liquid in a container of any shape (even a U-tube manometer!) is accelerated at constant rate, the liquid behaves as a rigid body with a tilted plane surface. The surface itself does not have to be continuous.
Patm a Rotation in a Cylindrical Container We know from experience that when a glass filled with water is rotated about its axis, the fluid is forced outward as a result of the so-called centrifugal force (but more properly explained in terms of centripetal accel eration), and the free surface of the liquid becomes concave. This is known as the forced vortex motion.
Δzs ax b Water tank 80 cm 휃 FIGURE 3–62 Schematic for Example 3–12.
cen96537_ch03_077-136.indd 110 14/01/17 2:23 pm 111 CHAPTER 3 Consider a vertical cylindrical container partially filled with a liquid. The container is now rotated about its axis at a constant angular velocity of ω, as shown in Fig. 3–63. After initial transients, the liquid will move as a rigid body together with the container. There is no deformation, and thus there can be no shear stress, and every fluid particle in the container moves with the same angular velocity. This problem is best analyzed in cylindrical coordinates (r, 𝜃, z), with z taken along the centerline of the container directed from the bottom toward the free surface, since the shape of the container is a cylinder, and the fluid particles undergo a circular motion. The centripetal acceleration of a fluid particle rotating with a constant angular velocity of ω at a distance r from the axis of rotation is rω2 and is directed radially toward the axis of rotation (negative r-direction). That is, ar = −rω2. There is symmetry about the z-axis, which is the axis of rotation, and thus there is no 𝜃 dependence. Then P = P(r, z) and a𝜃 = 0. Also, az = 0 since there is no motion in the z-direction. Then the equation of motion for accelerating fluids (Eq. 3–41) reduces to ∂P ∂r = ρr𝜔 2, ∂P ∂𝜃 = 0, and ∂P ∂z = −ρg (3–53) Then the total differential of P = P(r, z), which is dP = (∂P/∂r)dr + (∂P/∂z)dz, becomes dP = ρr𝜔 2 dr −ρg dz (3–54) The equation for surfaces of constant pressure is obtained by setting dP = 0 and replacing z by zisobar, which is the z-value (the vertical distance) of the surface as a function of r. It gives dzisobar dr = r𝜔 2 g (3–55) Integrating, the equation for the surfaces of constant pressure is determined to be Surfaces of constant pressure: zisobar = 𝜔 2 2g r2 + C1 (3–56) which is the equation of a parabola. Thus we conclude that the surfaces of constant pressure, including the free surface, are paraboloids of revolution (Fig. 3–64). The value of the integration constant C1 is different for different parabo loids of constant pressure (i.e., for different isobars). For the free surface, setting r = 0 in Eq. 3–56 gives zisobar(0) = C1 = hc, where hc is the distance of the free surface from the bottom of the container along the axis of rota tion (Fig. 3–63). Then the equation for the free surface becomes zs = 𝜔 2 2g r2 + hc (3–57) where zs is the distance of the free surface from the bottom of the con tainer at radius r. The underlying assumption in this analysis is that there is sufficient liquid in the container so that the entire bottom surface remains covered with liquid.
P7 Free surface P6 P5 P4 P3 P2 P1 Δzs, max ω FIGURE 3–64 Surfaces of constant pressure in a rotating liquid.
ho zs z Axis of rotation Free surface R r g hc ω FIGURE 3–63 Rigid-body motion of a liquid in a rotating vertical cylindrical container.
cen96537_ch03_077-136.indd 111 14/01/17 2:23 pm 112 pressure and fluid statics The volume of a cylindrical shell element of radius r, height zs, and thick ness dr is dV = 2πrzs dr. Then the volume of the paraboloid formed by the free surface is V = ∫ R r = 0 2𝜋zsr dr = 2𝜋 ∫ R r = 0 ( 𝜔 2 2g r2 + hc)r dr = 𝜋R2( 𝜔 2R2 4g + hc) (3–58) Since mass is conserved and density is constant, this volume must be equal to the original volume of the fluid in the container, which is V = 𝜋R2h0 (3–59) where h0 is the original height of the fluid in the container with no rotation. Setting these two volumes equal to each other, the height of the fluid along the centerline of the cylindrical container becomes hc = h0 −𝜔 2R2 4g (3–60) Then the equation of the free surface becomes Free surface: zs = h0 −𝜔 2 4g (R2 −2r2) (3–61) The paraboloid shape is independent of fluid properties, so the same free surface equation applies to any liquid. For example, spinning liquid mercury forms a parabolic mirror that is useful in astronomy (Fig. 3–65). The maximum vertical height occurs at the edge where r = R, and the maximum height difference between the edge and the center of the free sur face is determined by evaluating zs at r = R and also at r = 0, and taking their difference, Maximum height difference: Δzs, max = zs(R) −zs(0) = 𝜔 2 2g R2 (3–62) When 𝜌 = constant, the pressure difference between two points 1 and 2 in the fluid is determined by integrating dP = 𝜌rω2 dr − 𝜌g dz. This yields P2 −P1 = ρ𝜔 2 2 (r2 2 −r2 1) −ρg(z2 −z1) (3–63) Taking point 1 to be the origin (r = 0, z = 0) where the pressure is P0 and point 2 to be any point in the fluid (no subscript), the pressure at that point is expressed as Pressure at that point: P = P0 + ρ𝜔 2 2 r2 −ρgz (3–64) Note that at a fixed radius, the pressure varies hydrostatically in the vertical direction, as in a fluid at rest. For a fixed vertical distance z, the pressure varies with the square of the radial distance r, increasing from the centerline toward the outer edge. In any horizontal plane, the pressure difference between the center and edge of the container of radius R is ΔP = 𝜌ω2R2/2.
FIGURE 3–65 The 6-meter spinning liquid-mercury mirror of the Large Zenith Telescope located near Vancouver, British Columbia.
Photo courtesy of Paul Hickson, The University of British Columbia.
cen96537_ch03_077-136.indd 112 14/01/17 2:23 pm 113 CHAPTER 3 EXAMPLE 3–13 Rising of a Liquid During Rotation A 20-cm-diameter, 60-cm-high vertical cylindrical container, shown in Fig. 3–66, is partially filled with 50-cm-high liquid whose density is 850 kg/m3. Now the cylinder is rotated at a constant speed. Determine the rotational speed at which the liquid will start spilling from the edges of the container.
SOLUTION A vertical cylindrical container partially filled with a liquid is rotated. The angular speed at which the liquid will start spilling is to be determined.
Assumptions 1 The increase in the rotational speed is very slow so that the liq uid in the container always acts as a rigid body. 2 The bottom surface of the con tainer remains covered with liquid during rotation (no dry spots).
Analysis Taking the center of the bottom surface of the rotating vertical cylinder as the origin (r = 0, z = 0), the equation for the free surface of the liquid is given as zs = h0 −𝜔 2 4g (R2 −2r2) Then the vertical height of the liquid at the edge of the container where r = R becomes zs(R) = h0 + 𝜔 2R2 4g where h0 = 0.5 m is the original height of the liquid before rotation. Just before the liquid starts spilling, the height of the liquid at the edge of the container equals the height of the container, and thus zs(R) = H = 0.6 m. Solving the last equation for ω and substituting, the maximum rotational speed of the container is determined to be 𝜔 = √ 4g(H −h0) R2 = √ 4(9.81 m/s2)[(0.6 −0.5) m] (0.1 m)2 = 19.8 rad/s Noting that one complete revolution corresponds to 2π rad, the rotational speed of the container can also be expressed in terms of revolutions per minute (rpm) as n · = 𝜔 2𝜋= 19.8 rad/s 2𝜋 rad/rev( 60 s 1 min) = 189 rpm Therefore, the rotational speed of this container should be limited to 189 rpm to avoid any spill of liquid as a result of the centrifugal effect.
Discussion Note that the analysis is valid for any liquid since the result is inde pendent of density or any other fluid property. We should also verify that our assumption of no dry spots is valid. The liquid height at the center is zs(0) = h0 −𝜔 2R2 4g = 0.4 m Since zs(0) is positive, our assumption is validated.
h0 zs Free surface R H r z ω g FIGURE 3–66 Schematic for Example 3–13.
cen96537_ch03_077-136.indd 113 14/01/17 2:23 pm 114 pressure and fluid statics SUMMARY The normal force exerted by a fluid per unit area is called pres sure, and its SI unit is the pascal, 1 Pa ≡ 1 N/m2. The pressure relative to absolute vacuum is called the absolute pressure, and the difference between the absolute pressure and the local atmospheric pressure is called the gage pressure. Pressures below atmospheric pressure are sometimes called vacuum pres sures. The absolute, gage, and vacuum pressures are related by Pgage = Pabs −Patm Pvac = Patm −Pabs = −Pgage The pressure at a point in a fluid has the same magnitude in all directions. The variation of pressure with elevation in a fluid at rest is given by dP dz = −ρg where the positive z-direction is taken to be upward by con vention. When the density of the fluid is constant, the pres sure difference across a fluid layer of thickness Δz is Pbelow = Pabove + ρg|Δz| = Pabove + 𝛾s|Δz| The absolute and gage pressures in a static liquid open to the atmosphere at a depth h from the free surface are P = Patm + ρgh and Pgage = ρgh The pressure in a fluid at rest does not vary in the horizontal direction. Pascal’s law states that the pressure applied to a confined fluid increases the pressure throughout by the same amount. The atmospheric pressure can be measured by a barometer and is given by Patm = ρgh where h is the height of the liquid column. Fluid statics deals with problems associated with fluids at rest, and it is called hydrostatics when the fluid is a liquid. The magnitude of the resultant force acting on a plane surface of a completely submerged plate in a homogeneous fluid is equal to the product of the pressure PC at the centroid of the surface and the area A of the surface and is expressed as FR = (P0 + ρghC)A = PC A = PavgA where hC = yC sin 𝜃 is the vertical distance of the centroid from the free surface of the liquid. The pressure P0 is usually atmospheric pressure, which cancels out in most cases since it acts on both sides of the plate. The point of intersection of the line of action of the resultant force and the surface is the center of pressure. The vertical location of the line of action of the resultant force is given by yP = yC + Ixx, C [yC + P0 /(ρg sin 𝜃 )]A where Ixx, C is the second moment of area about the x-axis passing through the centroid of the area. A fluid exerts an upward force on a body immersed in it. This force is called the buoyant force and is expressed as FB = ρf gV where V is the volume of the body. This is known as Archimedes’ principle and is expressed as: the buoyant force acting on a body immersed in a fluid is equal to the weight of the fluid displaced by the body; it acts upward through the centroid of the displaced volume. In a fluid with constant density, the buoyant force is independent of the distance of the body from the free surface. For floating bodies, the sub merged volume fraction of the body is equal to the ratio of the average density of the body to the density of the fluid. The general equation of motion for a fluid that acts as a rigid body is ∇ › P + ρgk ›= −ρa › When gravity is aligned in the −z-direction, it is expressed in scalar form as ∂P ∂x = −ρax, ∂P ∂y = −ρay, and ∂P ∂z = −ρ(g + az) where ax, ay, and az are accelerations in the x-, y-, and z-directions, respectively. During linearly accelerating motion in the xz-plane, the pressure distribution is expressed as P = P0 −ρaxx −ρ(g + az)z The surfaces of constant pressure (including the free surface) in a liquid with constant acceleration in linear motion are parallel surfaces whose slope in some xz-plane is Slope = dzisobar dx = − ax g + az = −tan 𝜃 During rigid-body motion of a liquid in a rotating cylinder, the surfaces of constant pressure are paraboloids of revolu tion. The equation for the free surface is zs = h0 −𝜔 2 4g (R2 −2r2) where zs is the distance of the free surface from the bottom of the container at radius r and h0 is the original height of the fluid in the container with no rotation. The variation of pressure in the liquid is expressed as P = P0 + ρ𝜔 2 2 r2 −ρgz where P0 is the pressure at the origin (r = 0, z = 0). Pressure is a fundamental property, and it is hard to imag ine a significant fluid flow problem that does not involve pressure. Therefore, you will see this property in all chap ters in the rest of this book. The consideration of hydrostatic forces acting on plane or curved surfaces, however, is mostly limited to this chapter.
cen96537_ch03_077-136.indd 114 14/01/17 2:23 pm 115 CHAPTER 3 REFERENCES AND SUGGESTED READING 1. F. P. Beer, E. R. Johnston, Jr., E. R. Eisenberg, and G. H. Staab. Vector Mechanics for Engineers, Statics, 10th ed. New York: McGraw-Hill, 2012.
2. D. C. Giancoli. Physics, 6th ed. Upper Saddle River, NJ: Prentice Hall, 2012.
PROBLEMS Pressure, Manometer, and Barometer 3–1C What is the difference between gage pressure and absolute pressure?
3–2C A tiny steel cube is suspended in water by a string. If the lengths of the sides of the cube are very small, how would you compare the magnitudes of the pressures on the top, bottom, and side surfaces of the cube?
3–3C Explain why some people experience nose bleed ing and some others experience shortness of breath at high elevations.
3–4C Consider two identical fans, one at sea level and the other on top of a high mountain, running at identical speeds. How would you compare (a) the volume flow rates and (b) the mass flow rates of these two fans?
3–5C Someone claims that the absolute pressure in a liquid of constant density doubles when the depth is doubled. Do you agree? Explain.
3–6C Express Pascal’s law, and give a real-world example of it.
3–7 A pressure gage connected to a tank reads 500 kPa at a location where the atmospheric pressure is 94 kPa. Determine the absolute pressure in the tank.
3–8 A vacuum gage connected to a chamber reads 25 kPa at a location where the atmospheric pressure is 97 kPa. Determine the absolute pressure in the chamber.
3–9E The pressure at the exit of an air compressor is 150 psia. What is this pressure in kPa?
3–10 A diver’s watch resists an absolute pressure of 5.5 bar. At an ocean having density of 1025 kg/m3 and exposing an atmospheric pressure of 1 bar, what depth can maximally the diver dive to prevent water inlet into his watch? 1 bar = 105 Pa and g = 9.81 m/s2.
3–11E Show that 1 kgf/cm2 = 14.223 psi.
3–12E The pressure in a water line is 1500 kPa. What is the line pressure in (a) lbf/ft2 units and (b) Ibf/in2 (psi) units?
3–13 Blood pressure is usually measured by wrapping a closed air-filled jacket equipped with a pressure gage around the upper arm of a person at the level of the heart. Using a mercury manometer and a stethoscope, the systolic pressure (the maximum pressure when the heart is pumping) and the dia stolic pressure (the minimum pressure when the heart is resting) are measured in mmHg. The systolic and diastolic pressures of a healthy person are about 120 mmHg and 80 mmHg, respec tively, and are indicated as 120/80. Express both of these gage pressures in kPa, psi, and meter water column.
3–14 The maximum blood pressure in the upper arm of a healthy person is about 120 mmHg. If a vertical tube open to the atmosphere is connected to the vein in the arm of the per son, determine how high the blood will rise in the tube. Take the density of the blood to be 1040 kg/m3.
h FIGURE P3–14 3–15 Consider a 1.73-m-tall man standing vertically in water and completely submerged in a pool. Determine the difference between the pressures acting at the head and at the toes of this man, in kPa.
3–16E A manometer is used to measure the air pressure in a tank. The fluid used has a specific gravity of 1.40, and the differential height between the two arms of the manometer is 20 in. If the local atmospheric pressure is 12.7 psia, deter mine the absolute pressure in the tank for the cases of the Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and SI users can ignore them. Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software.
cen96537_ch03_077-136.indd 115 14/01/17 2:23 pm 116 pressure and fluid statics manometer arm with the (a) higher and (b) lower fluid level being attached to the tank.
3–17 The water in a tank is pressurized by air, and the pressure is measured by a multifluid manometer as shown in Fig. P3–17. Determine the gage pressure of air in the tank if h1 = 0.4 m, h2 = 0.6 m, and h3 = 0.8 m. Take the densities of water, oil, and mercury to be 1000 kg/m3, 850 kg/m3, and 13,600 kg/m3, respectively.
h1 h2 h3 Oil Mercury Water Air 1 2 FIGURE P3 –17 3–18 Determine the atmospheric pressure at a location where the barometric reading is 735 mmHg. Take the density of mercury to be 13,600 kg/m3.
3–19 The gage pressure in a liquid at a depth of 2.5 m is read to be 28 kPa. Determine the gage pressure in the same liquid at a depth of 9 m.
3–20 The absolute pressure in water at a depth of 8 m is read to be 175 kPa. Determine (a) the local atmospheric pressure, and (b) the absolute pressure at a depth of 8 m in a liquid whose specific gravity is 0.78 at the same location.
3–21E A 180-lbm man has a total foot imprint area of 68 in2. Determine the pressure this man exerts on the ground if (a) he stands on both feet and (b) he stands on one foot.
3–22 Consider a 55-kg woman who has a total foot imprint area of 400 cm2. She wishes to walk on the snow, but the snow cannot withstand pressures greater than 0.5 kPa. Deter mine the minimum size of the snowshoes needed (imprint area per shoe) to enable her to walk on the snow without sinking.
3–23 A vacuum gage connected to a tank reads 45 kPa at a location where the barometric reading is 755 mmHg. Determine the absolute pressure in the tank. Take 𝜌Hg = 13,590 kg/m3.
Answer: 55.6 kPa 3–24 The piston of a vertical piston-cylinder device containing a gas has a mass of 40 kg and a cross-sectional area of 0.012 m2 (Fig. P3–24). The local atmospheric pressure is 95 kPa, and the gravitational acceleration is 9.81 m/s2. (a) Determine the pressure inside the cylinder. (b) If some heat is transferred to the gas and its volume is doubled, do you expect the pressure inside the cylinder to change?
A = 0.012 m2 Patm = 95 kPa m = 40 kg FIGURE P3–24 3–25 The vacuum pressure of a condenser is given to be 65 kPa. If the atmospheric pressure is 98 kPa, what is the gage pressure and absolute pressure in kPa, kN/m2, lbf/in2, psi, and mmHg.
3–26 Water from a reservoir is raised in a vertical tube of internal diameter D = 30 cm under the influence of the pulling force F of a piston. Determine the force needed to raise the water to a height of h = 1.5 m above the free sur face. What would your response be for h = 3 m? Also, taking the atmospheric pressure to be 96 kPa, plot the absolute water pressure at the piston face as h varies from 0 to 3 m.
Water Air h D F FIGURE P3–26 3–27 The barometer of a mountain hiker reads 980 mbars at the beginning of a hiking trip and 790 mbars at the end. Neglecting the effect of altitude on local gravitational accel eration, determine the vertical distance climbed. Assume an average air density of 1.20 kg/m3. Answer: 1614 m 3–28 Determine the pressure exerted on a diver at 15 m below the free surface of the sea. Assume a baromet ric pressure of 101 kPa and a specific gravity of 1.03 for seawater. Answer: 253 kPa cen96537_ch03_077-136.indd 116 14/01/17 2:23 pm 117 CHAPTER 3 3–29 A gas is contained in a vertical, frictionless piston– cylinder device. The piston has a mass of 5 kg and a cross-sectional area of 35 cm2. A compressed spring above the piston exerts a force of 75 N on the piston. If the atmospheric pressure is 95 kPa, determine the pressure inside the cylinder.
Answer: 130 kPa A = 35 cm2 P = ?
Patm = 95 kPa 75 N mP = 5 kg FIGURE P3–29 3–30 Reconsider Prob. 3–29. Using appropriate soft ware, investigate the effect of the spring force in the range of 0 to 500 N on the pressure inside the cylinder. Plot the pressure against the spring force, and discuss the results.
3–31 The variation of pressure P in a gas with density 𝜌 is given by P = C𝜌n where C and n and are constants with P = P0 and 𝜌 = 𝜌0 at elevation z = 0. Obtain a relation for the variation of P with elevation in terms of z, g, n, P0 and 𝜌0.
3–32 Both a gage and a manometer are attached to a gas tank to measure its pressure. If the reading on the pressure gage is 65 kPa, determine the distance between the two fluid levels of the manometer if the fluid is (a) mercury (𝜌 = 13,600 kg/m3) or (b) water (𝜌 = 1000 kg/m3).
Gas h = ?
Pg = 65 kPa FIGURE P3–32 3–33 Reconsider Prob. 3–32. Using appropriate software, investigate the effect of the manom eter fluid density in the range of 800 to 13,000 kg/m3 on the differential fluid height of the manometer. Plot the dif ferential fluid height against the density, and discuss the results.
3–34 The system shown in the figure is used to accurately measure the pressure changes when the pressure is increased by ΔP in the water pipe. When Δh = 90 mm, what is the change in the pipe pressure?
FIGURE P3–34 Glycerin, SG = 1.26 D = 30 mm d = 3 mm Δh Water Pipe 3–35 The manometer shown in the figure is designed to measure pressures of up to a maximum of 100 Pa. If the reading error is estimated to be ±0.5 mm, what should the ratio of d/D be in order for the error associated with pressure measurement not to exceed 2.5% of the full scale.
FIGURE P3–35 D Scale 휃 = 30° L d P cen96537_ch03_077-136.indd 117 14/01/17 2:23 pm 118 pressure and fluid statics 3–36 A manometer containing oil (𝜌 = 850 kg/m3) is attached to a tank filled with air. If the oil-level difference between the two columns is 150 cm and the atmospheric pressure is 98 kPa, determine the absolute pressure of the air in the tank. Answer: 111 kPa 3–37 A mercury manometer (𝜌 = 13,600 kg/m3) is con nected to an air duct to measure the pressure inside. The difference in the manometer levels is 10 mm, and the atmo spheric pressure is 100 kPa. (a) Judging from Fig. P3–37, determine if the pressure in the duct is above or below the atmospheric pressure. (b) Determine the absolute pressure in the duct.
FIGURE P3–37 Air h = 10 mm P = ?
3–38 Repeat Prob. 3–37 for a differential mercury height of 25 mm.
3–39 Consider a U-tube whose arms are open to the atmo sphere. Now water is poured into the U-tube from one arm, and light oil (𝜌 = 790 kg/m3) from the other. One arm con tains 70-cm-high water, while the other arm contains both fluids with an oil-to-water height ratio of 6. Determine the height of each fluid in that arm.
70 cm Water Oil FIGURE P3–39 3–40 The hydraulic lift in a car repair shop has an output diameter of 45 cm and is to lift cars up to 2500 kg. Deter mine the fluid gage pressure that must be maintained in the reservoir.
3–41 Consider a double-fluid manometer attached to an air pipe shown in Fig. P3–41. If the specific gravity of one fluid is 13.55, determine the specific gravity of the other fluid for the indicated absolute pressure of air. Take the atmospheric pressure to be 100 kPa. Answer: 1.62 SG2 Air P = 76 kPa 22 cm 33 cm SG1 = 13.55 FIGURE P3–41 3–42E The pressure in a natural gas pipeline is measured by the manometer shown in Fig. P3–42E with one of the arms open to the atmosphere where the local atmospheric pressure is 14.2 psia. Determine the absolute pressure in the pipeline.
14 in 6 in 2 in 22 in Natural gas Water Air Mercury SG = 13.6 FIGURE P3–42E 3–43E Repeat Prob. 3–42E by replacing air by oil with a specific gravity of 0.69.
cen96537_ch03_077-136.indd 118 14/01/17 2:23 pm 119 CHAPTER 3 3–44 The gage pressure of the air in the tank shown in Fig. P3–44 is measured to be 50 kPa. Determine the differen tial height h of the mercury column.
Air 30 cm 75 cm h Mercury SG = 13.6 Water Oil SG = 0.72 50 kPa FIGURE P3–44 3–45 Repeat Prob. 3–44 for a gage pressure of 40 kPa.
3–46 The 500-kg load on the hydraulic lift shown in Fig. P3–46 is to be raised by pouring oil (𝜌 = 780 kg/m3) into a thin tube. Determine how high h should be in order to begin to raise the weight.
Load 500 kg h 1.2 m 1 cm FIGURE P3–46 3–47 Pressure is often given in terms of a liquid column and is expressed as “pressure head.” Express the standard atmospheric pressure in terms of (a) mercury (SG = 13.6), (b) water (SG = 1.0), and (c) glycerin (SG = 1.26) columns. Explain why we usually use mercury in manometers.
3–48 Freshwater and seawater flowing in parallel horizon tal pipelines are connected to each other by a double U-tube manometer, as shown in Fig. P3–48. Determine the pressure difference between the two pipelines. Take the density of sea water at that location to be 𝜌 = 1035 kg/m3. Can the air col umn be ignored in the analysis?
50 cm 10 cm 70 cm 30 cm Fresh water Sea-water Mercury Air FIGURE P3–48 3–49 Repeat Prob. 3–48 by replacing the air with oil whose specific gravity is 0.72.
3–50 The pressure difference between an oil pipe and water pipe is measured by a double-fluid manometer, as shown in Fig. P3–50. For the given fluid heights and specific gravities, calculate the pressure difference ΔP = PB − PA.
Oil SG = 0.88 Glycerin SG = 1.26 Water SG = 1.0 Mercury SG = 13.5 A B 20 cm 55 cm 10 cm 12 cm FIGURE P3–50 3–51 Consider the system shown in Fig. P3–51. If a change of 0.9 kPa in the pressure of air causes the brine-mercury interface in the right column to drop by 5 mm in the brine level in the right column while the pressure in the brine pipe remains constant, determine the ratio of A2/A1.
cen96537_ch03_077-136.indd 119 14/01/17 2:23 pm 120 pressure and fluid statics Mercury SG = 13.56 Water Air Area, A1 Area, A2 Brine pipe SG = 1.1 FIGURE P3–51 3–52 There is water at a height of 1 m in the tube open to the atmosphere (Patm = 100 kPa) connected to a tank with two sections. (a) Find the pressure readings (kPa) in Bourdon Type Manometer for the air in the A and B sections of the tank. (b) Find the absolute pressures (kPa) for the air in the A and B sections of the tank. Water density is 1000 kg/m3 and g = 9.79 m/s2.
Air A B Air 1 m 0.3 m Water Water Water 0.5 m PA PB Patm FIGURE P3–52 3–53 The top part of a water tank is divided into two com partments, as shown in Fig. P3–53. Now a fluid with an unknown density is poured into one side, and the water level rises a certain amount on the other side to compensate for this effect. Based on the final fluid heights shown on the fig ure, determine the density of the fluid added. Assume the liq uid does not mix with water.
FIGURE P3–53 45 cm 95 cm 75 cm Unknown liquid Water 3–54 A simple experiment has long been used to dem onstrate how negative pressure prevents water from being spilled out of an inverted glass. A glass that is fully filled by water and covered with a thin paper is inverted, as shown in Fig. P3–54. Determine the pressure at the bottom of the glass, and explain why water does not fall out.
Glass Water A piece of paper 12 cm FIGURE P3–54 3–55 A multifluid container is connected to a U-tube, as shown in Fig. P3–55. For the given specific gravities and fluid column heights, determine the gage pressure at A. Also determine the height of a mercury column that would create the same pressure at A. Answers: 0.415 kPa, 0.311 cm cen96537_ch03_077-136.indd 120 14/01/17 2:23 pm 121 CHAPTER 3 80 cm Oil SG = 0.90 Water Glycerin SG = 1.26 35 cm 18 cm 15 cm 90 cm A FIGURE P3–55 3–56 A hydraulic lift is to be used to lift a 2500 kg weight by putting a weight of 25 kg on a piston with a diameter of 10 cm. Determine the diameter of the piston on which the weight is to be placed.
FIGURE P3–56 F1 D2 F2 25 kg 10 cm Weight 2500 kg 3–57 On a day in which the local atmospheric pressure is 99.5 kPa, answer each of the following: (a) Calculate the column height of mercury in a mercury barometer in units of meters, feet, and inches.
(b) Francis is concerned about mercury poisoning, so he builds a water barometer to replace the mercury barometer. Calculate the column height of water in the water barometer in units of meters, feet, and inches.
(c) Explain why a water barometer is not very practical.
(d) Ignoring the practicality issue, which of the two (mercury or water) would be more precise? Explain. 3–58 A U-tube manometer is used to measure the pressure in a vacuum chamber as shown in the sketch. The fluid in the vacuum chamber is of density 𝜌1, and the manometer fluid is of density 𝜌2. The right side of the U-tube manometer is exposed to atmospheric pressure Patm, and elevations z1, z2, and zA are measured.
z1 Vacuum chamber A ρ z2 zA 2 ρ1 Patm FIGURE P3–58 (a) Generate an exact expression for the vacuum pressure at point A, i.e., generate an expression for PA,vac as a function of variables 𝜌1, 𝜌2, z1, z2, and zA.
(b) Simplify your expression for the case in which 𝜌1 ≪ 𝜌2.
Fluid Statics: Hydrostatic Forces on Plane and Curved Surfaces 3–59C Define the resultant hydrostatic force acting on a submerged surface, and the center of pressure.
3–60C You may have noticed that dams are much thicker at the bottom. Explain why dams are built that way.
3–61C Someone claims that she can determine the magni tude of the hydrostatic force acting on a plane surface sub merged in water regardless of its shape and orientation if she knew the vertical distance of the centroid of the surface from the free surface and the area of the surface. Is this a valid claim? Explain.
3–62C A submerged horizontal flat plate is suspended in water by a string attached at the centroid of its upper surface. Now the plate is rotated 45° about an axis that passes through its centroid. Discuss the change in the hydrostatic force act ing on the top surface of this plate as a result of this rotation. Assume the plate remains submerged at all times.
cen96537_ch03_077-136.indd 121 14/01/17 2:23 pm 122 pressure and fluid statics 3–63C Consider a submerged curved surface. Explain how you would determine the horizontal component of the hydro static force acting on this surface.
3–64C Consider a submerged curved surface. Explain how you would determine the vertical component of the hydro static force acting on this surface.
3–65C Consider a circular surface subjected to hydrostatic forces by a constant density liquid. If the magnitudes of the horizontal and vertical components of the resultant hydro static force are determined, explain how you would find the line of action of this force.
3–66E Consider a 200-ft-high, 1200-ft-wide dam filled to capacity. Determine (a) the hydrostatic force on the dam and (b) the force per unit area of the dam near the top and near the bottom.
3–67 A cylindrical tank is fully filled with water (Fig. P3–67). In order to increase the flow from the tank, an additional pressure is applied to the water surface by a compressor. For P0 = 0, P0 = 5 bar, and P0 = 10 bar, calculate the hydrostatic force on the surface A exerted by water.
FIGURE P3–67 Air, P0 Water level Water 60 cm A 3–68 Consider a 8-m-long, 8-m-wide, and 2-m-high aboveground swimming pool that is filled with water to the rim. (a) Determine the hydrostatic force on each wall and the distance of the line of action of this force from the ground. (b) If the height of the walls of the pool is doubled and the pool is filled, will the hydrostatic force on each wall double or quadruple? Why? Answer: (a) 157 kN 3–69 Consider a heavy car submerged in water in a lake with a flat bottom. The driver’s side door of the car is 1.1 m high and 0.9 m wide, and the top edge of the door is 10 m below the water surface. Determine the net force acting on the door (normal to its surface) and the location of the pres sure center if (a) the car is well-sealed and it contains air at atmospheric pressure and (b) the car is filled with water.
3–70 A room in the lower level of a cruise ship has a 40-cm-diameter circular window. If the midpoint of the window is 2 m below the water surface, determine the hydro static force acting on the window, and the pressure center. Take the specific gravity of seawater to be 1.025. Answers: 2527 N, 2.005 m Sea 2 m 40 cm FIGURE P3–70 3–71 The water side of the wall of a 70-m-long dam is a quarter circle with a radius of 7 m. Determine the hydro static force on the dam and its line of action when the dam is filled to the rim.
3–72 A water trough of semicircular cross section of radius 0.6 m consists of two symmetric parts hinged to each other at the bottom, as shown in Fig. P3–72. The two parts are held together by a cable and turnbuckle placed every 3 m along the length of the trough. Calculate the tension in each cable when the trough is filled to the rim.
1.2 m Cable Hinge FIGURE P3–72 3–73 Determine the resultant force acting on the 0.7-m-high and 0.7-m-wide triangular gate shown in Fig. P3–73 and its line of action.
FIGURE P3–73 Water 0.9 m 0.3 m 0.7 m 0.7 m β cen96537_ch03_077-136.indd 122 14/01/17 2:23 pm 123 CHAPTER 3 3–74 A 6-m-high, 5-m-wide rectangular plate blocks the end of a 5-m-deep freshwater channel, as shown in Fig. P3–74. The plate is hinged about a horizontal axis along its upper edge through a point A and is restrained from open ing by a fixed ridge at point B. Determine the force exerted on the plate by the ridge.
1 m 5 m A B FIGURE P3–74 3–75 Reconsider Prob. 3–74. Using appropriate soft ware, investigate the effect of water depth on the force exerted on the plate by the ridge. Let the water depth vary from 0 to 5 m in increments of 0.5 m. Tabulate and plot your results.
3–76E The flow of water from a reservoir is controlled by a 5-ft-wide L-shaped gate hinged at point A, as shown in Fig. P3–76E. If it is desired that the gate open when the water height is 12 ft, determine the mass of the required weight W. Answer: 30,900 lbm A B 8 ft 12 ft 15 ft Gate W FIGURE P3–76E 3–77E Repeat Prob. 3–76E for a water height of 6 ft.
3–78 For a gate width of 2 m into the paper (Fig. P3–78), determine the force required to hold the gate ABC at its location. Answer: 17.8 kN FIGURE P3–78 10 cm 45° Hinge 50 cm SG = 0.86 SG = 1.23 80 cm 40 cm C F B A 3–79E A long, solid cylinder of radius 2 ft hinged at point A is used as an automatic gate, as shown in Fig. P3–79E. When the water level reaches 12 ft, the cylindrical gate opens by turning about the hinge at point A. Determine (a) the hydrostatic force acting on the cylinder and its line of action when the gate opens and (b) the weight of the cylinder per ft length of the cylinder.
A 12 ft 2 ft FIGURE P3–79E 3–80 An open settling tank shown in the figure contains a liquid suspension. Determine the resultant force acting on the gate and its line of action if the liquid density is 850 kg/m3. The gate is parabolic as sketched, looking straight at the gate. Answers: 140 kN, 1.64 m from bottom FIGURE P3–80 y x y = 2x2 3 m 5 m θ = 60° cen96537_ch03_077-136.indd 123 14/01/17 2:23 pm 124 pressure and fluid statics 3–81 From Prob. 3–80, knowing that the density of the suspension depends on liquid depth and changes linearly from 800 kg/m3 to 900 kg/m3 in the vertical direction, deter mine the resultant force acting on the gate ABC, and its line of action.
3–82 The two sides of a V-shaped water trough are hinged to each other at the bottom where they meet, as shown in Fig. P3–82, making an angle of 45° with the ground from both sides. Each side is 0.75 m wide, and the two parts are held together by a cable and turnbuckle placed every 6 m along the length of the trough. Calculate the tension in each cable when the trough is filled to the rim. Answer: 5510 N 0.75 m 45° 45° Cable Hinge FIGURE P3–82 3–83 Repeat Prob. 3–82 for the case of a partially filled trough with a water height of 0.35 m directly above the hinge.
3–84 The bowl shown in the figure (the white volume) is to be cast in a pair of molding boxes. When liquid metal is poured in to the top, calculate the additional tensile force on each of 20 bolts located circumferentially. The specific gravity of the molten metal can be taken to be 7.8. 20 mm 40 cm 38 cm 20 bolts Upper molding box Lower molding box 20 cm 90 cm 20 mm FIGURE P3–84 3–85 A triangular-shaped gate is hinged at point A, as shown. Knowing that the weight of the gate is 100 N, deter mine the force needed to keep the gate at its position for unit width. The line of action of the weight of the gate is shown by the dashed line.
FIGURE P3–85 Water 2.4 m A 4.5 m 1 m F 1 m 3–86 Gate AB (0.6-m × 0.9-m) is located at the bottom of a tank filled with methyl alcohol (SG = 0.79), and hinged along its bottom edge A. Knowing that the weight of the gate is 300 N, determine the minimum force that must be applied to the cable (BCD) to open the gate. A B C 0.6 m 0.4 m 0.3 m 0.8 m F D FIGURE P3–86 3–87 Find the force applied by support BC to the gate AB. The width of the gate and support is 3 m and the weight of the gate is 1500 N. FIGURE P3–87 3 m Water 5 m 훼 = 50° 4 m F = ?
A, hinge B C 훽 cen96537_ch03_077-136.indd 124 14/01/17 2:23 pm 125 CHAPTER 3 3–88 A concrete block is attached to the gate as shown. If the water level is 1.3 m from the bottom of the container, there is no reaction force at A. What would be the reaction force for the level shown. Specific gravity of the concrete is 2.4. 40° 2 m 0.70 m 0.60 m A m FIGURE P3–88 3–89 The curved surface given in the figure is defined by y = 3√x. Determine the horizontal force and its line of action applied by water on the curved surface. The width of the gate is b = 2 m. Water y = 3√x 5 m x y FIGURE P3–89 3–90 A 4-m-long quarter-circular gate of radius 3 m and of negligible weight is hinged about its upper edge A, as shown in Fig. P3–90. The gate controls the flow of water over the ledge at B, where the gate is pressed by a spring. Determine the minimum spring force required to keep the gate closed when the water level rises to A at the upper edge of the gate.
A B 3 m Spring FIGURE P3–90 3–91 Repeat Prob. 3–90 for a radius of 2 m for the gate.
Answer: 78.5 kN Buoyancy 3–92C What is buoyant force? What causes it? What is the magnitude of the buoyant force acting on a submerged body whose volume is V? What are the direction and the line of action of the buoyant force?
3–93C Discuss the stability of (a) a submerged and (b) a floating body whose center of gravity is above the center of buoyancy.
3–94C Consider two 5-cm-diameter spherical balls—one made of aluminum, the other of iron—submerged in water. Will the buoyant forces acting on these two balls be the same or different? Explain.
3–95C Consider a 3-kg copper cube and a 3-kg copper ball submerged in a liquid. Will the buoyant forces acting on these two bodies be the same or different? Explain.
3–96C Consider two identical spherical balls submerged in water at different depths. Will the buoyant forces acting on these two balls be the same or different? Explain.
3–97 A 200-kg granite rock (𝜌 = 2700 kg/m3) is dropped into a lake. A man dives in and tries to lift the rock. Deter mine how much force the man needs to apply to lift it from the bottom of the lake. Do you think he can do it?
3–98 The hull of a boat has a volume of 180 m3, and the total mass of the boat when empty is 8560 kg. Determine how much load this boat can carry without sinking (a) in a lake and (b) in seawater with a specific gravity of 1.03.
3–99 The density of a liquid is to be determined by an old 1-cm-diameter cylindrical hydrometer whose division marks are completely wiped out. The hydrometer is first dropped in water, and the water level is marked. The hydrometer is then dropped into the other liquid, and it is observed that the mark for water has risen 0.3 cm above the liquid–air inter face (Fig. P3–99). If the height of the original water mark is 12.3 cm, determine the density of the liquid.
Mark for water Unknown liquid 0.3 cm 12 cm FIGURE P3–99 cen96537_ch03_077-136.indd 125 14/01/17 2:23 pm 126 pressure and fluid statics 3–100 It is said that Archimedes discovered his principle during a bath while thinking about how he could determine if King Hiero’s crown was actually made of pure gold. While in the bathtub, he conceived the idea that he could deter mine the average density of an irregularly shaped object by weighing it in air and also in water. If the crown weighed 3.55 kgf (= 34.8 N) in air and 3.25 kgf (= 31.9 N) in water, determine if the crown is made of pure gold. The density of gold is 19,300 kg/m3. Discuss how you can solve this prob lem without weighing the crown in water but by using an ordinary bucket with no calibration for volume. You may weigh anything in air.
3–101 It is estimated that 90 percent of an iceberg’s volume is below the surface, while only 10 percent is visible above the surface. For seawater with a density of 1025 kg/m3, esti mate the density of the iceberg.
FIGURE P3–101 ©Ralph Clevenger/Corbis 3–102 One of the common procedures in fitness programs is to determine the fat-to-muscle ratio of the body. This is based on the principle that the muscle tissue is denser than the fat tissue, and, thus, the higher the average density of the body, the higher is the fraction of muscle tissue. The aver age density of the body can be determined by weighing the person in air and also while submerged in water in a tank. Treating all tissues and bones (other than fat) as muscle with an equivalent density of 𝜌muscle, obtain a relation for the volume fraction of body fat xfat. Answer: xfat = (𝜌muscle − 𝜌avg)/ (𝜌muscle − 𝜌fat).
Submerged person Water tank Spring scale FIGURE P3–102 3–103 A cone floats in the glycerin (SG = 1.26), as shown in the figure. Find the mass of the cone. FIGURE P3–103 R r SG = 1.26 60 cm h = 30 cm 3–104 The weight of a body is usually measured by disregarding buoyancy force applied by the air. Consider a 20-cm-diameter spherical body of density 7800 kg/m3. What is the percentage error associated with the neglecting of air buoyancy?
Fluids in Rigid-Body Motion 3–105C Under what conditions can a moving body of fluid be treated as a rigid body?
3–106C Consider a vertical cylindrical container partially filled with water. Now the cylinder is rotated about its axis at a specified angular velocity, and rigid-body motion is established. Discuss how the pressure will be affected at the midpoint and at the edges of the bottom surface due to rotation.
3–107C Consider two identical glasses of water, one sta tionary and the other moving on a horizontal plane with con stant acceleration. Assuming no splashing or spilling occurs, which glass will have a higher pressure at the (a) front, (b) midpoint, and (c) back of the bottom surface?
cen96537_ch03_077-136.indd 126 14/01/17 2:23 pm 127 CHAPTER 3 3–108C Consider a glass of water. Compare the water pressures at the bottom surface for the following cases: the glass is (a) stationary, (b) moving up at constant velocity, (c) moving down at constant velocity, and (d) moving horizontally at constant velocity.
3–109 A water tank is being towed by a truck on a level road, and the angle the free surface makes with the horizontal is mea sured to be 8°. Determine the acceleration of the truck.
3–110 Consider two water tanks filled with water. The first tank is 8 m high and is stationary, while the second tank is 2 m high and is moving upward with an acceleration of 5 m/s2. Which tank will have a higher pressure at the bottom?
3–111 A water tank is being towed on an uphill road that makes 14° with the horizontal with a constant acceleration of 3.5 m/s2 in the direction of motion. Determine the angle the free surface of water makes with the horizontal. What would your answer be if the direction of motion were downward on the same road with the same acceleration?
3–112 The bottom quarter of a vertical cylindrical tank of total height 0.4 m and diameter 0.3 m is filled with a liquid (SG > 1, like glycerin) and the rest with water, as shown in the figure. The tank is now rotated about its vertical axis at a constant angular speed of ω. Determine (a) the value of the angular speed when the point P on the axis at the liquid-liquid interface touches the bottom of the tank and (b) the amount of water that would be spilled out at this angular speed.
FIGURE P3–112 P D = 0.3 m h = 0.1 m 3h 휔 3–113 A 3-m-diameter, 7-m-long cylindrical tank is com pletely filled with water. The tank is pulled by a truck on a level road with the 7-m-long axis being horizontal. Deter mine the pressure difference between the front and back ends of the tank along a horizontal line when the truck (a) acceler ates at 3 m/s2 and (b) decelerates at 4 m/s2.
3–114 A 30-cm-diameter, 90-cm-high vertical cylindrical container is partially filled with 60-cm-high water. Now the cylinder is rotated at a constant angular speed of 180 rpm. Determine how much the liquid level at the center of the cyl inder will drop as a result of this rotational motion.
3–115 A fish tank that contains 60-cm-high water is moved in the cabin of an elevator. Determine the pressure at the bot tom of the tank when the elevator is (a) stationary, (b) mov ing up with an upward acceleration of 3 m/s2, and (c) moving down with a downward acceleration of 3 m/s2.
3–116E A 15-ft-long, 6-ft-high rectangular tank open to the atmosphere is towed by a truck on a level road. The tank is filled with water to a depth of 5 ft. Determine the maximum acceleration or deceleration allowed if no water is to spill during towing.
3–117 Consider a tank of rectangular cross-section partially filled with a liquid placed on an inclined surface, as shown in the figure. When frictional effects are negligible, show that the slope of the liquid surface will be the same as the slope of the inclined surface when the tank is released. What can you say about the slope of the free surface when the friction is significant?
FIGURE P3–117 z y a α 3–118E A 3-ft-diameter vertical cylindrical tank open to the atmosphere contains 1-ft-high water. The tank is now rotated about the centerline, and the water level drops at the center while it rises at the edges. Determine the angu lar velocity at which the bottom of the tank will first be exposed. Also determine the maximum water height at this moment.
3–119 Milk with a density of 1020 kg/m3 is transported on a level road in a 9-m-long, 3-m-diameter cylindrical tanker. The tanker is completely filled with milk (no air space), and it accelerates at 4 m/s2. If the minimum pres sure in the tanker is 100 kPa, determine the maximum pressure difference and the location of the maximum pres sure. Answer: 66.7 kPa 9 m 3 m FIGURE P3–119 cen96537_ch03_077-136.indd 127 14/01/17 2:24 pm 128 pressure and fluid statics 3–120 Repeat Prob. 3–119 for a deceleration of 2.5 m/s2.
3–121 The distance between the centers of the two arms of a U-tube open to the atmosphere is 30 cm, and the U-tube contains 20-cm-high alcohol in both arms. Now the U-tube is rotated about the left arm at 3.5 rad/s. Deter mine the elevation difference between the fluid surfaces in the two arms.
30 cm 20 cm 휔 FIGURE P3–121 3–122 A 1.2-m-diameter, 3-m-high sealed vertical cylinder is completely filled with gasoline whose density is 740 kg/m3. The tank is now rotated about its vertical axis at a rate of 70 rpm. Determine (a) the difference between the pressures at the centers of the bottom and top surfaces and (b) the differ ence between the pressures at the center and the edge of the bottom surface.
3 m 1.20 m 휔 FIGURE P3–122 3–123 Reconsider Prob. 3–122. Using appropriate software, investigate the effect of rotational speed on the pressure difference between the center and the edge of the bottom surface of the cylinder. Let the rotational speed vary from 0 rpm to 500 rpm in increments of 50 rpm. Tabulate and plot your results.
3–124 A 4-m-diameter vertical cylindrical milk tank rotates at a constant rate of 15 rpm. If the pressure at the center of the bottom surface is 130 kPa, determine the pressure at the edge of the bottom surface of the tank. Take the density of the milk to be 1030 kg/m3.
3–125E An 8-ft-long tank open to the atmosphere initially contains 3-ft-high water. It is being towed by a truck on a level road. The truck driver applies the brakes and the water level at the front rises 0.75 ft above the initial level. Deter mine the deceleration of the truck. Answer: 6.04 ft/s2 8 ft Water 3 ft 0.75 ft FIGURE P3–125E 3–126 A 75-cm-high, 40-cm-diameter cylindrical water tank is being transported on a level road. The highest accel eration anticipated is 5 m/s2. Determine the allowable initial water height in the tank if no water is to spill out during acceleration. Answer: 64.8 cm 3–127 The rectangular tank is filled with heavy oil (like glycerin) at the bottom and water at the top, as shown in the figure. The tank is now moved to the right horizontally with a constant acceleration and ¼ of water is spilled out as a result from the back. Using geometrical considerations, determine how high the point A at the back of the tank on the oil-water interface will rise under this acceleration.
FIGURE P3–127 Water Oil a 0.75 m 1.0 m A L cen96537_ch03_077-136.indd 128 14/01/17 2:24 pm 129 CHAPTER 3 3–128 A centrifugal pump consists simply of a shaft and a few blades attached normally to the shaft. If the shaft is rotated at a constant rate of 1800 rpm, what would the theoretical pump head due to this rotation be? Take the impel ler diameter to be 35 cm and neglect the blade tip effects. Answer: 55.5 m 3–129 Two vertical and connected cylindrical tanks of diameter D are open to the atmosphere and they contain water at a height of h in the initial state, as shown in the figure. As the radial blades in the left tank are rotated about the cen terline of the tank, some water in the right tank flows into the left tank. Determine the angular velocity of radial blades 𝜔 such that half of water in the right tank flows into the left tank. Also, derive a general relation for angular velocity as a function of initial height of water in the tank. Assume water in the tanks would not spill. D = 2R = 45 cm, h = 40 cm and g = 9.81 m/s2.
h = 40 cm A B D = 45 cm Water 휔 FIGURE P3–129 3–130 The U-tube shown in the figure is subjected to an acceleration of a (m/s2) to left. If the difference between the free surfaces is ∆h = 20 cm and h = 0.4 m, calculate the acceleration. FIGURE P3–130 ∆h h a L = 0.5 m Liquid SG = 1.6 Water Review Problems 3–131E A pressure gage connected to a tank reads 60 psi at a location where the barometric reading is 29.1 inHg. Determine the absolute pressure in the tank. Take 𝜌Hg = 848.4 lbm/ft3. Answer: 74.3 psia 3–132 An air-conditioning system requires a 34-m-long section of 12-cm-diameter ductwork to be laid underwa ter. Determine the upward force the water will exert on the duct. Take the densities of air and water to be 1.3 kg/m3 and 1000 kg/m3, respectively.
3–133E Determine the pressure exerted on the surface of a submarine cruising 225 ft below the free surface of the sea. Assume that the barometric pressure is 14.7 psia and the spe cific gravity of seawater is 1.03.
3–134 A vertical, frictionless piston–cylinder device contains a gas at 600 kPa. The atmospheric pressure outside is 100 kPa, and the piston area is 30 cm2. Determine the mass of the piston.
3–135 If the rate of rotational speed of the 3-tube system shown in Fig. P3–135 is 𝜔 = 10 rad/s, determine the water heights in each tube leg. At what rotational speed will the middle tube be completely empty?
FIGURE P3–135 ω = 10 rad/s h = 15 cm 20 cm 10 cm 3–136 The average atmospheric pressure on earth is approximated as a function of altitude by the relation Patm = 101.325 (1 − 0.02256z)5.256, where Patm is the atmospheric pressure in kPa and z is the altitude in km with z = 0 at sea level. Determine the approximate atmospheric pressures at Atlanta (z = 306 m), Denver (z = 1610 m), Mexico City (z = 2309 m), and the top of Mount Everest (z = 8848 m).
3–137 Balloons are often filled with helium gas because it weighs only about one-seventh of what air weighs under identical conditions. The buoyancy force, which can be expressed as Fb = 𝜌airgVballoon, will push the balloon upward. If the balloon has a diameter of 12 m and carries two people, 70 kg each, determine the acceleration of the bal loon when it is first released. Assume the density of air is 𝜌 = 1.16 kg/m3, and neglect the weight of the ropes and the cage. Answer: 25.7 m/s2 cen96537_ch03_077-136.indd 129 14/01/17 2:24 pm 130 pressure and fluid statics Helium D = 12 m ρHe = ρair 1 7 m = 140 kg FIGURE P3–137 3–138 Reconsider Prob. 3–137. Using appropriate soft ware, investigate the effect of the number of people carried in the balloon on acceleration. Plot the accelera tion against the number of people, and discuss the results.
3–139 Determine the maximum amount of load, in kg, the balloon described in Prob. 3–137 can carry. Answer: 521 kg 3–140 The basic barometer can be used as an altitude- measuring device in airplanes. The ground control reports a barometric reading of 760 mmHg while the pilot’s reading is 420 mmHg. Estimate the altitude of the plane from ground level if the average air density is 1.20 kg/m3. Answer: 3853 m 3–141 The lower half of a 12-m-high cylindrical con tainer is filled with water (𝜌 = 1000 kg/m3) and the upper half with oil that has a specific gravity of 0.85. Determine the pressure difference between the top and bottom of the cylinder. Answer: 109 kPa Oil SG = 0.85 h = 12 m Water ρ = 1000 kg/m3 FIGURE P3–141 3–142 The 0.5-m-radius semi-circular gate shown in the figure is hinged through the top edge AB. Find the required force to be applied at the center of gravity to keep the gate closed. Answer: 11.3 kN FIGURE P3–142 Pair = 80 kPa (abs) Oil SG = 0.91 Glycerin SG = 1.26 4.74 m A B R CG F 3–143 A pressure cooker cooks a lot faster than an ordinary pan by maintaining a higher pressure and temperature inside. The lid of a pressure cooker is well sealed, and steam can escape only through an opening in the middle of the lid. A separate metal piece, the petcock, sits on top of this open ing and prevents steam from escaping until the pressure force overcomes the weight of the petcock. The periodic escape of the steam in this manner prevents any potentially danger ous pressure buildup and keeps the pressure inside at a con stant value. Determine the mass of the petcock of a pressure cooker whose operation pressure is 120 kPa gage and has an opening cross-sectional area of 3 mm2. Assume an atmo spheric pressure of 101 kPa, and draw the free-body diagram of the petcock. Answer: 36.7 g Petcock A = 3 mm2 Pressure cooker FIGURE P3–143 3–144 A glass tube is attached to a water pipe, as shown in Fig. P3–144. If the water pressure at the bottom of the tube is 110 kPa and the local atmospheric pressure is 98 kPa, deter mine how high the water will rise in the tube, in m. Assume g = 9.8 m/s2 at that location and take the density of water to be 1000 kg/m3.
cen96537_ch03_077-136.indd 130 14/01/17 2:24 pm 131 CHAPTER 3 Patm = 98 kPa h = ?
Water FIGURE P3–144 3–145 A system is equipped with two pressure gages and a manometer, as shown in Fig. P3–145. For Δh = 12 cm, deter mine the pressure difference ΔP = P2 − P1.
Air Manometer fluid SG = 2.15 Δh P1 P2 Oil SG = 0.87 FIGURE P3–145 3–146 An oil pipeline and a 1.3-m3 rigid air tank are connected to each other by a manometer, as shown in Fig. P3–146. If the tank contains 15 kg of air at 80°C, determine (a) the absolute pressure in the pipeline and (b) the change in Δh when the temperature in the tank drops to 20°C. Assume the pressure in the oil pipeline to remain constant, and the air volume in the manometer to be negligible relative to the volume of the tank.
1.3 m3 Air, 80°C Oil SG = 2.68 h = 75 cm D = 4 mm Mercury SG = 13.6 50° 3D Δh = 20 cm FIGURE P3–146 3–147 A 20-cm-diameter vertical cylindrical vessel is rotated about its vertical axis at a constant angular velocity of 70 rad/s. If the pressure at the midpoint of the inner top surface is atmo spheric pressure like the outer surface, determine the total upward force acting upon the entire top surface inside the cylinder.
3–148 An elastic air balloon having a diameter of 30 cm is attached to the base of a container partially filled with water at +4°C, as shown in Fig. P3–148. If the pressure of the air above the water is gradually increased from 100 kPa to 1.6 MPa, will the force on the cable change? If so, what is the percent change in the force? Assume the pressure on the free surface and the diameter of the balloon are related by P = CDn, where C is a constant and n = −2. The weight of the balloon and the air in it is negligible. Answer: 98.4 percent 50 cm Water 50 cm 20 cm P1 = 100 kPa D1 = 30 cm FIGURE P3–148 3–149 Reconsider Prob. 3–148. Using appropriate software, investigate the effect of air pressure above water on the cable force. Let this pressure vary from 0.5 MPa to 15 MPa. Plot the cable force versus the air pressure.
cen96537_ch03_077-136.indd 131 14/01/17 2:24 pm 132 pressure and fluid statics 3–150 A gasoline line is connected to a pressure gage through a double-U manometer, as shown in Fig. P3–150. If the reading of the pressure gage is 260 kPa, determine the gage pressure of the gasoline line.
45 cm 10 cm 22 cm 50 cm Mercury SG = 13.6 Gasoline SG = 0.70 Water Air Oil SG = 0.79 Pgage = 260 kPa Pipe FIGURE P3–150 3–151 Repeat Prob. 3–150 for a pressure gage reading of 330 kPa.
3–152E A water pipe is connected to a double-U manom eter as shown in Fig. P3–152E at a location where the local atmospheric pressure is 14.5 psia. Determine the absolute pressure at the center of the pipe.
Mercury SG = 13.6 Oil SG = 0.80 Oil SG = 0.80 Water pipe 15 in 40 in 60 in 35 in FIGURE P3–152E 3–153 Consider a U-tube filled with mercury as shown in Fig. P3–153. The diameter of the right arm of the U-tube is D = 1.5 cm, and the diameter of the left arm is twice that. Heavy oil with a specific gravity of 2.72 is poured into the left arm, forcing some mercury from the left arm into the right one. Determine the maximum amount of oil that can be added into the left arm. Answer: 0.0884 L Oil poured in here SG = 2.72 D = 1.5 cm 2D 12 cm Mercury SG = 13.6 FIGURE P3–153 3–154 The variation of pressure with density in a thick gas layer is given by P = C𝜌n, where C and n are constants. Noting that the pressure change across a differential fluid layer of thickness dz in the vertical z-direction is given as dP = −𝜌g dz, obtain a relation for pressure as a function of elevation z. Take the pressure and density at z = 0 to be P0 and 𝜌0, respectively.
3–155 A 3-m-high, 5-m-wide rectangular gate is hinged at the top edge at A and is restrained by a fixed ridge at B. Determine the hydrostatic force exerted on the gate by the 5-m-high water and the location of the pressure center.
2 m Water Gate 3 m A B FIGURE P3–155 3–156 Repeat Prob. 3–155 for a total water height of 2 m.
3–157E A semicircular 40-ft-diameter tunnel is to be built under a 150-ft-deep, 800-ft-long lake, as shown in Fig. P3–157E. Determine the total hydrostatic force acting on the roof of the tunnel.
cen96537_ch03_077-136.indd 132 14/01/17 2:24 pm 133 CHAPTER 3 Water 150 ft 40 ft Tunnel FIGURE P3–157E 3–158 A 30-ton, 4-m-diameter hemispherical dome on a level surface is filled with water, as shown in Fig. P3–158. Someone claims that he can lift this dome by making use of Pascal’s law by attaching a long tube to the top and filling it with water. Determine the required height of water in the tube to lift the dome. Disregard the weight of the tube and the water in it. Answer: 1.72 m 30 ton h 4 m Water FIGURE P3–158 3–159 The water in a 25-m-deep reservoir is kept inside by a 90-m-wide wall whose cross section is an equilateral triangle, as shown in Fig. P3–159. Determine (a) the total force (hydrostatic + atmospheric) acting on the inner surface of the wall and its line of action and (b) the magnitude of the horizontal component of this force. Take Patm = 100 kPa.
Water 60° 60° 25 m FIGURE P3–159 3–160 A 5-m-long, 4-m-high tank contains 2.5-m-deep water when not in motion and is open to the atmosphere through a vent in the middle. The tank is now accelerated to the right on a level surface at 2 m/s2. Determine the maxi mum pressure in the tank relative to the atmospheric pressure. Answer: 29.5 kPa 5 m Vent Water tank 2.5 m 1.5 m 2 m/s2 FIGURE P3–160 3–161 Reconsider Prob. 3–160. Using appropriate software, investigate the effect of acceleration on the slope of the free surface of water in the tank. Let the acceleration vary from 0 m/s2 to 15 m/s2 in increments of 1 m/s2. Tabulate and plot your results.
3–162 The density of a floating body can be determined by tying weights to the body until both the body and the weights are completely submerged, and then weighing them sepa rately in air. Consider a wood log that weighs 1400 N in air. If it takes 34 kg of lead (𝜌 = 11,300 kg/m3) to completely sink the log and the lead in water, determine the average den sity of the log. Answer: 822 kg/m3 3–163 A raft is made using a number of logs with 25 cm diameter and 2 m long, as shown in the figure. It is desired that a maximum 90 percent volume of each log will be sub merged when carrying two boys weighing 400 N each. Deter mine the minimum number of logs that must be used. The specific gravity of the wood is 0.75.
FIGURE P3–163 3–164 A prismatic timber is at equilibrium in a liquid, as shown in the figure. What is the specific gravity of the liquid?
cen96537_ch03_077-136.indd 133 14/01/17 2:24 pm 134 pressure and fluid statics 50° 3 m 1.8 m SG = ?
Cross section 10 cm by 2 cm FIGURE P3–164 3–165 The cylindrical tank containing water accelerates upward with az = 2 m/s2 while it rotates about its vertical axis by n . = 100 rpm. Determine the pressure at point A.
az 휔 1.2 m R = 0.3 m A FIGURE P3–165 3–166 A 30-cm-diameter, 10-cm-high vertical cylindrical vessel equipped with a vertical tube at the edge is rotated about its vertical axis at a constant angular velocity of 15 rad/s. If the water rise in the tube is 30 cm, determine the net vertical pressure force acting on the vessel.
A Water ω = 15 rad/s R = 15 cm Open end 30 cm 10 cm FIGURE P3–166 3–167 The 280-kg, 6-m-wide rectangular gate shown in Fig. P3–167 is hinged at B and leans against the floor at A making an angle of 45° with the horizontal. The gate is to be opened from its lower edge by applying a normal force at its center. Determine the minimum force F required to open the water gate. Answer: 626 kN 45° B A 0.5 m 3 m Water F FIGURE P3–167 3–168 Repeat Prob. 3–167 for a water height of 0.8 m above the hinge at B.
3–169 Determine the vertical force applied by water on the container.
FIGURE P3–169 D = 40 cm d = 15 cm 50 cm 60 cm 3–170 Knowing that the vertical component of force acting on a curved surface is given by dFV = PdAx, show that the net vertical force on a sphere submerged at a depth h in a liquid is equal to the buoyancy force on the same sphere.
h R FIGURE P3–170 cen96537_ch03_077-136.indd 134 14/01/17 2:24 pm 135 CHAPTER 3 3–171 In order to keep the cone-shaped plug closed as shown in the figure, what would be the maximum air pressure on the top of water? The tanks have the same width of b = 2 m.
FIGURE P3–171 Air SG = 1.6 Water 3 m 1 m D = 30 cm P = ?
2.4 m 60° Fundamentals of Engineering (FE) Exam Problems 3–172 The gage pressure in a pipe is measured by a manometer containing mercury (𝜌 = 13,600 kg/m3). The top of the mercury is open to the atmosphere and the atmospheric pressure is 100 kPa. If the mercury column height is 24 cm, the gage pressure in the pipe is (a) 32 kPa (b) 24 kPa (c) 76 kPa (d ) 124 kPa (e) 68 kPa 3–173 Which is of the highest value?
(a) 1 bar (b) 105 N/m2 (c) 1 atm (d) 100 kPa (e) None of these 3–174 The pressure in seawater where a submarine is sail ing is measured to be 1300 kPa. The submarine is in a water depth of (Take the density of water to be 1000 kg/m3.) (a) 130 m (b) 133 m (c) 0.133 m (d) 122 m (e) 0.122 m 3–175 The atmospheric pressure in a location is measured by a mercury (𝜌 = 13,600 kg/m3) barometer. If the height of the mercury column is 740 mm, the atmospheric pressure at that location is (a) 88.5 kPa (b) 93.9 kPa (c) 96.2 kPa (d ) 98.7 kPa (e) 101 kPa 3–176 A manometer is used to measure the pressure of a gas in a tank. The manometer fluid is water (𝜌 = 1000 kg/m3) and the manometer column height is 1.8 m. If the local atmo spheric pressure is 100 kPa, the absolute pressure within the tank is (a) 17,760 kPa (b) 100 kPa (c) 180 kPa (d ) 101 kPa (e) 118 kPa 3-177 Consider a hydraulic car jack with a piston area ratio of 50. A person can lift a 1000-kg car by applying a force of (a) 100 kgf (b) 10 kgf (c) 50 kgf (d) 20 kgf (e) 196 kgf 3–178 Consider the vertical rectangular wall of a water tank with a width of 5 m and a height of 8 m. The other side of the wall is open to the atmosphere. The resultant hydrostatic force on this wall is (a) 1570 kN (b) 2380 kN (c) 2505 kN (d ) 1410 kN (e) 404 kN 3–179 A vertical rectangular wall with a width of 20 m and a height of 12 m is holding a 7-m-deep water body. The resultant hydrostatic force acting on this wall is (a) 1370 kN (b) 4807 kN (c) 8240 kN (d ) 9740 kN (e) 11,670 kN 3–180 A vertical rectangular plate with a width of 16 m and a height of 12 m is located 4 m below a water surface. The resultant hydrostatic force on this plate is (a) 2555 kN (b) 3770 kN (c) 11,300 kN (d) 15,070 kN (e) 18,835 kN 3–181 A rectangular plate with a width of 16 m and a height of 12 m is located 4 m below a water surface. The plate is tilted and makes a 35° angle with the horizontal. The resultant hydrostatic force acting on the top surface of this plate is (a) 10,800 kN (b) 9745 kN (c) 8470 kN (d ) 6400 kN (e) 5190 kN 3–182 A vertical rectangular plate with a width of 16 m and a height of 12 m is located 4 m below a water surface. The line of action yp for the resultant hydrostatic force on this plate is (Neglect atmospheric pressure.) (a) 4 m (b) 5.3 m (c) 8 m (d) 11.2 m (e) 12 m 3–183 A 2-m-long and 3-m-wide horizontal rectangular plate is submerged in water. The distance of the top surface from the free surface is 5 m. The atmospheric pressure is 95 kPa. Considering atmospheric pressure, the hydrostatic force acting on the top surface of this plate is (a) 307 kN (b) 688 kN (c) 747 kN (d ) 864 kN (e) 2950 kN 3–184 Consider a 5-m-diameter spherical gate holding a body of water whose height is equal to the diameter of the gate. Atmospheric pressure acts on both sides of the gate. The horizontal component of the hydrostatic force acting on this curved surface is (a) 460 kN (b) 482 kN (c) 512 kN (d) 536 kN (e) 561 kN 3–185 Consider a 6-m-diameter spherical gate holding a body of water whose height is equal to the diameter of the gate. Atmospheric pressure acts on both sides of the gate. The vertical component of the hydrostatic force acting on this curved surface is (a) 89 kN (b) 270 kN (c) 327 kN (d ) 416 kN (e) 505 kN 3–186 A 0.75-cm-diameter spherical object is completely submerged in water. The buoyant force acting on this object is (a) 13,000 N (b) 9835 N (c) 5460 N (d ) 2167 N (e) 1267 N cen96537_ch03_077-136.indd 135 14/01/17 2:24 pm 136 pressure and fluid statics 3–187 A 3-kg object with a density of 7500 kg/m3 is placed in water. The weight of this object in water is (a) 29.4 N (b) 25.5 N (c) 14.7 N (d ) 30 N (e) 3 N 3–188 A 9-m-diameter hot air balloon is neither rising nor falling. The density of atmospheric air is 1.3 kg/m3. The total mass of the balloon including the people on board is (a) 496 kg (b) 458 kg (c) 430 kg (d ) 401 kg (e) 383 kg 3–189 A 10-kg object with a density of 900 kg/m3 is placed in a fluid with a density of 1100 kg/m3. The fraction of the volume of the object submerged in water is (a) 0.637 (b) 0.716 (c) 0.818 (d ) 0.90 (e) 1 3–190 Consider a cubical water tank with a side length of 3 m. The tank is half filled with water, and is open to the atmosphere. Now, a truck carrying this tank is accelerated at a rate of 5 m/s2. The maximum vertical rise of free surface of the water is (a) 1.5 m (b) 1.03 m (c) 1.34 m (d) 0.681 m (e) 0.765 m 3–191 A 20-cm-diameter, 40-cm-high vertical cylindrical container is partially filled with 25-cm-high water. Now the cylinder is rotated at a constant speed of 15 rad/s. The height of water at the edge of the cylinder is (a) 40 cm (b) 35.2 cm (c) 30.7 cm (d) 25 cm (e) 38.8 cm 3–192 A 30-cm-diameter, 40-cm-high vertical cylindrical container is partially filled with 25-cm-high water. Now the cylinder is rotated at a constant speed of 15 rad/s. The height of water at the center of the cylinder is (a) 28.0 cm (b) 24.2 cm (c) 20.5 cm (d) 16.5 cm (e) 12.1 cm Design and Essay Problems 3–193 Shoes are to be designed to enable people of up to 80 kg to walk on freshwater or seawater. The shoes are to be made of blown plastic in the shape of a sphere, a (American) football, or a loaf of French bread. Determine the equivalent diameter of each shoe and comment on the proposed shapes from the stability point of view. What is your assessment of the marketability of these shoes?
3–194 The volume of a rock is to be determined without using any volume measurement devices. Explain how you would do this with a waterproof spring scale.
3–195 There are the legends as C1 or Cl1, K1 or Kl1 on the gage scale of some Bourdon type metal manometers. Dis cuss what these legends mean.
3–196 Compare free vortex with forced vortex according to velocity and static pressure distribution. Think motion into hole of water in a bathtub. Discuss the reasons of for mation of free vortex motions rotating in different direc tions in north and south hemisphere in the world while no rotation motion is seen at equatorial plane. Investigate similar physical phenomena in universe.
3–197 The density of stainless steel is about 8000 kg/m3 (eight times denser than water), but a razor blade can float on water, even with some added weights. The water is at 20oC. The blade shown in the photograph is 4.3 cm long and 2.2 cm wide. For simplicity, the center cut-out area of the razor blade has been taped so that only the outer edges of the blade contribute to surface tension effects. Because the razor blade has sharp corners, the contact angle is not rele vant. Rather, the limiting case is when the water contacts the blade vertically as sketched (effective contact angle along the edge of the blade is 180°). (a) Considering surface ten sion alone, estimate (in grams) how much total mass (razor blade + weights placed on top of it) can be supported. (b) Refine your analysis by considering that the razor blade pushes the water down, and thus hydrostatic pressure effects are also present. Hint: You will also need to know that due to the curvature of the meniscus, the maximum possible depth is h = √ 2𝜎s ρg .
Added weights Pabove = Patm Pbelow h 휙 = 0 FIGURE P3–197 (Bottom) Photo by John M. Cimbala.
cen96537_ch03_077-136.indd 136 14/01/17 2:24 pm 4 CHAPTER 137 OBJECTIVES When you finish reading this chapter, you should be able to ■ ■ Understand the role of the material derivative in transforming between Lagrangian and Eulerian descriptions ■ ■ Distinguish between various types of flow visualizations and methods of plotting the characteristics of a fluid flow ■ ■ Appreciate the many ways that fluids move and deform ■ ■ Distinguish between rotational and irrotational regions of flow based on the flow property vorticity ■ ■ Understand the usefulness of the Reynolds transport theorem F LUI D KI N E MAT I CS F luid kinematics deals with describing the motion of fluids with out necessarily considering the forces and moments that cause the motion. In this chapter, we introduce several kinematic concepts related to flowing fluids. We discuss the material derivative and its role in transforming the conservation equations from the Lagrangian descrip tion of fluid flow (following a fluid particle) to the Eulerian description of fluid flow (pertaining to a flow field). We then discuss various ways to visualize flow fields—streamlines, streaklines, pathlines, timelines, optical methods schlieren and shadowgraph, and surface methods; and we describe three ways to plot flow data—profile plots, vector plots, and contour plots. We explain the four fundamental kinematic properties of fluid motion and deformation—rate of translation, rate of rotation, linear strain rate, and shear strain rate. The concepts of vorticity, rotationality, and irrotationality in fluid flows are then discussed. Finally, we discuss the Reynolds transport theorem (RTT ), emphasizing its role in transforming the equations of motion from those following a system to those pertaining to fluid flow into and out of a control volume. The analogy between material derivative for infinitesi mal fluid elements and RTT for finite control volumes is explained.
Satellite image of a hurricane near the Florida coast; water droplets move with the air, enabling us to visualize the counterclockwise swirling motion. However, the major portion of the hurricane is actually irrotational, while only the core (the eye of the storm) is rotational.
© StockTrek/Getty Images RF cen96537_ch04_137-188.indd 137 14/01/17 2:30 pm 138 Fluid Kinematics 4–1 ■ LAGRANGIAN AND EULERIAN DESCRIPTIONS The subject called kinematics concerns the study of motion. In fluid dynam ics, fluid kinematics is the study of how fluids flow and how to describe fluid motion. From a fundamental point of view, there are two distinct ways to describe motion. The first and most familiar method is the one you learned in high school physics—to follow the path of individual objects. For example, we have all seen physics experiments in which a ball on a pool table or a puck on an air hockey table collides with another ball or puck or with the wall (Fig. 4–1). Newton’s laws are used to describe the motion of such objects, and we can accurately predict where they go and how momentum and kinetic energy are exchanged from one object to another. The kinematics of such experiments involves keeping track of the position vector of each object, x → A, x → B, . . . , and the velocity vector of each object, V › A, V › B , . . . , as functions of time (Fig. 4–2). When this method is applied to a flowing fluid, we call it the Lagrangian description of fluid motion after the Italian mathematician Joseph Louis Lagrange (1736–1813). Lagrangian analysis is analogous to the (closed) system analysis that you learned in thermodynamics; namely, we fol low a mass of fixed identity. The Lagrangian description requires us to track the position and velocity of each individual fluid parcel, which we refer to as a fluid particle, and take to be a parcel of fixed identity. As you can imagine, this method of describing motion is much more dif ficult for fluids than for billiard balls! First of all we cannot easily define and identify fluid particles as they move around. Secondly, a fluid is a continuum (from a macroscopic point of view), so interactions between fluid particles are not as easy to describe as are interactions between distinct objects like billiard balls or air hockey pucks. Furthermore, the fluid par ticles continually deform as they move in the flow. From a microscopic point of view, a fluid is composed of billions of molecules that are continuously banging into one another, somewhat like billiard balls; but the task of following even a subset of these molecules is quite difficult, even for our fastest and largest computers. Nevertheless, there are many practical applications of the Lagrangian description, such as the tracking of passive scalars in a flow to model contaminant transport, rar efied gas dynamics calculations concerning reentry of a spaceship into the earth’s atmosphere, and the development of flow visualization and measure ment systems based on particle tracking (as discussed in Section 4–2). A more common method of describing fluid flow is the Eulerian description of fluid motion, named after the Swiss mathematician Leonhard Euler (1707– 1783). In the Eulerian description of fluid flow, a finite volume called a flow domain or control volume is defined, through which fluid flows in and out. Instead of tracking individual fluid particles, we define field variables, func tions of space and time, within the control volume. The field variable at a particular location at a particular time is the value of the variable for which ever fluid particle happens to occupy that location at that time. For example, the pressure field is a scalar field variable; for general unsteady three-dimensional fluid flow in Cartesian coordinates, Pressure field: P = P(x, y, z, t) (4–1) We define the velocity field as a vector field variable in similar fashion, Velocity field: V › = V › (x, y, z, t) (4–2) FIGURE 4–1 With a small number of objects, such as billiard balls on a pool table, individual objects can be tracked.
VB VC xA xB xC A B C VA FIGURE 4–2 In the Lagrangian description, we must keep track of the position and velocity of individual particles.
cen96537_ch04_137-188.indd 138 14/01/17 2:30 pm 139 CHAPTER 4 Likewise, the acceleration field is also a vector field variable, Acceleration field: a ›= a ›(x, y, z, t) (4–3) Collectively, these (and other) field variables define the flow field. The veloc ity field of Eq. 4–2 is expanded in Cartesian coordinates (x, y, z), (i → , j → , k → ) as V ›= (u, 𝜐 , w) = u(x, y, z, t) i ›+ 𝜐 (x, y, z, t) j ›+ w(x, y, z, t)k › (4–4) A similar expansion can be performed for the acceleration field of Eq. 4–3. In the Eulerian description, all such field variables are defined at any location (x, y, z) in the control volume and at any instant in time t (Fig. 4–3). In the Eulerian description we don’t really care what happens to individual fluid particles; rather we are concerned with the pressure, velocity, acceleration, etc., of whichever fluid particle happens to be at the location of interest at the time of interest. The difference between these two descriptions is made clearer by imagining a person standing beside a river, measuring its properties. In the Lagrangian approach, he throws in a probe that moves downstream with the water. In the Eulerian approach, he anchors the probe at a fixed location in the water. While there are many occasions in which the Lagrangian description is use ful, the Eulerian description is often more convenient for fluid mechanics appli cations. Furthermore, experimental measurements are generally more suited to the Eulerian description. In a wind tunnel, for example, velocity or pressure probes are usually placed at a fixed location in the flow, measuring V ›(x, y, z, t) or P(x, y, z, t). However, whereas the equations of motion in the Lagrangian description following individual fluid particles are well known (e.g., Newton’s second law), the equations of motion of fluid flow are not so readily apparent in the Eulerian description and must be carefully derived. We do this for control volume (integral) analysis via the Reynolds transport theorem at the end of this chapter. We derive the differential equations of motion in Chap. 9.
FIGURE 4–3 (a) In the Eulerian description, we define field variables, such as the pressure field and the velocity field, at any location and instant in time. (b) For example, the air speed probe mounted under the wing of an airplane measures the air speed at that location.
(Bottom) Photo by John M. Cimbala.
Control volume V(x, y, z, t) P(x, y, z, t) (x, y, z) (a) (b) EXAMPLE 4–1 A Steady Two-Dimensional Velocity Field A steady, incompressible, two-dimensional velocity field is given by V → = (u, 𝜐 ) = (0.5 + 0.8x) i → + (1.5 −0.8y) j → (1) where the x- and y-coordinates are in meters and the magnitude of velocity is in m/s. A stagnation point is defined as a point in the flow field where the veloc ity is zero. (a) Determine if there are any stagnation points in this flow field and, if so, where? (b) Sketch velocity vectors at several locations in the domain between x = −2 m to 2 m and y = 0 m to 5 m; qualitatively describe the flow field.
SOLUTION For the given velocity field, the location(s) of stagnation point(s) are to be determined. Several velocity vectors are to be sketched and the velocity field is to be described.
Assumptions 1 The flow is steady and incompressible. 2 The flow is two-dimensional, implying no z-component of velocity and no variation of u or 𝜐 with z.
Analysis (a) Since V › is a vector, all its components must equal zero in order for V › itself to be zero. Using Eq. 4–4 and setting Eq. 1 equal to zero, Stagnation point: u = 0.5 + 0.8x = 0 → x = −0.625 m 𝜐 = 1.5 −0.8y = 0 → y = 1.875 m Yes. There is one stagnation point located at x = −0.625 m, y = 1.875 m.
cen96537_ch04_137-188.indd 139 14/01/17 2:30 pm 140 Fluid Kinematics Acceleration Field As you should recall from your study of thermodynamics, the fundamen tal conservation laws (such as conservation of mass and the first law of thermodynamics) are expressed for a system of fixed identity (also called a closed system). In cases where analysis of a control volume (also called an open system) is more convenient than system analysis, it is necessary to rewrite these fundamental laws into forms applicable to the control volume. The same principle applies here. In fact, there is a direct analogy between systems versus control volumes in thermodynamics and Lagrangian versus Eulerian descriptions in fluid dynamics. The equations of motion for fluid flow (such as Newton’s second law) are written for a fluid particle, which we also call a material particle. If we were to follow a particular fluid par ticle as it moves around in the flow, we would be employing the Lagrangian description, and the equations of motion would be directly applicable. For example, we would define the particle’s location in space in terms of a material position vector (xparticle(t), yparticle(t), zparticle(t)). However, some mathematical manipulation is then necessary to convert the equations of motion into forms applicable to the Eulerian description. Consider, for example, Newton’s second law applied to our fluid particle, Newton’s second law: F → particle = mparticlea › particle (4–5) where F › particle is the net force acting on the fluid particle, mparticle is its mass, and a › particle is its acceleration (Fig. 4–6). By definition, the acceleration of the fluid particle is the time derivative of the particle’s velocity, Acceleration of a fluid particle: a › particle = dV → particle dt (4–6) However, at any instant in time t, the velocity of the particle is the same as the local value of the velocity field at the location (xparticle(t), yparticle(t), zparticle(t)) of the particle, since the fluid particle moves with the fluid by Scale: 5 4 3 2 y 1 0 –1 –3 –2 –1 0 x 1 2 3 10 m/s FIGURE 4–4 Velocity vectors (blue arrows) for the velocity field of Example 4–1. The scale is shown by the top arrow, and the solid black curves represent the approximate shapes of some streamlines, based on the calculated velocity vectors. The stagnation point is indicated by the blue circle. The shaded region represents a portion of the flow field that can approximate flow into an inlet (Fig. 4–5).
Region in which the velocity field is modeled Streamlines FIGURE 4–5 Flow field near the bell mouth inlet of a hydroelectric dam; a portion of the velocity field of Example 4–1 may be used as a first-order approximation of this physical flow field. The shaded region corresponds to that of Fig. 4–4.
(b) The x- and y-components of velocity are calculated from Eq. 1 for several (x, y) locations in the specified range. For example, at the point (x = 2 m, y = 3 m), u = 2.10 m/s and 𝜐 = −0.900 m/s. The magnitude of velocity (the speed) at that point is 2.28 m/s. At this and at an array of other locations, the velocity vector is constructed from its two components, the results of which are shown in Fig. 4–4. The flow can be described as stagnation point flow in which flow enters from the top and bottom and spreads out to the right and left about a horizontal line of symmetry at y = 1.875 m. The stagnation point of part (a) is indicated by the blue circle in Fig. 4–4. If we look only at the shaded portion of Fig. 4–4, this flow field models a converging, accelerating flow from the left to the right. Such a flow might be encountered, for example, near the submerged bell mouth inlet of a hydroelectric dam (Fig. 4–5). The useful portion of the given velocity field may be thought of as a first-order approximation of the shaded portion of the physical flow field of Fig. 4–5.
Discussion It can be verified from the material in Chap. 9 that this flow field is physically valid because it satisfies the differential equation for conservation of mass.
cen96537_ch04_137-188.indd 140 14/01/17 2:30 pm 141 CHAPTER 4 definition. In other words, V › particle(t) ≡ V ›(xparticle(t), yparticle(t), zparticle(t), t). To take the time derivative in Eq. 4–6, we must therefore use the chain rule, since the dependent variable (V ›) is a function of four independent variables (xparticle, yparticle, zparticle, and t), a › particle = dV › particle dt = dV › dt = dV › (xparticle, yparticle, zparticle, t) dt = ∂V › ∂t dt dt + ∂V › ∂xparticle dxparticle dt + ∂V › ∂yparticle dyparticle dt + ∂V › ∂zparticle dzparticle dt (4–7) In Eq. 4–7, ∂ is the partial derivative operator and d is the total derivative operator. Consider the second term on the right-hand side of Eq. 4–7. Since the acceleration is defined as that following a fluid particle (Lagrangian description), the rate of change of the particle’s x-position with respect to time is dxparticle/dt = u (Fig. 4–7), where u is the x-component of the veloc ity vector defined by Eq. 4–4. Similarly, dyparticle/dt = 𝜐 and dzparticle/dt = w. Furthermore, at any instant in time under consideration, the material position vector (xparticle, yparticle, zparticle) of the fluid particle in the Lagrangian frame is equal to the position vector (x, y, z) in the Eulerian frame. Equation 4–7 thus becomes a › particle(x, y, z, t) = dV › dt = ∂V › ∂t + u ∂V › ∂x + 𝜐 ∂V › ∂y + w ∂V › ∂z (4–8) where we have also used the (obvious) fact that dt/dt = 1. Finally, at any instant in time t, the acceleration field of Eq. 4–3 must equal the accelera tion of the fluid particle that happens to occupy the location (x, y, z) at that time t. Why? Because the fluid particle is by definition accelerating with the fluid flow. Hence, we may replace a › particle with a ›(x, y, z, t) in Eqs. 4–7 and 4–8 to transform from the Lagrangian to the Eulerian frame of refer ence. In vector form, Eq. 4–8 is written as Acceleration of a fluid particle expressed as a field variable: a › (x, y, z, t) = dV › dt = ∂V › ∂t + (V › · ∇ ›)V › (4–9) where ∇ › is the gradient operator or del operator, a vector operator that is defined in Cartesian coordinates as Gradient or del operator: ∇ ›= ( ∂ ∂x, ∂ ∂y, ∂ ∂z) = i ›∂ ∂x + j › ∂ ∂y + k › ∂ ∂z (4–10) In Cartesian coordinates then, the components of the acceleration vector are ax = ∂u ∂t + u ∂u ∂x + 𝜐 ∂u ∂y + w ∂u ∂z Cartesian coordinates: ay = ∂𝜐 ∂t + u ∂𝜐 ∂x + 𝜐 ∂𝜐 ∂y + w ∂𝜐 ∂z (4–11) az = ∂w ∂t + u ∂w ∂x + 𝜐 ∂w ∂y + w ∂w ∂z The first term on the right-hand side of Eq. 4–9, ∂V ›/∂t, is called the local acceleration and is nonzero only for unsteady flows. The second term, (V ›·∇ ›)V ›, is called the advective acceleration (sometimes the convective Vparticle V Fparticle aparticle (xparticle, yparticle, zparticle) Fluid particle at time t Fluid particle at time t + dt mparticle FIGURE 4–6 Newton’s second law applied to a fluid particle; the acceleration vector (purple arrow) is in the same direction as the force vector (green arrow), but the velocity vector (blue arrow) may act in a different direction.
Fluid particle at time t Fluid particle at time t + dt (xparticle, yparticle) (xparticle + dxparticle, yparticle + dyparticle) dyparticle dxparticle FIGURE 4–7 When following a fluid particle, the x-component of velocity, u, is defined as dxparticle/dt. Similarly, 𝜐 = dyparticle/dt and w = dzparticle/dt. Movement is shown here only in two dimensions for simplicity.
cen96537_ch04_137-188.indd 141 14/01/17 2:30 pm 142 Fluid Kinematics acceleration); this term can be nonzero even for steady flows. It accounts for the effect of the fluid particle moving (advecting or convecting) to a new location in the flow, where the velocity field is different. For example, consider steady flow of water through a garden hose nozzle (Fig. 4–8). We define steady in the Eulerian frame of reference to be when properties at any point in the flow field do not change with respect to time. Since the velocity at the exit of the nozzle is larger than that at the nozzle entrance, fluid particles clearly accelerate, even though the flow is steady. The accel eration is nonzero because of the advective acceleration terms in Eq. 4–9. Note that while the flow is steady from the point of view of a fixed observer in the Eulerian reference frame, it is not steady from the Lagrangian refer ence frame moving with a fluid particle that enters the nozzle and acceler ates as it passes through the nozzle.
EXAMPLE 4–2 Acceleration of a Fluid Particle through a Nozzle Nadeen is washing her car, using a nozzle similar to the one sketched in Fig. 4–8. The nozzle is 3.90 in (0.325 ft) long, with an inlet diameter of 0.420 in (0.0350 ft) and an outlet diameter of 0.182 in (see Fig. 4–9). The volume flow rate through the garden hose (and through the nozzle) is V . = 0.841 gal/min (0.00187 ft3/s), and the flow is steady. Estimate the magnitude of the acceleration of a fluid particle moving down the centerline of the nozzle.
SOLUTION The acceleration following a fluid particle down the center of a nozzle is to be estimated.
Assumptions 1 The flow is steady and incompressible. 2 The x-direction is taken along the centerline of the nozzle. 3 By symmetry, 𝜐 = w = 0 along the centerline, but u increases through the nozzle.
Analysis The flow is steady, so you may be tempted to say that the acceleration is zero. However, even though the local acceleration ∂V ›/∂t is identically zero for this steady flow field, the advective acceleration (V → ·∇ → )V → is not zero. We first cal culate the average x-component of velocity at the inlet and outlet of the nozzle by dividing volume flow rate by cross-sectional area: Inlet speed: uinlet ≅ V .
Ainlet = 4V .
𝜋 D2 inlet = 4(0.00187 ft3/s) 𝜋 (0.0350 ft)2 = 1.95 ft/s Similarly, the average outlet speed is uoutlet = 10.4 ft/s. We now calculate the accel eration in two ways, with equivalent results. First, a simple average value of accel eration in the x-direction is calculated based on the change in speed divided by an estimate of the residence time of a fluid particle in the nozzle, Δt = Δx/uavg (Fig. 4–10). By the fundamental definition of acceleration as the rate of change of veloc ity, Method A: ax ≅Δu Δt = uoutlet −uinlet Δx/uavg = uoutlet −uinlet 2 Δx/(uoutlet + uinlet) = uoutlet 2 −uinlet 2 2 Δx The second method uses the equation for acceleration field components in Carte sian coordinates, Eq. 4–11, FIGURE 4–8 Flow of water through the nozzle of a garden hose illustrates that fluid par ticles may accelerate, even in a steady flow. In this example, the exit speed of the water is much higher than the water speed in the hose, implying that fluid particles have accelerated even though the flow is steady.
Doutlet Dinlet uoutlet x Δx uinlet FIGURE 4–9 Flow of water through the nozzle of Example 4–2.
cen96537_ch04_137-188.indd 142 14/01/17 2:30 pm 143 CHAPTER 4 Material Derivative The total derivative operator d/dt in Eq. 4–9 is given a special name, the material derivative; it is assigned a special notation, D/Dt, in order to emphasize that it is formed by following a fluid particle as it moves through the flow field (Fig. 4–12). Other names for the material derivative include total, particle, Lagrangian, Eulerian, and substantial derivative.
Material derivative: D Dt = d dt = ∂ ∂t + (V › ·∇ › ) (4–12) When we apply the material derivative of Eq. 4–12 to the velocity field, the result is the acceleration field as expressed by Eq. 4–9, which is thus some times called the material acceleration, Material acceleration: a › (x, y, z, t) = DV › Dt = dV › dt = ∂V › ∂t + (V › ·∇ › )V › (4–13) Equation 4–12 can also be applied to other fluid properties besides velocity, both scalars and vectors. For example, the material derivative of pressure is written as Material derivative of pressure: DP Dt = dP dt = ∂P ∂t + (V › ·∇ › )P (4–14) Fluid particle at time t + Δt Δx Fluid particle at time t x FIGURE 4–10 Residence time Δt is defined as the time it takes for a fluid particle to travel through the nozzle from inlet to outlet (distance Δx).
Δq q Δx x ≈ dq dx Δq Δx FIGURE 4–11 A first-order finite difference approximation for derivative dq/dx is simply the change in dependent variable (q) divided by the change in independent variable (x).
t t + Δt t + 2Δt t + 3Δt FIGURE 4–12 The material derivative D/Dt is defined by following a fluid particle as it moves throughout the flow field. In this illustration, the fluid particle is accelerating to the right as it moves up and to the right.
Method B: ax = ∂u ∂t + u ∂u ∂x + 𝜐 ∂u ∂y + w ∂u ∂z ≅uavg Δu Δx Steady v = 0 along centerline w = 0 along centerline Here we see that only one advective term is nonzero. We approximate the average speed through the nozzle as the average of the inlet and outlet speeds, and we use a first-order finite difference approximation (Fig. 4–11) for the average value of derivative ∂u/∂x through the centerline of the nozzle: ax ≅ uoutlet + uinlet 2 uoutlet −uinlet Δx = uoutlet 2 −uinlet 2 2 Δx The result of method B is identical to that of method A. Substitution of the given values yields Axial acceleration: ax ≅ u2 outlet −u2 inlet 2 Δx = (10.4 ft/s)2 −(1.95 ft/s)2 2(0.325 ft) = 160 ft/s2 Discussion Fluid particles are accelerated through the nozzle at nearly five times the acceleration of gravity (almost five g’s)! This simple example clearly illustrates that the acceleration of a fluid particle can be nonzero, even in steady flow. Note that the acceleration is actually a point function, whereas we have estimated a sim ple average acceleration through the entire nozzle.
0 0 0 ⏟ ⏟ ⏟ ← ← ← cen96537_ch04_137-188.indd 143 14/01/17 2:30 pm 144 Fluid Kinematics Equation 4–14 represents the time rate of change of pressure follow ing a fluid particle as it moves through the flow and contains both local (unsteady) and advective components (Fig. 4–13).
Local Material derivative D Dt V · Δ Advective = + д дt FIGURE 4–13 The material derivative D/Dt is composed of a local or unsteady part and a convective or advective part.
Scale: 5 4 3 2 y 1 0 –1 –3 –2 –1 0 x 1 2 3 10 m/s2 FIGURE 4–14 Acceleration vectors (purple arrows) for the velocity field of Examples 4–1 and 4–3. The scale is shown by the top arrow, and the solid black curves represent the approximate shapes of some streamlines, based on the calculated velocity vectors (see Fig. 4–4). The stagnation point is indicated by the red circle.
EXAMPLE 4–3 Material Acceleration of a Steady Velocity Field Consider the steady, incompressible, two-dimensional velocity field of Example 4–1. (a) Calculate the material acceleration at the point (x = 2 m, y = 3 m). (b) Sketch the material acceleration vectors at the same array of x- and y-values as in Example 4–1.
SOLUTION For the given velocity field, the material acceleration vector is to be calculated at a particular point and plotted at an array of locations in the flow field.
Assumptions 1 The flow is steady and incompressible. 2 The flow is two-dimensional, implying no z-component of velocity and no variation of u or 𝜐 with z.
Analysis (a) Using the velocity field of Eq. 1 of Example 4–1 and the equation for material acceleration components in Cartesian coordinates (Eq. 4–11), we write expressions for the two nonzero components of the acceleration vector: ax = ∂u ∂t + u ∂u ∂x + 𝜐 ∂u ∂y + w ∂u ∂z = 0 + (0.5 + 0.8x)(0.8) + (1.5 −0.8y)(0) + 0 = (0.4 + 0.64x) m/s2 and ay = ∂𝜐 ∂t + u ∂𝜐 ∂x + 𝜐 ∂𝜐 ∂y + w ∂𝜐 ∂z = 0 + (0.5 + 0.8x)(0) + (1.5 −0.8y)(−0.8) + 0 = (−1.2 + 0.64y) m/s2 At the point (x = 2 m, y = 3 m), ax = 1.68 m/s2 and ay = 0.720 m/s2.
(b) The equations in part (a) are applied to an array of x- and y-values in the flow domain within the given limits, and the acceleration vectors are plotted in Fig. 4–14.
Discussion The acceleration field is nonzero, even though the flow is steady. Above the stagnation point (above y = 1.875 m), the acceleration vectors plotted in Fig. 4–14 point upward, increasing in magnitude away from the stagnation point. To the right of the stagnation point (to the right of x = −0.625 m), the acceleration vectors point to the right, again increasing in magnitude away from the stagnation point. This agrees qualitatively with the velocity vectors of Fig. 4–4 and the stream lines sketched in Fig. 4–14; namely, in the upper-right portion of the flow field, fluid particles are accelerated in the upper-right direction and therefore veer in the counterclockwise direction due to centripetal acceleration toward the upper right. The flow below y = 1.875 m is a mirror image of the flow above this symmetry line, and the flow to the left of x = −0.625 m is a mirror image of the flow to the right of this symmetry line.
cen96537_ch04_137-188.indd 144 14/01/17 2:30 pm 145 CHAPTER 4 4–2 ■ FLOW PATTERNS AND FLOW VISUALIZATION While quantitative study of fluid dynamics requires advanced mathemat ics, much can be learned from flow visualization—the visual exami nation of flow field features. Flow visualization is useful not only in physical experiments (Fig. 4–15), but in numerical solutions as well [computational fluid dynamics (CFD)]. In fact, the very first thing an engineer using CFD does after obtaining a numerical solution is simu late some form of flow visualization, so that he or she can see the “whole picture” rather than merely a list of numbers and quantitative data. Why? Because the human mind is designed to rapidly process an incredible amount of visual information; as they say, a picture is worth a thousand words. There are many types of flow patterns that can be visualized, both physically (experimentally) and/or computationally.
Streamlines and Streamtubes A streamline is a curve that is everywhere tangent to the instantaneous local velocity vector.
Streamlines are useful as indicators of the instantaneous direction of fluid motion throughout the flow field. For example, regions of recirculating flow and separation of a fluid off of a solid wall are easily identified by the streamline pattern. Streamlines cannot be directly observed experimentally except in steady flow fields, in which they are coincident with pathlines and streaklines, to be discussed next. Mathematically, however, we can write a simple expression for a streamline based on its definition. Consider an infinitesimal arc length d r › = dx i › + dy j › + dzk → along a streamline; dr › must be parallel to the local velocity vector V › = u i › + 𝜐 j › + wk → by definition of the streamline. By simple geometric arguments using simi lar triangles, we know that the components of dr → must be proportional to those of V › (Fig. 4–16). Hence, Equation for a streamline: dr V = dx u = dy 𝜐 = dz w (4–15) where dr is the magnitude of dr › and V is the speed, the magnitude of veloc ity vector V ›. Equation 4–15 is illustrated in two dimensions for simplicity in Fig. 4–16. For a known velocity field, we integrate Eq. 4–15 to obtain equations for the streamlines. In two dimensions, (x, y), (u, 𝜐), the following differential equation is obtained: Streamline in the xy-plane: ( dy dx) along a streamline = 𝜐 u (4–16) In some simple cases, Eq. 4–16 may be solvable analytically; in the general case, it must be solved numerically. In either case, an arbitrary constant of integration appears. Each chosen value of the constant represents a different streamline. The family of curves that satisfy Eq. 4–16 therefore represents streamlines of the flow field.
FIGURE 4–15 Spinning baseball. The late F. N. M. Brown devoted many years to developing and using smoke visualization in wind tunnels at the University of Notre Dame. Here the flow speed is about 77 ft/s and the ball is rotated at 630 rpm. Courtesy of Professor Thomas J. Mueller from the Collection of Professor F.N.M. Brown.
y x Point (x, y) Streamline Point (x + dx, y + dy) dx dy u υ V dr FIGURE 4–16 For two-dimensional flow in the xy-plane, arc length dr › = (dx, dy) along a streamline is everywhere tangent to the local instantaneous velocity vector V › = (u, 𝜐).
cen96537_ch04_137-188.indd 145 14/01/17 2:30 pm 146 Fluid Kinematics A streamtube consists of a bundle of streamlines (Fig. 4–18), much like a communications cable consists of a bundle of fiber-optic cables. Since streamlines are everywhere parallel to the local velocity, fluid cannot cross a streamline by definition. By extension, fluid within a streamtube must remain there and cannot cross the boundary of the streamtube. You must keep in mind that both streamlines and streamtubes are instantaneous quan tities, defined at a particular instant in time according to the velocity field at that instant. In an unsteady flow, the streamline pattern may change sig nificantly with time. Nevertheless, at any instant in time, the mass flow rate passing through any cross-sectional slice of a given streamtube must remain the same. For example, in a converging portion of an incompressible flow field, the diameter of the streamtube must decrease as the velocity increases in order to conserve mass (Fig. 4–19a). Likewise, the streamtube diameter increases in diverging portions of an incompressible flow (Fig. 4–19b).
Pathlines A pathline is the actual path traveled by an individual fluid particle over some time period.
5 4 3 2 y 1 0 –1 0 1 2 3 x 4 5 FIGURE 4–17 Streamlines (solid black curves) for the velocity field of Example 4–4; velocity vectors of Fig. 4–4 (blue arrows) are superimposed for comparison.
Streamlines Streamtube FIGURE 4–18 A streamtube consists of a bundle of individual streamlines.
EXAMPLE 4–4 Streamlines in the xy-Plane—An Analytical Solution For the steady, incompressible, two-dimensional velocity field of Example 4–1, plot several streamlines in the right half of the flow (x > 0) and compare to the velocity vectors plotted in Fig. 4–4.
SOLUTION An analytical expression for streamlines is to be generated and plot ted in the upper-right quadrant.
Assumptions 1 The flow is steady and incompressible. 2 The flow is two-dimensional, implying no z-component of velocity and no variation of u or 𝜐 with z.
Analysis Equation 4–16 is applicable here; thus, along a streamline, dy dx = 𝜐 u = 1.5 −0.8y 0.5 + 0.8x We solve this differential equation by separation of variables: dy 1.5 −0.8y = dx 0.5 + 0.8x → ∫ dy 1.5 −0.8y = ∫ dx 0.5 + 0.8x After some algebra, we solve for y as a function of x along a streamline, y = C 0.8(0.5 + 0.8x) + 1.875 where C is a constant of integration that can be set to various values in order to plot the streamlines. Several streamlines of the given flow field are shown in Fig. 4–17.
Discussion The velocity vectors of Fig. 4–4 are superimposed on the stream lines of Fig. 4–17; the agreement is excellent in the sense that the velocity vectors point everywhere tangent to the streamlines. Note that speed cannot be determined directly from the streamlines alone.
cen96537_ch04_137-188.indd 146 14/01/17 2:30 pm 147 CHAPTER 4 Pathlines are the easiest of the flow patterns to understand. A pathline is a Lagrangian concept in that we simply follow the path of an individual fluid particle as it moves around in the flow field (Fig. 4–20). Thus, a pathline is the same as the fluid particle’s material position vector (xparticle(t), yparticle(t), zparticle(t)), discussed in Section 4–1, traced out over some finite time inter val. In a physical experiment, you can imagine a tracer fluid particle that is marked somehow—either by color or brightness—such that it is easily dis tinguishable from surrounding fluid particles. Now imagine a camera with the shutter open for a certain time period, tstart < t < tend, in which the par ticle’s path is recorded; the resulting curve is called a pathline. An intrigu ing example is shown in Fig. 4–21 for the case of waves moving along the surface of water in a tank. Neutrally buoyant white tracer particles are sus pended in the water, and a time-exposure photograph is taken for one com plete wave period. The result is pathlines that are elliptical in shape, show ing that fluid particles bob up and down and forward and backward, but return to their original position upon completion of one wave period; there is no net forward motion. You may have experienced something similar while bobbing up and down on ocean waves at the beach.
(b) (a) FIGURE 4–19 In an incompressible flow field, a streamtube (a) decreases in diameter as the flow accelerates or converges and (b) increases in diameter as the flow decelerates or diverges.
Fluid particle at t = tstart Fluid particle at t = tend Fluid particle at some intermediate time Pathline FIGURE 4–20 A pathline is formed by following the actual path of a fluid particle.
FIGURE 4–21 Pathlines produced by white tracer particles suspended in water and captured by time-exposure photography; as waves pass horizontally, each particle moves in an elliptical path during one wave period.
Wallet, A & Ruellan, F. 1950, La Houille Blanche 5: 483–489. Used by permission. A modern experimental technique called particle image velocimetry (PIV) utilizes short segments of particle pathlines to measure the velocity field over an entire plane in a flow (Adrian, 1991). (Recent advances also extend the technique to three dimensions.) In PIV, tiny tracer particles are suspended in the fluid, much like in Fig. 4–21. However, the flow is illu minated by two flashes of light (usually a light sheet from a laser as in Fig. 4–22) to produce two bright spots (recorded by a camera) for each mov ing particle. Then, both the magnitude and direction of the velocity vector at each particle location can be inferred, assuming that the tracer particles are small enough that they move with the fluid. Modern digital photogra phy and fast computers have enabled PIV to be performed rapidly enough so that unsteady features of a flow field can also be measured. PIV is dis cussed in more detail in Chap. 8.
cen96537_ch04_137-188.indd 147 14/01/17 2:30 pm 148 Fluid Kinematics Pathlines can also be calculated numerically for a known velocity field. Specifically, the location of the tracer particle is integrated over time from some starting location x › start and starting time tstart to some later time t.
Tracer particle location at time t: x →= x → start + ∫ t tstart V › dt (4–17) When Eq. 4–17 is calculated for t between tstart and tend, a plot of x ›(t) is the pathline of the fluid particle during that time interval, as illustrated in Fig. 4–20. For some simple flow fields, Eq. 4–17 can be integrated analyti cally. For more complex flows, we must perform a numerical integration. If the velocity field is steady, individual fluid particles follow streamlines. Thus, for steady flow, pathlines are identical to streamlines.
Streaklines A streakline is the locus of fluid particles that have passed sequentially through a prescribed point in the flow.
Streaklines are the most common flow pattern generated in a physical experiment. If you insert a small tube into a flow and introduce a continu ous stream of tracer fluid (dye in a water flow or smoke in an airflow), the observed pattern is a streakline. Figure 4–23 shows a tracer being injected into a free-stream flow containing an object, such as a wing. The circles represent individual injected tracer fluid particles, released at a uniform time interval. As the particles are forced out of the way by the object, they accelerate around the shoulder of the object, as indicated by the increased distance between individual tracer particles in that region. The streakline is formed by connecting all the circles into a smooth curve. In physical experi ments in a wind or water tunnel, the smoke or dye is injected continuously, not as individual particles, and the resulting flow pattern is by definition a streakline. In Fig. 4–23, tracer particle 1 was released at an earlier time than tracer particle 2, and so on. The location of an individual tracer par ticle is determined by the surrounding velocity field from the moment of its injection into the flow until the present time. If the flow is unsteady, the surrounding velocity field changes, and we cannot expect the resulting streakline to resemble a streamline or pathline at any given instant in time. However, if the flow is steady, streamlines, pathlines, and streaklines are identical (Fig. 4–24). Streaklines are often confused with streamlines or pathlines. While the three flow patterns are identical in steady flow, they can be quite differ ent in unsteady flow. The main difference is that a streamline represents an instantaneous flow pattern at a given instant in time, while a streakline and a pathline are flow patterns that have some age and thus a time history associated with them. A streakline is an instantaneous snapshot of a time-integrated flow pattern. A pathline, on the other hand, is the time-exposed flow path of an individual particle over some time period. The time-integrative property of streaklines is vividly illustrated in an experiment by Cimbala et al. (1988), reproduced here as Fig. 4–25. The authors used a smoke wire for flow visualization in a wind tunnel. In opera tion, the smoke wire is a thin vertical wire that is coated with mineral oil. The oil breaks up into beads along the length of the wire due to surface V Streakline Object 8 7 6 5 4 3 2 1 Injected fluid particle Dye or smoke FIGURE 4–23 A streakline is formed by continuous introduction of dye or smoke from a point in the flow. Labeled tracer particles (1 through 8) were introduced sequentially.
–0.1 –0.05 0 0.05 z/c 0.88 0.9 0.92 0.94 0.96 0.98 1 1.02 ζ ζmin 0.15 0.1 0.05 0 –0.05 y c FIGURE 4–22 Stereo PIV measurements of the wing tip vortex in the wake of a NACA-66 airfoil at angle of attack. Color contours denote the local vorticity, normalized by the minimum value, as indicated in the color map. Vectors denote fluid mo tion in the plane of measurement. The black line denotes the location of the up stream wing trailling edge. Coordinates are normalized by the airfoil chord, and the origin is the wing root.
Photo by Michael H. Krane, ARL-Penn State.
cen96537_ch04_137-188.indd 148 14/01/17 2:30 pm 149 CHAPTER 4 tension effects. When an electric current heats the wire, each little bead of oil produces a streakline of smoke. In Fig. 4–25a, streaklines are introduced from a smoke wire located just downstream of a circular cylinder of diameter D aligned normal to the plane of view. (When multiple streaklines are intro duced along a line, as in Fig. 4–25, we refer to this as a rake of streaklines.) The Reynolds number of the flow is Re = 𝜌VD/𝜇 = 93. Because of unsteady vortices shed in an alternating pattern from the cylinder, the smoke collects into a clearly defined periodic pattern called a Kármán vortex street. A similar pattern can be seen at much larger scale in the air flow in the wake of an island (Fig. 4–26). From Fig. 4–25a alone, you may think that the shed vortices continue to exist to several hundred diameters downstream of the cylinder. However, the streakline pattern of this figure is misleading! In Fig. 4–25b, the smoke wire is placed 150 diameters downstream of the cylinder. The resulting streaklines are straight, indicating that the shed vortices have in reality disappeared by this downstream distance. The flow is steady and parallel at this location, and there are no more vortices; viscous diffusion has caused adjacent vorti ces of opposite sign to cancel each other out by around 100 cylinder diam eters. The patterns of Fig. 4–25a near x/D = 150 are merely remnants of the vortex street that existed upstream. The streaklines of Fig. 4–25b, however, show the correct features of the flow at that location. The streaklines gener ated at x/D = 150 are identical to streamlines or pathlines in that region of the flow—straight, nearly horizontal lines—since the flow is steady there. For a known velocity field, a streakline can be generated numerically. We need to follow the paths of a continuous stream of tracer particles from the time of their injection into the flow until the present time, using Eq. 4–17. Mathematically, the location of a tracer particle is integrated over time from the time of its injection tinject to the present time tpresent. Equation 4–17 becomes Integrated tracer particle location: x →= x → injection + ∫ tpresent tinject V › dt (4–18) FIGURE 4–26 Kármán vortices visible in the clouds in the wake of Alexander Selkirk Island in the southern Pacific Ocean.
Photo from Landsat 7 WRS Path 6 Row 83, center: -33.18, -79.99, 9/15/1999, earthobservatory.nasa.gov.
Courtesy of USGS EROS Data Center Satellite System Branch/NASA.
FIGURE 4–24 Streaklines produced by colored fluid introduced upstream; since the flow is steady, these streaklines are the same as streamlines and pathlines. Courtesy of ONERA. Photo by Werlé.
(a) (b) 0 50 Cylinder x/D 100 150 200 250 Cylinder FIGURE 4–25 Smoke streaklines introduced by a smoke wire at two different locations in the wake of a circular cylinder: (a) smoke wire just downstream of the cylinder and (b) smoke wire located at x/D = 150. The time-integrative nature of streaklines is clearly seen by comparing the two photographs.
Photos by John M. Cimbala.
cen96537_ch04_137-188.indd 149 14/01/17 2:30 pm 150 Fluid Kinematics In a complex unsteady flow, the time integration must be performed numer ically as the velocity field changes with time. When the locus of tracer par ticle locations at t = tpresent is connected by a smooth curve, the result is the desired streakline.
EXAMPLE 4–5 Comparison of Flow Patterns in an Unsteady Flow An unsteady, incompressible, two-dimensional velocity field is given by V ›= (u, 𝜐 ) = (0.5 + 0.8x) i ›+ (1.5 + 2.5 sin(𝜔 t) −0.8y) j › (1) where the angular frequency 𝜔 is equal to 2𝜋 rad/s (a physical frequency of 1 Hz). This velocity field is identical to that of Eq. 1 of Example 4–1 except for the additional periodic term in the 𝜐-component of velocity. In fact, since the period of oscillation is 1 s, when time t is any integral multi ple of 1 2 s (t = 0, 1 2, 1, 3 2, 2, . . . s), the sine term in Eq. 1 is zero and the veloc ity field is instantaneously identical to that of Example 4–1. Physically, we imagine flow into a large bell mouth inlet that is oscillating up and down at a frequency of 1 Hz. Consider two complete cycles of flow from t = 0 s to t = 2 s. Compare instantaneous streamlines at t = 2 s to pathlines and streaklines gener ated during the time period from t = 0 s to t = 2 s.
SOLUTION Streamlines, pathlines, and streaklines are to be generated and com pared for the given unsteady velocity field.
Assumptions 1 The flow is incompressible. 2 The flow is two-dimensional, implying no z-component of velocity and no variation of u or 𝜐 with z.
Analysis The instantaneous streamlines at t = 2 s are identical to those of Fig. 4–17, and several of them are replotted in Fig. 4–27. To simulate pathlines, we use the Runge–Kutta numerical integration technique to march in time from t = 0 s to t = 2 s, tracing the path of fluid particles released at three locations: (x = 0.5 m, y = 0.5 m), (x = 0.5 m, y = 2.5 m), and (x = 0.5 m, y = 4.5 m). These pathlines are shown in Fig. 4–27, along with the streamlines. Finally, streaklines are simulated by following the paths of many fluid tracer particles released at the given three locations at times between t = 0 s and t = 2 s, and connecting the locus of their positions at t = 2 s. These streaklines are also plotted in Fig. 4–27.
Discussion Since the flow is unsteady, the streamlines, pathlines, and streaklines are not coincident. In fact, they differ significantly from each other. Note that the streaklines and pathlines are wavy due to the undulating 𝜐-component of velocity. Two complete periods of oscillation have occurred between t = 0 s and t = 2 s, as verified by a careful look at the pathlines and streaklines. The streamlines have no such waviness since they have no time history; they represent an instantaneous snapshot of the velocity field at t = 2 s.
Timelines A timeline is a set of adjacent fluid particles that were marked at the same (earlier) instant in time.
Timelines are particularly useful in situations where the uniformity of a flow (or lack thereof) is to be examined. Figure 4–28 illustrates timelines in 5 4 3 2 y 1 0 –1 0 x 1 2 Streamlines at t = 2 s 3 4 5 Pathlines for 0 < t < 2 s Streaklines for 0 < t < 2 s FIGURE 4–27 Streamlines, pathlines, and streaklines for the oscillating velocity field of Example 4–5. The streaklines and pathlines are wavy because of their integrated time history, but the streamlines are not wavy since they represent an instantaneous snapshot of the velocity field.
Timeline at t = 0 Timeline at t = t1 Timeline at t = t2 Timeline at t = t3 Flow FIGURE 4–28 Timelines are formed by marking a line of fluid particles, and then watching that line move (and deform) through the flow field; timelines are shown at t = 0, t1, t2, and t3.
cen96537_ch04_137-188.indd 150 14/01/17 2:30 pm 151 CHAPTER 4 a channel flow between two parallel walls. Because of friction at the walls, the fluid velocity there is zero (the no-slip condition), and the top and bot tom of the timeline are anchored at their starting locations. In regions of the flow away from the walls, the marked fluid particles move at the local fluid velocity, deforming the timeline. In the example of Fig. 4–28, the speed near the center of the channel is fairly uniform, but small deviations tend to amplify with time as the timeline stretches. One very practical appli cation of timelines is that a velocity vector plot can be generated directly from a timeline (Fig. 4–29). Timelines can be generated experimentally in a water channel through use of a hydrogen bubble wire. When a short burst of electric current is sent through the cathode wire, electrolysis of the water occurs and tiny hydrogen gas bubbles form at the wire. Since the bubbles are so small, their buoyancy is nearly negligible, and the bubbles follow the water flow nicely (Fig. 4–30).
Refractive Flow Visualization Techniques Another category of flow visualization is based on the refractive property of light waves. As you recall from your study of physics, the speed of light through one material may differ somewhat from that in another material, or even in the same material if its density changes. As light travels through one fluid into a fluid with a different index of refraction, the light rays bend (they are refracted). There are two primary flow visualization techniques that utilize the fact that the index of refraction in air (or other gases) varies with density. They are the shadowgraph technique and the schlieren technique (Settles, 2001). Interferometry is a visualization technique that utilizes the related phase change of light as it passes through air of varying densities as the basis for flow visualization and is not discussed here (see Merzkirch, 1987). All these techniques are useful for flow visualization in flow fields where density changes from one location in the flow to another, such as nat ural convection flows (temperature differences cause the density variations), mixing flows (fluid species cause the density variations), and supersonic flows (shock waves and expansion waves cause the density variations). Unlike flow visualizations involving streaklines, pathlines, and timelines, the shadowgraph and schlieren methods do not require injection of a visible FIGURE 4–30 Timelines produced by a hydrogen bubble wire are used to visualize the boundary layer velocity profile shape along a flat plate. Flow is from left to right, and the hydrogen bubble wire is located to the left of the field of view. Bubbles near the wall reveal a flow instability that leads to turbulence. Bippes, H. 1972 Sitzungsber, heidelb. Akad. Wiss. Math. Naturwiss. Kl., no. 3, 103–180; NASA TM-75243, 1978.
Timeline at t = t2 FIGURE 4–29 Velocity vector plot generated from the timeline at t = t2 in Fig. 4–28. A suitable reference scale must be calcu lated so that the lengths of the arrows scale proportionally to that scale.
cen96537_ch04_137-188.indd 151 14/01/17 2:30 pm 152 Fluid Kinematics tracer (smoke or dye). Rather, density differences and the refractive property of light provide the necessary means for visualizing regions of activity in the flow field, allowing us to “see the invisible.” The image (a shadowgram) produced by the shadowgraph method is formed when the refracted rays of light rearrange the shadow cast onto a viewing screen or camera focal plane, causing bright or dark patterns to appear in the shadow. The dark patterns indicate the location where the refracted rays originate, while the bright pat terns mark where these rays end up, and can be misleading. As a result, the dark regions are less distorted than the bright regions and are more useful in the interpretation of the shadowgram. In the shadowgram of Fig. 4–31, for example, we can be confident of the shape and position of the bow shock wave (the dark band), but the refracted bright light has distorted the front of the sphere’s shadow. A shadowgram is not a true optical image; it is, after all, merely a shadow. A schlieren image, however, involves lenses (or mirrors) and a knife edge or other cutoff device to block the refracted light and is a true focused optical image. Schlieren imaging is more complicated to set up than is shadowgraphy (see Settles, 2001 for details) but has a number of advantages. For example, a schlieren image does not suffer from optical distortion by the refracted light rays. Schlieren imaging is also more sensitive to weak density gradients such as those caused by natural convection (Fig. 4–32) or by gradual phe nomena like expansion fans in supersonic flow. Color schlieren imaging tech niques have also been developed. Finally, one can adjust more components in a schlieren setup, such as the location, orientation, and type of the cutoff device, in order to produce an image that is most useful for the problem at hand.
Surface Flow Visualization Techniques Finally, we briefly mention some flow visualization techniques that are useful along solid surfaces. The direction of fluid flow immediately above a solid surface can be visualized with tufts—short, flexible strings glued to the sur face at one end that point in the flow direction. Tufts are especially useful for locating regions of flow separation, where the flow direction reverses. A technique called surface oil visualization can be used for the same purpose—oil placed on the surface forms streaks called friction lines that indicate the direction of flow. If it rains lightly when your car is dirty (espe cially in the winter when salt is on the roads), you may have noticed streaks along the hood and sides of the car, or even on the windshield. This is simi lar to what is observed with surface oil visualization. Lastly, there are pressure-sensitive and temperature-sensitive paints that enable researchers to observe the pressure or temperature distribution along solid surfaces.
4–3 ■ PLOTS OF FLUID FLOW DATA Regardless of how the results are obtained (analytically, experimentally, or computationally), it is usually necessary to plot flow data in ways that enable the reader to get a feel for how the flow properties vary in time and/or space. You are already familiar with time plots, which are especially useful in turbulent flows (e.g., a velocity component plotted as a function of time), and xy-plots (e.g., pressure as a function of radius). In this FIGURE 4–31 Color schlieren image of Mach 3.0 flow from left to right over a sphere. A curved shock wave called a bow shock forms in front of the sphere and curves downstream; its forward-most location appears as the thin red band to the left of the yellow band in this image. The yellow band is caused by the bow shock wrapping around the sphere. Shocks coming off the sphere downstream are due to boundary layer separation.
© G.S. Settles, Gas Dynamics Lab, Penn State University. Used with permission.
FIGURE 4–32 Schlieren image of natural convection due to a barbeque grill. © G.S. Settles, Gas Dynamics Lab, Penn State University. Used with permission.
cen96537_ch04_137-188.indd 152 14/01/17 2:30 pm 153 CHAPTER 4 section, we discuss three additional types of plots that are useful in fluid mechanics—profile plots, vector plots, and contour plots.
Profile Plots A profile plot indicates how the value of a scalar property varies along some desired direction in the flow field.
Profile plots are the simplest of the three to understand because they are like the common xy-plots that you have generated since grade school. Namely, you plot how one variable y varies as a function of a second vari able x. In fluid mechanics, profile plots of any scalar variable (pressure, temperature, density, etc.) can be created, but the most common one used in this book is the velocity profile plot. We note that since velocity is a vec tor quantity, we usually plot either the magnitude of velocity or one of the components of the velocity vector as a function of distance in some desired direction. For example, one of the timelines in the boundary layer flow of Fig. 4–30 is converted into a velocity profile plot by recognizing that at a given instant in time, the horizontal distance traveled by a hydrogen bubble at vertical location y is proportional to the local x-component of velocity u. We plot u as a function of y in Fig. 4–33. The values of u for the plot can also be obtained analytically (see Chaps. 9 and 10), experimentally using PIV or some kind of local velocity measurement device (see Chap. 8), or com putationally (see Chap. 15). Note that it is more physically meaningful in this example to plot u on the abscissa (horizontal axis) rather than on the ordinate (vertical axis) even though it is the dependent variable, since posi tion y is then in its proper orientation (up) rather than across. Finally, it is common to add arrows to velocity profile plots to make them more visually appealing, although no additional information is provided by the arrows. If more than one component of velocity is plotted by the arrow, the direction of the local velocity vector is indicated and the velocity profile plot becomes a velocity vector plot.
Vector Plots A vector plot is an array of arrows indicating the magnitude and direction of a vector property at an instant in time.
While streamlines indicate the direction of the instantaneous velocity field, they do not directly indicate the magnitude of the velocity (i.e., the speed). A useful flow pattern for both experimental and computational fluid flows is thus the vector plot, which consists of an array of arrows that indicate both magnitude and direction of an instantaneous vector property. We have already seen an example of a velocity vector plot in Fig. 4–4 and an accel eration vector plot in Fig. 4–14. These were generated analytically. Vector plots can also be generated from experimentally obtained data (e.g., from PIV measurements) or numerically from CFD calculations. To further illustrate vector plots, we generate a two-dimensional flow field consisting of free-stream flow impinging on a block of rectangular cross section. We perform CFD calculations, and the results are shown in Fig. 4–34. Note that this flow is by nature turbulent and unsteady, but only the long-time averaged results are calculated and displayed here. Streamlines y (a) u y (b) u FIGURE 4–33 Profile plots of the horizontal com ponent of velocity as a function of vertical distance; flow in the boundary layer growing along a horizontal flat plate: (a) standard profile plot and (b) profile plot with arrows.
cen96537_ch04_137-188.indd 153 14/01/17 2:30 pm 154 Fluid Kinematics are plotted in Fig. 4–34a; a view of the entire block and a large portion of its wake is shown. The closed streamlines above and below the symmetry plane indicate large recirculating eddies, one above and one below the line of symmetry. A velocity vector plot is shown in Fig. 4–34b. (Only the upper half of the flow is shown because of symmetry.) It is clear from this plot that the flow accelerates around the upstream corner of the block, so much so in fact that the boundary layer cannot negotiate the sharp corner and sep arates off the block, producing the large recirculating eddies downstream of the block. (Note that these velocity vectors are time-averaged values; the instantaneous vectors change in both magnitude and direction with time as vortices are shed from the body, similar to those of Fig. 4–25a.) A close-up view of the separated flow region is plotted in Fig. 4–34c, where we verify the reverse flow in the lower half of the large recirculating eddy. The vectors of Fig. 4–34 are colored by velocity magnitude, but with modern CFD codes and postprocessors, the vectors can be colored accord ing to some other flow property such as pressure (red for high pressure and blue for low pressure) or temperature (red for hot and blue for cold). In this manner, one can easily visualize not only the magnitude and direction of the flow, but other properties as well, simultaneously.
Contour Plots A contour plot shows curves of constant values of a scalar property (or mag nitude of a vector property) at an instant in time.
If you do any hiking, you are familiar with contour maps of mountain trails. The maps consist of a series of closed curves, each indicating a con stant elevation or altitude. Near the center of a group of such curves is the mountain peak or valley; the actual peak or valley is a point on the map showing the highest or lowest elevation. Such maps are useful in that not only do you get a bird’s-eye view of the streams and trails, etc., but you can also easily see your elevation and where the trail is flat or steep. In fluid mechanics, the same principle is applied to various scalar flow proper ties; contour plots (also called isocontour plots) are generated of pressure, temperature, velocity magnitude, species concentration, properties of turbu lence, etc. A contour plot can quickly reveal regions of high (or low) values of the flow property being studied. A contour plot may consist simply of curves indicating various levels of the property; this is called a contour line plot. Alternatively, the contours can be filled in with either colors or shades of gray; this is called a filled contour plot. An example of pressure contours is shown in Fig. 4–35 for the same flow as in Fig. 4–34. In Fig. 4–35a, filled contours are shown using color to identify regions of different pressure levels—blue regions indicate low pres sure and red regions indicate high pressure. It is clear from this figure that the pressure is highest at the front face of the block and lowest along the top of the block in the separated zone. The pressure is also low in the wake of the block, as expected. In Fig. 4–35b, the same pressure contours are shown, but as a contour line plot with labeled levels of gage pressure in units of pascal. In CFD, contour plots are often displayed in vivid colors with red usu ally indicating the highest value of the scalar and blue the lowest. A healthy (a) (c) Block Flow Recirculating eddy Symmetry plane Flow Block (b) Symmetry plane Block FIGURE 4–34 Results of CFD calculations of flow impinging on a block; (a) streamlines, (b) velocity vector plot of the upper half of the flow, and (c) velocity vector plot, close-up view revealing more details in the separated flow region.
cen96537_ch04_137-188.indd 154 14/01/17 2:30 pm 155 CHAPTER 4 human eye can easily spot a red or blue region and thus locate regions of high or low value of the flow property. Because of the pretty pictures produced by CFD, computational fluid dynamics is sometimes given the nickname “colorful fluid dynamics.” 4–4 ■ OTHER KINEMATIC DESCRIPTIONS Types of Motion or Deformation of Fluid Elements In fluid mechanics, as in solid mechanics, an element may undergo four fundamental types of motion or deformation, as illustrated in two dimen sions in Fig. 4–36: (a) translation, (b) rotation, (c) linear strain (some times called extensional strain), and (d) shear strain. The study of fluid dynamics is further complicated by the fact that all four types of motion or deformation usually occur simultaneously. Because fluid elements may be in constant motion, it is preferable in fluid dynamics to describe the motion and deformation of fluid elements in terms of rates. In particular, we discuss velocity (rate of translation), angular velocity (rate of rotation), linear strain rate (rate of linear strain), and shear strain rate (rate of shear strain). In order for these deformation rates to be useful in the calculation of fluid flows, we must express them in terms of velocity and derivatives of velocity. Translation and rotation are easily understood since they are commonly observed in the motion of solid particles such as billiard balls (Fig. 4–1). A vector is required in order to fully describe the rate of translation in three dimensions. The rate of translation vector is described mathematically as the velocity vector. In Cartesian coordinates, Rate of translation vector in Cartesian coordinates: V ›= u i ›+ 𝜐 j ›+ wk › (4–19) In Fig. 4–36a, the fluid element has moved in the positive horizontal (x) direction; thus u is positive, while 𝜐 (and w) are zero. Rate of rotation (angular velocity) at a point is defined as the average rotation rate of two initially perpendicular lines that intersect at that point. In Fig. 4–36b, for example, consider the point at the bottom-left corner of the initially square fluid element. The left edge and the bottom edge of the element intersect at that point and are initially perpendicular. Both of these lines rotate counterclockwise, which is the mathematically positive direc tion. The angle between these two lines (or between any two initially perpendicular lines on this fluid element) remains at 90° since solid body rotation is illustrated in the figure. Therefore, both lines rotate at the same rate, and the rate of rotation in the plane is simply the component of angular velocity in that plane. In the more general, but still two-dimensional case (Fig. 4–37), the fluid particle translates and deforms as it rotates, and the rate of rotation is cal culated according to the definition given in the previous paragraph. Namely, we begin at time t1 with two initially perpendicular lines (lines a and b in Fig. 4–37) that intersect at point P in the xy-plane. We follow these lines as they move and rotate in an infinitesimal increment of time dt = t2 − t1. (a) (b) 0 –10 –15 –20 –25 –30 –35 –40 –60 60 70 –50 10 20 40 Block Symmetry plane Symmetry plane Flow Block Flow FIGURE 4–35 Contour plots of the pressure field due to flow impinging on a block, as produced by CFD calculations; only the upper half is shown due to symmetry; (a) filled color contour plot and (b) contour line plot where pressure values are displayed in units of Pa (pascals) gage pressure.
cen96537_ch04_137-188.indd 155 14/01/17 2:30 pm 156 Fluid Kinematics At time t2, line a has rotated by angle 𝛼a, and line b has rotated by angle 𝛼b, and both lines have moved with the flow as sketched (both angle values are given in radians and are shown mathematically positive in the sketch). The average rotation angle is thus (𝛼a + 𝛼b)/2, and the rate of rotation or angular velocity in the xy-plane is equal to the time derivative of this average rota tion angle, Rate of rotation of fluid element about point P in Fig. 4–37: 𝜔 = d dt ( 𝛼a + 𝛼b 2 ) = 1 2 ( ∂𝜐 ∂x −∂u ∂y) (4–20) It is left as an exercise to prove the right side of Eq. 4–20 where we have writ ten 𝜔 in terms of velocity components u and 𝜐 in place of angles 𝛼a and 𝛼b. In three dimensions, we must define a vector for the rate of rotation at a point in the flow since its magnitude may differ in each of the three dimen sions. Derivation of the rate of rotation vector in three dimensions can be found in many fluid mechanics books such as Kundu and Cohen (2011) and White (2005). The rate of rotation vector is equal to the angular velocity vector and is expressed in Cartesian coordinates as Rate of rotation vector in Cartesian coordinates: 𝜔 →= 1 2 ( ∂w ∂y − ∂𝜐 ∂z) i ›+ 1 2 ( ∂u ∂z − ∂w ∂x ) j ›+ 1 2 ( ∂𝜐 ∂x − ∂u ∂y)k › (4–21) Linear strain rate is defined as the rate of increase in length per unit length. Mathematically, the linear strain rate of a fluid element depends on the initial orientation or direction of the line segment upon which we measure the linear strain. Thus, it cannot be expressed as a scalar or vector quantity. Instead, we define linear strain rate in some arbitrary direction, which we denote as the x𝛼-direction. For example, line segment PQ in Fig. 4–38 has an initial length of dx𝛼, and it grows to line segment P′Q′ as shown. From the given definition and using the lengths marked in Fig. 4–38, the linear strain rate in the x𝛼-direction is 𝜀𝛼𝛼= 1 dt ( PʹQʹ −PQ PQ ) (4–22) ≅1 dt ( (u𝛼+ ∂u𝛼 ∂x𝛼 dx𝛼) dt + dx𝛼−u𝛼 dt − dx𝛼 dx𝛼 ) = ∂u𝛼 ∂x𝛼 In Cartesian coordinates, we normally take the x𝛼-direction as that of each of the three coordinate axes, although we are not restricted to these directions.
Linear strain rate in Cartesian coordinates: 𝜀xx = ∂u ∂x 𝜀yy = ∂𝜐 ∂y 𝜀zz = ∂w ∂z (4–23) For the more general case, the fluid element moves and deforms as sketched in Fig. 4–37. It is left as an exercise to show that Eq. 4–23 is still valid for the general case.
(a) (c) (d) (b) FIGURE 4–36 Fundamental types of fluid element motion or deformation: (a) translation, (b) rotation, (c) linear strain, and (d) shear strain.
y x Fluid element at time t2 Fluid element at time t1 Line a Line b Line b Line a P′ u P υ 훼b π/2 훼a FIGURE 4–37 For a fluid element that translates and deforms as sketched, the rate of rotation at point P is defined as the average rotation rate of two initially perpendicular lines (lines a and b).
Length of PQ in the x𝛼-direction Length of PQ in the x𝛼-direction Length of P′Q′ in the x𝛼-direction cen96537_ch04_137-188.indd 156 14/01/17 2:30 pm 157 CHAPTER 4 Solid objects such as wires, rods, and beams stretch when pulled. You should recall from your study of engineering mechanics that when such an object stretches in one direction, it usually shrinks in direction(s) normal to that direction. The same is true of fluid elements. In Fig. 4–36c, the origi nally square fluid element stretches in the horizontal direction and shrinks in the vertical direction. The linear strain rate is thus positive horizontally and negative vertically. If the flow is incompressible, the net volume of the fluid element must remain constant; thus if the element stretches in one direction, it must shrink by an appropriate amount in other direction(s) to compensate. The volume of a compressible fluid element, however, may increase or decrease as its density decreases or increases, respectively. (The mass of a fluid element must remain constant, but since 𝜌 = m/V, density and volume are inversely proportional.) Consider for example a parcel of air in a cylinder being com pressed by a piston (Fig. 4–39); the volume of the fluid element decreases while its density increases such that the fluid element’s mass is conserved. The rate of increase of volume of a fluid element per unit volume is called its volumetric strain rate or bulk strain rate. This kinematic property is defined as positive when the volume increases. Another synonym of volu metric strain rate is rate of volumetric dilatation, which is easy to remem ber if you think about how the iris of your eye dilates (enlarges) when exposed to dim light. It turns out that the volumetric strain rate is the sum of the linear strain rates in three mutually orthogonal directions. In Carte sian coordinates (Eq. 4–23), the volumetric strain rate is thus Volumetric strain rate in Cartesian coordinates: 1 V DV Dt = 1 V dV dt = 𝜀xx + 𝜀yy + 𝜀zz = ∂u ∂x + ∂𝜐 ∂y + ∂w ∂z (4–24) In Eq. 4–24, the uppercase D notation is used to stress that we are talking about the volume following a fluid element, that is to say, the material vol ume of the fluid element, as in Eq. 4–12.
The volumetric strain rate is zero in an incompressible flow. Shear strain rate is a more difficult deformation rate to describe and understand. Shear strain rate at a point is defined as half of the rate of decrease of the angle between two initially perpendicular lines that intersect at the point. (The reason for the half will become clear later when we com bine shear strain rate and linear strain rate into one tensor.) In Fig. 4–36d, for example, the initially 90° angles at the lower-left corner and upper-right corner of the square fluid element decrease; this is by definition a positive shear strain. However, the angles at the upper-left and lower-right corners of the square fluid element increase as the initially square fluid element deforms; this is a negative shear strain. Obviously we cannot describe the shear strain rate in terms of only one scalar quantity or even in terms of one vector quantity for that matter. Rather, a full mathematical description of shear strain rate requires its specification in any two mutually perpendicular directions. In Cartesian coordinates, the axes themselves are the most obvi ous choice, although we are not restricted to these. Consider a fluid element in two dimensions in the xy-plane. The element translates and deforms with time as sketched in Fig. 4–40. Two initially mutually perpendicular lines y x xα uα P P′ Q′ Q uα + dxα дuα дxα uα + dxα дuα дxα dt ( ) uα dt dxα FIGURE 4–38 Linear strain rate in some arbitrary direction x𝛼 is defined as the rate of increase in length per unit length in that direction. Linear strain rate would be negative if the line segment length were to decrease. Here we follow the increase in length of line segment PQ into line segment P′Q′, which yields a positive linear strain rate. Velocity components and distances are truncated to first-order since dx𝛼 and dt are infinitesimally small.
Air parcel Time t1 Time t2 FIGURE 4–39 Air being compressed by a piston in a cylinder; the volume of a fluid element in the cylinder decreases, corresponding to a negative rate of volumetric dilatation.
cen96537_ch04_137-188.indd 157 14/01/17 2:30 pm 158 Fluid Kinematics (lines a and b in the x- and y-directions, respectively) are followed. The angle between these two lines decreases from 𝜋/2 (90°) to the angle marked 𝛼a-b at t2 in the sketch. It is left as an exercise to show that the shear strain rate at point P for initially perpendicular lines in the x- and y-directions is given by Shear strain rate, initially perpendicular lines in the x- and y-directions: 𝜀xy = −1 2 d dt 𝛼a-b = 1 2 ( ∂u ∂y + ∂𝜐 ∂x) (4–25) Equation 4–25 can be easily extended to three dimensions. The shear strain rate is thus Shear strain rate in Cartesian coordinates: 𝜀xy = 1 2 ( ∂u ∂y + ∂𝜐 ∂x) 𝜀zx = 1 2 ( ∂w ∂x + ∂u ∂z) 𝜀yz = 1 2 ( ∂𝜐 ∂z + ∂w ∂y ) (4–26) Finally, it turns out that we can mathematically combine linear strain rate and shear strain rate into one symmetric second-order tensor called the strain rate tensor, which is a combination of Eqs. 4–23 and 4–26: Strain rate tensor in Cartesian coordinates: 𝜀ij = ( 𝜀xx 𝜀yx 𝜀zx 𝜀xy 𝜀yy 𝜀zy 𝜀xz 𝜀yz 𝜀zz) = ( ∂u ∂x 1 2 ( ∂𝜐 ∂x + ∂u ∂y) 1 2 ( ∂w ∂x + ∂u ∂z) 1 2 ( ∂u ∂y + ∂𝜐 ∂x) ∂𝜐 ∂y 1 2 ( ∂w ∂y + ∂𝜐 ∂z) 1 2 ( ∂u ∂z + ∂w ∂x ) 1 2 ( ∂𝜐 ∂z + ∂w ∂y ) ∂w ∂z ) (4–27) The strain rate tensor obeys all the laws of mathematical tensors, such as tensor invariants, transformation laws, and principal axes. We use the nota tion 𝜀ij for the strain rate tensor to emphasize its nine components; this is also standard notation when using Cartesian tensor notation. Note that some authors use a double over-arrow instead, namely, 𝜀 → → This emphasizes that 𝜀 → → is a second-order tensor with 32 = 9 components, one mathematical step higher than a vector like V › which is a first-order tensor with 31 = 3 components. Figure 4–41 shows a general (although two-dimensional) situation in a compressible fluid flow in which all possible motions and deformations are present simultaneously. In particular, there is translation, rotation, lin ear strain, and shear strain. Because of the compressible nature of the fluid flow, there is also volumetric strain (dilatation). You should now have a bet ter appreciation of the inherent complexity of fluid dynamics, and the math ematical sophistication required to fully describe fluid motion.
αa-b at t2 Line a Line a u υ y x Fluid element at time t2 Fluid element at time t1 Line b Line b Pʹ P αa-b = π/2 FIGURE 4–40 For a fluid element that translates and deforms as sketched, the shear strain rate at point P is defined as half of the rate of decrease of the angle between two initially perpendicular lines (lines a and b).
C D A B C′ D′ A′ B′ FIGURE 4–41 A fluid element illustrating translation, rotation, linear strain, shear strain, and volumetric strain.
EXAMPLE 4–6 Calculation of Kinematic Properties in a Two-Dimensional Flow Consider the steady, two-dimensional velocity field of Example 4–1: V ›= (u, 𝜐 ) = (0.5 + 0.8 x) i ›+ (1.5 −0.8 y) j › (1) where lengths are in units of m, time in s, and velocities in m/s. There is a stagna tion point at (−0.625, 1.875) as shown in Fig. 4–42. Streamlines of the flow are also cen96537_ch04_137-188.indd 158 14/01/17 2:30 pm 159 CHAPTER 4 4 3 2 y 1 0 –1 –3 –2 –1 0 x 1 FIGURE 4–42 Streamlines for the velocity field of Example 4–6. The stagnation point is indicated by the red circle at x = −0.625 m and y = 1.875 m.
6 5 4 y 3 2 1 –1 0 1 2 x 3 FIGURE 4–43 Deformation of an initially square parcel of marked fluid subjected to the velocity field of Example 4–6 for a time period of 1.5 s. The stagnation point is indicated by the red circle at x = −0.625 m and y = 1.875 m, and several streamlines are plotted.
plotted in Fig. 4–42. Calculate the various kinematic properties, namely, the rate of translation, rate of rotation, linear strain rate, shear strain rate, and volumetric strain rate. Verify that this flow is incompressible.
SOLUTION We are to calculate several kinematic properties of a given velocity field and verify that the flow is incompressible.
Assumptions 1 The flow is steady. 2 The flow is two-dimensional, implying no z-component of velocity and no variation of u or 𝜐 with z.
Analysis By Eq. 4–19, the rate of translation is simply the velocity vector itself, given by Eq. 1. Thus, Rate of translation: u = 0.5 + 0.8x 𝜐 = 1.5 −0.8y w = 0 (2) The rate of rotation is found from Eq. 4–21. In this case, since w = 0 everywhere, and since neither u nor 𝜐 vary with z, the only nonzero component of rotation rate is in the z-direction. Thus, Rate of rotation: 𝜔 →= 1 2 ( ∂𝜐 ∂x −∂u ∂y)k ›= 1 2 (0 −0)k ›= 0 (3) In this case, we see that there is no net rotation of fluid particles as they move about. (This is a significant piece of information, to be discussed in more detail later in this chapter and also in Chap. 10.) Linear strain rates can be calculated in any arbitrary direction using Eq. 4–23. In the x-, y-, and z-directions, the linear strain rates are 𝜀xx = ∂u ∂x = 0.8 s−1 𝜀yy = ∂𝜐 ∂y = −0.8 s−1 𝜀zz = 0 (4) Thus, we predict that fluid particles stretch in the x-direction (positive lin ear strain rate) and shrink in the y-direction (negative linear strain rate). This is illustrated in Fig. 4–43, where we have marked an initially square parcel of fluid centered at (0.25, 4.25). By integrating Eqs. 2 with time, we calculate the location of the four corners of the marked fluid after an elapsed time of 1.5 s. Indeed this fluid parcel has stretched in the x-direction and has shrunk in the y-direction as predicted. Shear strain rate is determined from Eq. 4–26. Because of the two- dimensionality, nonzero shear strain rates can occur only in the xy-plane. Using lines parallel to the x- and y-axes as our initially perpendicular lines, we calculate 𝜀xy, 𝜀xy = 1 2 ( ∂u ∂y + ∂𝜐 ∂x) = 1 2 (0 + 0) = 0 (5) Thus, there is no shear strain in this flow, as also indicated by Fig. 4–43. Although the sample fluid particle deforms, it remains rectangular; its initially 90° corner angles remain at 90° throughout the time period of the calculation. Finally, the volumetric strain rate is calculated from Eq. 4–24: 1 V DV Dt = 𝜀xx + 𝜀yy + 𝜀zz = (0.8 −0.8 + 0) s−1 = 0 (6) Since the volumetric strain rate is zero everywhere, we can say definitively that fluid particles are neither dilating (expanding) nor shrinking (compressing) in vol ume. Thus, we verify that this flow is indeed incompressible. In Fig. 4–43, the area of the shaded fluid particle (and thus its volume since it is a 2-D flow) remains constant as it moves and deforms in the flow field.
cen96537_ch04_137-188.indd 159 14/01/17 2:30 pm 160 Fluid Kinematics 4–5 ■ VORTICITY AND ROTATIONALITY We have already defined the rate of rotation vector of a fluid element (see Eq. 4–21). A closely related kinematic property of great importance to the analysis of fluid flows is the vorticity vector, defined mathematically as the curl of the velocity vector V ›, Vorticity vector: 𝜁 ›= ∇ ›× V ›= curl(V ›) (4–28) Physically, you can tell the direction of the vorticity vector by using the right-hand rule for cross product (Fig. 4–44). The symbol 𝜁 used for vortic ity is the Greek letter zeta. You should note that this symbol for vorticity is not universal among fluid mechanics textbooks; some authors use the Greek letter omega (𝜔) while still others use uppercase omega (Ω). In this book, 𝜔 → is used to denote the rate of rotation vector (angular velocity vector) of a fluid element. It turns out that the rate of rotation vector is equal to half of the vorticity vector, Rate of rotation vector: 𝜔 › = 1 2 ∇ › × V ›= 1 2 curl(V ›) = 𝜁 › 2 (4–29) Thus, vorticity is a measure of rotation of a fluid particle. Specifically, Vorticity is equal to twice the angular velocity of a fluid particle (Fig. 4–45). If the vorticity at a point in a flow field is nonzero, the fluid particle that happens to occupy that point in space is rotating; the flow in that region is called rotational. Likewise, if the vorticity in a region of the flow is zero (or negligibly small), fluid particles there are not rotating; the flow in that region is called irrotational. Physically, fluid particles in a rotational region of flow rotate end over end as they move along in the flow. For example, fluid particles within the viscous boundary layer near a solid wall are rotational (and thus have nonzero vorticity), while fluid particles outside the boundary layer are irrotational (and their vorticity is zero). Both of these cases are illustrated in Fig. 4–46. Rotation of fluid elements is associated with wakes, boundary layers, flow through turbomachinery (fans, turbines, compressors, etc.), and flow with heat transfer. The vorticity of a fluid element cannot change except through the action of viscosity, nonuniform heating (temperature gradients), or other Discussion In this example it turns out that the linear strain rates (𝜀xx and 𝜀yy) are nonzero, while the shear strain rates (𝜀xy and its symmetric partner 𝜀yx) are zero. This means that the x- and y-axes of this flow field are the principal axes. The (two-dimensional) strain rate tensor in this orientation is thus 𝜀ij = ( 𝜀xx 𝜀yx 𝜀xy 𝜀yy) = ( 0.8 0 0 −0.8) s−1 (7) If we were to rotate the axes by some arbitrary angle, the new axes would not be principal axes, and all four elements of the strain rate tensor would be nonzero. You may recall rotating axes in your engineering mechanics classes through use of Mohr’s circles to determine principal axes, maximum shear strains, etc. Similar analyses are performed in fluid mechanics.
C = A × B A B FIGURE 4–44 The direction of a vector cross product is determined by the right-hand rule.
ζ ω FIGURE 4–45 The vorticity vector is equal to twice the angular velocity vector of a rotat ing fluid particle.
cen96537_ch04_137-188.indd 160 14/01/17 2:30 pm 161 CHAPTER 4 nonuniform phenomena. Thus if a flow originates in an irrotational region, it remains irrotational until some nonuniform process alters it. For example, air entering an inlet from quiescent (still) surroundings is irrotational and remains so unless it encounters an object in its path or is subjected to non uniform heating. If a region of flow can be approximated as irrotational, the equations of motion are greatly simplified, as you will see in Chap. 10. In Cartesian coordinates, (i →, j →, k →), (x, y, z), and (u, 𝜐, w), Eq. 4–28 is expanded as follows: Vorticity vector in Cartesian coordinates: 𝜁 ›= ( ∂w ∂y −∂𝜐 ∂z) i ›+ ( ∂u ∂z −∂w ∂x ) j ›+ ( ∂𝜐 ∂x −∂u ∂y) k › (4–30) If the flow is two-dimensional in the xy-plane, the z-component of velocity (w) is zero and neither u nor 𝜐 varies with z. Thus the first two components of Eq. 4–30 are identically zero and the vorticity reduces to Two-dimensional flow in Cartesian coordinates: 𝜁 → = ( ∂𝜐 ∂x −∂u ∂y)k › (4–31) Note that if a flow is two-dimensional in the xy-plane, the vorticity vector must point in either the z- or −z-direction (Fig. 4–47).
Fluid particles not rotating Velocity profile Irrotational outer flow region Rotational boundary layer region Fluid particles rotating Wall FIGURE 4–46 The difference between rotational and irrotational flow: fluid elements in a rotational region of the flow rotate, but those in an irrotational region of the flow do not. Note that the fluid elements would also distort as they move along in the flow, but in order to illustrate only particle rotation, such distortion is not shown here.
y z x ζ FIGURE 4–47 For two-dimensional flow in the xy-plane, the vorticity vector always points in the z- or −z-direction. In this illustration, the flag-shaped fluid particle rotates in the counterclockwise direction as it moves in the xy-plane; its vorticity points in the positive z-direction as shown.
Block Symmetry plane Flow FIGURE 4–48 Contour plot of the vorticity field 𝜁z due to flow impinging on a block, as produced by CFD calculations; only the upper half is shown due to symmetry. Blue regions represent large negative vorticity, and red regions represent large positive vorticity.
EXAMPLE 4–7 Vorticity Contours in a Two-Dimensional Flow Consider the CFD calculation of two-dimensional free-stream flow impinging on a block of rectangular cross section, as shown in Figs. 4–34 and 4–35. Plot vorticity contours and discuss.
SOLUTION We are to calculate the vorticity field for a given velocity field pro duced by CFD and then generate a contour plot of vorticity.
Analysis Since the flow is two-dimensional, the only nonzero component of vor ticity is in the z-direction, normal to the page in Figs. 4–34 and 4–35. A contour plot of the z-component of vorticity for this flow field is shown in Fig. 4–48. The blue region near the upper-left corner of the block indicates large negative values of vorticity, implying clockwise rotation of fluid particles in that region. This is due to the large velocity gradients encountered in this portion of the flow field; the boundary layer separates off the wall at the corner of the body and forms cen96537_ch04_137-188.indd 161 14/01/17 2:30 pm 162 Fluid Kinematics In cylindrical coordinates, (e → r, e → 𝜃, e → z), (r, 𝜃, z), and (ur, u𝜃, uz), Eq. 4–28 is expanded as Vorticity vector in cylindrical coordinates: 𝜁 → = ( 1 r ∂uz ∂𝜃− ∂u𝜃 ∂z )e → r + ( ∂ur ∂z − ∂uz ∂r )e → 𝜃+ 1 r( ∂(ru𝜃) ∂r − ∂ur ∂𝜃)e → z (4–32) a thin shear layer across which the velocity changes rapidly. The concentration of vorticity in the shear layer diminishes as vorticity diffuses downstream. The small red region near the top right corner of the block represents a region of positive vorticity (counterclockwise rotation)—a secondary flow pattern caused by the flow separation.
Discussion We expect the magnitude of vorticity to be highest in regions where spatial derivatives of velocity are high (see Eq. 4–30). Close examination reveals that the blue region in Fig. 4–48 does indeed correspond to large velocity gradients in Fig. 4–34. Keep in mind that the vorticity field of Fig. 4–48 is time-averaged. The instantaneous flow field is in reality turbulent and unsteady, and vortices are shed from the bluff body.
EXAMPLE 4–8 Determination of Rotationality in a Two-Dimensional Flow Consider the following steady, incompressible, two-dimensional velocity field: V ›= (u, 𝜐 ) = x2 i ›+ (−2xy −1) j › (1) Is this flow rotational or irrotational? Sketch some streamlines in the first quadrant and discuss.
SOLUTION We are to determine whether a flow with a given velocity field is rotational or irrotational, and we are to draw some streamlines in the first quadrant.
Analysis Since the flow is two-dimensional, Eq. 4–31 is applicable. Thus, Vorticity: 𝜁 → = ( ∂𝜐 ∂x −∂u ∂y)k ›= (−2y −0)k ›= −2yk › (2) Since the vorticity is nonzero, this flow is rotational. In Fig. 4–49 we plot several streamlines of the flow in the first quadrant; we see that fluid moves downward and to the right. The translation and deformation of a fluid parcel is also shown: at Δt = 0, the fluid parcel is square, at Δt = 0.25 s, it has moved and deformed, and at Δt = 0.50 s, the parcel has moved farther and is further deformed. In particular, the right-most portion of the fluid parcel moves faster to the right and faster downward compared to the left-most portion, stretching the parcel in the x-direction and squashing it in the vertical direction. It is clear that there is also a net clockwise rotation of the fluid parcel, which agrees with the result of Eq. 2.
Discussion From Eq. 4–29, individual fluid particles rotate at an angular velocity equal to 𝜔 → = −yk →, half of the vorticity vector. Since 𝜔 → is not constant, this flow is not solid-body rotation. Rather, 𝜔 → is a linear function of y. Further analysis reveals that this flow field is incompressible; the area (and volume) of the shaded regions representing the fluid parcel in Fig. 4–49 remains constant at all three instants in time.
4 3 y 2 1 0 0 1 2 3 x 4 Δt = 0.25 s Δt = 0 Δt = 0.50 s FIGURE 4–49 Deformation of an initially square fluid parcel subjected to the velocity field of Example 4–8 for a time period of 0.25 s and 0.50 s. Several streamlines are also plotted in the first quadrant. It is clear that this flow is rotational.
cen96537_ch04_137-188.indd 162 14/01/17 2:30 pm 163 CHAPTER 4 For two-dimensional flow in the r𝜃-plane, Eq. 4–32 reduces to Two-dimensional flow in cylindrical coordinates: 𝜁 → = 1 r( ∂(ru𝜃) ∂r − ∂ur ∂𝜃)k › (4–33) where k → is used as the unit vector in the z-direction in place of e → z. Note that if a flow is two-dimensional in the r𝜃-plane, the vorticity vector must point in either the z- or −z-direction (Fig. 4–50).
Comparison of Two Circular Flows Not all flows with circular streamlines are rotational. To illustrate this point, we consider two incompressible, steady, two-dimensional flows, both of which have circular streamlines in the r 𝜃-plane: Flow A—solid-body rotation: ur = 0 and u𝜃= 𝜔 r (4–34) Flow B—line vortex: ur = 0 and u𝜃= K r (4–35) where 𝜔 and K are constants. (Alert readers will note that u𝜃 in Eq. 4–35 is infinite at r = 0, which is of course physically impossible; we ignore the region close to the origin to avoid this problem.) Since the radial component of velocity is zero in both cases, the streamlines are circles about the origin. The velocity profiles for the two flows, along with their streamlines, are sketched in Fig. 4–51. We now calculate and compare the vorticity field for each of these flows, using Eq. 4–33.
Flow A—solid-body rotation: 𝜁 → = 1 r( ∂(𝜔 r2) ∂r −0)k ›= 2𝜔 k › (4–36) Flow B—line vortex: 𝜁 → = 1 r( ∂(K) ∂r −0)k ›= 0 (4–37) Not surprisingly, the vorticity for solid-body rotation is nonzero. In fact, it is a constant of magnitude twice the angular velocity and pointing in the same direction. (This agrees with Eq. 4–29.) Flow A is rotational. Physically, this means that individual fluid particles rotate as they revolve around the origin (Fig. 4–51a). By contrast, the vorticity of the line vortex is zero everywhere (except right at the origin, which is a mathematical singularity). Flow B is irrotational. Physically, fluid particles do not rotate as they revolve in cir cles about the origin (Fig. 4–51b). A simple analogy can be made between flow A and a merry-go-round or roundabout, and flow B and a Ferris wheel (Fig. 4–52). As children revolve around a roundabout, they also rotate at the same angular velocity as that of the ride itself. This is analogous to a rotational flow. In contrast, children on a Ferris wheel always remain oriented in an upright position as they trace out their circular path. This is analogous to an irrotational flow.
y z x r ζ θ FIGURE 4–50 For a two-dimensional flow in the r𝜃-plane, the vorticity vector always points in the z (or −z) direction. In this illustration, the flag-shaped fluid par ticle rotates in the clockwise direction as it moves in the r𝜃-plane; its vorticity points in the −z-direction as shown.
Flow A uθ uθ = ωr r (a) Flow B uθ r (b) uθ = r K FIGURE 4–51 Streamlines and velocity profiles for (a) flow A, solid-body rotation and (b) flow B, a line vortex. Flow A is rotational, but flow B is irrotational everywhere except at the origin. Note that the (oversized) fluid elements in flow B would also distort as they move, but in order to illustrate only particle ro tation, such distortion is not shown here.
cen96537_ch04_137-188.indd 163 14/01/17 2:30 pm 164 Fluid Kinematics 4–6 ■ THE REYNOLDS TRANSPORT THEOREM In thermodynamics and solid mechanics we often work with a system (also called a closed system), defined as a quantity of matter of fixed identity. In fluid dynamics, it is more common to work with a control volume (also EXAMPLE 4–9 Determination of Rotationality of a Line Sink A simple two-dimensional velocity field called a line sink is often used to simulate fluid being sucked into a line along the z-axis. Suppose the volume flow rate per unit length along the z-axis, V .
/L, is known, where V . is a negative quantity. In two dimensions in the r𝜃-plane, Line sink: ur = V .
2𝜋 L 1 r and u𝜃= 0 (1) Draw several streamlines of the flow and calculate the vorticity. Is this flow rota tional or irrotational?
SOLUTION Streamlines of the given flow field are to be sketched and the rota tionality of the flow is to be determined.
Analysis Since there is only radial flow and no tangential flow, we know imme diately that all streamlines must be rays into the origin. Several streamlines are sketched in Fig. 4–53. The vorticity is calculated from Eq. 4–33: 𝜁 → = 1 r( ∂(ru𝜃) ∂r −∂ ∂𝜃 ur)k → = 1 r(0 −∂ ∂𝜃 ( V .
2𝜋 L 1 r))k → = 0 (2) Since the vorticity vector is everywhere zero, this flow field is irrotational.
Discussion Many practical flow fields involving suction, such as flow into inlets and hoods, can be approximated quite accurately by assuming irrotational flow (Heinsohn and Cimbala, 2003).
y x Streamlines θ r FIGURE 4–53 Streamlines in the r𝜃-plane for the case of a line sink.
FIGURE 4–52 A simple analogy: (a) rotational circular flow is analogous to a roundabout, while (b) irrotational circular flow is analogous to a Ferris wheel.
(a) © McGraw-Hill Education/Mark Dierker, photographer (b) © DAJ/Getty Images RF (b) (b) (a) (a) cen96537_ch04_137-188.indd 164 14/01/17 2:31 pm 165 CHAPTER 4 called an open system), defined as a region in space chosen for study. The size and shape of a system may change during a process, but no mass crosses its boundaries. A control volume, on the other hand, allows mass to flow in or out across its boundaries, which are called the control surface. A control volume may also move and deform during a process, but many real-world applications involve fixed, nondeformable control volumes. Figure 4–54 illustrates both a system and a control volume for the case of deodorant being sprayed from a spray can. When analyzing the spraying pro cess, a natural choice for our analysis is either the moving, deforming fluid (a system) or the volume bounded by the inner surfaces of the can (a control volume). These two choices are identical before the deodorant is sprayed. When some contents of the can are discharged, the system approach consid ers the discharged mass as part of the system and tracks it (a difficult job indeed); thus the mass of the system remains constant. Conceptually, this is equivalent to attaching a flat balloon to the nozzle of the can and letting the spray inflate the balloon. The inner surface of the balloon now becomes part of the boundary of the system. The control volume approach, however, is not concerned at all with the deodorant that has escaped the can (other than its properties at the exit), and thus the mass of the control volume decreases during this process while its volume remains constant. Therefore, the system approach treats the spraying process as an expansion of the system’s vol ume, whereas the control volume approach considers it as a fluid discharge through the control surface of the fixed control volume. Most principles of fluid mechanics are adopted from solid mechanics, where the physical laws dealing with the time rates of change of extensive properties are expressed for systems. In fluid mechanics, it is usually more convenient to work with control volumes, and thus there is a need to relate the changes in a control volume to the changes in a system. The relationship between the time rates of change of an extensive property for a system and for a control volume is expressed by the Reynolds transport theorem (RTT), which provides the link between the system and control volume approaches (Fig. 4–55). RTT is named after the English engineer, Osborne Reynolds (1842–1912), who did much to advance its application in fluid mechanics. The general form of the Reynolds transport theorem can be derived by considering a system with an arbitrary shape and arbitrary interactions, but the derivation is rather involved. To help you grasp the fundamental mean ing of the theorem, we derive it first in a straightforward manner using a simple geometry and then generalize the results. Consider flow from left to right through a diverging (expanding) portion of a flow field as sketched in Fig. 4–56. The upper and lower bounds of the fluid under consideration are streamlines of the flow, and we assume uniform flow through any cross section between these two streamlines. We choose the control volume to be fixed between sections (1) and (2) of the flow field. Both (1) and (2) are normal to the direction of flow. At some ini tial time t, the system coincides with the control volume, and thus the system and control volume are identical (the greenish-shaded region in Fig. 4–56). During time interval Δt, the system moves in the flow direction at uniform speeds V1 at section (1) and V2 at section (2). The system at this later time is indicated by the hatched region. The region uncovered by the system during this motion is designated as section I (part of the CV), and the new region (a) Sprayed mass (b) System CV FIGURE 4–54 Two methods of analyzing the spray ing of deodorant from a spray can: (a) We follow the fluid as it moves and deforms. This is the system approach—no mass crosses the boundary, and the total mass of the system remains fixed. (b) We consider a fixed interior volume of the can. This is the control volume approach— mass crosses the boundary.
Control volume RTT System FIGURE 4–55 The Reynolds transport theorem (RTT) provides a link between the system approach and the control volume approach.
cen96537_ch04_137-188.indd 165 14/01/17 2:31 pm 166 Fluid Kinematics covered by the system is designated as section II (not part of the CV). There fore, at time t + Δt, the system consists of the same fluid, but it occupies the region CV − I + II. The control volume is fixed in space, and thus it remains as the shaded region marked CV at all times. Let B represent any extensive property (such as mass, energy, or momen tum), and let b = B/m represent the corresponding intensive property. Noting that extensive properties are additive, the extensive property B of the system at times t and t + Δt is expressed as Bsys, t = BCV, t (the system and CV coincide at time t) Bsys, t+Δt = BCV, t+Δt −BI, t+Δt + BII, t+Δt Subtracting the first equation from the second one and dividing by Δt gives Bsys, t+Δt −Bsys, t Δt = BCV, t+Δt −BCV, t Δt − BI, t+Δt Δt + BΠ, t+Δt Δt Taking the limit as Δt → 0, and using the definition of derivative, we get dBsys dt = dBCV dt −B .
in + B .
out (4–38) or dBsys dt = dBCV dt −b1ρ1V1A1 + b2ρ2V2A2 since BI, t+Δt = b1mI, t+Δt = b1ρ1V I, t+Δt = b1ρ1V1 Δt A1 BII, t+Δt = b2mII, t+Δt = b2ρ2V II, t+Δt = b2ρ2V2 Δt A2 and B .
in = B .
I = lim Δt→0 BI, t+Δt Δt = lim Δt→0 b1ρ1V1 Δt A1 Δt = b1ρ1V1 A1 B .
out = B .
II = lim Δt→0 BII, t+Δt Δt = lim Δt→0 b2ρ2V2 Δt A2 Δt = b2ρ2V2 A2 where A1 and A2 are the cross-sectional areas at locations 1 and 2. Equation 4–38 states that the time rate of change of the property B of the system is equal to the time rate of change of B of the control volume plus the net flux of B out of the control volume by mass crossing the control surface. This is the desired rela tion since it relates the change of a property of a system to the change of that property for a control volume. Note that Eq. 4–38 applies at any instant in time, where it is assumed that the system and the control volume occupy the same space at that particular instant in time. The influx B .
in and outflux B .
out of the property B in this case are easy to determine since there is only one inlet and one outlet, and the velocities are approximately normal to the surfaces at sections (1) and (2). In general, how ever, we may have several inlet and outlet ports, and the velocity may not be normal to the control surface at the point of entry. Also, the velocity may not be uniform. To generalize the process, we consider a differential surface area dA on the control surface and denote its unit outer normal by n →. The flow rate of property b through dA is 𝜌bV ›·n → dA since the dot product V ›·n → gives Control volume at time t + Δt (CV remains fixed in time) System at time t + Δt (hatched region) Inflow during Δt Outflow during Δt At time t: Sys = CV At time t + Δt: Sys = CV ‒ I + II II I V1 V2 System (material volume) and control volume at time t (shaded region) (1) (2) FIGURE 4–56 A moving system (hatched region) and a fixed control volume (shaded region) in a diverging portion of a flow field at times t and t + Δt. The upper and lower bounds are streamlines of the flow.
cen96537_ch04_137-188.indd 166 14/01/17 2:31 pm 167 CHAPTER 4 the normal component of the velocity. Then the net rate of outflow through the entire control surface is determined by integration to be (Fig. 4–57) B .
net = B .
out −B .
in = ∫CS ρbV ›·n → dA (inflow if negative) (4–39) An important aspect of this relation is that it automatically subtracts the inflow from the outflow, as explained next. The dot product of the velocity vector at a point on the control surface and the outer normal at that point is V › ·n →= ∣V ›∣∣n →∣ cos 𝜃= ∣V ›∣ cos 𝜃, where 𝜃 is the angle between the velocity vector and the outer normal, as shown in Fig. 4–58. For 𝜃 < 90°, cos 𝜃 > 0 and thus V ›·n → > 0 for outflow of mass from the control volume, and for 𝜃 > 90°, cos 𝜃 < 0 and thus V ›·n → < 0 for inflow of mass into the control volume. Therefore, the differential quantity 𝜌bV ›·n → dA is positive for mass flowing out of the control volume, and negative for mass flowing into the control volume, and its integral over the entire control surface gives the rate of net outflow of the property B by mass. The properties within the control volume may vary with position, in general. In such a case, the total amount of property B within the control volume must be determined by integration: BCV = ∫CV ρb dV (4–40) The term dBCV/dt in Eq. 4–38 is thus equal to d dt ∫CV ρb dV, and represents the time rate of change of the property B content of the control volume. A positive value for dBCV/dt indicates an increase in the B content, and a negative value indicates a decrease. Substituting Eqs. 4–39 and 4–40 into Eq. 4–38 yields the Reynolds transport theorem, also known as the system-to-control-volume transformation for a fixed control volume: RTT, fixed CV: dBsys dt = d dt ∫CV ρb dV + ∫CS ρbV ›·n → dA (4–41) Since the control volume is not moving or deforming with time, the time derivative on the right-hand side can be moved inside the integral, since the domain of integration does not change with time. (In other words, it is irrelevant whether we differentiate or integrate first.) But the time derivative in that case must be expressed as a partial derivative (∂/∂t) since density and the quantity b may depend not only on time, but also on the position within the control volume. Thus, an alternate form of the Reynolds transport theorem for a fixed control volume is Alternate RTT, fixed CV: dBsys dt = ∫CV ∂ ∂t (ρb) dV + ∫CS ρbV ›· n → dA (4–42) It turns out that Eq. 4–42 is also valid for the most general case of a mov ing and/or deforming control volume, provided that velocity vector V › is an absolute velocity (as viewed from a fixed reference frame). Next we consider yet another alternative form of the RTT. Equation 4–41 was derived for a fixed control volume. However, many practical systems such as turbine and propeller blades involve nonfixed control volumes. For tunately, Eq. 4–41 is also valid for moving and/or deforming control volumes Bnet = Bout ‒ Bin = ∫ CS ρbV · n dA · · · Control volume n Mass entering outward normal Mass leaving Mass leaving n n n n = FIGURE 4–57 The integral of b𝜌V ›· n → dA over the control surface gives the net amount of the property B flowing out of the control volume (into the control volume if it is negative) per unit time.
If θ < 90°, then cos θ > 0 (outflow).
If θ > 90°, then cos θ < 0 (inflow).
If θ = 90°, then cos θ = 0 (no flow).
n Outflow: θ < 90° dA n Inflow: θ > 90° dA · n = |V || n | cos θ = V cos θ V V V θ θ FIGURE 4–58 Outflow and inflow of mass across the differential area of a control surface.
cen96537_ch04_137-188.indd 167 14/01/17 2:31 pm 168 Fluid Kinematics provided that the absolute fluid velocity V › in the last term is replaced by the relative velocity V › r, Relative velocity: V › r = V ›−V › CS (4–43) where V › CS is the local velocity of the control surface (Fig. 4–59). The most general form of the Reynolds transport theorem is thus RTT, nonfixed CV: dBsys dt = d dt ∫CV ρb dV + ∫CS ρbV › r·n → dA (4–44) Note that for a control volume that moves and/or deforms with time, the time derivative is applied after integration in Eq. 4–44. As a simple example of a moving control volume, consider a toy car moving at a constant abso lute velocity V › car = 10 km/h to the right. A high-speed jet of water (absolute velocity = V › jet = 25 km/h to the right) strikes the back of the car and pro pels it (Fig. 4–60). If we draw a control volume around the car, the relative velocity is V › r = 25 − 10 = 15 km/h to the right. This represents the velocity at which an observer moving with the control volume (moving with the car) would observe the fluid crossing the control surface. In other words, V › r is the fluid velocity expressed relative to a coordinate system moving with the control volume. Finally, by application of the Leibniz theorem (to be discussed shortly), it can be shown that the Reynolds transport theorem for a general moving and/ or deforming control volume (Eq. 4–44) is equivalent to the form given by Eq. 4–42, which is repeated here: Alternate RTT, nonfixed CV: dBsys dt = ∫CV ∂ ∂t (ρb) dV + ∫CS ρbV ›· n → dA (4–45) In contrast to Eq. 4–44, the velocity vector V › in Eq. 4–45 must be taken as the absolute velocity (as viewed from a fixed reference frame) in order to apply to a nonfixed control volume. During steady flow, the amount of the property B within the control vol ume remains constant in time, and thus the time derivative in Eq. 4–44 becomes zero. Then the Reynolds transport theorem reduces to RTT, steady flow: dBsys dt = ∫CS ρbV › r·n → dA (4–46) Note that unlike the control volume, the property B content of the system may still change with time during a steady process. But in this case the change must be equal to the net property transported by mass across the control surface (an advective rather than an unsteady effect). In most practical engineering applications of the RTT, fluid crosses the boundary of the control volume at a finite number of well-defined inlets and outlets (Fig. 4–61). In such cases, it is convenient to cut the control sur face directly across each inlet and outlet and replace the surface integral in Eq. 4–44 with approximate algebraic expressions at each inlet and outlet based on the average values of fluid properties crossing the boundary. We define 𝜌avg, bavg, and Vr, avg as the average values of 𝜌, b, and Vr, respectively, across an inlet or outlet of cross-sectional area A, e.g., bavg = 1 A ∫A b dA. The CS –VCS VCS V Vr = V – VCS FIGURE 4–59 Relative velocity crossing a control surface is found by vector addition of the absolute velocity of the fluid and the negative of the local velocity of the control surface.
Control volume Absolute reference frame: Vjet Vcar Control volume Relative reference frame: V r = Vjet – Vcar FIGURE 4–60 Reynolds transport theorem applied to a control volume moving at constant velocity.
cen96537_ch04_137-188.indd 168 14/01/17 2:31 pm 169 CHAPTER 4 surface integrals in the RTT (Eq. 4–44), when applied over an inlet or outlet of cross-sectional area A, are then approximated by pulling property b out of the surface integral and replacing it with its average. This yields ∫A ρbV › r·n → dA ≅bavg ∫A ρV › r·n → dA = bavgm · r where m .
r is the mass flow rate through the inlet or outlet relative to the (mov ing) control surface. The approximation in this equation is exact when prop erty b is uniform over cross-sectional area A. Equation 4–44 thus becomes dBsys dt = d dt ∫CV ρb dV + ∑ out m · r bavg −∑ in m · r bavg (4–47) In some applications, we may wish to rewrite Eq. 4–47 in terms of volume (rather than mass) flow rate. In such cases, we make a further approxima tion that m · r ≈ρavgV .
r = ρavgVr, avg A. This approximation is exact when fluid density 𝜌 is uniform over A. Equation 4–47 then reduces to Approximate RTT for well-defined inlets and outlets: dBsys dt = d dt ∫CV ρb dV + ∑ out ρavgbavgVr, avg A −∑ in ρavgbavgVr,avg A (4–48) Note that these approximations simplify the analysis greatly but may not always be accurate, especially in cases where the velocity distribution across the inlet or outlet is not very uniform (e.g., pipe flows; Fig. 4–61). In par ticular, the control surface integral of Eq. 4–45 becomes nonlinear when property b contains a velocity term (e.g., when applying RTT to the linear momentum equation, b = V ›), and the approximation of Eq. 4–48 leads to errors. Fortunately we can eliminate the errors by including correction fac tors in Eq. 4–48, as discussed in Chaps. 5 and 6. Equations 4–47 and 4–48 apply to fixed or moving control volumes, but as discussed previously, the relative velocity must be used for the case of a nonfixed control volume. In Eq. 4–47 for example, the mass flow rate m .
r is relative to the (moving) control surface, hence the r subscript.
Alternate Derivation of the Reynolds Transport Theorem A more elegant mathematical derivation of the Reynolds transport theorem is possible through use of the Leibniz (sometimes Leibnitz) theorem (see Kundu and Cohen, 2011). You may be familiar with the one-dimensional version of this theorem, which allows you to differentiate an integral whose limits of integration are functions of the variable with which you need to dif ferentiate (Fig. 4–62): One-dimensional Leibniz theorem: d dt ∫ x=b(t) x=a(t) G(x, t) dx = ∫ b a ∂G ∂t dx + db dt G(b, t) −da dt G(a, t) (4–49) The Leibniz theorem takes into account the change of limits a(t) and b(t) with respect to time, as well as the unsteady changes of integrand G(x, t) with time.
CV 1 2 3 FIGURE 4–61 An example control volume in which there is one well-defined inlet (1) and two well-defined outlets (2 and 3). In such cases, the control surface integral in the RTT can be more conveniently written in terms of the average values of fluid properties crossing each inlet and outlet.
G(x, t) x b(t) a(t) x = b(t) x = a(t) ∫ G(x, t) dx FIGURE 4–62 The one-dimensional Leibniz theorem is required when calculating the time derivative of an integral (with respect to x) for which the limits of the integral are functions of time.
This section may be omitted without loss of continuity.
for each outlet for each inlet for each outlet for each inlet cen96537_ch04_137-188.indd 169 14/01/17 2:31 pm 170 Fluid Kinematics EXAMPLE 4–10 One-Dimensional Leibniz Integration Reduce the following expression as far as possible: F(t) = d dt ∫ x=Bt x=At e−2x2 dx SOLUTION F(t) is to be evaluated from the given expression.
Analysis The integral is F(t) = d dt ∫ x=Bt x=At e−2x2 dx (1) We could try integrating first, and then differentiating, but we can instead use the 1-D Leibniz theorem. Here, G(x, t) = e−2x2 (G is not a function of time in this sim ple example). The limits of integration are a(t) = At and b(t) = Bt. Thus, F(t) = ∫ b a ∂G ∂t dx + db dt G(b, t) −da dt G(a, t) (2) = 0 + Be−2b2 −Ae−2a2 or F(t) = Be−2B2t2−Ae−2A2t2 (3) Discussion You are welcome to try to obtain the same solution without using the Leibniz theorem. In three dimensions, the Leibniz theorem for a volume integral is Three-dimensional Leibniz theorem: d dt ∫V(t) G(x, y, z, t) dV = ∫V(t) ∂G ∂t dV + ∫A(t) GV › A·n → dA (4–50) where V(t) is a moving and/or deforming volume (a function of time), A(t) is its surface (boundary), and V › A is the absolute velocity of this (moving) surface (Fig. 4–63). Equation 4–50 is valid for any volume, moving and/or deforming arbitrarily in space and time. For consistency with the previous analyses, we set integrand G to 𝜌b for application to fluid flow, Three-dimensional Leibniz theorem applied to fluid flow: d dt ∫V(t) ρb dV = ∫V(t) ∂ ∂t (ρb) dV + ∫A(t) ρbV › A·n → dA (4–51) If we apply the Leibniz theorem to the special case of a material volume (a system of fixed identity moving with the fluid flow), then V › A = V › every where on the material surface since it moves with the fluid. Here V › is the local fluid velocity, and Eq. 4–51 becomes Leibniz theorem applied to a material volume: d dt ∫V(t) ρb dV = dBsys dt = ∫V(t) ∂ ∂t (ρb) dV + ∫A(t) ρbV ›·n → dA (4–52) A(t) V(t) G(x, y, z, t) dV G(x, y, z, t) VA v(t) ∫ FIGURE 4–63 The three-dimensional Leibniz theorem is required when calculating the time derivative of a volume integral for which the volume itself moves and/or deforms with time. It turns out that the three-dimensional form of the Leibniz theorem can be used in an alternative derivation of the Reynolds transport theorem.
cen96537_ch04_137-188.indd 170 14/01/17 2:31 pm 171 CHAPTER 4 Equation 4–52 is valid at any instant in time t. We define our control vol ume such that at this time t, the control volume and the system occupy the same space; in other words, they are coincident. At some later time t + Δt, the system has moved and deformed with the flow, but the control volume may have moved and deformed differently (Fig. 4–64). The key, however, is that at time t, the system (material volume) and control volume are one and the same. Thus, the volume integral on the right-hand side of Eq. 4–52 can be evaluated over the control volume at time t, and the surface integral can be evaluated over the control surface at time t. Hence, General RTT, nonfixed CV: dBsys dt = ∫CV ∂ ∂t (ρb) dV + ∫CS ρbV ›·n → dA (4–53) This expression is identical to that of Eq. 4–42 and is valid for an arbitrarily shaped, moving, and/or deforming control volume at time t. Keep in mind that V › in Eq. 4–53 is the absolute fluid velocity.
EXAMPLE 4–11 Reynolds Transport Theorem in Terms of Relative Velocity Beginning with the Leibniz theorem and the general Reynolds transport theorem for an arbitrarily moving and deforming control volume, Eq. 4–53, prove that Eq. 4–44 is valid.
SOLUTION Equation 4–44 is to be proven.
Analysis The general three-dimensional version of the Leibniz theorem, Eq. 4–50, applies to any volume. We choose to apply it to the control volume of interest, which can be moving and/or deforming differently than the material volume (Fig. 4–64). Setting G to 𝜌b, Eq. 4–50 becomes d dt ∫CV ρb dV = ∫CV ∂ ∂t (ρb) dV + ∫CS ρbV › CS · n → dA (1) We solve Eq. 4–53 for the control volume integral, ∫CV ∂ ∂t (ρb) dV = dBsys dt −∫CS ρbV ›·n› dA (2) Substituting Eq. 2 into Eq. 1, we get d dt ∫CV ρb dV = dBsys dt −∫CS ρbV ›·n› dA + ∫CS ρbV › CS·n› dA (3) Combining the last two terms and rearranging, dBsys dt = d dt ∫CV ρb dV + ∫CS ρb(V ›−V › CS) ·n› dA (4) But recall that the relative velocity is defined by Eq. 4–43. Thus, RTT in terms of relative velocity: dBsys dt = d dt ∫CV 𝞺b dV + ∫CS 𝞺bV › r · n› dA (5) Discussion Equation 5 is indeed identical to Eq. 4–44, and the power and elegance of the Leibniz theorem are demonstrated.
System (material volume) and control volume at time t System at time t + Δt Control volume at time t + Δt Flow FIGURE 4–64 The material volume (system) and control volume occupy the same space at time t (the greenish shaded area), but move and deform differently. At a later time they are not coincident.
cen96537_ch04_137-188.indd 171 14/01/17 2:31 pm 172 Fluid Kinematics Relationship between Material Derivative and RTT You may have noticed a similarity or analogy between the material derivative discussed in Section 4–1 and the Reynolds transport theorem discussed here. In fact, both analyses represent methods to transform from fundamentally Lagrangian concepts to Eulerian interpretations of those concepts. While the Reynolds transport theorem deals with finite-size control volumes and the material derivative deals with infinitesimal fluid particles, the same fundamental physical interpretation applies to both (Fig. 4–65). In fact, the Reynolds transport theorem can be thought of as the integral counterpart of the material derivative. In either case, the total rate of change of some property following an identified portion of fluid consists of two parts: There is a local or unsteady part that accounts for changes in the flow field with time (compare the first term on the right-hand side of Eq. 4–12 to that of Eq. 4–45). There is also an advective part that accounts for the movement of fluid from one region of the flow to another (compare the second term on the right-hand sides of Eqs. 4–12 and 4–45). Just as the material derivative can be applied to any fluid property, scalar or vector, the Reynolds transport theorem can be applied to any scalar or vector property as well. In Chaps. 5 and 6, we apply the Reynolds transport theorem to conservation of mass, energy, momentum, and angular momen tum by choosing parameter B to be mass, energy, momentum, and angular momentum, respectively. In this fashion we can easily convert from the fun damental system conservation laws (Lagrangian viewpoint) to forms that are valid and useful in a control volume analysis (Eulerian viewpoint).
Lagrangian description Eulerian description System analysis RTT Control volume analysis D Dt FIGURE 4–65 The Reynolds transport theorem for finite volumes (integral analysis) is analogous to the material derivative for infinitesimal volumes (differential analysis). In both cases, we transform from a Lagrangian or system viewpoint to an Eulerian or control volume viewpoint.
SUMMARY Fluid kinematics is concerned with describing fluid motion, without necessarily analyzing the forces responsible for such motion. There are two fundamental descriptions of fluid motion—Lagrangian and Eulerian. In a Lagrangian descrip tion, we follow individual fluid particles or collections of fluid particles, while in the Eulerian description, we define a control volume through which fluid flows in and out. We transform equations of motion from Lagrangian to Eulerian through use of the material derivative for infinitesimal fluid particles and through use of the Reynolds transport theo rem (RTT ) for systems of finite volume. For some extensive property B or its corresponding intensive property b, Material derivative: Db Dt = ∂b ∂t + (V ›·∇ › )b General RTT, nonfixed CV: dBsys dt = ∫CV ∂ ∂t (ρb) dV + ∫CS ρbV ›· n› dA In both equations, the total change of the property following a fluid particle or following a system is composed of two parts: a local (unsteady) part and an advective (movement) part. There are various ways to visualize and analyze flow fields—streamlines, streaklines, pathlines, timelines, surface imaging, shadowgraphy, schlieren imaging, profile plots, vector plots, and contour plots. We define each of these and provide examples in this chapter. In general unsteady flow, streamlines, streaklines, and pathlines differ, but in steady flow, streamlines, streaklines, and pathlines are coincident. Four fundamental rates of motion (deformation rates) are required to fully describe the kinematics of a fluid flow: veloc ity (rate of translation), angular velocity (rate of rotation), lin ear strain rate, and shear strain rate. Vorticity is a property of fluid flows that indicates the rotationality of fluid particles.
Vorticity vector: 𝜁 ›= ∇ ›× V ›= curl(V › ) = 2𝜔 › A region of flow is irrotational if the vorticity is zero in that region. The concepts learned in this chapter are used repeatedly throughout the rest of the book. We use the RTT to transform the conservation laws from closed systems to control volumes in Chaps. 5 and 6, and again in Chap. 9 in the derivation of the differential equations of fluid motion. The role of vortic ity and irrotationality is revisited in greater detail in Chap. 10 where we show that the irrotationality approximation leads to greatly reduced complexity in the solution of fluid flows. Finally, we use various types of flow visualization and data plots to describe the kinematics of example flow fields in nearly every chapter of this book.
cen96537_ch04_137-188.indd 172 14/01/17 2:31 pm 173 CHAPTER 4 Guest Author: Ganesh Raman, Illinois Institute of Technology Fluidic actuators are devices that use fluid logic circuits to produce oscilla tory velocity or pressure perturbations in jets and shear layers for delaying separation, enhancing mixing, and suppressing noise. Fluidic actuators are potentially useful for shear flow control applications for many reasons: they have no moving parts; they can produce perturbations that are controllable in frequency, amplitude, and phase; they can operate in harsh thermal environ ments and are not susceptible to electromagnetic interference; and they are easy to integrate into a functioning device. Although fluidics technology has been around for many years, recent advances in miniaturization and micro fabrication have made them very attractive candidates for practical use. The fluidic actuator produces a self-sustaining oscillatory flow using the principles of wall attachment and backflow that occur within miniature passages of the device.
Figure 4–66 demonstrates the application of a fluidic actuator for jet thrust vectoring. Fluidic thrust vectoring is important for future aircraft designs, since they can improve maneuverability without the complexity of additional sur faces near the nozzle exhaust. In the three images of Fig. 4–66, the primary jet exhausts from right to left and a single fluidic actuator is located at the top. Figure 4–66a shows the unperturbed jet. Figures 4–66b and c show the vectoring effect at two fluidic actuation levels. Changes to the primary jet are characterized using particle image velocimetry (PIV). A simplified explanation is as follows: In this technique tracer particles are introduced into the flow and illuminated by a thin laser light sheet that is pulsed to freeze particle motion. Laser light scattered by the particles is recorded at two instances in time using a digital camera. Using a spatial cross correlation, the local displacement vec tor is obtained. The results indicate that there exists the potential for integrat ing multiple fluidic sub-elements into aircraft components for improved per formance.
Figure 4–66 is actually a combination vector plot and contour plot. Velocity vectors are superimposed on contour plots of velocity magnitude (speed). The red regions represent high speeds, and the blue regions represent low speeds.
References Raman, G., Packiarajan, S., Papadopoulos, G., Weissman, C., and Raghu, S., “Jet Thrust Vectoring Using a Miniature Fluidic Oscillator,” ASME FEDSM 2001-18057, 2001.
Raman, G., Raghu, S., and Bencic, T. J., “Cavity Resonance Suppression Using Miniature Fluidic Oscillators,” AIAA Paper 99-1900, 1999.
APPLICATION SPOTLIGHT ■ Fluidic Actuators FIGURE 4–66 Time-averaged mean velocity field of a fluidic actuator jet. Results are from 150 PIV realizations, overlaid on an image of the seeded flow. Every seventh and second velocity vector is shown in the horizontal and vertical directions, respectively. The color levels denote the magnitude of the velocity field. (a) No actuation; (b) single actuator operating at 3 psig; (c) single actuator operating at 9 psig.
Copyright © Ganesh G. Raman (a) (b) (c) cen96537_ch04_137-188.indd 173 14/01/17 2:31 pm 174 Fluid Kinematics Guest Author: Rui Ni, Penn State University As G. I. Taylor recognized in his early seminal work, transport of particu late matter and small chemical volatiles should naturally be addressed in the Lagrangian framework, in which all information is encoded in the trajectories of the particles. One example of this application is the study of how humans smell food. The sense of smell arises from stimulation of olfactory receptor cells in the nasal cavity by odorized airflow (Fig. 4–67a). Food volatiles released from the back of the mouth can be transported up to the nasal cavity by exhaled airflow—so called retronasal olfaction—a process that is known to be important for humans to distinguish the delicate differences in food flavors. For a long time, however, it was not understood how those volatiles can be carried by exhaled air toward the nasal cavity rather than by inhaled air into the lungs, given that there is no directional valve in the human airway to promote such transport bias. To address this question, a model of human airway obtained from CT images has been created using a three-dimensional printer (Fig. 4–68). Water, seeded with tiny tracer particles, flowing through the channel with a matched Reynolds number was used to mimic the airflow carrying food volatiles in the human air way. Over a thousand particles were tracked simultaneously over time. Twenty-five of them from the same time window in both flow directions at Reynolds numbers close to 900 are shown in Fig. 4–67b. After being released from the back of the oral cavity, the particles, on average, were transported in the mean flow direction. However, during inhalation (red trajectories), tracers were trapped in a small area near the back of the mouth, and as a result, their overall displace ment was much smaller than those transported during exhalation (blue trajecto ries). This trapping mechanism during inhalation prevents food volatiles from being transported deeper into the respiratory system. However, depending on the flow conditions and Re, the fluctuations of the flow may occasionally drive the tracers out of the cavity further down to the trachea.
References Ni, R., Michalski, M. H., Brown, E., Doan, N., Zinter, J., Ouellette, N. T., and Shepherd, G. M., Optimal directional volatile transport in retronasal olfaction. Proceedings of the National Academy of Sciences, 112(47), 14700–14704, 2015.
Intagliata, C., Eat slowly and breathe smoothly to enhance taste. Scientific American, 2015.
APPLICATION SPOTLIGHT ■ Smelling Food; the Human Airway FIGURE 4–67 (a) Diagram of the human head. The green dashed line marks the region of the nasal cavity, nasopharynx, oropharynx, and trachea that is the subject of this study. (b) Side view of the three-dimensional model. Blue and red lines represent the trajectories of food volatiles released from the back of the oral cavity and transported with exhaled (Re = 883) and inhaled (Re = 1008) airflow, respectively.
(a) (b) Nasal cavity Back of oral cavity FIGURE 4–68 Flow passage created by a three-dimensional printer. A transparent cover (not shown) is mounted on the test piece for optical access.
© Rui Ni cen96537_ch04_137-188.indd 174 14/01/17 2:31 pm 175 CHAPTER 4 REFERENCES AND SUGGESTED READING 1. R. J. Adrian. “Particle-Imaging Technique for Experi mental Fluid Mechanics,” Annual Reviews in Fluid Mechanics, 23, pp. 261–304, 1991.
2. J. M. Cimbala, H. Nagib, and A. Roshko. “Large Struc ture in the Far Wakes of Two-Dimensional Bluff Bodies,” Journal of Fluid Mechanics, 190, pp. 265–298, 1988.
3. R. J. Heinsohn and J. M. Cimbala. Indoor Air Quality Engineering. New York: Marcel-Dekker, 2003.
4. P. K. Kundu and I. M. Cohen. Fluid Mechanics. Ed. 5, London, England: Elsevier Inc., 2011.
5. W. Merzkirch. Flow Visualization, 2nd ed. Orlando, FL: Academic Press, 1987.
6. G. S. Settles. Schlieren and Shadowgraph Techniques: Visualizing Phenomena in Transparent Media. Heidelberg: Springer-Verlag, 2001.
7. M. Van Dyke. An Album of Fluid Motion. Stanford, CA: The Parabolic Press, 1982.
8. F. M. White. Viscous Fluid Flow, 3rd ed. New York: McGraw-Hill, 2005.
PROBLEMS Introductory Problems 4–1C What does the word kinematics mean? Explain what the study of fluid kinematics involves.
4–2C Briefly discuss the difference between derivative operators d and ∂. If the derivative ∂u/∂x appears in an equa tion, what does this imply about variable u?
4–3 Consider the following steady, two-dimensional veloc ity field: V ›= (u, 𝜐 ) = (0.66 + 2.1x) i ›+ (−2.7 −2.1y) j › Is there a stagnation point in this flow field? If so, where is it?
Answer: Yes; x = −0.314, y = −1.29 4–4 Consider the following steady, two-dimensional veloc ity field: V ›= (u, 𝜐 ) = (a2 −(b −cx)2) i ›+ (−2cby + 2c2xy) j › Is there a stagnation point in this flow field? If so, where is it?
4–5 A steady, two-dimensional velocity field is given by V ›= (u, 𝜐 ) = (−0.781 −3.25x) i ›+ (−3.54 + 3.25y) j › Calculate the location of the stagnation point.
4–6 Consider steady flow of water through an axisymmet ric garden hose nozzle (Fig. P4–6). Along the centerline of the nozzle, the water speed increases from uentrance to uexit as sketched. Measurements reveal that the centerline water speed increases parabolically through the nozzle. Write an equation for centerline speed u(x), based on the parameters given here, from x = 0 to x = L.
Dexit Dentrance uexit uentrance u(x) x = L x = 0 FIGURE P4–6 Lagrangian and Eulerian Descriptions 4–7C What is the Eulerian description of fluid motion? How does it differ from the Lagrangian description?
4–8C Is the Lagrangian method of fluid flow analysis more similar to study of a system or a control volume? Explain.
4–9C A stationary probe is placed in a fluid flow and mea sures pressure and temperature as functions of time at one location in the flow (Fig. P4–9C). Is this a Lagrangian or an Eulerian measurement? Explain.
Flow Probe FIGURE P4–9C 4–10C A tiny neutrally buoyant electronic pressure probe is released into the inlet pipe of a water pump and transmits 2000 Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and SI users can ignore them. Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software.
cen96537_ch04_137-188.indd 175 14/01/17 2:31 pm 176 Fluid Kinematics pressure readings per second as it passes through the pump. Is this a Lagrangian or an Eulerian measurement? Explain.
4–11C Define a steady flow field in the Eulerian reference frame. In such a steady flow, is it possible for a fluid particle to experience a nonzero acceleration?
4–12C Is the Eulerian method of fluid flow analysis more similar to study of a system or a control volume? Explain.
4–13C A weather balloon is launched into the atmosphere by meteorologists. When the balloon reaches an altitude where it is neutrally buoyant, it transmits information about weather conditions to monitoring stations on the ground (Fig. P4–13C). Is this a Lagrangian or an Eulerian measurement? Explain.
Helium-filled weather balloon Transmitting instrumentation FIGURE P4–13C 4–14C A Pitot-static probe can often be seen protruding from the underside of an airplane (Fig. P4–14C). As the airplane flies, the probe measures relative wind speed. Is this a Lagrangian or an Eulerian measurement? Explain.
Probe FIGURE P4–14C 4–15C List at least three other names for the material deriv ative, and write a brief explanation about why each name is appropriate.
4–16 Consider steady, incompressible, two-dimensional flow through a converging duct (Fig. P4–16). A simple approximate velocity field for this flow is V ›= (u, 𝜐 ) = (U0 + bx) i ›−by j › where U0 is the horizontal speed at x = 0. Note that this equation ignores viscous effects along the walls but is a rea sonable approximation throughout the majority of the flow field. Calculate the material acceleration for fluid particles passing through this duct. Give your answer in two ways: (1) as acceleration components ax and ay and (2) as accelera tion vector a →.
y x U0 FIGURE P4–16 4–17 Converging duct flow is modeled by the steady, two- dimensional velocity field of Prob. 4–16. The pressure field is given by P = P0 −ρ 2 [2U0 bx + b2(x2 + y2)] where P0 is the pressure at x = 0. Generate an expression for the rate of change of pressure following a fluid particle.
4–18 A steady, incompressible, two-dimensional velocity field is given by the following components in the xy-plane: u = 1.85 + 2.05x + 0.656y 𝜐 = 0.754 −2.18x −2.05y Calculate the acceleration field (find expressions for accel eration components ax and ay), and calculate the acceleration at the point (x, y) = (−1, 3). Answers: ax = 1.51, ay = 2.74 4–19 A steady, incompressible, two-dimensional velocity field is given by the following components in the xy-plane: u = 0.205 + 0.97x + 0.851y 𝜐 = −0.509 + 0.953x −0.97y cen96537_ch04_137-188.indd 176 14/01/17 2:31 pm 177 CHAPTER 4 Calculate the acceleration field (find expressions for accel eration components ax and ay) and calculate the acceleration at the point (x, y) = (2, 1.5).
4–20 For the velocity field of Prob. 4–6, calculate the fluid acceleration along the nozzle centerline as a function of x and the given parameters.
4–21 Consider steady flow of air through the diffuser por tion of a wind tunnel (Fig. P4–21). Along the centerline of the diffuser, the air speed decreases from uentrance to uexit as sketched. Measurements reveal that the centerline air speed decreases parabolically through the diffuser. Write an equa tion for centerline speed u(x), based on the parameters given here, from x = 0 to x = L.
Dexit Dentrance uentrance u(x) x = L x = 0 uexit FIGURE P4–21 4–22 For the velocity field of Prob. 4–21, calculate the fluid acceleration along the diffuser centerline as a function of x and the given parameters. For L = 1.56 m, uentrance = 22.6 m/s, and uexit = 17.5 m/s, calculate the acceleration at x = 0 and x = 1.0 m. Answers: 0, −96.4 m/s2 4–23 A steady, incompressible, two-dimensional (in the xy-plane) velocity field is given by V ›= (0.523 −1.88x + 3.94y) i ›+ (−2.44 + 1.26x + 1.88y) j › Calculate the acceleration at the point (x, y) = (−1.55, 2.07).
4–24 The velocity field for a flow is given by V ›= ui ›+ 𝜐 j ›+ wk › where u = 3x, 𝜐 = −2y, w = 2z. Find the streamline that will pass through the point (1, 1, 0).
Flow Patterns and Flow Visualization 4–25C What is the definition of a pathline? What do path lines indicate?
4–26C What is the definition of a timeline? How can time lines be produced in a water channel? Name an application where timelines are more useful than streaklines.
4–27C What is the definition of a streamline? What do streamlines indicate?
4–28C What is the definition of a streakline? How do streaklines differ from streamlines?
4–29C Consider the visualization of flow over a 15° delta wing in Fig. P4–29C. Are we seeing streamlines, streaklines, pathlines, or timelines? Explain.
FIGURE P4–29C Visualization of flow over a 15° delta wing at a 20° angle of attack at a Reynolds number of 20,000. The visualiza-tion is produced by colored fluid injected into water from ports on the underside of the wing. Courtesy of ONERA. Photo by Werlé.
4–30C Consider the visualization of ground vortex flow in Fig. P4–30C. Are we seeing streamlines, streaklines, path lines, or timelines? Explain.
FIGURE P4–30C Visualization of ground vortex flow. A high-speed round air jet impinges on the ground in the presence of a free-stream flow of air from left to right. (The ground is at the bottom of the picture.) The portion of the jet that travels upstream forms a recirculating flow known as a ground vortex. The visualization is produced by a smoke wire mounted vertically to the left of the field of view. Photo by John M. Cimbala.
4–31C Consider the visualization of flow over a sphere in Fig. P4–31C. Are we seeing streamlines, streaklines, path lines, or timelines? Explain.
cen96537_ch04_137-188.indd 177 14/01/17 2:31 pm 178 Fluid Kinematics FIGURE P4–31C Visualization of flow over a sphere at a Reynolds number of 15,000. The visualization is produced by a time expo-sure of air bubbles in water. Courtesy of ONERA. Photo by Werlé.
4–32C Consider the visualization of flow over a 12° cone in Fig. P4–32C. Are we seeing streamlines, streaklines, path lines, or timelines? Explain.
FIGURE P4–32C Visualization of flow over a 12° cone at a 16° angle of attack at a Reynolds number of 15,000. The visualization is produced by colored fluid injected into water from ports in the body. Courtesy of ONERA. Photo by Werlé.
4–33C Consider a cross-sectional slice through an array of heat exchanger tubes (Fig. P4–33C). For each desired piece of information, choose which kind of flow visualization plot (vector plot or contour plot) would be most appropriate, and explain why.
(a) The location of maximum fluid speed is to be visualized.
(b) Flow separation at the rear of the tubes is to be visualized.
(c) The temperature field throughout the plane is to be visualized.
(d ) The distribution of the vorticity component normal to the plane is to be visualized.
Out In FIGURE P4–33C 4–34 A bird is flying in a room with a velocity field of V › = (u, 𝜐, w) = 0.6x + 0.2t – 1.4 (m/s). The room is heated by a heat pump so that the temperature distribution at steady state is T (x, y, z) = 400 – 0.4y – 0.6z – 0.2 (5 – x)2 (°C). Calculate the temperature change that the bird feels after 10 seconds of flight, as it flies through x = 1 m.
4–35E Converging duct flow is modeled by the steady, two-dimensional velocity field of Prob. 4–16. For the case in which U0 = 3.56 ft/s and b = 7.66 s−1, plot several stream lines from x = 0 ft to 5 ft and y = −2 ft to 2 ft. Be sure to show the direction of the streamlines.
4–36 The velocity field of a flow is described by V ›= (4x) i ›+ (5y + 3) j ›+ (3t2)k ›. What is the pathline of a particle at a location (1 m, 2 m, 4 m) at time t = 1 s?
4–37 Consider the following steady, incompressible, two-dimensional velocity field: V ›= (u, 𝜐 ) = (4.35 + 0.656x) i ›+ (−1.22 −0.656y) j › Generate an analytical expression for the flow streamlines and draw several streamlines in the upper-right quadrant from x = 0 to 5 and y = 0 to 6.
4–38 Consider the steady, incompressible, two-dimensional velocity field of Prob. 4–37. Generate a velocity vector plot in the upper-right quadrant from x = 0 to 5 and y = 0 to 6.
4–39 Consider the steady, incompressible, two-dimensional velocity field of Prob. 4–37. Generate a vector plot of the acceleration field in the upper-right quadrant from x = 0 to 5 and y = 0 to 6.
4–40 A steady, incompressible, two-dimensional velocity field is given by V ›= (u, 𝜐 ) = (1 + 2.5x + y)i ›+ (−0.5 −3x −2.5y) j › where the x- and y-coordinates are in m and the magnitude of velocity is in m/s.
(a) Determine if there are any stagnation points in this flow field, and if so, where they are.
(b) Sketch velocity vectors at several locations in the upper-right quadrant for x = 0 m to 4 m and y = 0 m to 4 m; quali tatively describe the flow field.
cen96537_ch04_137-188.indd 178 14/01/17 2:31 pm 179 CHAPTER 4 4–41 Consider the steady, incompressible, two-dimensional velocity field of Prob. 4–40.
(a) Calculate the material acceleration at the point (x = 2 m, y = 3 m). Answers: ax = 8.50 m/s2, ay = 8.00 m/s2 (b) Sketch the material acceleration vectors at the same array of x- and y-values as in Prob. 4–40.
4–42 The velocity field for solid-body rotation in the r𝜃-plane (Fig. P4–42) is given by ur = 0 u𝜃= 𝜔 r where 𝜔 is the magnitude of the angular velocity (𝜔 → points in the z-direction). For the case with 𝜔 = 1.5 s−1, plot a contour plot of velocity magnitude (speed). Specifically, draw curves of constant speed V = 0.5, 1.0, 1.5, 2.0, and 2.5 m/s. Be sure to label these speeds on your plot.
uθ uθ = ωr r FIGURE P4–42 4–43 The velocity field for a line source in the r𝜃-plane (Fig. P4–43) is given by ur = m 2𝜋 r u𝜃= 0 where m is the line source strength. For the case with m/(2𝜋) = 1.5 m2/s, plot a contour plot of velocity magnitude (speed). Specifically, draw curves of constant speed V = 0.5, 1.0, 1.5, 2.0, and 2.5 m/s. Be sure to label these speeds on your plot.
y x ur = m 2πr θ r FIGURE P4–43 4–44 A very small circular cylinder of radius Ri is rotating at angular velocity 𝜔i inside a much larger concentric cylinder of radius Ro that is rotating at angular velocity 𝜔o. A liquid of den sity 𝜌 and viscosity 𝜇 is confined between the two cylinders, as in Fig. P4–44. Gravitational and end effects can be neglected (the flow is two-dimensional into the page). If 𝜔i = 𝜔o and a long time has passed, generate an expression for the tangential velocity profile, u𝜃 as a function of (at most) r, 𝜔, Ri, Ro, 𝜌, and 𝜇, where 𝜔 = 𝜔i = 𝜔o. Also, calculate the torque exerted by the fluid on the inner cylinder and on the outer cylinder.
Liquid: ρ, μ Inner cylinder Outer cylinder Ro Ri ωo ωi FIGURE P4–44 4–45 Consider the same two concentric cylinders of Prob. 4–44. This time, however, the inner cylinder is rotat ing, but the outer cylinder is stationary. In the limit, as the outer cylinder is very large compared to the inner cylinder (imagine the inner cylinder spinning very fast while its radius gets very small), what kind of flow does this approximate? Explain. After a long time has passed, generate an expression for the tangential velocity profile, namely u𝜃 as a function of (at most) r, 𝜔i, Ri, Ro, 𝜌, and 𝜇. Hint: Your answer may con tain an (unknown) constant, which can be obtained by speci fying a boundary condition at the inner cylinder surface.
4–46 The velocity field for a line vortex in the r𝜃-plane (Fig. P4–46) is given by ur = 0 u𝜃= K r where K is the line vortex strength. For the case with K = 1.5 m/s2, plot a contour plot of velocity magnitude (speed). Specifically, draw curves of constant speed V = 0.5, 1.0, 1.5, 2.0, and 2.5 m/s. Be sure to label these speeds on your plot.
cen96537_ch04_137-188.indd 179 14/01/17 2:31 pm 180 Fluid Kinematics uθ r uθ = K r FIGURE P4–46 4–47 Converging duct flow (Fig. P4–16) is modeled by the steady, two-dimensional velocity field of Prob. 4–16. Gener ate an analytical expression for the flow streamlines.
Answer: y = C/(U0 + bx) Motion and Deformation of Fluid Elements; Vorticity and Rotationality 4–48C Name and briefly describe the four fundamental types of motion or deformation of fluid particles.
4–49C Explain the relationship between vorticity and rota tionality.
4–50 Converging duct flow is modeled by the steady, two-dimensional velocity field of Prob. 4–16. Use the equation for volumetric strain rate to verify that this flow field is incompressible.
4–51 Converging duct flow is modeled by the steady, two-dimensional velocity field of Prob. 4–16. A fluid particle (A) is located on the x-axis at x = xA at time t = 0 (Fig. P4–51). At some later time t, the fluid particle has moved down stream with the flow to some new location x = xA′, as shown in the figure. Since the flow is symmetric about the x-axis, the fluid particle remains on the x-axis at all times. Generate an analytical expression for the x-location of the fluid particle at some arbitrary time t in terms of its initial location xA and constants U0 and b. In other words, develop an expression for xA′. (Hint: We know that u = dxparticle/dt following a fluid particle. Plug in u, separate variables, and integrate.) Fluid particle at some later time t Fluid particle at time t = 0 y A x Aʹ FIGURE P4–51 4–52 Converging duct flow is modeled by the steady, two-dimensional velocity field of Prob. 4–16. Since the flow is symmetric about the x-axis, line segment AB along the x-axis remains on the axis, but stretches from length 𝜉 to length 𝜉 + Δ𝜉 as it flows along the channel center line (Fig. P4–52). Generate an analytical expression for the change in length of the line segment, Δ𝜉. (Hint: Use the result of Prob. 4–51.) Answer: (xB − xA)(ebt − 1) y A B Bʹ Aʹ x ξ ξ + Δξ FIGURE P4–52 4–53 Using the results from Prob. 4–52 and the funda mental definition of linear strain rate (the rate of increase in length per unit length), develop an expression for the linear strain rate in the x-direction (𝜀xx) of fluid particles located on the centerline of the channel. Compare your result to the gen eral expression for 𝜀xx in terms of the velocity field, i.e., 𝜀xx = ∂u/∂x. (Hint: Take the limit as time t → 0. You may need to apply a truncated series expansion for ebt.) Answer: b 4–54 Converging duct flow is modeled by the steady, two-dimensional velocity field of Prob. 4–16. A fluid par ticle (A) is located at x = xA and y = yA at time t = 0 (Fig. P4–54). At some later time t, the fluid particle has moved downstream with the flow to some new location x = xA′, y = yA′, as shown in the figure. Generate an analytical expres sion for the y-location of the fluid particle at arbitrary time t in terms of its initial y-location yA and constant b. In other words, develop an expression for yA′. (Hint: We know that 𝜐 = dyparticle/dt following a fluid particle. Substitute the equa tion for 𝜐, separate variables, and integrate.) Answer: yAe−bt Fluid particle at some later time t Fluid particle at time t = 0 y A Aʹ x FIGURE P4–54 cen96537_ch04_137-188.indd 180 14/01/17 2:31 pm 181 CHAPTER 4 4–55E Converging duct flow is modeled by the steady, two-dimensional velocity field of Prob. 4–16. For the case in which U0 = 5.0 ft/s and b = 4.6 s−1, consider an initially square fluid particle of edge dimension 0.5 ft, centered at x = 0.5 ft and y = 1.0 ft at t = 0 (Fig. P4–55E). Carefully calculate and plot where the fluid particle will be and what it will look like at time t = 0.2 s later. Comment on the fluid particle’s distortion. (Hint: Use the results of Probs. 4–51 and 4–54.) Initially square fluid particle at t = 0 Unknown shape and location of fluid particle at later time t y x ?
FIGURE P4–55E 4–56E Based on the results of Prob. 4–55E, verify that this converging duct flow field is indeed incompressible.
4–57 Converging duct flow is modeled by the steady, two-dimensional velocity field of Prob. 4–16. As vertical line segment AB moves downstream it shrinks from length 𝜂 to length 𝜂 + Δ𝜂 as sketched in Fig. P4–57. Generate an analyti cal expression for the change in length of the line segment, Δ𝜂. Note that the change in length, Δ𝜂, is negative. (Hint: Use the result of Prob. 4–54.) y A Aʹ Bʹ B x η η + Δη FIGURE P4–57 4–58 Using the results of Prob. 4–57 and the fundamental definition of linear strain rate (the rate of increase in length per unit length), develop an expression for the linear strain rate in the y-direction (𝜀yy) of fluid particles moving down the channel. Compare your result to the general expression for 𝜀yy in terms of the velocity field, i.e., 𝜀yy = ∂𝜐/∂y. (Hint: Take the limit as time t → 0. You may need to apply a trun cated series expansion for e−bt.) 4–59 Converging duct flow (Fig. P4–16) is modeled by the steady, two-dimensional velocity field of Prob. 4–16. Is this flow field rotational or irrotational? Show all your work. Answer: irrotational 4–60 A general equation for a steady, two-dimensional velocity field that is linear in both spatial directions (x and y) is V ›= (u, 𝜐 ) = (U + a1x + b1y) i ›+ (V + a2x + b2y) j › where U and V and the coefficients are constants. Their dimen sions are assumed to be appropriately defined. Calculate the x- and y-components of the acceleration field.
4–61 For the velocity field of Prob. 4–60, what relationship must exist between the coefficients to ensure that the flow field is incompressible? Answer: a1 + b2 = 0 4–62 For the velocity field of Prob. 4–60, calculate the lin ear strain rates in the x- and y-directions. Answers: a1, b2 4–63 For the velocity field of Prob. 4–60, calculate the shear strain rate in the xy-plane.
4–64 Combine your results from Probs. 4–62 and 4–63 to form the two-dimensional strain rate tensor 𝜀ij in the xy-plane, 𝜀ij = ( 𝜀xx 𝜀xy 𝜀yx 𝜀yy ) Under what conditions would the x- and y-axes be principal axes? Answer: b1 + a2 = 0 4–65 For the velocity field of Prob. 4–60, calculate the vorticity vector. In which direction does the vorticity vector point? Answer: (a2 − b1)k → in z − direction 4–66 Consider steady, incompressible, two-dimensional shear flow for which the velocity field is V ›= (u, 𝜐 ) = (a + by) i ›+ 0 j › where a and b are constants. Sketched in Fig. P4–66 is a small rectangular fluid particle of dimensions dx and dy at time t. The fluid particle moves and deforms with the flow such that at a later time (t + dt), the particle is no longer rect angular, as also shown in the figure. The initial location of each corner of the fluid particle is labeled in Fig. P4–66. The lower-left corner is at (x, y) at time t, where the x-component of velocity is u = a + by. At the later time, this corner moves to (x + u dt, y), or (x + (a + by) dt, y) (a) In similar fashion, calculate the location of each of the other three corners of the fluid particle at time t + dt.
cen96537_ch04_137-188.indd 181 14/01/17 2:31 pm 182 Fluid Kinematics (b) From the fundamental definition of linear strain rate (the rate of increase in length per unit length), calculate linear strain rates 𝜀xx and 𝜀yy. Answers: 0, 0 (c) Compare your results with those obtained from the equa tions for 𝜀xx and 𝜀yy in Cartesian coordinates, i.e., 𝜀xx = ∂u ∂x 𝜀yy = ∂𝜐 ∂y Particle at time t Particle at time t + dt y x (x + dx, y + dy) u = a + by dy (x + dx, y) (x, y + dy) (x, y) dx dx dx dx FIGURE P4–66 4–67 Use two methods to verify that the flow of Prob. 4–66 is incompressible: (a) by calculating the volume of the fluid particle at both times, and (b) by calculating the volumetric strain rate. Note that Prob. 4–66 should be completed before this problem.
4–68 Consider the steady, incompressible, two-dimensional flow field of Prob. 4–66. Using the results of Prob. 4–66(a), do the following: (a) From the fundamental definition of shear strain rate (half of the rate of decrease of the angle between two initially perpendicular lines that intersect at a point), calculate shear strain rate 𝜀xy in the xy-plane. (Hint: Use the lower edge and the left edge of the fluid particle, which intersect at 90° at the lower-left corner of the particle at the initial time.) (b) Compare your results with those obtained from the equa tion for 𝜀xy in Cartesian coordinates, i.e., 𝜀xy = 1 2 ( ∂u ∂y + ∂𝜐 ∂x) Answers: (a) b/2, (b) b/2 4–69 Consider the steady, incompressible, two-dimensional flow field of Prob. 4–66. Using the results of Prob. 4–66(a), do the following: (a) From the fundamental definition of the rate of rotation (average rotation rate of two initially perpendicular lines that intersect at a point), calculate the rate of rotation of the fluid particle in the xy-plane, 𝜔z. (Hint: Use the lower edge and the left edge of the fluid particle, which intersect at 90° at the lower-left corner of the particle at the initial time.) (b) Compare your results with those obtained from the equa tion for 𝜔z in Cartesian coordinates, i.e., 𝜔 z = 1 2 ( ∂𝜐 ∂x −∂u ∂y) Answers: (a) −b/2, (b) −b/2 4–70 From the results of Prob. 4–69, (a) Is this flow rotational or irrotational?
(b) Calculate the z-component of vorticity for this flow field.
4–71 A two-dimensional fluid element of dimensions dx and dy translates and distorts as shown in Fig. P4–71 during the infinitesimal time period dt = t2 − t1. The velocity com ponents at point P at the initial time are u and 𝜐 in the x- and y-directions, respectively. Show that the magnitude of the rate of rotation (angular velocity) about point P in the xy-plane is 𝜔 z = 1 2 ( ∂𝜐 ∂x −∂u ∂y) y x Fluid element at time t2 Fluid element at time t1 Line a Line b Line b Line a Pʹ Bʹ B dy dx A u P υ Aʹ αb αa π/2 FIGURE P4–71 4–72 A two-dimensional fluid element of dimensions dx and dy translates and distorts as shown in Fig. P4–71 dur ing the infinitesimal time period dt = t2 − t1. The velocity components at point P at the initial time are u and 𝜐 in the x- and y-directions, respectively. Consider the line segment PA in Fig. P4–71, and show that the magnitude of the linear strain rate in the x-direction is 𝜀xx = ∂u ∂x 4–73 A two-dimensional fluid element of dimensions dx and dy translates and distorts as shown in Fig. P4–71 during the infinitesimal time period dt = t2 − t1. The velocity cen96537_ch04_137-188.indd 182 14/01/17 2:31 pm 183 CHAPTER 4 components at point P at the initial time are u and 𝜐 in the x- and y-directions, respectively. Show that the magnitude of the shear strain rate about point P in the xy-plane is 𝜀xy = 1 2 ( ∂u ∂y + ∂𝜐 ∂x) 4–74 A cylindrical tank of water rotates in solid-body rota tion, counterclockwise about its vertical axis (Fig. P4–74) at angular speed n . = 175 rpm. Calculate the vorticity of fluid particles in the tank. Answer: 36.7 k → rad/s n · Liquid Free surface z rrim r FIGURE P4–74 4–75 A cylindrical tank of water rotates about its verti cal axis (Fig. P4–74). A PIV system is used to measure the vorticity field of the flow. The measured value of vorticity in the z-direction is −54.5 rad/s and is constant to within ±0.5 percent everywhere that it is measured. Calculate the angular speed of rotation of the tank in rpm. Is the tank rotat ing clockwise or counterclockwise about the vertical axis?
4–76 A cylindrical tank of radius rrim = 0.354 m rotates about its vertical axis (Fig. P4–74). The tank is partially filled with oil. The speed of the rim is 3.61 m/s in the counterclockwise direction (looking from the top), and the tank has been spin ning long enough to be in solid-body rotation. For any fluid particle in the tank, calculate the magnitude of the component of vorticity in the vertical z-direction. Answer: 20.4 rad/s 4–77 Consider a two-dimensional, incompressible flow field in which an initially square fluid particle moves and deforms. The fluid particle dimension is a at time t and is aligned with the x- and y-axes as sketched in Fig. P4–77. At some later time, the particle is still aligned with the x- and y-axes, but has deformed into a rectangle of horizontal length 2a. What is the vertical length of the rectangular fluid par ticle at this later time?
y x a a FIGURE P4–77 4–78 Consider a two-dimensional, compressible flow field in which an initially square fluid particle moves and deforms. The fluid particle dimension is a at time t and is aligned with the x- and y-axes as sketched in Fig. P4–77. At some later time, the particle is still aligned with the x- and y-axes but has deformed into a rectangle of horizontal length 1.08a and verti cal length 0.903a. (The particle’s dimension in the z-direction does not change since the flow is two-dimensional.) By what percentage has the density of the fluid particle increased or decreased?
4–79 Consider fully developed Couette flow—flow between two infinite parallel plates separated by distance h, with the top plate moving and the bottom plate stationary as illustrated in Fig. P4–79. The flow is steady, incompressible, and two-dimensional in the xy-plane. The velocity field is given by V ›= (u, 𝜐 ) = V y h i ›+ 0 j › Is this flow rotational or irrotational? If it is rotational, cal culate the vorticity component in the z-direction. Do fluid particles in this flow rotate clockwise or counterclockwise?
Answers: yes, −V/h, clockwise x y h u = V y h V FIGURE P4–79 4–80 For the Couette flow of Fig. P4–79, calculate the lin ear strain rates in the x- and y-directions, and calculate the shear strain rate 𝜀xy.
4–81 Combine your results from Prob. 4–80 to form the two-dimensional strain rate tensor 𝜀ij, 𝜀ij = ( 𝜀xx 𝜀xy 𝜀yx 𝜀yy ) Are the x- and y-axes principal axes?
cen96537_ch04_137-188.indd 183 14/01/17 2:31 pm 184 Fluid Kinematics 4–82 Consider a steady, two-dimensional, incompress ible flow field in the xy-plane. The linear strain rate in the x-direction is 1.75 s−1. Calculate the linear strain rate in the y-direction.
4–83 A steady, three-dimensional velocity field is given by V ›= (u, 𝜐 , w) = (2.49 + 1.36x −0.867y) i › + (1.95x −1.36y) j ›+ (− 0.458xy)k › Calculate the vorticity vector as a function of space variables (x, y, z).
4–84 Consider the following steady, three-dimensional veloc ity field: V ›= (u, 𝜐 , w) = (2.5 + 2.0x −y) i ›+ (2.0x −2.0y) j ›+ (0.8xy)k › Calculate the vorticity vector as a function of space (x, y, z).
4–85 A steady, two-dimensional velocity field is given by V ›= (u, 𝜐 ) = (2.85 + 1.26x −0.896y) i › + (3.45x + cx −1.26y) j › Calculate constant c such that the flow field is irrotational.
4–86 A steady, three-dimensional velocity field is given by V ›= (0.657 + 1.73x + 0.948y + az) i › + (2.61 + cx + 1.91y + bz) j › + (−2.73x −3.66y −3.64z)k › Calculate constants a, b, and c such that the flow field is irrotational.
Reynolds Transport Theorem 4–87C Briefly explain the purpose of the Reynolds trans port theorem (RTT). Write the RTT for extensive property B as a “word equation,” explaining each term in your own words.
4–88C Briefly explain the similarities and differences between the material derivative and the Reynolds transport theorem.
4–89C True or false: For each statement, choose whether the statement is true or false and discuss your answer briefly.
(a) The Reynolds transport theorem is useful for transform ing conservation equations from their naturally occurring control volume forms to their system forms.
(b) The Reynolds transport theorem is applicable only to nondeforming control volumes.
(c) The Reynolds transport theorem can be applied to both steady and unsteady flow fields.
(d ) The Reynolds transport theorem can be applied to both scalar and vector quantities.
4–90 Consider the integral d dt∫ 2t t x−2dx. Solve it two ways: (a) Take the integral first and then the time derivative.
(b) Use Leibniz theorem. Compare your results.
4–91 Solve the integral d dt∫ 2t t xxdx as far as you are able.
4–92 Consider the general form of the Reynolds transport theorem (RTT) given by dBsys dt = d dt ∫CV ρb dV + ∫CS ρbV › r·n › dA where V › r is the velocity of the fluid relative to the control surface. Let Bsys be the mass m of a closed system of fluid particles. We know that for a system, dm/dt = 0 since no mass can enter or leave the system by definition. Use the given equation to derive the equation of conservation of mass for a control volume.
4–93 Consider the general form of the Reynolds transport theorem (RTT) as stated in Prob. 4–92. Let Bsys be the linear momentum mV › of a system of fluid particles. We know that for a system, Newton’s second law is ∑F ›= ma ›= mdV › dt = d dt (mV ›)sys Use the RTT and Newton’s second law to derive the linear momentum equation for a control volume.
4–94 Consider the general form of the Reynolds transport theorem (RTT) as stated in Prob. 4–92. Let Bsys be the angu lar momentum H › = r → × mV › of a system of fluid particles, where r → is the moment arm. We know that for a system, con servation of angular momentum is ∑M ›= d dt H › sys where Σ M › is the net moment applied to the system. Use the RTT and the above equation to derive the equation of conser vation of angular momentum for a control volume.
Review Problems 4–95 Consider a steady, two-dimensional flow field in the xy-plane whose x-component of velocity is given by u = a + b(x −c)2 where a, b, and c are constants with appropriate dimensions. Of what form does the y-component of velocity need to be in order for the flow field to be incompressible? In other words, generate an expression for 𝜐 as a function of x, y, and the constants of the given equation such that the flow is incom pressible. Answer: −2b(x − c)y + f(x) 4–96 In a steady, two-dimensional flow field in the xy-plane, the x-component of velocity is u = ax + by + cx2 cen96537_ch04_137-188.indd 184 14/01/17 2:31 pm 185 CHAPTER 4 where a, b, and c are constants with appropriate dimensions. Generate a general expression for velocity component 𝜐 such that the flow field is incompressible.
4–97 The velocity field of a flow is given by V ›= k(x2 −y2) i ›−2kxy j › where k is a constant. If the radius of curvature of a streamline is R = [1 + yʹ2]3/2/∣yʺ∣, deter mine the normal acceleration of a particle (which is normal to the streamline) passing through the position x = l, y = 2.
4–98 The velocity field for an incompressible flow is given as V ›= 5x2 i ›−20 xy j ›+ 100t k ›. Determine if this flow is steady. Also determine the velocity and acceleration of a par ticle at (l, 3, 3) at t = 0.2 s.
4–99 Consider fully developed two-dimensional Poiseuille flow—flow between two infinite parallel plates separated by distance h, with both the top plate and bottom plate station ary, and a forced pressure gradient dP/dx driving the flow as illustrated in Fig. P4–99. (dP/dx is constant and negative.) The flow is steady, incompressible, and two-dimensional in the xy-plane. The velocity components are given by x y u(y) h FIGURE P4–99 u = 1 2𝜇 dP dx (y2 −hy) 𝜐 = 0 where 𝜇 is the fluid’s viscosity. Is this flow rotational or irro tational? If it is rotational, calculate the vorticity component in the z-direction. Do fluid particles in this flow rotate clock wise or counterclockwise?
4–100 For the two-dimensional Poiseuille flow of Prob. 4–99, calculate the linear strain rates in the x- and y-directions, and calculate the shear strain rate 𝜀xy.
4–101 Combine your results from Prob. 4–100 to form the two-dimensional strain rate tensor 𝜀ij in the xy-plane, 𝜀ij = ( 𝜀xx 𝜀xy 𝜀yx 𝜀yy ) Are the x- and y-axes principal axes?
4–102 Consider the two-dimensional Poiseuille flow of Prob. 4–99. The fluid between the plates is water at 40°C. Let the gap height h = 1.6 mm and the pressure gradient dP/dx = −230 N/m3. Calculate and plot seven path lines from t = 0 to t = 10 s. The fluid particles are released at x = 0 and at y = 0.2, 0.4, 0.6, 0.8, 1.0, 1.2, and 1.4 mm.
4–103 Consider the two-dimensional Poiseuille flow of Prob. 4–99. The fluid between the plates is water at 40°C. Let the gap height h = 1.6 mm and the pressure gradi ent dP/dx = −230 N/m3. Calculate and plot seven streak lines generated from a dye rake that introduces dye streaks at x = 0 and at y = 0.2, 0.4, 0.6, 0.8, 1.0, 1.2, and 1.4 mm (Fig. P4–103). The dye is introduced from t = 0 to t = 10 s, and the streaklines are to be plotted at t = 10 s.
x y u(y) Dye rake h FIGURE P4–103 4–104 Repeat Prob. 4–103 except that the dye is intro duced from t = 0 to t = 10 s, and the streak lines are to be plotted at t = 12 s instead of 10 s.
4–105 Compare the results of Probs. 4–103 and 4–104 and comment about the linear strain rate in the x-direction.
4–106 Consider the two-dimensional Poiseuille flow of Prob. 4–99. The fluid between the plates is water at 40°C. Let the gap height h = 1.6 mm and the pressure gradient dP/dx = −230 N/m3. Imagine a hydrogen bubble wire stretched vertically through the channel at x = 0 (Fig. P4–106). The wire is pulsed on and off such that bubbles are produced periodically to create timelines. Five distinct timelines are gen erated at t = 0, 2.5, 5.0, 7.5, and 10.0 s. Calculate and plot what these five timelines look like at time t = 12.5 s.
x u(y) H2 wire h y FIGURE P4–106 4–107 Consider fully developed axisymmetric Poiseuille flow—flow in a round pipe of radius R (diameter D = 2R), with a forced pressure gradient dP/dx driving the flow as illustrated in Fig. P4–107. (dP/dx is constant and negative.) The flow is steady, incompressible, and axisymmetric about the x-axis. The velocity components are given by u = 1 4𝜇 dP dx (r2 −R2) ur = 0 u𝜃= 0 where 𝜇 is the fluid’s viscosity. Is this flow rotational or irro tational? If it is rotational, calculate the vorticity component in the circumferential (𝜃) direction and discuss the sign of the rotation.
u(r) D R θ r x FIGURE P4–107 cen96537_ch04_137-188.indd 185 14/01/17 2:31 pm 186 Fluid Kinematics 4–108 For the axisymmetric Poiseuille flow of Prob. 4–107, calculate the linear strain rates in the x- and r-directions, and calculate the shear strain rate 𝜀xr. The strain rate tensor in cylindrical coordinates (r, 𝜃, x) and (ur, u𝜃, ux), is 𝜀ij = ( 𝜀rr 𝜀𝜃r 𝜀xr 𝜀r𝜃 𝜀𝜃𝜃 𝜀x𝜃 𝜀rx 𝜀𝜃x 𝜀xx) = ( ∂ur ∂r 1 2 ( r ∂ ∂r ( u𝜃 r ) + 1 r ∂ur ∂𝜃) 1 2 ( ∂ur ∂x + ∂ux ∂r ) 1 2 (r ∂ ∂r ( u𝜃 r ) + 1 r ∂ur ∂𝜃) 1 r ∂u𝜃 ∂𝜃+ ur r 1 2 ( 1 r ∂ux ∂𝜃+ ∂u𝜃 ∂x ) 1 2 ( ∂ur ∂x + ∂ux ∂r ) 1 2 ( 1 r ∂ux ∂𝜃+ ∂u𝜃 ∂x ) ∂ux ∂x ) 4–109 Combine your results from Prob. 4–108 to form the axisymmetric strain rate tensor 𝜀ij, 𝜀ij = ( 𝜀rr 𝜀xr 𝜀rx 𝜀xx) Are the x- and r-axes principal axes?
4–110 We approximate the flow of air into a vacuum cleaner attachment by the following velocity components in the centerplane (the xy-plane): u = −V .
x 𝜋 L x2 + y2 + b2 x4 + 2x2y2 + 2x2b2 + y4 −2y2b2 + b4 and 𝜐 = −V .
y 𝜋 L x2 + y2 −b2 x4 + 2x2y2 + 2x2b2 + y4 −2y2b2 + b4 where b is the distance of the attachment above the floor, L is the length of the attachment, and V . is the volume flow rate of air being sucked up into the hose (Fig. P4–110). Deter mine the location of any stagnation point(s) in this flow field.
Answer: at the origin Floor y L z x b V FIGURE P4–110 4–111 Consider the vacuum cleaner of Prob. 4–110. For the case where b = 2.0 cm, L = 35 cm, and V . = 0.1098 m3/s, cre ate a velocity vector plot in the upper half of the xy-plane from x = −3 cm to 3 cm and from y = 0 cm to 2.5 cm. Draw as many vectors as you need to get a good feel of the flow field. Note: The velocity is infinite at the point (x, y) = (0, 2.0 cm), so do not attempt to draw a velocity vector at that point.
4–112 Consider the approximate velocity field given for the vacuum cleaner of Prob. 4–110. Calculate the flow speed along the floor. Dust particles on the floor are most likely to be sucked up by the vacuum cleaner at the location of maximum speed. Where is that location? Do you think the vacuum cleaner will do a good job at sucking up dust directly below the inlet (at the origin)? Why or why not?
4–113 There are numerous occasions in which a fairly uni form free-stream flow encounters a long circular cylinder aligned normal to the flow (Fig. P4–113). Examples include air flowing around a car antenna, wind blowing against a flag pole or telephone pole, wind hitting electrical wires, and ocean currents impinging on the submerged round beams that support oil platforms. In all these cases, the flow at the rear of the cylinder is separated and unsteady, and usually turbulent. However, the flow in the front half of the cylinder is much more steady and predictable. In fact, except for a very thin boundary layer near the cylinder surface, the flow field may be approximated by the following steady, two-dimensional velocity components in the xy- or r𝜃-plane: ur = V cos 𝜃(1 −a2 r2) u𝜃= −V sin 𝜃(1 + a2 r2) Is this flow field rotational or irrotational? Explain.
V y r = a r θ x FIGURE P4–113 4–114 Consider the flow field of Prob. 4–113 (flow over a circular cylinder). Consider only the front half of the flow (x < 0). There is one stagnation point in the front half of the flow field. Where is it? Give your answer in both cylindrical (r, 𝜃) coordinates and Cartesian (x, y) coordinates.
4–115 Consider the upstream half (x < 0) of the flow field of Prob. 4–113 (flow over a circular cyl inder). We introduce a parameter called the stream function 𝜓, which is constant along streamlines in two-dimensional flows such as the one being considered here (Fig. P4–115). The velocity field of Prob. 4–113 corresponds to a stream function given by 𝜓= V sin 𝜃(r −a2 r ) cen96537_ch04_137-188.indd 186 14/01/17 2:31 pm 187 CHAPTER 4 (a) Setting 𝜓 to a constant, generate an equation for a stream line. (Hint: Use the quadratic rule to solve for r as a function of 𝜃.) Streamlines y x ψ4 ψ3 ψ2 ψ1 FIGURE P4–115 (b) For the particular case in which V = 1.00 m/s and cylinder radius a = 10.0 cm, plot several streamlines in the upstream half of the flow (90° < 𝜃 < 270°). For consistency, plot in the range −0.4 m < x < 0 m, −0.2 m < y < 0.2 m, with stream function values evenly spaced between −0.16 m2/s and 0.16 m2/s.
4–116 Consider the flow field of Prob. 4–113 (flow over a circular cylinder). Calculate the two linear strain rates in the r𝜃-plane; i.e., calculate 𝜀rr and 𝜀𝜃𝜃. Discuss whether fluid line segments stretch (or shrink) in this flow field. (Hint: The strain rate tensor in cylindrical coordinates is given in Prob. 4–108.) 4–117 Based on your results of Prob. 4–116, discuss the compressibility (or incompressibility) of this flow. Answer: flow is incompressible 4–118 Consider the flow field of Prob. 4–113 (flow over a circular cylinder). Calculate 𝜀r𝜃, the shear strain rate in the r𝜃-plane. Discuss whether fluid particles in this flow deform with shear or not. (Hint: The strain rate tensor in cylindrical coordinates is given in Prob. 4–108.) 4–119 In a steady, two-dimensional flow field in the xy-plane, the x-component of velocity is u = ax + by + cx2 −dxy where a, b, c, and d are constants with appropriate dimen sions. Generate a general expression for velocity component 𝜐 such that the flow field is incompressible.
4–120 A steady, two-dimensional velocity field in the xy-plane is given by V › = (a + bx)i › + (c + dy)j › + 0k ›.
(a) What are the primary dimensions (m, L, t, T, . . .) of coef ficients a, b, c, and d? (b) What relationship between the coefficients is necessary in order for this flow to be incompressible? (c) What relationship between the coefficients is necessary in order for this flow to be irrotational? (d ) Write the strain rate tensor for this flow. (e) For the simplified case of d = −b, derive an equation for the streamlines of this flow, namely, y = function(x, a, b, c).
4–121 A velocity field is given by u = 5y2, v = 3x, w = 0. (Do not concern yourself with units in this problem.) (a) Is this flow steady or unsteady? Is it two- or three-dimensional? (b) At (x,y,z) = (3,2,–3), compute the velocity vector. (c) At (x,y,z) = (3,2,–3), compute the local (i.e., unsteady part) of the acceleration vector. (d ) At (x,y,z) = (3,2,–3), compute the convective (or advec tive) part of the acceleration vector. (e) At (x,y,z) = (3,2,–3), compute the (total) acceleration vector.
Fundamentals of Engineering (FE) Exam Problems 4–122 The actual path traveled by an individual fluid par ticle over some period is called a (a) Pathline (b) Streamtube (c) Streamline (d ) Streakline (e) Timeline 4–123 The locus of fluid particles that have passed sequen tially through a prescribed point in the flow is called a (a) Pathline (b) Streamtube (c) Streamline (d ) Streakline (e) Timeline 4–124 A curve that is everywhere tangent to the instanta neous local velocity vector is called a (a) Pathline (b) Streamtube (c) Streamline (d ) Streakline (e) Timeline 4–125 A steady, incompressible, two-dimensional velocity field is given by V ›= (u, 𝜐 ) = (2.5 −1.6x) i ›+ (0.7 + 1.6y) j › where the x- and y-coordinates are in meters and the magni tude of velocity is in m/s. The values of x and y at the stagna tion point, respectively, are (a) 0.9375 m, 0.375 m (b) 1.563 m, −0.4375 m (c) 2.5 m, 0.7 m (d ) 0.731 m, 1.236 m (e) −1.6 m, 0.8 m 4–126 Water is flowing in a 3-cm-diameter garden hose at a rate of 25 L/min. A 20-cm nozzle is attached to the hose which decreases the diameter to 1.2 cm. The magnitude of the acceleration of a fluid particle moving down the center line of the nozzle is (a) 9.81 m/s2 (b) 17.3 m/s2 (c) 28.6 m/s2 (d ) 33.1 m/s2 (e) 42.4 m/s2 4–127 A steady, incompressible, two-dimensional velocity field is given by V ›= (u, 𝜐 ) = (0.65 + 1.7x) i ›+ (1.3 −1.7y) j › where the x- and y-coordinates are in meters and the magni tude of velocity is in m/s. The y-component of the accelera tion vector ay is (a) l.7y (b) −l.7y (c) 2.89y − 2.21 (d ) 3.0x − 2.73 (e) 0.84y + 1.42 4–128 A steady, incompressible, two-dimensional velocity field is given by V ›= (u, 𝜐 ) = (2.5 −1.6x) i ›+ (0.7 + 1.6y) j › cen96537_ch04_137-188.indd 187 14/01/17 2:31 pm 188 Fluid Kinematics where the x- and y-coordinates are in meters and the magni tude of velocity is in m/s. The x- and y-component of mate rial acceleration ax and ay at the point (x = 1 m, y = 1 m), respectively, in m/s2, are (a) −1.44, 3.68 (b) −1.6, 1.5 (c) 3.1, −1.32 (d ) 2.56, −4 (e) −0.8, 1.6 4–129 A steady, incompressible, two-dimensional velocity field is given by V ›= (u, 𝜐 ) = (2.5 −1.6x) i ›+ (0.7 + 1.6y) j › where the x- and y-coordinates are in meters and the magni tude of velocity is in m/s. The x-component of the accelera tion vector ax is (a) 0.8y (b) −1.6x (c) 2.5x − 1.6 (d ) 2.56x − 4 (e) 2.56x + 0.8y 4–130 A steady, incompressible, two-dimensional velocity field is given by V ›= (u, 𝜐 ) = (0.65 + 1.7x) i ›+ (1.3 −1.7y) j › where the x- and y-coordinates are in meters and the magni tude of velocity is in m/s. The x- and y-component of mate rial acceleration ax and ay at the point (x = 0 m, y = 0 m), respectively, in m/s2, are (a) 0.37, −1.85 (b) −1.7, 1.7 (c) 1.105, −2.21 (d ) 1.7, −1.7 (e) 0.65, 1.3 4–131 A steady, incompressible, two-dimensional velocity field is given by V ›= (u, 𝜐 ) = (0.8 + 1.7x) i ›+ (1.5 −1.7y) j › where the x- and y-coordinates are in meters and the magnitude of velocity is in m/s. The x- and y-component of velocity u and 𝜐 at the point (x = 1 m, y = 2 m), respectively, in m/s, are (a) 0.54, −2.31 (b) −1.9, 0.75 (c) 0.598, −2.21 (d ) 2.5, −1.9 (e) 0.8, 1.5 4–132 An array of arrows indicating the magnitude and direction of a vector property at an instant in time is called a (a) Profiler plot (b) Vector plot (c) Contour plot (d ) Velocity plot (e) Time plot 4–133 Which one is not a fundamental type of motion or deformation an element may undergo in fluid mechanics?
(a) Rotation (b) Converging (c) Translation (d ) Linear strain (e) Shear strain 4–134 A steady, incompressible, two-dimensional velocity field is given by V ›= (u, 𝜐 ) = (2.5 −1.6 x) i ›+ (0.7 + 1.6y) j › where the x- and y-coordinates are in meters and the magnitude of velocity is in m/s. The linear strain rate in the x-direction in s−1 is (a) −1.6 (b) 0.8 (c) 1.6 (d ) 2.5 (e) −0.875 4-135 A steady, incompressible, two-dimensional velocity field is given by V ›= (u, 𝜐 ) = (2.5 −1.6 x) i ›+ (0.7 + 1.6y) j › where the x- and y-coordinates are in meters and the magni tude of velocity is in m/s. The shear strain rate in s−1 is (a) −1.6 (b) 1.6 (c) 2.5 (d ) 0.7 (e) 0 4–136 A steady, two-dimensional velocity field is given by V ›= (u, 𝜐 ) = (2.5 −1.6 x) i ›+ (0.7 + 0.8y) j › where the x- and y-coordinates are in meters and the magni tude of velocity is in m/s. The volumetric strain rate in s−1 is (a) 0 (b) 3.2 (c) −0.8 (d ) 0.8 (e) −1.6 4–137 If the vorticity in a region of the flow is zero, the flow is (a) Motionless (b) Incompressible (c) Compressible (d ) Irrotational (e) Rotational 4–138 The angular velocity of a fluid particle is 20 rad/s. The vorticity of this fluid particle is (a) 20 rad/s (b) 40 rad/s (c) 80 rad/s (d ) 10 rad/s (e) 5 rad/s 4–139 A steady, incompressible, two-dimensional velocity field is given by V ›= (u, 𝜐 ) = (0.75 + 1.2 x) i ›+ (2.25 −1.2y) j › where the x- and y-coordinates are in meters and the magni tude of velocity is in m/s. The vorticity of this flow is (a) 0 (b) 1.2yk › (c) −1.2yk › (d ) yk › (e) −1.2xyk › 4–140 A steady, incompressible, two-dimensional velocity field is given by V ›= (u, 𝜐 ) = (2xy + 1) i ›+ (−y2 −0.6) j › where the x- and y-coordinates are in meters and the magni tude of velocity is in m/s. The angular velocity of this flow is (a) 0 (b) −2yk › (c) 2yk › (d ) −2xk › (e) −xk › 4–141 A cart is moving at a constant absolute velocity V › cart = 3 km/h to the right. A high-speed jet of water at an absolute velocity of V › jet = 15 km/h to the right strikes the back of the car. The relative velocity of the water is (a) 0 km/h (b) 3 km/h (c) 12 km/h (d ) 15 km/h (e) 18 km/h cen96537_ch04_137-188.indd 188 14/01/17 2:32 pm 5 CHAPTER 189 BERNOULLI AND ENERGY EQUATIONS T his chapter deals with three equations commonly used in fluid mechan ics: the mass, Bernoulli, and energy equations. The mass equation is an expression of the conservation of mass principle. The Bernoulli equa tion is concerned with the conservation of kinetic, potential, and flow energies of a fluid stream and their conversion to each other in regions of flow where net viscous forces are negligible and where other restrictive conditions apply. The energy equation is a statement of the conservation of energy principle. In fluid mechanics, it is convenient to separate mechanical energy from thermal energy and to consider the conversion of mechanical energy to thermal energy as a result of frictional effects as mechanical energy loss. Then the energy equation becomes the mechanical energy balance. We start this chapter with an overview of conservation principles and the conservation of mass relation. This is followed by a discussion of various forms of mechanical energy and the efficiency of mechanical work devices such as pumps and turbines. Then we derive the Bernoulli equation by applying Newton’s second law to a fluid element along a streamline and demonstrate its use in a variety of applications. We continue with the devel opment of the energy equation in a form suitable for use in fluid mechanics and introduce the concept of head loss. Finally, we apply the energy equa tion to various engineering systems.
OBJECTIVES When you finish reading this chapter, you should be able to ■ ■ Apply the conservation of mass equation to balance the incom ing and outgoing flow rates in a flow system ■ ■ Recognize various forms of mechanical energy, and work with energy conversion efficiencies ■ ■ Understand the use and limita tions of the Bernoulli equation, and apply it to solve a variety of fluid flow problems ■ ■ Work with the energy equation expressed in terms of heads, and use it to determine turbine power output and pumping power requirements Wind turbine “farms” are being constructed all over the world to extract kinetic energy from the wind and convert it to electrical energy. The mass, energy, momentum, and angular momentum balances are utilized in the design of a wind turbine. The Bernoulli equation is also useful in the preliminary design stage.
© J. Luke/PhotoLink/Getty Images RF cen96537_ch05_189-248.indd 189 14/01/17 2:38 pm 190 BERNOULLI AND ENERGY EQUATIONS 5–1 ■ INTRODUCTION You are already familiar with numerous conservation laws such as the laws of conservation of mass, conservation of energy, and conservation of momentum. Historically, the conservation laws are first applied to a fixed quantity of matter called a closed system or just a system, and then extended to regions in space called control volumes. The conservation relations are also called balance equations since any conserved quantity must balance during a process. We now give a brief description of the conservation of mass and energy relations, and the linear momentum equation (Fig. 5–1).
Conservation of Mass The conservation of mass relation for a closed system undergoing a change is expressed as msys = constant or dmsys/dt = 0, which is the statement that the mass of the system remains constant during a process. For a control vol ume (CV), mass balance is expressed in rate form as Conservation of mass: m · in −m · out = dmCV dt (5–1) where m .
in and m .
out are the total rates of mass flow into and out of the con trol volume, respectively, and dmCV/dt is the rate of change of mass within the control volume boundaries. In fluid mechanics, the conservation of mass relation written for a differential control volume is usually called the conti nuity equation. Conservation of mass is discussed in Section 5–2.
The Linear Momentum Equation The product of the mass and the velocity of a body is called the linear momentum or just the momentum of the body, and the momentum of a rigid body of mass m moving with a velocity V › is mV ›. Newton’s second law states that the acceleration of a body is proportional to the net force acting on it and is inversely proportional to its mass, and that the rate of change of the momentum of a body is equal to the net force acting on the body. Therefore, the momentum of a system remains constant only when the net force acting on it is zero, and thus the momentum of such systems is conserved. This is known as the conservation of momentum principle. In fluid mechanics, Newton’s second law is usually referred to as the linear momentum equation, which is discussed in Chap. 6 together with the angular momentum equation.
Conservation of Energy Energy can be transferred to or from a closed system by heat or work, and the conservation of energy principle requires that the net energy transfer to or from a system during a process be equal to the change in the energy con tent of the system. Control volumes involve energy transfer via mass flow also, and the conservation of energy principle, also called the energy balance, is expressed as Conservation of energy: E · in −E · out = dECV dt (5–2) where E .
in and E .
out are the total rates of energy transfer into and out of the control volume, respectively, and dECV/dt is the rate of change of energy within the control volume boundaries. In fluid mechanics, we usually limit FIGURE 5–1 Many fluid flow devices such as this Pelton wheel hydraulic turbine are analyzed by applying the conservation of mass and energy principles, along with the linear momentum equation.
Courtesy of Hydro Tasmania, www.hydro.com.au. Used by permission.
cen96537_ch05_189-248.indd 190 14/01/17 2:38 pm 191 CHAPTER 5 our consideration to mechanical forms of energy only. Conservation of energy is discussed in Section 5–6.
5–2 ■ CONSERVATION OF MASS The conservation of mass principle is one of the most fundamental princi ples in nature. We are all familiar with this principle, and it is not difficult to understand. A person does not have to be a rocket scientist to figure out how much vinegar-and-oil dressing will be obtained by mixing 100 g of oil with 25 g of vinegar. Even chemical equations are balanced on the basis of the conservation of mass principle. When 16 kg of oxygen reacts with 2 kg of hydrogen, 18 kg of water is formed (Fig. 5–2). In an electrolysis process, the water separates back to 2 kg of hydrogen and 16 kg of oxygen. Technically, mass is not exactly conserved. It turns out that mass m and energy E can be converted to each other according to the well-known for mula proposed by Albert Einstein (1879–1955): E = mc2 (5–3) where c is the speed of light in a vacuum, which is c = 2.9979 × 108 m/s. This equation suggests that there is equivalence between mass and energy. All physical and chemical systems exhibit energy interactions with their sur roundings, but the amount of energy involved is equivalent to an extremely small mass compared to the system’s total mass. For example, when 1 kg of liquid water is formed from oxygen and hydrogen at normal atmospheric conditions, the amount of energy released is 15.8 MJ, which corresponds to a mass of only 1.76 × 10−10 kg. However, even in nuclear reactions, the mass equivalence of the amount of energy interacted is a very small fraction of the total mass involved. Therefore, in most engineering analyses, we con sider both mass and energy as conserved quantities. For closed systems, the conservation of mass principle is implicitly used by requiring that the mass of the system remain constant during a process. For control volumes, however, mass can cross the boundaries, and so we must keep track of the amount of mass entering and leaving the control volume.
Mass and Volume Flow Rates The amount of mass flowing through a cross section per unit time is called the mass flow rate and is denoted by m . . The dot over a symbol is used to indicate time rate of change. A fluid flows into or out of a control volume, usually through pipes or ducts. The differential mass flow rate of fluid flowing across a small area element dAc in a cross section of a pipe is proportional to dAc itself, the fluid density 𝜌, and the component of the flow velocity normal to dAc, which we denote as Vn, and is expressed as (Fig. 5–3) 𝛿 m · = ρVn dAc (5–4) Note that both 𝛿 and d are used to indicate differential quantities, but 𝛿 is typically used for quantities (such as heat, work, and mass transfer) that are path functions and have inexact differentials, while d is used for quantities 2 kg H2 + 16 kg O2 18 kg H2O FIGURE 5–2 Mass is conserved even during chemical reactions.
dAc Vn V n Control surface FIGURE 5–3 The normal velocity Vn for a surface is the component of velocity perpendicular to the surface.
cen96537_ch05_189-248.indd 191 14/01/17 2:38 pm 192 BERNOULLI AND ENERGY EQUATIONS (such as properties) that are point functions and have exact differentials. For flow through an annulus of inner radius r1 and outer radius r2, for example, ∫ 2 1 dAc = Ac2 −Ac1 = 𝜋(r 2 2 −r1 2) but ∫ 2 1 𝛿 m · = m · total (total mass flow rate through the annulus), not m .
2 − m .
1. For specified values of r1 and r2, the value of the integral of dAc is fixed (thus the names point function and exact differential), but this is not the case for the integral of 𝛿m . (thus the names path function and inexact differential). The mass flow rate through the entire cross-sectional area of a pipe or duct is obtained by integration: m · = ∫Ac 𝛿 m · = ∫Ac ρVn dAc (kg/s) (5–5) While Eq. 5–5 is always valid (in fact it is exact), it is not always practical for engineering analyses because of the integral. We would like instead to express mass flow rate in terms of average values over a cross section of the pipe. In a general compressible flow, both 𝜌 and Vn vary across the pipe. In many practical applications, however, the density is essentially uniform over the pipe cross section, and we can take 𝜌 outside the integral of Eq. 5–5. Velocity, however, is never uniform over a cross section of a pipe because of the no-slip condition at the walls. Rather, the velocity varies from zero at the walls to some maximum value at or near the centerline of the pipe. We define the average velocity Vavg as the average value of Vn across the entire cross section of the pipe (Fig. 5–4), Average velocity: Vavg = 1 Ac∫Ac Vn dAc (5–6) where Ac is the area of the cross section normal to the flow direction. Note that if the speed were Vavg all through the cross section, the mass flow rate would be identical to that obtained by integrating the actual velocity pro file. Thus for incompressible flow or even for compressible flow where 𝜌 is approximated as uniform across Ac, Eq. 5–5 becomes m · = ρVavg Ac (kg/s) (5–7) For compressible flow, we can think of 𝜌 as the bulk average density over the cross section, and then Eq. 5–7 can be used as a reasonable approximation. For simplicity, we drop the subscript on the average velocity. Unless otherwise stated, V denotes the average velocity in the flow direction. Also, Ac denotes the cross-sectional area normal to the flow direction. The volume of the fluid flowing through a cross section per unit time is called the volume flow rate V . (Fig. 5–5) and is given by V .
= ∫Ac Vn dAc = Vavg Ac = VAc (m3/s) (5–8) An early form of Eq. 5–8 was published in 1628 by the Italian monk Bene detto Castelli (circa 1577–1644). Note that many fluid mechanics textbooks use Q instead of V . for volume flow rate. We use V . to avoid confusion with heat transfer. The mass and volume flow rates are related by m · = ρV .
= V .
v (5–9) Vavg Cross section Ac V = VavgAc FIGURE 5–5 The volume flow rate is the volume of fluid flowing through a cross section per unit time.
Vavg FIGURE 5–4 Average velocity Vavg is defined as the average speed through a cross section.
cen96537_ch05_189-248.indd 192 14/01/17 2:38 pm 193 CHAPTER 5 where v is the specific volume. This relation is analogous to m = 𝜌V = V/v, which is the relation between the mass and the volume of a fluid in a container.
Conservation of Mass Principle The conservation of mass principle for a control volume can be expressed as: The net mass transfer to or from a control volume during a time interval Δt is equal to the net change (increase or decrease) of the total mass within the control volume during Δt. That is, ( Total mass entering the CV during Δt ) −( Total mass leaving the CV during Δt ) = ( Net change of mass within the CV during Δt ) or min −mout = ΔmCV (kg) (5–10) where ΔmCV = mfinal – minitial is the change in the mass of the control volume during the process (Fig. 5–6). It can also be expressed in rate form as m · in −m · out = dmCV/dt (kg/s) (5–11) where m .
in and m .
out are the total rates of mass flow into and out of the con trol volume, and dmCV/dt is the rate of change of mass within the control volume boundaries. Equations 5–10 and 5–11 are often referred to as the mass balance and are applicable to any control volume undergoing any kind of process. Consider a control volume of arbitrary shape, as shown in Fig. 5–7. The mass of a differential volume dV within the control volume is dm = 𝜌 dV. The total mass within the control volume at any instant in time t is deter mined by integration to be Total mass within the CV: mCV = ∫CV ρ dV (5–12) Then the time rate of change of the amount of mass within the control volume is expressed as Rate of change of mass within the CV: dmCV dt = d dt∫CV ρ dV (5–13) For the special case of no mass crossing the control surface (i.e., the con trol volume is a closed system), the conservation of mass principle reduces to dmCV/dt = 0. This relation is valid whether the control volume is fixed, moving, or deforming. Now consider mass flow into or out of the control volume through a dif ferential area dA on the control surface of a fixed control volume. Let n → be the outward unit vector of dA normal to dA and V › be the flow velocity at dA relative to a fixed coordinate system, as shown in Fig. 5–7. In general, the velocity may cross dA at an angle 𝜃 off the normal of dA, and the mass flow rate is proportional to the normal component of velocity V › n = V › cos 𝜃 ranging from a maximum outflow of V › for 𝜃 = 0 (flow is normal to dA) to a minimum of zero for 𝜃 = 90° (flow is tangent to dA) to a maximum inflow of V › for 𝜃 = 180° (flow is normal to dA but in the opposite direction). Control volume (CV) Control surface (CS) dV dm dA n V 휃 FIGURE 5–7 The differential control volume dV and the differential control surface dA used in the derivation of the conservation of mass relation.
Water Δmbathtub = min – mout = 20 kg min = 50 kg mout = 30 kg FIGURE 5–6 Conservation of mass principle for an ordinary bathtub.
cen96537_ch05_189-248.indd 193 14/01/17 2:39 pm 194 BERNOULLI AND ENERGY EQUATIONS Making use of the concept of dot product of two vectors, the magnitude of the normal component of velocity is Normal component of velocity: Vn = V cos 𝜃 = V › ·n › (5–14) The mass flow rate through dA is proportional to the fluid density 𝜌, normal velocity Vn, and the flow area dA, and is expressed as Differential mass flow rate: 𝛿 m · = ρVn dA = ρ(V cos 𝜃 ) dA = ρ(V › ·n › ) dA (5–15) The net flow rate into or out of the control volume through the entire con trol surface is obtained by integrating 𝛿m . over the entire control surface, Net mass flow rate: m · net = ∫ CS 𝛿 m · = ∫ CS ρV n dA = ∫ CS ρ(V › ·n › ) dA (5–16) Note that V n = V ›·n → = V cos 𝜃 is positive for 𝜃 < 90° (outflow) and nega tive for 𝜃 > 90° (inflow). Therefore, the direction of flow is automatically accounted for, and the surface integral in Eq. 5–16 directly gives the net mass flow rate. A positive value for m .
net indicates a net outflow of mass and a negative value indicates a net inflow of mass. Rearranging Eq. 5–11 as dmCV/dt + m .
out − m .
in = 0, the conservation of mass relation for a fixed control volume is then expressed as General conservation of mass: d dt ∫ CV ρ dV + ∫ CS ρ(V › ·n › ) dA = 0 (5–17) It states that the time rate of change of mass within the control volume plus the net mass flow rate through the control surface is equal to zero. The general conservation of mass relation for a control volume can also be derived using the Reynolds transport theorem (RTT) by taking the prop erty B to be the mass m (Chap. 4). Then we have b = 1 since dividing mass by mass to get the property per unit mass gives unity. Also, the mass of a closed system is constant, and thus its time derivative is zero. That is, dmsys/dt = 0. Then the Reynolds transport equation reduces immediately to Eq. 5–17, as shown in Fig. 5–8, and thus illustrates that the Reynolds trans port theorem is a very powerful tool indeed. Splitting the surface integral in Eq. 5–17 into two parts—one for the out going flow streams (positive) and one for the incoming flow streams (negative)—the general conservation of mass relation can also be expressed as d dt ∫ CV ρ dV + ∑ out ρ ∣Vn ∣A −∑ in ρ ∣Vn ∣A = 0 (5–18) where A represents the area for an inlet or outlet, and the summation signs are used to emphasize that all the inlets and outlets are to be considered. Using the definition of mass flow rate, Eq. 5–18 can also be expressed as d dt ∫ CV ρ dV = ∑ in m · −∑ out m · or dmCV dt = ∑ in m · −∑ out m · (5–19) There is considerable flexibility in the selection of a control volume when solving a problem. Many control volume choices are available, but some are more convenient to work with. A control volume should not introduce any unnecessary complications. A wise choice of a control volume can make the solution of a seemingly complicated problem rather easy. A simple rule in selecting a control volume is to make the control surface normal to the flow = + B = m b = 1 b = 1 dBsys dt V d dt CV ∫ 휌b( · n ) dA CS ∫ = + dmsys dt V d dt CV ∫ 휌( · n ) dA CS ∫ 휌dV 휌bdV FIGURE 5–8 The conservation of mass equation is obtained by replacing B in the Reynolds transport theorem by mass m, and b by 1 (m per unit mass = 1).
cen96537_ch05_189-248.indd 194 14/01/17 2:39 pm 195 CHAPTER 5 at all locations where it crosses the fluid flow, whenever possible. This way the dot product V ›·n → simply becomes the magnitude of the velocity, and the integral ∫ A ρ(V › ·n › ) dA becomes simply 𝜌VA (Fig. 5–9).
Moving or Deforming Control Volumes Equations 5–17 and 5–19 are also valid for moving control volumes pro vided that the absolute velocity V › is replaced by the relative velocity V › r, which is the fluid velocity relative to the control surface (Chap. 4). In the case of a moving but nondeforming control volume, relative velocity is the fluid velocity observed by a person moving with the control volume and is expressed as V › r = V › − V › CS, where V › is the fluid velocity and V › CS is the velocity of the control surface, both relative to a fixed point outside. Note that this is a vector subtraction. Some practical problems (such as the injection of medication through the needle of a syringe by the forced motion of the plunger) involve deforming control volumes. The conservation of mass relations developed can still be used for such deforming control volumes provided that the velocity of the fluid crossing a deforming part of the control surface is expressed relative to the control surface (that is, the fluid velocity should be expressed relative to a reference frame attached to the deforming part of the control surface). The relative velocity in this case at any point on the control surface is expressed again as V › r = V › − V › CS, where V › CS is the local velocity of the control surface at that point relative to a fixed point outside the control volume.
Mass Balance for Steady-Flow Processes During a steady-flow process, the total amount of mass contained within a control volume does not change with time (mCV = constant). Then the con servation of mass principle requires that the total amount of mass entering a control volume equal the total amount of mass leaving it. For a garden hose nozzle in steady operation, for example, the amount of water entering the nozzle per unit time is equal to the amount of water leaving it per unit time. When dealing with steady-flow processes, we are not interested in the amount of mass that flows in or out of a device over time; instead, we are interested in the amount of mass flowing per unit time, that is, the mass flow rate m . . The conservation of mass principle for a general steady-flow system with multiple inlets and outlets is expressed in rate form as (Fig. 5–10) Steady flow: ∑ in m · = ∑ out m · (kg/s) (5–20) It states that the total rate of mass entering a control volume is equal to the total rate of mass leaving it. Many engineering devices such as nozzles, diffusers, turbines, compres sors, and pumps involve a single stream (only one inlet and one outlet). For these cases, we typically denote the inlet state by the subscript 1 and the outlet state by the subscript 2, and drop the summation signs. Then Eq. 5–20 reduces, for single-stream steady-flow systems, to Steady flow (single stream): m · 1 = m · 2 → ρ1V1A1 = ρ2V2A2 (5–21) n A m = 휌V A V FIGURE 5–9 A control surface should always be selected normal to the flow at all locations where it crosses the fluid flow to avoid complications, even though the result is the same.
V 휃 n Vn = V cos 휃 A/cos 휃 A m = 휌(V cos 휃 )(A/cos 휃 ) = 휌VA (a) Control surface at an angle to the flow (b) Control surface normal to the flow m CV 1 = 2 kg/s m 2 = 3 kg/s m3 = m1 + m2 = 5 kg/s FIGURE 5–10 Conservation of mass principle for a two-inlet–one-outlet steady-flow system.
cen96537_ch05_189-248.indd 195 14/01/17 2:39 pm 196 BERNOULLI AND ENERGY EQUATIONS Special Case: Incompressible Flow The conservation of mass relations can be simplified even further when the fluid is incompressible, which is usually the case for liquids. Canceling the density from both sides of the general steady-flow relation gives Steady, incompressible flow: ∑ in V · = ∑ out V · (m3/s) (5–22) For single-stream steady-flow systems Eq. 5–22 becomes Steady, incompressible flow (single stream): V · 1 = V · 2 →V1A1 = V2A2 (5–23) It should always be kept in mind that there is no such thing as a “conserva tion of volume” principle. Therefore, the volume flow rates into and out of a steady-flow device may be different. The volume flow rate at the outlet of an air compressor is much less than that at the inlet even though the mass flow rate of air through the compressor is constant (Fig. 5–11). This is due to the higher density of air at the compressor exit. For steady flow of liq uids, however, the volume flow rates remain nearly constant since liquids are essentially incompressible (constant-density) substances. Water flow through the nozzle of a garden hose is an example of the latter case. The conservation of mass principle requires every bit of mass to be accounted for during a process. If you can balance your checkbook (by keeping track of deposits and withdrawals, or by simply observing the “con servation of money” principle), you should have no difficulty applying the conservation of mass principle to engineering systems.
m1 = 2 kg/s Air compressor m 2 = 2 kg/s V2 = 0.8 m3/s V1 = 1.4 m3/s FIGURE 5–11 During a steady-flow process, volume flow rates are not necessarily conserved although mass flow rates are.
FIGURE 5–12 Schematic for Example 5–1.
Photo by John M. Cimbala EXAMPLE 5–1 Water Flow through a Garden Hose Nozzle A garden hose attached with a nozzle is used to fill a 10-gal bucket. The inner diameter of the hose is 2 cm, and it reduces to 0.8 cm at the nozzle exit (Fig. 5–12). If it takes 50 s to fill the bucket with water, determine (a) the volume and mass flow rates of water through the hose, and (b) the average velocity of water at the nozzle exit.
SOLUTION A garden hose is used to fill a water bucket. The volume and mass flow rates of water and the exit velocity are to be determined.
Assumptions 1 Water is a nearly incompressible substance. 2 Flow through the hose is steady. 3 There is no waste of water by splashing.
Properties We take the density of water to be 1000 kg/m3 = 1 kg/L.
Analysis (a) Noting that 10 gal of water are discharged in 50 s, the volume and mass flow rates of water are V · = ΔV Δt = 10 gal 50 s ( 3.7854 L 1 gal ) = 0.757 L/s m · = ρV · = (1 kg/L)(0.757 L/s) = 0.757 kg/s (b) The cross-sectional area of the nozzle exit is Ae = 𝜋r 2 e = 𝜋(0.4 cm)2 = 0.5027 cm2 = 0.5027 × 10−4 m2 cen96537_ch05_189-248.indd 196 14/01/17 2:39 pm 197 CHAPTER 5 The volume flow rate through the hose and the nozzle is constant. Then the average velocity of water at the nozzle exit becomes Ve = V · Ae = 0.757 L/s 0.5027 × 10−4 m2 ( 1 m 3 1000 L) = 15.1 m/s Discussion It can be shown that the average velocity in the hose is 2.4 m/s. Therefore, the nozzle increases the water velocity by over six times.
EXAMPLE 5–2 Discharge of Water from a Tank A 4-ft-high, 3-ft-diameter cylindrical water tank whose top is open to the atmosphere is initially filled with water. Now the discharge plug near the bottom of the tank is pulled out, and a water jet whose diameter is 0.5 in streams out (Fig. 5–13). The average velocity of the jet is approximated as V = √2gh, where h is the height of water in the tank measured from the center of the hole (a variable) and g is the gravitational acceleration. Determine how long it takes for the water level in the tank to drop to 2 ft from the bottom.
SOLUTION The plug near the bottom of a water tank is pulled out. The time it takes for half of the water in the tank to empty is to be determined.
Assumptions 1 Water is a nearly incompressible substance. 2 The distance between the bottom of the tank and the center of the hole is negligible compared to the total water height. 3 The gravitational acceleration is 32.2 ft/s2.
Analysis We take the volume occupied by water as the control volume. The size of the control volume decreases in this case as the water level drops, and thus this is a variable control volume. (We could also treat this as a fixed control volume that consists of the interior volume of the tank by disregarding the air that replaces the space vacated by the water.) This is obviously an unsteady-flow problem since the properties (such as the amount of mass) within the control volume change with time. The conservation of mass relation for a control volume undergoing any process is given in rate form as m · in −m · out = dmCV dt (1) During this process no mass enters the control volume (m .
in = 0), and the mass flow rate of discharged water is m · out = (ρVA)out = ρ√2ghAjet (2) where Ajet = 𝜋D2 jet/4 is the cross-sectional area of the jet, which is constant. Noting that the density of water is constant, the mass of water in the tank at any time is mCV = ρV = ρAtankh (3) where Atank = 𝜋D2 tank/4 is the base area of the cylindrical tank. Substituting Eqs. 2 and 3 into the mass balance relation (Eq. 1) gives −ρ√2ghAjet = d(ρAtankh) dt →−ρ√2gh(𝜋D2 jet/4) = ρ(𝜋D2 tank/4)dh dt Water Air 0 Dtank Djet h2 h0 h FIGURE 5–13 Schematic for Example 5–2.
cen96537_ch05_189-248.indd 197 14/01/17 2:39 pm 198 BERNOULLI AND ENERGY EQUATIONS Canceling the densities and other common terms and separating the variables give dt = − D2 tank D2 jet dh √2gh Integrating from t = 0 at which h = h0 to t = t at which h = h2 gives ∫ t 0 dt = − D2 tank D2 jet√2g ∫ h2 h0 dh √h → t = √h0 −√h2 √g/2 ( Dtank Djet ) 2 Substituting, the time of discharge is determined to be t = √4 ft −√2 ft √32.2/2 ft/s2 ( 3 × 12 in 0.5 in ) 2 = 757 s = 12.6 min Therefore, it takes 12.6 min after the discharge hole is unplugged for half of the tank to be emptied.
Discussion Using the same relation with h2 = 0 gives t = 43.1 min for the discharge of the entire amount of water in the tank. Therefore, emptying the bottom half of the tank takes much longer than emptying the top half. This is due to the decrease in the average discharge velocity of water with decreasing h.
5–3 ■ MECHANICAL ENERGY AND EFFICIENCY Many fluid systems are designed to transport a fluid from one location to another at a specified flow rate, velocity, and elevation difference, and the system may generate mechanical work in a turbine or it may consume mechanical work in a pump or fan during this process (Fig. 5–14). These systems do not involve the conversion of nuclear, chemical, or thermal energy to mechanical energy. Also, they do not involve heat transfer in any significant amount, and they operate essentially at constant tempera ture. Such systems can be analyzed conveniently by considering only the mechanical forms of energy and the frictional effects that cause the mechan ical energy to be lost (i.e., to be converted to thermal energy that usually cannot be used for any useful purpose). The mechanical energy is defined as the form of energy that can be con verted to mechanical work completely and directly by an ideal mechanical device such as an ideal turbine. Kinetic and potential energies are the famil iar forms of mechanical energy. Thermal energy is not mechanical energy, however, since it cannot be converted to work directly and completely (the second law of thermodynamics). A pump transfers mechanical energy to a fluid by raising its pressure, and a turbine extracts mechanical energy from a fluid by dropping its pres sure. Therefore, the pressure of a flowing fluid is also associated with its mechanical energy. In fact, the pressure unit Pa is equivalent to Pa = N/m2 = N·m/m3 = J/m3, which is energy per unit volume, and the product Pv or its equivalent P/𝜌 has the unit J/kg, which is energy per unit mass. Note that pressure itself is not a form of energy; rather, it can be thought of as a measure of stored potential energy per unit volume. But a pressure force acting on a fluid through a distance produces work, called flow work, in the amount of P/𝜌 per unit mass. Flow work is expressed in terms of fluid FIGURE 5–14 Mechanical energy is a useful concept for flows that do not involve significant heat transfer or energy conversion, such as the flow of gasoline from an underground tank into a car.
© Corbis RF cen96537_ch05_189-248.indd 198 14/01/17 2:39 pm 199 CHAPTER 5 properties, and it is convenient to view it as part of the energy of a flowing fluid and call it flow energy. Therefore, the mechanical energy of a flowing fluid can be expressed on a unit-mass basis as emech = P ρ + V 2 2 + gz where P/𝜌 is the flow energy, V 2/2 is the kinetic energy, and gz is the poten tial energy of the fluid, all per unit mass. Then the mechanical energy change of a fluid during incompressible flow becomes Δemech = P2 −P1 ρ + V 2 2 −V 2 1 2 + g(z2 −z1) (kJ/kg) (5–24) Therefore, the mechanical energy of a fluid does not change during flow if its pressure, density, velocity, and elevation remain constant. In the absence of any irreversible losses, the mechanical energy change represents the mechanical work supplied to the fluid (if ∆emech > 0) or extracted from the fluid (if ∆emech < 0). The maximum (ideal) power generated by a turbine, for example, is W .
max = m . Δemech, as shown in Fig. 5–15. Consider a container of height h filled with water, as shown in Fig. 5–16, with the reference level selected at the bottom surface. The gage pressure and the potential energy per unit mass are, respectively, Pgage, A = 0 and peA = gh at point A at the free surface, and Pgage, B = 𝜌gh and peB = 0 at point B at the bottom of the container. An ideal hydraulic turbine at the bottom elevation would produce the same work per unit mass wturbine = gh whether it receives water (or any other fluid with constant density) from the top or from the bottom of the container. Note that we are assuming ideal flow (no irreversible losses) through the pipe leading from the tank to the turbine and negligible kinetic energy at the turbine outlet. Therefore, the total avail able mechanical energy of water at the bottom is equivalent to that at the top. The transfer of mechanical energy is usually accomplished by a rotating shaft, and thus mechanical work is often referred to as shaft work. A pump or a fan receives shaft work (usually from an electric motor) and transfers it to the fluid as mechanical energy (less frictional losses). A turbine, on the other hand, converts the mechanical energy of a fluid to shaft work. Because of irreversibilities such as friction, mechanical energy cannot be converted entirely from one mechanical form to another, and the mechanical efficiency of a device or process is defined as 𝜂mech = Mechanical energy output Mechanical energy input = Emech, out Emech, in = 1 − Emech, loss Emech, in (5–25) A conversion efficiency of less than 100 percent indicates that conver sion is less than perfect and some losses have occurred during conversion. A mechanical efficiency of 74 percent indicates that 26 percent of the mechanical energy input is converted to thermal energy as a result of fric tional heating (Fig. 5–17), and this manifests itself as a slight rise in the temperature of the fluid. In fluid systems, we are usually interested in increasing the pressure, veloc ity, and/or elevation of a fluid. This is done by supplying mechanical energy to the fluid by a pump, a fan, or a compressor (we refer to all of them as pumps). Or we are interested in the reverse process of extracting mechanical energy FIGURE 5–15 Mechanical energy is illustrated by an ideal hydraulic turbine coupled with an ideal generator. In the absence of irreversible losses, the maximum produced power is proportional to (a) the change in water surface elevation from the upstream to the downstream reservoir or (b) (close-up view) the drop in water pressure from just upstream to just downstream of the turbine.
Generator Turbine W 2 3 4 1 Generator Turbine h W W ˙ max = m ˙ ∆emech = m ˙ g(z1 − z4) = m ˙ gh since P 1 ≈ P 4 = P atm and V 1 = V 4 ≈ 0 (a) W ˙ max = m ˙ ∆emech = m ˙ P2 −P3 ρ = m ˙ ΔP ρ since V2 ≈ V3 and z2 ≈ z3 (b) cen96537_ch05_189-248.indd 199 14/01/17 2:39 pm 200 BERNOULLI AND ENERGY EQUATIONS from a fluid by a turbine and producing mechanical power in the form of a rotating shaft that can drive a generator or any other rotary device. The degree of perfection of the conversion process between the mechanical work supplied or extracted and the mechanical energy of the fluid is expressed by the pump efficiency and turbine efficiency. In rate form, these are defined as 𝜂pump = Mechanical power increase of the fluid Mechanical power input = ΔE .
mech, fluid W .
shaft, in = W .
pump, u W .
pump (5–26) where ∆E .
mech, fluid = E .
mech, out − E .
mech, in is the rate of increase in the mechani cal energy of the fluid, which is equivalent to the useful pumping power W .
pump, u supplied to the fluid, and 𝜂turbine = Mechanical power output Mechanical power decrease of the fluid = W .
shaft, out ∣ΔE .
mech, fluid∣ = W .
turbine W .
turbine, e (5–27) where │ΔE .
mech, fluid│ = E .
mech, in − E .
mech, out is the rate of decrease in the mechanical energy of the fluid, which is equivalent to the mechanical power extracted from the fluid by the turbine W .
turbine, e, and we use the absolute value sign to avoid negative values for efficiencies. A pump or turbine efficiency of 100 percent indicates perfect conversion between the shaft work and the mechanical energy of the fluid, and this value can be approached (but never attained) as the frictional effects are minimized. The mechanical efficiency should not be confused with the motor efficiency and the generator efficiency, which are defined as Motor: 𝜂motor = Mechanical power output Electric power input = W .
shaft, out W .
elect, in (5–28) and Generator: 𝜂generator = Electric power output Mechanical power input = W .
elect, out W .
shaft, in (5–29) A pump is usually packaged together with its motor, and a turbine with its generator. Therefore, we are usually interested in the combined or overall efficiency of pump–motor and turbine–generator combinations (Fig. 5–18), which are defined as 𝜂pump−motor = 𝜂pump 𝜂motor = W .
pump, u W .
elect, in = ΔE .
mech, fluid W .
elect, in (5–30) and 𝜂turbine−gen = 𝜂turbine 𝜂generator = W .
elect, out W .
turbine, e = W .
elect, out ∣ΔE .
mech, fluid∣ (5–31) All the efficiencies just defined range between 0 and 100 percent. The lower limit of 0 percent corresponds to the conversion of the entire mechanical or electric energy input to thermal energy, and the device in this case functions like a resistance heater. The upper limit of 100 percent corresponds to the case of perfect conversion with no friction or other irre versibilities, and thus no conversion of mechanical or electric energy to thermal energy (no losses).
m = 0.506 kg/s Fan 50.0 W 1 2 = = ≈ ≈ ≈ = 0.741 mech, fan = (0.506 kg/s)(12.1 m/s)2/2 50.0 W mV 2 2/2 Wshaft, in 0, V1 = 12.1 m/s = z2 z1 Patm P1 Patm and P2 V2 ∆Emech, fluid Wshaft, in η FIGURE 5–17 The mechanical efficiency of a fan is the ratio of the rate of increase of the mechanical energy of the air to the mechanical power input.
m 0 z h pe = gh Pgage = 0 Pgage = ρgh pe = 0 A B m Wmax = mgh Wmax = mgh FIGURE 5–16 The available mechanical energy of water at the bottom of a container is equal to the available mechanical energy at any depth including the free surface of the container.
cen96537_ch05_189-248.indd 200 14/01/17 2:39 pm 201 CHAPTER 5 EXAMPLE 5–3 Pumping Water from a Lake to a Storage Tank Water is pumped from a lake to a storage tank 18 m above at a rate of 70 L/s while consuming 20.4 kW of electric power (Fig. 5–19). The top of the tank is open to the atmosphere. Disregarding any frictional losses in the pipes and any changes in kinetic energy, determine (a) the overall efficiency of the pump–motor unit and (b) the pressure difference between the inlet and the exit of the pump.
SOLUTION Water is pumped from a lake to a storage tank at a specified rate. The overall efficiency of the pump–motor unit and the pressure difference between the inlet and the exit of the pump are to be determined.
Assumptions 1 The elevations of the tank and the lake remain constant. 2 Frictional losses in the pipes are negligible. 3 The changes in kinetic energy are negligible. 4 The elevation difference across the pump is negligible. Properties We take the density of water to be 𝜌 = 1000 kg/m3.
Analysis (a) We take the free surface of the lake to be point 1 and the free surfaces of the storage tank to be point 2. We also take the lake surface as the reference level (z1 = 0), and thus the potential energy at points 1 and 2 are pe1 = 0 and pe2 = gz2. The flow energy at both points is zero since both 1 and 2 are open to the atmosphere (P1 = P2 = Patm). Further, the kinetic energy at both points is zero (ke1 = ke2 = 0) since the water at both locations is essentially stationary. The mass flow rate of water and its potential energy at point 2 are m · = ρV · = (1000 kg/m3)(0.070 m3/s) = 70 kg/s pe2 = gz2 = (9.81 m/s2)(18 m)( 1 kJ/kg 1000 m2/s2) = 0.177 kJ/kg Then the rate of increase of the mechanical energy of water becomes ΔE · mech,fluid = m ·(emech,out −emech,in) = m ·(pe2 −0) = m ·pe2 = (70 kg/s)(0.177 kJ/kg) = 12.4 kW The overall efficiency of the combined pump–motor unit is determined from its definition, 𝜂pump−motor = ΔE .
mech,fluid W .
elect,in = 12.4 kW 20.4 kW = 0.606 or 60.6% (b) Now we consider the pump. The change in the mechanical energy of water as it flows through the pump consists of the change in the flow energy only since the elevation difference across the pump and the change in the kinetic energy are negligible. Also, this change must be equal to the useful mechanical energy supplied by the pump, which is 12.4 kW: ΔE · mech,fluid = m ·(emech,out −emech,in) = m · P2 −P1 ρ = V · ΔP Solving for ΔP and substituting, ΔP = ΔE · mech,fluid V · = 12.4 kJ/s 0.070 m3/s( 1 kPa·m3 1 kJ ) = 177 kPa Pump Storage tank 18 m 1 2 FIGURE 5–19 Schematic for Example 5–3.
Welect. out Turbine Generator ηturbine = 0.75 ηgenerator = 0.97 0.73 0.75 × 0.97 ηturbine–gen = = = ηturbine ηgenerator FIGURE 5–18 The overall efficiency of a turbine– generator is the product of the efficiency of the turbine and the efficiency of the generator, and represents the fraction of the mechanical power of the fluid converted to electrical power.
cen96537_ch05_189-248.indd 201 14/01/17 2:39 pm 202 BERNOULLI AND ENERGY EQUATIONS EXAMPLE 5–4 Conservation of Energy for an Oscillating Steel Ball The motion of a steel ball in a hemispherical bowl of radius h shown in Fig. 5–20 is to be analyzed. The ball is initially held at the highest location at point A, and then it is released. Obtain relations for the conservation of energy of the ball for the cases of frictionless and actual motions.
SOLUTION A steel ball is released in a bowl. Relations for the energy balance are to be obtained.
Assumptions For the frictionless case, friction between the ball, the bowl, and the air is negligible.
Analysis When the ball is released, it accelerates under the influence of gravity, reaches a maximum velocity (and minimum elevation) at point B at the bottom of the bowl, and moves up toward point C on the opposite side. In the ideal case of frictionless motion, the ball will oscillate between points A and C. The actual motion involves the conversion of the kinetic and potential energies of the ball to each other, together with overcoming resistance to motion due to friction (doing frictional work). The general energy balance for any system undergoing any process is Ein −Eout = ΔEsystem Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc., energies Then the energy balance (per unit mass) for the ball for a process from point 1 to point 2 becomes −wfriction = (ke2 + pe2) −(ke1 + pe1) or V 2 1 2 + gz1 = V 2 2 2 + gz2 + wfriction since there is no energy transfer by heat or mass and no change in the internal energy of the ball (the heat generated by frictional heating is dissipated to the surrounding air). The frictional work term wfriction is often expressed as eloss to represent the loss (conversion) of mechanical energy into thermal energy. For the idealized case of frictionless motion, the last relation reduces to V 2 1 2 + gz1 = V 2 2 2 + gz2 or V 2 2 + gz = C = constant Therefore, the pump must boost the pressure of water by 177 kPa in order to raise its elevation by 18 m. Discussion Note that only two-thirds of the electric energy consumed by the pump–motor is converted to the mechanical energy of water; the remaining one-third is wasted because of the inefficiencies of the pump and the motor.
Steel ball 0 z h A B C 1 2 FIGURE 5–20 Schematic for Example 5–4.
cen96537_ch05_189-248.indd 202 14/01/17 2:39 pm 203 CHAPTER 5 Most processes encountered in practice involve only certain forms of energy, and in such cases it is more convenient to work with the simplified versions of the energy balance. For systems that involve only mechanical forms of energy and its transfer as shaft work, the conservation of energy principle can be expressed conveniently as Emech, in −Emech, out = ΔEmech, system + Emech, loss (5–32) where Emech, loss represents the conversion of mechanical energy to ther mal energy due to irreversibilities such as friction. For a system in steady operation, the rate of mechanical energy balance becomes E .
mech, in = E .
mech, out + E .
mech, loss (Fig. 5–21).
5–4 ■ THE BERNOULLI EQUATION The Bernoulli equation is an approximate relation between pressure, velocity, and elevation, and is valid in regions of steady, incompressible flow where net frictional forces are negligible (Fig. 5–22). Despite its sim plicity, it has proven to be a very powerful tool in fluid mechanics. In this section, we derive the Bernoulli equation by applying the conservation of linear momentum principle, and we demonstrate both its usefulness and its limitations. The key approximation in the derivation of the Bernoulli equation is that viscous effects are negligibly small compared to inertial, gravitational, and pressure effects. Since all fluids have viscosity (there is no such thing as an “inviscid fluid”), this approximation cannot be valid for an entire flow field of practical interest. In other words, we cannot apply the Bernoulli equation everywhere in a flow, no matter how small the fluid’s viscosity. However, it turns out that the approximation is reasonable in certain regions of many practical flows. We refer to such regions as inviscid regions of flow, and we stress that they are not regions where the fluid itself is inviscid or friction less, but rather they are regions where net viscous or frictional forces are negligibly small compared to other forces acting on fluid particles. Care must be exercised when applying the Bernoulli equation since it is an approximation that applies only to inviscid regions of flow. In general, fric tional effects are always important very close to solid walls (boundary layers) and directly downstream of bodies (wakes). Thus, the Bernoulli approxima tion is typically useful in flow regions outside of boundary layers and wakes, where the fluid motion is governed by the combined effects of pressure and gravity forces.
where the value of the constant is C = gh. That is, when the frictional effects are negligible, the sum of the kinetic and potential energies of the ball remains constant.
Discussion This is certainly a more intuitive and convenient form of the conservation of energy equation for this and other similar processes such as the swinging motion of a pendulum. The relation obtained is analogous to the Bernoulli equation derived in Section 5–4.
Bernoulli equation valid Bernoulli equation not valid FIGURE 5–22 The Bernoulli equation is an approximate equation that is valid only in inviscid regions of flow where net viscous forces are negligibly small compared to inertial, gravitational, or pressure forces. Such regions occur outside of boundary layers and wakes.
z2 = z1 + h P1 = P2 = Patm Emech, in = Emech, out + Emech, loss Wpump + mgz1 = mgz2 + Emech, loss Wpump = mgh + Emech, loss h 2 Steady flow 1 V1 = V2 ≈ 0 Wpump FIGURE 5–21 Many fluid flow problems involve mechanical forms of energy only, and such problems are conveniently solved by using a rate of mechanical energy balance.
cen96537_ch05_189-248.indd 203 14/01/17 2:39 pm 204 BERNOULLI AND ENERGY EQUATIONS Acceleration of a Fluid Particle The motion of a particle and the path it follows are described by the velocity vector as a function of time and space coordinates and the initial position of the particle. When the flow is steady (no change with time at a speci fied location), all particles that pass through the same point follow the same path (which is the streamline), and the velocity vectors remain tangent to the path at every point. Often it is convenient to describe the motion of a particle in terms of its distance s along a streamline together with the radius of curvature along the streamline. The speed of the particle is related to the distance by V = ds/dt, which may vary along the streamline. In two-dimensional flow, the acceleration can be decomposed into two components: streamwise acceleration as along the streamline and normal acceleration an in the direction normal to the stream line, which is given as an = V 2/R. Note that streamwise acceleration is due to a change in speed along a streamline, and normal acceleration is due to a change in direction. For particles that move along a straight path, an = 0 since the radius of curvature is infinity and thus there is no change in direction. The Bernoulli equation results from a force balance along a streamline. One may be tempted to think that acceleration is zero in steady flow since acceleration is the rate of change of velocity with time, and in steady flow there is no change with time. Well, a garden hose nozzle tells us that this understanding is not correct. Even in steady flow and thus constant mass flow rate, water accelerates through the nozzle (Fig. 5–23 as discussed in Chap. 4). Steady simply means no change with time at a specified location, but the value of a quantity may change from one location to another. In the case of a nozzle, the velocity of water remains constant at a specified point, but it changes from the inlet to the exit (water accelerates along the nozzle). Mathematically, this can be expressed as follows: We take the velocity V of a fluid particle to be a function of s and t. Taking the total differential of V(s, t) and dividing both sides by dt yield dV = ∂V ∂s ds + ∂V ∂t dt and dV dt = ∂V ∂s ds dt + ∂V ∂t (5–33) In steady flow 𝜕V/𝜕t = 0 and thus V = V(s), and the acceleration in the s-direction becomes as = dV dt = ∂V ∂s ds dt = ∂V ∂s V = V dV ds (5–34) where V = ds/dt if we are following a fluid particle as it moves along a streamline. Therefore, acceleration in steady flow is due to the change of velocity with position.
Derivation of the Bernoulli Equation Consider the motion of a fluid particle in a flow field in steady flow. Apply ing Newton’s second law (which is referred to as the linear momentum equation in fluid mechanics) in the s-direction on a particle moving along a streamline gives ∑Fs = mas (5–35) FIGURE 5–23 During steady flow, a fluid may not accelerate in time at a fixed point, but it may accelerate in space.
cen96537_ch05_189-248.indd 204 14/01/17 2:39 pm 205 CHAPTER 5 In regions of flow where net frictional forces are negligible, there is no pump or turbine, and there is no heat transfer along the streamline, the significant forces acting in the s-direction are the pressure (acting on both sides) and the component of the weight of the particle in the s-direction (Fig. 5–24). There fore, Eq. 5–35 becomes P dA −(P + dP) dA −W sin 𝜃 = mV dV ds (5–36) where 𝜃 is the angle between the normal of the streamline and the vertical z-axis at that point, m = 𝜌V = 𝜌 dA ds is the mass, W = mg = 𝜌g dA ds is the weight of the fluid particle, and sin 𝜃 = dz/ds. Substituting, −dP dA −ρg dA ds dz ds = ρ dA ds V dV ds (5–37) Canceling dA from each term and simplifying, −dP −ρg dz = ρV dV (5–38) Noting that V dV = 1 2 d(V 2) and dividing each term by 𝜌 gives dP ρ + 1 2 d(V2) + g dz = 0 (5–39) Integrating, Steady flow: ∫ dP ρ + V 2 2 + gz = constant (along a streamline) (5–40) since the last two terms are exact differentials. In the case of incompressible flow, the first term also becomes an exact differential, and integration gives Steady, incompressible flow: P ρ + V2 2 + gz = constant (along a streamline) (5–41) This is the famous Bernoulli equation (Fig. 5–25), which is commonly used in fluid mechanics for steady, incompressible flow along a stream line in inviscid regions of flow. The Bernoulli equation was first stated in words by the Swiss mathematician Daniel Bernoulli (1700–1782) in a text written in 1738 when he was working in St. Petersburg, Russia. It was later derived in equation form by his associate Leonhard Euler (1707–1783) in 1755. The value of the constant in Eq. 5–41 can be evaluated at any point on the streamline where the pressure, density, velocity, and elevation are known. The Bernoulli equation can also be written between any two points on the same streamline as Steady, incompressible flow: P1 ρ + V 2 1 2 + gz1 = P2 ρ + V 2 2 2 + gz2 (5–42) We recognize V 2/2 as kinetic energy, gz as potential energy, and P/𝜌 as flow energy, all per unit mass. Therefore, the Bernoulli equation can be viewed z x W s n P dA (P + dP) dA Steady flow along a streamline dx dz ds 휃 휃 ds g FIGURE 5–24 The forces acting on a fluid particle along a streamline.
General: (For steady flow along a streamline) The Bernoulli Equation Incompressible flow (ρ = constant): ∫ + + + gz = constant dP 2 + gz = constant P 2 V2 V2 ρ ρ FIGURE 5–25 The incompressible Bernoulli equation is derived assuming incompressible flow, and thus it should not be used for flows with significant compressibility effects.
cen96537_ch05_189-248.indd 205 14/01/17 2:39 pm 206 BERNOULLI AND ENERGY EQUATIONS as an expression of mechanical energy balance and can be stated as follows (Fig. 5–26): The sum of the kinetic, potential, and flow energies of a fluid particle is constant along a streamline during steady flow when compressibility and frictional effects are negligible. The kinetic, potential, and flow energies are the mechanical forms of energy, as discussed in Section 5–3, and the Bernoulli equation can be viewed as the “conservation of mechanical energy principle.” This is equivalent to the general conservation of energy principle for systems that do not involve any conversion of mechanical energy and thermal energy to each other, and thus the mechanical energy and thermal energy are conserved separately. The Bernoulli equation states that during steady, incompressible flow with negli gible friction, the various forms of mechanical energy are converted to each other, but their sum remains constant. In other words, there is no dissipation of mechanical energy during such flows since there is no friction that con verts mechanical energy to sensible thermal (internal) energy. Recall that energy is transferred to a system as work when a force is applied to the system through a distance. In the light of Newton’s second law of motion, the Bernoulli equation can also be viewed as: The work done by the pressure and gravity forces on the fluid particle is equal to the increase in the kinetic energy of the particle. The Bernoulli equation is obtained from Newton’s second law for a fluid particle moving along a streamline. It can also be obtained from the first law of thermodynamics applied to a steady-flow system, as shown in Section 5–6. Despite the highly restrictive approximations used in its derivation, the Bernoulli equation is commonly used in practice since a variety of practical fluid flow problems can be analyzed to reasonable accuracy with it. This is because many flows of practical engineering interest are steady (or at least steady in the mean), compressibility effects are relatively small, and net frictional forces are negligible in some regions of interest in the flow.
Force Balance across Streamlines It is left as an exercise to show that a force balance in the direction n normal to the streamline yields the following relation applicable across the stream lines for steady, incompressible flow: P ρ + ∫ V 2 R dn + gz = constant (across streamlines) (5–43) where R is the local radius of curvature of the streamline. For flow along curved streamlines (Fig. 5–27a), the pressure decreases towards the center of curvature, and fluid particles experience a corresponding centripetal force and centripetal acceleration due to this pressure gradient. For flow along a straight line, R → ∞ and Eq. 5–43 reduces to P/𝜌 + gz = constant or P = −𝜌gz + constant, which is an expression for the variation of hydrostatic pressure with vertical distance for a stationary fluid body. There fore, the variation of pressure with elevation in steady, incompressible flow along a straight line in an inviscid region of flow is the same as that in the stationary fluid (Fig. 5–27b).
+ gz = constant Flow energy P 2 Potential energy Kinetic energy V2 ρ + FIGURE 5–26 The Bernoulli equation states that the sum of the kinetic, potential, and flow energies (all per unit mass) of a fluid particle is constant along a streamline during steady flow.
Stationary fluid A (a) (b) z z B C D PB – PA = PD – PC Flowing fluid A B PA > PB FIGURE 5–27 Pressure decreases towards the center of curvature when streamlines are curved (a), but the variation of pressure with elevation in steady, incompressible flow along a straight line (b) is the same as that in stationary fluid.
cen96537_ch05_189-248.indd 206 14/01/17 2:39 pm 207 CHAPTER 5 Unsteady, Compressible Flow Similarly, using both terms in the acceleration expression (Eq. 5–33), it can be shown that the Bernoulli equation for unsteady, compressible flow is Unsteady, compressible flow: ∫ dP ρ + ∫ ∂V ∂t ds + V 2 2 + gz = constant (5–44) Static, Dynamic, and Stagnation Pressures The Bernoulli equation states that the sum of the flow, kinetic, and poten tial energies of a fluid particle along a streamline is constant. Therefore, the kinetic and potential energies of the fluid can be converted to flow energy (and vice versa) during flow, causing the pressure to change. This phenom enon can be made more visible by multiplying the Bernoulli equation by the density 𝜌, P + ρ V 2 2 + ρgz = constant (along a streamline) (5–45) Each term in this equation has pressure units, and thus each term represents some kind of pressure: • P is the static pressure (it does not incorporate any dynamic effects); it represents the actual thermodynamic pressure of the fluid. This is the same as the pressure used in thermodynamics and property tables.
• 𝜌V 2/2 is the dynamic pressure; it represents the pressure rise when the fluid in motion is brought to a stop isentropically.
• 𝜌gz is the hydrostatic pressure term, which is not pressure in a real sense since its value depends on the reference level selected; it accounts for the elevation effects, i.e., fluid weight on pressure. (Be careful of the sign— unlike hydrostatic pressure 𝜌gh which increases with fluid depth h, the hydrostatic pressure term 𝜌gz decreases with fluid depth.) The sum of the static, dynamic, and hydrostatic pressures is called the total pressure. Therefore, the Bernoulli equation states that the total pressure along a streamline is constant. The sum of the static and dynamic pressures is called the stagnation pressure, and it is expressed as Pstag = P + ρ V 2 2 (kPa) (5–46) The stagnation pressure represents the pressure at a point where the fluid is brought to a complete stop isentropically. The static, dynamic, and stagna tion pressure heads are shown in Fig. 5–28, where head is an equivalent col umn height of the fluid. When static and stagnation pressures are measured at a specified location, the fluid velocity at that location is calculated from V = √ 2(P stag −P) ρ (5–47) Equation 5–47 is useful in the measurement of flow velocity when a combina tion of a static pressure tap and a Pitot tube is used, as illustrated in Fig. 5–28. A static pressure tap is simply a small hole drilled into a wall such that the Proportional to static pressure, P Proportional to stagnation pressure, Pstag Stagnation point 2(Pstag – P) Proportional to dynamic pressure Pitot tube Piezometer 2g V V = V2 ρ √ FIGURE 5–28 The static, dynamic, and stagnation pressure heads measured using piezometer tubes.
cen96537_ch05_189-248.indd 207 14/01/17 2:39 pm 208 BERNOULLI AND ENERGY EQUATIONS plane of the hole is parallel to the flow direction. It measures the static pres sure. A Pitot tube is a small tube with its open end aligned into the flow so as to sense the full impact pressure of the flowing fluid. It measures the stag nation pressure. In situations in which the static and stagnation pressure of a flowing liquid are greater than atmospheric pressure, a vertical transparent tube called a piezometer tube (or simply a piezometer) can be attached to the pres sure tap and to the Pitot tube, as sketched in Fig. 5–28. The liquid rises in the piezometer tube to a column height (head) that is proportional to the pressure being measured. If the pressures to be measured are below atmospheric, or if measuring pressures in gases, piezometer tubes do not work. However, the static pressure tap and Pitot tube can still be used, but they must be connected to some other kind of pressure measurement device such as a U-tube manometer or a pressure transducer (Chap. 3). Sometimes it is convenient to integrate static pressure holes on a Pitot probe. The result is a Pitot-static probe (also called a Pitot-Darcy probe), as shown in Fig. 5–29 and discussed in more detail in Chap. 8. A Pitot-static probe connected to a pressure transducer or a manometer measures the dynamic pressure (and thus infers the fluid velocity) directly. When the static pressure is measured by drilling a hole in the tube wall, care must be exercised to ensure that the opening of the hole is flush with the wall surface, with no extrusions before or after the hole (Fig. 5–30). Otherwise the reading would incorporate some dynamic effects, and thus it would be in error. When a stationary body is immersed in a flowing stream, the fluid is brought to a stop at the nose of the body (the stagnation point). The flow streamline that extends from far upstream to the stagnation point is called the stagnation streamline (Fig. 5–31). For a two-dimensional flow in the xy-plane, the stag nation point is actually a line parallel to the z-axis, and the stagnation stream line is actually a surface that separates fluid that flows over the body from fluid that flows under the body. In an incompressible flow, the fluid deceler ates nearly isentropically from its free-stream velocity to zero at the stagnation point, and the pressure at the stagnation point is thus the stagnation pressure.
Limitations on the Use of the Bernoulli Equation The Bernoulli equation (Eq. 5–41) is one of the most frequently used and misused equations in fluid mechanics. Its versatility, simplicity, and ease of use make it a very valuable tool for use in analysis, but the same attributes also make it very tempting to misuse. Therefore, it is important to under stand the restrictions on its applicability and observe the limitations on its use, as explained here: 1. Steady flow The first limitation on the Bernoulli equation is that it is applicable to steady flow. Therefore, it should not be used during the transient start-up and shut-down periods, or during periods of change in the flow conditions. Note that there is an unsteady form of the Bernoulli equation (Eq. 5–44), discussion of which is beyond the scope of the present text (see Panton, 2005).
2. Negligible viscous effects Every flow involves some friction, no matter how small, and frictional effects may or may not be negligible. The situation is complicated even more by the amount of error that can be tolerated. In general, frictional effects are negligible for short flow sections with large cross sections, especially at low flow velocities. Stagnation pressure hole Static pressure holes FIGURE 5–29 Close-up of a Pitot-static probe, showing the stagnation pressure hole and two of the five static circumferential pressure holes.
Photo by Po-Ya Abel Chuang. Used by permission.
High Correct Low FIGURE 5–30 Careless drilling of the static pressure tap may result in an erroneous reading of the static pressure head.
Stagnation streamline FIGURE 5–31 Streaklines produced by colored fluid introduced upstream of an airfoil; since the flow is steady, the streaklines are the same as streamlines and pathlines. The stagnation streamline is marked.
Courtesy of ONERA. Photo by Werlé.
cen96537_ch05_189-248.indd 208 14/01/17 2:39 pm 209 CHAPTER 5 Frictional effects are usually significant in long and narrow flow pas sages, in the wake region downstream of an object, and in diverging flow sections such as diffusers because of the increased possibility of the fluid separating from the walls in such geometries. Frictional effects are also significant near solid surfaces, and thus the Bernoulli equation is usually applicable along a streamline in the core region of the flow, but not along a streamline close to the surface (Fig. 5–32). A component that disturbs the streamlined structure of flow and thus causes considerable mixing and backflow such as a sharp entrance of a tube or a partially closed valve in a flow section can make the Ber noulli equation inapplicable.
3. No shaft work The Bernoulli equation was derived from a force balance on a particle moving along a streamline. Therefore, the Bernoulli equation is not applicable in a flow section that involves a pump, turbine, fan, or any other machine or impeller since such devices disrupt the streamlines and carry out energy interactions with the fluid particles. When the flow section considered involves any of these devices, the energy equation should be used instead to account for the shaft work input or output. However, the Bernoulli equation can still be applied to a flow section prior to or past a machine (assuming, of course, that the other restrictions on its use are satisfied). In such cases, the Bernoulli constant changes from upstream to downstream of the device.
4. Incompressible flow One of the approximations used in the derivation of the Bernoulli equation is that 𝜌 = constant and thus the flow is in compressible. This condition is satisfied by liquids and also by gases at Mach numbers less than about 0.3 since compressibility effects and thus density variations of gases are negligible at such relatively low veloci ties. Note that there is a compressible form of the Bernoulli equation (Eqs. 5–40 and 5–44).
5. Negligible heat transfer The density of a gas is inversely proportional to temperature, and thus the Bernoulli equation should not be used for flow sections that involve significant temperature change such as heating or cooling sections.
6. Flow along a streamline Strictly speaking, the Bernoulli equation P/𝜌 + V2/2 + gz = C is applicable along a streamline, and the value of the constant C is generally different for different streamlines. However, when a region of the flow is irrotational and there is no vorticity in the flow field, the value of the constant C remains the same for all streamlines, and the Bernoulli equation becomes applicable across streamlines as well (Fig. 5–33). Therefore, we do not need to be concerned about the streamlines when the flow is irrotational, and we can apply the Bernoulli equation between any two points in the irrotational region of the flow (Chap. 10). We derived the Bernoulli equation by considering two-dimensional flow in the xz-plane for simplicity, but the equation is valid for general three-dimensional flow as well, as long as it is applied along the same streamline. We should always keep in mind the approximations used in the derivation of the Bernoulli equation and make sure that they are valid before applying it.
A fan A sudden expansion A long narrow tube A heating section Flow through a valve 2 2 2 2 1 1 1 1 1 2 A boundary layer A wake FIGURE 5–32 Frictional effects, heat transfer, and components that disturb the streamlined structure of flow make the Bernoulli equation invalid. It should not be used in any of the flows shown here.
cen96537_ch05_189-248.indd 209 14/01/17 2:39 pm 210 BERNOULLI AND ENERGY EQUATIONS Hydraulic Grade Line (HGL) and Energy Grade Line (EGL) It is often convenient to represent the level of mechanical energy graphically using heights to facilitate visualization of the various terms of the Bernoulli equation. This is done by dividing each term of the Bernoulli equation by g to give P ρg + V 2 2g + z = H = constant (along a streamline) (5–48) Each term in this equation has the dimension of length and represents some kind of “head” of a flowing fluid as follows: • P/𝜌g is the pressure head; it represents the height of a fluid column that produces the static pressure P.
• V 2/2g is the velocity head; it represents the elevation needed for a fluid to reach the velocity V during frictionless free fall.
• z is the elevation head; it represents the potential energy of the fluid.
Also, H is the total head for the flow. Therefore, the Bernoulli equation is expressed in terms of heads as: The sum of the pressure, velocity, and elevation heads along a streamline is constant during steady flow when compressibility and frictional effects are negligible (Fig. 5–34). If a piezometer (which measures static pressure) is tapped into a pressur ized pipe, as shown in Fig. 5–35, the liquid would rise to a height of P/𝜌g above the pipe center. The hydraulic grade line (HGL) is obtained by doing this at several locations along the pipe and drawing a curve through the liq uid levels in the piezometers. The vertical distance above the pipe center is a measure of pressure within the pipe. Similarly, if a Pitot tube (measures static + dynamic pressure) is tapped into a pipe, the liquid would rise to a height of P/𝜌g + V 2/2g above the pipe center, or a distance of V 2/2g above the HGL. The energy grade line (EGL) is obtained by doing this at several locations along the pipe and drawing a curve through the liquid levels in the Pitot tubes. Noting that the fluid also has elevation head z (unless the reference level is taken to be the centerline of the pipe), the HGL and EGL are defined as fol lows: The line that represents the sum of the static pressure and the elevation heads, P/𝜌g + z, is called the hydraulic grade line. The line that represents + + z = H = constant Pressure head P 2g Elevation head Velocity head Total head V2 ρg FIGURE 5–34 An alternative form of the Bernoulli equation is expressed in terms of heads as: The sum of the pressure, velocity, and elevation heads is constant along a streamline.
Streamlines 1 2 + + gz1 = P1 2 + + gz2 P2 ρ ρ 1 V 2 2 2 V 2 FIGURE 5–33 When the flow is irrotational, the Bernoulli equation becomes applicable between any two points along the flow (not just on the same streamline).
FIGURE 5–35 The hydraulic grade line (HGL) and the energy grade line (EGL) for free discharge from a reservoir through a horizontal pipe with a diffuser.
/2g 2 V 2 /2g 1 V 2 Diffuser Arbitrary reference plane (z = 0) HGL EGL 1 z 0 2 3 cen96537_ch05_189-248.indd 210 14/01/17 2:39 pm 211 CHAPTER 5 the total head of the fluid, P/𝜌g + V 2/2g + z, is called the energy grade line. The difference between the heights of EGL and HGL is equal to the dynamic head, V 2/2g. We note the following about the HGL and EGL: • For stationary bodies such as reservoirs or lakes, the EGL and HGL coin cide with the free surface of the liquid. The elevation of the free surface z in such cases represents both the EGL and the HGL since the velocity is zero and the static (gage) pressure is zero.
• The EGL is always a distance V 2/2g above the HGL. These two curves approach each other as the velocity decreases, and they diverge as the velocity increases. The height of the HGL decreases as the velocity in creases, and vice versa.
• In an idealized Bernoulli-type flow, EGL is horizontal and its height remains constant. This would also be the case for HGL when the flow velocity is constant (Fig. 5–36).
• For open-channel flow, the HGL coincides with the free surface of the liq uid, and the EGL is a distance V 2/2g above the free surface.
• At a pipe exit, the pressure head is zero (atmospheric pressure) and thus the HGL coincides with the pipe outlet (location 3 on Fig. 5–35).
• The mechanical energy loss due to frictional effects (conversion to thermal energy) causes the EGL and HGL to slope downward in the direction of flow. The slope is a measure of the head loss in the pipe (discussed in detail in Chap. 8). A component that generates significant frictional effects such as a valve causes a sudden drop in both EGL and HGL at that location.
• A steep jump occurs in EGL and HGL whenever mechanical energy is added to the fluid (by a pump, for example). Likewise, a steep drop occurs in EGL and HGL whenever mechanical energy is removed from the fluid (by a turbine, for example), as shown in Fig. 5–37.
• The gage pressure of a fluid is zero at locations where the HGL intersects the fluid. The pressure in a flow section that lies above the HGL is negative, and the pressure in a section that lies below the HGL is positive (Fig. 5–38). Therefore, an accurate drawing of a piping system overlaid with the HGL can be used to determine the regions where the gage pressure in the pipe is negative (below atmospheric pressure). The last remark enables us to avoid situations in which the pressure drops below the vapor pressure of the liquid (which may cause cavitation, as dis cussed in Chap. 2). Proper consideration is necessary in the placement of a liquid pump to ensure that the suction side pressure does not fall too low, especially at elevated temperatures where vapor pressure is higher than it is at low temperatures. Now we examine Fig. 5–35 more closely. At point 0 (at the liquid surface), EGL and HGL are even with the liquid surface since there is no flow there. HGL decreases rapidly as the liquid accelerates into the pipe; however, EGL decreases very slowly through the well-rounded pipe inlet. EGL declines con tinually along the flow direction due to friction and other irreversible losses in the flow. EGL cannot increase in the flow direction unless energy is sup plied to the fluid. HGL can rise or fall in the flow direction, but can never exceed EGL. HGL rises in the diffuser section as the velocity decreases, and the static pressure recovers somewhat; the total pressure does not recover, Reference level 0 (Horizontal) EGL z HGL P g V2/2g ρ FIGURE 5–36 In an idealized Bernoulli-type flow, EGL is horizontal and its height remains constant. But this is not the case for HGL when the flow velocity varies along the flow.
Pump Turbine EGL HGL Wpump Wturbine FIGURE 5–37 A steep jump occurs in EGL and HGL whenever mechanical energy is added to the fluid by a pump, and a steep drop occurs whenever mechanical energy is removed from the fluid by a turbine.
Negative P P = 0 P = 0 HGL Positive P Positive P FIGURE 5–38 The gage pressure of a fluid is zero at locations where the HGL intersects the fluid, and the gage pressure is negative (vacuum) in a flow section that lies above the HGL.
cen96537_ch05_189-248.indd 211 14/01/17 2:39 pm 212 BERNOULLI AND ENERGY EQUATIONS however, and EGL decreases through the diffuser. The difference between EGL and HGL is V 2 1 /2g at point 1, and V 2 2 /2g at point 2. Since V1 > V2, the difference between the two grade lines is larger at point 1 than at point 2. The downward slope of both grade lines is larger for the smaller diameter section of pipe since the frictional head loss is greater. Finally, HGL decays to the liquid surface at the outlet since the pressure there is atmospheric. However, EGL is still higher than HGL by the amount V 2 2 /2g since V3 = V2 at the outlet.
Applications of the Bernoulli Equation So far, we have discussed the fundamental aspects of the Bernoulli equa tion. Now, we demonstrate its use in a wide range of applications through examples.
EXAMPLE 5–5 Spraying Water into the Air Water is flowing from a garden hose (Fig. 5–39). A child places his thumb to cover most of the hose outlet, causing a thin jet of high-speed water to emerge. The pressure in the hose just upstream of his thumb is 400 kPa. If the hose is held upward, what is the maximum height that the jet could achieve?
SOLUTION Water from a hose attached to the water main is sprayed into the air. The maximum height the water jet can rise is to be determined.
Assumptions 1 The flow exiting into the air is steady, incompressible, and irrotational (so that the Bernoulli equation is applicable). 2 The surface ten sion effects are negligible. 3 The friction between the water and air is negligible. 4 The irreversibilities that occur at the outlet of the hose due to abrupt contraction are not taken into account.
Properties We take the density of water to be 1000 kg/m3.
Analysis This problem involves the conversion of flow, kinetic, and potential energies to each other without involving any pumps, turbines, and wasteful compo nents with large frictional losses, and thus it is suitable for the use of the Bernoulli equation. The water height will be maximum under the stated assumptions. The velocity inside the hose is negligibly small compared to that of the jet (V1 2 ≪ Vj 2, see magnified portion of Fig. 5–39) and we take the elevation just below the hose outlet as the reference level (z1 = 0). At the top of the water trajectory V2 = 0, and atmospheric pressure pertains. Then the Bernoulli equation along a streamline from 1 to 2 simplifies to P1 ρg + V 2 1 2g + z1 = P2 ρg + V 2 2 2g + z2 → P1 ρg = Patm ρg + z2 Solving for z2 and substituting, z2 = P1 −Patm ρg = P1, gage ρg = 400 kPa (1000 kg/m3)(9.81 m/s2) ( 1000 N/m2 1 kPa ) ( 1 kg·m/s2 1 N ) = 40.8 m Therefore, the water jet can rise as high as 40.8 m into the sky in this case.
Discussion The result obtained by the Bernoulli equation represents the upper limit and should be interpreted accordingly. It tells us that the water cannot possibly rise more than 40.8 m, and, in all likelihood, the rise will be much less than 40.8 m due to irreversible losses that we neglected. ignore 0 0 ⟶ ⟶ Water jet 2 Hose 0 z Vj V12 ≪Vj2 1 1 Magnifying glass FIGURE 5–39 Schematic for Example 5–5. Inset shows a magnified view of the hose outlet region.
⟶ cen96537_ch05_189-248.indd 212 14/01/17 2:39 pm 213 CHAPTER 5 EXAMPLE 5–6 Water Discharge from a Large Tank A large tank open to the atmosphere is filled with water to a height of 5 m from the outlet tap (Fig. 5–40). A tap near the bottom of the tank is now opened, and water flows out from the smooth and rounded outlet. Determine the maximum water velocity at the outlet.
SOLUTION A tap near the bottom of a tank is opened. The maximum exit velocity of water from the tank is to be determined.
Assumptions 1 The flow is incompressible and irrotational (except very close to the walls). 2 The water drains slowly enough that the flow can be approxi mated as steady (actually quasi-steady when the tank begins to drain). 3 Irreversible losses in the tap region are neglected.
Analysis This problem involves the conversion of flow, kinetic, and potential energies to each other without involving any pumps, turbines, and wasteful com ponents with large frictional losses, and thus it is suitable for the use of the Bernoulli equation. We take point 1 to be at the free surface of water so that P1 = Patm (open to the atmosphere), V1 is negligibly small compared to V2 (the tank diameter is very large relative to the outlet diameter), z1 = 5 m, and z2 = 0 (we take the reference level at the center of the outlet). Also, P2 = Patm (water discharges into the atmosphere). For flow along a streamline from 1 to 2, the Bernoulli equation simplifies to P1 ρg + V 2 1 2g + z1 = P2 ρg + V 2 2 2g + z2 → z1 = V 2 2 2g Solving for V2 and substituting, V2 = √2gz1 = √2(9.81 m/s2)(5 m) = 9.9 m/s The relation V = √2gz is called the Torricelli equation. Therefore, the water leaves the tank with an initial maximum velocity of 9.9 m/s. This is the same velocity that would manifest if a solid were dropped a distance of 5 m in the absence of air friction drag. (What would the velocity be if the tap were at the bottom of the tank instead of on the side?) Discussion If the orifice were sharp-edged instead of rounded, then the flow would be disturbed, and the average exit velocity would be less than 9.9 m/s. Care must be exercised when attempting to apply the Bernoulli equation to situations where abrupt expansions or contractions occur since the friction and flow disturbance in such cases may not be negligible. From conservation of mass, (V1 /V2)2 = (D2/D1)4. So, for example, if D2/D1 = 0.1, then (V1 /V2)2 = 0.0001, and our approximation that V1 2 ≪ V2 2 is justified. ignore 0 ⟶ ⟶ Water 5 m 0 z 1 2 V2 FIGURE 5–40 Schematic for Example 5–6.
EXAMPLE 5–7 Siphoning Out Gasoline from a Fuel Tank During a trip to the beach (Patm = 1 atm = 101.3 kPa), a car runs out of gasoline, and it becomes necessary to siphon gas out of the car of a Good Samaritan (Fig. 5–41). The siphon is a small-diameter hose, and to start the siphon it is necessary to insert one siphon end in the full gas tank, fill the hose with gasoline via suction, and then place the other end in a gas can below the level of the gas tank. The dif ference in pressure between point 1 (at the free surface of the gasoline in the tank) and point 2 (at the outlet of the tube) causes the liquid to flow from the higher to 0 z z3 z1 z2 Gas tank 0.75 m 2 m Gasoline siphoning tube 1 2 3 Gas can FIGURE 5–41 Schematic for Example 5–7.
cen96537_ch05_189-248.indd 213 14/01/17 2:39 pm 214 BERNOULLI AND ENERGY EQUATIONS the lower elevation. Point 2 is located 0.75 m below point 1 in this case, and point 3 is located 2 m above point 1. The siphon diameter is 5 mm, and frictional losses in the siphon are to be disregarded. Determine (a) the minimum time to withdraw 4 L of gasoline from the tank to the can and (b) the pressure at point 3. The density of gasoline is 750 kg/m3.
SOLUTION Gasoline is to be siphoned from a tank. The minimum time it takes to withdraw 4 L of gasoline and the pressure at the highest point in the system are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 Even though the Bernoulli equation is not valid through the pipe because of frictional losses, we employ the Bernoulli equation anyway in order to obtain a best-case estimate. 3 The change in the gasoline surface level inside the tank is negligible compared to elevations z1 and z2 during the siphoning period.
Properties The density of gasoline is given to be 750 kg/m3.
Analysis (a) We take point 1 to be at the free surface of gasoline in the tank so that P1 = Patm (open to the atmosphere), V1 ≅ 0 (the tank is large relative to the tube diameter), and z2 = 0 (point 2 is taken as the reference level). Also, P2 = Patm (gasoline discharges into the atmosphere). Then the Bernoulli equation simplifies to P1 ρg + V 2 1 2g + z1 = P2 ρg + V 2 2 2g + z2 → z1 = V 2 2 2g Solving for V2 and substituting, V2 = √2gz1 = √2(9.81 m/s2)(0.75 m) = 3.84 m/s The cross-sectional area of the tube and the flow rate of gasoline are A = 𝜋D2/4 = 𝜋(5 × 10−3 m)2/4 = 1.96 × 10−5 m2 V · = V2A = (3.84 m/s)(1.96 × 10−5 m2) = 7.53 × 10−5 m3/s = 0.0753 L/s Then the time needed to siphon 4 L of gasoline becomes Δt = V V · = 4 L 0.0753 L/s = 53.1 s (b) The pressure at point 3 is determined by writing the Bernoulli equation along a streamline between points 3 and 2. Noting that V2 = V3 (conservation of mass), z2 = 0, and P2 = Patm, P2 ρg + V 2 2 2g + z2 = P3 ρg + V 2 3 2g + z3 → Patm ρg = P3 ρg + z3 Solving for P3 and substituting, P3 = Patm −ρgz3 = 101.3 kPa −(750 kg/m3)(9.81 m/s2)(2.75 m)( 1 N 1 kg·m/s2) ( 1 kPa 1000 N/m2) = 81.1 kPa ≈0 0 ⟶ ⟶ 0 ⟶ cen96537_ch05_189-248.indd 214 14/01/17 2:39 pm 215 CHAPTER 5 Discussion The siphoning time is determined by neglecting frictional effects, and thus this is the minimum time required. In reality, the time will be longer than 53.1 s because of friction between the gasoline and the tube surface, along with other irreversible losses, as discussed in Chap. 8. Also, the pressure at point 3 is below the atmospheric pressure. If the elevation difference between points 1 and 3 is too high, the pressure at point 3 may drop below the vapor pressure of gasoline at the gasoline temperature, and some gasoline may evaporate (cavitate). The vapor then may form a pocket at the top and halt the flow of gasoline.
h3 = 12 cm h2 = 7 cm h1 = 3 cm Stagnation point Water 1 2 V1 FIGURE 5–42 Schematic for Example 5–8.
EXAMPLE 5–8 Velocity Measurement by a Pitot Tube A piezometer and a Pitot tube are tapped into a horizontal water pipe, as shown in Fig. 5–42, to measure static and stagnation (static + dynamic) pressures. For the indicated water column heights, determine the velocity at the center of the pipe.
SOLUTION The static and stagnation pressures in a horizontal pipe are meas ured. The velocity at the center of the pipe is to be determined.
Assumptions 1 The flow is steady and incompressible. 2 Points 1 and 2 are close enough together that the irreversible energy loss between these two points is negligible, and thus we can use the Bernoulli equation.
Analysis We take points 1 and 2 along the streamline at the centerline of the pipe, with point 1 directly under the piezometer and point 2 at the tip of the Pitot tube. This is a steady flow with straight and parallel streamlines, and the gage pressures at points 1 and 2 can be expressed as P1 = ρg(h1 + h2) P2 = ρg(h1 + h2 + h3) Noting that z1 = z2, and point 2 is a stagnation point and thus V2 = 0, the applica tion of the Bernoulli equation between points 1 and 2 gives P1 ρg + V 2 1 2g + z1 = P2 ρg + V 2 2 2g + z2 → V 2 1 2g = P2 −P1 ρg Substituting the P1 and P2 expressions gives V 2 1 2g = P2 −P1 ρg = ρg(h1 + h2 + h3) −ρg(h1 + h2) ρg = h3 Solving for V1 and substituting, V1 = √2gh3 = √2(9.81 m/s2)(0.12 m) = 1.53 m/s Discussion Note that to determine the flow velocity, all we need is to measure the height of the excess fluid column in the Pitot tube compared to that in the pie zometer tube. 0 ⟶ cen96537_ch05_189-248.indd 215 14/01/17 2:39 pm 216 BERNOULLI AND ENERGY EQUATIONS Calm ocean level Ocean Hurricane Eye 1 2 3 h3 h2 A B FIGURE 5–43 Schematic for Example 5–9. The vertical scale is greatly exaggerated.
FIGURE 5–44 The eye of hurricane Linda (1997 in the Pacific Ocean near Baja California) is clearly visible in this satellite photo.
© Brand X Pictures/PunchStock RF EXAMPLE 5–9 The Rise of the Ocean Due to a Hurricane A hurricane is a tropical storm formed over the ocean by low atmospheric pressures. As a hurricane approaches land, inordinate ocean swells (very high tides) accompany the hurricane. A Class-5 hurricane features winds in excess of 155 mph, although the wind velocity at the center “eye” is very low. Figure 5–43 depicts a hurricane hovering over the ocean swell below. The atmospheric pressure 200 mi from the eye is 30.0 in Hg (at point 1, generally normal for the ocean) and the winds are calm. The atmo spheric pressure at the eye of the storm is 22.0 in Hg. Estimate the ocean swell at (a) the eye of the hurricane at point 3 and (b) point 2, where the wind velocity is 155 mph. Take the density of seawater and mercury to be 64 lbm/ft3 and 848 lbm/ft3, respec tively, and the density of air at normal sea-level temperature and pressure to be 0.076 lbm/ft3.
SOLUTION A hurricane is moving over the ocean. The amount of ocean swell at the eye and at active regions of the hurricane are to be determined.
Assumptions 1 The airflow within the hurricane is steady, incompressible, and irrotational (so that the Bernoulli equation is applicable). (This is certainly a very questionable assumption for a highly turbulent flow, but it is justified in the discus sion.) 2 The effect of water sucked into the air is negligible.
Properties The densities of air at normal conditions, seawater, and mercury are given to be 0.076 lbm/ft3, 64.0 lbm/ft3, and 848 lbm/ft3, respectively.
Analysis (a) Reduced atmospheric pressure over the water causes the water to rise. Thus, decreased pressure at point 2 relative to point 1 causes the ocean water to rise at point 2. The same is true at point 3, where the storm air velocity is negligible. The pressure difference given in terms of the mercury column height is expressed in terms of the seawater column height by ΔP = (ρgh)Hg = (ρgh)sw → hsw = ρHg ρsw hHg Then the pressure difference between points 1 and 3 in terms of the seawater col umn height becomes h3 = ρHg ρsw hHg = ( 848 lbm/ft3 64.0 lbm/ft3)(30 −22) in Hg = 8.83 ft which is equivalent to the storm surge at the eye of the hurricane (Fig. 5–44) since the wind velocity there is negligible and there are no dynamic effects.
(b) To determine the additional rise of ocean water at point 2 due to the high winds at that point, we write the Bernoulli equation between points A and B, which are on top of points 2 and 3, respectively. Noting that VB ≅ 0 (the eye region of the hur ricane is relatively calm) and zA = zB (both points are on the same horizontal line), the Bernoulli equation simplifies to PA ρg + V 2 A 2g + zA = PB ρg + V 2 B 2g + zB → PB −PA ρg = V 2 A 2g Substituting, PB −PA ρg = V 2 A 2g = (155 mph)2 2(32.2 ft/s2) ( 1.4667 ft/s 1 mph ) 2 = 803 ft 0 ⟶ cen96537_ch05_189-248.indd 216 14/01/17 2:39 pm 217 CHAPTER 5 where 𝜌 is the density of air in the hurricane. Noting that the density of an ideal gas at constant temperature is proportional to absolute pressure and the density of air at the normal atmospheric pressure of 14.7 psia ≅ 30 in Hg is 0.076 lbm/ft3, the density of air in the hurricane is ρair = Pair Patm air ρatm air = ( 22 in Hg 30 in Hg)(0.076 lbm/ft3) = 0.056 lbm/ft3 Using the relation developed above in part (a), the seawater column height equiva lent to 803 ft of air column height is determined to be hdynamic = ρair ρsw hair = ( 0.056 lbm/ft3 64 lbm/ft3 )(803 ft) = 0.70 ft Therefore, the pressure at point 2 is 0.70 ft seawater column lower than the pressure at point 3 due to the high wind velocities, causing the ocean to rise an additional 0.70 ft. Then the total storm surge at point 2 becomes h2 = h3 + hdynamic = 8.83 + 0.70 = 9.53 ft Discussion This problem involves highly turbulent flow and the intense break down of the streamlines, and thus the applicability of the Bernoulli equation in part (b) is questionable. Furthermore, the flow in the eye of the storm is not irrotational, and the Bernoulli equation constant changes across streamlines (see Chap. 10). The Bernoulli analysis can be thought of as the limiting, ideal case, and shows that the rise of seawater due to high-velocity winds cannot be more than 0.70 ft. The wind power of hurricanes is not the only cause of damage to coastal areas. Ocean flooding and erosion from excessive tides is just as serious, as are high waves generated by the storm turbulence and energy.
EXAMPLE 5–10 Bernoulli Equation for Compressible Flow Derive the Bernoulli equation when the compressibility effects are not negligi ble for an ideal gas undergoing (a) an isothermal process and (b) an isentropic process.
SOLUTION The Bernoulli equation for compressible flow is to be obtained for an ideal gas for isothermal and isentropic processes.
Assumptions 1 The flow is steady and frictional effects are negligible. 2 The fluid is an ideal gas, so the relation P = 𝜌RT is applicable. 3 The specific heats are constant so that P/𝜌k = constant during an isentropic process.
Analysis (a) When the compressibility effects are significant and the flow cannot be assumed to be incompressible, the Bernoulli equation is given by Eq. 5–40 as ∫ dP ρ + V 2 2 + gz = constant (along a streamline) (1) cen96537_ch05_189-248.indd 217 14/01/17 2:39 pm 218 BERNOULLI AND ENERGY EQUATIONS The compressibility effects can be properly accounted for by performing the integration ∫dP/𝜌 in Eq. 1. But this requires a relation between P and 𝜌 for the process. For the isothermal expansion or compression of an ideal gas, the inte gral in Eq. 1 is performed easily by noting that T = constant and substituting 𝜌 = P/RT, ∫ dP ρ = ∫ dP P/RT = RT ln P Substituting into Eq. 1 gives the desired relation, Isothermal process: RT ln P + V 2 2 + gz = constant (2) (b) A more practical case of compressible flow is the isentropic flow of ideal gases through equipment that involves high-speed fluid flow such as nozzles, diffusers, and the passages between turbine blades (Fig. 5–45). Isentropic (i.e., reversible and adiabatic) flow is closely approximated by these devices, and it is characterized by the relation P/𝜌k = C = constant, where k is the specific heat ratio of the gas. Solving for 𝜌 from P/𝜌k = C gives 𝜌 = C −1/kP1/k. Performing the integration, ∫ dP ρ = ∫ C1/kP−1/k dP = C1/k P−1/k+1 −1/k + 1 = P1/k ρ P−1/k+1 −1/k + 1 = ( k k −1) P ρ (3) Substituting, the Bernoulli equation for steady, isentropic, compressible flow of an ideal gas becomes Isentropic flow: ( k k −1) P 𝞺+ V 2 2 + gz = constant (4a) or ( k k −1) P1 ρ1 + V 2 1 2 + gz1 = ( k k −1) P2 ρ2 + V 2 2 2 + gz2 (4b) A common practical situation involves the acceleration of a gas from rest (stagnation conditions at state 1) with negligible change in elevation. In that case we have z1 = z2 and V1 = 0. Noting that 𝜌 = P/RT for ideal gases, P/𝜌k = constant for isentropic flow, and the Mach number is defined as Ma = V/c where c = √kRT is the local speed of sound for ideal gases, Eq. 4b simplifies to P1 P2 = [1 + ( k −1 2 )Ma 2 2 ] k /(k−1) (4c) where state 1 is the stagnation state and state 2 is any state along the flow.
Discussion It can be shown that the results obtained using the compressible and incompressible equations deviate no more than 2 percent when the Mach number is less than 0.3. Therefore, the flow of an ideal gas can be considered to be incompressible when Ma ≲ 0.3. For atmospheric air at normal conditions, this corresponds to a flow speed of about 100 m/s or 360 km/h.
FIGURE 5–45 Compressible flow of a gas through turbine blades is often modeled as isentropic, and the compressible form of the Bernoulli equation is a reasonable approximation.
© Corbis RF cen96537_ch05_189-248.indd 218 14/01/17 2:39 pm 219 CHAPTER 5 5–5 ■ GENERAL ENERGY EQUATION One of the most fundamental laws in nature is the first law of thermodynamics, also known as the conservation of energy principle, which provides a sound basis for studying the relationships among the various forms of energy and energy interactions. It states that energy can be neither created nor destroyed during a process; it can only change forms. Therefore, every bit of energy must be accounted for during a process. A rock falling off a cliff, for example, picks up speed as a result of its potential energy being converted to kinetic energy (Fig. 5–46). Experimen tal data show that the decrease in potential energy equals the increase in kinetic energy when the air resistance is negligible, thus confirming the conservation of energy principle. The conservation of energy principle also forms the backbone of the diet industry: a person who has a greater energy input (food) than energy output (exercise) will gain weight (store energy in the form of fat), and a person who has a smaller energy input than out put will lose weight. The change in the energy content of a system is equal to the difference between the energy input and the energy output, and the conservation of energy principle for any system can be expressed simply as Ein − Eout = ∆E. The transfer of any quantity (such as mass, momentum, and energy) is recognized at the boundary as the quantity crosses the boundary. A quantity is said to enter a system (or control volume) if it crosses the boundary from the outside to the inside, and to exit the system if it moves in the reverse direction. A quantity that moves from one location to another within a sys tem is not considered as a transferred quantity in an analysis since it does not enter or exit the system. Therefore, it is important to specify the system and thus clearly identify its boundaries before an engineering analysis is performed. The energy content of a fixed quantity of mass (a closed system) can be changed by two mechanisms: heat transfer Q and work transfer W. Then the conservation of energy for a fixed quantity of mass can be expressed in rate form as (Fig. 5–47) Q .
net in + W .
net in = dE sys dt or Q .
net in + W .
net in = d dt ∫ sys ρe dV (5–49) where the overdot stands for time rate of change, and Q · net in = Q · in − Q · out is the net rate of heat transfer to the system (negative, if from the system), W · net in = W · in − W · out is the net power input to the system in all forms (negative, if power output), and dEsys/dt is the rate of change of the total energy con tent of the system. For simple compressible systems, total energy consists of internal, kinetic, and potential energies, and it is expressed on a unit-mass basis as (see Chap. 2) e = u + ke + pe = u + V 2 2 + gz (5–50) Note that total energy is a property, and its value does not change unless the state of the system changes.
PE1 = 10 kJ m KE1 = 0 PE2 = 7 kJ m KE2 = 3 kJ Δz FIGURE 5–46 Energy cannot be created or destroyed during a process; it can only change forms.
Wshaft, in = 6 kJ = 18 kJ Qout = 3 kJ Qin = 15 kJ ΔE = (15 – 3) + 6 FIGURE 5–47 The energy change of a system during a process is equal to the net work and heat transfer between the system and its surroundings.
cen96537_ch05_189-248.indd 219 14/01/17 2:39 pm 220 BERNOULLI AND ENERGY EQUATIONS Energy Transfer by Heat, Q In daily life, we frequently refer to the sensible and latent forms of internal energy as heat, and talk about the heat content of bodies. Scientifically the more correct name for these forms of energy is thermal energy. For single-phase substances, a change in the thermal energy of a given mass results in a change in temperature, and thus temperature is a good representative of thermal energy. Thermal energy tends to move naturally in the direction of decreasing temperature. The transfer of energy from one system to another as a result of a temperature difference is called heat transfer. The warming up of a canned drink in a warmer room, for example, is due to heat transfer (Fig. 5–48). The time rate of heat transfer is called heat transfer rate and is denoted by Q .. The direction of heat transfer is always from the higher-temperature body to the lower-temperature one. Once temperature equality is established, heat transfer stops. There cannot be any net heat transfer between two systems (or a system and its surroundings) that are at the same temperature. A process during which there is no heat transfer is called an adiabatic process. There are two ways a process can be adiabatic: Either the system is well insulated so that only a negligible amount of heat can pass through the system boundary, or both the system and the surroundings are at the same temperature and therefore there is no driving force (temperature difference) for net heat transfer. An adiabatic process should not be confused with an isothermal process. Even though there is no net heat transfer during an adiabatic process, the energy content and thus the temperature of a system can still be changed by other means such as work transfer.
Energy Transfer by Work, W An energy interaction is work if it is associated with a force acting through a distance. A rising piston, a rotating shaft, and an electric wire crossing the system boundary are all associated with work interactions. The time rate of doing work is called power and is denoted by W .
. Car engines and hydraulic, steam, and gas turbines produce power (W .
shaft, in < 0); compressors, pumps, fans, and mixers consume power (W .
shaft, in > 0). Work-consuming devices transfer energy to the fluid, and thus increase the energy of the fluid. A fan in a room, for example, mobilizes the air and increases its kinetic energy. The electric energy a fan consumes is first con verted to mechanical energy by its motor that forces the shaft of the blades to rotate. This mechanical energy is then transferred to the air, as evidenced by the increase in air velocity. This energy transfer to air has nothing to do with a temperature difference, so it cannot be heat transfer. Therefore, it must be work. Air discharged by the fan eventually comes to a stop and thus loses its mechanical energy as a result of friction between air particles of different velocities. But this is not a “loss” in the real sense; it is simply the conversion of mechanical energy to an equivalent amount of thermal energy (which is of limited value, and thus the term loss) in accordance with the conservation of energy principle. If a fan runs a long time in a sealed room, we can sense the buildup of this thermal energy by a rise in air temperature.
Room air 25°C No heat transfer Heat Heat 25°C 8 J/s 16 J/s 15°C 5°C FIGURE 5–48 Temperature difference is the driving force for heat transfer. The larger the temperature difference, the higher is the rate of heat transfer. Condensation of water vapor from the room is shown for the coldest can.
cen96537_ch05_189-248.indd 220 14/01/17 2:39 pm 221 CHAPTER 5 A system may involve numerous forms of work, and the total work can be expressed as Wtotal = W shaft + W pressure + W viscous + W other (5–51) where Wshaft is the work transmitted by a rotating shaft, Wpressure is the work done by the pressure forces on the control surface, Wviscous is the work done by the normal and shear components of viscous forces on the control sur face, and Wother is the work done by other forces such as electric, magnetic, and surface tension, which are insignificant for simple compressible systems and are not considered in this text. We do not consider Wviscous either, since moving walls (such as fan blades or turbine runners) are usually inside the control volume and are not part of the control surface. But it should be kept in mind that the work done by shear forces as the blades shear through the fluid may need to be considered in a refined analysis of turbomachinery.
Shaft Work Many flow systems involve a machine such as a pump, a turbine, a fan, or a compressor whose shaft protrudes through the control surface, and the work transfer associated with all such devices is simply referred to as shaft work Wshaft. The power transmitted via a rotating shaft is proportional to the shaft torque Tshaft and is expressed as W .
shaft = 𝜔 Tshaft = 2𝜋 n · Tshaft (5–52) where 𝜔 is the angular speed of the shaft in rad/s and n . is the number of revolutions of the shaft per unit time, often expressed in rev/min or rpm.
Work Done by Pressure Forces Consider a gas being compressed in the piston-cylinder device shown in Fig. 5–49a. When the piston moves down a differential distance ds under the influence of the pressure force PA, where A is the cross-sectional area of the piston, the boundary work done on the system is 𝛿Wboundary = PA ds. Dividing both sides of this relation by the differential time interval dt gives the time rate of boundary work (i.e., power), 𝛿 W · pressure = 𝛿 W · boundary = PAV piston where Vpiston = ds/dt is the piston speed, which is the speed of the moving boundary at the piston face. Now consider a material chunk of fluid (a system) of arbitrary shape that moves with the flow and is free to deform under the influence of pressure, as shown in Fig. 5–49b. Pressure always acts inward and normal to the sur face, and the pressure force acting on a differential area dA is PdA. Again noting that work is force times distance and distance traveled per unit time is velocity, the time rate at which work is done by pressure forces on this differential part of the system is 𝛿 W .
pressure = −P dA Vn = −P dA(V › ·n ›) (5–53) since the normal component of velocity through the differential area dA is Vn = V cos 𝜃 = V ›·n →. Note that n → is the outward normal of dA, and thus the quantity V ›·n → is positive for expansion and negative for compression. The negative sign in Eq. 5–53 ensures that work done by pressure forces System System boundary, A dV dm dA P n 휃 V (b) (a) ds P A Vpiston System (gas in cylinder) FIGURE 5–49 The pressure force acting on (a) the moving boundary of a system in a piston-cylinder device, and (b) the differential surface area of a system of arbitrary shape.
cen96537_ch05_189-248.indd 221 14/01/17 2:39 pm 222 BERNOULLI AND ENERGY EQUATIONS is positive when it is done on the system, and negative when it is done by the system, which agrees with our sign convention. The total rate of work done by pressure forces is obtained by integrating 𝛿W .
pressure over the entire surface A, W .
pressure, net in = −∫ A P(V › ·n ›)dA = −∫ A P ρ ρ(V › ·n ›)dA (5–54) In light of these discussions, the net power transfer can be expressed as W .
net in = W .
shaft, net in + W .
pressure, net in = W .
shaft, net in −∫ A P (V › ·n ›) dA (5–55) Then the rate form of the conservation of energy relation for a closed system becomes Q .
net in + W .
shaft, net in + W .
pressure, net in = dEsys dt (5–56) To obtain a relation for the conservation of energy for a control volume, we apply the Reynolds transport theorem by replacing B with total energy E, and b with total energy per unit mass e, which is e = u + ke + pe = u + V 2/2 + gz (Fig. 5–50). This yields dEsys dt = d dt ∫ CV eρ dV + ∫ CS eρ (V r ›·n ›)A (5–57) Substituting the left-hand side of Eq. 5–56 into Eq. 5–57, the general form of the energy equation that applies to fixed, moving, or deforming control volumes becomes Q .
net in + W .
shaft, net in + W .
pressure, net in = d dt ∫ CV eρ dV + ∫ CS eρ(V r ›·n ›) dA (5–58) which is stated in words as ( The net rate of energy transfer into a CV by heat and work transfer ) = ( The time rate of change of the energy content of the CV ) + ( The net flow rate of energy out of the control surface by mass flow ) Here V › r = V › − V › CS is the fluid velocity relative to the control surface, and the product 𝜌(V › r·n →) dA represents the mass flow rate through area element dA into or out of the control volume. Again noting that n → is the outward nor mal of dA, the quantity V › r·n → and thus mass flow is positive for outflow and negative for inflow. Substituting the surface integral for the rate of pressure work from Eq. 5–54 into Eq. 5–58 and combining it with the surface integral on the right give Q .
net in + W .
shaft, net in = d dt ∫CV eρ dV + ∫ CS ( P ρ + e)ρ (V r ›·n ›)dA (5–59) This is a convenient form for the energy equation since pressure work is now combined with the energy of the fluid crossing the control surface and we no longer have to deal with pressure work.
= + bρdV B = E b = e b = e dBsys dt V d dt CV bρ( r · n ) dA CS = + eρdV dEsys dt V d dt CV eρ( r · n ) dA CS ∫ ∫ ∫ ∫ FIGURE 5–50 The conservation of energy equation is obtained by replacing B in the Reynolds transport theorem by energy E and b by e.
cen96537_ch05_189-248.indd 222 14/01/17 2:39 pm 223 CHAPTER 5 The term P/𝜌 = Pv = wflow is the flow work, which is the work per unit mass associated with pushing a fluid into or out of a control volume. Note that the fluid velocity at a solid surface is equal to the velocity of the solid surface because of the no-slip condition. As a result, the pressure work along the portions of the control surface that coincide with nonmoving solid surfaces is zero. Therefore, pressure work for fixed control volumes can exist only along the imaginary part of the control surface where the fluid enters and leaves the control volume, i.e., inlets and outlets. For a fixed control volume (no motion or deformation of the control vol ume), V › r = V › and the energy equation Eq. 5–59 becomes Fixed CV: Q .
net in + W .
shaft, net in = d dt ∫ CV eρ dV + ∫ CS ( P ρ + e)ρ(V › ·n ›) dA (5–60) This equation is not in a convenient form for solving practical engineer ing problems because of the integrals, and thus it is desirable to rewrite it in terms of average velocities and mass flow rates through inlets and outlets. If P/𝜌 + e is nearly uniform across an inlet or outlet, we can simply take it outside the integral. Noting that m · = ∫ Ac ρ(V › ·n ›) dAc is the mass flow rate across an inlet or outlet, the rate of inflow or outflow of energy through the inlet or outlet can be approximated as m .(P/𝜌 + e). Then the energy equation becomes (Fig. 5–51) Q .
net in + W .
shaft, net in = d dt ∫CV eρ dV + ∑ out m · ( P ρ + e) −∑ in m · ( P ρ + e) (5–61) where e = u + V 2/2 + gz (Eq. 5–50) is the total energy per unit mass for both the control volume and flow streams. Then, (5–62) or Q .
net in + W .
shaft, net in = d dt ∫ CV eρ dV + ∑ out m · (h + V 2 2 + gz) −∑ in m · (h + V 2 2 + gz) (5–63) where we used the definition of specific enthalpy h = u + Pv = u + P/𝜌. The last two equations are fairly general expressions of conservation of energy, but their use is still limited to fixed control volumes, uniform flow at inlets and outlets, and negligible work due to viscous forces and other effects. Also, the subscript “net in” stands for “net input,” and thus any heat or work transfer is positive if to the system and negative if from the system.
5–6 ■ ENERGY ANALYSIS OF STEADY FLOWS For steady flows, the time rate of change of the energy content of the con trol volume is zero, and Eq. 5–63 simplifies to Q .
net in + W .
shaft, net in = ∑ out m · (h + V 2 2 + gz) −∑ in m · (h + V 2 2 + gz) (5–64) Q .
net in + W .
shaft, net in = d dt ∫ CV eρ dV + ∑ out m · ( P ρ + u + V2 2 + gz) −∑ in m · ( P ρ + u + V2 2 + gz) min , energyin In mout , Out mout , Out Wshaft, net in mout , energyout energyout min , energyin energyout Qnet in In Out Fixed control volume FIGURE 5–51 In a typical engineering problem, the control volume may contain many inlets and outlets; energy flows in at each inlet, and energy flows out at each outlet. Energy also enters the control volume through net heat transfer and net shaft work.
cen96537_ch05_189-248.indd 223 14/01/17 2:39 pm 224 BERNOULLI AND ENERGY EQUATIONS It states that during steady flow the net rate of energy transfer to a control volume by heat and work transfers is equal to the difference between the rates of outgoing and incoming energy flows by mass flow. Many practical problems involve just one inlet and one outlet (Fig. 5–52). The mass flow rate for such single-stream devices is the same at the inlet and outlet, and Eq. 5–64 reduces to Q .
net in + W .
shaft, net in = m · (h2 −h1 + V 2 2 −V 2 1 2 + g(z2 −z1)) (5–65) where subscripts 1 and 2 refer to the inlet and outlet, respectively. The steady-flow energy equation on a unit-mass basis is obtained by dividing Eq. 5–65 by the mass flow rate m ., qnet in + wshaft, net in = h2 −h1 + V 2 2 −V 2 1 2 + g(z2 −z1) (5–66) where qnet in = Q .
net in/m . is the net heat transfer to the fluid per unit mass and wshaft, net in = W .
shaft, net in/m . is the net shaft work input to the fluid per unit mass. Using the definition of enthalpy h = u + P/𝜌 and rearranging, the steady-flow energy equation can also be expressed as wshaft, net in + P1 ρ1 + V 2 1 2 + gz1 = P2 ρ2 + V 2 2 2 + gz2 + (u2 −u1 −qnet in) (5–67) where u is the internal energy, P/𝜌 is the flow energy, V 2/2 is the kinetic energy, and gz is the potential energy of the fluid, all per unit mass. These relations are valid for both compressible and incompressible flows. The left side of Eq. 5–67 represents the mechanical energy input, while the first three terms on the right side represent the mechanical energy output. If the flow is ideal with no irreversibilities such as friction, the total mechanical energy must be conserved, and the term in parentheses (u2 − u1 − qnet in) must equal zero. That is, Ideal flow (no mechanical energy loss): qnet in = u2 −u1 (5–68) Any increase in u2 − u1 above qnet in is due to the irreversible conversion of mechanical energy to thermal energy, and thus u2 − u1 − qnet in represents the mechanical energy loss per unit mass (Fig. 5–53). That is, Real flow (with mechanical energy loss): emech, loss = u2 −u1 −qnet in (5–69) For single-phase fluids (a gas or a liquid), u2 − u1 = cv(T2 − T1) where cv is the constant-volume specific heat. The steady-flow energy equation on a unit-mass basis can be written con veniently as a mechanical energy balance, emech, in = emech, out + emech, loss (5–70) or wshaft, net in + P1 ρ1 + V 2 1 2 + gz1 = P2 ρ2 + V 2 2 2 + gz2 + emech, loss (5–71) 15.2°C 15.0°C Water 0.7 kg/s Δu = 0.84 kJ/kg ΔT = 0.2°C 2 kW pump = 0.70 η FIGURE 5–53 The lost mechanical energy in a fluid flow system results in an increase in the internal energy of the fluid and thus in a rise of fluid temperature.
+ + h1 Qnet in + Wshaft, net in gz1 2 1 m 2 V 2 2 V In Out Fixed control volume 2 1 ( ) + + h2 gz2 2 m( ) FIGURE 5–52 A control volume with only one inlet and one outlet and energy interactions.
cen96537_ch05_189-248.indd 224 14/01/17 2:39 pm 225 CHAPTER 5 Noting that wshaft, net in = wpump − wturbine, the mechanical energy balance can be written more explicitly as P1 ρ1 + V 2 1 2 + gz1 + wpump = P2 ρ2 + V 2 2 2 + gz2 + wturbine + emech, loss (5–72) where wpump is the mechanical work input (due to the presence of a pump, fan, compressor, etc.) and wturbine is the mechanical work output (due to a tur bine). When the flow is incompressible, either absolute or gage pressure can be used for P since Patm/𝜌 would appear on both sides and would cancel out. Multiplying Eq. 5–72 by the mass flow rate m . gives m · ( P1 ρ1 + V 2 1 2 + gz1)+ W .
pump = m · ( P2 ρ2 + V 2 2 2 + gz2)+ W .
turbine + E .
mech, loss (5–73) where W .
pump is the shaft power input through the pump’s shaft, W .
turbine is the shaft power output through the turbine’s shaft, and E .
mech, loss is the total mechanical power loss, which consists of pump and turbine losses as well as the frictional losses in the piping network. That is, E .
mech, loss = E .
mech loss, pump + E .
mech loss, turbine + E .
mech loss, piping By convention, irreversible pump and turbine losses are treated separately from irreversible losses due to other components of the piping system (Fig. 5–54). Thus, the energy equation is expressed in its most common form in terms of heads by dividing each term in Eq. 5 –73 by m .g. The result is P1 ρ1g + V 2 1 2g + z1 + hpump, u = P2 ρ2g + V 2 2 2g + z2 + hturbine, e + hL (5–74) where • hpump, u = wpump, u g = W .
pump, u m · g = 𝜂pump W .
pump m · g is the useful head delivered to the fluid by the pump. Because of irreversible losses in the pump, hpump, u is less than W .
pump/m .g by the factor 𝜂pump.
• hturbine, e = wturbine, e g = W .
turbine, e m · g = W .
turbine 𝜂turbinem · g is the extracted head removed from the fluid by the turbine. Because of irreversible losses in the turbine, hturbine, e is greater than W .
turbine/m .g by the factor 𝜂turbine. • hL = emech loss, piping g = E .
mech loss, piping m · g is the irreversible head loss between 1 and 2 due to all components of the piping system other than the pump or turbine.
Note that the head loss hL represents the frictional losses associated with fluid flow in piping, and it does not include the losses that occur within the pump or turbine due to the inefficiencies of these devices—these losses are taken into account by 𝜂pump and 𝜂turbine. Equation 5–74 is illustrated sche matically in Fig. 5–55. The pump head is zero if the piping system does not involve a pump, a fan, or a compressor, and the turbine head is zero if the system does not involve a turbine.
FIGURE 5–54 A typical power plant has numerous pipes, elbows, valves, pumps, and turbines, all of which have irreversible losses.
© Brand X Pictures/PunchStock RF cen96537_ch05_189-248.indd 225 14/01/17 2:39 pm 226 BERNOULLI AND ENERGY EQUATIONS Special Case: Incompressible Flow with No Mechanical Work Devices and Negligible Friction When piping losses are negligible, there is negligible dissipation of mechan ical energy into thermal energy, and thus hL = emech loss, piping/g ≅ 0, as shown later in Example 5–11. Also, hpump, u = hturbine, e = 0 when there are no mechanical work devices such as fans, pumps, or turbines. Then Eq. 5–74 reduces to P1 ρg + V 2 1 2g + z1 = P2 ρg + V 2 2 2g + z2 or P ρg + V 2 2g + z = constant (5–75) which is the Bernoulli equation derived earlier using Newton’s second law of motion. Thus, the Bernoulli equation can be thought of as a degenerate form of the energy equation.
Kinetic Energy Correction Factor, α The average flow velocity Vavg was defined such that the relation 𝜌VavgA gives the actual mass flow rate. Therefore, there is no such thing as a correction factor for mass flow rate. However, as Gaspard Coriolis (1792–1843) showed, the kinetic energy of a fluid stream obtained from V2/2 is not the same as the actual kinetic energy of the fluid stream since the square of a sum is not equal to the sum of the squares of its components (Fig. 5–56). This error can be corrected by replacing the kinetic energy terms V 2/2 in the energy equation by 𝛼V 2 avg/2, where 𝛼 is the kinetic energy correction factor. By using equations for the variation of velocity with the radial distance, it can be shown that the correction factor is 2.0 for fully developed laminar pipe flow, and it ranges between 1.04 and 1.11 for fully developed turbulent flow in a round pipe. The kinetic energy correction factors are often ignored (i.e., 𝛼 is set equal to 1) in an elementary analysis since (1) most flows encountered in prac tice are turbulent, for which the correction factor is near unity, and (2) the kinetic energy terms are often small relative to the other terms in the energy equation, and multiplying them by a factor less than 2.0 does not make much difference. When the velocity and thus the kinetic energy are high, the flow turns turbulent, and a unity correction factor is more appropriate. However, you should keep in mind that you may encounter some situations Control volume Emech loss, pump Wpump, u Wpump hpump, u hturbine, e Emech fluid, out Wturbine, e Wturbine Emech loss, turbine hL + + P2 z2 2g Emech loss, piping Emech fluid, in ρg 2 V 2 P1 z1 + + 2g ρg 1 V 2 FIGURE 5–55 Mechanical energy flow chart for a fluid flow system that involves a pump and a turbine. Vertical dimensions show each energy term expressed as an equivalent column height of fluid, i.e., head, corresponding to each term of Eq. 5–74.
KEact = ∫ ke훿m =∫A A [V(r)] 2 [ρV(r) dA] m = ρVavg A, ρ = constant V(r) A 1 2 ρAV 3 1 2 avg = avg mV 2 1 2 KEact 훼 = = 3 dA KEavg V(r) Vavg 1 A ρ A[V(r)] 3 dA = 1 2 KEavg = ( ) ∫ ∫ FIGURE 5–56 The determination of the kinetic energy correction factor using the actual velocity distribution V(r) and the average velocity Vavg at a cross section.
cen96537_ch05_189-248.indd 226 14/01/17 2:39 pm 227 CHAPTER 5 for which these factors are significant, especially when the flow is laminar. Therefore, we recommend that you always include the kinetic energy cor rection factor when analyzing fluid flow problems. When the kinetic energy correction factors are included, the energy equations for steady incompressible flow (Eqs. 5–73 and 5–74) become m · ( P1 ρ + 𝛼1 V1 2 2 + gz1) + W · pump = m · ( P2 ρ + 𝛼2 V2 2 2 + gz2) + W · turbine + E · mech, loss (5–76) P1 ρg + 𝛼1 V1 2 2g + z1 + hpump, u = P2 ρg + 𝛼2 V2 2 2g + z2 + hturbine, e + hL (5–77) If the flow at an inlet or outlet is fully developed turbulent pipe flow, we recommend using 𝛼 = 1.05 as a reasonable estimate of the correction factor. This leads to a more conservative estimate of head loss, and it does not take much additional effort to include 𝛼 in the equations.
EXAMPLE 5–11 Effect of Friction on Fluid Temperature and Head Loss Show that during steady and incompressible flow of a fluid in an adiabatic flow section (a) the temperature remains constant and there is no head loss when fric tion is ignored and (b) the temperature increases and some head loss occurs when frictional effects are considered. Discuss if it is possible for the fluid temperature to decrease during such flow (Fig. 5–57).
SOLUTION Steady and incompressible flow through an adiabatic section is considered. The effects of friction on the temperature and the heat loss are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The flow section is adiabatic and thus there is no heat transfer, qnet in = 0.
Analysis The density of a fluid remains constant during incompressible flow and the entropy change is Δs = cv ln T2 T1 This relation represents the entropy change of the fluid per unit mass as it flows through the flow section from state 1 at the inlet to state 2 at the outlet. Entropy change is caused by two effects: (1) heat transfer and (2) irrevers i bilities. There fore, in the absence of heat transfer, entropy change is due to irre vers i bilities only, whose effect is always to increase entropy.
(a) The entropy change of the fluid in an adiabatic flow section (qnet in = 0) is zero when the process does not involve any irreversibilities such as friction and swirling turbulent eddies, and thus for adiabatic reversible flow we have Temperature change: Δs = cv ln T2 T1 = 0 → T2 = T1 1 2 T1 u1 T2 u2 ρ = constant (adiabatic) FIGURE 5–57 Schematic for Example 5–11.
cen96537_ch05_189-248.indd 227 14/01/17 2:39 pm 228 BERNOULLI AND ENERGY EQUATIONS EXAMPLE 5–12 Pumping Power and Frictional Heating in a Pump The pump of a water distribution system is powered by a 15-kW electric motor whose efficiency is 90 percent (Fig. 5–58). The water flow rate through the pump is 50 L/s. The diameters of the inlet and outlet pipes are the same, and the elevation difference across the pump is negligible. If the absolute pressures at the inlet and outlet of the pump are measured to be 100 kPa and 300 kPa, respectively, determine (a) the mechanical efficiency of the pump and (b) the temperature rise of water as it flows through the pump due to mechanical inefficiencies.
SOLUTION The pressures across a pump are measured. The mechanical effi ciency of the pump and the temperature rise of water are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The pump is driven by an external motor so that the heat generated by the motor is dis sipated to the atmosphere. 3 The elevation difference between the inlet and out let of the pump is negligible, z1 ≅ z2. 4 The inlet and outlet diameters are the same and thus the average inlet and outlet velocities are equal, V1 = V2. 5 The kinetic energy correction factors are equal, 𝛼1 = 𝛼2.
Properties We take the density of water to be 1 kg/L = 1000 kg/m3 and its spe cific heat to be 4.18 kJ/kg·°C.
Analysis (a) The mass flow rate of water through the pump is m · = ρV · = (1 kg/L)(50 L/s) = 50 kg/s Mechanical energy loss: emech loss, piping = u2 −u1 −qnet in = cv(T2 −T1) −qnet in = 0 −0 = 0 Head loss: hL = emech loss, piping/g = 0 Thus we conclude that when heat transfer and frictional effects are negligible, (1) the temperature of the fluid remains constant, (2) no mechanical energy is converted to thermal energy, and (3) there is no irreversible head loss.
(b) When irreversibilities such as friction are taken into account, the entropy change is positive and thus we have: Temperature change: Δs = cv ln T2 T1 > 0 → T2 > T1 Mechanical energy loss: emech loss, piping = u2 −u1 −qnet in = cv(T2 −T1) > 0 Head loss: hL = emech loss, piping/g > 0 Thus we conclude that when the flow is adiabatic and irreversible, (1) the tem perature of the fluid increases, (2) some mechanical energy is converted to thermal energy, and (3) some irreversible head loss occurs.
Discussion It is impossible for the fluid temperature to decrease during steady, incompressible, adiabatic flow since this would require the entropy of an adiabatic system to decrease, which would be a violation of the second law of thermodynamics.
300 kPa Water Motor 15 kW ηmotor = 90% 50 L/s 100 kPa 1 2 Wpump FIGURE 5–58 Schematic for Example 5–12.
cen96537_ch05_189-248.indd 228 14/01/17 2:39 pm 229 CHAPTER 5 The motor draws 15 kW of power and is 90 percent efficient. Thus the mechanical (shaft) power it delivers to the pump is W .
pump, shaft = 𝜂motorW .
electric = (0.90)(15 kW) = 13.5 kW To determine the mechanical efficiency of the pump, we need to know the increase in the mechanical energy of the fluid as it flows through the pump, which is ΔE · mech, fluid = E · mech, out −E · mech, in = m · ( P2 ρ + 𝛼2 V 2 2 2 + gz2)−m · ( P1 ρ +𝛼1 V 2 1 2 +gz1) Simplifying it for this case and substituting the given values, ΔE · mech, fluid = m · ( P2 −P1 ρ ) = (50 kg/s)( (300 −100) kPa 1000 kg/m3 ) ( 1 kJ 1 kPa · m3)= 10.0 kW Then the mechanical efficiency of the pump becomes 𝜂pump = W · pump, u W · pump, shaft = ΔE · mech, fluid W · pump, shaft = 10.0 kW 13.5 kW = 0.741 or 74.1% (b) Of the 13.5-kW mechanical power supplied by the pump, only 10.0 kW is imparted to the fluid as mechanical energy. The remaining 3.5 kW is converted to thermal energy due to frictional effects, and this “lost” mechanical energy manifests itself as a heating effect in the fluid, E · mech, loss = W .
pump,shaft −ΔE · mech, fluid = 13.5 −10.0 = 3.5 kW The temperature rise of water due to this mechanical inefficiency is determined from the thermal energy balance, E .
mech, loss = m . (u2 − u1) = m . c∆T. Solving for ∆T, ΔT = E · mech, loss m · c = 3.5 kW (50 kg/s)(4.18 kJ/ kg·° C) = 0.017° C Therefore, the water experiences a temperature rise of 0.017°C which is very small, due to mechanical inefficiency, as it flows through the pump.
Discussion In an actual application, the temperature rise of water would probably be less since part of the heat generated would be transferred to the casing of the pump and from the casing to the surrounding air. If the entire pump and motor were submerged in water, then the 1.5 kW dissipated due to motor inefficiency would also be transferred to the surrounding water as heat.
EXAMPLE 5–13 Hydroelectric Power Generation from a Dam In a hydroelectric power plant, 100 m3/s of water flows from an elevation of 120 m to a turbine, where electric power is generated (Fig. 5–59). The total irreversible head loss in the piping system from point 1 to point 2 (excluding the turbine unit) is determined to be 35 m. If the overall efficiency of the turbine–generator is 80 percent, estimate the electric power output.
SOLUTION The available head, flow rate, head loss, and efficiency of a hydroelectric turbine are given. The electric power output is to be determined. ηturbine–gen = 80% 100 m3/s hL = 35 m 2 120 m 1 Generator Turbine FIGURE 5–59 Schematic for Example 5–13.
cen96537_ch05_189-248.indd 229 14/01/17 2:39 pm 230 BERNOULLI AND ENERGY EQUATIONS Assumptions 1 The flow is steady and incompressible. 2 Water levels at the res ervoir and the discharge site remain constant.
Properties We take the density of water to be 1000 kg/m3.
Analysis The mass flow rate of water through the turbine is m · = ρV · = (1000 kg/m3)(100 m3/s) = 105 kg/s We take point 2 as the reference level, and thus z2 = 0. Also, both points 1 and 2 are open to the atmosphere (P1 = P2 = Patm) and the flow velocities are negligible at both points (V1 = V2 = 0). Then the energy equation for steady, incompressible flow reduces to P1 ρg + 𝛼1 V 2 1 2g + z1 + hpump, u = P2 ρg + 𝛼2 V 2 2 2g + z2 + hturbine, e + hL or hturbine, e = z1 −hL Substituting, the extracted turbine head and the corresponding turbine power are hturbine, e = z1 − hL = 120 − 35 = 85 m W .
turbine, e = m ·ghturbine, e = (105 kg/s)(9.81 m/s2)(85 m)( 1 kJ/kg 1000 m2/s2)= 83,400 kW Therefore, a perfect turbine–generator would generate 83,400 kW of electricity from this resource. The electric power generated by the actual unit is W .
electric = 𝜂turbine–genW .
turbine, e = (0.80)(83.4 MW) = 66.7 MW Discussion Note that the power generation would increase by almost 1 MW for each percentage point improvement in the efficiency of the turbine– generator unit. You will learn how to determine hL in Chap. 8. 0 0 ↗ ⟶ EXAMPLE 5–14 Fan Selection for Air Cooling of a Computer A fan is to be selected to cool a computer case whose dimensions are 12 cm × 40 cm × 40 cm (Fig. 5–60). Half of the volume in the case is expected to be filled with components and the other half to be air space. A 5-cm-diameter hole is available at the back of the case for the installation of the fan that is to replace the air in the void spaces of the case once every second. Small low-power fan–motor combined units are available in the market and their efficiency is estimated to be 30 percent. Determine (a) the wattage of the fan–motor unit to be purchased and (b) the pressure difference across the fan. Take the air density to be 1.20 kg/m3.
SOLUTION A fan is to cool a computer case by completely replacing the air inside once every second. The power of the fan and the pressure difference across it are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 Losses other than those due to the inefficiency of the fan–motor unit are negligible. 3 The flow at the outlet is fairly uniform except near the center (due to the wake of the fan motor), and the kinetic energy correction factor at the outlet is 1.10.
Fan Casing Streamline V2 1 2 3 4 Welect FIGURE 5–60 Schematic for Example 5–14.
cen96537_ch05_189-248.indd 230 14/01/17 2:39 pm 231 CHAPTER 5 Properties The density of air is given to be 1.20 kg/m3.
Analysis (a) Noting that half of the volume of the case is occupied by the com ponents, the air volume in the computer case is ΔVair = (Void fraction)(Total case volume) = 0.5(12 cm × 40 cm × 40 cm) = 9600 cm3 Therefore, the volume and mass flow rates of air through the case are V · = ΔVair Δt = 9600 cm3 1 s = 9600 cm3/s = 9.6 × 10−3 m3/s m · = ρV · = (1.20 kg/m3)(9.6 × 10−3 m3/s) = 0.0115 kg/s The cross-sectional area of the opening in the case and the average air velocity through the outlet are A = 𝜋D2 4 = 𝜋(0.05 m)2 4 = 1.96 × 10−3 m2 V = V · A = 9.6 × 10−3 m3/s 1.96 × 10−3 m2 = 4.90 m/s We draw the control volume around the fan such that both the inlet and the outlet are at atmospheric pressure (P1 = P2 = Patm), as shown in Fig. 5–60, where the inlet section 1 is large and far from the fan so that the flow velocity at the inlet section is negligible (V1 ≅ 0). Noting that z1 = z2 and frictional losses in the flow are dis regarded, the mechanical losses consist of fan losses only and the energy equation (Eq. 5–76) simplifies to m · ( P1 ρ + 𝛼1 V 2 1 2 + gz2) + W .
fan = m · ( P2 ρ + 𝛼2 V 2 2 2 + gz2) + W .
turbine + E · mech loss, fan Solving for W .
fan − E .
mech loss, fan = W .
fan, u and substituting, W .
fan, u = m ·𝛼2 V 2 2 2 = (0.0115 kg/s)(1.10) (4.90 m/s)2 2 ( 1 N 1 kg·m/s2) = 0.152 W Then the required electric power input to the fan is determined to be W .
elect = W .
fan, u 𝜂fan−motor = 0.152 W 0.3 = 0.506 W Therefore, a fan–motor rated at about a half watt is adequate for this job (Fig. 5–61). (b) To determine the pressure difference across the fan unit, we take points 3 and 4 to be on the two sides of the fan on a horizontal line. This time z3 = z4 again and V3 = V4 since the fan is a narrow cross section, and the energy equation reduces to m · P3 ρ + W .
fan = m · P4 ρ + E · mech loss, fan → W .
fan, u = m · P4 −P3 ρ Solving for P4 − P3 and substituting, P4 −P3 = ρW .
fan, u m · = (1.2 kg/m3)(0.152 W) 0.0115 kg/s ( 1 Pa·m3 1 Ws ) = 15.8 Pa Therefore, the pressure rise across the fan is 15.8 Pa. 0 0 ↗ —⟶ FIGURE 5–61 The cooling fans used in computers and computer power supplies are typically small and consume only a few watts of electrical power.
© PhotoDisc/Getty RF cen96537_ch05_189-248.indd 231 14/01/17 2:40 pm 232 BERNOULLI AND ENERGY EQUATIONS EXAMPLE 5–15 Pumping Water from a Lake to a Pool A submersible pump with a shaft power of 5 kW and an efficiency of 72 percent is used to pump water from a lake to a pool through a constant diameter pipe (Fig. 5–62). The free surface of the pool is 25 m above the free surface of the lake. If the irreversible head loss in the piping system is 4 m, determine the dis charge rate of water and the pressure difference across the pump.
SOLUTION Water from a lake is pumped to a pool at a given elevation. For a given head loss, the flow rate and the pressure difference across the pump are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 Both the lake and pool are large enough that their surface elevations remain fixed.
Properties We take the density of water to be 1 kg/L = 1000 kg/m3.
Analysis The pump delivers 5 kW of shaft power and is 72 percent efficient. The useful mechanical power it imparts to the water is W .
pump u = 𝜂pumpW .
shaft = (0.72)(5 kW) = 3.6 kW We take point 1 at the free surface of the lake, which is also taken as the reference level (z1 = 0), and point 2 at the free surface of the pool. Also, both points 1 and 2 are open to the atmosphere (P1 = P2 = Patm), and the velocities are negligible there (V1 ≅ V2 ≅ 0). Then the energy equation for steady, incompressible flow through a control volume between these two surfaces that includes the pump is expressed as m · ( P1 ρ + 𝛼1 V 2 1 2 + gz1) + W .
pump, u = m · ( P2 ρ + 𝛼2 V 2 2 2 + gz2) + W .
turbine, e + E · mech loss, piping Under the stated assumptions, the energy equation reduces to W .
pump, u = m . gz2 + E .
mech loss, piping Noting that E .
mech loss, piping= m .ghL, the mass and volume flow rates of water become m · = W .
pump, u gz2 + ghL = W .
pump, u g(z2 + hL) = 3.6 kJ/s (9.81 m/s2)(25 + 4 m) ( 1000 m2/s2 1 kJ )= 12.7 kg/s V · = m · ρ = 12.7 kg/s 1000 kg/m3 = 12.7 × 10−3 m3/s = 12.7 L/s Discussion The efficiency of the fan–motor unit is given to be 30 percent, which means 30 percent of the electric power W .
electric consumed by the unit is converted to useful mechanical energy while the rest (70 percent) is “lost” and converted to thermal energy. Also, a more powerful fan is required in an actual system to overcome frictional losses inside the computer case. Note that if we had ignored the kinetic energy correction factor at the outlet, the required electrical power and pressure rise would have been 10 percent lower in this case (0.460 W and 14.4 Pa, respectively).
25 m Pool 2 1 FIGURE 5–62 Schematic for Example 5–15.
cen96537_ch05_189-248.indd 232 14/01/17 2:40 pm 233 CHAPTER 5 We now take the pump as the control volume. Assuming that the elevation differ ence and the kinetic energy change across the pump are negligible, the energy equa tion for this control volume yields ΔP = P out −Pin = W .
pump, u V .
= 3.6 kJ/s 12.7 × 10−3 m3/s ( 1 kN·m 1 kJ )( 1 kPa 1 kN/m2) = 283 kPa Discussion It can be shown that in the absence of head loss (hL = 0) the flow rate of water would be 14.7 L/s, which is an increase of 16 percent. Therefore, frictional losses in pipes should be minimized since they always cause the flow rate to decrease.
SUMMARY This chapter deals with the mass, Bernoulli, and energy equations and their applications. The amount of mass flow ing through a cross section per unit time is called the mass flow rate and is expressed as m · = ρVAc = ρV · where 𝜌 is the density, V is the average velocity, V . is the volume flow rate of the fluid, and Ac is the cross-sectional area normal to the flow direction. The conservation of mass relation for a control volume is expressed as d dt ∫CV ρ dV + ∫CS ρ(V › ·n ›) dA = 0 It states that the time rate of change of the mass within the control volume plus the net mass flow rate out of the control surface is equal to zero. In simpler terms, dmCV dt = ∑ in m · −∑ out m · For steady-flow devices, the conservation of mass princi ple is expressed as Steady flow: ∑ in m · = ∑ out m · Steady flow (single stream): m · 1 = m · 2 → ρ1V1A1 = ρ2V2A2 Steady, incompressible flow: ∑ in V · = ∑ out V · Steady, incompressible flow (single stream): V · 1 = V · 2 → V1 A1 = V2 A2 Mechanical energy is the form of energy associated with the velocity, elevation, and pressure of the fluid, and it can be converted to mechanical work completely and directly by an ideal mechanical device. The efficiencies of various real devices are defined as 𝜂pump = ΔE · mech, fluid W .
shaft, in = W .
pump, u W .
pump 𝜂turbine = W .
shaft, out ∣ΔE · mech, fluid∣ = W .
turbine W .
turbine, e 𝜂motor = Mechanical power output Electric power input = W .
shaft, out W .
elect, in 𝜂generator = Electric power output Mechanical power input = W .
elect, out W .
shaft, in 𝜂pump−motor = 𝜂pump𝜂motor = ΔE .
mech, fluid W .
elect, in = W .
pump, u W .
elect, in 𝜂turbine–gen = 𝜂turbine𝜂generator = W .
elect, out ∣ΔE .
mech, fluid∣ = W .
elect, out W .
turbine, e cen96537_ch05_189-248.indd 233 14/01/17 2:40 pm 234 BERNOULLI AND ENERGY EQUATIONS REFERENCES AND SUGGESTED READING 1. R. C. Dorf, ed. in chief. The Engineering Handbook, 2nd ed. Boca Raton, FL: CRC Press, 2004.
2. R. L. Panton. Incompressible Flow, 3rd ed. New York: Wiley, 2005.
3. M. Van Dyke. An Album of Fluid Motion. Stanford, CA: The Parabolic Press, 1982. The Bernoulli equation is a relation between pressure, velocity, and elevation in steady, incompressible flow, and is expressed along a streamline and in regions where net vis cous forces are negligible as P ρ + V2 2 + gz = constant It can also be expressed between any two points on a stream line as P1 ρ + V 2 1 2 + gz1 = P2 ρ + V 2 2 2 + gz2 The Bernoulli equation is an expression of mechanical energy balance and can be stated as: The sum of the kinetic, potential, and flow energies of a fluid particle is constant along a streamline during steady flow when the compress ibility and frictional effects are negligible. Multiplying the Bernoulli equation by density gives P + ρ V 2 2 + ρgz = constant where P is the static pressure, which represents the actual pressure of the fluid; 𝜌V2/2 is the dynamic pressure, which represents the pressure rise when the fluid in motion is brought to a stop isentropically; and 𝜌gz is the hydrostatic pressure, which accounts for the effects of fluid weight on pressure. The sum of the static, dynamic, and hydrostatic pressures is called the total pressure. The Bernoulli equa tion states that the total pressure along a streamline is con stant. The sum of the static and dynamic pressures is called the stagnation pressure, which represents the pressure at a point where the fluid is brought to a complete stop in an isentropic manner. The Bernoulli equation can also be repre sented in terms of “heads” by dividing each term by g, P ρg + V 2 2g + z = H = constant where P/𝜌g is the pressure head, which represents the height of a fluid column that produces the static pressure P; V 2/2g is the velocity head, which represents the eleva tion needed for a fluid to reach the velocity V during fric tionless free fall; and z is the elevation head, which repre sents the potential energy of the fluid. Also, H is the total head for the flow. The curve that represents the sum of the static pressure and the elevation heads, P/𝜌g + z, is called the hydraulic grade line (HGL), and the curve that represents the total head of the fluid, P/𝜌g + V 2/2g + z, is called the energy grade line (EGL). The energy equation for steady, incompressible flow is P1 ρg + 𝛼1 V 2 1 2g + z1 + hpump, u = P2 ρg + 𝛼2 V 2 2 2g + z2 + hturbine, e + hL where hpump, u = wpump, u g = W .
pump, u m · g = 𝜂pumpW .
pump m · g hturbine, e = wturbine, e g = W .
turbine, e m · g = W .
turbine 𝜂turbinem · g hL = emech loss, piping g = E .
mech loss, piping m · g emech, loss = u2 −u1 −qnet in The mass, Bernoulli, and energy equations are three of the most fundamental relations in fluid mechanics, and they are used extensively in the chapters that follow. In Chap. 6, either the Bernoulli equation or the energy equation is used together with the mass and momentum equations to determine the forces and torques acting on fluid systems. In Chaps. 8 and 14, the mass and energy equations are used to determine the pumping power requirements in fluid systems and in the design and analysis of turbomachinery. In Chaps. 12 and 13, the energy equation is also used to some extent in the analy sis of compressible flow and open-channel flow.
cen96537_ch05_189-248.indd 234 14/01/17 2:40 pm 235 CHAPTER 5 PROBLEMS Conservation of Mass 5–1C Does the amount of mass entering a control volume have to be equal to the amount of mass leaving during an unsteady-flow process?
5–2C Define mass and volume flow rates. How are they related to each other?
5–3C Name four physical quantities that are conserved and two quantities that are not conserved during a process.
5–4C When is the flow through a control volume steady?
5–5C Consider a device with one inlet and one outlet. If the volume flow rates at the inlet and at the outlet are the same, is the flow through this device necessarily steady? Why?
5–6 A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls the air in and forces it through the resistors where it is heated. If the density of air is 1.20 kg/m3 at the inlet and 1.05 kg/m3 at the exit, determine the percent increase in the velocity of air as it flows through the hair dryer.
1.05 kg/m3 1.20 kg/m3 FIGURE P5–6 5–7E A garden hose attached with a nozzle is used to fill a 20-gal bucket. The inner diameter of the hose is 1 in and it reduces to 0.4 in at the nozzle exit. If the average veloc ity in the hose is 6 ft/s, determine (a) the volume and mass flow rates of water through the hose, (b) how long it will take to fill the bucket with water, and (c) the average velocity of water at the nozzle exit.
5–8E Air whose density is 0.082 lbm/ft3 enters the duct of an air-conditioning system at a volume flow rate of 450 ft3/min. If the diameter of the duct is 16 in, determine the velocity of the air at the duct inlet and the mass flow rate of air.
5–9 A 0.75-m3 rigid tank initially contains air whose den sity is 1.18 kg/m3. The tank is connected to a high-pressure supply line through a valve. The valve is opened, and air is allowed to enter the tank until the density in the tank rises to 4.95 kg/m3. Determine the mass of air that has entered the tank. Answer: 2.83 kg 5–10 Consider the flow of an incompressible Newtonian fluid between two parallel plates that are 4 mm apart. If the upper plate moves to right with u1 = 5 m/s while the bottom one moves to the left with u2 = 1.5 m/s, what would be the net flow rate at a cross-section between two plates? Take the plate width to be 5 cm.
5–11 A desktop computer is to be cooled by a fan whose flow rate is 0.30 m3/min. Determine the mass flow rate of air through the fan at an elevation of 3400 m where the air density is 0.7 kg/m3. Also, if the average velocity of air is not to exceed 95 m/min, determine the minimum diameter of the casing of the fan. Answers: 0.00350 kg/s, 0.0634 m 5–12 The minimum fresh air requirement of a residen tial building is specified to be 0.35 air changes per hour (ASHRAE, Standard 62, 1989). That is, 35 percent of the entire air contained in a residence should be replaced by fresh outdoor air every hour. If the ventilation requirement of a 2.7-m-high, 200-m2 residence is to be met entirely by a fan, determine the flow capacity in L/min of the fan that needs to be installed. Also determine the minimum diameter of the duct if the average air velocity is not to exceed 5 m/s.
5–13 The ventilating fan of the bathroom of a build ing (Fig. P5–13) has a volume flow rate of 50 L/s and runs continuously. If the density of air inside is 1.20 kg/m3, deter mine the mass of air vented out in one day.
Bathroom 22°C Fan 50 L/s FIGURE P5–13 Problems designated by a “C” are concept questions, and stu dents are encouraged to answer them all. Problems designated by an “E” are in English units, and the SI users can ignore them. Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software.
cen96537_ch05_189-248.indd 235 14/01/17 2:40 pm 236 BERNOULLI AND ENERGY EQUATIONS 5–14 Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is 60 cm2, determine (a) the mass flow rate through the nozzle, and (b) the exit area of the nozzle. Answers: (a) 0.265 kg/s, (b) 23.2 cm2 5–15 Air at 40°C flow steadily through the pipe shown in Fig. P5–15. If P1 = 40 kPa (gage), P2 = 10 kPa (gage), D = 3d, P atm ≅100 kPa, the average velocity at section 2 is V2 = 25 m/s, and air temperature remains nearly constant, determine the average speed at section 1.
FIGURE P5–15 P1 P2 d D 1 2 5–16 In climates with low night-time temperatures, an energy-efficient way of cooling a house is to install a fan in the ceiling that draws air from the interior of the house and discharges it to a ventilated attic space. Consider a house whose interior air volume is 720 m3. If air in the house is to be exchanged once every 20 minutes, determine (a) the required flow rate of the fan and (b) the average discharge speed of air if the fan diameter is 0.5 m.
Mechanical Energy and Efficiency 5–17C What is mechanical energy? How does it differ from thermal energy? What are the forms of mechanical energy of a fluid stream?
5–18C Define turbine efficiency, generator efficiency, and combined turbine–generator efficiency.
5–19C What is mechanical efficiency? What does a mechanical efficiency of 100 percent mean for a hydraulic turbine?
5–20C How is the combined pump–motor efficiency of a pump and motor system defined? Can the combined pump– motor efficiency be greater than either the pump or the motor efficiency?
5–21 At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy of air per unit mass and the power generation potential of a wind turbine with 70-m-diameter blades at that location. Also determine the actual electric power generation assuming an overall effi ciency of 30 percent. Take the air density to be 1.25 kg/m3.
5–22 Reconsider Prob. 5–21. Using appropriate soft ware, investigate the effect of wind velocity and the blade span diameter on wind power generation. Let the velocity vary from 5 to 20 m/s in increments of 5 m/s, and the diameter to vary from 20 to 80 m in increments of 20 m. Tabulate the results, and discuss their significance.
5–23E A differential thermocouple with sensors at the inlet and exit of a pump indicates that the temperature of water rises 0.048°F as it flows through the pump at a rate of 1.5 ft3/s. If the shaft power input to the pump is 23 hp and the heat loss to the surrounding air is negligible, determine the mechanical efficiency of the pump. Answer: 72.4 percent ΔT = 0.048°F Pump FIGURE P5–23E 5–24 Electric power is to be generated by installing a hydraulic turbine–generator at a site 110 m below the free surface of a large water reservoir that can supply water steadily at a rate of 900 kg/s. If the mechanical power out put of the turbine is 800 kW and the electric power gen eration is 750 kW, determine the turbine efficiency and the combined turbine–generator efficiency of this plant. Neglect losses in the pipes.
5–25 Consider a river flowing toward a lake at an average speed of 3 m/s at a rate of 500 m3/s at a location 55 m above the lake surface. Determine the total mechanical energy of the river water per unit mass and the power generation potential of the entire river at that location. Answer: 272 MW River 3 m/s 55 m FIGURE P5–25 Bernoulli Equation 5–26C Express the Bernoulli equation in three different ways using (a) energies, (b) pressures, and (c) heads.
5–27C What are the three major assumptions used in the derivation of the Bernoulli equation?
5–28C Define static, dynamic, and hydrostatic pressure. Under what conditions is their sum constant for a flow stream?
5–29C What is streamwise acceleration? How does it differ from normal acceleration? Can a fluid particle accelerate in steady flow?
cen96537_ch05_189-248.indd 236 14/01/17 2:40 pm 237 CHAPTER 5 5–30C What is stagnation pressure? Explain how it can be measured.
5–31C Define pressure head, velocity head, and elevation head for a fluid stream and express them for a fluid stream whose pressure is P, velocity is V, and elevation is z.
5–32C How is the location of the hydraulic grade line deter mined for open-channel flow? How is it determined at the outlet of a pipe discharging to the atmosphere?
5–33C In a certain application, a siphon must go over a high wall. Can water or oil with a specific gravity of 0.8 go over a higher wall? Why?
5–34C What is the hydraulic grade line? How does it differ from the energy grade line? Under what conditions do both lines coincide with the free surface of a liquid?
5–35C A glass manometer with oil as the working fluid is connected to an air duct as shown in Fig. P5–35C. Will the oil levels in the manometer be as in Fig. P5–35Ca or b? Explain. What would your response be if the flow direction is reversed?
Flow (a) (b) Flow FIGURE P5–35C 5–36C The velocity of a fluid flowing in a pipe is to be measured by two different Pitot-type mercury manometers shown in Fig. P5–36C. Would you expect both manometers to predict the same velocity for flowing water? If not, which would be more accurate? Explain. What would your response be if air were flowing in the pipe instead of water?
Flow Flow 1 2 FIGURE P5–36C 5–37C The water level of a tank on a building roof is 20 m above the ground. A hose leads from the tank bottom to the ground. The end of the hose has a nozzle, which is pointed straight up. What is the maximum height to which the water could rise? What factors would reduce this height?
5–38C Explain how and why a siphon works. Someone proposes siphoning cold water over a 7-m-high wall. Is this feasible? Explain.
5–39C A student siphons water over a 8.5-m-high wall at sea level. She then climbs to the summit of Mount Shasta (elevation 4390 m, Patm = 58.5 kPa) and attempts the same experiment. Comment on her prospects for success.
5–40 In a hydroelectric power plant, water enters the tur bine nozzles at 800 kPa absolute with a low velocity. If the nozzle outlets are exposed to atmospheric pressure of 100 kPa, determine the maximum velocity to which water can be accelerated by the nozzles before striking the turbine blades.
5–41 A Pitot-static probe is used to measure the speed of an aircraft flying at 3000 m. If the differential pressure reading is 3 kPa, determine the speed of the aircraft.
5–42 The air velocity in the duct of a heating system is to be measured by a Pitot-static probe inserted into the duct par allel to the flow. If the differential height between the water columns connected to the two outlets of the probe is 3.2 cm, determine (a) the flow velocity and (b) the pressure rise at the tip of the probe. The air temperature and pressure in the duct are 45°C and 98 kPa, respectively.
5–43 A piezometer and a Pitot tube are tapped into a 4-cm-diameter horizontal water pipe, and the height of the water columns are measured to be 26 cm in the piezometer and 35 cm in the Pitot tube (both measured from the top sur face of the pipe). Determine the velocity at the center of the pipe.
5–44 The diameter of a cylindrical water tank is Do and its height is H. The tank is filled with water, which is open to the atmosphere. An orifice of diameter D with a smooth entrance (i.e., negligible losses) is open at the bottom. Develop a rela tion for the time required for the tank (a) to empty halfway and (b) to empty completely.
5–45E A siphon pumps water from a large reservoir to a lower tank that is initially empty. The tank also has a rounded orifice 20 ft below the reservoir surface where the water leaves the tank. Both the siphon and the orifice diameters are 2 in. Ignoring frictional losses, determine to what height the water will rise in the tank at equilibrium.
5–46E Water flows through a horizontal pipe at a rate of 2.4 gal/s. The pipe consists of two sections of diameters 4 in and 2 in with a smooth reducing section. The pressure dif ference between the two pipe sections is measured by a mer cury manometer. Neglecting frictional effects, determine the differential height of mercury between the two pipe sections. Answer: 3.0 in cen96537_ch05_189-248.indd 237 14/01/17 2:40 pm 238 BERNOULLI AND ENERGY EQUATIONS 4 in 2 in h FIGURE P5–46E 5–47 An airplane is flying at an altitude of 10,500 m. Determine the gage pressure at the stagnation point on the nose of the plane if the speed of the plane is 450 km/h. How would you solve this problem if the speed were 1050 km/h? Explain.
5–48 While traveling on a dirt road, the bottom of a car hits a sharp rock and a small hole develops at the bottom of its gas tank. If the height of the gasoline in the tank is 30 cm, deter mine the initial velocity of the gasoline at the hole. Discuss how the velocity will change with time and how the flow will be affected if the lid of the tank is closed tightly. Answer: 2.43 m/s 5–49 The water in an 8-m-diameter, 3-m-high above-ground swimming pool is to be emptied by unplugging a 3-cm-diameter, 25-m-long horizontal pipe attached to the bottom of the pool. Determine the maximum discharge rate of water through the pipe. Also, explain why the actual flow rate will be less.
5–50 Reconsider Prob. 5–49. Determine how long it will take to empty the swimming pool completely. Answer: 15.4 h 5–51 Reconsider Prob. 5–50. Using appropriate soft ware, investigate the effect of the discharge pipe diameter on the time required to empty the pool completely. Let the diameter vary from 1 to 10 cm in increments of 1 cm. Tabulate and plot the results.
5–52 Air at 105 kPa and 37°C flows upward through a 6-cm-diameter inclined duct at a rate of 65 L/s. The duct diameter is then reduced to 4 cm through a reducer. The pressure change across the reducer is measured by a water manometer. The elevation difference between the two points on the pipe where the two arms of the manometer are attached is 0.20 m. Determine the differential height between the fluid levels of the two arms of the manometer.
Air h FIGURE P5–52 5–53 Water at 20°C is siphoned from a reservoir as shown in Fig. P5–53. For d = 8 cm and D = 16 cm, determine (a) the minimum flow rate that can be achieved without cavitation occurring in the piping system and (b) the maximum elevation of the highest point of the piping system to avoid cavitation. (c) Also, discuss the ways of increasing the maximum elevation of the highest point of the piping system to avoid cavitation.
1 m 4 m 2 m 7 m T = 20°C d D 1 2 3 4 FIGURE P5–53 5–54 The water pressure in the mains of a city at a particu lar location is 270 kPa gage. Determine if this main can serve water to neighborhoods that are 25 m above this location.
5–55 A pressurized tank of water has a 10-cm-diameter orifice at the bottom, where water discharges to the atmos phere. The water level is 2.5 m above the outlet. The tank air pressure above the water level is 250 kPa (absolute) while the atmospheric pressure is 100 kPa. Neglecting frictional effects, determine the initial discharge rate of water from the tank. Answer: 0.147 m3/s 2.5 m 10 cm Air 250 kPa FIGURE P5–55 5–56 Reconsider Prob. 5–55. Using appropriate software, investigate the effect of water height in the tank on the discharge velocity. Let the water height vary from 0 to 5 m in increments of 0.5 m. Tabulate and plot the results.
5–57E Air is flowing through a venturi meter whose diameter is 2.6 in at the entrance part (location 1) and 1.8 in at the throat (location 2). The gage pressure is measured cen96537_ch05_189-248.indd 238 14/01/17 2:40 pm 239 CHAPTER 5 to be 12.2 psia at the entrance and 11.8 psia at the throat. Neglecting frictional effects, show that the volume flow rate can be expressed as V · = A2√ 2(P1 −P2) ρ(1 −A 2 2 /A 2 1 ) and determine the flow rate of air. Take the air density to be 0.075 lbm/ft3.
2.6 in 12.2 psia Air 1.8 in 11.8 psia FIGURE P5–57E 5–58 The water level in a tank is 20 m above the ground. A hose is connected to the bottom of the tank, and the nozzle at the end of the hose is pointed straight up. The tank cover is airtight, and the air pressure above the water surface is 2 atm gage. The system is at sea level. Determine the maximum height to which the water stream could rise. Answer: 40.7 m 20 m 2 atm h FIGURE P5–58 5–59E The air velocity in a duct is measured by a Pitot-static probe connected to a differential pressure gage. If the air is at 13.4 psia absolute and 70°F and the reading of the dif ferential pressure gage is 0.15 psi, determine the air velocity. Answer: 143 ft/s 5–60 In cold climates, water pipes may freeze and burst if proper precautions are not taken. In such an occurrence, the exposed part of a pipe on the ground ruptures, and water shoots up to 55 m. Estimate the gage pressure of water in the pipe. State your assumptions and discuss if the actual pres sure is more or less than the value you predicted.
5–61 Water enters a tank of diameter DT steadily at a mass flow rate of m .
in. An orifice at the bottom with diameter Do allows water to escape. The orifice has a rounded entrance, so the frictional losses are negligible. If the tank is initially empty, (a) determine the maximum height that the water will reach in the tank and (b) obtain a relation for water height z as a function of time.
Do z DT in m FIGURE P5–61 5–62 A fluid of density 𝜌 and viscosity 𝜇 flows through a section of horizontal converging–diverging duct. The duct cross-sectional areas Ainlet, Athroat, and Aoutlet are known at the inlet, throat (minimum area), and outlet, respectively. Average pressure Poutlet is measured at the outlet, and average veloc ity Vinlet is measured at the inlet. (a) Neglecting any irrevers ibilities such as friction, generate expressions for the average velocity and average pressure at the inlet and the throat in terms of the given variables. (b) In a real flow (with irrevers ibilities), do you expect the actual pressure at the inlet to be higher or lower than the prediction? Explain.
5–63 What is the minimum diameter at section (1) to avoid cavitation at that point? Take D2 = 15 cm.
Water (3) (1) (2) 5 m FIGURE P5–63 Energy Equation 5–64C What is irreversible head loss? How is it related to the mechanical energy loss?
5–65C What is useful pump head? How is it related to the power input to the pump?
cen96537_ch05_189-248.indd 239 14/01/17 2:40 pm 240 BERNOULLI AND ENERGY EQUATIONS 5–66C Consider the steady adiabatic flow of an incom pressible fluid. Can the temperature of the fluid decrease dur ing flow? Explain.
5–67C Consider the steady adiabatic flow of an incom pressible fluid. If the temperature of the fluid remains con stant during flow, is it accurate to say that the frictional effects are negligible?
5–68C What is the kinetic energy correction factor? Is it significant?
5–69C The water level in a tank is 20 m above the ground. A hose is connected to the bottom of the tank, and the noz zle at the end of the hose is pointed straight up. The water stream from the nozzle is observed to rise 25 m above the ground. Explain what may cause the water from the hose to rise above the tank level.
5–70C A 3-m-high tank filled with water has a discharge valve near the bottom and another near the top. (a) If these two valves are opened, will there be any difference between the discharge velocities of the two water streams? (b) If a hose whose discharge end is left open on the ground is first connected to the lower valve and then to the higher valve, will there be any difference between the discharge rates of water for the two cases? Disregard any frictional effects.
5–71C A person is filling a knee-high bucket with water using a garden hose and holding it such that water discharges from the hose at the level of his waist. Someone suggests that the bucket will fill faster if the hose is lowered such that water discharges from the hose at the knee level. Do you agree with this suggestion? Explain. Disregard any frictional effects.
5–72 Water is being pumped from a large lake to a reservoir 25 m above at a rate of 25 L/s by a 10-kW (shaft) pump. If the irreversible head loss of the piping system is 5 m, determine the mechanical efficiency of the pump. Answer: 73.6 percent 5–73 Reconsider Prob. 5–72. Using appropriate soft ware, investigate the effect of irreversible head loss on the mechanical efficiency of the pump. Let the head loss vary from 0 to 15 m in increments of 1 m. Plot the results, and discuss them.
5–74 A 15-hp (shaft) pump is used to raise water to a 45-m higher elevation. If the mechanical efficiency of the pump is 82 percent, determine the maximum volume flow rate of water.
5–75 Water flows at a rate of 0.040 m3/s in a horizontal pipe whose diameter is reduced from 15 cm to 8 cm by a reducer. If the pressure at the centerline is measured to be 480 kPa and 440 kPa before and after the reducer, respectively, determine the irreversible head loss in the reducer. Take the kinetic energy correction factors to be 1.05. Answer: 0.963 m 5–76 The water level in a tank is 20 m above the ground. A hose is connected to the bottom of the tank, and the noz zle at the end of the hose is pointed straight up. The tank is at sea level, and the water surface is open to the atmosphere. In the line leading from the tank to the nozzle is a pump, which increases the pressure of water. If the water jet rises to a height of 27 m from the ground, determine the minimum pressure rise supplied by the pump to the water line.
20 m 27 m FIGURE P5–76 5–77 A hydraulic turbine has 50 m of head available at a flow rate of 1.30 m3/s, and its overall turbine–generator effi ciency is 78 percent. Determine the electric power output of this turbine.
5–78E In a hydroelectric power plant, water flows from an elevation of 400 ft to a turbine, where electric power is gen erated. For an overall turbine–generator efficiency of 85 per cent, determine the minimum flow rate required to generate 100 kW of electricity. Answer: 217 lbm/s 5–79E Reconsider Prob. 5–78E. Determine the flow rate of water if the irreversible head loss of the piping system between the free surfaces of the inlet and the outlet is 36 ft.
5–80 A fan is to be selected to ventilate a bathroom whose dimensions are 2 m × 3 m × 3 m. The air velocity is not to exceed 7 m/s to minimize vibration and noise. The combined efficiency of the fan–motor unit to be used can be taken to be 50 percent. If the fan is to replace the entire volume of air in 15 min, determine (a) the wattage of the fan–motor unit to be purchased, (b) the diameter of the fan casing, and (c) the pressure difference across the fan. Take the air density to be 1.25 kg/m3 and disregard the effect of the kinetic energy cor rection factors.
Air 7 m/s Exhaust fan FIGURE P5–80 5–81 Water flows at a rate of 20 L/s through a horizontal pipe whose diameter is constant at 3 cm. The pressure drop across a valve in the pipe is measured to be 2 kPa, as shown in Fig. P5–81. Determine the irreversible head loss of the cen96537_ch05_189-248.indd 240 14/01/17 2:40 pm 241 CHAPTER 5 valve, and the useful pumping power needed to overcome the resulting pressure drop. Answers: 0.204 m, 40 W Water 20 L/s ΔP = 2 kPa FIGURE P5–81 5–82E The water level in a tank is 34 ft above the ground. A hose is connected to the bottom of the tank at the ground level and the nozzle at the end of the hose is pointed straight up. The tank cover is airtight, but the pressure over the water surface is unknown. Determine the minimum tank air pres sure (gage) that will cause a water stream from the nozzle to rise 72 ft from the ground.
5–83 A large tank is initially filled with water 4 m above the center of a sharp-edged 10-cm-diameter orifice. The tank water surface is open to the atmosphere, and the orifice drains to the atmosphere. If the total irreversible head loss in the system is 0.2 m, determine the initial discharge velocity of water from the tank. Take the kinetic energy correction factor at the orifice to be 1.2.
5–84 Water enters a hydraulic turbine through a 30-cm-diameter pipe at a rate of 0.6 m3/s and exits through a 25-cm-diameter pipe. The pressure drop in the turbine is measured by a mercury manometer to be 1.2 m. For a com bined turbine– generator efficiency of 83 percent, determine the net electric power output. Disregard the effect of the kinetic energy correction factors.
W e 30 cm Turbine ΔP = 1.2 m Hg Generator 25 cm FIGURE P5–84 5–85E A 78-percent efficient 12-hp pump is pumping water from a lake to a nearby pool at a rate of 1.2 ft3/s through a constant-diameter pipe. The free surface of the pool is 32 ft above that of the lake. Determine the irreversible head loss of the piping system, in ft, and the mechanical power used to overcome it.
5–86 Water is pumped from a lower reservoir to a higher reservoir by a pump that provides 23 kW of useful mechani cal power to the water. The free surface of the upper reservoir is 57 m higher than the surface of the lower reservoir. If the flow rate of water is measured to be 0.03 m3/s, determine the irreversible head loss of the system and the lost mechanical power during this process.
0.03 m3/s 57 m Pump 23 kW FIGURE P5–86 5–87 Water in a partially filled large tank is to be supplied to the roof top, which is 8 m above the water level in the tank, through a 2.5-cm-internal-diameter pipe by maintaining a constant air pressure of 300 kPa (gage) in the tank. If the head loss in the piping is 2 m of water, determine the dis charge rate of the supply of water to the roof top.
5–88 Underground water is to be pumped by a 78 percent efficient 5-kW submerged pump to a pool whose free surface is 30 m above the underground water level. The diameter of the pipe is 7 cm on the intake side and 5 cm on the discharge side. Determine (a) the maximum flow rate of water and (b) the pressure difference across the pump. Assume the eleva tion difference between the pump inlet and the outlet and the effect of the kinetic energy correction factors to be negligible.
30 m Pool FIGURE P5–88 cen96537_ch05_189-248.indd 241 14/01/17 2:40 pm 242 BERNOULLI AND ENERGY EQUATIONS 5–89 Reconsider Prob. 5–88. Determine the flow rate of water and the pressure difference across the pump if the irre versible head loss of the piping system is 4 m.
5–90 The velocity profile for turbulent flow in a circular pipe is approximated as u(r) = umax(1 − r/R)1/n, where n = 9. Determine the kinetic energy correction factor for this flow. Answer: 1.04 5–91 The demand for electric power is usually much higher during the day than it is at night, and utility companies often sell power at night at much lower prices to encourage con sumers to use the available power generation capacity and to avoid building new expensive power plants that will be used only a short time during peak periods. Utilities are also will ing to purchase power produced during the day from private parties at a high price.
Pump– turbine Lake 50 m Reservoir FIGURE P5–91 Suppose a utility company is selling electric power for $0.06/kWh at night and is willing to pay $0.13/kWh for power produced during the day. To take advantage of this opportunity, an entrepreneur is considering building a large reservoir 50 m above the lake level, pumping water from the lake to the reservoir at night using cheap power, and letting the water flow from the reservoir back to the lake during the day, producing power as the pump–motor operates as a turbine–generator during reverse flow. Preliminary analysis shows that a water flow rate of 2 m3/s can be used in either direction, and the irreversible head loss of the piping system is 4 m. The combined pump–motor and turbine–generator efficiencies are expected to be 75 percent each. Assuming the system operates for 10 h each in the pump and turbine modes during a typical day, determine the potential revenue this pump–turbine system can generate per year.
5–92 A fireboat is to fight fires at coastal areas by drawing seawater with a density of 1030 kg/m3 through a 10-cm-diameter pipe at a rate of 0.04 m3/s and discharg ing it through a hose nozzle with an exit diameter of 5 cm. The total irreversible head loss of the system is 3 m, and the position of the nozzle is 3 m above sea level. For a pump efficiency of 70 percent, determine the required shaft power input to the pump and the water discharge velocity. Answers: 39.2 kW, 20.4 m/s 3 m FIGURE P5–92 Review Problems 5–93 Consider a fully filled tank of semi-circular cross section tank with radius R and width of b into the page, as shown in Fig. P5–93. If the water is pumped out of the tank at flow rate of V · = Kh2, where K is a positive constant and h is the water depth at time t. Determine the time needed to drop the water level to a specified h value of ho in terms of R, K, and ho.
Water R h FIGURE P5–93 5–94 The velocity of a liquid flowing in a circular pipe of radius R varies from zero at the wall to a maximum at the pipe center. The velocity distribution in the pipe can be represented as V(r), where r is the radial distance from the pipe center. Based on the definition of mass flow rate m ., obtain a relation for the average velocity in terms of V(r), R, and r.
5–95 Air at 2.50 kg/m3 enters a nozzle that has an inlet-to-exit area ratio of 2:1 with a velocity of 120 m/s and leaves with a velocity of 330 m/s. Determine the density of air at the exit. Answer: 1.82 kg/m3 5–96 The air in a 5-m × 5-m × 3-m hospital room is to be completely replaced by conditioned air every 15 min. If the average air velocity in the circular air duct leading to the room is not to exceed 5 m/s, determine the minimum diameter of the duct.
5–97E The water level in a tank is 70 ft above the ground. A hose is connected to the bottom of the tank, and the nozzle at the end of the hose is pointed straight up. The tank is at sea level, and the water surface is open to the atmosphere. In the line leading from the tank to the cen96537_ch05_189-248.indd 242 14/01/17 2:40 pm 243 CHAPTER 5 nozzle is a pump, which increases the water pressure by 15 psia. Determine the maximum height to which the water stream could rise.
5–98 A pressurized 2-m-diameter tank of water has a 10-cm-diameter orifice at the bottom, where water discharges to the atmosphere. The water level initially is 3 m above the outlet. The tank air pressure above the water level is maintained at 450 kPa absolute and the atmos pheric pressure is 100 kPa. Neglecting frictional effects, determine (a) how long it will take for half of the water in the tank to be discharged and (b) the water level in the tank after 10 s.
5–99 Underground water is being pumped into a pool whose cross section is 3 m × 4 m while water is discharged through a 5-cm-diameter orifice at a constant average veloc ity of 5 m/s. If the water level in the pool rises at a rate of 1.5 cm/min, determine the rate at which water is supplied to the pool, in m3/s.
5–100 Air at 98 kPa and 20°C flows in a horizontal duct of variable cross section. The water column in the manometer that measures the difference between two sections has a verti cal displacement of 5 cm. If the velocity in the first section is low and the friction is negligible, determine the velocity at the second section. Also, if the manometer reading has a pos sible error of ±2 mm, conduct an error analysis to estimate the range of validity for the velocity found.
5–101 A very large tank contains air at 102 kPa at a loca tion where the atmospheric air is at 100 kPa and 20°C. Now a 2-cm-diameter tap is opened. Determine the maximum flow rate of air through the hole. What would your response be if air is discharged through a 2-m-long, 4-cm-diameter tube with a 2-cm-diameter nozzle? Would you solve the prob lem the same way if the pressure in the storage tank were 300 kPa?
2 cm Air 102 kPa 100 kPa 20°C 2 cm 4 cm FIGURE P5–101 5–102 Water is flowing through a Venturi meter whose diameter is 7 cm at the entrance part and 4 cm at the throat. The pressure is measured to be 380 kPa at the entrance and 200 kPa at the throat. Neglecting frictional effects, determine the flow rate of water. Answer: 0.0252 m3/s 5–103 Water flows at a rate of 0.011 m3/s in a horizontal pipe whose diameter increases from 6 to 11 cm by an enlargement section. If the head loss across the enlargement section is 0.65 m and the kinetic energy correction factor at both the inlet and the outlet is 1.05, determine the pressure change.
5–104 Air flows through a pipe at a rate of 120 L/s. The pipe consists of two sections of diameters 22 cm and 10 cm with a smooth reducing section that connects them. The pres sure difference between the two pipe sections is measured by a water manometer. Neglecting frictional effects, determine the differential height of water between the two pipe sections. Take the air density to be 1.20 kg/m3. Answer: 1.37 cm 22 cm Air 120 L/s 10 cm h FIGURE P5–104 5–105 A 3-m-high large tank is initially filled with water. The tank water surface is open to the atmosphere, and a sharp-edged 10-cm-diameter orifice at the bottom drains to the atmosphere through a horizontal 80-m-long pipe. If the total irreversible head loss of the system is determined to be 1.5 m, determine the initial velocity of the water from the tank. Disregard the effect of the kinetic energy correction factors. Answer: 5.42 m/s 3 m Water 10 cm 80 m FIGURE P5–105 5–106 Reconsider Prob. 5–105. Using appropriate software, investigate the effect of the tank height on the initial discharge velocity of water from the completely filled tank. Let the tank height vary from 2 to 15 m in increments of 1 m, and assume the irreversible head loss to remain constant. Tabu late and plot the results.
5–107 Reconsider Prob. 5–105. In order to drain the tank faster, a pump is installed near the tank exit. Determine the pump head input necessary to establish an average water velocity of 6.5 m/s when the tank is full.
cen96537_ch05_189-248.indd 243 14/01/17 2:40 pm 244 BERNOULLI AND ENERGY EQUATIONS 5–108 A D0 = 12-m-diameter tank is initially filled with water 2 m above the center of a D = 10-cm-diameter valve near the bottom. The tank surface is open to the atmosphere, and the tank drains through a L = 95-m-long pipe connected to the valve. The friction factor of the pipe is given to be f = 0.015, and the discharge velocity is expressed as V = √ 2gz 1.5 + fL/D where z is the water height above the center of the valve. Determine (a) the initial discharge velocity from the tank and (b) the time required to empty the tank. The tank can be considered to be empty when the water level drops to the center of the valve.
5–109 An oil pump is drawing 18 kW of electric power while pumping oil with 𝜌 = 860 kg/m3 at a rate of 0.1 m3/s. The inlet and outlet diameters of the pipe are 8 cm and 12 cm, respectively. If the pressure rise of oil in the pump is measured to be 250 kPa and the motor efficiency is 95 percent, deter mine the mechanical efficiency of the pump. Take the kinetic energy correction factor to be 1.05.
18 kW ΔP = 250 kPa 0.1 m3/s Motor 8 cm 12 cm Oil Pump FIGURE P5–109 5–110 A wind tunnel draws atmospheric air at 20°C and 101.3 kPa by a large fan located near the exit of the tunnel. If the air velocity in the tunnel is 80 m/s, determine the pressure in the tunnel.
20°C 101.3 kPa Wind tunnel 80 m/s FIGURE P5–110 5–111 Consider a spherical tank containing compressed air. It is known from the elementary compressible theory that if a hole is opened on the tank, the compressed air will leave the tank at a mass flow rate of m . = k𝜌 where k is a constant and 𝜌 is the density at any time. Assuming an initial density 𝜌0 and pressure P0, derive an expression for P(t).
ρ0, P0 m = k휌 D/2 FIGURE P5–111 5–112 A tank with openings 1, 2, and 3 is moving to left at a speed of 25 km/h. Knowing that D1 = D2 = 20 cm and D3 = 10 cm, find the volume flow rate at each opening. Assume a frictionless incompressible flow. Air velocity far from the tank is assumed to be zero. 1 2 3 V 60° FIGURE P5–112 5–113 Two dimensionally identical containers are connected to each other by a pipe with a diameter of 3 cm. Container A initially contains water but container B is empty. The cross-sectional area of each container is 4 m2. If the viscous effects are negligible, determine the time needed to empty the con tainer A completely.
A B 0.7 m 2.5 m 0.5 m FIGURE P5–113 cen96537_ch05_189-248.indd 244 14/01/17 2:40 pm 245 CHAPTER 5 5–114 A circular thin plate is placed on the top of a tube, as shown in the figure. (a) Find the exit velocity from the gap. (b) Find the velocity and pressure distributions in the gap between plate and tube’s tip for d/2 < r < D/2. Use 𝜌air = 1.2 kg/m3 and Patm = 101,325 Pa. Air flow D = 10 cm d = 3 mm s = 1.5 mm Air exit V = 3 m/s r FIGURE P5–114 5–115 A pump-storage plant uses a turbine to generate hydropower during the day, letting water from the upper res ervoir flow through the turbine into the lower reservoir as sketched. The plant then pumps water back up to the upper reservoir during the night. In a typical pump-storage plant, the same turbomachine is used as both the pump and the tur bine, and is called a pump–turbine. The plant is profitable because the demand for electric power is much higher during the day than it is at night, and utility companies sell power at much lower prices at night to encourage customers to use the available power generation capacity and to avoid build ing new expensive power plants that would be used only a short time during peak periods. Utilities are also willing to purchase power produced during the day from private parties at a high price. Suppose a utility company is selling electric power for $0.030/kWh at night and is willing to pay $0.120/ kWh for power produced during the day. A pump-storage plant has a gross head of 90.0 m, and can handle a volume flow rate of 4.6 m3/s in either direction. The irreversible head losses in the system are estimated as 5.0 m in either direction at this flow rate. The efficiency of the combined pump–motor is 88 percent and that of the combined turbine–generator is 92 percent. The plant runs in pump mode for 10 hours each night and runs in turbine mode for 10 hours each day. It oper ates 340 days in a year. How much net revenue ($) does this pump-storage plant generate in one year?
Pump– turbine Hgross Generator/motor FIGURE P5–115 5–116 A diffuser in a pipe flow is basically a slow expan sion of the pipe diameter, which slows down the fluid veloc ity and increases the pressure (the Bernoulli effect). Water at room temperature flows with a volume flow rate of 0.0250 m3/s through a horizontal diffuser in which the pipe diameter increases gradually from D1 = 6.00 to D2 = 11.00 cm. The irreversible head loss through the diffuser is estimated to be 0.450 m. The flow is turbulent, and the kinetic energy cor rection factors at both the inlet and outlet of the diffuser are assumed to be 1.05.
V1 V2 P1 P2 D1 D2 1 2 휃 FIGURE P5–116 (a) Calculate the pressure difference P2 – P1 in units of kPa using the energy equation. (b) Repeat using the Bernoulli equation (ignore irreversible head losses and ignore kinetic energy correction factors—in other words, set the kinetic energy correction factors to 1). Calculate the percentage error in the result due to the Ber noulli approximation, and explain why (or why not) Bernoulli is applicable here.
(c) It may be surprising that the answer to part (a) is posi tive, i.e., the pressure rises downstream. How is this possi ble? Explain by calculating the change i n energy grade line ΔEGL and the change in hydraulic grade line ΔHGL from the upstream to the downstream location. In particular, does EGL go up or down, and does HGL go up or down?
Fundamentals of Engineering (FE) Exam Problems 5–117 Water flows in a 10-cm-diameter pipe at a velocity of 0.75 m/s. The mass flow rate of water in the pipe is (a) 353 kg/min (b) 209 kg/min (c) 88.4 kg/min (d ) 44.5 kg/min (e) 5.9 kg/min cen96537_ch05_189-248.indd 245 14/01/17 2:40 pm 246 BERNOULLI AND ENERGY EQUATIONS 5–118 Water flows in a 3-cm-diameter pipe at a velocity of 0.55 m/s. The volume flow rate of water in the pipe is (a) 23.3 L/min (b) 0.39 L/min (c) 1400 L/min (d) 55 L/min (e) 70.7 L/min 5–119 Cold water at a rate of 25 L/min is mixed with hot water at 40 L/min in a continuous manner in a mixing cham ber. The rate of water output from the mixing chamber is (a) 0.65 kg/s (b) 1.08 kg/s (c) 15 kg/s (d) 32.5 kg/s (e) 65 kg/s 5–120 Air enters a steady-flow compressor at 1 atm and 25°C at a rate of 0.35 m3/s and leaves at a rate of 0.12 m3/s. The density of air at the compressor exit is (a) 1.2 kg/m3 (b) 1.63 kg/m3 (c) 2.48 kg/m3 (d) 3.45 kg/m3 (e) 4.57 kg/m3 5–121 A 75-m-high water body that is open to the atmo sphere is available. The mechanical energy of this water body per unit mass is (a) 736 kJ/kg (b) 0.736 kJ/kg (c) 0.75 kJ/kg (d) 75 kJ/kg (e) 150 kJ/kg 5–122 Water enters a motor-pump unit at a pressure of 95 kPa at a rate of 115 kg/min. If the motor consumes 0.8 kW of electricity, the maximum water pressure at the exit of the pump is (a) 408 kPa (b) 512 kPa (c) 816 kPa (d) 1150 kPa (e) 1020 kPa 5–123 A pump is used to increase the pressure of water from 100 kPa to 900 kPa at a rate of 160 L/min. If the shaft power input to the pump is 3 kW, the efficiency of the pump is (a) 0.532 (b) 0.660 (c) 0.711 (d ) 0.747 (e) 0.855 5–124 A hydraulic turbine is used to generate power by using the water in a dam. The elevation difference between the free surfaces upstream and downstream of the dam is 130 m. The water is supplied to the turbine at a rate of 150 kg/s. If the shaft power output from the turbine is 155 kW, the effi ciency of the turbine is (a) 0.79 (b) 0.81 (c) 0.83 (d ) 0.85 (e) 0.88 5–125 The efficiency of a hydraulic turbine-generator unit is specified to be 85 percent. If the generator efficiency is 96 percent, the turbine efficiency is (a) 0.816 (b) 0.850 (c) 0.862 (d ) 0.885 (e) 0.960 5–126 Which one is not an assumption involved with the Bernoulli equation? (a) No elevation change (b) Incompressible flow (c) Steady flow (d) No shaft work (e) No friction 5–127 Consider incompressible, frictionless flow of a fluid in a horizontal piping. The pressure and velocity of a fluid is measured to be 150 kPa and 1.25 m/s at a specified point. The density of the fluid is 700 kg/m3. If the pressure is 140 kPa at another point, the velocity of the fluid at that point is (a) 1.26 m/s (b) 1.34 m/s (c) 3.75 m/s (d ) 5.49 m/s (e) 7.30 m/s 5–128 Consider incompressible, frictionless flow of water in a vertical piping. The pressure is 240 kPa at 2 m from the ground level. The velocity of water does not change during this flow. The pressure at 15 m from the ground level is (a) 227 kPa (b) 174 kPa (c) 127 kPa (d ) 120 kPa (e) 113 kPa 5–129 Consider water flow in a piping network. The pres sure, velocity, and elevation at a specified point (point 1) of the flow are 300 kPa, 2.4 m/s, and 5 m. The velocity and ele vation at point 2 are 1.9 m/s and 18 m, respectively. Take the correction factor to be 1. If the irreversible head loss between point 1 and point 2 of the pipe is 2 m, the pressure of water at point 2 is (a) 286 kPa (b) 230 kPa (c) 179 kPa (d) 154 kPa (e) 101 kPa 5–130 The static and stagnation pressures of a fluid in a pipe are measured by a piezometer and a pitot tube to be 200 kPa and 210 kPa, respectively. If the density of the fluid is 550 kg/m3, the velocity of the fluid is (a) 10 m/s (b) 6.03 m/s (c) 5.55 m/s (d ) 3.67 m/s (e) 0.19 m/s 5–131 The static and stagnation pressures of a fluid in a pipe are measured by a piezometer and a pitot tube. The heights of the fluid in the piezometer and pitot tube are measured to be 2.2 m and 2.0 m, respectively. If the density of the fluid is 5000 kg/m3, the velocity of the fluid in the pipe is (a) 0.92 m/s (b) 1.43 m/s (c) 1.65 m/s (d ) 1.98 m/s (e) 2.39 m/s 5–132 The difference between the heights of energy grade line (EGL) and hydraulic grade line (HGL) is equal to (a) z (b) P/ρg (c) V2/2g (d ) z + P/ρg (e) z + V 2/2g 5–133 Water at 120 kPa (gage) is flowing in a horizontal pipe at a velocity of 1.15 m/s. The pipe makes a 90° angle at the exit and the water exits the pipe vertically into the air. The maximum height the water jet can rise is (a) 6.9 m (b) 7.8 m (c) 9.4 m (d ) 11.5 m (e) 12.3 m 5–134 Water is withdrawn at the bottom of a large tank open to the atmosphere. The water velocity is 9.5 m/s. The minimum height of the water in the tank is (a) 2.22 m (b) 3.54 m (c) 4.60 m (d ) 5.23 m (e) 6.07 m 5–135 Water at 80 kPa (gage) enters a horizontal pipe at a velocity of 1.7 m/s. The pipe makes a 90° angle at the exit and the water exits the pipe vertically into the air. Take the correction factor to be 1. If the irreversible head loss between the inlet and exit of the pipe is 3 m, the height the water jet can rise is (a) 3.4 m (b) 5.3 m (c) 8.2 m (d ) 10.5 m (e) 12.3 m 5–136 Liquid ethanol (𝜌 = 783 kg/m3) at a pressure of 230 kPa enters a 10-cm-diameter pipe at a rate of 2.8 kg/s. Ethanol leaves the pipe at 100 kPa at 15 m above the inlet level. Take the correction factor to be 1. If the diameter of cen96537_ch05_189-248.indd 246 14/01/17 2:40 pm 247 CHAPTER 5 the pipe at the exit is 12 cm, the irreversible head loss in this pipe is (a) 0.95 m (b) 1.93 m (c) 1.23 m (d ) 4.11 m (e) 2.86 m 5–137 Seawater is to be pumped into a large tank at a rate of 165 kg/min. The tank is open to the atmosphere and the water enters the tank from a 80-m-height. The overall effi ciency of the motor-pump unit is 75 percent and the motor consumes electricity at a rate of 3.2 kW. Take the correction factor to be 1. If the irreversible head loss in the piping is 7 m, the velocity of the water at the tank inlet is (a) 2.34 m/s (b) 4.05 m/s (c) 6.21 m/s (d ) 8.33 m/s (e) 10.7 m/s 5–138 An adiabatic pump is used to increase the pressure of water from 100 kPa to 500 kPa at a rate of 400 L/min. If the efficiency of the pump is 75 percent, the maximum tempera ture rise of the water across the pump is (a) 0.096°C (b) 0.058°C (c) 0.035°C (d ) 1.52°C (e) 1.27°C 5–139 The shaft power from a 90 percent-efficient tur bine is 500 kW. If the mass flow rate through the turbine is 440 kg/s, the extracted head removed from the fluid by the turbine is (a) 44.0 m (b) 49.5 (c) 142 m (d ) 129 m (e) 98.5 m Design and Essay Problems 5–140 Using a large bucket whose volume is known and measuring the time it takes to fill the bucket with water from a garden hose, determine the mass flow rate and the average velocity of water through the hose.
5–141 Your company is setting up an experiment that involves the measurement of airflow rate in a duct, and you are to come up with proper instrumentation. Research the available techniques and devices for airflow rate measure ment, discuss the advantages and disadvantages of each tech nique, and make a recommendation.
5–142 Computer-aided designs, the use of better materi als, and better manufacturing techniques have resulted in a tremendous increase in the efficiency of pumps, turbines, and electric motors. Contact one or more pump, turbine, and motor manufacturers and obtain information about the effi ciency of their products. In general, how does efficiency vary with rated power of these devices?
5–143 Using a handheld bicycle pump to generate an air jet, a soda can as the water reservoir, and a straw as the tube, design and build an atomizer. Study the effects of various parameters such as the tube length, the diameter of the exit hole, and the pumping speed on performance.
5–144 Using a flexible drinking straw and a ruler, explain how you would measure the water flow velocity in a river.
5–145 The power generated by a wind turbine is propor tional to the cube of the wind velocity. Inspired by the accel eration of a fluid in a nozzle, someone proposes to install a reducer casing to capture the wind energy from a larger area and accelerate it before the wind strikes the turbine blades, as shown in Fig. P5–145. Evaluate if the proposed modification should be given a consideration in the design of new wind turbines.
Wind FIGURE P5–145 cen96537_ch05_189-248.indd 247 14/01/17 2:40 pm This page intentionally left blank 6 CHAPTER 249 M O M E N T UM AN ALYSI S O F F LOW SYST E M S W hen dealing with engineering problems, it is desirable to obtain fast and accurate solutions at minimal cost. Most engineering problems, including those associated with fluid flow, can be analyzed using one of three basic approaches: differential, experimental, and control volume. In differential approaches, the problem is formulated accu rately using differential quantities, but the solution of the resulting differ ential equations is difficult, usually requiring the use of numerical methods with extensive computer codes. Experimental approaches complemented with dimensional analysis are highly accurate, but they are typically time consuming and expensive. The finite control volume approach described in this chapter is remarkably fast and simple and usually gives answers that are sufficiently accurate for most engineering purposes. Therefore, despite the approximations involved, the basic finite control volume analysis performed with paper and pencil has always been an indispensable tool for engineers.
In Chap. 5, the control volume mass and energy analysis of fluid flow systems was presented. In this chapter, we present the finite control volume momentum analysis of fluid flow problems. First we give an overview of Newton’s laws and the conservation relations for linear and angular momen tum. Then using the Reynolds transport theorem, we develop the linear momentum and angular momentum equations for control volumes and use them to determine the forces and torques associated with fluid flow.
OBJECTIVES When you finish reading this chapter, you should be able to ■ ■ Identify the various kinds of forces and moments acting on a control volume ■ ■ Use control volume analysis to determine the forces associ ated with fluid flow ■ ■ Use control volume analysis to determine the moments caused by fluid flow and the torque transmitted Steady swimming of the jellyfish Aurelia aurita. Fluorescent dye placed directly upstream of the animal is drawn underneath the bell as the body relaxes and forms vortex rings below the animal as the body contracts and ejects fluid. The vortex rings simultaneously induce flows for both feeding and propulsion.
Adapted from Dabiri et al., J. Exp. Biol. 208: 1257–1265. Photo by Sean P. Colin and John H. Costello.
cen96537_ch06_249-296.indd 249 14/01/17 2:46 pm 250 MOMENTUM ANALYSIS OF FLOW SYSTEMS 6–1 ■ NEWTON’S LAWS Newton’s laws are relations between motions of bodies and the forces acting on them. Newton’s first law states that a body at rest remains at rest, and a body in motion remains in motion at the same velocity in a straight path when the net force acting on it is zero. Therefore, a body tends to preserve its state of inertia. Newton’s second law states that the acceleration of a body is proportional to the net force acting on it and is inversely propor tional to its mass. Newton’s third law states that when a body exerts a force on a second body, the second body exerts an equal and opposite force on the first. Therefore, the direction of an exposed reaction force depends on the body taken as the system. For a rigid body of mass m, Newton’s second law is expressed as Newton’s second law: F ›= ma ›= m dV › dt = d(mV ›) dt (6–1) where F › is the net force acting on the body and a → is the acceleration of the body under the influence of F ›. The product of the mass and the velocity of a body is called the linear momentum or just the momentum of the body. The momentum of a rigid body of mass m moving with velocity V › is mV › (Fig. 6–1). Then Newton’s second law expressed in Eq. 6–1 can also be stated as the rate of change of the momentum of a body is equal to the net force acting on the body (Fig. 6–2). This statement is more in line with Newton’s original statement of the second law, and it is more appropriate for use in fluid mechanics when studying the forces generated as a result of velocity changes of fluid streams. Therefore, in fluid mechanics, Newton’s second law is usually referred to as the linear momentum equation. The momentum of a system remains constant only when the net force act ing on it is zero, and thus the momentum of such a system is conserved. This is known as the conservation of momentum principle. This principle has proven to be a very useful tool when analyzing collisions such as those between balls; between balls and rackets, bats, or clubs; and between atoms or subatomic particles; and explosions such as those that occur in rockets, missiles, and guns. In fluid mechanics, however, the net force acting on a system is typically not zero, and we prefer to work with the linear momentum equation rather than the conservation of momentum principle. Note that force, acceleration, velocity, and momentum are vector quantities, and as such they have direction as well as magnitude. Also, momentum is a constant multiple of velocity, and thus the direction of momentum is the direction of velocity as shown in Fig. 6–1. Any vector equation can be written in scalar form for a specified direction using magnitudes, e.g., Fx = max = d(mVx)/dt in the x-direction. The counterpart of Newton’s second law for rotating rigid bodies is ex -pressed as M › = I𝛼 →, where M › is the net moment or torque applied on the body, I is the moment of inertia of the body about the axis of rotation, and 𝛼 → is the angular acceleration. It can also be expressed in terms of the rate of change of angular momentum d H ›/dt as Angular momentum equation: M ›= I 𝛼 ›= I d 𝜔 › dt = d(I 𝜔 ›) dt = d H › dt (6–2) V mV m m FIGURE 6–1 Linear momentum is the product of mass and velocity, and its direction is the direction of velocity.
Net force Rate of change of momentum F = ma = m V d dt dt = d(mV ) FIGURE 6–2 Newton’s second law is also expressed as the rate of change of the momentum of a body is equal to the net force acting on it.
cen96537_ch06_249-296.indd 250 14/01/17 2:46 pm 251 CHAPTER 6 where 𝜔 → is the angular velocity. For a rigid body rotating about a fixed x-axis, the angular momentum equation is written in scalar form as Angular momentum about x-axis: Mx = Ix d𝜔x dt = dHx dt (6–3) The angular momentum equation can be stated as the rate of change of the angular momentum of a body is equal to the net torque acting on it (Fig. 6–3). The total angular momentum of a rotating body remains constant when the net torque acting on it is zero, and thus the angular momentum of such systems is conserved. This is known as the conservation of angular momen tum principle and is expressed as I𝜔 = constant. Many interesting phenom ena such as ice skaters spinning faster when they bring their arms close to their bodies and divers rotating faster when they curl after the jump can be explained easily with the help of the conservation of angular momentum principle (in both cases, the moment of inertia I is decreased and thus the angular velocity 𝜔 is increased as the outer parts of the body are brought closer to the axis of rotation).
6–2 ■ CHOOSING A CONTROL VOLUME We now briefly discuss how to wisely select a control volume. A control volume can be selected as any arbitrary region in space through which fluid flows, and its bounding control surface can be fixed, moving, and even deforming during flow. The application of a basic conservation law is a systematic procedure for bookkeeping or accounting of the quantity under consideration, and thus it is extremely important that the boundaries of the control volume are well defined during an analysis. Also, the flow rate of any quantity into or out of a control volume depends on the flow velocity relative to the control surface, and thus it is essential to know if the control volume remains at rest during flow or if it moves. Many flow systems involve stationary hardware firmly fixed to a station ary surface, and such systems are best analyzed using fixed control volumes. When determining the reaction force acting on a tripod holding the nozzle of a hose, for example, a natural choice for the control volume is one that passes perpendicularly through the nozzle exit flow and through the bottom of the tripod legs (Fig. 6–4a). This is a fixed control volume, and the water velocity relative to a fixed point on the ground is the same as the water velocity relative to the nozzle exit plane. When analyzing flow systems that are moving or deforming, it is usu ally more convenient to allow the control volume to move or deform. When determining the thrust developed by the jet engine of an airplane cruising at constant velocity, for example, a wise choice of control volume is one that encloses the airplane and cuts through the nozzle exit plane (Fig. 6–4b). The control volume in this case moves with velocity V › CV, which is identical to the cruising velocity of the airplane relative to a fixed point on earth. When determining the flow rate of exhaust gases leaving the nozzle, the proper velocity to use is the velocity of the exhaust gases relative to the nozzle exit plane, that is, the relative velocity V › r. Since the entire control volume moves at velocity V › CV, the relative velocity becomes V › r = V › − V › CV, where V › is the α d(Iω) M = I = I dω dt dt = dH dt = Net torque Rate of change of angular momentum FIGURE 6–3 The rate of change of the angular momentum of a body is equal to the net torque acting on it.
cen96537_ch06_249-296.indd 251 14/01/17 2:46 pm 252 MOMENTUM ANALYSIS OF FLOW SYSTEMS absolute velocity of the exhaust gases, i.e., the velocity relative to a fixed point on earth. Note that V › r is the fluid velocity expressed relative to a coor dinate system moving with the control volume. Also, this is a vector equa tion, and velocities in opposite directions have opposite signs. For example, if the airplane is cruising at 500 km/h to the left, and the velocity of the exhaust gases is 800 km/h to the right relative to the ground, the velocity of the exhaust gases relative to the nozzle exit is V › r = V ›−V › CV = 800 i ›−(−500 i › ) = 1300 i › km/h That is, the exhaust gases leave the nozzle at 1300 km/h to the right rela tive to the nozzle exit (in the direction opposite to that of the airplane); this is the velocity that should be used when evaluating the outflow of exhaust gases through the control surface (Fig. 6–4b). Note that the exhaust gases would appear motionless to an observer on the ground if the relative veloc ity were equal in magnitude to the airplane velocity. When analyzing the purging of exhaust gases from a reciprocating inter nal combustion engine, a wise choice for the control volume is one that comprises the space between the top of the piston and the cylinder head (Fig. 6–4c). This is a deforming control volume, since part of the control surface moves relative to other parts. The relative velocity for an inlet or outlet on the deforming part of a control surface (there are no such inlets or outlets in Fig. 6–4c) is then given by V › r = V › − V › CS where V › is the absolute fluid velocity and V › CS is the control surface velocity, both relative to a fixed point outside the control volume. Note that V › CS = V › CV for moving but nondeforming control volumes, and V › CS = V › CV = 0 for fixed ones.
6–3 ■ FORCES ACTING ON A CONTROL VOLUME The forces acting on a control volume consist of body forces that act throughout the entire body of the control volume (such as gravity, electric, and magnetic forces) and surface forces that act on the control surface (such as pressure and viscous forces and reaction forces at points of contact). Only external forces are considered in the analysis. Internal forces (such as the pressure force between a fluid and the inner surfaces of the flow section) are not considered in a control volume analysis unless they are exposed by passing the control surface through that area. In control volume analysis, the sum of all forces acting on the control vol ume at a particular instant in time is represented by Σ F › and is expressed as Total force acting on control volume: ∑F ›= ∑F › body + ∑F › surface (6–4) Body forces act on each volumetric portion of the control volume. The body force acting on a differential element of fluid of volume dV within the con trol volume is shown in Fig. 6–5, and we must perform a volume integral to account for the net body force on the entire control volume. Surface forces act on each portion of the control surface. A differential surface element of area dA and unit outward normal n → on the control surface is shown in Fig. 6–5, along with the surface force acting on it. We must perform an area integral to obtain the net surface force acting on the entire control surface. As sketched, the surface force may act in a direction independent of that of the outward normal vector.
V V (a) (b) (c) CV V V V CV r r Moving control volume Deforming control volume Fixed control volume x x y VCS FIGURE 6–4 Examples of (a) fixed, (b) moving, and (c) deforming control volumes.
cen96537_ch06_249-296.indd 252 14/01/17 2:46 pm 253 CHAPTER 6 The most common body force is that of gravity, which exerts a down ward force on every differential element of the control volume. While other body forces, such as electric and magnetic forces, may be important in some analyses, we consider only gravitational forces here. The differential body force d F › body = d F › gravity acting on the small fluid ele ment shown in Fig. 6–6 is simply its weight, Gravitational force acting on a fluid element: dF › gravity = ρg› dV (6–5) where 𝜌 is the average density of the element and g → is the gravitational vector. In Cartesian coordinates we adopt the convention that g → acts in the negative z-direction, as in Fig. 6–6, so that Gravitational vector in Cartesian coordinates: g ›= −gk › (6–6) Note that the coordinate axes in Fig. 6–6 are oriented so that the gravity vector acts downward in the −z-direction. On earth at sea level, the gravita tional constant g is equal to 9.807 m/s2. Since gravity is the only body force being considered, integration of Eq. 6–5 yields Total body force acting on control volume: ∑F › body = ∫CV ρg › dV = mCVg › (6–7) Surface forces are not as simple to analyze since they consist of both normal and tangential components. Furthermore, while the physical force acting on a surface is independent of orientation of the coordinate axes, the description of the force in terms of its coordinate components changes with orientation (Fig. 6–7). In addition, we are rarely fortunate enough to have each of the control surfaces aligned with one of the coordinate axes. While not desiring to delve too deeply into tensor algebra, we are forced to define a second-order tensor called the stress tensor 𝜎ij in order to adequately describe the surface stresses at a point in the flow, Stress tensor in Cartesian coordinates: 𝜎ij = ( 𝜎xx 𝜎yx 𝜎zx 𝜎xy 𝜎yy 𝜎zy 𝜎xz 𝜎yz 𝜎zz) (6–8) The diagonal components of the stress tensor, 𝜎xx, 𝜎yy, and 𝜎zz, are called normal stresses; they are composed of pressure (which always acts inwardly normal) and viscous stresses. Viscous stresses are discussed in more detail in Chap. 9. The off-diagonal components, 𝜎xy, 𝜎zx, etc., are called shear stresses; since pressure can act only normal to a surface, shear stresses are composed entirely of viscous stresses. When the face is not parallel to one of the coordinate axes, mathematical laws for axes rotation and tensors can be used to calculate the normal and tangential components acting at the face. In addition, an alternate notation called tensor notation is convenient when working with tensors but is usu ally reserved for graduate studies. (For a more in-depth analysis of tensors and tensor notation see, for example, Kundu and Cohen, 2011.) In Eq. 6–8, 𝜎ij is defined as the stress (force per unit area) in the j-direction acting on a face whose normal is in the i-direction. Note that i and j are merely indices of the tensor and are not the same as unit vectors i › and j ›. For example, 𝜎xy is defined as positive for the stress pointing in the y-direction on a face whose outward normal is in the x-direction. This component of the body Control volume (CV) Control surface (CS) n dFbody dFsurface dA dV g FIGURE 6–5 The total force acting on a control volume is composed of body forces and surface forces; body force is shown on a differential volume element, and surface force is shown on a differential surface element.
g dFbody = dFgravity = ρg dV z, k y, j x, i dy dz dx dV,ρ FIGURE 6–6 The gravitational force acting on a differential volume element of fluid is equal to its weight; the axes are oriented so that the gravity vector acts downward in the negative z-direction.
cen96537_ch06_249-296.indd 253 14/01/17 2:46 pm 254 MOMENTUM ANALYSIS OF FLOW SYSTEMS stress tensor, along with the other eight components, is shown in Fig. 6–8 for the case of a differential fluid element aligned with the axes in Carte sian coordinates. All the components in Fig. 6–8 are shown on positive faces (right, top, and front) and in their positive orientation by definition. Positive stress components on the opposing faces of the fluid element (not shown) point in exactly opposite directions. The dot product of a second-order tensor and a vector yields a second vector; this operation is often called the contracted product or the inner product of a tensor and a vector. In our case, it turns out that the inner product of the stress tensor 𝜎ij and the unit outward normal vector n → of a differential surface element yields a vector whose magnitude is the force per unit area acting on the surface element and whose direction is the direction of the surface force itself. Mathematically we write Surface force acting on a differential surface element: dF › surface = 𝜎ij · n › dA (6–9) Finally, we integrate Eq. 6–9 over the entire control surface, Total surface force acting on control surface: ∑F › surface = ∫ CS 𝜎ij · n › dA (6–10) Substitution of Eqs. 6–7 and 6–10 into Eq. 6–4 yields ∑F ›= ∑F › body + ∑F › surface = ∫ CV ρg › dV + ∫ CS 𝜎ij · n › dA (6–11) This equation turns out to be quite useful in the derivation of the differ ential form of conservation of linear momentum, as discussed in Chap. 9. For practical control volume analysis, however, it is rare that we need to use Eq. 6–11, especially the cumbersome surface integral that it contains. A careful selection of the control volume enables us to write the total force acting on the control volume, Σ F ›, as the sum of more readily avail able quantities like weight, pressure, and reaction forces. We recommend the following for control volume analysis: Total force: ∑F ›= ∑F › gravity + ∑F › pressure + ∑F › viscous + ∑F › other (6–12) total force body force surface forces The first term on the right-hand side of Eq. 6–12 is the body force weight, since gravity is the only body force we are considering. The other three terms combine to form the net surface force; they are pressure forces, vis cous forces, and “other” forces acting on the control surface. Σ F › other is com posed of reaction forces required to turn the flow; forces at bolts, cables, struts, or walls through which the control surface cuts; etc. All these surface forces arise as the control volume is isolated from its surroundings for analysis, and the effect of any detached object is accounted for by a force at that location. This is similar to drawing a free-body dia gram in your statics and dynamics classes. We should choose the control volume such that forces that are not of interest remain internal, and thus they do not complicate the analysis. A well-chosen control volume exposes only the forces that are to be determined (such as reaction forces) and a minimum number of other forces.
Control surface y x (a) (b) dFsurface dFsurface, y dFsurface, x dFsurface, normal n dFsurface, tangential dA Control surface y x dFsurface, y dFsurface, x dFsurface, normal dFsurface, tangential dA dFsurface n FIGURE 6–7 When coordinate axes are rotated (a) to (b), the components of the surface force change, even though the force itself remains the same; only two dimensions are shown here.
cen96537_ch06_249-296.indd 254 14/01/17 2:46 pm 255 CHAPTER 6 A common simplification in the application of Newton’s laws of motion is to subtract the atmospheric pressure and work with gage pressures. This is because atmospheric pressure acts in all directions, and its effect cancels out in every direction (Fig. 6–9). This means we can also ignore the pressure forces at outlet sections where the fluid is discharged at subsonic velocities to the atmosphere since the discharge pressures in such cases are very near atmospheric pressure. As an example of how to wisely choose a control volume, consider con trol volume analysis of water flowing steadily through a faucet with a par tially closed gate valve spigot (Fig. 6–10). It is desired to calculate the net force on the flange to ensure that the flange bolts are strong enough. There are many possible choices for the control volume. Some engineers restrict their control volumes to the fluid itself, as indicated by CV A (the purple control volume) in Fig. 6–10. With this control volume, there are pressure forces that vary along the control surface, there are viscous forces along the pipe wall and at locations inside the valve, and there is a body force, namely, the weight of the water in the control volume. Fortunately, to cal culate the net force on the flange, we do not need to integrate the pressure and viscous stresses all along the control surface. Instead, we can lump the unknown pressure and viscous forces together into one reaction force, repre senting the net force of the walls on the water. This force, plus the weight of the faucet and the water, is equal to the net force on the flange. (We must be very careful with our signs, of course.) When choosing a control volume, you are not limited to the fluid alone. Often it is more convenient to slice the control surface through solid objects such as walls, struts, or bolts as illustrated by CV B (the red control volume) in Fig. 6–10. A control volume may even surround an entire object, like the one shown here. Control volume B is a wise choice because we are not con cerned with any details of the flow or even the geometry inside the control volume. For the case of CV B, we assign a net reaction force acting at the portions of the control surface that slice through the flange bolts. Then, the only other things we need to know are the gage pressure of the water at the flange (the inlet to the control volume) and the weights of the water and the faucet assembly. The pressure everywhere else along the control surface is atmospheric (zero gage pressure) and cancels out. This problem is revis ited in Section 6–4, Example 6–7.
6–4 ■ THE LINEAR MOMENTUM EQUATION Newton’s second law for a system of mass m subjected to net force Σ F › is expressed as ∑F ›= ma ›= m dV › dt = d dt (mV ›) (6–13) where mV › is the linear momentum of the system. Noting that both the density and velocity may change from point to point within the system, Newton’s second law can be expressed more generally as ∑F ›= d dt∫sys ρV › dV (6–14) dy dz dx σxz σxx σxy σyz σyy σyx σzy σzx σzz y x z FIGURE 6–8 Components of the stress tensor in Cartesian coordinates on the right, top, and front faces.
FR P1 W Patm Patm P1 (gage) With atmospheric pressure considered With atmospheric pressure cancelled out FR W FIGURE 6–9 Atmospheric pressure acts in all directions, and thus it can be ignored when performing force balances since its effect cancels out in every direction.
Wfaucet Wwater CV B Out Spigot In Bolts x z CV A FIGURE 6–10 Cross section through a faucet assembly, illustrating the importance of choosing a control volume wisely; CV B is much easier to work with than CV A.
cen96537_ch06_249-296.indd 255 14/01/17 2:46 pm 256 MOMENTUM ANALYSIS OF FLOW SYSTEMS where 𝜌V › dV is the momentum of a differential element dV, which has mass 𝛿m = 𝜌 dV. Therefore, Newton’s second law can be stated as the sum of all external forces acting on a system is equal to the time rate of change of linear momentum of the system. This statement is valid for a coordinate system that is at rest or moves with a constant velocity, called an inertial coordinate system or inertial reference frame. Accelerating systems such as aircraft during takeoff are best analyzed using noninertial (or accel erating) coordinate systems fixed to the aircraft. Note that Eq. 6–14 is a vector relation, and thus the quantities F › and V › have direction as well as magnitude. Equation 6–14 is for a given mass of a solid or fluid and is of limited use in fluid mechanics since most flow systems are analyzed using control vol umes. The Reynolds transport theorem developed in Section 4–6 provides the necessary tools to shift from the system formulation to the control vol ume formulation. Setting b = V › and thus B = mV ›, the Reynolds transport theorem is expressed for linear momentum as (Fig. 6–11) d(mV ›)sys dt = d dt ∫CV ρV › dV + ∫CS ρV › (V › r · n › ) dA (6–15) The left-hand side of this equation is, from Eq. 6–13, equal to Σ F ›. Substi tuting, the general form of the linear momentum equation that applies to fixed, moving, or deforming control volumes is General: ∑F ›= d dt ∫CV ρV › dV + ∫CS ρV › (V › r· n › ) dA (6–16) which is stated in words as ( The sum of all external forces acting on a CV) = ( The time rate of change of the linear momentum of the contents of the CV) + ( The net flow rate of linear momentum out of the control surface by mass flow) Here V › r = V › − V › CS is the fluid velocity relative to the control surface (for use in mass flow rate calculations at all locations where the fluid crosses the control surface), and V › is the fluid velocity as viewed from an inertial refer ence frame. The product 𝜌(V › r·n →) dA represents the mass flow rate through area element dA into or out of the control volume. For a fixed control volume (no motion or deformation of the control volume), V › r = V › and the linear momentum equation becomes Fixed CV: ∑F ›= d dt ∫CV ρV › dV + ∫CS ρV › (V › · n › ) dA (6–17) Note that the momentum equation is a vector equation, and thus each term should be treated as a vector. Also, the components of this equation can be resolved along orthogonal coordinates (such as x, y, and z in the Cartesian coordinate system) for convenience. The sum of forces Σ F › in most cases consists of weights, pressure forces, and reaction forces (Fig. 6–12). The momentum equation is commonly used to calculate the forces (usually on support systems or connectors) induced by the flow.
= + ρb dV B = mV dBsys dt d dt CV ∫ ρb(V r · n ) dA CS ∫ ∫ ∫ = + ρV dV d(mV )sys dt V d dt CV ρV( r · n ) dA CS b = V b = V FIGURE 6–11 The linear momentum equation is obtained by replacing B in the Reynolds transport theorem by the momentum mV ›, and b by the momentum per unit mass V ›.
FR1 FR2 P2,gageA2 P1,gageA1 A2 An 180° elbow supported by the ground (Pressure force) CS (Reaction force) (Reaction force) A1 W (Weight) FIGURE 6–12 In most flow systems, the sum of forces Σ F › consists of weights, pressure forces, and reaction forces. Gage pressures are used here since atmospheric pressure cancels out on all sides of the control surface.
cen96537_ch06_249-296.indd 256 14/01/17 2:46 pm 257 CHAPTER 6 Special Cases Most momentum problems considered in this text are steady. During steady flow, the amount of momentum within the control volume remains constant, and thus the time rate of change of linear momentum of the contents of the control volume (the second term of Eq. 6–16) is zero. Thus, Steady flow: ∑F ›= ∫CS ρV › (V › r· n › ) dA (6–18) For a case in which a non-deforming control volume moves at constant velocity (an inertial reference frame), the first V › in Eq. 6–18 may also be taken relative to the moving control surface. While Eq. 6–17 is exact for fixed control volumes, it is not always con venient when solving practical engineering problems because of the inte grals. Instead, as we did for conservation of mass, we would like to rewrite Eq. 6–17 in terms of average velocities and mass flow rates through inlets and outlets. In other words, our desire is to rewrite the equation in algebraic rather than integral form. In many practical applications, fluid crosses the boundaries of the control volume at one or more inlets and one or more out lets, and carries with it some momentum into or out of the control volume. For simplicity, we always draw our control surface such that it slices normal to the inflow or outflow velocity at each such inlet or outlet (Fig. 6–13). The mass flow rate m . into or out of the control volume across an inlet or outlet at which 𝜌 is nearly constant is Mass flow rate across an inlet or outlet: m · = ∫Ac ρ(V › ·n › ) dAc = ρVavgAc (6–19) Comparing Eq. 6–19 to Eq. 6–17, we notice an extra velocity in the control surface integral of Eq. 6–17. If V › were uniform (V › = V › avg) across the inlet or outlet, we could simply take it outside the integral. Then we could write the rate of inflow or outflow of momentum through the inlet or outlet in simple algebraic form, Momentum flow rate across a uniform inlet or outlet: ∫Ac ρV › (V › · n › ) dAc = ρVavg AcV › avg = m · V › avg (6–20) The uniform flow approximation is reasonable at some inlets and outlets, e.g., the well-rounded entrance to a pipe, the flow at the entrance to a wind tunnel test section, and a slice through a water jet moving at nearly uniform speed through air (Fig. 6–14). At each such inlet or outlet, Eq. 6–20 can be applied directly.
Momentum-Flux Correction Factor, β Unfortunately, the velocity across most inlets and outlets of practical engineering interest is not uniform. Nevertheless, it turns out that we can still convert the control surface integral of Eq. 6–17 into algebraic form, but a dimensionless correction factor 𝛽, called the momentum-flux correction factor, is required, Vavg,2 m2, ⋅ m1, Vavg,1 ⋅ In In Out Out Out Fixed control volume m4, Vavg,4 ⋅ m3, Vavg,3 ⋅ m5, Vavg,5 ⋅ FIGURE 6–13 In a typical engineering problem, the control volume may contain multiple inlets and outlets; at each inlet or outlet we define the mass flow rate m . and the average velocity V › avg.
cen96537_ch06_249-296.indd 257 14/01/17 2:46 pm 258 MOMENTUM ANALYSIS OF FLOW SYSTEMS EXAMPLE 6–1 Momentum-Flux Correction Factor for Laminar Pipe Flow Consider laminar flow through a very long straight section of round pipe. It is shown in Chap. 8 that the velocity profile through a cross-sectional area of the pipe is parabolic (Fig. 6–15), with the axial velocity component given by V = 2Vavg(1 −r2 R2) (1) where R is the radius of the inner wall of the pipe and Vavg is the average velocity. Calculate the momentum-flux correction factor through a cross section of the pipe for the case in which the pipe flow represents an outlet of the control volume, as sketched in Fig. 6–15.
SOLUTION For a given velocity distribution we are to calculate the momentum-flux correction factor.
as first shown by the French scientist Joseph Boussinesq (1842–1929). The algebraic form of Eq. 6–17 for a fixed control volume is then written as ∑F ›= d dt ∫CV ρV › dV + ∑ out 𝛽m · V › avg −∑ in 𝛽m · V › avg (6–21) where a unique value of momentum-flux correction factor is applied to each inlet and outlet in the control surface. Note that 𝛽 = 1 for the case of uniform flow over an inlet or outlet, as in Fig. 6–14. For the general case, we define 𝛽 such that the integral form of the momentum flux into or out of the con trol surface at an inlet or outlet of cross-sectional area Ac can be expressed in terms of mass flow rate m . through the inlet or outlet and average velocity V › avg through the inlet or outlet, Momentum flux across an inlet or outlet: ∫Ac ρV › (V › · n › ) dAc = 𝛽m · V › avg (6–22) For the case in which density is uniform over the inlet or outlet and V › is in the same direction as V › avg over the inlet or outlet, we solve Eq. 6–22 for 𝛽, 𝛽= ∫Ac ρV(V › ·n › ) dAc m · Vavg = ∫Ac ρV(V › · n › ) dAc ρVavg AcVavg (6–23) where we have substituted 𝜌Vavg Ac for m · in the denominator. The densi ties cancel and since Vavg is constant, it can be brought inside the integral. Furthermore, if the control surface slices normal to the inlet or outlet area, (V ›·n →) dAc = V dAc. Then, Eq. 6–23 simplifies to Momentum-flux correction factor: 𝛽= 1 Ac ∫Ac ( V Vavg) 2 dAc (6–24) It may be shown that 𝛽 is always greater than or equal to unity.
FIGURE 6–14 Examples of inlets or outlets in which the uniform flow approximation is reasonable: (a) the well-rounded entrance to a pipe, (b) the entrance to a wind tunnel test section, and (c) a slice through a free water jet in air.
CV (a) V ≈ Vavg CV (b) V ≈ Vavg CV Nozzle (c) V ≈ Vavg cen96537_ch06_249-296.indd 258 14/01/17 2:46 pm 259 CHAPTER 6 Assumptions 1 The flow is incompressible and steady. 2 The control volume slices through the pipe normal to the pipe axis, as sketched in Fig. 6–15.
Analysis We substitute the given velocity profile for V in Eq. 6–24 and integrate, noting that dAc = 2𝜋r dr, 𝛽= 1 Ac ∫ Ac ( V Vavg) 2 dAc = 4 𝜋R2 ∫ R 0 (1 −r2 R2) 2 2𝜋r dr (2) Defining a new integration variable y = 1 − r 2/R2 and thus dy = −2r dr/R 2 (also, y = 1 at r = 0, and y = 0 at r = R) and performing the integration, the momentum-flux correction factor for fully developed laminar flow becomes Laminar flow: 𝛽= −4 ∫ 0 1 y2 dy = −4[ y3 3 ] 0 1 = 4 3 (3) Discussion We have calculated 𝛽 for an outlet, but the same result would have been obtained if we had considered the cross section of the pipe as an inlet to the control volume. From Example 6–1 we see that 𝛽 is not very close to unity for fully devel oped laminar pipe flow, and ignoring 𝛽 could potentially lead to significant error. If we were to perform the same kind of integration as in Example 6–1 but for fully developed turbulent rather than laminar pipe flow, we would find that 𝛽 ranges from about 1.01 to 1.04. Since these values are so close to unity, many practicing engineers completely disregard the momentum-flux correction factor. While the neglect of 𝛽 in turbulent flow calculations may have an insignificant effect on the final results, it is wise to keep it in our equations. Doing so not only improves the accuracy of our calculations, but reminds us to include the momentum-flux correction factor when solving laminar flow control volume problems.
For turbulent flow 𝛽 may have an insignificant effect at inlets and outlets, but for laminar flow 𝛽 may be important and should not be neglected. It is wise to include 𝛽 in all momentum control volume problems.
Steady Flow If the flow is also steady, the time derivative term in Eq. 6–21 vanishes and we are left with Steady linear momentum equation: ∑F ›= ∑ out 𝛽m · V ›−∑ in 𝛽 m · V › (6–25) where we have dropped the subscript “avg” from average velocity. Equa tion 6–25 states that the net force acting on the control volume during steady flow is equal to the difference between the rates of outgoing and incoming momentum flows. This statement is illustrated in Fig. 6–16. It can also be expressed for any direction, since Eq. 6–25 is a vector equation.
Vavg V R r CV FIGURE 6–15 Velocity profile over a cross section of a pipe in which the flow is fully developed and laminar.
In In Out out in Fixed control volume Out Out V 2 β2m2 ⋅ V 1 β1m1 ⋅ ⋅ ⋅ V 3 β3m3 ⋅ V 4 β4m4 ⋅ V 5 β5m5 ⋅ ΣF = ΣβmV – ΣβmV ΣF FIGURE 6–16 The net force acting on the control volume during steady flow is equal to the difference between the outgoing and the incoming momentum fluxes.
cen96537_ch06_249-296.indd 259 14/01/17 2:46 pm 260 MOMENTUM ANALYSIS OF FLOW SYSTEMS Steady Flow with One Inlet and One Outlet Many practical engineering problems involve just one inlet and one outlet (Fig. 6–17). The mass flow rate for such single-stream systems remains constant, and Eq. 6–25 reduces to One inlet and one outlet: ∑F ›= m · (𝛽2V › 2 −𝛽1V › 1) (6–26) where we have adopted the usual convention that subscript 1 implies the inlet and subscript 2 the outlet, and V › 1 and V › 2 denote the average velocities across the inlet and outlet, respectively. We emphasize again that all the preceding relations are vector equations, and thus all the additions and subtractions are vector additions and subtractions. Recall that subtracting a vector is equivalent to adding it after reversing its direction (Fig. 6–18). When writing the momentum equation for a specified coordinate direction (such as the x-axis), we use the projections of the vec tors on that axis. For example, Eq. 6–26 is written along the x-coordinate as Along x-coordinate: ∑Fx = m · (𝛽2V2, x −𝛽1V1, x) (6–27) where ΣFx is the vector sum of the x-components of the forces, and V2, x and V1, x are the x-components of the outlet and inlet velocities of the fluid stream, respectively. The force or velocity components in the positive x-direction are positive quantities, and those in the negative x-direction are negative quantities. Also, it is good practice to take the direction of unknown forces in the positive directions (unless the problem is very straightforward). A negative value obtained for an unknown force indicates that the assumed direction is wrong and should be reversed.
Flow with No External Forces An interesting situation arises when there are no external forces (such as weight, pressure, and reaction forces) acting on the body in the direction of motion—a common situation for space vehicles and satellites. For a control volume with multiple inlets and outlets, Eq. 6–21 reduces in this case to No external forces: 0 = d(mV › )CV dt + ∑ out 𝛽m · V ›−∑ in 𝛽m · V › (6–28) This is an expression of the conservation of momentum principle, which is stated in words as in the absence of external forces, the rate of change of the momentum of a control volume is equal to the difference between the rates of incoming and outgoing momentum flow rates. When the mass m of the control volume remains nearly constant, the first term of Eq. 6–28 becomes simply mass times acceleration, since d(mV ›)CV dt = mCV dV › CV dt = (ma › )CV = mCVa › Therefore, the control volume in this case can be treated as a solid body (a fixed-mass system) with a net thrusting force (or just thrust) of Thrust: F › thrust = mbodya ›=∑ in 𝛽m · V ›−∑ out 𝛽m · V › (6– 29) acting on the body. In Eq 6–29, fluid velocities are relative to an inertial reference frame—that is, a coordinate system that is fixed in space or is In Out Fixed control volume 2 1 V 1 β1m ⋅ ⋅ V 2 β2m ⋅ ΣF = m(β2V2 – β1V ) → ΣF FIGURE 6–17 A control volume with only one inlet and one outlet.
(Reaction force) Support Water flow CS Note: V2 ≠ V1 even if |V2| = |V1| θ θ FR FR V 1 β1m ⋅ –β1mV 1 ⋅ V 2 β2m ⋅ β2mV 2 ⋅ FIGURE 6–18 The determination by vector addition of the reaction force on the support caused by a change of direction of water.
cen96537_ch06_249-296.indd 260 14/01/17 2:46 pm 261 CHAPTER 6 moving uniformly at constant velocity on a straight path. When analyzing the motion of bodies moving at constant velocity on a straight path, it is conve nient to choose an inertial reference frame that moves with the body at the same velocity on the same path. In this case the velocities of fluid streams relative to the inertial reference frame are identical to the velocities relative to the moving body, which are much easier to apply. This approach, while not strictly valid for noninertial reference frames, can also be used to calculate the initial acceleration of a space vehicle when its rocket is fired (Fig. 6–19). Recall that thrust is a mechanical force typically generated through the reaction of an accelerating fluid. In the jet engine of an aircraft, for example, hot exhaust gases are accelerated by the action of expansion and outflow of gases through the back of the engine, and a thrusting force is produced by a reaction in the opposite direction. The generation of thrust is based on Newton’s third law of motion, which states that for every action at a point there is an equal and opposite reaction. In the case of a jet engine, if the engine exerts a force on exhaust gases, then the exhaust gases exert an equal force on the engine in the opposite direction. That is, the pushing force exerted on the departing gases by the engine is equal to the thrusting force the departing gases exert on the remaining mass of the aircraft in the opposite direction F › thrust = −F › push. On the free-body diagram of an aircraft, the effect of outgoing exhaust gases is accounted for by the insertion of a force in the opposite direction of motion of the exhaust gases.
L = 2 m V0 = 2000 m/s FIGURE 6–19 The thrust needed to lift the space shuttle is generated by the rocket engines as a result of momentum change of the fuel as it is accelerated from about zero to an exit speed of about 2000 m/s after combustion.
NASA FRz FRx Patm 30° 30 cm P1,gage z x CV 1 2 mV1 · mV2 · FIGURE 6–20 Schematic for Example 6–2.
EXAMPLE 6–2 The Force to Hold a Deflector Elbow in Place A reducing elbow is used to deflect water flow at a rate of 14 kg/s in a horizontal pipe upward 30° while accelerating it (Fig. 6–20). The elbow discharges water into the atmosphere. The cross-sectional area of the elbow is 113 cm2 at the inlet and 7 cm2 at the outlet. The elevation difference between the centers of the outlet and the inlet is 30 cm. The weight of the elbow and the water in it is considered to be negligible. Determine (a) the gage pressure at the center of the inlet of the elbow and (b) the anchoring force needed to hold the elbow in place.
SOLUTION A reducing elbow deflects water upward and discharges it to the atmosphere. The pressure at the inlet of the elbow and the force needed to hold the elbow in place are to be determined.
Assumptions 1 The flow is steady, and the frictional effects are negligible. 2 The weight of the elbow and the water in it is negligible. 3 The water is dis charged to the atmosphere, and thus the gage pressure at the outlet is zero. 4 The flow is turbulent and fully developed at both the inlet and outlet of the control volume, and we take the momentum-flux correction factor to be 𝛽 = 1.03 (as a conservative estimate) at both the inlet and the outlet.
Properties We take the density of water to be 1000 kg/m3.
Analysis (a) We take the elbow as the control volume and designate the inlet by 1 and the outlet by 2. We also take the x- and z-coordinates as shown. The continuity equation for this one-inlet, one-outlet, steady-flow system is m .
1 = m .
2 = m . = 14 kg/s. Noting that m . = 𝜌AV, the inlet and outlet velocities of water are V1 = m · ρA1 = 14 kg/s (1000 kg/m3)(0.0113 m2) = 1.24 m/s cen96537_ch06_249-296.indd 261 14/01/17 2:46 pm 262 MOMENTUM ANALYSIS OF FLOW SYSTEMS V2 = m · ρA2 = 14 kg/s (1000 kg/m3)(7 × 10−4 m2) = 20.0 m/s We use the Bernoulli equation (Chap. 5) as a first approximation to calculate the pressure. In Chap. 8 we will learn how to account for frictional losses along the walls. Taking the center of the inlet cross section as the reference level (z1 = 0) and noting that P2 = Patm, the Bernoulli equation for a streamline going through the cen ter of the elbow is expressed as P1 ρg + V 2 1 2g + z1 = P2 ρg + V 2 2 2g + z2 P1 −P2 = ρg( V 2 2 −V 2 1 2g + z2 −z1) P1 −Patm = (1000 kg/m3)(9.81 m/s2) × ( (20 m/s)2 −(1.24 m/s)2 2(9.81 m/s2) + 0.3 −0) ( 1 kN 1000 kg·m/s2) P1, gage = 202.2 kN/m2 = 202.2 kPa (gage) (b) The momentum equation for steady flow is ∑F ›= ∑ out 𝛽m · V ›−∑ in 𝛽m · V › We let the x- and z-components of the anchoring force of the elbow be FRx and FRz, and assume them to be in the positive direction. We also use gage pressure since the atmospheric pressure acts on the entire control surface. Then the momentum equa tions along the x- and z-axes become FRx + P1, gage A1 = 𝛽m · V2 cos 𝜃−𝛽m · V1 FRz = 𝛽m · V2 sin 𝜃 where we have set 𝛽 = 𝛽1 = 𝛽2. Solving for FRx and FRz, and substituting the given values, FRx = 𝛽m · (V2 cos 𝜃−V1) −P1, gageA1 = 1.03(14 kg/s)(20 cos 30° −1.24) m/s −(202,200 N/m2)(0.0113 m2) = 232 −2285 = −2053 N FRz = 𝛽m · V2 sin 𝜃= (1.03)(14 kg/s)(20 sin 30° m/s)( 1 N 1 kg·m/s2) = 144 N The negative result for FRx indicates that the assumed direction is wrong, and it should be reversed. Therefore, FRx acts in the negative x-direction.
Discussion There is a nonzero pressure distribution along the inside walls of the elbow, but since the control volume is outside the elbow, these pressures do not appear in our analysis. The weight of the elbow and the water in it could be added to the vertical force for better accuracy. The actual value of P1, gage will be higher than that calculated here because of frictional and other irreversible losses in the elbow.
cen96537_ch06_249-296.indd 262 14/01/17 2:46 pm 263 CHAPTER 6 FRz FRx Patm P1,gage 1 2 mV1 · mV2 · CV FIGURE 6–21 Schematic for Example 6–3.
35 m/s 10 m/s FRx Water jet FIGURE 6–22 Schematic for Example 6–4.
EXAMPLE 6–3 The Force to Hold a Reversing Elbow in Place The deflector elbow in Example 6–2 is replaced by a reversing elbow such that the fluid makes a 180° U-turn before it is discharged, as shown in Fig. 6–21. The elevation difference between the centers of the inlet and the exit sections is still 0.3 m. Determine the anchoring force needed to hold the elbow in place.
SOLUTION The inlet and the outlet velocities and the pressure at the inlet of the elbow remain the same, but the vertical component of the anchoring force at the connection of the elbow to the pipe is zero in this case (FRz = 0) since there is no other force or momentum flux in the vertical direction (we are neglecting the weight of the elbow and the water). The horizontal component of the anchor ing force is determined from the momentum equation written in the x-direction. Noting that the outlet velocity is negative since it is in the negative x-direction, we have FRx + P1, gage A1 = 𝛽2m · (−V2) −𝛽1m · V1 = −𝛽m · (V2 + V1) Solving for FRx and substituting the known values, FRx = −𝛽m · (V2 + V1) −P1, gage A1 = −(1.03)(14 kg/s)(20 + 1.24) m/s−(202,200 N/m2)(0.0113 m2) = −306 −2285 = −2591 N Therefore, the horizontal force on the flange is 2591 N acting in the negative x-direction (the elbow is trying to separate from the pipe). This force is equivalent to the weight of about 260 kg mass, and thus the connectors (such as bolts) used must be strong enough to withstand this force.
Discussion The reaction force in the x-direction is larger than that of Example 6–2 since the walls turn the water over a much greater angle. If the reversing elbow is replaced by a straight nozzle (like one used by firefighters) such that water is discharged in the positive x-direction, the momentum equation in the x-direction becomes FRx + P1, gage A1 = 𝛽m · V2 −𝛽m · V1 → FRx = 𝛽m · (V2 −V1) −P1, gage A1 since both V1 and V2 are in the positive x-direction. This shows the importance of using the correct sign (positive if in the positive direction and negative if in the opposite direction) for velocities and forces.
EXAMPLE 6–4 Water Jet Striking a Moving Cart Water accelerated by a nozzle to 35 m/s strikes the vertical back surface of a cart moving horizontally at a constant velocity of 10 m/s in the flow direction (Fig. 6–22). The mass flow rate of water through the stationary nozzle is 30 kg/s. After the strike, the water stream splatters off in all directions in the plane of the back surface. (a) Determine the force that needs to be applied by the brakes of the cart to prevent it from accelerating. (b) If this force were used to generate power instead of wasting it on the brakes, determine the maximum amount of power that could ideally be generated. (c) If the mass of the cart is 400 kg and cen96537_ch06_249-296.indd 263 14/01/17 2:46 pm 264 MOMENTUM ANALYSIS OF FLOW SYSTEMS the brakes fail, determine the acceleration of the cart when the water first strikes it. Assume the mass of water that wets the back surface is negligible.
SOLUTION Water accelerated by a nozzle strikes the back surface of a cart moving horizontally at a constant velocity. The braking force, the power wasted by the brakes, and the acceleration of the cart if the brakes fail are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The water splatters off the sides of the plate in all directions in the plane of the back surface. 3 The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the splattered water is the atmospheric pressure which is disregarded since it acts on all surfaces. 4 Fiction during motion is negligible. 5 The motions of the water jet and the cart are horizontal. 6 Jet flow is nearly uniform and thus the effect of the momentum-flux correction factor is negligible, 𝛽 ≅ 1.
Analysis We take the cart as the control volume, and the direction of flow as the positive direction of x-axis. The relative velocity between the cart and the jet is Vr = Vjet −Vcart = 35 − 10 = 25 m/s Therefore, we can view the cart as being stationary and the jet moving with a velocity of 25 m/s. Noting that water leaves the nozzle at 20 m/s and the corresponding mass flow rate relative to nozzle exit is 30 kg/s, the mass flow rate of water striking the cart corresponding to a water jet velocity of 25 m/s relative to the cart is m .
r = Vr Vjet m .
jet = 25 m/s 35 m/s (30 kg/s) = 21.43 kg/s The momentum equation for steady flow in the x (flow)-direction reduces in this case to ∑F ›= ∑ out 𝛽m · V ›−∑ in 𝛽m · V › → FRx = −m .
iVi → Fbrake = −m .
rVr We note that the brake force acts in the opposite direction to flow, and we should not forget the negative sign for forces and velocities in the negative x-direction. Substituting the given values, Fbrake = −m .
rVr = −(21.43 kg/s)(+25 m/s)( 1 N 1 kg·m/s2) = −535.8 N ≅ −536 N The negative sign indicates that the braking force acts in the opposite direction to motion, as expected. Just as the water jet here imparts a force to the cart, the air jet from a helicopter (downwash) imparts a force on the surface of the water (Fig. 6–23). Noting that work is force times distance and the distance traveled by the cart per unit time is the cart velocity, the power wasted by the brakes is W .
= FbrakeVcart = (535.8 N)(10 m/s)( 1 W 1 N·m/s) = 5358 W ≅ 5.36 kW FIGURE 6–23 The downwash of a helicopter is similar to the jet discussed in Example 6–4. The jet impinges on the surface of the water in this case, causing circular waves as seen here.
© Purestock/Superstock RF cen96537_ch06_249-296.indd 264 14/01/17 2:47 pm 265 CHAPTER 6 Note that the power wasted is equivalent to the maximum power that can be gener ated as the cart velocity is maintained constant. (c) When the brakes fail, the braking force will propel the cart forward, and the acceleration will be a = F mcart = 535.8 N 400 kg ( 1 kg·m/s2 1 N ) = 1.34 m/s2 Discussion This is the acceleration at the moment the brakes fail. The accelera tion will decrease as the relative velocity between the water jet and the cart (and thus the force) decreases.
EXAMPLE 6–5 Power Generation and Wind Loading of a Wind Turbine A wind generator with a 30-ft-diameter blade span has a cut-in wind speed (minimum speed for power generation) of 7 mph, at which velocity the turbine generates 0.4 kW of electric power (Fig. 6–24). Determine (a) the efficiency of the wind turbine–genera tor unit and (b) the horizontal force exerted by the wind on the supporting mast of the wind turbine. What is the effect of doubling the wind velocity to 14 mph on power generation and the force exerted? Assume the efficiency remains the same, and take the density of air to be 0.076 lbm/ft3.
SOLUTION The power generation and loading of a wind turbine are to be ana lyzed. The efficiency and the force exerted on the mast are to be determined, and the effects of doubling the wind velocity are to be investigated.
Assumptions 1 The wind flow is steady and incompressible. 2 The effi ciency of the turbine–generator is independent of wind speed. 3 The frictional effects are negligible, and thus none of the incoming kinetic energy is con verted to thermal energy. 4 The average velocity of air through the wind turbine is the same as the wind velocity (actually, it is considerably less—see Chap. 14). 5 The wind flow is nearly uniform upstream and downstream of the wind turbine and thus the momentum-flux correction factor is 𝛽 = 𝛽1 = 𝛽2 ≅ 1.
Properties The density of air is given to be 0.076 lbm/ft3.
Analysis Kinetic energy is a mechanical form of energy, and thus it can be con verted to work entirely. Therefore, the power potential of the wind is proportional to its kinetic energy, which is V 2/2 per unit mass, and thus the maximum power is m . V 2/2 for a given mass flow rate: V1 = (7 mph)( 1.4667 ft/s 1 mph ) = 10.27 ft/s m · = ρ1V1A1 = ρ1V1 𝜋D2 4 = (0.076 lbm/ft3)(10.27 ft/s) 𝜋(30 ft)2 4 = 551.7 lbm/s W · max = m · ke1 = m · V 2 1 2 = (551.7 lbm/s) (10.27 ft/s)2 2 ( 1 lbf 32.2 lbm·ft/s2) ( 1 kW 737.56 lbf·ft/s) = 1.225 kW 1 2 Patm Patm mV1 · FR Streamline x mV2 · CV FIGURE 6–24 Schematic for Example 6–5.
cen96537_ch06_249-296.indd 265 14/01/17 2:47 pm 266 MOMENTUM ANALYSIS OF FLOW SYSTEMS Therefore, the available power to the wind turbine is 1.225 kW at the wind velocity of 7 mph. Then the turbine–generator efficiency becomes 𝜂wind turbine = W · act W · max = 0.4 kW 1.225 kW = 0.327 (or 32.7%) (b) The frictional effects are assumed to be negligible, and thus the portion of incoming kinetic energy not converted to electric power leaves the wind turbine as outgoing kinetic energy. Noting that the mass flow rate remains constant, the exit velocity is determined to be m · ke2 = m · ke1(1 −𝜂wind turbine) → m · V 2 2 2 = m · V 2 1 2 (1 −𝜂wind turbine) (1) or V2 = V1√1 −𝜂wind turbine = (10.27 ft/s)√1 −0.327 = 8.43 ft/s To determine the force on the mast (Fig. 6–25), we draw a control volume around the wind turbine such that the wind is normal to the control surface at the inlet and the outlet and the entire control surface is at atmospheric pressure (Fig. 6–23). The momentum equation for steady flow is given as ∑F ›= ∑ out 𝛽m · V ›−∑ in 𝛽m · V › (2) Writing Eq. 2 along the x-direction and noting that 𝛽 = 1, V1, x = V1, and V2, x = V2 give FR = m · V2 −m · V1 = m · (V2 −V1) (3) Substituting the known values into Eq. 3 gives FR = m · (V2 −V1) = (551.7 lbm/s)(8.43 −10.27 ft/s) ( 1 lbf 32.2 lbm·ft/s2) = −31.5 lbf The negative sign indicates that the reaction force acts in the negative x-direction, as expected. Then the force exerted by the wind on the mast becomes Fmast = −FR = 31.5 lbf. The power generated is proportional to V 3 since the mass flow rate is p roportional to V and the kinetic energy to V 2. Therefore, doubling the wind velocity to 14 mph will increase the power generation by a factor of 23 = 8 to 0.4 × 8 = 3.2 kW. The force exerted by the wind on the support mast is proportional to V 2. Therefore, doubling the wind velocity to 14 mph will increase the wind force by a factor of 22 = 4 to 31.5 × 4 = 126 lbf.
Discussion Wind turbines are treated in more detail in Chap. 14.
800 m/s 80 kg/s 3000 m/s x FIGURE 6–26 Schematic for Example 6–6.
© Brand X Pictures/PureStock RF FIGURE 6–25 Forces and moments on the supporting mast of a modern wind turbine can be substantial, and increase like V 2; thus the mast is typically quite large and strong.
© Ingram Publishing/SuperStock RF EXAMPLE 6–6 Deceleration of a Spacecraft A spacecraft with a mass of 12,000 kg is dropping vertically towards a planet at a constant speed of 800 m/s (Fig. 6–26). To slow down the spacecraft, a solid-fuel rocket at the bottom is fired, and combustion gases leave the rocket at a constant rate of 80 kg/s and at a velocity of 3000 m/s relative to the spacecraft in the direction of motion of the spacecraft for a period of 5 s. Disregarding the cen96537_ch06_249-296.indd 266 14/01/17 2:47 pm 267 CHAPTER 6 small change in the mass of the spacecraft, determine (a) the deceleration of the spacecraft during this period, (b) the change of velocity of the spacecraft, and (c) the thrust exerted on the spacecraft.
SOLUTION The rocket of a spacecraft is fired in the direction of motion. The deceleration, the velocity change, and the thrust are to be determined. Assumptions 1 The flow of combustion gases is steady and one-dimensional during the firing period, but the flight of the spacecraft is unsteady. 2 There are no external forces acting on the spacecraft, and the effect of pressure force at the nozzle outlet is negligible. 3 The mass of discharged fuel is negligible relative to the mass of the spacecraft, and thus, the spacecraft may be treated as a solid body with a constant mass. 4 The nozzle is well designed such that the effect of the momentum-flux correction factor is negligible, and thus, 𝛽≅ 1.
Analysis (a) For convenience, we choose an inertial reference frame that moves with the spacecraft at the same initial velocity. Then the velocities of the fluid stream relative to an inertial reference frame become simply the velocities relative to the spacecraft. We take the direction of motion of the spacecraft as the positive direction along the x-axis. There are no external forces acting on the spacecraft, and its mass is essentially constant. Therefore, the spacecraft can be treated as a solid body with constant mass, and the momentum equation in this case is, from Eq. 6–29, F › thrust = m spacecraft a › spacecraft = ∑ in 𝛽m · V ›−∑ out 𝛽m · V › where the fluid stream velocities relative to the inertial reference frame in this case are identical to the velocities relative to the spacecraft. Noting that the motion is on a straight line and the discharged gases move in the positive x-direction, we write the momentum equation using magnitudes as mspacecraftaspacecraft = mspacecraft dVspacecraft dt = −m · gasVgas Noting that gases leave in the positive x-direction and substituting, the acceleration of the spacecraft during the first 5 seconds is determined to be aspacecraft = dVspacecraft dt = − m .
gas mspacecraft Vgas = −80 kg/s 12,000 kg (+3000 m/s) = −20 m/s2 The negative value confirms that the spacecraft is decelerating in the positive x direction at a rate of 20 m/s2.
(b) Knowing the deceleration, which is constant, the velocity change of the spacecraft during the first 5 seconds is determined from the definition of accelera tion to be dVspacecraft = aspacecraftdt → ΔVspacecraft = aspacecraftΔt = (−20 m/s2)(5 s) = −100 m/s (c) The thrusting force exerted on the space aircraft is, from Eq. 6–29, Fthrust = 0 −m .
gasV gas = 0 −(80 kg/s)(+3000 m/s)( 1 kN 1000 kg·m/s2)= −240 kN cen96537_ch06_249-296.indd 267 14/01/17 2:47 pm 268 MOMENTUM ANALYSIS OF FLOW SYSTEMS The negative sign indicates that the trusting force due to firing of the rocket acts on the aircraft in the negative x-direction. Discussion Note that if this fired rocket were attached somewhere on a test stand, it would exert a force of 240 kN (equivalent to the weight of about 24 tons of mass) to its support in the opposite direction of the discharged gases.
EXAMPLE 6–7 Net Force on a Flange Water flows at a rate of 18.5 gal/min through a flanged faucet with a partially closed gate valve spigot (Fig. 6–27). The inner diameter of the pipe at the location of the flange is 0.780 in (= 0.0650 ft), and the pressure at that location is measured to be 13.0 psig. The total weight of the faucet assembly plus the water within it is 12.8 lbf. Calculate the net force on the flange.
SOLUTION Water flow through a flanged faucet is considered. The net force acting on the flange is to be calculated.
Assumptions 1 The flow is steady and incompressible. 2 The flow at the inlet and at the outlet is turbulent and fully developed so that the momentum-flux correc tion factor is about 1.03. 3 The pipe diameter at the outlet of the faucet is the same as that at the flange.
Properties The density of water at room temperature is 62.3 lbm/ft3.
Analysis We choose the faucet and its immediate surroundings as the control volume, as shown in Fig. 6–27 along with all the forces acting on it. These forces include the weight of the water and the weight of the faucet assembly, the gage pressure force at the inlet to the control volume, and the net force of the flange on the control volume, which we call F → R. We use gage pressure for convenience since the gage pressure on the rest of the control surface is zero (atmospheric pressure). Note that the pressure through the outlet of the control volume is also atmospheric since we are assuming incompressible flow; hence, the gage pressure is also zero through the outlet. We now apply the control volume conservation laws. Conservation of mass is trivial here since there is only one inlet and one outlet; namely, the mass flow rate into the control volume is equal to the mass flow rate out of the control volume. Also, the outflow and inflow average velocities are identical since the inner diam eter is constant and the water is incompressible, and are determined to be V2 = V1=V = V · Ac = V · 𝜋D2/4 = 18.5 gal/min 𝜋(0.065 ft)2/4 ( 0.1337 ft3 1 gal ) ( 1 min 60 s )=12.42 ft/s Also, m · = ρV · = (62.3 lbm/ft3)(18.5 gal/min)( 0.1337 ft3 1 gal ) ( 1 min 60 s ) = 2.568 lbm/s Next we apply the momentum equation for steady flow, ∑F ›= ∑ out 𝛽m · V ›−∑ in 𝛽m · V › (1) We let the x- and z-components of the force acting on the flange be FRx and FRz, and assume them to be in the positive directions. The magnitude of the velocity Wfaucet Wwater P1,gage CV Out Spigot Flange x z In FR FIGURE 6–27 Control volume for Example 6–7 with all forces shown; gage pressure is used for convenience.
cen96537_ch06_249-296.indd 268 14/01/17 2:47 pm 269 CHAPTER 6 6–5 ■ REVIEW OF ROTATIONAL MOTION AND ANGULAR MOMENTUM The motion of a rigid body can be considered to be the combination of translational motion of its center of mass and rotational motion about its center of mass. The translational motion is analyzed using the linear momentum equation, Eq. 6–1. Now we discuss the rotational motion—a motion during which all points in the body move in circles about the axis of rotation. Rotational motion is described with angular quantities such as angular distance 𝜃, angular velocity → 𝜔, and angular acceleration → 𝛼.
in the x-direction is +V1 at the inlet, but zero at the outlet. The magnitude of the velocity in the z-direction is zero at the inlet, but −V2 at the outlet. Also, the weight of the faucet assembly and the water within it acts in the −z-direction as a body force. No pressure or viscous forces act on the chosen (wise) control volume in the z-direction. The components of Eq. 1 along the x- and z-directions become FRx + P1, gageA1 = 0 −m · (+V1) FRz −Wfaucet −Wwater = m · (−V2) −0 Solving for FRx and FRz, and substituting the given values, FRx = −m · V1 −P1, gageA1 = −(2.568 lbm/s)(12.42 ft/s)( 1 lbf 32.2 lbm·ft/s2) −(13 lbf/in2) 𝜋(0.780 in)2 4 = −7.20 lbf FRz = −m · V2 + Wfaucet+water = −(2.568 lbm/s)(12.42 ft/s)( 1 lbf 32.2 lbm·ft/s2) + 12.8 lbf = 11.8 lbf Then the net force of the flange on the control volume is expressed in vector form as F › R = FRx i ›+ FRzk ›= −7.20 i ›+ 11.8k › lbf From Newton’s third law, the force the faucet assembly exerts on the flange is the negative of F → R, F › faucet on flange = −F › R = 7.20 i ›−11.8 k › lbf Discussion The faucet assembly pulls to the right and down; this agrees with our intuition. Namely, the water exerts a high pressure at the inlet, but the outlet pressure is atmospheric. In addition, the momentum of the water at the inlet in the x-direction is lost in the turn, causing an additional force to the right on the pipe walls. The faucet assembly weighs much more than the momentum effect of the water, so we expect the force to be downward. Note that labeling forces such as “faucet on flange” clarifies the direction of the force.
cen96537_ch06_249-296.indd 269 14/01/17 2:47 pm 270 MOMENTUM ANALYSIS OF FLOW SYSTEMS The amount of rotation of a point in a body is expressed in terms of the angle 𝜃 swept by a line of length r that connects the point to the axis of rota tion and is perpendicular to the axis. The angle 𝜃 is expressed in radians (rad), which is the arc length corresponding to 𝜃 on a circle of unit radius. Noting that the circumference of a circle of radius r is 2𝜋r, the angular distance traveled by any point in a rigid body during a complete rotation is 2𝜋 rad.
The physical distance traveled by a point along its circular path is l = 𝜃r, where r is the normal distance of the point from the axis of rotation and 𝜃 is the angular distance in rad. Note that 1 rad corresponds to 360/(2𝜋) ≅ 57.3°. The magnitude of angular velocity 𝜔 is the angular distance traveled per unit time, and the magnitude of angular acceleration 𝛼 is the rate of change of angular velocity. They are expressed as (Fig. 6–28), 𝜔= d𝜃 dt = d(l/r) dt = 1 r dl dt = V r and 𝛼= d𝜔 dt = d 2𝜃 dt 2 = 1 r dV dt = at r (6–30) or V = r𝜔 and at = r𝛼 (6–31) where V is the linear velocity and at is the linear acceleration in the tangen tial direction for a point located at a distance r from the axis of rotation. Note that 𝜔 and 𝛼 are the same for all points of a rotating rigid body, but V and at are not (they are proportional to r). Newton’s second law requires that there must be a force acting in the tangential direction to cause angular acceleration. The strength of the rotat ing effect, called the moment or torque, is proportional to the magnitude of the force and its distance from the axis of rotation. The perpendicular distance from the axis of rotation to the line of action of the force is called the moment arm, and the magnitude of torque M acting on a point mass m at normal distance r from the axis of rotation is expressed as M = rFt = rmat = mr2𝛼 (6–32) The total torque acting on a rotating rigid body about an axis is determined by integrating the torque acting on differential mass 𝛿m over the entire body to give Magnitude of torque: M = ∫mass r2𝛼 𝛿m = [∫mass r2 𝛿m]𝛼= I𝛼 (6–33) where I is the moment of inertia of the body about the axis of rotation, which is a measure of the inertia of a body against rotation. The relation M = I𝛼 is the counterpart of Newton’s second law, with torque replacing force, moment of inertia replacing mass, and angular acceleration replacing linear accelera tion (Fig. 6–29). Note that unlike mass, the rotational inertia of a body also depends on the distribution of the mass of the body with respect to the axis of rotation. Therefore, a body whose mass is closely packed about its axis of rotation has a small resistance against angular acceleration, while a body whose mass is concentrated at its periphery has a large resistance against angular acceleration. A flywheel is a good example of the latter. The linear momentum of a body of mass m having velocity V › is mV ›, and the direction of linear momentum is identical to the direction of velocity. ω ω = = dθ dt θ r r V = rω V r FIGURE 6–28 The relations between angular distance 𝜃, angular velocity 𝜔, and linear velocity V in a plane.
Linear velocity, V Angular velocity, ω Force, F Torque, M mV Iω Linear momentum Angular momentum F = ma M = Iα M = r × F H = r × mV Mass, m Moment of inertia, I Moment of force, M Moment of momentum, H Linear acceleration, a Angular acceleration, α FIGURE 6–29 Analogy between corresponding linear and angular quantities.
cen96537_ch06_249-296.indd 270 14/01/17 2:47 pm 271 CHAPTER 6 Noting that the moment of a force is equal to the product of the force and the normal distance, the magnitude of the moment of momentum, called the angular momentum, of a point mass m about an axis is expressed as H = rmV = r2m𝜔, where r is the normal distance from the axis of rotation to the line of action of the momentum vector (Fig. 6–30). Then the total angular momentum of a rotating rigid body is determined by integration to be Magnitude of angular momentum: H = ∫mass r2𝜔 𝛿m = [∫mass r2 𝛿m]𝜔= I𝜔 (6–34) where again I is the moment of inertia of the body about the axis of rota tion. It can also be expressed more generally in vector form as H ›= I 𝜔 › (6–35) Note that the angular velocity 𝜔 → is the same at every point of a rigid body. Newton’s second law F › = ma → was expressed in terms of the rate of change of linear momentum in Eq. 6–1 as F › = d(mV ›)/dt. Likewise, the counter part of Newton’s second law for rotating bodies M › = I𝛼 → is expressed in Eq. 6–2 in terms of the rate of change of angular momentum as Angular momentum equation: M ›= I 𝛼 ›= I d 𝜔 › dt = d(I 𝜔 › ) dt = d H › dt (6–36) where M › is the net torque applied on the body about the axis of rotation. The angular velocity of rotating machinery is typically expressed in rpm (number of revolutions per minute) and denoted by n .. Noting that veloc ity is distance traveled per unit time and the angular distance traveled during each revolution is 2𝜋, the angular velocity of rotating machinery is 𝜔 = 2𝜋n . rad/min or Angular velocity versus rpm: 𝜔= 2𝜋n · (rad/min) = 2𝜋n · 60 (rad/s) (6–37) Consider a constant force F acting in the tangential direction on the outer surface of a shaft of radius r rotating at an rpm of n .. Noting that work W is force times distance, and power W . is work done per unit time and thus force times velocity, we have W .
shaft = FV = Fr𝜔 = M𝜔. Therefore, the power transmitted by a shaft rotating at an rpm of n . under the influence of an applied torque M is (Fig. 6–31) Shaft power: W .
shaft = 𝜔M = 2𝜋n ·M (6–38) The kinetic energy of a body of mass m during translational motion is KE = 1 2mV 2. Noting that V = r𝜔, the rotational kinetic energy of a body of mass m at a distance r from the axis of rotation is KE = 1 2mr2𝜔2. The total rotational kinetic energy of a rotating rigid body about an axis is determined by integrating the rotational kinetic energies of differential masses dm over the entire body to give Rotational kinetic energy: KEr = 1 2 I𝜔2 (6–39) H = rmV = rm(rω) = r2mω = Iω ω r m mV = mrω V = rω FIGURE 6–30 Angular momentum of point mass m rotating at angular velocity 𝜔 at distance r from the axis of rotation.
Wshaft = ωM = 2πnM ω = 2πn ⋅ ⋅ ⋅ FIGURE 6–31 The relations between angular velocity, rpm, and the power transmitted through a rotating shaft.
cen96537_ch06_249-296.indd 271 14/01/17 2:47 pm 272 MOMENTUM ANALYSIS OF FLOW SYSTEMS where again I is the moment of inertia of the body and 𝜔 is the angular velocity. During rotational motion, the direction of velocity changes even when its magnitude remains constant. Velocity is a vector quantity, and thus a change in direction constitutes a change in velocity with time, and thus accelera tion. This is called centripetal acceleration. Its magnitude is ar = V2 r = r𝜔2 Centripetal acceleration is directed toward the axis of rotation (opposite direction of radial acceleration), and thus the radial acceleration is negative. Noting that acceleration is a constant multiple of force, centripetal accelera tion is the result of a force acting on the body toward the axis of rotation, known as the centripetal force, whose magnitude is Fr = mV 2/r. Tangential and radial accelerations are perpendicular to each other (since the radial and tangential directions are perpendicular), and the total linear acceleration is determined by their vector sum, a → = a → t + a → r. For a body rotating at constant angular velocity, the only acceleration is the centripetal acceleration. The centripetal force does not produce torque since its line of action intersects the axis of rotation.
6–6 ■ THE ANGULAR MOMENTUM EQUATION The linear momentum equation discussed in Section 6–4 is useful for deter mining the relationship between the linear momentum of flow streams and the resultant forces. Many engineering problems involve the moment of the linear momentum of flow streams, and the rotational effects caused by them. Such problems are best analyzed by the angular momentum equa tion, also called the moment of momentum equation. An important class of fluid devices, called turbomachines, which include centrifugal pumps, tur bines, and fans, is analyzed by the angular momentum equation. The moment of a force F › about a point O is the vector (or cross) product (Fig. 6–32) Moment of a force: M ›= r›× F › (6–40) where r → is the position vector from point O to any point on the line of action of F ›. The vector product of two vectors is a vector whose line of action is normal to the plane that contains the crossed vectors (r → and F › in this case) and whose magnitude is Magnitude of the moment of a force: M = Fr sin 𝜃 (6–41) where 𝜃 is the angle between the lines of action of the vectors r → and F ›.
Therefore, the magnitude of the moment about point O is equal to the magnitude of the force multiplied by the normal distance of the line of action of the force from the point O. The sense of the moment vector M › is determined by the right-hand rule: when the fingers of the right hand are curled in the direction that the force tends to cause rotation, the thumb points the direction of the moment vector (Fig. 6–33). Note that a force whose line of action passes through point O produces zero moment about point O.
Direction of rotation O r F M = r × F M = Fr sin θ r sin θ θ FIGURE 6–32 The moment of a force F › about a point O is the vector product of the position vector r → and F ›.
Sense of the moment ω Axis of rotation r F M = r × F FIGURE 6–33 The determination of the direction of the moment by the right-hand rule.
cen96537_ch06_249-296.indd 272 14/01/17 2:47 pm 273 CHAPTER 6 The vector product of r → and the momentum vector mV › gives the moment of momentum, also called the angular momentum, about a point O as Moment of momentum: H ›= r›× mV › (6–42) Therefore, r → × V › represents the angular momentum per unit mass, and the angular momentum of a differential mass 𝛿m = 𝜌 dV is d H › = (r → × V › )𝜌 dV. Then the angular momentum of a system is determined by integration to be Moment of momentum (system): H › sys = ∫sys ( r›× V ›)ρ dV (6–43) The rate of change of the moment of momentum is Rate of change of moment of momentum: dH › sys dt = d dt ∫sys (r›× V ›)ρ dV (6–44) The angular momentum equation for a system was expressed in Eq. 6–2 as ∑M ›= dH › sys dt (6–45) where Σ M › = Σ(r → × F ›) is the net torque or moment applied on the sys tem, which is the vector sum of the moments of all forces acting on the system, and d H › sys/dt is the rate of change of the angular momentum of the system. Equation 6–45 is stated as the rate of change of angular momentum of a system is equal to the net torque acting on the system. This equation is valid for a fixed quantity of mass and an inertial reference frame, i.e., a refer ence frame that is fixed or moves with a constant velocity in a straight path. The general control volume formulation of the angular momentum equa tion is obtained by setting b = r → × V › and thus B = H › in the general Reyn olds transport theorem. It gives (Fig. 6–34) dH › sys dt = d dt ∫CV ( r›× V › )ρ dV + ∫CS ( r›× V › )ρ(V › r· n › ) dA (6–46) The left-hand side of this equation is, from Eq. 6–45, equal to Σ M ›. Substi tuting, the angular momentum equation for a general control volume (sta tionary or moving, fixed shape or distorting) is General: ∑M ›= d dt ∫CV ( r›× V › )ρ dV + ∫CS ( r›× V › )ρ(V › r· n › ) dA (6–47) which is stated in words as ( The sum of all external moments acting on a CV ) = ( The time rate of change of the angular momentum of the contents of the CV) + ( The net flow rate of angular momentum out of the control surface by mass flow) Again, V › r = V › − V › CS is the fluid velocity relative to the control surface (for use in mass flow rate calculations at all locations where the fluid crosses the control surface), and V › is the fluid velocity as viewed from a fixed refer ence frame. The product 𝜌(V › r·n →) dA represents the mass flow rate through dA into or out of the control volume, depending on the sign.
= dBsys dt d dt ρb dV + V ρb( r · n ) dA B = H b = r × V b = r × V (r × V)ρ dV = dHsys dt d dt (r × V) ρ(Vr · n) dA + ∫ CV ∫ CV ∫ CS ∫ CS FIGURE 6–34 The angular momentum equation is obtained by replacing B in the Reynolds transport theorem by the angular momentum H ›, and b by the angular momentum per unit mass r → × V ›.
cen96537_ch06_249-296.indd 273 14/01/17 2:47 pm 274 MOMENTUM ANALYSIS OF FLOW SYSTEMS For a fixed control volume (no motion or deformation of the control volume), V › r = V › and the angular momentum equation becomes Fixed CV: ∑M ›= d dt ∫CV ( r ›× V › )ρ dV + ∫CS ( r›× V › )ρ( V › ·n › ) dA (6–48) Also, note that the forces acting on the control volume consist of body forces that act throughout the entire body of the control volume such as grav ity, and surface forces that act on the control surface such as the pressure and reaction forces at points of contact. The net torque consists of the moments of these forces as well as the torques applied on the control volume.
Special Cases During steady flow, the amount of angular momentum within the con trol volume remains constant, and thus the time rate of change of angular momentum of the contents of the control volume is zero. Then, Steady flow: ∑M ›= ∫CS ( r›× V › )ρ( V › r· n › ) dA (6–49) In many practical applications, the fluid crosses the boundaries of the control volume at a certain number of inlets and outlets, and it is convenient to replace the area integral by an algebraic expression written in terms of the average prop erties over the cross-sectional areas where the fluid enters or leaves the control volume. In such cases, the angular momentum flow rate can be expressed as the difference in the angular momentum of outgoing and incoming streams. Furthermore, in many cases the moment arm r → is either constant along the inlet or outlet (as in radial flow turbomachines) or is large compared to the diameter of the inlet or outlet pipe (as in rotating lawn sprinklers, Fig. 6–35). In such cases, the average value of r → is used throughout the cross-sectional area of the inlet or outlet. Then, an approximate form of the angular momen tum equation in terms of average properties at inlets and outlets becomes ∑M ›≅ d dt ∫CV (r›× V › )ρ dV + ∑ out (r›× m · V ›) −∑ in (r›× m · V ›) (6–50) You may be wondering why we don’t introduce a correction factor into Eq. 6–50, like we did for conservation of energy (Chap. 5) and for conserva tion of linear momentum (Section 6–4). The reason is that the cross product of r → and m . V › is dependent on problem geometry, and thus, such a correction factor would vary from problem to problem. Therefore, whereas we can readily calculate a kinetic energy flux correction factor and a momentum flux correction factor for fully developed pipe flow that can be applied to various problems, we cannot do so for angular momentum. Fortunately, in many problems of practical engineering interest, the error associated with using average values of radius and velocity is small, and the approximation of Eq. 6–50 is reasonable. If the flow is steady, Eq. 6–50 further reduces to (Fig. 6–36) Steady flow: ∑M ›= ∑ out (r›× m · V › ) −∑ in (r›× m · V › ) (6–51) FIGURE 6–35 A rotating lawn sprinkler is a good example of application of the angular momentum equation.
© John A. Rizzo/Getty Images RF FIGURE 6–36 The net torque acting on a control volume during steady flow is equal to the difference between the outgoing and incoming angular momentum flow rates.
Σ M = Σ r × m V – Σ r × m V out in ⋅ ⋅ cen96537_ch06_249-296.indd 274 14/01/17 2:47 pm 275 CHAPTER 6 Equation 6–51 states that the net torque acting on the control volume during steady flow is equal to the difference between the outgoing and incoming angular momentum flow rates. This statement can also be expressed for any specified direction. Note that velocity V › in Eq. 6–51 is the velocity relative to an inertial coordinate system. In many problems, all the significant forces and momentum flows are in the same plane, and thus all give rise to moments in the same plane and about the same axis. For such cases, Eq. 6–51 can be expressed in scalar form as ∑M = ∑ out rm .V −∑ in rm .V (6–52) where r represents the average normal distance between the point about which moments are taken and the line of action of the force or velocity, provided that the sign convention for the moments is observed. That is, all moments in the counterclockwise direction are positive, and all moments in the clockwise direction are negative.
Flow with No External Moments When there are no external moments applied, the angular momentum equa tion Eq. 6–50 reduces to No external moments: 0 = dH › CV dt + ∑ out (r›× m · V › ) −∑ in (r›× m · V › ) (6–53) This is an expression of the conservation of angular momentum principle, which can be stated as in the absence of external moments, the rate of change of the angular momentum of a control volume is equal to the differ ence between the incoming and outgoing angular momentum fluxes. When the moment of inertia I of the control volume remains constant, the first term on the right side of Eq. 6–53 becomes simply moment of inertia times angular acceleration, I𝛼 →. Therefore, the control volume in this case can be treated as a solid body, with a net torque of M › body = Ibody 𝛼 ›= ∑ in (r›× m · V › ) −∑ out (r›× m · V › ) (6–54) (due to a change of angular momentum) acting on it. This approach can be used to determine the angular acceleration of space vehicles and aircraft when a rocket is fired in a direction different than the direction of motion.
Radial-Flow Devices Many rotary-flow devices such as centrifugal pumps and fans involve flow in the radial direction normal to the axis of rotation and are called radial-flow devices (Chap. 14). In a centrifugal pump, for example, the fluid enters the device in the axial direction through the eye of the impeller, turns out ward as it flows through the passages between the blades of the impel ler, collects in the scroll, and is discharged in the tangential direction, as shown in Fig. 6–37. Axial-flow devices are easily analyzed using the linear momentum equation. But radial-flow devices involve large changes in angu lar momentum of the fluid and are best analyzed with the help of the angu lar momentum equation. cen96537_ch06_249-296.indd 275 14/01/17 2:47 pm 276 MOMENTUM ANALYSIS OF FLOW SYSTEMS To analyze a centrifugal pump, we choose the annular region that encloses the impeller section as the control volume, as shown in Fig. 6–38. Note that the average flow velocity, in general, has normal and tangential components at both the inlet and the outlet of the impeller section. Also, when the shaft rotates at angular velocity 𝜔, the impeller blades have tangential velocity 𝜔r1 at the inlet and 𝜔r2 at the outlet. For steady, incompressible flow, the conser vation of mass equation is written as V · 1 = V · 2 = V · → (2𝜋r1b1)V1, n = (2𝜋r2b2)V2, n (6–55) where b1 and b2 are the flow widths at the inlet where r = r1 and at the outlet where r = r2, respectively. (Note that the actual circumferential cross-sectional area is somewhat less than 2𝜋rb since the blade thickness is not zero.) Then the average normal components V1, n and V2, n of abso lute velocity can be expressed in terms of the volumetric flow rate V . as V1, n = V · 2𝜋r1b1 and V2, n = V · 2𝜋r2b2 (6–56) The normal velocity components V1, n and V2, n as well as pressure acting on the inner and outer circumferential areas pass through the shaft center, and thus they do not contribute to torque about the origin. Then only the tan gential velocity components contribute to torque, and the application of the angular momentum equation ∑M = ∑ out rm · V −∑ in rm · V to the control volume gives Euler’s turbine equation: Tshaft = m · (r2V2, t −r1V1, t) (6–57) which is known as Euler’s turbine equation. When the angles 𝛼1 and 𝛼2 between the direction of absolute flow velocities and the radial direction are known, Eq. 6–57 becomes Tshaft = m · (r2V2 sin 𝛼2 −r1V1 sin 𝛼1) (6–58) In the idealized case of the tangential fluid velocity being equal to the blade angular velocity both at the inlet and the exit, we have V1, t = 𝜔r1 and V2, t = 𝜔r2, and the torque becomes Tshaft, ideal = m · 𝜔(r 2 2 −r 2 1 ) (6–59) where 𝜔 = 2𝜋n . is the angular velocity of the blades. When the torque is known, the shaft power is determined from W .
shaft = 𝜔Tshaft = 2𝜋n .Tshaft.
Scroll Casing Shaft Eye Side view Frontal view Impeller blade Impeller shroud r1 r2 b2 b1 In Out In ω ω FIGURE 6–37 Side and frontal views of a typical centrifugal pump.
V V V α2 O 2 2, n 2, t 1, n 1, t Control volume Tshaft r1 r2 V V V 1 ω α1 FIGURE 6–38 An annular control volume that encloses the impeller section of a centrifugal pump.
cen96537_ch06_249-296.indd 276 14/01/17 2:47 pm 277 CHAPTER 6 EXAMPLE 6–8 Bending Moment Acting at the Base of a Water Pipe Underground water is pumped through a 10-cm- diameter pipe that consists of a 2-m-long vertical and 1-m-long horizontal section, as shown in Fig. 6–39. Water discharges to atmospheric air at an average velocity of 3 m/s, and the mass of the horizontal pipe section when filled with water is 12 kg per meter length. The pipe is anchored on the ground by a concrete base. Determine the bending moment act ing at the base of the pipe (point A) and the required length of the horizontal sec tion that would make the moment at point A zero.
SOLUTION Water is pumped through a piping section. The moment acting at the base and the required length of the horizontal section to make this moment zero is to be determined.
Assumptions 1 The flow is steady. 2 The water is discharged to the atmosphere, and thus the gage pressure at the outlet is zero. 3 The pipe diameter is small com pared to the moment arm, and thus we use average values of radius and velocity at the outlet.
Properties We take the density of water to be 1000 kg/m3.
Analysis We take the entire L-shaped pipe as the control volume, and designate the inlet by 1 and the outlet by 2. We also take the x- and z-coordinates as shown. The control volume and the reference frame are fixed. The conservation of mass equation for this one-inlet, one-outlet, steady-flow sys tem is m .
1 = m .
2 = m . , and V1 = V2 = V since Ac = constant. The mass flow rate and the weight of the horizontal section of the pipe are m · = ρAcV = (1000 kg/m3)𝜋(0.10 m)2/4 = 23.56 kg/s W = mg = (12 kg/m)(1 m)(9.81 m/s2)( 1 N 1 kg·m/s2) = 117.7 N To determine the moment acting on the pipe at point A, we need to take the moment of all forces and momentum flows about that point. This is a steady-flow problem, and all forces and momentum flows are in the same plane. Therefore, the angular momentum equation in this case is expressed as ∑M = ∑ out rm · V −∑ in rm · V where r is the average moment arm, V is the average speed, all moments in the counterclockwise direction are positive, and all moments in the clockwise direction are negative. The free-body diagram of the L-shaped pipe is given in Fig. 6–39. Noting that the moments of all forces and momentum flows passing through point A are zero, the only force that yields a moment about point A is the weight W of the horizontal pipe section, and the only momentum flow that yields a moment is the outlet stream (both are negative since both moments are in the clockwise direction). Then the angular momentum equation about point A becomes MA −r1W = −r2m · V2 2 m 1 m 3 m/s 10 cm A FIGURE 6–39 Schematic for Example 6–8 and the free-body diagram.
V r1 = 0.5 m MA x z m ⋅ 2 V m ⋅ 1 r2 = 2 m W WA FR P1,gage A A cen96537_ch06_249-296.indd 277 14/01/17 2:47 pm 278 MOMENTUM ANALYSIS OF FLOW SYSTEMS Solving for MA and substituting give MA = r1W −r2m · V2 = (0.5 m)(118 N) −(2 m)(23.56 kg/s)(3 m/s)( 1 N 1 kg·m/s2) = −82.5 N·m The negative sign indicates that the assumed direction for MA is wrong and should be reversed. Therefore, a moment of 82.5 N·m acts at the stem of the pipe in the clockwise direction. That is, the concrete base must apply a 82.5 N·m moment on the pipe stem in the clockwise direction to counteract the excess moment caused by the exit stream. The weight of the horizontal pipe is w = W/L = 117.7 N per m length. Therefore, the weight for a length of Lm is Lw with a moment arm of r1 = L/2. Setting MA = 0 and substituting, the length L of the horizontal pipe that would cause the moment at the pipe stem to vanish is determined to be 0 = r1W −r2m · V2 → 0 = (L/2)Lw −r2m · V2 or L = √ 2r2m · V2 w = √ 2(2 m)(23.56 kg/s)(3 m/s) 117.7 N/m ( N kg·m/s2) = 1.55 m Discussion Note that the pipe weight and the momentum of the exit stream cause opposing moments at point A. This example shows the importance of accounting for the moments of momentums of flow streams when performing a dynamic analysis and evaluating the stresses in pipe materials at critical cross sections.
EXAMPLE 6–9 Power Generation from a Sprinkler System A large lawn sprinkler (Fig. 6–40) with four identical arms is to be converted into a turbine to generate electric power by attaching a generator to its rotating head, as shown in Fig. 6–41. Water enters the sprinkler from the base along the axis of rotation at a rate of 20 L/s and leaves the nozzles in the tangential direction. The sprinkler rotates at a rate of 300 rpm in a horizontal plane. The diameter of each jet is 1 cm, and the normal distance between the axis of rotation and the center of each nozzle is 0.6 m. Estimate the electric power produced.
SOLUTION A four-armed sprinkler is used to generate electric power. For a specified flow rate and rotational speed, the power produced is to be determined.
Assumptions 1 The flow is cyclically steady (i.e., steady from a frame of reference rotating with the sprinkler head). 2 The water is discharged to the atmosphere, and thus the gage pressure at the nozzle exit is zero. 3 Generator losses and air drag of rotating components are neglected. 4 The nozzle diameter is small compared to the moment arm, and thus we use average values of radius and velocity at the outlet.
Properties We take the density of water to be 1000 kg/m3 = 1 kg/L.
FIGURE 6–40 Lawn sprinklers often have rotating heads to spread the water over a large area.
© Andy Sotiriou/Getty Images RF cen96537_ch06_249-296.indd 278 14/01/17 2:47 pm 279 CHAPTER 6 Analysis We take the disk that encloses the sprinkler arms as the control volume, which is a stationary control volume. The conservation of mass equation for this steady-flow system is m .
1 = m .
2 = m .
total. Noting that the four nozzles are identical, we have m .
nozzle = m .
total/4 or V .
nozzle = V .
total/4 since the density of water is constant. The average jet exit velocity relative to the rotating nozzle is Vjet,r = V · nozzle Ajet = 5 L/s [𝜋(0.01 m)2/4] ( 1 m3 1000 L) = 63.66 m/s The angular and tangential velocities of the nozzles are 𝜔= 2𝜋n · = 2𝜋(300 rev/min) ( 1 min 60 s ) = 31.42 rad/s Vnozzle = r𝜔= (0.6 m)(31.42 rad/s) = 18.85 m/s Note that water in the nozzle is also moving at an average velocity of 18.85 m/s in the opposite direction when it is discharged. The average absolute velocity of the water jet (velocity relative to a fixed location on earth) is the vec tor sum of its relative velocity (jet velocity relative to the nozzle) and the absolute nozzle velocity, V › jet = V › jet, r + V › nozzle All of these three velocities are in the tangential direction, and taking the direc tion of jet flow as positive, the vector equation can be written in scalar form using magnitudes as Vjet = Vjet,r −Vnozzle = 63.66 −18.85 = 44.81 m/s Noting that this is a cyclically steady-flow problem, and all forces and momentum flows are in the same plane, the angular momentum equation is approximated as ∑M = ∑ out rm · V −∑ in rm · V, where r is the moment arm, all moments in the counterclockwise direction are positive, and all moments in the clockwise direction are negative. The free-body diagram of the disk that contains the sprinkler arms is given in Fig. 6–41. Note that the moments of all forces and momentum flows passing through the axis of rotation are zero. The momentum flows via the water jets leav ing the nozzles yield a moment in the clockwise direction and the effect of the generator on the control volume is a moment also in the clockwise direction (thus both are negative). Then the angular momentum equation about the axis of rotation becomes −Tshaft = −4rm · nozzleVjet or Tshaft = rm · totalVjet Substituting, the torque transmitted through the shaft is Tshaft = rm · totalVjet = (0.6 m)(20 kg/s)(44.81 m/s)( 1 N 1 kg·m/s2) = 537.7 N·m since m .
total = 𝜌V .
total = (1 kg/L)(20 L/s) = 20 kg/s. Then the power generated becomes W .
= 𝜔Tshaft = (31.42 rad/s)(537.7 N·m)( 1 kW 1000 N·m/s) = 16.9 kW FIGURE 6–41 Schematic for Example 6–9 and the free-body diagram.
mtotal Electric generator ω ⋅ Vjet Vjet Vjet Vjet r = 0.6 m Tshaft mnozzleVjet ⋅ mnozzleVjet ⋅ mnozzleVjet ⋅ mnozzleVjet ⋅ cen96537_ch06_249-296.indd 279 14/01/17 2:47 pm 280 MOMENTUM ANALYSIS OF FLOW SYSTEMS 20 0 5 10 15 200 0 400 600 Power produced, kW rpm 800 1000 1200 FIGURE 6–42 The variation of power produced with angular speed for the turbine of Example 6–9.
FIGURE 6–43 The manta ray is the largest of the rays, reaching up to 8 m in span. They swim with a motion that is a combination of flapping and undulation of their large pectoral fins. © Frank & Joyce Burek/Getty Images RF Therefore, this sprinkler-type turbine has the potential to produce 16.9 kW of power.
Discussion To put the result obtained in perspective, we consider two limiting cases. In the first limiting case, the sprinkler is stuck, and thus, the angular velocity is zero. The torque developed is maximum in this case, since Vnozzle = 0. Thus Vjet = Vjet, r = 63.66 m/s, giving Tshaft, max = 764 N⋅m. The power generated is zero since the generator shaft does not rotate. In the second limiting case, the sprinkler shaft is disconnected from the generator (and thus both the useful torque and power generation are zero), and the shaft accelerates until it reaches an equilibrium velocity. Setting Tshaft = 0 in the angular momentum equation gives the absolute water-jet velocity (jet velocity relative to an observer on earth) to be zero, Vjet = 0. Therefore, the relative velocity Vjet, r and absolute velocity Vnozzle are equal but in opposite direction. So, the absolute tangential velocity of the jet (and thus torque) is zero, and the water mass drops straight down like a waterfall under gravity with zero angular momentum (around the axis of rotation). The angular speed of the sprinkler in this case is n · = 𝜔 2𝜋= Vnozzle 2𝜋r = 63.66 m/s 2𝜋(0.6 m)( 60 s 1 min) = 1013 rpm Of course, the Tshaft = 0 case is possible only for an ideal, frictionless nozzle (i.e., 100 percent nozzle efficiency, as a no-load ideal turbine). Otherwise, there would be a resisting torque due to friction of the water, shaft, and surrounding air. The variation of power produced with angular speed is plotted in Fig. 6–42. Note that the power produced increases with increasing rpm, reaches a maximum (at about 500 rpm in this case), and then decreases. The actual power produced would be less than this due to generator inefficiency (Chap. 5) and other irreversible losses such as fluid friction within the nozzle (Chap. 8), shaft friction, and aerodynamic drag (Chap. 11).
Guest Authors: Alexander Smits, Keith Moored and Peter Dewey, Princeton University Aquatic animals propel themselves using a wide variety of mechanisms. Most fish flap their tail to produce thrust, and in doing so they shed two single vortices per flapping cycle, creating a wake that resembles a reverse von Kármán vortex street. The non-dimensional number that describes this vortex shedding is the Strouhal number St, where St = fA/U∞, where f is the frequency of actuation, A is the peak-to-peak amplitude of the trailing edge motion at the half-span, and U∞ is the steady swimming veloc ity. Remarkably, a wide variety of fish and mammals swim in the range 0.2 < St < 0.35. In manta rays (Fig. 6–43), propulsion is achieved by combining oscillatory and undulatory motions of flexible pectoral fins. That is, as the manta ray flaps its fins, it is also generating a traveling wave motion along the chord, APPLICATION SPOTLIGHT ■ Manta Ray Swimming cen96537_ch06_249-296.indd 280 14/01/17 2:47 pm 281 CHAPTER 6 opposite to the direction of its motion. This wave motion is not readily apparent because the wavelength is 6 to 10 times greater than the chord length. A similar undulation is observed in sting rays, but there it is more obvious because the wavelength is less than the chord length. Field observa tions indicate that many species of manta ray are migratory, and that they are very efficient swimmers. They are difficult to study in the laboratory because they are a protected and somewhat fragile creature. However, it is possible to study many aspects of their swimming behavior by mimicking their propulsive techniques using robots or mechanical devices such as that shown in Fig. 6–44. The flow field generated by such a fin displays the vor tex shedding seen in other fish studies, and when time-averaged displays a high momentum jet that contributes to the thrust (Fig. 6–45). The thrust and efficiencies can also be measured directly, and it appears that the undulatory motion due to the traveling wave is most important to thrust production at high efficiency in the manta ray.
References Clark, R. P. and Smits, A. J., “Thrust Production and Wake Structure of a Batoid-Inspired Oscillating Fin.” Journal of Fluid Mechanics, 562, 415–429, 2006.
Dewey, P. A., Carriou, A., and Smits, A. J. “On the Relationship between Effi ciency and Wake Structure of a Batoid-Inspired Oscillating Fin.” Journal of Fluid Mechanics, Vol. 691, pp. 245–266, 2011.
Moored, K. W., Dewey, P. A., Leftwich, M. C., Bart-Smith, H., and Smits, A. J., “Bio-Inspired Propulsion Mechanisms Based on Lamprey and Manta Ray Locomotion.” The Marine Technology Society Journal, Vol. 45(4), pp. 110–118, 2011.
Triantafyllou, G. S., Triantafyllou, M. S., and Grosenbaugh, M. A., “Optimal Thrust Development in Oscillating Foils with Application to Fish Propulsion.” J. Fluid. Struct., 7:205–224, 1993.
1.2 10 5 –5 –10 0 2 1.5 0.5 0 1 1 0.8 0.6 0.4 0.2 –0.4 –0.2 0 0.2 0.4 ζ, s–1 U/U∞ y –0.4 –0.2 0 0.2 0.4 y x 1.2 1 0.8 0.6 0.4 0.2 x U∞ FIGURE 6–44 Manta ray fin mechanism, showing the vortex pattern produced in the wake when it is swimming in a range where two single vortices are shed into the wake per flapping cycle. The artificial flexible fin is actuated by four rigid spars; by changing the relative phase differences between adjacent actuators, undulations of varying wavelength can be produced. FIGURE 6–45 Measurements of the wake of the manta ray fin mechanism, with the flow going from bottom to top. On the left, we see the vortices shed in the wake, alternating between positive vorticity (red) and negative vorticity (blue). The induced velocities are shown by the black arrows, and in this case we see that thrust is being produced. On the right, we see the time-averaged velocity field. The unsteady velocity field induced by the vortices produces a high velocity jet in the time-averaged field. The momentum flux associated with this jet contributes to the total thrust on the fin.
Image courtesy of Peter Dewey, Keith Moored and Alexander Smits. Used by permission.
cen96537_ch06_249-296.indd 281 14/01/17 2:47 pm MOMENTUM ANALYSIS OF FLOW SYSTEMS 282 MOMENTUM ANALYSIS OF FLOW SYSTEMS SUMMARY This chapter deals mainly with the conservation of momen tum for finite control volumes. The forces acting on the con trol volume consist of body forces that act throughout the entire body of the control volume (such as gravity, electric, and magnetic forces) and surface forces that act on the con trol surface (such as the pressure forces and reaction forces at points of contact). The sum of all forces acting on the control volume at a particular instant in time is represented by ΣF ›and is expressed as ∑F ›= ∑F › gravity + ∑F › pressure + ∑F › viscous + ∑F › other total force body force surface forces Newton’s second law can be stated as the sum of all external forces acting on a system is equal to the time rate of change of linear momentum of the system. Setting b = V › and thus B = mV › in the Reynolds transport theorem and utilizing Newton’s second law gives the linear momentum equation for a control volume as ∑F ›= d dt ∫CV ρV › dV + ∫CS ρV › (V › r ·n ›) dA which reduces to the following special cases: Steady flow: ∑F ›= ∫CS ρV › (V › r·n ›) dA Unsteady flow (algebraic form): ∑F ›= d dt ∫CV ρV › dV + ∑ out 𝛽m · V ›−∑ in 𝛽m · V › Steady flow (algebraic form): ∑F ›= ∑ out 𝛽m · V ›−∑ in 𝛽m · V › No external forces: 0 = d(mV › )CV dt + ∑ out 𝛽m · V ›−∑ in 𝛽m · V › where 𝛽 is the momentum-flux correction factor. A A con trol volume whose mass m remains constant can be treated as a solid body (a fixed-mass system) with a net thrusting force (also called simply the thrust) of F › thrust = mCVa ›= ∑ in 𝛽m · V ›−∑ out 𝛽m · V › acting on the body. Newton’s second law can also be stated as the rate of change of angular momentum of a system is equal to the net torque acting on the system. Setting b = r → × V › and thus B = H › in the general Reynolds transport theorem gives the angular momentum equation as ∑M ›= d dt ∫CV (r›× V ›)ρ dV + ∫CS (r›× V ›)ρ(V › r·n › ) dA which reduces to the following special cases: Steady flow: ∑M ›= ∫CS (r›× V ›)ρ(V › r·n › ) dA Unsteady flow (algebraic form): ∑M ›= d dt ∫CV (r›× V ›)ρ dV + ∑ out r›× m · V ›−∑ in r›× m · V › Steady and uniform flow: ∑M ›= ∑ out r›× m · V ›−∑ in r›× m · V › Scalar form for one direction: ∑M = ∑ out rm · V −∑ in rm · V No external moments: 0 = dH › CV dt + ∑ out r›× m · V ›−∑ in r›× m · V › A control volume whose moment of inertia I remains constant can be treated as a solid body (a fixed-mass system), with a net torque of M › CV = ICV𝛼 › = ∑ in r›× m · V ›−∑ out r›× m · V › acting on the body. This relation is used to determine the angular acceleration of a spacecraft when a rocket is fired. The linear and angular momentum equations are of funda mental importance in the analysis of turbomachinery and are used extensively in Chap. 14.
REFERENCES AND SUGGESTED READING 1. Kundu, P. K., Cohen, I. M., and Dowling, D. R., Fluid Mechanics, ed. 5. San Diego, CA: Academic Press, 2011.
2. Terry Wright, Fluid Machinery: Performance, Analysis, and Design, Boca Raton, FL: CRC Press, 1999.
cen96537_ch06_249-296.indd 282 14/01/17 2:47 pm 283 CHAPTER 6 PROBLEMS Newton’s Laws and Conservation of Momentum 6–1C Express Newton’s second law of motion for rotat ing bodies. What can you say about the angular velocity and angular momentum of a rotating nonrigid body of constant mass if the net torque acting on it is zero? 6–2C Express Newton’s first, second, and third laws.
6–3C Is momentum a vector? If so, in what direction does it point?
6–4C Express the conservation of momentum principle. What can you say about the momentum of a body if the net force acting on it is zero?
Linear Momentum Equation 6–5C How do surface forces arise in the momentum analy sis of a control volume? How can we minimize the number of surface forces exposed during analysis?
6–6C Explain the importance of the Reynolds transport theorem in fluid mechanics, and describe how the linear momentum equation is obtained from it.
6–7C What is the importance of the momentum-flux correction factor in the momentum analysis of flow systems? For which type(s) of flow is it significant and must it be con sidered in analysis: laminar flow, turbulent flow, or jet flow?
6–8C Write the momentum equation for steady one-dimensional flow for the case of no external forces and explain the physical significance of its terms.
6–9C In the application of the momentum equation, explain why we can usually disregard the atmospheric pressure and work with gage pressures only.
6–10C Two firefighters are fighting a fire with identical water hoses and nozzles, except that one is holding the hose straight so that the water leaves the nozzle in the same direc tion it comes, while the other holds it backward so that the water makes a U-turn before being discharged. Which fire fighter will experience a greater reaction force?
6–11C A rocket in space (no friction or resistance to motion) can expel gases relative to itself at some high veloc ity V. Is V the upper limit to the rocket’s ultimate velocity?
6–12C Describe in terms of momentum and airflow how a helicopter is able to hover.
Problems designated by a “C” are concept questions, and stu dents are encouraged to answer them all. Problems designated by an “E” are in English units, and SI users can ignore them. Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software.
FIGURE P6–12C © JupiterImages/Thinkstock/Alamy RF 6–13C Does it take more, equal, or less power for a helicopter to hover at the top of a high mountain than it does at sea level? Explain.
6–14C In a given location, would a helicopter require more energy in summer or winter to achieve a specified perfor mance? Explain.
6–15C Describe body forces and surface forces, and explain how the net force acting on a control volume is determined. Is fluid weight a body force or surface force? How about pressure? 6–16C A constant-velocity horizontal water jet from a sta tionary nozzle impinges normally on a vertical flat plate that rides on a nearly frictionless track. As the water jet hits the plate, it begins to move due to the water force. Will the accel eration of the plate remain constant or change? Explain.
FIGURE P6–16C Nozzle Water jet 6–17C A horizontal water jet of constant velocity V from a stationary nozzle impinges normally on a vertical flat plate that rides on a nearly frictionless track. As the water jet hits the plate, it begins to move due to the water force. What is the highest velocity the plate can attain? Explain.
6–18C A horizontal water jet from a nozzle of constant exit cross section impinges normally on a stationary vertical flat plate. A certain force F is required to hold the plate against the water stream. If the water velocity is doubled, will the necessary holding force also be doubled? Explain.
6–19 A 2.5-cm-diameter horizontal water jet with a speed of Vj = 40 m/s relative to the ground is deflected by a 60° stationary cone whose base diameter is 25 cm. Water veloc ity along the cone varies linearly from zero at the cone sur face to the incoming jet speed of 40 m/s at the free surface. cen96537_ch06_249-296.indd 283 14/01/17 2:47 pm MOMENTUM ANALYSIS OF FLOW SYSTEMS 284 MOMENTUM ANALYSIS OF FLOW SYSTEMS Disregarding the effect of gravity and the shear forces, deter mine the horizontal force F needed to hold the cone stationary.
FIGURE P6–19 θ = 60° Dc = 25 cm F Water jet, Vj Vj 6–20 A 90° elbow in a horizontal pipe is used to direct water flow upward at a rate of 40 kg/s. The diameter of the entire elbow is 10 cm. The elbow discharges water into the atmosphere, and thus the pressure at the exit is the local atmospheric pressure. The elevation difference between the centers of the exit and the inlet of the elbow is 50 cm. The weight of the elbow and the water in it is considered to be negligible. Determine (a) the gage pressure at the center of the inlet of the elbow and (b) the anchoring force needed to hold the elbow in place. Take the momentum-flux correction factor to be 1.03 at both the inlet and the outlet.
FIGURE P6–20 Water 40 kg/s 50 cm 6–21 Repeat Prob. 6–20 for the case of another (identical) elbow attached to the existing elbow so that the fluid makes a U-turn. Answers: (a) 9.81 kPa, (b) 497 N 6–22E A horizontal water jet impinges against a vertical flat plate at 30 ft/s and splashes off the sides in the verti cal plane. If a horizontal force of 500 lbf is required to hold the plate against the water stream, determine the volume flow rate of the water.
6–23 Water enters a 7-cm-diameter pipe steadily with a uniform velocity of 2 m/s and exits with the turbulent flow velocity distribution given by u = umax (1 − r/R)1/7. If the pressure drop along the pipe is 10 kPa, determine the drag force exerted on the pipe by water flow.
6–24 A reducing elbow in a horizontal pipe is used to deflect water flow by an angle 𝜃 = 45° from the flow direction while accelerating it. The elbow discharges water into the atmo sphere. The cross- sectional area of the elbow is 150 cm2 at the inlet and 25 cm2 at the exit. The elevation difference between the centers of the exit and the inlet is 40 cm. The mass of the elbow and the water in it is 50 kg. Determine the anchoring force needed to hold the elbow in place. Take the momentum-flux correction factor to be 1.03 at both the inlet and outlet.
FIGURE P6–24 150 cm2 40 cm 45° 25 cm2 Water 30.0 kg/s 6–25 Repeat Prob. 6–24 for the case of 𝜃 = 125°.
6–26E A 100-ft3/s water jet is moving in the positive x-direction at 18 ft/s. The stream hits a stationary splitter, such that half of the flow is diverted upward at 45° and the other half is directed downward, and both streams have a final average speed of 18 ft/s. Disregarding gravitational effects, determine the x- and z-components of the force required to hold the splitter in place against the water force.
FIGURE P6–26E 100 ft3/s 18 ft/s Splitter 45° 45° x z 6–27E Reconsider Prob. 6–26E. Using appropriate software, investigate the effect of the splitter angle on the force exerted on the splitter in the incoming flow direction. Let the half splitter angle vary from 0° to 180° in increments of 10°. Tabulate and plot your results, and draw some conclusions.
6–28 Commercially available large wind turbines have blade span diameters larger than 100 m and generate over 3 MW of electric power at peak design conditions. Consider a wind turbine with a 75-m blade span subjected cen96537_ch06_249-296.indd 284 14/01/17 2:47 pm 285 CHAPTER 6 to 25-km/h steady winds. If the combined turbine–generator efficiency of the wind turbine is 32 percent, determine (a) the power generated by the turbine and (b) the horizontal force exerted by the wind on the supporting mast of the turbine. Take the density of air to be 1.25 kg/m3, and disregard fric tional effects on mast.
FIGURE P6–28 25 km/h 75 m 6–29E A fan with 24-in-diameter blades moves 2000 cfm (cubic feet per minute) of air at 70°F at sea level. Determine (a) the force required to hold the fan and (b) the minimum power input required for the fan. Choose a control volume sufficiently large to contain the fan, with the inlet sufficiently far upstream so that the gage pressure at the inlet is nearly zero. Assume air approaches the fan through a large area with negligible veloc ity and air exits the fan with a uniform velocity at atmospheric pressure through an imaginary cylinder whose diameter is the fan blade diameter. Answers: (a) 0.820 lbf, (b) 5.91 W 6–30E A 3-in-diameter horizontal jet of water, with veloc ity 140 ft/s, strikes a bent plate, which deflects the water by 135° from its original direction. How much force is required to hold the plate against the water stream and what is its direction? Disregard frictional and gravitational effects.
6–31 Firefighters are holding a nozzle at the end of a hose while trying to extinguish a fire. If the nozzle exit diameter is 8 cm and the water flow rate is 12 m3/min, determine (a) the average water exit velocity and (b) the horizontal resistance force required of the firefighters to hold the nozzle. Answers: (a) 39.8 m/s, (b) 7958 N FIGURE P6–31 12 m3/min 6–32 A 5-cm-diameter horizontal jet of water with a velocity of 30 m/s relative to the ground strikes a flat plate that is moving in the same direction as the jet at a velocity of 20 m/s. The water splatters in all directions in the plane of the plate. How much force does the water stream exert on the plate?
6–33 Reconsider Prob. 6–32. Using appropriate soft ware, investigate the effect of the plate velocity on the force exerted on the plate. Let the plate velocity vary from 0 to 30 m/s, in increments of 3 m/s. Tabulate and plot your results.
6–34E A 3-in-diameter horizontal water jet having a velocity of 90 ft/s strikes a curved plate, which deflects the water 180° at the same speed. Ignoring the frictional effects, determine the force required to hold the plate against the water stream.
FIGURE P6–34E 90 ft/s 90 ft/s 3 in Water jet 6–35 An unloaded helicopter of mass 12,000 kg hovers at sea level while it is being loaded. In the unloaded hover mode, the blades rotate at 550 rpm. The horizontal blades above the heli copter cause a 18-m-diameter air mass to move downward at an average velocity proportional to the overhead blade rotational velocity (rpm). A load of 14,000 kg is loaded onto the helicop ter, and the helicopter slowly rises. Determine (a) the volumet ric airflow rate downdraft that the helicopter generates during unloaded hover and the required power input and (b) the rpm of the helicopter blades to hover with the 14,000-kg load and the required power input. Take the density of atmospheric air to be 1.18 kg/m3. Assume air approaches the blades from the top through a large area with negligible velocity and air is forced by the blades to move down with a uniform velocity through an imaginary cylinder whose base is the blade span area.
FIGURE P6–35 Load 14,000 kg 18 m cen96537_ch06_249-296.indd 285 14/01/17 2:47 pm MOMENTUM ANALYSIS OF FLOW SYSTEMS 286 MOMENTUM ANALYSIS OF FLOW SYSTEMS 6–36 Reconsider the helicopter in Prob. 6–35, except that it is hovering on top of a 2200-m-high mountain where the air density is 0.987 kg/m3. Noting that the unloaded heli copter blades must rotate at 550 rpm to hover at sea level, determine the blade rotational velocity to hover at the higher altitude. Also determine the percent increase in the required power input to hover at 2200-m altitude relative to that at sea level. Answers: 601 rpm, 9.3 percent 6–37 Water is flowing through a 10-cm-diameter water pipe at a rate of 0.1 m3/s. Now a diffuser with an outlet diameter of 20 cm is bolted to the pipe in order to slow down water, as shown in Fig. P6–37. Disregarding fric tional effects, determine the force exerted on the bolts due to the water flow.
FIGURE P6–37 d = 10 cm D = 20 cm Diffuser 6–38 Water flowing in a horizontal 25-cm-diameter pipe at 8 m/s and 300 kPa gage enters a 90° bend reducing section, which connects to a 15-cm-diameter vertical pipe. The inlet of the bend is 50 cm above the exit. Neglecting any frictional and gravitational effects, determine the net resultant force exerted on the reducer by the water. Take the momentum-flux correction factor to be 1.04.
6–39 A horizontal 4-cm-diameter water jet with a veloc ity of 18 m/s impinges normally upon a vertical plate of mass 750 kg. The plate rides on a nearly frictionless track and is initially stationary. When the jet strikes the plate, the plate begins to move in the direction of the jet. The water always splatters in the plane of the retreating plate. Determine (a) the acceleration of the plate when the jet first strikes it (time = 0), (b) the time it takes for the plate to reach a velocity of 9 m/s, and (c) the plate velocity 20 s after the jet first strikes the plate. For simplicity, assume the velocity of the jet is increased as the cart moves such that the impulse force exerted by the water jet on the plate remains constant.
6–40 Water enters a centrifugal pump axially at atmo spheric pressure at a rate of 0.09 m3/s and at a velocity of 5 m/s, and leaves in the normal direction along the pump cas ing, as shown in Fig. P6–40. Determine the force acting on the shaft (which is also the force acting on the bearing of the shaft) in the axial direction.
FIGURE P6–40 n ⋅ Blade Shaft 0.09 m3/S Impeller shroud 6–41 An incompressible fluid of density 𝜌 and viscosity 𝜇 flows through a curved duct that turns the flow 180°. The duct cross-sectional area remains constant. The average velocity, momentum flux correction factor, and gage pressure are known at the inlet (1) and outlet (2), as in Fig. P6–41. (a) Write an expression for the horizontal force Fx of the fluid on the walls of the duct in terms of the given variables. (b) Verify your expres sion by plugging in the following values: 𝜌 = 998.2 kg/m3, 𝜇 = 1.003 × 10−3 kg/m·s, A1 = A2 = 0.025 m2, 𝛽1 = 1.01, 𝛽2 = 1.03, V1 = 10 m/s, P1,gage = 78.47 kPa, and P2,gage = 65.23 kPa. Answer: (b) Fx = 8680 N to the right FIGURE P6–41 V1 V2 P1,gage P2,gage + Fx A1 A2 β1 β2 6–42 Consider the curved duct of Prob. 6–41, except allow the cross-sectional area to vary along the duct (A1 ≠ A2). (a) Write an expression for the horizontal force Fx of the fluid on the walls of the duct in terms of the given variables. (b) Verify your expres sion by plugging in the following values: 𝜌 = 998.2 kg/m3, A1 = 0.025 m2, A2 = 0.015 m2, 𝛽1 = 1.02, 𝛽2 = 1.04, V1 = 20 m/s, P1,gage = 88.34 kPa, and P2,gage = 67.48 kPa. Answer: (b) Fx = 30,700 N to the right 6–43 As a follow-up to Prob. 6 –41, it turns out that for a large enough area ratio A2/A1, the inlet pressure is actu ally smaller than the outlet pressure! Explain how this can be true in light of the fact that there is friction and other cen96537_ch06_249-296.indd 286 14/01/17 2:47 pm 287 CHAPTER 6 irreversibilities due to turbulence, and pressure must be lost along the axis of the duct to overcome these irreversibilities.
6–44 Water of density 𝜌 = 998.2 kg/m3 flows through a fire man’s nozzle—a converging section of pipe that accelerates the flow. The inlet diameter is d1 = 0.100 m, and the outlet diameter is d2 = 0.050 m. The average velocity, momentum flux correction factor, and gage pressure are known at the inlet (1) and outlet (2), as in Fig. P6–44. (a) Write an expres sion for the horizontal force Fx of the fluid on the walls of the nozzle in terms of the given variables. (b) Verify your expression by plugging in the following values: 𝛽1 = 1.03, 𝛽2 = 1.02, V1 = 3 m/s, P1,gage = 137,000 Pa, and P2,gage = 0 Pa.
Answer: (b) Fx = 861 N to the right FIGURE P6–44 L V1 A1 P1 V2 r x A2 F P2 d1 6–45 The weight of a water tank open to the atmosphere is balanced by a counterweight, as shown in Fig. P6–45. There is a 4-cm hole at the bottom of the tank with a discharge coefficient of 0.90, and water level in the tank is maintained constant at 50 cm by water entering the tank horizontally. Determine how much mass must be added to or removed from the counterweight to maintain balance when the hole at the bottom is opened.
FIGURE P6–45 Water h = 50 cm W Hole, d = 4 cm 6–46 A sluice gate, which controls flow rate in a channel by simply raising or lowering a vertical plate, is commonly used in irrigation systems. A force is exerted on the gate due to the difference between the water heights y1 and y2 and the flow velocities V1 and V2 upstream and downstream from the gate, respectively. Take the width of the sluice gate (into the page) to be w. Wall shear stresses along the channel walls may be ignored, and for simplicity, we assume steady, uni form flow at locations 1 and 2. Develop a relationship for the force FR acting on the sluice gate as a function of depths y1 and y2, mass flow rate m ., gravitational constant g, gate width w, and water density 𝜌.
FIGURE P6–46 y1 V1 Sluice gate V2 y2 6–47 A room is to be ventilated using a centrifugal fan, mounted as shown in the figure. The fan discharge pipe has a cross-sectional area of 150 cm2 while the ventilating duct cross-sectional area is 500 cm2. If the fan capacity is 0.40 m3/s and a minimum ventilating rate of V · room = 0.30 m3/s is required, determine the proper installation angle 𝛽. Assume a local loss 0.5V1 2/2g for the duct entering from the room.
FIGURE P6–47 Room β 1 2 Angular Momentum Equation 6–48C How is the angular momentum equation obtained from Reynolds transport equations?
6–49C Consider two rigid bodies having the same mass and angular speed. Do you think these two bodies must have the same angular momentum? Explain.
6–50C Express the angular momentum equation in scalar form about a specified axis of rotation for a fixed control vol ume for steady and uniform flow. 6–51C Express the unsteady angular momentum equation in vector form for a control volume that has a constant moment of inertia I, no external moments applied, one outgoing uni form flow stream of velocity V → , and mass flow rate m ..
cen96537_ch06_249-296.indd 287 14/01/17 2:47 pm MOMENTUM ANALYSIS OF FLOW SYSTEMS 288 MOMENTUM ANALYSIS OF FLOW SYSTEMS 6–52E A large lawn sprinkler with two identical arms is used to generate electric power by attaching a generator to its rotating head. Water enters the sprinkler from the base along the axis of rotation at a rate of 5 gal/s and leaves the nozzles in the tangential direction. The sprinkler rotates at a rate of 180 rpm in a horizontal plane. The diameter of each jet is 0.5 in, and the normal distance between the axis of rotation and the center of each nozzle is 2 ft. Determine the maxi mum possible electrical power produced.
6–53E Reconsider the lawn sprinkler in Prob. 6–52E. If the rotating head is somehow stuck, determine the moment act ing on the head. 6–54 The impeller of a centrifugal pump has inner and outer diameters of 15 and 35 cm, respectively, and a flow rate of 0.15 m3/s at a rotational speed of 1400 rpm. The blade width of the impeller is 8 cm at the inlet and 3.5 cm at the outlet. If water enters the impeller in the radial direction and exits at an angle of 60° from the radial direction, deter mine the minimum power requirement for the pump.
6–55 Water is flowing through a 15-cm-diameter pipe that consists of a 3-m-long vertical and 2-m-long horizontal section with a 90° elbow at the exit to force the water to be discharged downward, as shown in Fig. P6–55, in the vertical direction. Water discharges to atmospheric air at a velocity of 5 m/s, and the mass of the pipe section when filled with water is 17 kg per meter length. Determine the moment acting at the intersection of the vertical and horizontal sections of the pipe (point A). What would your answer be if the flow were discharged upward instead of downward?
FIGURE P6–55 2 m 15 cm 3 m A 5 m/s 6–56 Water enters vertically and steadily at a rate of 35 L/s into the sprinkler shown in Fig. P6–56 with unequal arms and unequal discharge areas. The smaller jet has a discharge area of 3 cm2 and a normal distance of 50 cm from the axis of rotation. The larger jet has a discharge area of 5 cm2 and a normal distance of 35 cm from the axis of rotation. Dis regarding any frictional effects, determine (a) the rotational speed of the sprinkler in rpm and (b) the torque required to prevent the sprinkler from rotating.
FIGURE P6–56 50 cm 35 cm Water jet Water jet 6–57 Repeat Prob. 6–56 for a water flow rate of 60 L/s.
6–58 Consider a centrifugal blower that has a radius of 20 cm and a blade width of 8.2 cm at the impeller inlet, and a radius of 45 cm and a blade width of 5.6 cm at the outlet. The blower delivers air at a rate of 0.70 m3/s at a rotational speed of 700 rpm. Assuming the air to enter the impeller in the radial direction and to exit at an angle of 50° from the radial direction, determine the minimum power consumption of the blower. Take the density of air to be 1.25 kg/m3.
FIGURE P6–58 V = 50° α2 1 2 Impeller region r1 r2 V ω 6–59 Reconsider Prob. 6–58. For the specified flow rate, investigate the effect of discharge angle 𝛼2 on the minimum power input requirements. Assume the air to enter the impeller in the radial direction (𝛼1 = 0°), and vary 𝛼2 from 0° to 85° in increments of 5°. Plot the variation of power input versus 𝛼2, and discuss your results. 6–60E Water enters the impeller of a centrifugal pump radially at a rate of 45 cfm (cubic feet per minute) when the shaft is rotating at 500 rpm. The tangential component of absolute velocity of water at the exit of the 2-ft outer diameter impeller is 110 ft/s. Determine the torque applied cen96537_ch06_249-296.indd 288 14/01/17 2:47 pm 289 CHAPTER 6 to the impeller and the minimum power input to the pump. Answers: 160 lbf⋅ft, 11.3 kW 6–61 A lawn sprinkler with three identical arms is used to water a garden by rotating in a horizontal plane by the impulse caused by water flow. Water enters the sprinkler along the axis of rotation at a rate of 45 L/s and leaves the 1.5-cm-diameter nozzles in the tangential direction. The bearing applies a retarding torque of T0 = 40 N·m due to friction at the anticipated operating speeds. For a normal distance of 40 cm between the axis of rotation and the center of the nozzles, determine the angular velocity of the sprin kler shaft. 6–62 Pelton wheel turbines are commonly used in hydro electric power plants to generate electric power. In these turbines, a high-speed jet at a velocity of Vj impinges on buckets, forcing the wheel to rotate. The buckets reverse the direction of the jet, and the jet leaves the bucket mak ing an angle 𝛽 with the direction of the jet, as shown in Fig. P6–62. Show that the power produced by a Pelton wheel of radius r rotating steadily at an angular velocity of 𝜔 is W . shaft = 𝜌𝜔rV .
(Vj − 𝜔r)(1 − cos 𝛽), where 𝜌 is the density and V . is the volume flow rate of the fluid. Obtain the numerical value for 𝜌 = 1000 kg/m3, r = 2 m, V . =10 m3/s, n . = 150 rpm, 𝛽 = 160°, and Vj = 50 m/s.
FIGURE P6–62 Vj – rω ω β Vj r Nozzle Shaft rω 6–63 Reconsider Prob. 6–62. The maximum efficiency of the turbine occurs when 𝛽 = 180°, but this is not practical. Investigate the effect of 𝛽 on the power genera tion by allowing it to vary from 0° to 180°. Do you think we are wasting a large fraction of power by using buckets with a 𝛽 of 160°?
6–64 The impeller of a centrifugal blower has a radius of 18 cm and a blade width of 6.1 cm at the inlet, and a radius of 30 cm and a blade width of 3.4 cm at the outlet. The blower delivers atmospheric air at 20°C and 95 kPa. Disregarding any losses and assuming the tangential components of air velocity at the inlet and the outlet to be equal to the impel ler velocity at respective locations, determine the volumetric flow rate of air when the rotational speed of the shaft is 900 rpm and the power consumption of the blower is 120 W. Also determine the normal components of velocity at the inlet and outlet of the impeller.
FIGURE P6–64 Outlet ω Inlet Review Problems 6–65 An 8-cm-diameter horizontal water jet having a veloc ity of 35 m/s strikes a vertical stationary flat plate. The water splatters in all directions in the plane of the plate. How much force is required to hold the plate against the water stream? Answer: 6110 N 6–66 Water flowing steadily at a rate of 0.16 m3/s is deflected downward by an angled elbow as shown in Fig. P6–66. For D = 30 cm, d = 10 cm, and h = 50 cm, determine the force acting on the flanges of the elbow and the angle its line of action makes with the horizontal. Take the internal vol ume of the elbow to be 0.03 m3 and disregard the weight of the elbow material and the frictional effects. FIGURE P6–66 Flange Bolts 60° h d D 6–67 Repeat Prob. 6–66 by taking into consideration the weight of the elbow whose mass is 5 kg.
cen96537_ch06_249-296.indd 289 14/01/17 2:47 pm MOMENTUM ANALYSIS OF FLOW SYSTEMS 290 MOMENTUM ANALYSIS OF FLOW SYSTEMS 6–68 A 16-cm diameter horizontal water jet with a speed of Vj = 20 m/s relative to the ground is deflected by a 40° cone moving to the left at Vc = 10 m/s. Determine the external force, F, needed to maintain the motion of the cone. Disregard the gravity and surface shear effects and assume the cross-sectional area of water jet normal to the direction of motion remains constant throughout the flow. Answer: 4230 N FIGURE P6–68 θ = 40° Vc = 10 m/s F Water jet, Vj 6–69 Water enters vertically and steadily at a rate of 10 L/s into the sprinkler shown in Fig. P6–69. Both water jets have a diameter of 1.2 cm. Disregarding any frictional effects, determine (a) the rotational speed of the sprinkler in rpm and (b) the torque required to prevent the sprinkler from rotating.
FIGURE P6–69 40 cm 40 cm 60° 60° 6–70 Repeat Prob. 6–69 for the case of unequal arms—the left one being 60 cm and the right one 20 cm from the axis of rotation.
6–71 A tripod holding a nozzle, which directs a 5-cm-diameter stream of water from a hose, is shown in Fig. P6–71. The nozzle mass is 10 kg when filled with water. The tripod is rated to provide 1800 N of holding force. A firefighter was standing 60 cm behind the nozzle and was hit by the noz zle when the tripod suddenly failed and released the nozzle. You have been hired as an accident reconstructionist and, after testing the tripod, have determined that as water flow rate increased, it did collapse at 1800 N. In your final report you must state the water velocity and the flow rate consistent with the failure and the nozzle velocity when it hit the fire fighter. For simplicity, ignore pressure and momentum effects in the upstream portion of the hose. Answers: 30.3 m/s, 0.0595 m3/s, 14.7 m/s FIGURE P6–71 Nozzle Tripod D = 5 cm 6–72 Consider an airplane with a jet engine attached to the tail section that expels combustion gases at a rate of 18 kg/s with a velocity of V = 300 m/s relative to the plane. During landing, a thrust reverser (which serves as a brake for the air craft and facilitates landing on a short runway) is lowered in the path of the exhaust jet, which deflects the exhaust from rearward to 120°. Determine (a) the thrust (forward force) that the engine produces prior to the insertion of the thrust reverser and (b) the braking force produced after the thrust reverser is deployed.
FIGURE P6–72 300 m/s 120° Thrust reverser Thrust reverser 6–73 Reconsider Prob. 6–72. Using appropriate soft ware, investigate the effect of thrust reverser angle on the braking force exerted on the airplane. Let the reverser angle vary from 0° (no reversing) to 180° (full reversing) in increments of 10°. Tabulate and plot your results and draw conclusions.
6–74E A spacecraft cruising in space at a constant veloc ity of 2000 ft/s has a mass of 25,000 lbm. To slow down the spacecraft, a solid fuel rocket is fired, and the combus tion gases leave the rocket at a constant rate of 150 lbm/s at a velocity of 5000 ft/s in the same direction as the space craft for a period of 5 s. Assuming the mass of the spacecraft remains constant, determine (a) the deceleration of the space craft during this 5-s period, (b) the change of velocity of the spacecraft during this time period, and (c) the thrust exerted on the spacecraft.
cen96537_ch06_249-296.indd 290 14/01/17 2:47 pm 291 CHAPTER 6 6–75 A 60-kg ice skater is standing on ice with ice skates (negligible friction). She is holding a flexible hose (essen tially weightless) that directs a 2-cm-diameter stream of water horizontally parallel to her skates. The water velocity at the hose outlet is 10 m/s relative to the skater. If she is initially standing still, determine (a) the velocity of the skater and the distance she travels in 5 s and (b) how long it will take to move 5 m and the velocity at that moment. Answers: (a) 2.62 m/s, 6.54 m, (b) 4.4 s, 2.3 m/s FIGURE P6–75 10 m/s Ice skater D = 2 cm 6–76 A 5-cm-diameter horizontal jet of water, with velocity 30 m/s, strikes the tip of a horizontal cone, which deflects the water by 60° from its original direction. How much force is required to hold the cone against the water stream?
6–77 Water is flowing into and discharging from a pipe U-section as shown in Fig. P6–77. At flange (1), the total absolute pressure is 200 kPa, and 55 kg/s flows into the pipe. At flange (2), the total pressure is 150 kPa. At loca tion (3), 15 kg/s of water discharges to the atmosphere, which is at 100 kPa. Determine the total x- and z-forces at the two flanges connecting the pipe. Discuss the significance of grav ity force for this problem. Take the momentum-flux correc tion factor to be 1.03 throughout the pipes.
FIGURE P6–77 10 cm 3 cm 15 kg/s 40 kg/s 55 kg/s g 5 cm 1 2 3 x z 6–78 Indiana Jones needs to ascend a 10-m-high building. There is a large hose filled with pressurized water hanging down from the building top. He builds a square platform and mounts four 4-cm-diameter nozzles pointing down at each corner. By connecting hose branches, a water jet with 15-m/s velocity can be produced from each nozzle. Jones, the platform, and the nozzles have a combined mass of 150 kg. Determine (a) the minimum water jet veloc ity needed to raise the system, (b) how long it takes for the system to rise 10 m when the water jet velocity is 18 m/s and the velocity of the platform at that moment, and (c) how much higher will the momentum raise Jones if he shuts off the water at the moment the platform reaches 10 m above the ground. How much time does he have to jump from the platform to the roof? Answers: (a) 17.1 m/s, (b) 4.37 s, 4.57 m/s, (c) 1.07 m, 0.933 s FIGURE P6–78 D = 4 cm 18 m/s 6–79E An engineering student considers using a fan as a levitation demonstration. She plans to face the box-enclosed fan so the air blast is directed face down through a 2-ft-diameter blade span area. The system weighs 3 lbf, and the student will secure the system from rotating. By increasing the power to the fan, she plans to increase the blade rpm and air exit velocity until the exhaust provides sufficient upward force to cause the box fan to hover in the air. Determine (a) the air exit velocity to produce 3 lbf, (b) the volumetric flow rate needed, and (c) the minimum mechanical power that must be supplied to the airstream. Take the air density to be 0.078 lbm/ft3.
cen96537_ch06_249-296.indd 291 14/01/17 2:48 pm MOMENTUM ANALYSIS OF FLOW SYSTEMS 292 MOMENTUM ANALYSIS OF FLOW SYSTEMS FIGURE P6–79E 6–80 A walnut with a mass of 50 g requires a force of 200 N applied continuously for 0.002 s to be cracked. If wal nuts are to be cracked by dropping them from a high place onto a hard surface, determine the minimum height required. Disregard air friction.
6–81 A 7-cm diameter vertical water jet is injected upwards by a nozzle at a speed of 15 m/s. Determine the maximum weight of a flat plate that can be supported by this water jet at a height of 2 m from the nozzle.
6–82 Repeat Prob. 6–81 for a height of 8 m from the nozzle.
6–83 A horizontal water jet of constant velocity V impinges normally on a vertical flat plate and splashes off the sides in the vertical plane. The plate is moving toward the oncoming water jet with velocity 1 2V. If a force F is required to maintain the plate stationary, how much force is required to move the plate toward the water jet?
FIGURE P6–83 Water jet 1 2 V V 6–84 Show that the force exerted by a liquid jet on a stationary nozzle as it leaves with a velocity V is proportional to V 2 or, alternatively, to m . 2. Assume the jet stream is perpendicular to the incoming liquid flow line.
6–85 Consider steady developing laminar flow of water in a constant-diameter horizontal discharge pipe attached to a tank. The fluid enters the pipe with nearly uniform velocity V and pressure P1. The velocity profile becomes parabolic after a certain distance with a momentum correction factor of 2 while the pressure drops to P2. Obtain a relation for the horizontal force acting on the bolts that hold the pipe attached to the tank.
FIGURE P6–85 z r 6–86 A soldier jumps from a plane and opens his parachute when his velocity reaches the terminal velocity VT. The para chute slows him down to his landing velocity of VF. After the parachute is deployed, the air resistance is proportional to the velocity squared (i.e., F = kV2). The soldier, his parachute, and his gear have a total mass of m. Show that k = mg/V 2 F and develop a relation for the soldier’s velocity after he opens the parachute at time t = 0.
Answer: V = VF VT + VF + (VT −VF)e−2gt/VF VT + VF −(VT −VF)e−2gt/VF FIGURE P6–86 © Corbis RF 6–87 Water enters a mixed flow pump axially at a rate of 0.25 m3/s and at a velocity of 5 m/s, and is discharged to the atmosphere at an angle of 75° from the horizontal, as shown in Fig. P6–87. If the discharge flow area is half the inlet area, determine the force acting on the shaft in the axial direction.
FIGURE P6–87 n Blade Shaft 0.25 m3/S 75° 6–88 Water accelerated by a nozzle enters the impeller of a turbine through its outer edge of diameter D with a velocity of V making an angle 𝛼 with the radial direction at a mass flow rate of m .. Water leaves the impeller in the radial direc tion. If the angular speed of the turbine shaft is n ., show that the maximum power that can be generated by this radial tur bine is W .
shaft = 𝜋n .m .DV sin 𝛼.
cen96537_ch06_249-296.indd 292 14/01/17 2:48 pm 293 CHAPTER 6 6–89 Water enters a two-armed lawn sprinkler along the vertical axis at a rate of 75 L/s, and leaves the sprinkler nozzles as 2-cm diameter jets at an angle of 𝜃 from the tan gential direction, as shown in Fig. P6–89. The length of each sprinkler arm is 0.52 m. Disregarding any frictional effects, determine the rate of rotation n . of the sprinkler in rev/min for (a) 𝜃 = 0°, (b) 𝜃 = 30°, and (c) 𝜃 = 60°.
FIGURE P6–89 r = 0.52 m θ θ 6–90 Reconsider Prob. 6–89. For the specified flow rate, investigate the effect of discharge angle 𝜃 on the rate of rotation n . by varying 𝜃 from 0° to 90° in incre ments of 10°. Plot the rate of rotation versus 𝜃, and discuss your results.
6–91 A stationary water tank of diameter D is mounted on wheels and is placed on a nearly frictionless level surface. A smooth hole of diameter Do near the bottom of the tank allows water to jet horizontally and rearward and the water jet force propels the system forward. The water in the tank is much heavier than the tank-and-wheel assembly, so only the mass of water remaining in the tank needs to be consid ered in this problem. Considering the decrease in the mass of water with time, develop relations for (a) the acceleration, (b) the velocity, and (c) the distance traveled by the system as a function of time.
6–92 Nearly frictionless vertical guide rails maintain a plate of mass mp in a horizontal position, such that it can slide freely in the vertical direction. A nozzle directs a water stream of area A against the plate underside. The water jet splatters in the plate plane, applying an upward force against the plate. The water flow rate m . (kg/s) can be controlled. Assume that distances are short, so the velocity of the ris ing jet can be considered constant with height. (a) Determine the minimum mass flow rate m .
min necessary to just levitate the plate and obtain a relation for the steady-state velocity of the upward moving plate for m . > m .
min. (b) At time t = 0, the plate is at rest, and the water jet with m . > m .
min is suddenly turned on. Apply a force balance to the plate and obtain the integral that relates velocity to time (do not solve).
FIGURE P6–92 mp Nozzle m ⋅ Guide rails 6–93 A horizontal water jet with a flow rate of V . and cross-sectional area of A drives a covered cart of mass mc along a level and nearly frictionless path. The jet enters a hole at the rear of the cart and all water that enters the cart is retained, increasing the system mass. The relative velocity between the jet of constant velocity VJ and the cart of variable velocity V is VJ − V. If the cart is initially empty and stationary when the jet action is initiated, develop a relation (integral form is acceptable) for cart velocity versus time.
FIGURE P6–93 V Cart mc A VJ 6–94 Water is discharged from a pipe through a 1.2-m long 5-mm wide rectangular slit underneath of the pipe. Water dis charge velocity profile is parabolic, varying from 3 m/s on one end of the slit to 7 m/s on the other, as shown in Fig. P6–94. Determine (a) the rate of discharge through the slit and (b) the vertical force acting on the pipe due to this discharge process.
FIGURE P6–94 Parabolic velocity distribution 1.2 m V1 = 3 m/s Slit width = 5 mm V2 = 7 m/s 6–95 A water jet strikes a moving plate at velocity Vjet = 10 m/s as the plate moves at a velocity of U = 2 m/s, as shown in the figure. (a) Determine the force needed to hold the plate in its place.
(b) What would be the force if U is reversed?
cen96537_ch06_249-296.indd 293 14/01/17 2:48 pm MOMENTUM ANALYSIS OF FLOW SYSTEMS 294 MOMENTUM ANALYSIS OF FLOW SYSTEMS FIGURE P6–95 U = 2 m/s Djet = 12 cm Vjet = 10 m/s 6–96 Water flows at mass flow rate m . through a 90º ver tically oriented elbow of elbow radius R (to the centerline) and inner pipe diameter D as sketched. The outlet is exposed to the atmosphere. (Hint: This means that the pressure at the outlet is atmospheric pressure.) The pressure at the inlet must obviously be higher than atmospheric in order to push the water through the elbow and to raise the elevation of the water. The irreversible head loss through the elbow is hL. Assume that the kinetic energy flux correction factor 𝛼 is not unity, but is the same at the inlet and outlet of the elbow (𝛼1 = 𝛼2). Assume that the same thing applies to the momen tum flux correction factor 𝛽 (i.e., 𝛽1 = 𝛽2).
(a) Using the head form of the energy equation, derive an expression for the gage pressure Pgage,1 at the center of the inlet as a function of the other variables as needed.
(b) Plug in these numbers and solve for Pgage,1: 𝜌 = 998.0 kg/m3, D = 10.0 cm, R = 35.0 cm, hL = 0.259 m (of equivalent water column height), 𝛼1 = 𝛼2 = 1.05, 𝛽1 = 𝛽2 = 1.03, and m . = 25.0 kg/s. Use g = 9.807 m/s2 for consistency. Your answer should lie between 5 and 6 kPa.
(c) Neglecting the weight of the elbow itself and the weight of the water in the elbow, calculate the x and z components of the anchoring force required to hold the elbow in place. Your final answer for the anchoring force should be given as a vec tor, F › = Fx i › + Fz k ›. Your answer for Fx should lie between −120 and −140 N, and your answer for Fz should lie between 80 and 90 N.
(d) Repeat Part (c) without neglecting the weight of the water in the elbow. Is it reasonable to neglect the weight of the water in this problem?
FIGURE P6–96 Inner dia. D In, 1 Out, 2 m g Fx,anchoring Fz,anchoring · R 6–97 A cart with frictionless wheels and a large tank shoots water at a deflector plate, turning it by angle 𝜃 as sketched. The cart tries to move to the left, but a cable prevents it from doing so. At the exit of the deflector, the water jet area Ajet, its average velocity Vjet, and its momentum flux correction factor 𝛽jet are known. Generate an expression for the tension T in the cable in terms of the given variables. FIGURE P6–97 z x Vjet, Ajet, βjet Tank Cart Cable θ 6–98 Water shoots out of a large tank sitting on a cart with frictionless wheels. The water jet velocity is Vj = 7.00 m/s, its cross-sectional area is Aj = 20.0 mm2, and the momentum flux correction factor of the jet is 1.04. The water is deflected 135º as shown (𝜃 = 45º), and all of the water flows back into the tank. The density of the water is 1000 kg/m3. Calculate the hor izontal force F (in units of N) required to hold the cart in place. FIGURE P6–98 Vj, Aj, βj Tank F Cart θ Fundamentals of Engineering (FE) Exam Problems 6–99 When determining the thrust developed by a jet engine, a wise choice of control volume is (a) Fixed control volume (b) Moving control volume (c) Deforming control volume (d) Moving or deforming control volume (e) None of these 6–100 Consider an airplane cruising at 1000 km/h to the right. If the velocity of exhaust gases is 700 km/h to the left relative to the ground, the velocity of the exhaust gases rela tive to the nozzle exit is (a) 1700 km/h (b) 1000 km/h (c) 700 km/h (d) 300 km/h (e) 150 km/h 6–101 A water jet strikes a stationary vertical plate hori zontally at a rate of 7 kg/s with a velocity of 35 km/h. Assume the water stream moves in the vertical direction after the strike. The force needed to prevent the plate from moving horizontally is (a) 24.3 N (b) 35.0 N (c) 48.6 N (d) 68.1 N (e) 79.3 N cen96537_ch06_249-296.indd 294 14/01/17 2:48 pm 295 CHAPTER 6 6–102 Consider water flow through a horizontal, short garden hose at a rate of 30 kg/min. The velocity at the inlet is 1.5 m/s and that at the outlet is 14.5 m/s. Disregard the weight of the hose and water. Taking the momentum-flux correction factor to be 1.04 at both the inlet and the outlet, the anchoring force required to hold the hose in place is (a) 2.8 N (b) 8.6 N (c) 17.5 N (d) 27.9 N (e) 43.3 N 6–103 Consider water flow through a horizontal, short gar den hose at a rate of 30 kg/min. The velocity at the inlet is 1.5 m/s and that at the outlet is 11.5 m/s. The hose makes a 180° turn before the water is discharged. Disregard the weight of the hose and water. Taking the momentum-flux correction factor to be 1.04 at both the inlet and the outlet, the anchor ing force required to hold the hose in place is (a) 7.6 N (b) 28.4 N (c) 16.6 N (d) 34.1 N (e) 11.9 N 6–104 Consider water flow through a horizontal, short gar den hose at a rate of 40 kg/min. The velocity at the inlet is 1.5 m/s and that at the outlet is 16 m/s. The hose makes a 90° turn to a vertical direction before the water is discharged. Disregard the weight of the hose and water. Taking the momentum-flux correction factor to be 1.04 at both the inlet and the outlet, the reaction force in the vertical direction required to hold the hose in place is (a) 11.1 N (b) 10.1 N (c) 9.3 N (d) 27.2 N (e) 28.9 N 6–105 Consider water flow through a horizontal, short pipe at a rate of 80 kg/min. The velocity at the inlet is 1.5 m/s and that at the outlet is 16.5 m/s. The pipe makes a 90° turn to a vertical direction before the water is discharged. Disregard the weight of the pipe and water. Taking the momentum-flux correction factor to be 1.04 at both the inlet and the outlet, the reaction force in the horizontal direction required to hold the pipe in place is (a) 73.7 N (b) 97.1 N (c) 99.2 N (d) 122 N (e) 153 N 6–106 A water jet strikes a stationary horizontal plate verti cally at a rate of 18 kg/s with a velocity of 20 m/s. The mass of the plate is 10 kg. Assume the water stream moves in the horizontal direction after the strike. The force needed to pre vent the plate from moving vertically is (a) 186 N (b) 262 N (c) 334 N (d) 410 N (e) 522 N 6–107 The velocity of wind at a wind turbine is measured to be 6 m/s. The blade span diameter is 24 m and the effi ciency of the wind turbine is 29 percent. The density of air is 1.22 kg/m3. The horizontal force exerted by the wind on the supporting mast of the wind turbine is (a) 2524 N (b) 3127 N (c) 3475 N (d) 4138 N (e) 4313 N 6–108 The velocity of wind at a wind turbine is measured to be 8 m/s. The blade span diameter is 12 m. The density of air is 1.2 kg/m3. If the horizontal force exerted by the wind on the supporting mast of the wind turbine is 1620 N, the efficiency of the wind turbine is (a) 27.5% (b) 31.7% (c) 29.5% (d) 35.1% (e) 33.8% 6–109 A 3-cm-diameter horizontal pipe attached to a sur face makes a 90° turn to a vertical upward direction before the water is discharged at a velocity of 9 m/s. The horizon tal section is 5 m long and the vertical section is 4 m long. Neglecting the mass of the water contained in the pipe, the bending moment acting on the base of the pipe on the wall is (a) 286 N·m (b) 229 N·m (c) 207 N·m (d) 175 N·m (e) 124 N·m 6–110 A 3-cm-diameter horizontal pipe attached to a sur face makes a 90° turn to a vertical upward direction before the water is discharged at a velocity of 6 m/s. The horizon tal section is 5 m long and the vertical section is 4 m long. Neglecting the mass of the pipe and considering the weight of the water contained in the pipe, the bending moment act ing on the base of the pipe on the wall is (a) 11.9 N·m (b) 46.7 N·m (c) 127 N·m (d) 104 N·m (e) 74.8 N·m 6–111 A large lawn sprinkler with four identical arms is to be converted into a turbine to generate electric power by attaching a generator to its rotating head. Water enters the sprinkler from the base along the axis of rotation at a rate of 10 kg/s and leaves the nozzles in the tangential direction at a velocity of 50 m/s relative to the rotating nozzle. The sprinkler rotates at a rate of 400 rpm in a horizontal plane. The normal distance between the axis of rotation and the center of each nozzle is 30 cm. Estimate the electric power produced.
(a) 4704 W (b) 5855 W (c) 6496 W (d) 7051 W (e) 7840 W 6–112 Consider the impeller of a centrifugal pump with a rotational speed of 900 rpm and a flow rate of 95 kg/min. The impeller radii at the inlet and outlet are 7 cm and 16 cm, respectively. Assuming that the tangential fluid velocity is equal to the blade angular velocity both at the inlet and the exit, the power requirement of the pump is (a) 83 W (b) 291 W (c) 409 W (d) 756 W (e) 1125 W 6–113 Water enters the impeller of a centrifugal pump radially at a rate of 450 L/min when the shaft is rotating at 400 rpm. The tangential component of absolute velocity of water at the exit of the 70-cm outer diameter impeller is 55 m/s. The torque applied to the impeller is (a) 144 N·m (b) 93.6 N·m (c) 187 N·m (d) 112 N·m (e) 235 N·m 6–114 The shaft of a turbine rotates at a speed of 600 rpm. If the torque of the shaft is 3500 N·m, the shaft power is (a) 207 kW (b) 220 kW (c) 233 kW (d) 246 kW (e) 350 kW Design and Essay Problem 6–115 Visit a fire station and obtain information about flow rates through hoses and discharge diameters. Using this infor mation, calculate the impulse force to which the firefighters are subjected when holding a fire hose.
cen96537_ch06_249-296.indd 295 14/01/17 2:48 pm cen96537_ch06_249-296.indd 296 14/01/17 2:48 pm This page intentionally left blank 7 CHAPTER 297 OBJECTIVES When you finish reading this chapter, you should be able to ■ ■ Develop a better understanding of dimensions, units, and dimensional homo geneity of equations ■ ■ Understand the numerous benefits of dimensional analysis ■ ■ Know how to use the method of repeating variables to identify nondimensional parameters ■ ■ Understand the concept of dynamic similarity and how to apply it to experimental modeling D I ME N S I O N AL AN ALYSI S AN D M O DE L I N G I n this chapter, we first review the concepts of dimensions and units. We then review the fundamental principle of dimensional homogeneity, and show how it is applied to equations in order to nondimensionalize them and to identify dimensionless groups. We discuss the concept of similarity between a model and a prototype. We also describe a powerful tool for engineers and scientists called dimensional analysis, in which the combina tion of dimensional variables, nondimensional variables, and dimensional constants into nondimensional parameters reduces the number of necessary independent parameters in a problem. We present a step-by-step method for obtaining these nondimensional parameters, called the method of repeating variables, which is based solely on the dimensions of the variables and con stants. Finally, we apply this technique to several practical problems to illus trate both its utility and its limitations.
A 1:46.6 scale model of an Arleigh Burke class U.S. Navy fleet destroyer being tested in the 100-m-long towing tank at the University of Iowa. The model is 3.048 m long. In tests like this, the Froude number is the most important nondimensional parameter. Photo courtesy of IIHR-Hydroscience & Engineering, University of Iowa. Used by permission.
cen96537_ch07_297-350.indd 297 29/12/16 5:22 pm 298 dimensional ANALYSIS and modeling 7–1 ■ DIMENSIONS AND UNITS A dimension is a measure of a physical quantity (without numerical val ues), while a unit is a way to assign a number to that dimension. For exam ple, length is a dimension that is measured in units such as microns (µm), feet (ft), centimeters (cm), meters (m), kilometers (km), etc. (Fig. 7–1). There are seven primary dimensions (also called fundamental or basic dimensions)—mass, length, time, temperature, electric current, amount of light, and amount of matter.
All nonprimary dimensions can be formed by some combination of the seven primary dimensions.
For example, force has the same dimensions as mass times acceleration (by Newton’s second law). Thus, in terms of primary dimensions, Dimensions of force: {Force} = {Mass Length Time2 } = {mL/t2} (7–1) where the brackets indicate “the dimensions of” and the abbreviations are taken from Table 7–1. You should be aware that some authors prefer force instead of mass as a primary dimension—we do not follow that practice.
3.2 cm 1 2 3 cm Length FIGURE 7–1 A dimension is a measure of a physical quantity without numerical values, while a unit is a way to assign a number to the dimension. For example, length is a dimension, but centimeter is a unit.
FIGURE 7–2 Schematic for Example 7–1.
H = r × mV Point O mV Fluid particle r TABLE 7–1 Primary dimensions and their associated primary SI and English units Dimension Symbol SI Unit English Unit Mass Length Time† Temperature Electric current Amount of light Amount of matter m L t T I C N kg (kilogram) m (meter) s (second) K (kelvin) A (ampere) cd (candela) mol (mole) lbm (pound-mass) ft (foot) s (second) R (rankine) A (ampere) cd (candela) mol (mole) We italicize symbols for variables, but not symbols for dimensions.
† Note that some authors use the symbol T for the time dimension and the symbol 𝜃 for the temperature dimension. We do not follow this convention to avoid confusion between time and temperature.
EXAMPLE 7–1 Primary Dimensions of Angular Momentum Angular momentum, also called moment of momentum (H ›), is formed by the cross product of a moment arm (r ›) and the linear momentum (mV ›) of a fluid particle, as sketched in Fig. 7–2. What are the primary dimensions of angular momentum? List the units of angular momentum in primary SI units and in primary English units.
SOLUTION We are to write the primary dimensions of angular momentum and list its units. Analysis Angular momentum is the product of length, mass, and velocity, Primary dimensions of angular momentum: {H › } = {length × mass × length time } = { mL2 t } (1) or, in exponent form, {H ›} = {m1 L2 t−1} cen96537_ch07_297-350.indd 298 29/12/16 5:22 pm 299 CHAPTER 7 7–2 ■ DIMENSIONAL HOMOGENEITY We’ve all heard the old saying, You can’t add apples and oranges (Fig. 7–3). This is actually a simplified expression of a far more global and fundamen tal mathematical law for equations, the law of dimensional homogeneity, stated as Every additive term in an equation must have the same dimensions.
Consider, for example, the change in total energy of a simple compressible closed system from one state and/or time (1) to another (2), as illustrated in Fig. 7–4. The change in total energy of the system (ΔE) is given by Change of total energy of a system: ΔE = ΔU + ΔKE + ΔPE (7–2) where E has three components: internal energy (U), kinetic energy (KE), and potential energy (PE). These components can be written in terms of the system mass (m); measurable quantities and thermodynamic properties at each of the two states, such as speed (V), elevation (z), and specific internal energy (u); and the gravitational acceleration constant (g), ΔU = m(u2 −u1) ΔKE = 1 2 m (V 2 2 −V 2 1 ) ΔPE = mg(z2 −z1) (7–3) It is straightforward to verify that the left side of Eq. 7–2 and all three additive terms on the right side of Eq. 7–2 have the same dimensions—energy. Using the definitions of Eq. 7–3, we write the primary dimensions of each term, {ΔE} = {Energy} = {Force × Length} → {ΔE} = {mL2/t2} {ΔU} = {Mass Energy Mass } = {Energy} → {ΔU} = {mL2/t2} {ΔKE} = {Mass Length2 Time2 } → {ΔKE} = {mL2/t2} {ΔPE} = {Mass Length Time2 Length} → {ΔPE} = {mL2/t2} If at some stage of an analysis we find ourselves in a position in which two additive terms in an equation have different dimensions, this would be a clear indication that we have made an error at some earlier stage in the analysis (Fig. 7–5). In addition to dimensional homogeneity, calculations are valid only when the units are also homogeneous in each additive term. For example, units of energy in the above terms may be J, N·m, or kg·m2/s2, all of which are equivalent. Suppose, however, that kJ were used in place of J for one of the terms. This term would be off by a factor of 1000 compared to We write the primary SI units according to Eq. 1, Primary SI units: Units of angular momentum = kg·m2 s and in primary English units, Primary English units: Units of angular momentum = lbm·ft2 s Discussion Primary units are not required for dimensional analysis, but are often useful for unit conversions and for verification of proper units when solving a problem.
+ + = ?
FIGURE 7–3 You can’t add apples and oranges!
System at state 2 E2 = U2 + KE2 + PE2 System at state 1 E1 = U1 + KE1 + PE1 FIGURE 7–4 Total energy of a system at state 1 and at state 2.
FIGURE 7–5 An equation that is not dimensionally homogeneous is a sure sign of an error.
cen96537_ch07_297-350.indd 299 29/12/16 5:22 pm 300 dimensional ANALYSIS and modeling the other terms. It is wise to write out all units when performing mathemati cal calculations in order to avoid such errors.
EXAMPLE 7–2 Dimensional Homogeneity of the Bernoulli Equation Probably the most well-known (and most misused) equation in fluid mechanics is the Bernoulli equation (Fig. 7–6), discussed in Chap. 5. One standard form of the Bernoulli equation for incompressible irrotational fluid flow is Bernoulli equation: P + 1 2ρV 2 + ρgz = C (1) (a) Verify that each additive term in the Bernoulli equation has the same dimen sions. (b) What are the dimensions of the constant C?
SOLUTION We are to verify that the primary dimensions of each additive term in Eq. 1 are the same, and we are to determine the dimensions of constant C.
Analysis (a) Each term is written in terms of primary dimensions, {P} = {Pressure} = { Force Area} = {Mass Length Time2 1 Length2} = { m t2L} { 1 2ρV 2 } = { Mass Volume ( Length Time ) 2 } = { Mass × Length2 Length3 × Time2} = { m t2L} {ρgz} = { Mass Volume Length Time2 Length} = { Mass × Length2 Length3 × Time2} = { m t2L} Indeed, all three additive terms have the same dimensions.
(b) From the law of dimensional homogeneity, the constant must have the same dimensions as the other additive terms in the equation. Thus, Primary dimensions of the Bernoulli constant: {C} = { m t2L} Discussion If the dimensions of any of the terms were different from the others, it would indicate that an error was made somewhere in the analysis.
Nondimensionalization of Equations The law of dimensional homogeneity guarantees that every additive term in an equation has the same dimensions. It follows that if we divide each term in the equation by a collection of variables and constants whose product has those same dimensions, the equation is rendered nondimensional (Fig. 7–7). If, in addition, the nondimensional terms in the equation are of order unity, the equation is called normalized. Normalization is thus more restrictive than nondimensionalization, even though the two terms are some times (incorrectly) used interchangeably.
Each term in a nondimensional equation is dimensionless.
In the process of nondimensionalizing an equation of motion, nondimen sional parameters often appear—most of which are named after a notable scientist or engineer (e.g., the Reynolds number and the Froude number). This process is referred to by some authors as inspectional analysis.
Equation of the Day The Bernoulli equation P + ρV2 + ρgz = C 1 2 FIGURE 7–6 The Bernoulli equation is a good example of a dimensionally homoge neous equation. All additive terms, including the constant, have the same dimensions, namely that of pressure. In terms of primary dimensions, each term has dimensions {m/(t2L)}.
The nondimensionalized Bernoulli equation P ρV2 ρgz C P∞ 2P∞ P∞ P∞ {1} {1} {1} {1} + + = FIGURE 7–7 A nondimensionalized form of the Bernoulli equation is formed by dividing each additive term by a pressure (here we use P∞). Each resulting term is dimensionless (dimensions of {1}).
cen96537_ch07_297-350.indd 300 29/12/16 5:22 pm 301 CHAPTER 7 As a simple example, consider the equation of motion describing the ele vation z of an object falling by gravity through a vacuum (no air drag), as in Fig. 7–8. The initial location of the object is z0 and its initial velocity is w0 in the z-direction. From high school physics, Equation of motion: d 2z dt2 = −g (7–4) Dimensional variables are defined as dimensional quantities that change or vary in the problem. For the simple differential equation given in Eq. 7–4, there are two dimensional variables: z (dimension of length) and t (dimension of time). Nondimensional (or dimensionless) variables are defined as quan tities that change or vary in the problem, but have no dimensions; an exam ple is angle of rotation, measured in degrees or radians which are dimension less units. Gravitational constant g, while dimensional, remains constant and is called a dimensional constant. Two additional dimensional constants are relevant to this particular problem, initial location z0 and initial vertical speed w0. While dimensional constants may change from problem to problem, they are fixed for a particular problem and are thus distinguished from dimensional variables. We use the term parameters for the combined set of dimensional variables, nondimensional variables, and dimensional constants in the problem. Equation 7–4 is easily solved by integrating twice and applying the initial conditions. The result is an expression for elevation z at any time t: Dimensional result: z = z0 + w0t −1 2 gt2 (7–5) The constant 1 2 and the exponent 2 in Eq. 7–5 are dimensionless results of the integration. Such constants are called pure constants. Other common examples of pure constants are 𝜋 and e. To nondimensionalize Eq. 7–4, we need to select scaling parameters, based on the primary dimensions contained in the original equation. In fluid flow problems there are typically at least three scaling parameters, e.g., L, V, and P0 − P∞ (Fig. 7–9), since there are at least three primary dimensions in the general problem (e.g., mass, length, and time). In the case of the falling object being discussed here, there are only two primary dimensions, length and time, and thus we are limited to selecting only two scaling parameters. We have some options in the selection of the scaling parameters since we have three available dimensional constants g, z0, and w0. We choose z0 and w0. You are invited to repeat the analysis with g and z0 and/or with g and w0. With these two chosen scaling parameters we nondimensionalize the dimen sional variables z and t. The first step is to list the primary dimensions of all dimensional variables and dimensional constants in the problem, Primary dimensions of all parameters: {z} = {L} {t} = {t} {z0} = {L} {w0} = {L/t} {g} = {L/t2} The second step is to use our two scaling parameters to nondimensionalize z and t (by inspection) into nondimensional variables z and t, Nondimensionalized variables: z = z z0 t = w0t z0 (7–6) w = component of velocity in the z-direction z = vertical distance g = gravitational acceleration in the negative z-direction FIGURE 7–8 Object falling in a vacuum. Vertical velocity is drawn positively, so w < 0 for a falling object.
L P0 V, P∞ FIGURE 7–9 In a typical fluid flow problem, the scaling parameters usually include a characteristic length L, a characteristic velocity V, and a reference pressure difference P0 − P∞. Other parameters and fluid properties such as density, viscosity, and gravitational accelera tion enter the problem as well.
cen96537_ch07_297-350.indd 301 29/12/16 5:22 pm 302 dimensional ANALYSIS and modeling Substitution of Eq. 7–6 into Eq. 7–4 gives d 2z dt 2 = d2(z0z) d(z0t/w0)2 = w 2 0 z0 d2z dt2 = −g → w 2 0 gz0 d2z dt2 = −1 (7–7) which is the desired nondimensional equation. The grouping of dimensional constants in Eq. 7–7 is the square of a well-known nondimensional param eter or dimensionless group called the Froude number, Froude number: Fr = w0 √gz0 (7–8) The Froude (pronounced “Frude”) number also appears as a nondimen sional parameter in free-surface flows (Chap. 13), and can be thought of as the ratio of inertial force to gravitational force (Fig. 7–10). You should note that in some older textbooks, Fr is defined as the square of the parameter shown in Eq. 7–8. Substitution of Eq. 7–8 into Eq. 7–7 yields Nondimensionalized equation of motion: d 2z dt2 = −1 Fr 2 (7–9) In dimensionless form, only one parameter remains, namely the Froude number. Equation 7–9 is easily solved by integrating twice and applying the initial conditions. The result is an expression for dimensionless elevation z as a function of dimensionless time t: Nondimensional result: z = 1 + t − 1 2Fr2 t2 (7–10) Comparison of Eqs. 7–5 and 7–10 reveals that they are equivalent. In fact, for practice, substitute Eqs. 7–6 and 7–8 into Eq. 7–5 to verify Eq. 7–10. It seems that we went through a lot of extra algebra to generate the same final result. What then is the advantage of nondimensionalizing the equation? Before answering this question, we note that the advantages are not so clear in this simple example because we were able to analytically integrate the dif ferential equation of motion. In more complicated problems, the differential equation (or more generally the coupled set of differential equations) cannot be integrated analytically, and engineers must either integrate the equations numerically, or design and conduct physical experiments to obtain the needed results, both of which can incur considerable time and expense. In such cases, the nondimensional parameters generated by nondimensionalizing the equations are extremely useful and can save much effort and expense in the long run. There are two key advantages of nondimensionalization (Fig. 7–11). First, it increases our insight about the relationships between key parameters. Equation 7–8 reveals, for example, that doubling w0 has the same effect as decreasing z0 by a factor of 4. Second, it reduces the number of parameters in the problem. For example, the original problem contains one dependent variable, z; one independent variable, t; and three additional dimensional constants, g, w0, and z0. The nondimensionalized problem contains one dependent parameter, z; one independent parameter, t; and only one additional parameter, namely the dimensionless Froude number, Fr. The number of additional parameters has been reduced from three to one! Example 7–3 further illustrates the advantages of nondimensionalization.
Sluice gate y1 V 1 2 V y2 FIGURE 7–10 The Froude number is important in free-surface flows such as flow in open channels. Shown here is flow through a sluice gate. The Froude number upstream of the sluice gate is Fr1 = V1 /√gy1, and it is Fr2 = V2 /√gy2 downstream of the sluice gate.
Relationships between key parameters in the problem are identified.
The number of parameters in a nondimensionalized equation is less than the number of parameters in the original equation.
FIGURE 7–11 The two key advantages of non dimensionalization of an equation.
cen96537_ch07_297-350.indd 302 29/12/16 5:22 pm 303 CHAPTER 7 If you are still not convinced that nondimensionalizing the equations and the parameters has many advantages, consider this: In order to reasonably docu ment the trajectories of Example 7–3 for a range of all three of the dimensional EXAMPLE 7–3 Illustration of the Advantages of Nondimensionalization Your little brother’s high school physics class conducts experiments in a large vertical pipe whose inside is kept under vacuum conditions. The students are able to remotely release a steel ball at initial height z0 between 0 and 15 m (measured from the bottom of the pipe), and with initial vertical speed w0 between 0 and 10 m/s. A computer coupled to a network of photosensors along the pipe enables students to plot the trajectory of the steel ball (height z plotted as a function of time t) for each test. The students are unfamiliar with dimensional analysis or nondimensionalization techniques, and therefore conduct several “brute force” experiments to determine how the trajectory is affected by initial conditions z0 and w0. First they hold w0 fixed at 4 m/s and conduct experiments at five different values of z0: 3, 6, 9, 12, and 15 m. The experimental results are shown in Fig. 7–12a. Next, they hold z0 fixed at 10 m and conduct experiments at five different values of w0: 2, 4, 6, 8, and 10 m/s. These results are shown in Fig. 7–12b. Later that evening, your brother shows you the data and the trajectory plots and tells you that they plan to conduct more experiments at different values of z0 and w0. You explain to him that by first nondimensionalizing the data, the problem can be reduced to just one parameter, and no further experiments are required. Prepare a nondimensional plot to prove your point and discuss.
SOLUTION A nondimensional plot is to be generated from all the available trajectory data. Specifically, we are to plot z as a function of t.
Assumptions The inside of the pipe is subjected to strong enough vacuum pressure that aerodynamic drag on the ball is negligible.
Properties The gravitational constant is 9.81 m/s2.
Analysis Equation 7–4 is valid for this problem, as is the nondimensionalization that resulted in Eq. 7–9. As previously discussed, this problem combines three of the original dimensional parameters (g, z0, and w0) into one nondimensional parameter, the Froude number. After converting to the dimensionless variables of Eq. 7–6, the 10 trajectories of Fig. 7–12a and b are replotted in dimension less format in Fig. 7–13. It is clear that all the trajectories are of the same fam ily, with the Froude number as the only remaining parameter. Fr2 varies from about 0.041 to about 1.0 in these experiments. If any more experiments are to be conducted, they should include combinations of z0 and w0 that produce Froude numbers outside of this range. A large number of additional experiments would be unnecessary, since all the trajectories would be of the same family as those plotted in Fig. 7–13.
Discussion At low Froude numbers, gravitational forces are much larger than inertial forces, and the ball falls to the floor in a relatively short time. At large values of Fr on the other hand, inertial forces dominate initially, and the ball rises a significant distance before falling; it takes much longer for the ball to hit the ground. The students are obviously not able to adjust the gravitational constant, but if they could, the brute force method would require many more experiments to document the effect of g. If they nondimensionalize first, however, the dimension less trajectory plots already obtained and shown in Fig. 7–13 would be valid for any value of g; no further experiments would be required unless Fr were outside the range of tested values.
FIGURE 7–12 Trajectories of a steel ball falling in a vacuum: (a) w0 fixed at 4 m/s, and (b) z0 fixed at 10 m (Example 7–3). (a) (b) z, m 0 0 0.5 1 1.5 2 t, s 2.5 3 2 4 6 8 10 12 14 16 15 m z0 = wo = 4 m/s z0 = 10 m 12 m 9 m 6 m 3 m z, m 0 0 0.5 1 1.5 2 t, s 2.5 3 2 4 6 8 10 12 14 16 10 m/s w0 = 8 m/s 6 m/s 4 m/s 2 m/s Fr2 Fr2 = 1.0 z 0 0 0.5 1 1.5 2 t 2.5 3 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 Fr2 = 0.041 FIGURE 7–13 Trajectories of a steel ball falling in a vacuum. Data of Fig. 7–12a and b are nondimensionalized and combined onto one plot.
cen96537_ch07_297-350.indd 303 29/12/16 5:22 pm 304 dimensional ANALYSIS and modeling parameters g, z0, and w0, the brute force method would require several (say a minimum of four) additional plots like Fig. 7–12a at various values (levels) of w0, plus several additional sets of such plots for a range of g. A complete data set for three parameters with five levels of each parameter would require 53 = 125 experiments! Nondimensionalization reduces the number of param eters from three to one—a total of only 51 = 5 experiments are required for the same resolution. (For five levels, only five dimensionless trajectories like those of Fig. 7–13 are required, at carefully chosen values of Fr.) Another advantage of nondimensionalization is that extrapolation to untested values of one or more of the dimensional parameters is possible. For example, the data of Example 7–3 were taken at only one value of grav itational acceleration. Suppose you wanted to extrapolate these data to a dif ferent value of g. Example 7–4 shows how this is easily accomplished via the dimensionless data.
EXAMPLE 7–4 Extrapolation of Nondimensionalized Data The gravitational constant at the surface of the moon is only about one-sixth of that on earth. An astronaut on the moon throws a baseball at an ini tial speed of 21.0 m/s at a 5° angle above the horizon and at 2.0 m above the moon’s surface (Fig. 7–14). (a) Using the dimensionless data of Example 7–3 shown in Fig. 7–13, predict how long it takes for the baseball to fall to the ground. (b) Do an exact calculation and compare the result to that of part (a).
SOLUTION Experimental data obtained on earth are to be used to predict the time required for a baseball to fall to the ground on the moon.
Assumptions 1 The horizontal velocity of the baseball is irrelevant. 2 The surface of the moon is perfectly flat near the astronaut. 3 There is no aerodynamic drag on the ball since there is no atmosphere on the moon. 4 Moon gravity is one-sixth that of earth.
Properties The gravitational constant on the moon is gmoon ≅ 9.81/6 = 1.63 m/s2.
Analysis (a) The Froude number is calculated based on the value of gmoon and the vertical component of initial speed, w0 = (21.0 m/s) sin (5°) = 1.830 m/s from which Fr2 = w 2 0 gmoon z0 = (1.830 m/s)2 (1.63 m/s2) (2.0 m) = 1.03 This value of Fr2 is nearly the same as the largest value plotted in Fig. 7–13. Thus, in terms of dimensionless variables, the baseball strikes the ground at t ≅ 2.75, as determined from Fig. 7–13. Converting back to dimensional variables using Eq. 7–6, Estimated time to strike the ground: t = tz0 w0 = 2.75 (2.0 m) 1.830 m/s = 3.01 s (b) An exact calculation is obtained by setting z equal to zero in Eq. 7–5 and solv ing for time t (using the quadratic formula), FIGURE 7–14 Throwing a baseball on the moon (Example 7–4).
cen96537_ch07_297-350.indd 304 29/12/16 5:22 pm 305 CHAPTER 7 The differential equations of motion for fluid flow are derived and dis cussed in Chap. 9. In Chap. 10 you will find an analysis similar to that pre sented here, but applied to the differential equations for fluid flow. It turns out that the Froude number also appears in that analysis, as do three other important dimensionless parameters—the Reynolds number, Euler number, and Strouhal number (Fig. 7–15).
7–3 ■ DIMENSIONAL ANAL YSIS AND SIMILARITY Nondimensionalization of an equation by inspection is useful only when we know the equation to begin with. However, in many cases in real-life engi neering, the equations are either not known or too difficult to solve; often times experimentation is the only method of obtaining reliable information. In most experiments, to save time and money, tests are performed on a geo metrically scaled model, rather than on the full-scale prototype. In such cases, care must be taken to properly scale the results. We introduce here a powerful technique called dimensional analysis. While typically taught in fluid mechanics, dimensional analysis is useful in all disciplines, especially when it is necessary to design and conduct experiments. You are encouraged to use this powerful tool in other subjects as well, not just in fluid mechanics. The three primary purposes of dimensional analysis are • To generate nondimensional parameters that help in the design of experiments (physical and/or numerical) and in the reporting of experimental results • To obtain scaling laws so that prototype performance can be predicted from model performance • To (sometimes) predict trends in the relationship between parameters Before discussing the technique of dimensional analysis, we first explain the underlying concept of dimensional analysis—the principle of similarity. There are three necessary conditions for complete similarity between a model and a prototype. The first condition is geometric similarity—the model must be the same shape as the prototype, but may be scaled by some constant scale factor. The second condition is kinematic similarity, which means that the velocity at any point in the model flow must be proportional Exact time to strike the ground: t = w0 + √w 2 0 + 2z0g g = 1.830 m/s + √(1.830 m/s)2 + 2(2.0 m)(1.63 m/s2) 1.63 m/s2 = 3.05 s Discussion If the Froude number had landed between two of the trajectories of Fig. 7–13, interpolation would have been required. Since some of the num bers are precise to only two significant digits, the small difference between the results of part (a) and part (b) is of no concern. The final result is t = 3.0 s to two significant digits.
ρ, μ g P∞ P0 L f V Re = ρVL μ St = fL V Eu = P0 – P∞ ρV2 Fr = V gL FIGURE 7–15 In a general unsteady fluid flow prob lem with a free surface, the scaling parameters include a characteristic length L, a characteristic velocity V, a characteristic frequency f, and a reference pressure difference P0 − P∞. Nondimensionalization of the differential equations of fluid flow produces four dimensionless parameters: the Reynolds number, Froude number, Strouhal number, and Euler number (see Chap. 10).
cen96537_ch07_297-350.indd 305 29/12/16 5:22 pm 306 dimensional ANALYSIS and modeling (by a constant scale factor) to the velocity at the corresponding point in the prototype flow (Fig. 7–16). Specifically, for kinematic similarity the velocity at corresponding points must scale in magnitude and must point in the same relative direction. You may think of geometric similarity as length-scale equivalence and kinematic similarity as time-scale equivalence. Geometric similarity is a prerequisite for kinematic similarity. Just as the geometric scale factor can be less than, equal to, or greater than one, so can the velocity scale factor. In Fig. 7–16, for example, the geometric scale factor is less than one (model smaller than prototype), but the velocity scale is greater than one (velocities around the model are greater than those around the prototype). You may recall from Chap. 4 that streamlines are kinematic phenomena; hence, the streamline pattern in the model flow is a geometrically scaled copy of that in the prototype flow when kinematic similarity is achieved. The third and most restrictive similarity condition is that of dynamic similarity. Dynamic similarity is achieved when all forces in the model flow scale by a constant factor to corresponding forces in the prototype flow (force-scale equivalence). As with geometric and kinematic similarity, the scale factor for forces can be less than, equal to, or greater than one. In Fig. 7–16 for example, the force-scale factor is less than one since the force on the model building is less than that on the prototype. Kinematic similar ity is a necessary but insufficient condition for dynamic similarity. It is thus possible for a model flow and a prototype flow to achieve both geomet ric and kinematic similarity, yet not dynamic similarity. All three similarity conditions must exist for complete similarity to be ensured.
In a general flow field, complete similarity between a model and prototype is achieved only when there is geometric, kinematic, and dynamic similarity. We let uppercase Greek letter Pi (Π) denote a nondimensional parameter. In Section 7–2, we have already discussed one Π, namely the Froude num ber, Fr. In a general dimensional analysis problem, there is one Π that we call the dependent Π, giving it the notation Π1. The parameter Π1 is in general a function of several other Π’s, which we call independent Π’s. The functional relationship is Functional relationship between Π’s: Π1 = f (Π2, Π3, … , Πk) (7–11) where k is the total number of Π’s. Consider an experiment in which a scale model is tested to simulate a prototype flow. To ensure complete similarity between the model and the prototype, each independent Π of the model (subscript m) must be identical to the corresponding independent Π of the prototype (subscript p), i.e., Π2, m = Π2, p, Π3, m = Π3, p, . . . , Πk, m = Πk, p.
To ensure complete similarity, the model and prototype must be geometrically similar, and all independent Π groups must match between model and prototype.
Under these conditions the dependent Π of the model (Π1, m) is guaranteed to also equal the dependent Π of the prototype (Π1, p). Mathematically, we write a conditional statement for achieving similarity, If Π2, m = Π2, p and Π3, m = Π3, p … and Πk, m = Πk, p, then Π1, m = Π1, p (7–12) Prototype: Model: Vp Vm FD, m FD, p FIGURE 7–16 Kinematic similarity is achieved when, at all locations, the speed in the model flow is proportional to that at corresponding locations in the prototype flow, and points in the same direction.
cen96537_ch07_297-350.indd 306 29/12/16 5:23 pm 307 CHAPTER 7 Consider, for example, the design of a new sports car, the aerodynamics of which is to be tested in a wind tunnel. To save money, it is desirable to test a small, geometrically scaled model of the car rather than a full-scale prototype of the car (Fig. 7–17). In the case of aerodynamic drag on an automobile, it turns out that if the flow is approximated as incompressible, there are only two Π’s in the problem, Π1 = f (Π2) where Π1 = FD ρV 2L2 and Π2 = ρVL 𝜇 (7–13) The procedure used to generate these Π’s is discussed in Section 7–4. In Eq. 7–13, FD is the magnitude of the aerodynamic drag on the car, 𝜌 is the air density, V is the car’s speed (or the speed of the air in the wind tunnel), L is the length of the car, and 𝜇 is the viscosity of the air. Π1 is a nonstand ard form of the drag coefficient, and Π2 is the Reynolds number, Re. You will find that many problems in fluid mechanics involve a Reynolds number (Fig. 7–18).
The Reynolds number is the most well known and useful dimensionless parameter in all of fluid mechanics. In the problem at hand there is only one independent Π, and Eq. 7–12 ensures that if the independent Π’s match (the Reynolds numbers match: Π2, m = Π2, p), then the dependent Π’s also match (Π1, m = Π1, p). This enables engineers to measure the aerodynamic drag on the model car and then use this value to predict the aerodynamic drag on the prototype car.
Prototype car Model car Vp μp, ρp Lp Vm μm, ρm Lm FIGURE 7–17 Geometric similarity between a prototype car of length Lp and a model car of length Lm.
Re = = ρVL μ ρ, μ VL ν V L FIGURE 7–18 The Reynolds number Re is formed by the ratio of density, characteristic speed, and characteristic length to viscosity. Alternatively, it is the ratio of characteristic speed and length to kinematic viscosity, defined as 𝜈 = 𝜇/𝜌.
EXAMPLE 7–5 Similarity between Model and Prototype Cars The aerodynamic drag of a new sports car is to be predicted at a speed of 50.0 mi/h at an air temperature of 25°C. Automotive engineers build a one-fifth scale model of the car to test in a wind tunnel. It is winter and the wind tunnel is located in an unheated building; the temperature of the wind tunnel air is only about 5°C. Deter mine how fast the engineers should run the wind tunnel in order to achieve similarity between the model and the prototype.
SOLUTION We are to utilize the concept of similarity to determine the speed of the wind tunnel.
Assumptions 1 Compressibility of the air is negligible (the validity of this approximation is discussed later). 2 The wind tunnel walls are far enough away so as to not interfere with the aerodynamic drag on the model car. 3 The model is geometrically similar to the prototype. 4 The wind tunnel has a moving belt to simulate the ground under the car, as in Fig. 7–19. (The moving belt is necessary in order to achieve kinematic similarity everywhere in the flow, in particular underneath the car.) Properties For air at atmospheric pressure and at T = 25°C, 𝜌 = 1.184 kg/m3 and 𝜇 = 1.849 × 10−5 kg/m·s. Similarly, at T = 5°C, 𝜌 = 1.269 kg/m3 and 𝜇 = 1.754 × 10−5 kg/m·s.
Analysis Since there is only one independent Π in this problem, the similarity equa tion (Eq. 7–12) holds if Π2, m = Π2, p, where Π2 is given by Eq. 7–13, and we call it the Reynolds number. Thus, we write Π2, m = Rem = ρmVmLm 𝜇m = Π2, p = Rep = ρpVpLp 𝜇p cen96537_ch07_297-350.indd 307 29/12/16 5:23 pm 308 dimensional ANALYSIS and modeling Once we are convinced that complete similarity has been achieved between the model tests and the prototype flow, Eq. 7–12 can be used again to predict the performance of the prototype based on measurements of the performance of the model. This is illustrated in Example 7–6.
which we solve for the unknown wind tunnel speed for the model tests, Vm, Vm = Vp( 𝜇m 𝜇p ) ( ρp ρm) ( Lp Lm) = (50.0 mi/h) ( 1.754 × 10−5 kg/m·s 1.849 × 10−5 kg/m·s) ( 1.184 kg/m3 1.269 kg/m3)(5) = 221 mi/h Thus, to ensure similarity, the wind tunnel should be run at 221 mi/h (to three significant digits). Note that we were never given the actual length of either car, but the ratio of Lp to Lm is known because the prototype is five times larger than the scale model. When the dimensional parameters are rearranged as nondimensional ratios (as done here), the unit system is irrelevant. Since the units in each numerator cancel those in each denominator, no unit conversions are necessary.
Discussion This speed is quite high (about 100 m/s), and the wind tunnel may not be able to run at that speed. Furthermore, the incompressible approximation may come into question at this high speed (we discuss this in more detail in Example 7–8).
EXAMPLE 7–6 Prediction of Aerodynamic Drag Force on a Prototype Car This example is a follow-up to Example 7–5. Suppose the engineers run the wind tunnel at 221 mi/h to achieve similarity between the model and the prototype. The aerodynamic drag force on the model car is measured with a drag balance (Fig. 7–19). Several drag readings are recorded, and the average drag force on the model is 21.2 lbf. Predict the aerodynamic drag force on the prototype (at 50 mi/h and 25°C).
SOLUTION Because of similarity, the model results are to be scaled up to predict the aerodynamic drag force on the prototype.
Analysis The similarity equation (Eq. 7–12) shows that since Π2, m = Π2, p, Π1, m = Π1, p, where Π1 is given for this problem by Eq. 7–13. Thus, we write Π1, m = FD, m ρmV 2 m L 2 m = Π1, p = FD, p ρpV 2 p L 2 p which we solve for the unknown aerodynamic drag force on the prototype car, FD, p, FD, p = FD, m( ρp ρm) ( Vp Vm) 2 ( Lp Lm) 2 = (21.2 lbf)( 1.184 kg/m3 1.269 kg/m3) ( 50.0 mi/h 221 mi/h ) 2 (5)2 = 25.3 lbf Model Moving belt Wind tunnel test section Drag balance FD V FIGURE 7–19 A drag balance is a device used in a wind tunnel to measure the aero dynamic drag of a body. When testing automobile models, a moving belt is often added to the floor of the wind tunnel to simulate the moving ground (from the car’s frame of reference).
cen96537_ch07_297-350.indd 308 29/12/16 5:23 pm 309 CHAPTER 7 The power of using dimensional analysis and similarity to supplement experimental analysis is further illustrated by the fact that the actual values of the dimensional parameters (density, velocity, etc.) are irrelevant. As long as the corresponding independent Π’s are set equal to each other, similarity is achieved—even if different fluids are used. This explains why automobile or aircraft performance can be simulated in a water tunnel, and the perfor mance of a submarine can be simulated in a wind tunnel (Fig. 7–21). Sup pose, for example, that the engineers in Examples 7–5 and 7–6 use a water tunnel instead of a wind tunnel to test their one-fifth scale model. Using the properties of water at room temperature (20°C is assumed), the water tunnel speed required to achieve similarity is easily calculated as Vm = Vp( 𝜇m 𝜇p ) ( ρp ρm) ( Lp Lm) = (50.0 mi/h)( 1.002 × 10−3 kg/m·s) 1.849 × 10−5 kg/m·s ) ( 1.184 kg/m3 998.0 kg/m3)(5) = 16.1 mi/h As can be seen, one advantage of a water tunnel is that the required water tunnel speed is much lower than that required for a wind tunnel using the same size model.
7–4 ■ THE METHOD OF REPEATING VARIABLES AND THE BUCKINGHAM PI THEOREM We have seen several examples of the usefulness and power of dimensional analysis. Now we are ready to learn how to generate the nondimensional parameters, i.e., the Π’s. There are several methods that have been developed for this purpose, but the most popular (and simplest) method is the method of repeating variables, popularized by Edgar Buckingham (1867–1940). The method was first published by the Russian scientist Dimitri Ria bou chinsky (1882–1962) in 1911. We can think of this method as a step-by-step procedure or “recipe” for obtaining nondimensional parameters. There are six steps, listed concisely in Fig. 7–22, and in more detail in Table 7–2. These steps are explained in further detail as we work through a number of example problems. As with most new procedures, the best way to learn is by example and practice. As a simple first example, consider a ball falling in a vacuum as discussed in Section 7–2. Let us pretend that we do not know that Eq. 7–4 is appropriate for this problem, nor do we know much physics concerning falling objects. In fact, suppose that all we know is that the instantaneous Discussion By arranging the dimensional parameters as nondimensional ratios, the units cancel nicely even though they are a mixture of SI and English units. Because both velocity and length are squared in the equation for Π1, the higher speed in the wind tunnel nearly compensates for the model’s smaller size, and the drag force on the model is nearly the same as that on the prototype. In fact, if the density and viscosity of the air in the wind tunnel were identical to those of the air flowing over the prototype, the two drag forces would be identical as well (Fig. 7–20).
FIGURE 7–21 Similarity can be achieved even when the model fluid is different than the prototype fluid. Here a submarine model is tested in a wind tunnel.
Courtesy NASA Langley Research Center.
Prototype Model Vp μp, ρp FD, p Lp Vm = Vp μm = μp ρm = ρp F D, m = F D, p Lm Lp Lm FIGURE 7–20 For the special case in which the wind tunnel air and the air flowing over the prototype have the same properties (𝜌m = 𝜌p, 𝜇m = 𝜇p), and under simi larity conditions (Vm = VpLp/Lm), the aerodynamic drag force on the prototype is equal to that on the scale model. If the two fluids do not have the same properties, the two drag forces are not necessarily the same, even under dynamically similar conditions.
cen96537_ch07_297-350.indd 309 29/12/16 5:23 pm 310 dimensional ANALYSIS and modeling elevation z of the ball must be a function of time t, initial vertical speed w0, initial elevation z0, and gravitational constant g (Fig. 7–23). The beauty of dimensional analysis is that the only other thing we need to know is the pri mary dimensions of each of these quantities. As we go through each step of the method of repeating variables, we explain some of the subtleties of the technique in more detail using the falling ball as an example.
Step 1 There are five parameters (dimensional variables, nondimensional variables, and dimensional constants) in this problem; n = 5. They are listed in func tional form, with the dependent variable listed as a function of the inde pendent variables and constants: List of relevant parameters: z = f(t, w0, z0, g) n = 5 The Method of Repeating Variables Step 1: List the parameters in the problem and count their total number n.
Step 2: List the primary dimensions of each of the n parameters.
Step 5: Construct the k II’s, and manipulate as necessary.
Step 6: Write the final functional relationship and check your algebra.
Step 4: Choose j repeating parameters.
Step 3: Set the reduction j as the number of primary dimensions. Calculate k, the expected number of II’s, k = n – j FIGURE 7–22 A concise summary of the six steps that comprise the method of repeating variables.
w0 = initial vertical speed z = 0 (datum plane) z0 = initial elevation g = gravitational acceleration in the negative z-direction z = elevation of ball = f (t, w0, z0, g) FIGURE 7–23 Setup for dimensional analysis of a ball falling in a vacuum. Elevation z is a function of time t, initial verti cal speed w0, initial elevation z0, and gravitational constant g.
TABLE 7–2 Detailed description of the six steps that comprise the method of repeating variables Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 List the parameters (dimensional variables, nondimensional variables, and dimensional constants) and count them. Let n be the total number of parameters in the problem, including the dependent variable. Make sure that any listed independent parameter is indeed independent of the others, i.e., it cannot be expressed in terms of them. (For example, don’t include radius r and area A = 𝜋r 2, since r and A are not independent.) List the primary dimensions for each of the n parameters.
Guess the reduction j. As a first guess, set j equal to the number of primary dimensions represented in the problem. The expected number of Π’s (k) is equal to n minus j, according to the Buckingham Pi theorem, The Buckingham Pi theorem: k = n − j (7–14) If at this step or during any subsequent step, the analysis does not work out, verify that you have included enough parameters in step 1. Otherwise, go back and reduce j by one and try again.
Choose j repeating parameters that will be used to construct each Π. Since the repeating parameters have the potential to appear in each Π, be sure to choose them wisely (Table 7–3).
Generate the Π’s one at a time by grouping the j repeating parameters with one of the remaining parameters, forcing the product to be dimensionless. In this way, construct all k Π’s. By convention the first Π, designated as Π1, is the dependent Π (the one on the left side of the list). Manipulate the Π’s as necessary to achieve established dimensionless groups (Table 7–5).
Check that all the Π’s are indeed dimensionless. Write the final functional relationship in the form of Eq. 7–11.
This is a step-by-step method for finding the dimensionless Π groups when performing a dimensional analysis.
cen96537_ch07_297-350.indd 310 29/12/16 5:23 pm 311 CHAPTER 7 Step 2 The primary dimensions of each parameter are listed here. We recommend writing each dimension with exponents since this helps with later algebra.
z t w0 z0 g {L1} {t1} {L1t−1} {L1} {L1t−2} Step 3 As a first guess, j is set equal to 2, the number of primary dimensions repre sented in the problem (L and t).
Reduction: j = 2 If this value of j is correct, the number of Π’s predicted by the Buckingham Pi theorem is Number of expected Π’s: k = n −j = 5 −2 = 3 Step 4 We need to choose two repeating parameters since j = 2. Since this is often the hardest (or at least the most mysterious) part of the method of repeating variables, several guidelines about choosing repeating parameters are listed in Table 7–3. Following the guidelines of Table 7–3 on the next page, the wisest choice of two repeating parameters is w0 and z0.
Repeating parameters: w0 and z0 Step 5 Now we combine these repeating parameters into products with each of the remaining parameters, one at a time, to create the Π’s. The first Π is always the dependent Π and is formed with the dependent variable z.
Dependent Π: Π1 = zw a1 0 z b1 0 (7–15) where a1 and b1 are constant exponents that need to be determined. We apply the primary dimensions of step 2 into Eq. 7–15 and force the Π to be dimensionless by setting the exponent of each primary dimension to zero: Dimensions of Π1: {Π1} = {L0t0} = {zw a1 0 z b1 0 } = {L1(L1t−1)a1Lb1} Since primary dimensions are by definition independent of each other, we equate the exponents of each primary dimension independently to solve for exponents a1 and b1 (Fig. 7–24).
Time: {t0} = {t−a1} 0 = −a1 a1 = 0 Length: {L0} = {L1La1Lb1} 0 = 1 + a1 + b1 b1 = −1 −a1 b1 = −1 Equation 7–15 thus becomes Π1 = z z0 (7–16) Multiplication: Add exponents xaxbx2c = xa+b+2c Division: Subtract exponents × = xa–b–2c xa xb 1 x2c FIGURE 7–24 The mathematical rules for adding and subtracting exponents during multiplication and division, respectively.
cen96537_ch07_297-350.indd 311 29/12/16 5:23 pm 312 dimensional ANALYSIS and modeling In similar fashion we create the first independent Π (Π2) by combining the repeating parameters with independent variable t.
First independent Π: Π2 = tw a2 0 z b2 0 Dimensions of Π2: {Π2} = {L0t0} = {tw a2 0 z b2 0 } = {t(L1t−1)a2Lb2} TABLE 7–3 Guidelines for choosing repeating parameters in step 4 of the method of repeating variables Guideline Comments and Application to Present Problem 1. Never pick the dependent variable. In the present problem we cannot choose z, but we must choose from among Otherwise, it may appear in all the the remaining four parameters. Therefore, we must choose two of the following Π’s, which is undesirable. parameters: t, w0, z0, and g.
2. The chosen repeating parameters In the present problem, any two of the independent parameters would be valid must not by themselves be able according to this guideline. For illustrative purposes, however, suppose we have to form a dimensionless group. to pick three instead of two repeating parameters. We could not, for example, Otherwise, it would be impossible choose t, w0, and z0, because these can form a Π all by themselves (tw0/z0). to generate the rest of the Π’s. 3. The chosen repeating parameters Suppose for example that there were three primary dimensions (m, L, and t) and must represent all the primary two repeating parameters were to be chosen. You could not choose, say, a length dimensions in the problem. and a time, since primary dimension mass would not be represented in the dimensions of the repeating parameters. An appropriate choice would be a density and a time, which together represent all three primary dimensions in the problem.
4. Never pick parameters that are Suppose an angle 𝜃 were one of the independent parameters. We could not choose already dimensionless. These are 𝜃 as a repeating parameter since angles have no dimensions (radian and degree Π’s already, all by themselves. are dimensionless units). In such a case, one of the Π’s is already known, namely 𝜃.
5. Never pick two parameters with In the present problem, two of the parameters, z and z0, have the same the same dimensions or with dimensions (length). We cannot choose both of these parameters. dimensions that differ by only (Note that dependent variable z has already been eliminated by guideline 1.) an exponent. Suppose one parameter has dimensions of length and another parameter has dimensions of volume. In dimensional analysis, volume contains only one primary dimension (length) and is not dimensionally distinct from length—we cannot choose both of these parameters.
6. Whenever possible, choose If we choose time t as a repeating parameter in the present problem, it would dimensional constants over appear in all three Π’s. While this would not be wrong, it would not be wise dimensional variables so that since we know that ultimately we want some nondimensional height as a only one Π contains the function of some nondimensional time and other nondimensional parameter(s). dimensional variable. From the original four independent parameters, this restricts us to w0, z0, and g.
7. Pick common parameters since In fluid flow problems we generally pick a length, a velocity, and a mass or they may appear in each of the Π’s. density (Fig. 7–25). It is unwise to pick less common parameters like viscosity 𝜇 or sur face tension 𝜎s, since we would in general not want 𝜇 or 𝜎s to appear in each of the Π’s. In the present problem, w0 and z0 are wiser choices than g.
8. Pick simple parameters over It is better to pick parameters with only one or two basic dimensions (e.g., complex parameters whenever a length, a time, a mass, or a velocity) instead of parameters that are composed possible. of several basic dimensions (e.g., an energy or a pressure).
These guidelines, while not infallible, help you to pick repeating parameters that usually lead to established nondimensional Π groups with minimal effort.
cen96537_ch07_297-350.indd 312 29/12/16 5:23 pm 313 CHAPTER 7 Equating exponents, Time: {t0} = {t1t−a2} 0 = 1 −a2 a2 = 1 Length: {L0} = {La2Lb2} 0 = a2 + b2 b2 = −a2 b2 = −1 Π2 is thus Π2 = w0t z0 (7–17) Finally we create the second independent Π (Π3) by combining the repeat ing parameters with g and forcing the Π to be dimensionless (Fig. 7–26).
Second independent Π: Π3 = gw a3 0 z b3 0 Dimensions of Π3: {Π3} = {L0t0} = {gw a3 0 z b3 0 } = {L1t−2(L1t−1)a3Lb3} Equating exponents, Time: {t0} = {t−2t−a3} 0 = −2 −a3 a3 = −2 Length: {L0} = {L1La3Lb3} 0 = 1 + a3 + b3 b3 = −1 −a3 b3 = 1 Π3 is thus Π3 = gz0 w 2 0 (7–18) All three Π’s have been found, but at this point it is prudent to examine them to see if any manipulation is required. We see immediately that Π1 and Π2 are the same as the nondimensionalized variables z and t defined by Eq. 7–6—no manipulation is necessary for these. However, we recognize that the third Π must be raised to the power of −1 2 to be of the same form as an established dimensionless parameter, namely the Froude number of Eq. 7–8: Modified Π3: Π3, modified = ( gz0 w2 0 ) −1/2 = w0 √gz0 = Fr (7–19) Such manipulation is often necessary to put the Π’s into proper estab lished form. The Π of Eq. 7–18 is not wrong, and there is certainly no mathematical advantage of Eq. 7–19 over Eq. 7–18. Instead, we like to say that Eq. 7–19 is more “socially acceptable” than Eq. 7–18, since it is a named, established nondimensional parameter that is commonly used in the literature. Table 7–4 lists some guidelines for manipulation of nondi mensional Π groups into established nondimensional parameters. Table 7–5 lists some established nondimensional parameters, most of which are named after a notable scientist or engineer (see Fig. 7–27 and the Historical Spotlight on p. 317). This list is by no means exhaustive. When ever possible, you should manipulate your Π’s as necessary in order to con vert them into established nondimensional parameters.
Hint of the Day A wise choice of repeating parameters for most fluid flow problems is a length, a velocity, and a mass or density.
FIGURE 7–25 It is wise to choose common parameters as repeating parameters since they may appear in each of your dimensionless Π groups.
{II1} = {m0L0t0T0I0C0N0} = {1} {II2} = {m0L0t0T0I0C0N0} = {1} {IIk} = {m0L0t0T0I0C0N0} = {1} • • • FIGURE 7–26 The Π groups that result from the method of repeating variables are guaranteed to be dimensionless because we force the overall exponent of all seven primary dimensions to be zero.
cen96537_ch07_297-350.indd 313 29/12/16 5:23 pm 314 dimensional ANALYSIS and modeling Step 6 We should double-check that the Π’s are indeed dimensionless (Fig. 7–28). You can verify this on your own for the present example. We are finally ready to write the functional relationship between the nondimen sional parameters. Combining Eqs. 7–16, 7–17, and 7–19 into the form of Eq. 7–11, Relationship between Π’s: Π1 = f (Π2, Π3) → z z0 = f ( w0t z0 , w0 √gz0) Or, in terms of the nondimensional variables z and t defined previously by Eq. 7–6 and the definition of the Froude number, Final result of dimensional analysis: z = f(t , Fr) (7–20) It is useful to compare the result of dimensional analysis, Eq. 7–20, to the exact analytical result, Eq. 7–10. The method of repeating variables properly predicts the functional relationship between dimensionless groups. However, The method of repeating variables cannot predict the exact mathematical form of the equation.
This is a fundamental limitation of dimensional analysis and the method of repeating variables. For some simple problems, however, the form of the equation can be predicted to within an unknown constant, as is illustrated in Example 7–7.
Wow!
Aaron, you've made it! They named a nondimensional parameter after you!
FIGURE 7–27 Established nondimensional parameters are usually named after a notable scientist or engineer.
TABLE 7–4 Guidelines for manipulation of the Π’s resulting from the method of repeating variables Guideline Comments and Application to Present Problem 1. We may impose a constant We can raise a Π to any exponent n (changing it to Πn) without changing the (dimensionless) exponent on dimensionless stature of the Π. For example, in the present problem, we a Π or perform a functional imposed an exponent of −1/2 on Π3. Similarly we can perform the functional operation on a Π. operation sin(Π), exp(Π), etc., without influencing the dimensions of the Π.
2. We may multiply a Π by a Sometimes dimensionless factors of 𝜋, 1/2, 2, 4, etc., are included in a Π for pure (dimensionless) constant. convenience. This is perfectly okay since such factors do not influence the dimen sions of the Π.
3. We may form a product (or quotient) We could replace Π3 by Π3Π1, Π3/Π2, etc. Sometimes such manipulation of any Π with any other Π in the is necessary to convert our Π into an established Π. In many cases, the problem to replace one of the Π’s. established Π would have been produced if we would have chosen different repeating parameters. 4. We may use any of guidelines In general, we can replace any Π with some new Π such as AΠ3 B sin(Π1 C), 1 to 3 in combination. where A, B, and C are pure constants.
5. We may substitute a dimensional For example, the Π may contain the square of a length or the cube of a parameter in the Π with other length, for which we may substitute a known area or volume, respectively, parameter(s) of the same dimensions. in order to make the Π agree with established conventions.
These guidelines are useful in step 5 of the method of repeating variables and are listed to help you convert your nondimensional Π groups into standard, established nondimensional parameters, many of which are listed in Table 7–5.
cen96537_ch07_297-350.indd 314 29/12/16 5:23 pm 315 CHAPTER 7 TABLE 7–5 Some common established nondimensional parameters or Π’s encountered in fluid mechanics and heat transfer Name Definition Ratio of Significance Archimedes number Ar = ρsgL3 𝜇2 (ρs −ρ) Gravitational force Viscous force Aspect ratio AR = L W or L D Length Width or Length Diameter Biot number Bi = hL k Internal thermal resistance Surface thermal resistance Bond number Bo = g(ρf −ρv)L2 𝜎s Gravitational force Surface tension force Cavitation number Ca (sometimes 𝜎c) = P −Pv ρV 2 Pressure −Vapor pressure Inertial pressure (sometimes 2(P −Pv) ρV 2 ) Darcy friction factor f = 8𝜏w ρV2 Wall friction force Inertial force Drag coefficient CD = FD 1 2ρV 2A Drag force Dynamic force Eckert number Ec = V 2 cpT Kinetic energy Enthalpy Euler number Eu = ΔP ρV 2 (sometimes ΔP 1 2ρV 2) Pressure difference Dynamic pressure Fanning friction factor Cf = 2𝜏w ρV2 Wall friction force Inertial force Fourier number Fo (sometimes 𝜏) = 𝛼 t L2 Physical time Thermal diffusion time Froude number Fr = V √gL (sometimes V 2 gL) Inertial force Gravitational force Grashof number Gr = g𝛽 ∣ΔT ∣L3ρ2 𝜇2 Buoyancy force Viscous force Jakob number Ja = cp(T −Tsat) hfg Sensible energy Latent energy Knudsen number Kn = 𝜆 L Mean free path length Characteristic length Lewis number Le = k ρcpDAB = 𝛼 DAB Thermal diffusion Species diffusion Lift coefficient CL = FL 1 2ρV2A Lift force Dynamic force FIGURE 7–28 A quick check of your algebra is always wise.
ARE YOUR PI’S DIMENSIONLESS?
(continued) cen96537_ch07_297-350.indd 315 29/12/16 5:24 pm 316 dimensional ANALYSIS and modeling TABLE 7–5 (Continued) Name Definition Ratio of Significance Mach number Ma (sometimes M) = V c Flow speed Speed of sound Nusselt number Nu = Lh k Convection heat transfer Conduction heat transfer Peclet number Pe = ρLVcp k = LV 𝛼 Bulk heat transfer Conduction heat transfer Power number NP = W .
ρD5𝜔 3 Power Rotational inertia Prandtl number Pr = 𝜈 𝛼 = 𝜇cp k Viscous diffusion Thermal diffusion Pressure coefficient Cp = P −P∞ 1 2ρV 2 Static pressure difference Dynamic pressure Rayleigh number Ra = g𝛽 ∣ΔT∣L3ρ2cp k𝜇 Buoyancy force Viscous force Reynolds number Re = ρVL 𝜇 = VL v Inertial force Viscous force Richardson number Ri = L5gΔρ ρV · 2 Buoyancy force Inertial force Schmidt number Sc = 𝜇 ρDAB = 𝜈 DAB Viscous diffusion Species diffusion Sherwood number Sh = VL DAB Overall mass diffusion Species diffusion Specific heat ratio k (sometimes 𝛾) = cp cv Enthalpy Internal energy Stanton number St = h ρcpV Convection heat transfer Thermal capacity Stokes number Stk (sometimes St) = ρpD 2 p V 18𝜇L Particle relaxation time Characteristic flow time Strouhal number St (sometimes S or Sr) = f L V Characteristic flow time Period of oscillation Weber number We = ρV2L 𝜎s Inertial force Surface tension force A is a characteristic area, D is a characteristic diameter, f is a characteristic frequency (Hz), L is a characteristic length, t is a characteristic time, T is a characteristic (absolute) temperature, V is a characteristic velocity, W is a characteristic width, W . is a characteristic power, 𝜔 is a characteristic angular velocity (rad/s). Other parameters and fluid properties in these Π’s include: c = speed of sound, cp, cv = specific heats, Dp = particle diameter, DAB = species diffusion coefficient, h = convective heat transfer coefficient, hfg = latent heat of evaporation, k = thermal conductivity, P = pressure, Tsat = saturation temperature, V · = volume flow rate, 𝛼 = thermal diffusivity, 𝛽 = coefficient of thermal expansion, 𝜆 = mean free path length, 𝜇 = viscosity, 𝜈 = kinematic viscosity, 𝜌 = fluid density, 𝜌f = liquid density, 𝜌p = particle density, 𝜌s = solid density, 𝜌v = vapor density, 𝜎s = surface tension, and 𝜏w = shear stress along a wall.
cen96537_ch07_297-350.indd 316 29/12/16 5:24 pm 317 CHAPTER 7 Guest Author: Glenn Brown, Oklahoma State University Commonly used, established dimensionless numbers have been given names for convenience, and to honor persons who have contributed in the development of science and engineering. In many cases, the namesake was not the first to define the number, but usually he/she used it or a similar parameter in his/her work. The following is a list of some, but not all, such persons. Also keep in mind that some numbers may have more than one name.
Archimedes (287–212 bc) Greek mathematician who defined buoyant forces.
Biot, Jean-Baptiste (1774–1862) French mathematician who did pioneering work in heat, electricity, and elasticity. He also helped measure the arc of the meridian as part of the metric system development.
Darcy, Henry P. G. (1803–1858) French engineer who per formed extensive experiments on pipe flow and the first quantifiable filtration tests.
Eckert, Ernst R. G. (1904–2004) German–American engineer and student of Schmidt who did early work in boundary layer heat transfer.
Euler, Leonhard (1707–1783) Swiss mathematician and associate of Daniel Bernoulli who formulated equations of fluid motion and introduced the concept of centrifugal machinery.
Fanning, John T. (1837–1911) American engineer and textbook author who published in 1877 a modified form of Weisbach’s equation with a table of resistance values computed from Darcy’s data.
Fourier, Jean B. J. (1768–1830) French mathematician who did pioneering work in heat transfer and several other topics.
Froude, William (1810–1879) English engineer who developed naval modeling methods and the transfer of wave and boundary resistance from model to prototype.
Grashof, Franz (1826–1893) German engineer and educa tor known as a prolific author, editor, corrector, and dispatcher of publications.
Jakob, Max (1879–1955) German–American physicist, engineer, and textbook author who did pioneering work in heat transfer.
Knudsen, Martin (1871–1949) Danish physicist who helped develop the kinetic theory of gases.
Lewis, Warren K. (1882–1975) American engineer who researched distillation, extraction, and fluidized bed reactions.
Mach, Ernst (1838–1916) Austrian physicist who was first to realize that bodies traveling faster than the speed of sound would drastically alter the properties of the fluid. His ideas had great influence on twentieth-century thought, both in physics and in philosophy, and influenced Einstein’s development of the theory of relativity.
Nusselt, Wilhelm (1882–1957) German engineer who was the first to apply similarity theory to heat transfer.
Peclet, Jean C. E. (1793–1857) French educator, physicist, and industrial researcher.
Prandtl, Ludwig (1875–1953) German engineer and developer of boundary layer theory who is considered the founder of modern fluid mechanics.
Lord Raleigh, John W. Strutt (1842–1919) English scientist who investigated dynamic similarity, cavitation, and bubble collapse.
Reynolds, Osborne (1842–1912) English engineer who investigated flow in pipes and developed viscous flow equations based on mean velocities.
Richardson, Lewis F. (1881–1953) English mathematician, physicist, and psychologist who was a pioneer in the application of fluid mechanics to the modeling of atmospheric turbulence.
Schmidt, Ernst (1892–1975) German scientist and pioneer in the field of heat and mass transfer. He was the first to measure the velocity and temperature field in a free convection boundary layer.
Sherwood, Thomas K. (1903–1976) American engineer and educator. He researched mass transfer and its interaction with flow, chemical reactions, and industrial process operations.
Stanton, Thomas E. (1865–1931) English engineer and student of Reynolds who contributed to a number of areas of fluid flow.
Stokes, George G. (1819–1903) Irish scientist who devel oped equations of viscous motion and diffusion.
Strouhal, Vincenz (1850–1922) Czech physicist who showed that the period of oscillations shed by a wire are related to the velocity of the air passing over it.
Weber, Moritz (1871–1951) German professor who applied similarity analysis to capillary flows.
HISTORICAL SPOTLIGHT ■ Persons Honored by Nondimensional Parameters cen96537_ch07_297-350.indd 317 29/12/16 5:24 pm 318 dimensional ANALYSIS and modeling EXAMPLE 7–7 Pressure in a Soap Bubble Some children are playing with soap bubbles, and you become curious as to the rela tionship between soap bubble radius and the pressure inside the soap bubble (Fig. 7–29). You reason that the pressure inside the soap bubble must be greater than atmospheric pressure, and that the shell of the soap bubble is under tension, much like the skin of a balloon. You also know that the property surface tension must be important in this problem. Not knowing any other physics, you decide to approach the problem using dimensional analysis. Establish a relationship between pressure dif ference ΔP = Pinside − Poutside, soap bubble radius R, and the surface tension 𝜎s of the soap film.
SOLUTION The pressure difference between the inside of a soap bubble and the outside air is to be analyzed by the method of repeating variables.
Assumptions 1 The soap bubble is neutrally buoyant in the air, and gravity is not relevant. 2 No other variables or constants are important in this problem.
Analysis The step-by-step method of repeating variables is employed.
Step 1 There are three variables and constants in this problem; n = 3. They are listed in functional form, with the dependent variable listed as a function of the independent variables and constants: List of relevant parameters: ΔP = f (R, 𝜎s) n = 3 Step 2 The primary dimensions of each parameter are listed. The dimensions of surface tension are obtained from Example 7–1, and those of pressure from Example 7–2. ΔP R 𝜎s {m1L−1t−2} {L1} {m1t−2} Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t).
Reduction (first guess): j = 3 If this value of j is correct, the expected number of Π’s is k = n − j = 3 − 3 = 0. But how can we have zero Π’s? Something is obviously not right (Fig. 7–30). At times like this, we need to first go back and make sure that we are not neglecting some important variable or constant in the problem. Since we are confident that the pressure difference should depend only on soap bubble radius and surface tension, we reduce the value of j by one, Reduction (second guess): j = 2 If this value of j is correct, k = n − j = 3 − 2 = 1. Thus we expect one Π, which is more physically realistic than zero Π’s.
Step 4 We need to choose two repeating parameters since j = 2. Following the guidelines of Table 7–3, our only choices are R and 𝜎s, since ΔP is the dependent variable.
Step 5 We combine these repeating parameters into a product with the dependent variable ΔP to create the dependent Π, Dependent Π: Π1 = ΔPRa1𝜎b1 s (1) Soap film Pinside Poutside σs σs R FIGURE 7–29 The pressure inside a soap bubble is greater than that surrounding the soap bubble due to surface tension in the soap film.
What happens if k = n – j = 0?
Do the following: • Check your list of parameters.
• Check your algebra.
• If all else fails, reduce j by one.
FIGURE 7–30 If the method of repeating variables indicates zero Π’s, we have either made an error, or we need to reduce j by one and start over.
cen96537_ch07_297-350.indd 318 29/12/16 5:24 pm 319 CHAPTER 7 We apply the primary dimensions of step 2 into Eq. 1 and force the Π to be dimensionless.
Dimensions of Π1: {Π1} = {m0L0t0} = {ΔPR a1𝜎b1 s } = {(m1L−1t−2)La1(m1t−2)b1} We equate the exponents of each primary dimension to solve for a1 and b1: Time: {t0} = {t−2t−2b1} 0 = −2 −2b1 b1 = −1 Mass: {m0} = {m1mb1} 0 = 1 + b1 b1 = −1 Length: {L0} = {L−1La1} 0 = −1 + a1 a1 = 1 Fortunately, the first two results agree with each other, and Eq. 1 thus becomes Π1 = ΔPR 𝜎s (2) From Table 7–5, the established nondimensional parameter most similar to Eq. 2 is the Weber number, defined as a pressure (𝜌V 2) times a length divided by surface tension. There is no need to further manipulate this Π.
Step 6 We write the final functional relationship. In the case at hand, there is only one Π, which is a function of nothing. This is possible only if the Π is constant. Putting Eq. 2 into the functional form of Eq. 7–11, Relationship between Π’s: Π1 = ΔPR 𝜎s = f(nothing) = constant → ΔP = constant 𝞼s R (3) Discussion This is an example of how we can sometimes predict trends with dimensional analysis, even without knowing much of the physics of the problem. For example, we know from our result that if the radius of the soap bubble doubles, the pressure difference decreases by a factor of 2. Similarly, if the value of surface tension doubles, ΔP increases by a factor of 2. Dimensional analysis cannot predict the value of the constant in Eq. 3; further analysis (or one experiment) reveals that the constant is equal to 4 (Chap. 2).
EXAMPLE 7–8 Lift on a Wing Some aeronautical engineers are designing an airplane and wish to predict the lift produced by their new wing design (Fig. 7–31). The chord length Lc of the wing is 1.12 m, and its planform area A (area viewed from the top when the wing is at zero angle of attack) is 10.7 m2. The prototype is to fly at V = 52.0 m/s close to the ground where T = 25°C. They build a one-tenth scale model of the wing to test in a pressurized wind tunnel. The wind tunnel can be pressurized to a maximum of 5 atm. At what speed and pressure should they run the wind tunnel in order to achieve dynamic similarity?
SOLUTION We are to determine the speed and pressure at which to run the wind tunnel in order to achieve dynamic similarity.
Lc FL V ρ, μ, c α FIGURE 7–31 Lift FL on a wing of chord length Lc at angle of attack 𝛼 in a flow of free-stream speed V with density 𝜌, viscosity 𝜇, and speed of sound c. The angle of attack 𝛼 is measured relative to the free-stream flow direction.
cen96537_ch07_297-350.indd 319 29/12/16 5:24 pm 320 dimensional ANALYSIS and modeling Assumptions 1 The prototype wing flies through the air at standard atmospheric pressure. 2 The model is geometrically similar to the prototype.
Analysis First, the step-by-step method of repeating variables is employed to obtain the nondimensional parameters. Then, the dependent Π’s are matched between prototype and model.
Step 1 There are seven parameters (variables and constants) in this problem; n = 7. They are listed in functional form, with the dependent variable listed as a function of the independent parameters: List of relevant parameters: FL = f (V, Lc, ρ, 𝜇, c, 𝛼 ) n = 7 where FL is the lift force on the wing, V is the fluid speed, Lc is the chord length, 𝜌 is the fluid density, 𝜇 is the fluid viscosity, c is the speed of sound in the fluid, and 𝛼 is the angle of attack of the wing.
Step 2 The primary dimensions of each parameter are listed; angle 𝛼 is dimensionless: FL V Lc ρ 𝜇 c 𝛼 {m1L1t−2} {L1t−1} {L1} {m1L−3} {m1L−1t−1} {L1t−1} {1} Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t).
Reduction: j = 3 If this value of j is correct, the expected number of Π’s is k = n − j = 7 − 3 = 4.
Step 4 We need to choose three repeating parameters since j = 3. Following the guidelines listed in Table 7–3, we cannot pick the dependent variable FL. Nor can we pick 𝛼 since it is already dimensionless. We cannot choose both V and c since their dimensions are identical. It would not be desirable to have 𝜇 appear in all the Π’s. The best choice of repeating parameters is thus either V, Lc, and 𝜌 or c, Lc, and 𝜌. Of these, the former is the better choice since the speed of sound appears in only one of the established nondimensional parameters of Table 7–5, whereas the velocity scale is more “common” and appears in several of the parameters (Fig. 7–32).
Repeating parameters: V, Lc, and ρ Step 5 The dependent Π is generated: Π1 = FLVa1L b1 c ρc1 → {Π1} = {(m1L1t−2)(L1t−1)a1(L1)b1(m1L−3)c1} The exponents are calculated by forcing the Π to be dimensionless (algebra not shown). We get a1 = −2, b1 = −2, and c1 = −1. The dependent Π is thus Π1 = FL ρV 2Lc 2 From Table 7–5, the established nondimensional parameter most similar to our Π1 is the lift coefficient, defined in terms of planform area A rather than the square of chord length, and with a factor of 1/2 in the denominator. Thus, we may manipulate this Π according to the guidelines listed in Table 7–4 as follows: Modified Π1: Π1, modified = FL 1 2ρV 2A = Lift coefficient = CL FIGURE 7–32 Oftentimes when performing the method of repeating variables, the most difficult part of the procedure is choosing the repeating parameters. With practice, however, you will learn to choose these parameters wisely.
cen96537_ch07_297-350.indd 320 29/12/16 5:24 pm 321 CHAPTER 7 Similarly, the first independent Π is generated: Π2 = 𝜇V a2L b2 c ρc2 → {Π2} = {(m1L−1t−1)(L1t−1)a2(L1)b2(m1L−3)c2} from which a2 = −1, b2 = −1, and c2 = −1, and thus Π2 = 𝜇 ρVLc We recognize this Π as the inverse of the Reynolds number. So, after inverting, Modified Π2: Π2, modified = ρVLc 𝜇 = Reynolds number = Re The third Π is formed with the speed of sound, the details of which are left for you to generate on your own. The result is Π3 = V c = Mach number = Ma Finally, since the angle of attack 𝛼 is already dimensionless, it is a dimensionless Π group all by itself (Fig. 7–33). You are invited to go through the algebra; you will find that all the exponents turn out to be zero, and thus Π4 = 𝛼 = Angle of attack Step 6 We write the final functional relationship as CL = F L 1 2 ρV2A = f (Re, Ma, 𝞪) (1) To achieve dynamic similarity, Eq. 7–12 requires that all three of the dependent nondimensional parameters in Eq. 1 match between the model and the prototype. While it is trivial to match the angle of attack, it is not so simple to simultaneously match the Reynolds number and the Mach number. For example, if the wind tunnel were run at the same temperature and pressure as those of the prototype, such that 𝜌, 𝜇, and c of the air flowing over the model were the same as 𝜌, 𝜇, and c of the air flowing over the prototype, Reynolds number similarity would be achieved by setting the wind tunnel air speed to 10 times that of the prototype (since the model is one-tenth scale). But then the Mach numbers would differ by a factor of 10. At 25°C, c is approximately 346 m/s, and the Mach number of the prototype airplane wing is Map = 52.0/346 = 0.150—subsonic. At the required wind tunnel speed, Mam would be 1.50—supersonic! This is clearly unacceptable since the physics of the flow changes dramatically from subsonic to supersonic conditions. At the other extreme, if we were to match Mach numbers, the Reynolds number of the model would be 10 times too small. What should we do? A common rule of thumb is that for Mach numbers less than about 0.3, as is the fortunate case here, compressibility effects are practi cally negligible. Thus, it is not necessary to exactly match the Mach number; rather, as long as Mam is kept below about 0.3, approximate dynamic similarity can be achieved by matching the Reynolds number. Now the problem shifts to one of how to match Re while maintaining a low Mach number. This is where the pressurization feature of the wind tunnel comes in. At constant tempera ture, density is proportional to pressure, while viscosity and speed of sound are very weak functions of pressure. If the wind tunnel pressure could be pumped to 10 atm, we could run the model test at the same speed as the prototype and A parameter that is already dimensionless becomes a П parameter all by itself.
FIGURE 7–33 A parameter that is dimensionless (like an angle) is already a nondimensional Π all by itself— we know this Π without doing any further algebra.
cen96537_ch07_297-350.indd 321 29/12/16 5:24 pm 322 dimensional ANALYSIS and modeling Recall that in Examples 7–5 and 7–6 the air speed of the prototype car is 50.0 mi/h, and that of the wind tunnel is 221 mi/h. At 25°C, this corre sponds to a prototype Mach number of Map = 0.065, and at 5°C, the Mach number of the wind tunnel is 0.29—on the borderline of the incompress ible limit. In hindsight, we should have included the speed of sound in our dimensional analysis, which would have generated the Mach number as an additional Π. Another way to match the Reynolds number while keeping the Mach number low is to use a liquid such as water, since liquids are nearly incompressible, even at fairly high speeds.
achieve a nearly perfect match in both Re and Ma. However, at the maximum wind tunnel pressure of 5 atm, the required wind tunnel speed would be twice that of the prototype, or 104 m/s. The Mach number of the wind tunnel model would thus be Mam = 104/346 = 0.301—approximately at the limit of incom pressibility according to our rule of thumb. In summary, the wind tunnel should be run at approximately 100 m/s, 5 atm, and 25°C.
Discussion This example illustrates one of the (frustrating) limitations of dimensional analysis; namely, You may not always be able to match all the dependent Π’s simultaneously in a model test. Compromises must be made in which only the most important Π’s are matched. In many practical situations in fluid mechanics, the Reynolds number is not critical for dynamic similarity, pro vided that Re is high enough. If the Mach number of the prototype were sig nificantly larger than about 0.3, we would be wise to precisely match the Mach number rather than the Reynolds number in order to ensure reasonable results. Furthermore, if a different gas were used to test the model, we would also need to match the specific heat ratio (k), since compressible flow behavior is strongly dependent on k (Chap. 12). We discuss such model testing problems in more detail in Section 7–5.
EXAMPLE 7–9 Friction in a Pipe Consider flow of an incompressible fluid of density 𝜌 and viscosity 𝜇 through a long, horizontal section of round pipe of diameter D. The velocity profile is sketched in Fig. 7–34; V is the average speed across the pipe cross section, which by conservation of mass remains constant down the pipe. For a very long pipe, the flow eventually becomes hydrodynamically fully developed, which means that the velocity profile also remains uniform down the pipe. Because of frictional forces between the fluid and the pipe wall, there exists a shear stress 𝜏w on the inside pipe wall as sketched. The shear stress is also constant down the pipe in the fully developed region. We assume some constant average roughness height 𝜀 along the inside wall of the pipe. In fact, the only parameter that is not constant down the length of pipe is the pressure, which must decrease (linearly) down the pipe in order to “push” the fluid through the pipe to overcome friction. Develop a nondimensional relationship between shear stress 𝜏w and the other parameters in the problem.
SOLUTION We are to generate a nondimensional relationship between shear stress and other parameters.
τw V D ε ρ, μ FIGURE 7–34 Friction on the inside wall of a pipe. The shear stress 𝜏w on the pipe walls is a function of average fluid speed V, average wall roughness height 𝜀, fluid density 𝜌, fluid viscosity 𝜇, and inside pipe diameter D.
cen96537_ch07_297-350.indd 322 29/12/16 5:24 pm 323 CHAPTER 7 Assumptions 1 The flow is hydrodynamically fully developed. 2 The fluid is incompressible. 3 No other parameters are significant in the problem.
Analysis The step-by-step method of repeating variables is employed to obtain the nondimensional parameters.
Step 1 There are six variables and constants in this problem; n = 6. They are listed in functional form, with the dependent variable listed as a function of the independent variables and constants: List of relevant parameters: 𝜏w = f(V, 𝜀 , ρ, 𝜇, D) n = 6 Step 2 The primary dimensions of each parameter are listed. Note that shear stress is a force per unit area, and thus has the same dimensions as pressure.
𝜏w V 𝜀 ρ 𝜇 D {m1L−1t−2} {L1t−1} {L1} {m1L−3} {m1L−1t−1} {L1} Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the problem (m, L, and t). Reduction: j = 3 If this value of j is correct, the expected number of Π’s is k = n − j = 6 − 3 = 3.
Step 4 We choose three repeating parameters since j = 3. Following the guidelines of Table 7–3, we cannot pick the dependent variable 𝜏w. We cannot choose both 𝜀 and D since their dimensions are identical, and it would not be desirable to have 𝜇 or 𝜀 appear in all the Π’s. The best choice of repeating parameters is thus V, D, and 𝜌.
Repeating parameters: V, D, and ρ Step 5 The dependent Π is generated: Π1 = 𝜏wV a1Db1ρc1 → {Π1} = {(m1L−1t−2)(L1t−1)a1(L1)b1(m1L−3)c1} from which a1 = −2, b1 = 0, and c1 = −1, and thus the dependent Π is Π1 = 𝜏w ρV 2 From Table 7–5, the established nondimensional parameter most similar to this Π1 is the Darcy friction factor, defined with a factor of 8 in the numerator (Fig. 7–35). Thus, we manipulate this Π according to the guidelines listed in Table 7–4 as follows: Modified Π1: Π1, modified = 8𝜏w ρV 2 = Darcy friction factor = f Similarly, the two independent Π’s are generated, the details of which are left for you to do on your own: Π2 = 𝜇V a2 Db2 ρc2 → Π2 = ρVD 𝜇 = Reynolds number = Re Π3 = 𝜀 V a3 Db3 ρc3 → Π3 = 𝜀 D = Roughness ratio Darcy friction factor: Fanning friction factor: τw ρ V 8τw ρV2 f = 2τw ρV2 Cf = FIGURE 7–35 Although the Darcy friction factor for pipe flows is most common, you should be aware of an alternative, less common friction factor called the Fanning friction factor. The relationship between the two is f = 4Cf.
cen96537_ch07_297-350.indd 323 29/12/16 5:24 pm 324 dimensional ANALYSIS and modeling To verify the validity of Eq. 1 of Example 7–9, we use computational fluid dynamics (CFD) to predict the velocity profiles and the values of wall shear stress for two physically different but dynamically similar pipe flows: • Air at 300 K flowing at an average speed of 14.5 ft/s through a pipe of inner diameter 1.00 ft and average roughness height 0.0010 ft.
• Water at 300 K flowing at an average speed of 3.09 m/s through a pipe of inner diameter 0.0300 m and average roughness height 0.030 mm.
The two pipes are clearly geometrically similar since they are both round pipes. They have the same average roughness ratio (𝜀/D = 0.0010 in both cases). We have carefully chosen the values of average speed and diameter such that the two flows are also dynamically similar. Specifically, the other independent Π (the Reynolds number) also matches between the two flows.
Reair = ρairVair Dair 𝜇air = (1.225 kg/m3)(14.5 ft/s)(1.00 ft) 1.789 × 10−5 kg/m·s ( 0.3048 m ft ) 2 = 9.22 × 104 where the fluid properties are those built into the CFD code, and Rewater = ρwaterVwater Dwater 𝜇water = (998.2 kg/m3)(3.09 m/s)(0.0300 m) 0.001003 kg/m·s = 9.22 × 104 Hence by Eq. 7–12, we expect that the dependent Π’s should match between the two flows as well. We generate a computational mesh for each of the two flows, and use a commercial CFD code to generate the velocity profile, from which the shear stress is calculated. Fully developed, time- averaged, turbulent velocity profiles near the far end of both pipes are compared. Although the pipes are of different diameters and the fluids are vastly dif ferent, the velocity profile shapes look quite similar. In fact, when we plot normalized axial velocity (u/V) as a function of normalized radius (r/R), we find that the two profiles fall on top of each other (Fig. 7–36). Wall shear stress is also calculated from the CFD results for each flow, a comparison of which is shown in Table 7–6. There are several reasons why the wall shear stress in the water pipe is orders of magnitude larger than that in the air pipe. Namely, water is over 800 times as dense as air and over 50 times as viscous. Furthermore, shear stress is proportional to the gradient of velocity, and the water pipe diameter is less than one-tenth that of the air Step 6 We write the final functional relationship as f = 8𝜏w 𝞺V2 = f(Re, 𝞮 D) (1) Discussion The result applies to both laminar and turbulent fully developed pipe flow; it turns out, however, that the second independent Π (roughness ratio 𝜀/D) is not nearly as important in laminar pipe flow as in turbulent pipe flow. This problem presents an interesting connection between geometric similarity and dimensional analysis. Namely, it is necessary to match 𝜀/D since it is an independent Π in the problem. From a different perspective, thinking of roughness as a geometric property, it is necessary to match 𝜀/D to ensure geometric similarity between two pipes.
r/R 0 0 0.5 1 1.5 u/V 0.2 0.4 0.6 0.8 1 1.2 FIGURE 7–36 Normalized axial velocity profiles for fully developed flow through a pipe as predicted by CFD; profiles of air (circles) and water (crosses) are shown on the same plot.
cen96537_ch07_297-350.indd 324 29/12/16 5:24 pm 325 CHAPTER 7 pipe, leading to steeper velocity gradients. In terms of the nondimensional ized wall shear stress, f, however, Table 7–6 shows that the results are iden tical due to dynamic similarity between the two flows. Note that although the values are reported to three significant digits, the reliability of turbu lence models in CFD is accurate to at most two significant digits (Chap. 15).
7–5 ■ EXPERIMENTAL TESTING, MODELING, AND INCOMPLETE SIMILARITY One of the most useful applications of dimensional analysis is in designing physical and/or numerical experiments, and in reporting the results of such experiments. In this section we discuss both of these applications, and point out situations in which complete dynamic similarity is not achievable.
Setup of an Experiment and Correlation of Experimental Data As a generic example, consider a problem in which there are five origi nal parameters (one of which is the dependent parameter). A complete set of experiments (called a full factorial test matrix) is conducted by testing every possible combination of several levels of each of the four independent parameters. A full factorial test with five levels of each of the four inde pendent parameters would require 54 = 625 experiments. While experimen tal design techniques (fractional factorial test matrices; see Montgomery, 2013) can significantly reduce the size of the test matrix, the number of required experiments would still be large. However, assuming that three pri mary dimensions are represented in the problem, we can reduce the number of parameters from five to two (k = 5 − 3 = 2 nondimensional Π groups), and the number of independent parameters from four to one. Thus, for the same resolution (five tested levels of each independent parameter) we would then need to conduct a total of only 51 = 5 experiments. You don’t have to be a genius to realize that replacing 625 experiments by 5 experiments is cost effective. You can see why it is wise to perform a dimensional analysis before conducting an experiment. Continuing our discussion of this generic example (a two-Π problem), once the experiments are complete, we plot the dependent dimensionless para meter (Π1) as a function of the independent dimensionless parameter (Π2), as in Fig. 7–37. We then determine the functional form of the relationship by Π1 (a) Π2 Π1 (b) Π2 FIGURE 7–37 For a two-Π problem, we plot dependent dimensionless parameter (Π1) as a function of independent dimensionless parameter (Π2). The resulting plot can be (a) linear or (b) nonlinear. In either case, regression and curve-fitting techniques are available to determine the relationship between the Π’s.
TABLE 7–6 Comparison of wall shear stress and nondimensionalized wall shear stress for fully developed flow through an air pipe and a water pipe as predicted by CFD Parameter Air Flow Water Flow Wall shear stress 𝜏w, air = 0.0557 N/m2 𝜏w, water = 22.2 N/m2 Dimensionless wall shear stress (Darcy friction factor) fair = 8𝜏w, air ρair V2 air = 0.0186 fwater = 8𝜏w, water ρwater V 2 water = 0.0186 Data obtained with ANSYS-FLUENT using the standard k-𝜀 turbulence model with wall functions.
cen96537_ch07_297-350.indd 325 29/12/16 5:24 pm 326 dimensional ANALYSIS and modeling performing a regression analysis on the data. If we are lucky, the data may correlate linearly. If not, we can try linear regression on log–linear or log– log coordinates, polynomial curve fitting, etc., to establish an approximate relationship between the two Π’s. See Holman (2001) for details about these curve-fitting techniques. If there are more than two Π’s in the problem (e.g., a three-Π problem or a four-Π problem), we need to set up a test matrix to determine the relation ship between the dependent Π and the independent Π’s. In many cases we discover that one or more of the dependent Π’s has negligible effect and can be removed from the list of necessary dimensionless parameters. As we have seen (Example 7–7), dimensional analysis sometimes yields only one Π. In a one-Π problem, we know the form of the relationship between the original parameters to within some unknown constant. In such a case, only one experiment is needed to determine that constant.
Incomplete Similarity We have shown several examples in which the nondimensional Π groups are easily obtained with paper and pencil through straightforward use of the method of repeating variables. In fact, after sufficient practice, you should be able to obtain the Π’s with ease—sometimes in your head or on the “back of an envelope.” Unfortunately, it is often a much different story when we go to apply the results of our dimensional analysis to experimental data. The problem is that it is not always possible to match all the Π’s of a model to the corresponding Π’s of the prototype, even if we are careful to achieve geometric similarity. This situation is called incomplete similarity. Fortunately, in some cases of incomplete similarity, we are still able to extrapolate model test data to obtain reasonable full-scale predictions.
Wind Tunnel Testing We illustrate incomplete similarity with the problem of measuring the aero dynamic drag force on a model truck in a wind tunnel (Fig. 7–38). Sup pose we purchase a one-sixteenth scale die-cast model of a tractor-trailer rig (18-wheeler). The model is geometrically similar to the prototype—even in the details such as side mirrors, mud flaps, etc. The model truck is 0.991 m long, corresponding to a full-scale prototype length of 15.9 m. The model truck is to be tested in a wind tunnel that has a maximum speed of 70 m/s. The wind tunnel test section is 1.0 m tall and 1.2 m wide—big enough to accommodate the model without needing to worry about wall interference or blockage effects. The air in the wind tunnel is at the same temperature and pressure as the air flowing around the prototype. We want to simulate flow at Vp = 60 mi/h (26.8 m/s) over the full-scale prototype truck. The first thing we do is match the Reynolds numbers, Rem = ρmVm Lm 𝜇m = Rep = ρpVp Lp 𝜇p which can be solved for the required wind tunnel speed for the model tests Vm, Vm = Vp( 𝜇m 𝜇p)( ρp ρm)( Lp Lm) = (26.8 m/s)(1)(1)( 16 1 ) = 429 m/s Model Moving belt Drag balance FD V Wind tunnel test section FIGURE 7–38 Measurement of aerodynamic drag on a model truck in a wind tunnel equipped with a drag balance and a moving belt ground plane.
cen96537_ch07_297-350.indd 326 29/12/16 5:24 pm 327 CHAPTER 7 Thus, to match the Reynolds number between model and prototype, the wind tunnel should be run at 429 m/s (to three significant digits). We obvi ously have a problem here, since this speed is more than six times greater than the maximum achievable wind tunnel speed. Moreover, even if we could run the wind tunnel that fast, the flow would be supersonic, since the speed of sound in air at room temperature is about 346 m/s. While the Mach number of the prototype truck moving through the air is 26.8/335 = 0.080, that of the wind tunnel air moving over the model would be 429/335 = 1.28 (if the wind tunnel could go that fast). It is clearly not possible to match the model Reynolds number to that of the prototype with this model and wind tunnel facility. What do we do? There are several options: • If we had a bigger wind tunnel, we could test with a larger model. Auto mobile manufacturers typically test with three-eighths scale model cars and with one-eighth scale model trucks and buses in very large wind tunnels. Some wind tunnels are even large enough for full-scale automobile tests (Fig. 7–39a). As you can imagine, however, the bigger the wind tunnel and the model the more expensive the tests. We must also be careful that the model is not too big for the wind tunnel. A useful rule of thumb is that the blockage (ratio of the model frontal area to the cross-sectional area of the test section) should be less than 7.5 percent. Otherwise, the wind tunnel walls adversely affect both geometric and kinematic similarity.
• We could use a different fluid for the model tests. For example, water tunnels can achieve higher Reynolds numbers than can wind tunnels of the same size, but they are much more expensive to build and operate (Fig. 7–39b).
• We could pressurize the wind tunnel and/or adjust the air temperature to increase the maximum Reynolds number capability. While these techniques can help, the increase in the Reynolds number is limited.
• If all else fails, we could run the wind tunnel at several speeds near the maximum speed, and then extrapolate our results to the full-scale Reynolds number.
Fortunately, it turns out that for many wind tunnel tests the last option is quite viable. While drag coefficient CD is a strong function of the Reynolds number at low values of Re, CD often levels off for Re above some value. In other words, for flow over many objects, especially “bluff” objects like trucks, buildings, etc., the flow is Reynolds number independent above some threshold value of Re (Fig. 7–40), typically when the boundary layer and the wake are both fully turbulent.
EXAMPLE 7–10 Model Truck Wind Tunnel Measurements A one-sixteenth scale model tractor-trailer truck (18-wheeler) is tested in a wind tunnel as sketched in Fig. 7–38. The model truck is 0.991 m long, 0.257 m tall, and 0.159 m wide. During the tests, the moving ground belt speed is adjusted so as to always match the speed of the air moving through the test section. Aero dynamic drag force FD is measured as a function of wind tunnel speed; the exper CD Re Unreliable data at low Re Re independence FIGURE 7–40 For many objects, the drag coefficient levels off at Reynolds numbers above some threshold value. This fortunate situation is called Reynolds number independence. It enables us to extrapolate to prototype Reynolds numbers that are outside of the range of our experimental facility.
FIGURE 7–39 (a) The Langley full-scale wind tunnel (LFST) is large enough that full-scale vehicles can be tested. (b) For the same scale model and speed, water tunnels achieve higher Reynolds numbers than wind tunnels.
(a) NASA/Ames/Dominic Hart (b) NASA/Eric James (a) (b) cen96537_ch07_297-350.indd 327 29/12/16 5:24 pm 328 dimensional ANALYSIS and modeling imental results are listed in Table 7–7. Plot the drag coefficient CD as a function of the Reynolds number Re, where the area used for the calculation of CD is the frontal area of the model truck (the area you see when you look at the model from upstream), and the length scale used for calculation of Re is truck width W. Have we achieved dynamic similarity? Have we achieved Reynolds number independence in our wind tunnel test? Estimate the aerodynamic drag force on the prototype truck traveling on the highway at 26.8 m/s. Assume that both the wind tunnel air and the air flowing over the prototype car are at 25°C and standard atmospheric pressure.
SOLUTION We are to calculate and plot CD as a function of Re for a given set of wind tunnel measurements and determine if dynamic similarity and/or Reynolds number independence have been achieved. Finally, we are to estimate the aerody namic drag force acting on the prototype truck.
Assumptions 1 The model truck is geometrically similar to the proto type truck. 2 The aerodynamic drag on the strut(s) holding the model truck is negligible.
Properties For air at atmospheric pressure and at T = 25°C, 𝜌 = 1.184 kg/m3 and 𝜇 = 1.849 × 10−5 kg/m·s.
Analysis We calculate CD and Re for the last data point listed in Table 7–7 (at the fastest wind tunnel speed), CD, m = FD, m 1 2 ρmV 2 m Am = 89.9 N 1 2 (1.184 kg/m3) (70 m/s)2 (0.159 m) (0.257 m) ( 1 kg·m/s2 1 N ) = 0.758 and Rem = ρmVmWm 𝜇m = (1.184 kg/m3) (70 m/s) (0.159 m) 1.849 × 10−5 kg/m·s = 7.13 × 105 (1) We repeat these calculations for all the data points in Table 7–7, and we plot CD versus Re in Fig. 7–41. Have we achieved dynamic similarity? Well, we have geometric similarity between model and prototype, but the Reynolds number of the prototype truck is Rep = ρpVpWp 𝜇p = (1.184 kg/m3) (26.8 m/s)[16(0.159 m)] 1.849 × 10−5 kg/m·s = 4.37 × 106 (2) where the width of the prototype is specified as 16 times that of the model. Com parison of Eqs. 1 and 2 reveals that the prototype Reynolds number is more than six times larger than that of the model. Since we cannot match the independent Π’s in the problem, dynamic similarity has not been achieved. Have we achieved Reynolds number independence? From Fig. 7–41 we see that Reynolds number independence has indeed been achieved—at Re greater than about 5 × 105, CD has leveled off to a value of about 0.76 (to two s ignificant digits). Since we have achieved Reynolds number independence, we can extrapolate to the full-scale prototype, assuming that CD remains constant as Re is increased to that of the full-scale prototype.
TABLE 7–7 Wind tunnel data: aerodynamic drag force on a model truck as a function of wind tunnel speed V, m/s FD, N 20 12.4 25 19.0 30 22.1 35 29.0 40 34.3 45 39.9 50 47.2 55 55.5 60 66.0 65 77.6 70 89.9 CD 0.6 2 7 6 5 4 3 8 Re × 10–5 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 FIGURE 7–41 Aerodynamic drag coefficient as a function of the Reynolds number. The values are calculated from wind tunnel test data on a model truck (Table 7–7).
cen96537_ch07_297-350.indd 328 29/12/16 5:24 pm 329 CHAPTER 7 Flows with Free Surfaces For the case of model testing of flows with free surfaces (boats and ships, floods, river flows, aqueducts, hydroelectric dam spillways, interaction of waves with piers, soil erosion, etc.), complications arise that preclude com plete similarity between model and prototype. For example, if a model river is built to study flooding, the model is often several hundred times smaller than the prototype due to limited lab space. If the vertical dimensions of the model were scaled proportionately, the depth of the model river would be so small that surface tension effects (and the Weber number) would become important, and would perhaps even dominate the model flow, even though surface tension effects are negligible in the prototype flow. In addition, although the flow in the actual river may be turbulent, the flow in the model river may be laminar, especially if the slope of the riverbed is geometrically similar to that of the prototype. To avoid these problems, researchers often use a distorted model in which the vertical scale of the model (e.g., river depth) is exaggerated in comparison to the horizontal scale of the model (e.g., river width). In addition, the model riverbed slope is often made pro portionally steeper than that of the prototype. These modifications result in incomplete similarity due to lack of geometric similarity. Model tests are still useful under these circumstances, but other tricks (like deliberately roughening the model surfaces) and empirical corrections and correlations are required to properly scale up the model data. In many practical problems involving free surfaces, both the Reynolds number and Froude number appear as relevant independent Π groups in the dimensional analysis (Fig. 7–42). It is difficult (often impossible) to match both of these dimensionless parameters simultaneously. For a free-surface flow with length scale L, velocity scale V, and kinematic viscosity 𝜈, the Reynolds number is matched between model and prototype when Rep = Vp Lp 𝜈p = Rem = Vm Lm 𝜈m (7–21) The Froude number is matched between model and prototype when Fr p = Vp √gLp = Frm = Vm √gLm (7–22) Predicted aerodynamic drag on the prototype: FD, p = 1 2 ρpV 2 p Ap CD, p = 1 2(1.184 kg/m3)(26.8 m/s)2162 (0.159 m) (0.257 m) ( 1 N 1 kg·m/s2) = 3400 N Discussion We give our final result to two significant digits. More than that cannot be justified. As always, we must exercise caution when performing an extrapolation, since we have no guarantee that the extrapolated results are correct.
g L V ρ, μ Re = = ν μ ρVL VL Fr = gL V √ FIGURE 7–42 In many flows involving a liquid with a free surface, both the Reynolds number and Froude number are relevant nondimensional parameters. Since it is not always possible to match both Re and Fr between model and prototype, we are sometimes forced to settle for incomplete similarity. cen96537_ch07_297-350.indd 329 29/12/16 5:24 pm 330 dimensional ANALYSIS and modeling To match both Re and Fr, we solve Eqs. 7–21 and 7–22 simultaneously for the required length scale factor Lm/Lp, Lm Lp = 𝜈m 𝜈p Vp Vm = ( Vm Vp ) 2 (7–23) Eliminating the ratio Vm/Vp from Eq. 7–23, we see that Required ratio of kinematic viscosities to match both Re and Fr: 𝜈m 𝜈p = ( Lm Lp ) 3/2 (7–24) Thus, to ensure complete similarity (assuming geometric similarity is achievable without unwanted surface tension effects as discussed previ ously), we would need to use a liquid whose kinematic viscosity satisfies Eq. 7–24. Although it is sometimes possible to find an appropriate liquid for use with the model, in most cases it is either impractical or impossible, as Example 7–11 illustrates. In such cases, it is more important to match Froude number than Reynolds number (Fig. 7–43).
EXAMPLE 7–11 Model Lock and River In the late 1990s the U.S. Army Corps of Engineers designed an experiment to model the flow of the Tennessee River downstream of the Kentucky Lock and Dam (Fig. 7–44). Because of laboratory space restrictions, they built a scale model with a length scale factor of Lm/Lp = 1/100. Suggest a liquid that would be appropriate for the experiment.
SOLUTION We are to suggest a liquid to use in an experiment involving a one-hundredth scale model of a lock, dam, and river.
Assumptions 1 The model is geometrically similar to the prototype. 2 The model river is deep enough that surface tension effects are not significant.
Properties For water at atmospheric pressure and at T = 20°C, the prototype kin ematic viscosity is 𝜈p = 1.002 × 10−6 m2/s.
Analysis From Eq. 7–24, Required kinematic viscosity of model liquid: 𝜈m = 𝜈p( Lm Lp ) 3/2 = (1.002 × 10−6 m2/s)( 1 100) 3/2 = 1.00 × 10−9 m2/s (1) Thus, we need to find a liquid that has a viscosity of 1.00 × 10−9 m2/s. A quick glance through the appendices yields no such liquid. Hot water has a lower kinematic viscosity than cold water, but only by a factor of about 3.
Liquid mercury has a very small kinematic viscosity, but it is of order 10−7 m2/s—still two orders of magnitude too large to satisfy Eq. 1. Even if liquid mercury would work, it would be too expensive and too hazard ous to use in such a test. What do we do? The bottom line is that we can not match both the Froude number and the Reynolds number in this model test. FIGURE 7–43 A NACA 0024 airfoil being tested in a towing tank at Fr = (a) 0.19, (b) 0.37, and (c) 0.55. In tests like this, the Froude number is the most important nondimensional parameter. Photo courtesy of IIHR-Hydroscience & Engineering, University of Iowa. Used by permission.
(a) (b) (c) cen96537_ch07_297-350.indd 330 29/12/16 5:24 pm 331 CHAPTER 7 In closing this section on experiments and incomplete similarity, we mention the importance of similarity in the production of Hollywood movies in which model boats, trains, airplanes, buildings, monsters, etc., are blown up or burned. Movie producers must pay attention to dynamic similarity in order to make the small-scale fires and explosions appear as realistic as possible. You may recall some low-budget movies where the special effects are unconvincing. In most cases this is due to lack of dynamic similarity between the small model and the full-scale prototype. If the model’s Froude number and/or Reynolds number differ too much from those of the prototype, the special effects don’t look right, even to the untrained eye. The next time you watch a movie, be on the alert for incomplete similarity!
In other words, it is impossible to achieve complete similarity between model and prototype in this case. Instead, we do the best job we can under conditions of incomplete similarity. Water is typically used in such tests for con venience.
Discussion It turns out that for this kind of experiment, Froude number matching is more critical than Reynolds number matching. As discussed previ ously for wind tunnel testing, Reynolds number independence is achieved at high enough values of Re. Even if we are unable to achieve Reynolds number inde pendence, we can often extrapolate our low Reynolds number model data to pre dict full-scale Reynolds number behavior (Fig. 7–45). A high level of confidence in using this kind of extrapolation comes only after much laboratory experience with similar problems.
FIGURE 7–44 A 1:100 scale model constructed to investigate navigation conditions in the lower lock approach for a distance of 2 mi downstream of the dam. The model includes a scaled version of the spillway, powerhouse, and existing lock. In addition to navigation, the model was used to evaluate environmental issues associated with the new lock and required railroad and highway bridge relocations. The view here is looking upstream toward the lock and dam. At this scale, 52.8 ft on the model represents 1 mi on the prototype. A (real, full-scale) pickup truck in the background gives you a feel for the model scale.
Photo courtesy of the U.S. Army Corps of Engi neers, US Army Engineer Research and Develop ment Center (USACE-ERDC), Nashville.
Measured parameter Re Rep Range of Rem Extrapolated result FIGURE 7–45 In many experiments involving free surfaces, we cannot match both the Froude number and the Reynolds number. However, we can often extrapolate low Re model test data to predict high Re prototype behavior.
cen96537_ch07_297-350.indd 331 29/12/16 5:24 pm 332 dimensional ANALYSIS and modeling Guest Author: Michael Dickinson, California Institute of Technology An interesting application of dimensional analysis is in the study of how insects fly. The small size and fast wing speed of an insect, such as a tiny fruit fly, make it difficult to directly measure the forces or visualize the air motion cre ated by the fly’s wings. However, using principles of dimensional analysis, it is possible to study insect aerodynamics on a larger-scale, slowly moving model— a mechanical robot. The forces created by a hovering fly and flapping robot are dynamically similar if the Reynolds number is the same for each case. For a flapping wing, Re is calculated as 2ΦRLc𝜔/𝜈, where Φ is the angular amplitude of the wing stroke, R is the wing length, Lc is the average wing width (chord length), 𝜔 is the angular frequency of the stroke, and 𝜈 is the kinematic vis cosity of the surrounding fluid. A fruit fly flaps its 2.5-mm-long, 0.7-mm-wide wings 200 times per second over a 2.8-rad stroke in air with a kinematic viscos ity of 1.5 × 10−5 m2/s. The resulting Reynolds number is approximately 130. By choosing mineral oil with a kinematic viscosity of 1.15 × 10−4 m2/s, it is possible to match this Reynolds number on a robotic fly that is 100 times larger, flapping its wings over 1000 times more slowly! If the fly is not stationary, but rather moving through the air, it is necessary to match another dimensionless parameter to ensure dynamic similarity, the reduced frequency, 𝜎 = 2ΦR𝜔/V, which measures the ratio of the flapping velocity of the wing tip (2ΦR𝜔) to the forward velocity of the body (V). To simulate forward flight, a set of motors tows Robofly through its oil tank at an appropriately scaled speed. Dynamically scaled robots have helped show that insects use a variety of dif ferent mechanisms to produce forces as they fly. During each back-and-forth stroke, insect wings travel at high angles of attack, generating a prominent leading-edge vortex. The low pressure of this large vortex pulls the wings upward. Insects can further augment the strength of the leading-edge vortex by rotating their wings at the end of each stroke. After the wing changes direction, it can also generate forces by quickly running through the wake of the previous stroke. Figure 7–46a shows a real fly flapping its wings, and Fig. 7–46b shows Robofly flapping its wings. Because of the larger length scale and shorter time scale of the model, measurements and flow visualizations are possible. Experi ments with dynamically scaled model insects continue to teach researchers how insects manipulate wing motion to steer and maneuver.
References Dickinson, M. H., Lehmann, F.-O., and Sane, S., “Wing Rotation and the Aerody namic Basis of Insect Flight,” Science, 284, p. 1954, 1999.
Dickinson, M. H., “Solving the Mystery of Insect Flight,” Scientific American, 284, No. 6, pp. 35–41, June 2001.
Fry, S. N., Sayaman, R., and Dickinson, M. H., “The Aerodynamics of Free-flight Maneuvers in Drosophila,” Science, 300, pp. 495–498, 2003.
APPLICATION SPOTLIGHT ■ How a Fly Flies (a) FIGURE 7–46 (a) The fruit fly, Drosophila melanogaster, flaps its tiny wings back and forth 200 times a second, creating a blurred image of the stroke plane. (b) The dynamically scaled model, Robofly, flaps its wings once every 5 s in 2 tons of mineral oil. Sensors at the base of the wings record aerodynamic forces, while fine bubbles are used to visualize the flow. The size and speed of the robot, as well as the properties of the oil, were carefully chosen to match the Reynolds number of a real fly.
Photos © Courtesy of Michael Dickinson, CALTECH.
(b) cen96537_ch07_297-350.indd 332 29/12/16 5:24 pm 333 CHAPTER 7 SUMMARY There is a difference between dimensions and units; a dimen sion is a measure of a physical quantity (without numerical values), while a unit is a way to assign a number to that dimen sion. There are seven primary dimensions—not just in fluid mechanics, but in all fields of science and engineering. They are mass, length, time, temperature, electric current, amount of light, and amount of matter. All other dimensions can be formed by a combination of these seven primary dimensions. All mathematical equations must be dimensionally homo geneous; this fundamental principle can be applied to equa tions in order to nondimensionalize them and to identify dimensionless groups, also called nondimensional param eters. A powerful tool to reduce the number of necessary independent parameters in a problem is called dimensional analysis. The method of repeating variables is a step-by-step procedure for finding the nondimensional parameters, or Π’s, based simply on the dimensions of the variables and constants in the problem. The six steps in the method of repeating variables are summarized here.
Step 1 List the n parameters (variables and constants) in the problem.
Step 2 List the primary dimensions of each parameter.
Step 3 Guess the reduction j, usually equal to the num ber of primary dimensions in the problem. If the analy sis does not work out, reduce j by one and try again. The expected number of Π’s (k) is equal to n minus j.
Step 4 Wisely choose j repeating parameters for con struction of the Π’s.
Step 5 Generate the k Π’s one at a time by grouping the j repeating parameters with each of the remaining variables or constants, forcing the product to be dimen sionless, and manipulating the Π’s as necessary to achieve established nondimensional parameters.
Step 6 Check your work and write the final functional relationship. When all the dimensionless groups match between a model and a prototype, dynamic similarity is achieved, and we are able to directly predict prototype performance based on model experiments. However, it is not always possible to match all the Π groups when trying to achieve similarity between a model and a prototype. In such cases, we run the model tests under conditions of incomplete similarity, match ing the most important Π groups as best we can, and then extrapolating the model test results to prototype conditions. We use the concepts presented in this chapter throughout the remainder of the book. For example, dimensional analysis is applied to fully developed pipe flows in Chap. 8 (friction factors, loss coefficients, etc.). In Chap. 10, we normalize the differential equations of fluid flow derived in Chap. 9, produc ing several dimensionless parameters. Drag and lift coeffi cients are used extensively in Chap. 11, and dimensionless parameters also appear in the chapters on compressible flow and open-channel flow (Chaps. 12 and 13). We learn in Chap. 14 that dynamic similarity is often the basis for design and test ing of pumps and turbines. Finally, dimensionless parameters are also used in computations of fluid flows (Chap. 15).
REFERENCES AND SUGGESTED READING 1. D. C. Montgomery. Design and Analysis of Experiments, 8th ed. New York: Wiley, 2013.
2. J. P. Holman. Experimental Methods for Engineers, 7th ed. New York: McGraw-Hill, 2001.
PROBLEMS Dimensions and Units, Primary Dimensions 7–1C What is the difference between a dimension and a unit? Give three examples of each.
7–2C List the seven primary dimensions. What is signifi cant about these seven?
7–3 Write the primary dimensions of the universal ideal gas constant Ru. (Hint: Use the ideal gas law, PV = nRuT where P is pressure, V is volume, T is absolute temperature, and n is the number of moles of the gas.) Answer: {m1L2t−2T−1N−1} 7–4 Write the primary dimensions of each of the following variables from the field of thermodynamics, showing all your work: (a) energy E; (b) specific energy e = E/m; (c) power W .
.
Answers: (a) {m1L2t−2}, (b) {L2t−2}, (c) {m1L2t−3} Problems designated by a “C” are concept questions, and stu dents are encouraged to answer them all. Problems designated by an “E” are in English units, and SI users can ignore them. Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software.
cen96537_ch07_297-350.indd 333 29/12/16 5:24 pm 334 dimensional ANALYSIS and modeling 7–5 What are the primary dimensions of electric voltage (E)? (Hint: Make use of the fact that electric power is equal to voltage times current.) 7–6 When performing a dimensional analysis, one of the first steps is to list the primary dimensions of each relevant param eter. It is handy to have a table of parameters and their primary dimensions. We have started such a table for you (Table P7–6), in which we have included some of the basic parameters com monly encountered in fluid mechanics. As you work through homework problems in this chapter, add to this table. You should be able to build up a table with dozens of parameters.
TABLE P7–6 Parameter Parameter Primary Name Symbol Dimensions Acceleration a L1t−2 Angle 𝜃, 𝜙, etc. 1 (none) Density 𝜌 m1L−3 Force F m1L1t−2 Frequency f t−1 Pressure P m1L−1t−2 Surface tension 𝜎s m1t−2 Velocity V L1t−1 Viscosity 𝜇 m1L−1t−1 Volume flow rate V · L3t−1 7–7 Consider the table of Prob. 7–6 where the primary dimensions of several variables are listed in the mass– length–time system. Some engineers prefer the force–length– time system (force replaces mass as one of the primary dimensions). Write the primary dimensions of three of these (density, surface tension, and viscosity) in the force– length–time system.
7–8 On a periodic chart of the elements, molar mass (M), also called atomic weight, is often listed as though it were a dimensionless quantity (Fig. P7–8). In reality, atomic weight is the mass of 1 mol of the element. For example, the atomic weight of nitrogen Mnitrogen = 14.0067. We interpret this as 14.0067 g/mol of elemental nitrogen, or in the English sys tem, 14.0067 lbm/lbmol of elemental nitrogen. What are the primary dimensions of atomic weight?
7 N 14.0067 8 O 15.9994 6 C 12.011 15 P 30.9738 16 S 32.060 14 Si 28.086 FIGURE P7–8 7–9 We define the specific ideal gas constant Rgas for a par ticular gas as the ratio of the universal gas constant and the molar mass (also called molecular weight) of the gas, Rgas = Ru/M. For a particular gas, then, the ideal gas law is written as follows: PV = mRgasT or P = ρRgasT where P is pressure, V is volume, m is mass, T is absolute temperature, and 𝜌 is the density of the particular gas. What are the primary dimensions of Rgas? For air, Rair = 287.0 J/kg·K in standard SI units. Verify that these units agree with your result.
7–10 The moment of force ( M ›) is formed by the cross product of a moment arm (r›) and an applied force (F ›), as sketched in Fig. P7–10. What are the primary dimensions of moment of force? List its units in primary SI units and in primary English units.
F M = r × F Point O r FIGURE P7–10 7–11 You are probably familiar with Ohm’s law for elec tric circuits (Fig. P7–11), where ΔE is the voltage difference or potential across the resistor, I is the electric current pass ing through the resistor, and R is the electrical resistance. What are the primary dimensions of electrical resistance? Answer: {m1L2t−3I−2} I R ΔE = IR FIGURE P7–11 7–12 Write the primary dimensions of each of the follow ing variables, showing all your work: (a) acceleration a; (b) angular velocity 𝜔; (c) angular acceleration 𝛼.
7–13 Write the primary dimensions of each of the following variables, showing all your work: (a) specific heat at constant pressure cp; (b) specific weight 𝜌g; (c) specific enthalpy h.
7–14 Thermal conductivity k is a measure of the ability of a material to conduct heat (Fig. P7–14). For conduction heat transfer in the x-direction through a surface normal to the x-direction, Fourier’s law of heat conduction is expressed as Q .
conduction = −kA dT dx cen96537_ch07_297-350.indd 334 29/12/16 5:24 pm 335 CHAPTER 7 where Q .
conduction is the rate of heat transfer and A is the area normal to the direction of heat transfer. Determine the pri mary dimensions of thermal conductivity (k). Look up a value of k in the appendices and verify that its SI units are consistent with your result. In particular, write the primary SI units of k.
T1 T2 A k x Qconduction • FIGURE P7–14 7–15 Write the primary dimensions of each of the follow ing variables from the study of convection heat transfer (Fig. P7–15), showing all your work: (a) heat generation rate g . (Hint: rate of conversion of thermal energy per unit volume); (b) heat flux q . (Hint: rate of heat transfer per unit area); (c) heat transfer coefficient h (Hint: heat flux per unit temperature difference).
q • g • h = q • Ts – T∞ T∞ Ts FIGURE P7–15 7–16 Thumb through the appendices of your thermodynam ics book, and find three properties or constants not mentioned in Probs. 7–1 to 7–15. List the name of each property or con stant and its SI units. Then write out the primary dimensions of each property or constant.
Dimensional Homogeneity 7–17C Explain the law of dimensional homogeneity in sim ple terms.
7–18 Cold water enters a pipe, where it is heated by an external heat source (Fig. P7–18). The inlet and outlet water temperatures are Tin and Tout, respectively. The total rate of heat transfer Q . from the surroundings into the water in the pipe is Q .
= m · cp(Tout −Tin) where m . is the mass flow rate of water through the pipe, and cp is the specific heat of the water. Write the primary dimen sions of each additive term in the equation, and verify that the equation is dimensionally homogeneous. Show all your work.
Q = mcp(Tout – Tin) m Tin Tout ⋅ ⋅ ⋅ FIGURE P7–18 7–19 The Reynolds transport theorem (RTT) is discussed in Chap. 4. For the general case of a moving and/or deforming control volume, we write the RTT as follows: dBsys dt = d dt ∫ CV ρb dV + ∫ CS ρbV › r·n› dA where V › r is the relative velocity, i.e., the velocity of the fluid relative to the control surface. Write the primary dimensions of each additive term in the equation, and verify that the equation is dimensionally homogeneous. Show all your work. (Hint: Since B can be any property of the flow—scalar, vec tor, or even tensor—it can have a variety of dimensions. So, just let the dimensions of B be those of B itself, {B}. Also, b is defined as B per unit mass.) 7–20 An important application of fluid mechanics is the study of room ventilation. In particular, suppose there is a source S (mass per unit time) of air pollution in a room of volume V (Fig. P7–20). Examples include carbon monoxide from cigarette smoke or an unvented kerosene heater, gases like ammonia from household cleaning products, and vapors given off by evaporation of volatile organic compounds (VOCs) from an open container. We let c represent the mass concentration (mass of contaminant per unit volume of air). V · is the volume flow rate of fresh air entering the room. If the room air is well mixed so that the mass concentration c is uniform throughout the room, but varies with time, the differ ential equation for mass concentration in the room as a func tion of time is V dc dt = S −V ·c −cAskw Supply Exhaust c(t) As kw S V V .
FIGURE P7–20 where kw is an adsorption coefficient and As is the surface area of walls, floors, furniture, etc., that adsorb some of the cen96537_ch07_297-350.indd 335 29/12/16 5:24 pm 336 dimensional ANALYSIS and modeling contaminant. Write the primary dimensions of the first three terms in the equation (including the term on the left side), and verify that those terms are dimensionally homogeneous. Then determine the dimensions of kw. Show all your work.
7–21 In Chap. 9, we discuss the differential equation for conservation of mass, the continuity equation. In cylindrical coordinates, and for steady flow, 1 r ∂(rur) ∂r + 1 r ∂u𝜃 ∂𝜃 + ∂uz ∂z = 0 Write the primary dimensions of each additive term in the equation, and verify that the equation is dimensionally homo geneous. Show all your work.
7–22 In Chap. 4 we defined volumetric strain rate as the rate of increase of volume of a fluid element per unit volume (Fig. P7–22). In Cartesian coordinates we write the volumet ric strain rate as 1 V DV Dt = ∂u ∂x + ∂𝜐 ∂y + ∂w ∂z Write the primary dimensions of each additive term, and ver ify that the equation is dimensionally homogeneous. Show all your work.
Time = t1 Time = t2 Volume = V2 Volume = V1 FIGURE P7–22 7–23 In Chap. 4, we defined the material acceleration, which is the acceleration following a fluid particle, a ›(x, y, z, t) = ∂V › ∂t + (V › · ∇ › )V › (a) What are the primary dimensions of the gradient operator ∇ →? (b) Verify that each additive term in the equation has the same dimensions. Answers: (a) {L−1}; (b) {L1t−2} V = V(x, y, z, t) F (x, y, z) Fluid particle at time t Fluid particle at time t + dt a = a(x, y, z, t) m FIGURE P7–23 7–24 Newton’s second law is the foundation for the differ ential equation of conservation of linear momentum (to be discussed in Chap. 9). In terms of the material acceleration following a fluid particle (Fig. P7–23), we write Newton’s second law as follows: F ›= ma› = m( ∂V › ∂t + (V ›·∇ › )V › ) Or, dividing both sides by the mass m of the fluid particle, F › m = ∂V › ∂t + (V › ·∇ › )V › Write the primary dimensions of each additive term in the (second) equation, and verify that the equation is dimension ally homogeneous. Show all your work.
Nondimensionalization of Equations 7–25C What is the primary reason for nondimensional izing an equation?
7–26 Recall from Chap. 4 that the volumetric strain rate is zero for a steady incompressible flow. In Cartesian coordi nates we express this as ∂u ∂x + ∂𝜐 ∂y + ∂w ∂z = 0 Suppose the characteristic speed and characteristic length for a given flow field are V and L, respectively (Fig. P7–26). Define the following dimensionless variables, x = x L , y = y L , z = z L , u = u V , 𝜐 = 𝜐 V , and w = w V Nondimensionalize the equation, and identify any established (named) dimensionless parameters that may appear. Discuss.
V L FIGURE P7–26 7–27 In Chap. 9, we define the stream function 𝜓 for two-dimensional incompressible flow in the xy-plane, u = ∂𝜓 ∂y 𝜐 = −∂𝜓 ∂x cen96537_ch07_297-350.indd 336 29/12/16 5:24 pm 337 CHAPTER 7 where u and 𝜐 are the velocity components in the x- and y-directions, respectively. (a) What are the primary dimen sions of 𝜓? (b) Suppose a certain two-dimensional flow has a characteristic length scale L and a characteristic time scale t. Define dimensionless forms of variables x, y, u, 𝜐, and 𝜓. (c) Rewrite the equations in nondimensional form, and identify any established dimensionless parameters that may appear.
7–28 In an oscillating incompressible flow field the force per unit mass acting on a fluid particle is obtained from Newton’s second law in intensive form (see Prob. 7–19), F › m = ∂V › ∂t + (V ›·∇ › )V › Suppose the characteristic speed and characteristic length for a given flow field are V∞ and L, respectively. Also suppose that 𝜔 is a characteristic angular frequency (rad/s) of the oscillation (Fig. P7–28). Define the following nondimension alized variables, t = 𝜔 t, x › = x › L , ∇ › = L∇ › , and V › = V › V∞ Since there is no given characteristic scale for the force per unit mass acting on a fluid particle, we assign one, noting that {F ›/m} = {L/t2}. Namely, we let (F ›/m) = 1 𝜔 2L F › /m Nondimensionalize the equation of motion and identify any established (named) dimensionless parameters that may appear.
V F/m a m V∞ ω L FIGURE P7–28 7–29 A wind tunnel is used to measure the pressure distri bution in the airflow over an airplane model (Fig. P7–29). The air speed in the wind tunnel is low enough that compressible effects are negligible. As discussed in Chap. 5, the Bernoulli equation approximation is valid in such a flow situation eve rywhere except very close to the body surface or wind tunnel wall surfaces and in the wake region behind the model. Far away from the model, the air flows at speed V∞ and pressure P∞, and the air density 𝜌 is approximately constant. Gravita tional effects are generally negligible in airflows, so we write the Bernoulli equation as P ∞, ρ V ∞ Wind tunnel test section Model Traverse Crank Pressure probe Strut FIGURE P7–29 P + 1 2 ρV 2 = P∞+ 1 2 ρV 2 ∞ Nondimensionalize the equation, and generate an expression for the pressure coefficient Cp at any point in the flow where the Bernoulli equation is valid. Cp is defined as Cp = P −P∞ 1 2ρV 2 ∞ Answer: Cp = 1 − V2/V ∞ 2 7–30 Consider ventilation of a well-mixed room as in Fig. P7–21. The differential equation for mass concentration in the room as a function of time is given in Prob. 7–21 and is repeated here for convenience, V dc dt = S −V ·c −cAskw There are three characteristic parameters in such a situation: L, a characteristic length scale of the room (assume L = V1/3); V ·, the volume flow rate of fresh air into the room, and climit, the maximum mass concentration that is not harmful. (a) Using these three characteristic parameters, define dimensionless forms of all the variables in the equation. (Hint: For example, define c = c/climit.) (b) Rewrite the equation in dimension less form, and identify any established dimensionless groups that may appear.
7–31 In an oscillating compressible flow field the volumet ric strain rate is not zero, but varies with time following a fluid particle. In Cartesian coordinates we express this as 1 V DV Dt = ∂u ∂x + ∂v ∂y + ∂w ∂z Suppose the characteristic speed and characteristic length for a given flow field are V and L, respectively. Also sup pose that f is a characteristic frequency of the oscillation (Fig. P7–31). Define the following dimensionless variables, t = ft, V = V L3 , x = x L , y = y L , z = z L , u = u V , 𝜐 = 𝜐 V , and w = w V cen96537_ch07_297-350.indd 337 29/12/16 5:24 pm 338 dimensional ANALYSIS and modeling Nondimensionalize the equation and identify any established (named) dimensionless parameters that may appear.
ƒ = frequency of oscillation L V V Time t1 Time t2 Time t3 FIGURE P7–31 Dimensional Analysis and Similarity 7–32C List the three primary purposes of dimensional analysis.
7–33C List and describe the three necessary conditions for complete similarity between a model and a prototype.
7–34 A student team is to design a human-powered subma rine for a design competition. The overall length of the proto type submarine is 4.85 m, and its student designers hope that it can travel fully submerged through water at 0.440 m/s. The water is freshwater (a lake) at T = 15°C. The design team builds a one-fifth scale model to test in their university’s wind tunnel (Fig. P7–34). A shield surrounds the drag balance strut so that the aerodynamic drag of the strut itself does not influence the measured drag. The air in the wind tunnel is at 25°C and at one standard atmosphere pressure. At what air speed do they need to run the wind tunnel in order to achieve similarity? Answer: 30.2 m/s P∞, ρ FD V Wind tunnel test section Model Shield Drag balance Strut FIGURE P7–34 7–35 Repeat Prob. 7–34 with all the same conditions except that the only facility available to the students is a much smaller wind tunnel. Their model submarine is a one-twenty-seventh scale model instead of a one-fifth scale model. At what air speed do they need to run the wind tunnel in order to achieve similarity? Do you notice anything disturbing or suspicious about your result? Discuss your results.
7–36 This is a follow-up to Prob. 7–34. The students meas ure the aerodynamic drag on their model submarine in the wind tunnel (Fig. P7–34). They are careful to run the wind tunnel at conditions that ensure similarity with the prototype submarine. Their measured drag force is 5.70 N. Estimate the drag force on the prototype submarine at the conditions given in Prob. 7–34. Answer: 25.5 N 7–37E The aerodynamic drag of a new sports car is to be predicted at a speed of 60.0 mi/h at an air temperature of 25°C. Automotive engineers build a one-third scale model of the car (Fig. P7–37E) to test in a wind tunnel. The temperature of the wind tunnel air is also 25°C. The drag force is measured with a drag balance, and the moving belt is used to simulate the moving ground (from the car’s frame of reference). Deter mine how fast the engineers should run the wind tunnel to achieve similarity between the model and the prototype.
Moving belt Drag balance FD, m Vm Lm ρm, μm Wind tunnel test section FIGURE P7–37E 7–38E This is a follow-up to Prob. 7–37E. The aerody namic drag on the model in the wind tunnel (Fig. P7–37E) is measured to be 33.5 lbf when the wind tunnel is operated at the speed that ensures similarity with the prototype car. Esti mate the drag force (in lbf) on the prototype car at the condi tions given in Prob. 7–37E.
7–39 Consider the common situation in which a researcher is trying to match the Reynolds number of a large prototype vehicle with that of a small-scale model in a wind tunnel. Is it better for the air in the wind tunnel to be cold or hot? Why? Support your argument by comparing wind tunnel air at 10°C and at 45°C, all else being equal.
7–40 Some wind tunnels are pressurized. Discuss why a research facility would go through all the extra trouble and expense to pressurize a wind tunnel. If the air pressure in the tunnel increases by a factor of 1.8, all else being equal (same wind speed, same model, etc.), by what factor will the Reyn olds number increase?
7–41E Some students want to visualize flow over a spin ning baseball. Their fluids laboratory has a nice water tun nel into which they can inject multicolored dye streaklines, so they decide to test a spinning baseball in the water tun nel (Fig. P7–41E). Similarity requires that they match both cen96537_ch07_297-350.indd 338 29/12/16 5:24 pm 339 CHAPTER 7 the Rey n olds number and the Strouhal number between their model test and the actual baseball that moves through the air at 90 mi/h and spins at 300 rpm. Both the air and the water are at 68°F. At what speed should they run the water in the water tunnel, and at what rpm should they spin their base ball? Answers: 5.96 mi/h, 19.9 rpm V ρ, μ Water tunnel test section Spinning baseball Shield Motor Dye injection Strut n FIGURE P7–41E 7–42E A lightweight parachute is being designed for mili tary use (Fig. P7–42E). Its diameter D is 20 ft and the total weight W of the falling payload, parachute, and equipment is 145 lbf. The design terminal settling speed Vt of the parachute at this weight is 18 ft/s. A one-twelfth scale model of the par achute is tested in a wind tunnel. The wind tunnel temperature and pressure are the same as those of the prototype, namely 60°F and standard atmospheric pressure. (a) Calculate the drag coefficient of the prototype. (Hint: At terminal settling speed, weight is balanced by aerodynamic drag.) (b) At what wind tunnel speed should the wind tunnel be run in order to achieve dynamic similarity? (c) Estimate the aerodynamic drag of the model parachute in the wind tunnel (in lbf).
Payload V t D FIGURE P7–42E Dimensionless Parameters and the Method of Repeating Variables 7–43 Using primary dimensions, verify that the Archime des number (Table 7–5) is indeed dimensionless.
7–44 Using primary dimensions, verify that the Grashof number (Table 7–5) is indeed dimensionless.
7–45 Using primary dimensions, verify that the Rayleigh number (Table 7–5) is indeed dimensionless. What other established nondimensional parameter is formed by the ratio of Ra and Gr? Answer: the Prandtl number 7–46 The Richardson number is defined as Ri = L5g Δρ ρV · 2 Miguel is working on a problem that has a characteristic length scale L, a characteristic velocity V, a characteristic density difference Δ𝜌, a characteristic (average) density 𝜌, and of course the gravitational constant g, which is always available. He wants to define a Richardson number, but does not have a characteristic volume flow rate. Help Miguel define a characteristic volume flow rate based on the param eters available to him, and then define an appropriate Rich ardson number in terms of the given parameters.
7–47 A periodic Kármán vortex street is formed when a uniform stream flows over a circular cylinder (Fig. P7–47). Use the method of repeating variables to generate a dimen sionless relationship for Kármán vortex shedding frequency fk as a function of free-stream speed V, fluid density 𝜌, fluid viscosity 𝜇, and cylinder diameter D. Show all your work. Answer: St = f (Re) ρ, μ V D fk FIGURE P7–47 7–48 Repeat Prob. 7–47, but with an additional independ ent parameter included, namely, the speed of sound c in the fluid. Use the method of repeating variables to generate a dimensionless relationship for Kármán vortex shedding fre quency fk as a function of free-stream speed V, fluid density 𝜌, fluid viscosity 𝜇, cylinder diameter D, and speed of sound c. Show all your work.
7–49 A stirrer is used to mix chemicals in a large tank (Fig. P7–49). The shaft power W . supplied to the stirrer blades is a function of stirrer diameter D, liquid density 𝜌, liquid vis cosity 𝜇, and the angular velocity 𝜔 of the spinning blades. cen96537_ch07_297-350.indd 339 29/12/16 5:24 pm 340 dimensional ANALYSIS and modeling W ρ, μ ω .
D FIGURE P7–49 Use the method of repeating variables to generate a dimen sionless relationship between these parameters. Show all your work and be sure to identify your Π groups, modifying them as necessary. Answer: Np = f (Re) 7–50 Repeat Prob. 7–49 except do not assume that the tank is large. Instead, let tank diameter Dtank and average liquid depth htank be additional relevant parameters.
7–51 Albert Einstein is pondering how to write his (soon-to-be-famous) equation. He knows that energy E is a function of mass m and the speed of light c, but he doesn't know the functional relationship (E = m2c? E = mc4?). Pretend that Albert knows nothing about dimensional analysis, but since you are taking a fluid mechanics class, you help Albert come up with his equation. Use the step-by-step method of repeating variables to generate a dimensionless relationship between these parameters, showing all of your work. Com pare this to Einstein's famous equation—does dimensional analysis give you the correct form of the equation?
FIGURE P7–51 7–52 Consider fully developed Couette flow—flow between two infinite parallel plates separated by distance h, with the top plate moving and the bottom plate stationary as illustrated in Fig. P7–52. The flow is steady, incompressible, and two-dimensional in the xy-plane. Use the method of repeat ing variables to generate a dimensionless relationship for the x-component of fluid velocity u as a function of fluid viscosity 𝜇, top plate speed V, distance h, fluid density 𝜌, and distance y. Show all your work. Answer: u/ V = f (Re, y/h) ρ, μ h y u V x FIGURE P7–52 7–53 Consider developing Couette flow—the same flow as Prob. 7–52 except that the flow is not yet steady-state, but is developing with time. In other words, time t is an additional parameter in the problem. Generate a dimensionless relation ship between all the variables.
7–54 The speed of sound c in an ideal gas is known to be a function of the ratio of specific heats k, absolute temperature T, and specific ideal gas constant Rgas (Fig. P7–54). Showing all your work, use dimensional analysis to find the functional relationship between these parameters.
c k, T, Rgas FIGURE P7–54 7–55 Repeat Prob. 7–54, except let the speed of sound c in an ideal gas be a function of absolute temperature T, univer sal ideal gas constant Ru, molar mass (molecular weight) M of the gas, and ratio of specific heats k. Showing all your work, use dimensional analysis to find the functional relationship between these parameters.
7–56 Repeat Prob. 7–54, except let the speed of sound c in an ideal gas be a function only of absolute temperature T and specific ideal gas constant Rgas. Showing all your work, use dimensional analysis to find the functional relationship between these parameters. Answer: c/√Rgas T = constant 7–57 Repeat Prob. 7–54, except let speed of sound c in an ideal gas be a function only of pressure P and gas density 𝜌. Showing all your work, use dimensional analysis to find the functional relationship between these parameters. Verify that your results are consistent with the equation for speed of sound in an ideal gas, c = √kRgas T .
7–58 When small aerosol particles or microorganisms move through air or water, the Reynolds number is very small (Re ≪ 1). Such flows are called creeping flows. The aero dynamic drag on an object in creeping flow is a function cen96537_ch07_297-350.indd 340 29/12/16 5:24 pm 341 CHAPTER 7 only of its speed V, some characteristic length scale L of the object, and fluid viscosity 𝜇 (Fig. P7–58). Use dimensional analysis to generate a relationship for FD as a function of the independent variables.
V L FD μ FIGURE P7–58 7–59 A tiny aerosol particle of density 𝜌p and character istic diameter Dp falls in air of density 𝜌 and viscosity 𝜇 (Fig. P7–59). If the particle is small enough, the creeping flow approximation is valid, and the terminal settling speed of the particle V depends only on Dp, 𝜇, gravitational con stant g, and the density difference (𝜌p − 𝜌). Use dimensional analysis to generate a relationship for V as a function of the independent variables. Name any established dimensionless parameters that appear in your analysis.
V g ρ, μ Dp ρp FIGURE P7–59 7–60 Combine the results of Probs. 7–58 and 7–59 to gener ate an equation for the settling speed V of an aerosol particle falling in air (Fig. P7–59). Verify that your result is consistent with the functional relationship obtained in Prob. 7–59. For consistency, use the notation of Prob. 7–59. (Hint: For a particle falling at constant settling speed, the particle’s net weight must equal its aerodynamic drag. Your final result should be an equa tion for V that is valid to within some unknown constant.) 7–61 You will need the results of Prob. 7–60 to do this problem. A tiny aerosol particle falls at steady settling speed V. The Reynolds number is small enough that the creeping flow approximation is valid. If the particle size is doubled, all else being equal, by what factor will the settling speed go up? If the density difference (𝜌p − 𝜌) is doubled, all else being equal, by what factor will the settling speed go up?
7–62 An incompressible fluid of density 𝜌 and viscosity 𝜇 flows at average speed V through a long, horizontal section of round pipe of length L, inner diameter D, and inner wall roughness height 𝜀 (Fig. P7–62). The pipe is long enough that the flow is fully developed, meaning that the velocity profile does not change down the pipe. Pressure decreases (linearly) down the pipe in order to “push” the fluid through the pipe to overcome friction. Using the method of repeating variables, develop a nondimensional relationship between pres sure drop ΔP = P1 − P2 and the other parameters in the prob lem. Be sure to modify your Π groups as necessary to achieve established nondimensional parameters, and name them. (Hint: For consistency, choose D rather than L or 𝜀 as one of your repeating parameters.) Answer: Eu = f (Re, 𝜀/D, L/D) D P1 P2 V ρ, μ ε L FIGURE P7–62 7–63 Consider laminar flow through a long section of pipe, as in Fig. P7–62. For laminar flow it turns out that wall roughness is not a relevant parameter unless 𝜀 is very large. The volume flow rate V · through the pipe is a function of pipe diameter D, fluid viscosity 𝜇, and axial pressure gradient dP/dx. If pipe diameter is doubled, all else being equal, by what factor will volume flow rate increase? Use dimensional analysis.
7–64 In the study of turbulent flow, turbulent viscous dis sipation rate 𝜀 (rate of energy loss per unit mass) is known to be a function of length scale l and velocity scale u′ of the large-scale turbulent eddies. Using dimensional analysis (Buckingham pi and the method of repeating variables) and showing all of your work, generate an expression for 𝜀 as a function of l and u′.
7–65 Bill is working on an electrical circuit problem. He remembers from his electrical engineering class that voltage drop ΔE is a function of electrical current I and electrical resistance R. Unfortunately, he does not recall the exact form of the equation for ΔE. However, he is taking a fluid mechan ics class and decides to use his newly acquired knowledge about dimensional analysis to recall the form of the equation. Help Bill develop the equation for ΔE using the method of repeating variables, showing all of your work. Compare this to Ohm’s law—does dimensional analysis give you the correct form of the equation?
7–66 A boundary layer is a thin region (usually along a wall) in which viscous forces are significant and within which the flow is rotational. Consider a boundary layer growing along a thin flat plate (Fig. P7–66). The flow is steady. The boundary layer thickness 𝛿 at any downstream distance x is a function of x, free-stream velocity V∞, and fluid properties 𝜌 (density) and 𝜇 (viscosity). Use the method of repeating cen96537_ch07_297-350.indd 341 29/12/16 5:24 pm 342 dimensional ANALYSIS and modeling x y V δ(x) ρ, μ FIGURE P7–66 variables to generate a dimensionless relationship for 𝛿 as a function of the other parameters. Show all your work.
7–67 A liquid of density 𝜌 and viscosity 𝜇 is pumped at volume flow rate V · through a pump of diameter D. The blades of the pump rotate at angular velocity 𝜔. The pump supplies a pressure rise ΔP to the liquid. Using dimensional analysis, generate a dimensionless relationship for ΔP as a function of the other parameters in the problem. Identify any established nondimensional parameters that appear in your result. Hint: For consistency (and whenever possible), it is wise to choose a length, a density, and a velocity (or angular velocity) as repeating variables.
7–68 A propeller of diameter D rotates at angular veloc ity 𝜔 in a liquid of density 𝜌 and viscosity 𝜇. The required torque T is determined to be a function of D, 𝜔, 𝜌, and 𝜇. Using dimensional analysis, generate a dimensionless rela tionship. Identify any established nondimensional parameters that appear in your result. Hint: For consistency (and when ever possible), it is wise to choose a length, a density, and a velocity (or angular velocity) as repeating variables.
7–69 Repeat Prob. 7–68 for the case in which the propeller operates in a compressible gas instead of a liquid.
7–70 Jen is working on a spring–mass–damper system, as shown in Fig. P7–70. She remembers from her dynamic systems class that the damping ratio 𝜁 is a nondimensional property of such systems and that 𝜁 is a function of spring constant k, mass m, and damping coefficient c. Unfortunately, she does not recall the exact form of the equation for 𝜁. How ever, she is taking a fluid mechanics class and decides to use her newly acquired knowledge about dimensional analysis to recall the form of the equation. Help Jen develop the equation for 𝜁 using the method of repeating variables, showing all of your work. (Hint: Typical units for k are N/m and those for c are N·s/m.) Spring (k) Mass (m) Damper (c) FIGURE P7–70 7–71 The rate of heat transfer to water flowing in a pipe was analyzed in Prob. 7–18. Let us approach that same problem, but now with dimensional analysis. Cold water enters a pipe, where it is heated by an external heat source (Fig. P7–71). The inlet and outlet water temperatures are Tin and Tout, respec tively. The total rate of heat transfer Q . from the surroundings into the water in the pipe is known to be a function of mass flow rate m ., the specific heat cp of the water, and the tem perature difference between the incoming and outgoing water. Showing all your work, use dimensional analysis to find the functional relationship between these parameters, and com pare to the analytical equation given in Prob. 7–18. (Note: We are pretending that we do not know the analytical equation.) Q m Tin cp = specific heat of the water Tout ⋅ ⋅ FIGURE P7–71 7–72 Consider a liquid in a cylindrical container in which both the container and the liquid are rotating as a rigid body (solid-body rotation). The elevation difference h between the center of the liquid surface and the rim of the liquid surface is a function of angular velocity 𝜔, fluid density 𝜌, gravitational acceleration g, and radius R (Fig. P7–72). Use the method of repeating variables to find a dimensionless relationship between the parameters. Show all your work. Answer: h/R = f (Fr) Liquid Free surface g R ω ρ h FIGURE P7–72 7–73 Consider the case in which the container and liquid of Prob. 7–72 are initially at rest. At t = 0 the container begins to rotate. It takes some time for the liquid to rotate as a rigid body, and we expect that the liquid’s viscosity is an addi tional relevant parameter in the unsteady problem. Repeat cen96537_ch07_297-350.indd 342 29/12/16 5:24 pm 343 CHAPTER 7 Prob. 7–72, but with two additional independent parameters included, namely, fluid viscosity 𝜇 and time t. (We are inter ested in the development of height h as a function of time and the other parameters.) 7–74 One of the first things you learn in physics class is the law of universal gravitation, F = G m1m2 r2 , where F is the attractive force between two bodies, m1 and m2 are the masses of the two bodies, r is the distance between the two bodies, and G is the universal gravitational constant equal to (6.67428 ± 0.00067) × 10−11 [the units of G are not given here]. (a) Calculate the SI units of G. For consistency, give your answer in terms of kg, m, and s. (b) Suppose you don’t remember the law of universal gravitation, but you are clever enough to know that F is a function of G, m1, m2, and r. Use dimensional analysis and the method of repeating variables (show all your work) to generate a nondimensional expres sion for F = F(G, m1, m2, r). Give your answer as Π1 = func tion of (Π2, Π3, …). (c) Dimensional analysis cannot yield the exact form of the function. However, compare your result to the law of universal gravitation to find the form of the func tion (e.g., Π1 = Π2 2 or some other functional form).
Experimental Testing and Incomplete Similarity 7–75C Although we usually think of a model as being smaller than the prototype, describe at least three situations in which it is better for the model to be larger than the prototype.
7–76C Discuss the purpose of a moving ground belt in wind tunnel tests of flow over model automobiles. Think of an alternative if a moving ground belt is unavailable.
7–77C Define wind tunnel blockage. What is the rule of thumb about the maximum acceptable blockage for a wind tunnel test? Explain why there would be measurement errors if the blockage were significantly higher than this value.
7–78C Consider again the model truck example discussed in Section 7–5, except that the maximum speed of the wind tunnel is only 50 m/s. Aerodynamic force data are taken for wind tunnel speeds between V = 20 and 50 m/s—assume the same data for these speeds as those listed in Table 7–7. Based on these data alone, can the researchers be confident that they have reached Reynolds number independence?
7–79C What is the rule of thumb about the Mach number limit in order that the incompressible flow approximation is reasonable? Explain why wind tunnel results would be incor rect if this rule of thumb were violated.
7–80 A one-sixteenth scale model of a new sports car is tested in a wind tunnel. The prototype car is 4.37 m long, 1.30 m tall, and 1.69 m wide. During the tests, the moving ground belt speed is adjusted so as to always match the speed of the air moving through the test section. Aerodynamic drag force FD is measured as a function of wind tunnel speed; the experimental results are listed in Table P7–80. Plot drag coef ficient CD as a function of the Reynolds number Re, where the area used for calculation of CD is the frontal area of the model car (assume A = width × height), and the length scale used for calculation of Re is car width W. Have we achieved dynamic similarity? Have we achieved Reynolds number independence in our wind tunnel test? Estimate the aerody namic drag force on the prototype car traveling on the high way at 24.6 m/s (55 mi/h). Assume that both the wind tunnel air and the air flowing over the prototype car are at 25°C and atmospheric pressure. Answers: no, yes, 252 N TABLE P7–80 V, m/s FD, N 10 0.29 15 0.64 20 0.96 25 1.41 30 1.55 35 2.10 40 2.65 45 3.28 50 4.07 55 4.91 7–81 Water at 20°C flows through a long, straight pipe. The pressure drop is measured along a section of the pipe of length L = 1.3 m as a function of average velocity V through the pipe (Table P7–81). The inner diameter of the pipe is D = 10.4 cm. (a) Nondimensionalize the data and plot the Euler number as a function of the Reynolds number. Has the exper iment been run at high enough speeds to achieve Reyn olds number independence? (b) Extrapolate the experimen tal data to predict the pressure drop at an average speed of 80 m/s. Answer: 1,940,000 N/m2 TABLE P7–81 V, m/s ΔP, N/m2 0.5 77.0 1 306 2 1218 4 4865 6 10,920 8 19,440 10 30,340 15 68,330 20 121,400 25 189,800 30 273,200 35 372,100 40 485,300 45 614,900 50 758,700 cen96537_ch07_297-350.indd 343 29/12/16 5:24 pm 344 dimensional ANALYSIS and modeling 7–82 In the model truck example discussed in Section 7–5, the wind tunnel test section is 3.5 m long, 0.85 m tall, and 0.90 m wide. The one-sixteenth scale model truck is 0.991 m long, 0.257 m tall, and 0.159 m wide. What is the wind tun nel blockage of this model truck? Is it within acceptable limits according to the standard rule of thumb?
7–83 Use dimensional analysis to show that in a problem involving shallow water waves (Fig. P7–83), both the Froude number and the Reynolds number are relevant dimensionless parameters. The wave speed c of waves on the surface of a liquid is a function of depth h, gravitational acceleration g, fluid density 𝜌, and fluid viscosity 𝜇. Manipulate your Π’s to get the parameters into the following form: Fr = c √gh = f (Re) where Re = ρch 𝜇 ρ, μ h c g FIGURE P7–83 7–84E A small wind tunnel in a university’s undergradu ate fluid flow laboratory has a test section that is 20 by 20 in in cross section and is 4.0 ft long. Its maximum speed is 145 ft/s. Some students wish to build a model 18-wheeler to study how aerodynamic drag is affected by rounding off the back of the trailer. A full-size (prototype) tractor-trailer rig is 52 ft long, 8.33 ft wide, and 12 ft high. Both the air in the wind tunnel and the air flowing over the prototype are at 80°F and atmospheric pressure. (a) What is the largest scale model they can build to stay within the rule-of-thumb guidelines for blockage? What are the dimen sions of the model truck in inches? (b) What is the maximum model truck Reynolds number achievable by the students? (c) Are the students able to achieve Reynolds number inde pendence? Discuss.
Review Problems 7–85C There are many established nondimensional param eters besides those listed in Table 7–5. Do a literature search or an Internet search and find at least three established, named nondimensional parameters that are not listed in Table 7–5. For each one, provide its definition and its ratio of significance, following the format of Table 7–5. If your equation contains any variables not identified in Table 7–5, be sure to identify those variables.
7–86C Think about and describe a prototype flow and a corresponding model flow that have geometric similarity, but not kinematic similarity, even though the Reynolds numbers match. Explain.
7–87C For each statement, choose whether the statement is true or false and discuss your answer briefly. (a) Kinematic similarity is a necessary and sufficient condi tion for dynamic similarity.
(b) Geometric similarity is a necessary condition for dynamic similarity. (c) Geometric similarity is a necessary condition for kin ematic similarity. (d ) Dynamic similarity is a necessary condition for kinematic similarity.
7–88 What are the primary dimensions of electric charge q, the units of which are coulombs (C)? (Hint: Look up the fun damental definition of electric current.) 7–89 What are the primary dimensions of electrical capaci tance C, the units of which are farads? (Hint: Look up the fundamental definition of electrical capacitance.) 7–90 Write the primary dimensions of each of the follow ing variables from the field of solid mechanics, showing all your work: (a) moment of inertia I; (b) modulus of elastic ity E, also called Young’s modulus; (c) strain 𝜀; (d) stress 𝜎. (e) Finally, show that the relationship between stress and strain (Hooke’s law) is a dimensionally homogeneous equation.
7–91 Some authors prefer to use force as a primary dimen sion in place of mass. In a typical fluid mechanics problem, then, the four represented primary dimensions m, L, t, and T are replaced by F, L, t, and T. The primary dimension of force in this system is {force} = {F}. Using the results of Prob. 7–3, rewrite the primary dimensions of the universal gas constant in this alternate system of primary dimensions.
7–92 From fundamental electronics, the current flowing through a capacitor at any instant of time is equal to the capacitance times the rate of change of voltage (electromotive force) across the capacitor, I = C dE dt Write the primary dimensions of both sides of this equation, and verify that the equation is dimensionally homogeneous. Show all your work.
7–93 Force F is applied at the tip of a cantilever beam of length L and moment of inertia I (Fig. P7–93). The modu lus of elasticity of the beam material is E. When the force is applied, the tip deflection of the beam is zd. Use dimensional analysis to generate a relationship for zd as a function of the independent variables. Name any established dimensionless parameters that appear in your analysis.
F L E, I zd FIGURE P7–93 7–94 An explosion occurs in the atmosphere when an anti-aircraft missile meets its target (Fig. P7–94). A shock wave (also called a blast wave) spreads out radially from the explosion. The pressure difference across the blast wave ΔP cen96537_ch07_297-350.indd 344 29/12/16 5:24 pm 345 CHAPTER 7 and its radial distance r from the center are functions of time t, speed of sound c, and the total amount of energy E released by the explosion. (a) Generate dimensionless relationships between ΔP and the other parameters and between r and the other parameters. (b) For a given explosion, if the time t since the explosion doubles, all else being equal, by what factor will ΔP decrease?
Blast wave Pow!
E c r ΔP FIGURE P7–94 7–95 The Archimedes number listed in Table 7–5 is appro priate for buoyant particles in a fluid. Do a literature search or an Internet search and find an alternative definition of the Archimedes number that is appropriate for buoyant fluids (e.g., buoyant jets and buoyant plumes, heating and air-conditioning applications). Provide its definition and its ratio of significance, following the format of Table 7–5. If your equa tion contains any variables not identified in Table 7–5, be sure to identify those variables. Finally, look through the established dimensionless parameters listed in Table 7–5 and find one that is similar to this alternate form of the Archi medes number.
7–96 Consider steady, laminar, fully developed, two-dimensional Poiseuille flow—flow between two infinite paral lel plates separated by distance h, with both the top plate and bottom plate stationary, and a forced pressure gradient dP/dx driving the flow as illustrated in Fig. P7–96. (dP/dx is con stant and negative.) The flow is steady, incompressible, and two-dimensional in the xy-plane. The flow is also fully devel oped, meaning that the velocity profile does not change with downstream distance x. Because of the fully developed nature of the flow, there are no inertial effects and density does not enter the problem. It turns out that u, the velocity component in the x-direction, is a function of distance h, pressure gradient dP/dx, fluid viscosity 𝜇, and vertical coordinate y. Perform a dimensional analysis (showing all your work), and generate a dimensionless relationship between the given variables.
x u(y) umax μ h y FIGURE P7–96 7–97 Consider the steady, laminar, fully developed, two-dimensional Poiseuille flow of Prob. 7–96. The maximum velocity umax occurs at the center of the channel. (a) Generate a dimensionless relationship for umax as a function of distance between plates h, pressure gradient dP/dx, and fluid viscos ity 𝜇. (b) If the plate separation distance h is doubled, all else being equal, by what factor will umax change? (c) If the pressure gradient dP/dx is doubled, all else being equal, by what factor will umax change? (d) How many experiments are required to describe the complete relationship between umax and the other parameters in the problem?
7–98 Compare the primary dimensions of each of the fol lowing properties in the mass-based primary dimension sys tem (m, L, t, T, I, C, N) to those in the force-based primary dimension system (F, L, t, T, I, C, N): (a) pressure or stress; (b) moment or torque; (c) work or energy. Based on your results, explain when and why some authors prefer to use force as a primary dimension in place of mass.
7–99 Oftentimes it is desirable to work with an established dimensionless parameter, but the characteristic scales avail able do not match those used to define the parameter. In such cases, we create the needed characteristic scales based on dimensional reasoning (usually by inspection). Suppose for example that we have a characteristic velocity scale V, char acteristic area A, fluid density 𝜌, and fluid viscosity 𝜇, and we wish to define a Reynolds number. We create a length scale L = √A, and define Re = ρV√A 𝜇 In similar fashion, define the desired established dimension less parameter for each case: (a) Define a Froude number, given V ·′ = volume flow rate per unit depth, length scale L, and gravitational constant g. (b) Define a Reynolds number, given V ·′ = volume flow rate per unit depth and kinematic viscosity 𝜈. (c) Define a Richardson number (see Table 7–5), given V ·′ = volume flow rate per unit depth, length scale L, characteristic density difference Δ𝜌, characteristic density 𝜌, and gravitational constant g.
7–100 A liquid of density 𝜌 and viscosity 𝜇 flows by gravity through a hole of diameter d in the bottom of a tank of diameter D (Fig. P7–100). At the start of the experi ment, the liquid surface is at height h above the bottom of the tank, as sketched. The liquid exits the tank as a jet with average velocity V straight down as also sketched. Using dimensional analysis, generate a dimensionless relationship for V as a function of the other parameters in the problem. Identify any established nondimensional parameters that appear in your result. (Hint: There are three length scales in this problem. For consistency, choose h as your length scale.) cen96537_ch07_297-350.indd 345 29/12/16 5:25 pm 346 dimensional ANALYSIS and modeling h D d ρ, μ g V FIGURE P7–100 7–101 Repeat Prob. 7–100 except for a different dependent parameter, namely, the time required to empty the tank tempty. Generate a dimensionless relationship for tempty as a function of the following independent parameters: hole diameter d, tank diameter D, density 𝜌, viscosity 𝜇, initial liquid surface height h, and gravitational acceleration g.
7–102 A liquid delivery system is being designed such that ethylene glycol flows out of a hole in the bottom of a large tank, as in Fig. P7–100. The designers need to predict how long it will take for the ethylene glycol to completely drain. Since it would be very expensive to run tests with a full-scale prototype using ethylene glycol, they decide to build a one-quarter scale model for experimental testing, and they plan to use water as their test liquid. The model is geometrically similar to the prototype (Fig. P7–102). (a) The temperature of the ethylene glycol in the prototype tank is 60°C, at which 𝜈 = 4.75 × 10−6 m2/s. At what temperature should the water in the model experiment be set in order to ensure complete similarity between model and prototype? (b) The experiment is run with water at the proper temperature as calculated in part (a). It takes 4.12 min to drain the model tank. Predict how long it will take to drain the ethylene glycol from the prototype tank. Answers: (a) 45.8°C, (b) 8.24 min hp Dp Prototype ρp, μp dp Dm hm Model ρm, μm dm g FIGURE P7–102 7–103 Liquid flows out of a hole in the bottom of a tank as in Fig. P7–100. Consider the case in which the hole is very small compared to the tank (d ≪ D). Experiments reveal that average jet velocity V is nearly independent of d, D, 𝜌, or 𝜇. In fact, for a wide range of these parameters, it turns out that V depends only on liquid surface height h and gravitational acceleration g. If the liquid surface height is doubled, all else being equal, by what factor will the average jet velocity increase? Answer: √2 7–104 An aerosol particle of characteristic size Dp moves in an airflow of characteristic length L and characteristic velocity V. The characteristic time required for the particle to adjust to a sudden change in air speed is called the particle relaxation time 𝜏p, 𝜏p = ρp D 2 p 18𝜇 ρ, μ V L Dp ρp FIGURE P7–104 Verify that the primary dimensions of 𝜏p are time. Then cre ate a dimensionless form of 𝜏p, based on some characteris tic velocity V and some characteristic length L of the airflow (Fig. P7–104). What established dimensionless parameter do you create?
7–105 The pressure drop ΔP = P1 − P2 through a long section of round pipe can be written in terms of the shear stress 𝜏w along the wall. Shown in Fig. P7–105 is the shear stress acting by the wall on the fluid. The shaded region is a control volume composed of the fluid in the pipe between axial locations 1 and 2. There are two dimensionless param eters related to the pressure drop: the Euler number Eu and the Darcy friction factor f. (a) Using the control volume sketched in Fig. P7–105, generate a relationship for f in terms of Eu (and any other properties or parameters in the problem as needed). (b) Using the experimental data and conditions of Prob. 7–81 (Table P7–81), plot the Darcy friction factor as a function of Re. Does f show Reynolds number independence at large values of Re? If so, what is the value of f at very high Re?
Answers: (a) f = 2 D LEu; (b) yes, 0.0487 cen96537_ch07_297-350.indd 346 29/12/16 5:25 pm 347 CHAPTER 7 1 2 P1 CV P2 τw L D V ρ, μ FIGURE P7–105 7–106 The Stanton number is listed as a named, established nondimensional parameter in Table 7–5. However, careful analysis reveals that it can actually be formed by a combi nation of the Reynolds number, Nusselt number, and Prandtl number. Find the relationship between these four dimen sionless groups, showing all your work. Can you also form the Stanton number by some combination of only two other established dimensionless parameters?
7–107 Consider a variation of the fully developed Couette flow problem of Prob. 7–52—flow between two infinite paral lel plates separated by distance h, with the top plate moving at speed Vtop and the bottom plate moving at speed Vbottom as illustrated in Fig. P7–107. The flow is steady, incompressible, and two-dimensional in the xy-plane. Generate a dimension less relationship for the x-component of fluid velocity u as a function of fluid viscosity 𝜇, plate speeds Vtop and Vbottom, dis tance h, fluid density 𝜌, and distance y. (Hint: Think carefully about the list of parameters before rushing into the algebra.) ρ, μ h y u x Vbottom Vtop FIGURE P7–107 7–108 In many electronic circuits in which some kind of time scale is involved, such as filters and time-delay circuits (Fig. P7–108—a low-pass filter), you often see a resistor (R) and a capacitor (C) in series. In fact, the product of R and C is called the electrical time constant, RC. Showing all your work, what are the primary dimensions of RC? Using dimensional reasoning alone, explain why a resistor and capacitor are often found together in timing circuits.
C R Eout Ein FIGURE P7–108 7–109 In Example 7–7, the mass-based system of primary dimensions was used to establish a relationship for the pres sure difference ΔP = Pinside − Poutside between the inside and outside of a soap bubble as a function of soap bubble radius R and surface tension 𝜎s of the soap film (Fig. P7–109). Repeat the dimensional analysis using the method of repeating vari ables, but use the force-based system of primary dimensions instead. Show all your work. Do you get the same result?
Soap film Pinside Poutside σs σs R FIGURE P7–109 7–110 When a capillary tube of small diameter D is inserted into a container of liquid, the liquid rises to height h inside the tube (Fig. P7–110). h is a function of liquid density 𝜌, tube diameter D, gravitational constant g, contact angle 𝜙, and the surface tension 𝜎s of the liquid. (a) Generate a dimen sionless relationship for h as a function of the given param eters. (b) Compare your result to the exact analytical equation for h given in Chap. 2. Are your dimensional analysis results consistent with the exact equation? Discuss.
h D ϕ ρ, σs g FIGURE P7–110 7–111 Repeat part (a) of Prob. 7–110, except instead of height h, find a functional relationship for the time scale trise needed for the liquid to climb up to its final height in the cap illary tube. (Hint: Check the list of independent parameters in Prob. 7–110. Are there any additional relevant parameters?) cen96537_ch07_297-350.indd 347 29/12/16 5:25 pm 348 dimensional ANALYSIS and modeling 7–112 Sound intensity I is defined as the acoustic power per unit area emanating from a sound source. We know that I is a function of sound pressure level P (dimensions of pres sure) and fluid properties 𝜌 (density) and speed of sound c. (a) Use the method of repeating variables in mass-based pri mary dimensions to generate a dimensionless relationship for I as a function of the other parameters. Show all your work. What happens if you choose three repeating variables? Dis cuss. (b) Repeat part (a), but use the force-based primary dimension system. Discuss.
7–113 Repeat Prob. 7–112, but with the distance r from the sound source as an additional independent parameter.
7–114 Engineers at MIT have developed a mechanical model of a tuna fish to study its locomotion. The “Robotuna” shown in Fig. P7–114 is 1.0 m long and swims at speeds up to 2.0 m/s. Real bluefin tuna can exceed 3.0 m in length and have been clocked at speeds greater than 13 m/s. How fast would the 1.0-m Robotuna need to swim in order to match the Reynolds number of a real tuna that is 2.0 m long and swims at 10 m/s?
FIGURE P7–114 Photo by David Barrett MIT, used by permission.
7–115 Experiments are being designed to measure the horizontal force F on a fireman’s nozzle, as shown in Fig. P7–115. Force F is a function of velocity V1, pressure drop ΔP = P1 − P2, density 𝜌, viscosity 𝜇, inlet area A1, out let area A2, and length L. Perform a dimensional analysis for F = f (V1, ΔP, 𝜌, 𝜇, A1, A2, L). For consistency, use V1, A1, and 𝜌 as the repeating parameters and generate a dimension less relationship. Identify any established nondimensional parameters that appear in your result.
L V1 A1 P1 V2 r x A2 F P2 d1 FIGURE P7–115 7–116 Many of the established nondimensional parameters listed in Table 7–5 can be formed by the product or ratio of two other established nondimensional parameters. For each pair of nondimensional parameters listed, find a third estab lished nondimensional parameter that is formed by some manipulation of the two given parameters: (a) Reynolds num ber and Prandtl number; (b) Schmidt number and Prandtl number; (c) Reynolds number and Schmidt number.
7–117 A common device used in various applications to clean particle-laden air is the reverse-flow cyclone (Fig. P7–117). Dusty air (volume flow rate V · and density 𝜌) enters tangentially through an opening in the side of the cyclone and swirls around in the tank. Dust particles are flung out ward and fall out the bottom, while clean air is drawn out the top. The reverse-flow cyclones being studied are all geomet rically similar; hence, diameter D represents the only length scale required to fully specify the entire cyclone geometry. Engineers are concerned about the pressure drop 𝛿P through the cyclone. (a) Generate a dimensionless relationship between the pressure drop through the cyclone and the given param eters. Show all your work. (b) If the cyclone size is dou bled, all else being equal, by what factor will the pressure drop change? (c) If the volume flow rate is doubled, all else being equal, by what factor will the pressure drop change?
Answers: (a) D4𝛿P/𝜌V · 2 = constant, (b) 1/16, (c) 4 D Dusty air in Dust and bleed air out Clean air out V, ρ ⋅ FIGURE P7–117 7–118 An electrostatic precipitator (ESP) is a device used in various applications to clean particle-laden air. First, the dusty air passes through the charging stage of the ESP, where dust particles are given a positive charge qp (cou lombs) by charged ionizer wires (Fig. P7–118). The dusty air then enters the collector stage of the device, where it flows between two oppositely charged plates. The applied electric field strength between the plates is Ef (voltage difference per unit distance). Shown in Fig. P7–118 is a charged dust par ticle of diameter Dp. It is attracted to the negatively charged plate and moves toward that plate at a speed called the drift velocity w. If the plates are long enough, the dust particle impacts the negatively charged plate and adheres to it. Clean air exits the device. It turns out that for very small particles cen96537_ch07_297-350.indd 348 29/12/16 5:25 pm 349 CHAPTER 7 the drift velocity depends only on qp, Ef, Dp, and air viscos ity 𝜇. (a) Generate a dimensionless relationship between the drift velocity through the collector stage of the ESP and the given parameters. Show all your work. (b) If the electric field strength is doubled, all else being equal, by what factor will the drift velocity change? (c) For a given ESP, if the particle diameter is doubled, all else being equal, by what factor will the drift velocity change?
Dusty air in Dust particle, diameter Dp Ionizer wire Charging stage + + + + + + + + – – – – Collector stage Clean air out Ef w V qp μ FIGURE P7–118 7–119 The output power W . of a spinning shaft is a function of torque T and angular velocity 𝜔. Use dimensional analysis to express the relationship between W . , T, and 𝜔 in dimen sionless form. Compare your result to what you know from physics and discuss briefly.
FIGURE P7–119 T W ω ⋅ 7–120 The radius R of a mushroom cloud generated by a nuclear bomb grows in time. We expect that R is a function of time t, initial energy of the explosion E, and average air density 𝜌. Use dimensional analysis to express the relation ship between R, t, E, and 𝜌 in dimensionless form.
FIGURE P7–120 © Galerie Bilderwelt/Getty Images Fundamentals of Engineering (FE) Exam Problems 7–121 Which one is not a primary dimension?
(a) Velocity (b) Time (c) Electric current (d ) Temperature (e) Mass 7–122 The primary dimensions of the universal gas constant Ru are (a) m·L/t2·T (b) m2·L/N (c) m·L2/t2·N·T (d ) L2/t2·T (e) N/m·t 7–123 The thermal conductivity of a substance may be defined as the rate of heat transfer per unit length per unit temperature difference. The primary dimensions of thermal conductivity are (a) m2·L/t2·T (b) m2·L2/t·T (c) L2/m·t2·T (d ) m·L/t3·T (e) m·L2/t3·T 7–124 The primary dimensions of the gas constant over the universal gas constant R/Ru are (a) L2/t2·T (b) m·L/N (c) m/t·N·T (d ) m/L3 (e) N/m 7–125 The primary dimensions of kinematic viscosity are (a) m·L/t2 (b) m/L·t (c) L2/t (d ) L2/m·t (e) L/m·t2 7–126 There are four additive terms in an equation, and their units are given below. Which one is not consistent with this equation?
(a) J (b) W/m (c) kg·m2/s2 (d) Pa·m3 (e) N·m 7–127 The drag coefficient CD is a nondimensional param eter and is a function of drag force FD, density 𝜌, velocity V, and area A. The drag coefficient is expressed as (a) FDV2 2ρA (b) 2FD ρVA (c) ρVA2 FD (d) FD A ρV (e) 2FD ρV2A 7–128 The nondimensional heat transfer coefficient is a function of convection coefficient h (W/m2·K), thermal con ductivity k (W/m·K), and characteristic length L. This nondi mensional parameter is expressed as (a) hL/k (b) h/kL (c) L/hk (d ) hk/L (e) kL/h 7–129 The heat transfer coefficient is a nondimensional parameter which is a function of viscosity 𝜇, specific heat cp (kJ/kg·K), and thermal conductivity k (W/m·K). This nondi mensional parameter is expressed as (a) cp/𝜇k (b) k/𝜇cp (c) 𝜇/cpk (d ) 𝜇cp/k (e) cpk/𝜇 7–130 A one-third scale model of a car is to be tested in a wind tunnel. The conditions of the actual car are V = 75 km/h and T = 0°C and the air temperature in the wind tunnel is 20°C.
The properties of air at 1 atm and 0°C: 𝜌 = 1.292 kg/m3, 𝜈 = 1.338 × 10−5 m2/s.
The properties of air at 1 atm and 20°C: 𝜌 = 1.204 kg/m3, 𝜈 = 1.516 × 10−5 m2/s.
In order to achieve similarity between the model and the pro totype, the wind tunnel velocity should be (a) 255 km/h (b) 225 km/h (c) 147 km/h (d ) 75 km/h (e) 25 km/h cen96537_ch07_297-350.indd 349 29/12/16 5:25 pm 350 dimensional ANALYSIS and modeling 7–131 A one-fourth scale model of an airplane is to be tested in water. The airplane has a velocity of 700 km/h in air at −50°C. The water temperature in the test section is 10°C. In order to achieve similarity between the model and the pro totype, the test is done at a water velocity of 393 km/h.
The properties of air at 1 atm and −50°C: 𝜌 = 1.582 kg/m3, 𝜇 = 1.474 × 10−5 kg/m·s.
The properties of water at 1 atm and 10°C: 𝜌 = 999.7 kg/m3, 𝜇 = 1.307 × 10−3 kg/m·s.
If the average drag force on the model is measured to be 13,800 N, the drag force on the prototype is (a) 590 N (b) 862 N (c) 1109 N (d ) 4655 N (e) 3450 N 7–132 A one-third scale model of an airplane is to be tested in water. The airplane has a velocity of 900 km/h in air at −50°C. The water temperature in the test section is 10°C.
The properties of air at 1 atm and −50°C: 𝜌 = 1.582 kg/m3, 𝜇 = 1.474 × 10−5 kg/m·s.
The properties of water at 1 atm and 10°C: 𝜌 = 999.7 kg/m3, 𝜇 = 1.307 × 10−3 kg/m·s.
In order to achieve similarity between the model and the pro totype, the water velocity on the model should be (a) 97 km/h (b) 186 km/h (c) 263 km/h (d ) 379 km/h (e) 450 km/h 7–133 A one-fourth scale model of a car is to be tested in a wind tunnel. The conditions of the actual car are V = 45 km/h and T = 0°C and the air temperature in the wind tunnel is 20°C. In order to achieve similarity between the model and the proto type, the wind tunnel is run at 180 km/h.
The properties of air at 1 atm and 0°C: 𝜌 = 1.292 kg/m3, 𝜈 = 1.338 × 10−5 m2/s.
The properties of air at 1 atm and 20°C: 𝜌 = 1.204 kg/m3, 𝜈 = 1.516 × 10−5 m2/s.
If the average drag force on the model is measured to be 70 N, the drag force on the prototype is (a) 66.5 N (b) 70 N (c) 75.1 N (d ) 80.6 N (e) 90 N 7–134 Consider a boundary layer growing along a thin flat plate. This problem involves the following parameters: boundary layer thickness 𝛿, downstream distance x, free-stream velocity V, fluid density 𝜌, and fluid viscosity 𝜇. The dependent parameter is 𝛿. If we choose three repeating parameters as x, 𝜌, and V, the dependent Π is (a) 𝛿x2/V (b) 𝛿V2/x𝜌 (c) 𝛿𝜌/xV (d ) x/𝛿V (e) 𝛿/x 7–135 Consider a boundary layer growing along a thin flat plate. This problem involves the following parameters: boundary layer thickness 𝛿, downstream distance x, free-stream velocity V, fluid density 𝜌, and fluid viscosity 𝜇. The number of primary dimensions represented in this problem is (a) 1 (b) 2 (c) 3 (d ) 4 (e) 5 7–136 Consider a boundary layer growing along a thin flat plate. This problem involves the following parameters: boundary layer thickness 𝛿, downstream distance x, free-stream velocity V, fluid density 𝜌, and fluid viscosity 𝜇. The number of expected nondimensional parameters Πs for this problem is (a) 5 (b) 4 (c) 3 (d ) 2 (e) 1 cen96537_ch07_297-350.indd 350 29/12/16 5:25 pm 8 CHAPTER 351 OBJECTIVES When you finish reading this chapter, you should be able to ■ ■ Have a deeper understanding of laminar and turbulent flow in pipes and the analysis of fully developed flow ■ ■ Calculate the major and minor losses associated with pipe flow in piping networks and determine the pumping power requirements ■ ■ Understand various velocity and flow rate measurement techniques and learn their advantages and disadvantages I NT E R N A L F LOW F luid flow is classified as external or internal, depending on whether the fluid is forced to flow over a surface or in a conduit. Internal and external flows exhibit very different characteristics. In this chapter we consider internal flow where the conduit is completely filled with the fluid, and the flow is driven primarily by a pressure difference. This should not be confused with open-channel flow (Chap. 13) where the conduit is partially filled by the fluid and thus the flow is partially bounded by solid surfaces, as in an irrigation ditch, and the flow is driven by gravity alone.
We start this chapter with a general physical description of internal flow through pipes and ducts including the entrance region and the fully devel oped region. We continue with a discussion of the dimen sion less Reynolds number and its physical significance. We then introduce the pressure drop correlations associated with pipe flow for both laminar and turbulent flows. Then, we discuss minor losses and determine the pressure drop and pump ing power requirements for real-world piping systems. Finally, we present a brief overview of flow measurement devices.
Internal flows through pipes, elbows, tees, valves, etc., as in this oil refinery, are found in nearly every industry.
© Corbis RF cen96537_ch08_351-442.indd 351 14/01/17 2:58 pm 352 internal flow 8–1 ■ INTRODUCTION Liquid or gas flow through pipes or ducts is commonly used in heating and cooling applications and fluid distribution networks. The fluid in such appli cations is usually forced to flow by a fan or pump through a flow section. We pay particular attention to friction, which is directly related to the pressure drop and head loss during flow through pipes and ducts. The pressure drop is then used to determine the pumping power requirement. A typical piping system involves pipes of different diameters connected to each other by var ious fittings or elbows to route the fluid, valves to control the flow rate, and pumps to pressurize the fluid.
The terms pipe, duct, and conduit are usually used interchangeably for flow sections. In general, flow sections of circular cross section are referred to as pipes (especially when the fluid is a liquid), and flow sections of non circular cross section as ducts (especially when the fluid is a gas). Small-diameter pipes are usually referred to as tubes. Given this uncertainty, we will use more descriptive phrases (such as a circular pipe or a rectangular duct) whenever necessary to avoid any misunderstandings.
You have probably noticed that most fluids, especially liquids, are trans ported in circular pipes. This is because pipes with a circular cross section can withstand large pressure differences between the inside and the outside without undergoing significant distortion. Noncircular pipes are usually used in applications such as the heating and cooling systems of buildings where the pressure difference is relatively small, the manufacturing and installation costs are lower, and the available space is limited for ductwork (Fig. 8–1).
Although the theory of fluid flow is reasonably well understood, theoretical solutions are obtained only for a few simple cases such as fully developed laminar flow in a circular pipe. Therefore, we must rely on experimental results and empirical relations for most fluid flow problems rather than closed-form analytical solutions. Noting that the experimental results are obtained under carefully controlled laboratory conditions and that no two systems are exactly alike, we must not be so naive as to view the results obtained as “exact.” An error of 10 percent (or more) in friction factors calculated using the relations in this chapter is the “norm” rather than the “exception.” The fluid velocity in a pipe changes from zero at the wall because of the no-slip condition to a maximum at the pipe center. In fluid flow, it is con venient to work with an average velocity Vavg, which remains constant in incompressible flow when the cross-sectional area of the pipe is constant (Fig. 8–2). The average velocity in heating and cooling applications may change somewhat because of changes in density with temperature. But, in practice, we evaluate the fluid properties at some average temperature and treat them as constants. The convenience of working with constant proper ties usually more than justifies the slight loss in accuracy.
Also, the friction between the fluid particles in a pipe does cause a slight rise in fluid temperature as a result of the mechanical energy being con verted to sensible thermal energy. But this temperature rise due to frictional heating is usually too small to warrant any consideration in calculations and thus is disregarded. For example, in the absence of any heat transfer, no Circular pipe Rectangular duct Water 50 atm Air 1.2 atm FIGURE 8–1 Circular pipes can withstand large pressure differences between the inside and the outside without undergoing any significant distortion, but noncircular pipes cannot.
Vavg Vmax FIGURE 8–2 Average velocity Vavg is defined as the average speed through a cross section. For fully developed laminar pipe flow, Vavg is half of the maximum velocity.
cen96537_ch08_351-442.indd 352 14/01/17 2:58 pm 353 CHAPTER 8 noticeable difference can be detected between the inlet and outlet tempera tures of water flowing in a pipe. The primary consequence of friction in fluid flow is pressure drop, and thus any significant temperature change in the fluid is due to heat transfer. The value of the average velocity Vavg at some streamwise cross section is determined from the requirement that the conservation of mass principle be satisfied (Fig. 8–2). That is, m · = ρVavg Ac = ∫Ac ρu(r) dAc (8–1) where m . is the mass flow rate, 𝜌 is the density, Ac is the cross-sectional area, and u(r) is the velocity profile. Then the average velocity for incompressible flow in a circular pipe of radius R is expressed as Vavg = ∫Ac ρu(r) dAc ρAc = ∫ R 0 ρu(r)2𝜋r dr ρ𝜋R2 = 2 R2 ∫ R 0 u(r)r dr (8–2) Therefore, when we know the flow rate or the velocity profile, the average velocity can be determined easily.
8–2 ■ LAMINAR AND TURBULENT FLOWS If you have been around smokers, you probably noticed that the cigarette smoke rises in a smooth plume for the first few centimeters and then starts fluctuating randomly in all directions as it continues its rise. Other plumes behave similarly (Fig. 8–3). Likewise, a careful inspection of flow in a pipe reveals that the fluid flow is streamlined at low velocities but turns chaotic as the velocity is increased above a critical value, as shown in Fig. 8–4. The flow regime in the first case is said to be laminar, characterized by smooth streamlines and highly ordered motion, and turbulent in the sec ond case, where it is characterized by velocity fluctuations and highly dis ordered motion. The transition from laminar to turbulent flow does not occur suddenly; rather, it occurs over some region in which the flow fluctu ates between laminar and turbulent flows before it becomes fully turbulent. Most flows encountered in practice are turbulent. Laminar flow is encoun tered when highly viscous fluids such as oils flow in small pipes or narrow passages. We can verify the existence of these laminar, transitional, and turbulent flow regimes by injecting some dye streaks into the flow in a glass pipe, as the British engineer Osborne Reynolds (1842–1912) did over a century ago. We observe that the dye streak forms a straight and smooth line at low velocities when the flow is laminar (we may see some blurring because of molecular diffusion), has bursts of fluctuations in the transitional regime, and zigzags rapidly and disorderly when the flow becomes fully turbulent. These zigzags and the dispersion of the dye are indicative of the fluctuations in the main flow and the rapid mixing of fluid particles from adjacent layers. The intense mixing of the fluid in turbulent flow as a result of rapid fluctu ations enhances momentum transfer between fluid particles, which increases the friction force on the pipe wall and thus the required pumping power. The friction factor reaches a maximum when the flow becomes fully turbulent.
Laminar flow Turbulent flow FIGURE 8–3 Laminar and turbulent flow regimes of a candle smoke plume.
Dye trace (b) Turbulent flow Dye trace Dye injection Vavg Vavg (a) Laminar flow Dye injection FIGURE 8–4 The behavior of colored fluid injected into the flow in (a) laminar and (b) turbulent flow in a pipe.
cen96537_ch08_351-442.indd 353 14/01/17 2:58 pm 354 internal flow Reynolds Number The transition from laminar to turbulent flow depends on the geometry, sur face roughness, flow velocity, surface temperature, and type of fluid, among other things. After exhaustive experiments in the 1880s, Osborne Reyn olds discovered that the flow regime depends mainly on the ratio of iner tial forces to viscous forces in the fluid (Fig. 8–5). This ratio is called the Reynolds number and is expressed for internal flow in a circular pipe as Re = Inertial forces Viscous forces = VavgD 𝜈 = ρVavgD 𝜇 (8–3) where Vavg = average flow velocity (m/s), D = characteristic length of the geometry (diameter in this case, in m), and 𝜈 = 𝜇/𝜌 = kinematic viscosity of the fluid (m2/s). Note that the Reynolds number is a dimensionless quantity (Chap. 7). Also, kinematic viscosity has units m2/s, and can be viewed as viscous diffusivity or diffusivity for momentum.
At large Reynolds numbers, the inertial forces, which are proportional to the fluid density and the square of the fluid velocity, are large relative to the viscous forces, and thus the viscous forces cannot prevent the random and rapid fluctuations of the fluid. At small or moderate Reynolds numbers, however, the viscous forces are large enough to suppress these fluctuations and to keep the fluid “in line.” Thus the flow is turbulent in the first case and laminar in the second.
The Reynolds number at which the flow becomes turbulent is called the critical Reynolds number, Recr. The value of the critical Reynolds number is different for different geometries and flow conditions. For internal flow in a circular pipe, the generally accepted value of the critical Reynolds number is Recr = 2300.
For flow through noncircular pipes, the Reynolds number is based on the hydraulic diameter Dh defined as (Fig. 8–6) Hydraulic diameter: Dh = 4Ac p (8–4) where Ac is the cross-sectional area of the pipe and p is its wetted perimeter. The hydraulic diameter is defined such that it reduces to ordinary diameter D for circular pipes, Circular pipes: Dh = 4Ac p = 4(𝜋D2/4) 𝜋D = D It certainly is desirable to have precise values of Reynolds numbers for laminar, transitional, and turbulent flows, but this is not the case in practice. It turns out that the transition from laminar to turbulent flow also depends on the degree of disturbance of the flow by surface roughness, pipe vibrations, and fluctuations in the upstream flow. Under most practical conditions, the flow in a circular pipe is laminar for Re ≲ 2300, turbulent for Re ≳ 4000, and transitional in between. That is, Re ≲ 2300 laminar flow 2300 ≲ Re ≲ 4000 transitional flow Re ≳4000 turbulent flow μ ρV avg L Inertial forces Viscous forces Re = = = = V avg L v ρV2 avg L 2 μV avg L L Vavg FIGURE 8–5 The Reynolds number can be viewed as the ratio of inertial forces to viscous forces acting on a fluid element.
Dh = = D 4(πD2/4) πD Dh = = a 4a2 4a Dh = = 4ab 2(a + b) 2ab a + b Circular tube: Rectangular duct: Square duct: a b D a a Dh = 4ab 2a + b Channel: a b FIGURE 8–6 The hydraulic diameter Dh = 4Ac/p is defined such that it reduces to ordi nary diameter for circular tubes. When there is a free surface, such as in open-channel flow, the wetted perimeter includes only the walls in contact with the fluid.
cen96537_ch08_351-442.indd 354 14/01/17 2:58 pm 355 CHAPTER 8 In transitional flow, the flow switches between laminar and turbulent in a disorderly fashion (Fig. 8–7). It should be kept in mind that laminar flow can be maintained at much higher Reynolds numbers in very smooth pipes by avoiding flow disturbances and pipe vibrations. In such carefully controlled laboratory experiments, laminar flow has been maintained at Reynolds num bers of up to 100,000.
8–3 ■ THE ENTRANCE REGION Consider a fluid entering a circular pipe at a uniform velocity. Because of the no-slip condition, the fluid particles in the layer in contact with the wall of the pipe come to a complete stop. This layer also causes the fluid parti cles in the adjacent layers to slow down gradually as a result of friction. To make up for this velocity reduction, the velocity of the fluid at the midsec tion of the pipe has to increase to keep the mass flow rate through the pipe constant. As a result, a velocity gradient develops along the pipe. The region of the flow in which the effects of the viscous shearing forces caused by fluid viscosity are felt is called the velocity bound ary layer or just the boundary layer. The hypothetical boundary surface divides the flow in a pipe into two regions: the boundary layer region, in which the viscous effects and the velocity changes are significant, and the irrotational (core) flow region, in which the frictional effects are negligi ble and the velocity remains essentially constant in the radial direction. The thickness of this boundary layer increases in the flow direction until the boundary layer reaches the pipe center and thus fills the entire pipe, as shown in Fig. 8–8, and the velocity becomes fully developed a little farther downstream. The region from the pipe inlet to the point at which the veloc ity profile is fully developed is called the hydrodynamic entrance region, and the length of this region is called the hydrodynamic entry length Lh. Flow in the entrance region is called hydrodynamically developing flow since this is the region where the velocity profile develops. The region beyond the entrance region in which the velocity profile is fully developed and remains unchanged is called the hydrodynamically fully developed region. The flow is said to be fully developed when the normalized temperature profile remains unchanged as well. Hydrodynamically fully developed flow is equivalent to fully developed flow when the fluid in the pipe is not heated or cooled since the fluid temperature in this case remains essentially constant throughout. The velocity profile in the fully developed region is parabolic in laminar flow and x r Hydrodynamic entrance region Hydrodynamically fully developed region Velocity boundary layer Developing velocity profile Fully developed velocity profile Irrotational (core) flow region Vavg Vavg Vavg Vavg Vavg FIGURE 8–8 The development of the velocity boundary layer in a pipe. (The developed average velocity profile is parabolic in laminar flow, as shown, but much flatter or fuller in turbulent flow.) Laminar Turbulent Vavg Dye trace Dye injection FIGURE 8–7 In the transitional flow region of 2300 ≤ Re ≤ 4000, the flow switches between laminar and turbulent somewhat randomly.
cen96537_ch08_351-442.indd 355 14/01/17 2:58 pm 356 internal flow much flatter (or fuller) in turbulent flow due to eddy motion and more vigor ous mixing in the radial direction. The time-averaged velocity profile remains unchanged when the flow is fully developed, and thus Hydrodynamically fully developed: ∂u(r, x) ∂x = 0 → u = u(r) (8–5) The shear stress at the pipe wall 𝜏w is related to the slope of the velocity profile at the surface. Noting that the velocity profile remains unchanged in the hydrodynamically fully developed region, the wall shear stress also remains constant in that region (Fig. 8–9). Consider fluid flow in the hydrodynamic entrance region of a pipe. The wall shear stress is the highest at the pipe inlet where the thickness of the boundary layer is smallest, and decreases gradually to the fully developed value, as shown in Fig. 8–10. Therefore, the pressure drop is higher in the entrance regions of a pipe, and the effect of the entrance region is always to increase the average friction factor for the entire pipe. This increase may be significant for short pipes but is negligible for long ones.
Entry Lengths The hydrodynamic entry length is usually taken to be the distance from the pipe entrance to where the wall shear stress (and thus the friction fac tor) reaches within about 2 percent of the fully developed value. In laminar flow, the nondimensional hydrodynamic entry length is given approximately as [see Kays and Crawford (2004) and Shah and Bhatti (1987)] Lh, laminar D ≅0.05Re (8–6) For Re = 20, the hydrodynamic entry length is about the size of the diameter, but increases linearly with velocity. In the limiting laminar case of Re = 2300, the hydrodynamic entry length is 115D.
τw τw τw τw τw τw Lh x r x Fully developed region Fully developed region Entrance region Entrance region τw τw Vavg Vavg FIGURE 8–10 The variation of wall shear stress in the flow direction for flow in a pipe from the entrance region into the fully developed region.
τw τw τw τw FIGURE 8–9 In the fully developed flow region of a pipe, the velocity profile does not change downstream, and thus the wall shear stress remains constant as well.
cen96537_ch08_351-442.indd 356 14/01/17 2:58 pm 357 CHAPTER 8 In turbulent flow, the intense mixing during random fluctuations usually overshadows the effects of molecular diffusion. The nondimensional hydro dynamic entry length for turbulent flow is approximated as [see Bhatti and Shah (1987) and Zhi-qing (1982)] Lh, turbulent D = 1.359Re1/4 (8–7) Compared to high Reynolds number laminar flow, the entry length is much shorter in turbulent flow, as expected, and its dependence on the Reynolds number is weaker. In many pipe flows of practical engineering interest, the entrance effects become insignificant beyond a pipe length of about 10 diam eters, and the nondimensional hydrodynamic entry length is approximated as Lh, turbulent D ≈ 10 (8–8) Precise correlations for calculating the frictional head losses in entrance regions are available in the literature. However, the pipes used in practice are usually several times the length of the entrance region, and thus the flow through the pipes is often assumed to be fully developed for the entire length of the pipe. This simplistic approach gives reasonable results for long pipes but sometimes poor results for short ones since it underpredicts the wall shear stress and thus the friction factor.
8–4 ■ LAMINAR FLOW IN PIPES We mentioned in Section 8–2 that flow in pipes is laminar for Re ≲ 2300, and that the flow is fully developed if the pipe is sufficiently long (relative to the entry length) so that the entrance effects are negligible. In this section, we consider the steady, laminar, incompressible flow of fluid with constant properties in the fully developed region of a straight circular pipe. We obtain the momentum equation by applying a momentum balance to a differential volume element, and we obtain the velocity profile by solving it. Then we use it to obtain a relation for the friction factor. An important aspect of the analysis here is that it is one of the few available for viscous flow. In fully developed laminar flow, each fluid particle moves at a constant axial velocity along a streamline and the velocity profile u(r) remains unchanged in the flow direction. There is no motion in the radial direction, and thus the velocity component in the direction normal to the pipe axis is everywhere zero. There is no acceleration since the flow is steady and fully developed. Now consider a ring-shaped differential volume element of radius r, thick ness dr, and length dx oriented coaxially with the pipe, as shown in Fig. 8–11. The volume element involves only pressure and viscous effects and thus the pressure and shear forces must balance each other. The pressure force acting on a submerged plane surface is the product of the pressure at the centroid of the surface and the surface area. A force balance on the volume element in the flow direction gives (2𝜋r dr P)x −(2𝜋r dr P)x+dx + (2𝜋r dx 𝜏)r −(2𝜋r dx 𝜏)r+dr = 0 (8–9) which indicates that in fully developed flow in a horizontal pipe, the viscous and pressure forces balance each other. Dividing by 2𝜋drdx and rearranging, r Px+dx −P x dx + (r𝜏)r+dr −(r𝜏)r dr = 0 (8–10) u(r) umax x dx dr r R Px Px + dx τr τr + dr FIGURE 8–11 Free-body diagram of a ring-shaped differential fluid element of radius r, thickness dr, and length dx oriented coaxially with a horizontal pipe in fully developed laminar flow. (The size of the fluid element is greatly exaggerated for clarity.) cen96537_ch08_351-442.indd 357 14/01/17 2:58 pm 358 internal flow But we recognize that the two numerators in Eq. 8–10 are dP and d(r𝜏), respectively. Thus, r dP dx + d(r𝜏) dr = 0 (8–11) Substituting 𝜏 = −𝜇(du/dr), dividing by r, and taking 𝜇 = constant gives the desired equation, 𝜇 r d dr(r du dr) = dP dx (8–12) The quantity du/dr is negative in pipe flow, and the negative sign is included to obtain positive values for 𝜏. (Or, du/dr = −du/dy if we define y = R − r.) The left side of Eq. 8–12 is a function of r, and the right side is a function of x. The equality must hold for any value of r and x, and an equal ity of the form f(r) = g(x) can be satisfied only if both f (r) and g(x) are equal to the same constant. Thus we conclude that dP/dx = constant. This is verified by writing a force balance on a volume element of radius R and thickness dx (a slice of the pipe as in Fig. 8–12), which gives dP dx = −2𝜏w R (8–13) Here 𝜏w is constant since the viscosity and the velocity profile are constants in the fully developed region. Therefore, dP/dx = constant. Equation 8–12 is solved by rearranging and integrating it twice to give u(r) = r 2 4𝜇( dP dx ) + C1 ln r + C2 (8–14) The velocity profile u(r) is obtained by applying the boundary conditions 𝜕u/𝜕r = 0 at r = 0 (because of symmetry about the centerline) and u = 0 at r = R (the no-slip condition at the pipe wall), u(r) = −R2 4𝜇( dP dx ) (1 −r 2 R2) (8–15) Therefore, the velocity profile in fully developed laminar flow in a pipe is parabolic with a maximum at the centerline and a minimum (zero) at the pipe wall. Also, the axial velocity u is positive for any r, and thus the axial pressure gradient dP/dx must be negative (i.e., pressure must decrease in the flow direction because of viscous effects—it takes pressure to push the fluid through the pipe). The average velocity is determined from its definition by substituting Eq. 8–15 into Eq. 8–2, and performing the integration, yielding Vavg = 2 R2 ∫ R 0 u(r)r dr = −2 R2 ∫ R 0 R2 4𝜇( dP dx ) (1 −r 2 R2)r dr = −R2 8𝜇( dP dx ) (8–16) Combining the last two equations, the velocity profile is rewritten as u(r) = 2Vavg(1 −r 2 R2) (8–17) τw πR2P – πR2(P + dP) – 2πR dx τw = 0 – = dP dx R r x 2πR dx τw πR2(P + dP) 2 πR2P R Force balance: Simplifying: dx FIGURE 8–12 Free-body diagram of a fluid disk element of radius R and length dx in fully developed laminar flow in a horizontal pipe.
cen96537_ch08_351-442.indd 358 14/01/17 2:58 pm 359 CHAPTER 8 This is a convenient form for the velocity profile since Vavg can be deter mined easily from the flow rate information.
The maximum velocity occurs at the centerline and is determined from Eq. 8–17 by substituting r = 0, umax = 2Vavg (8–18) Therefore, the average velocity in fully developed laminar pipe flow is one-half of the maximum velocity.
Pressure Drop and Head Loss A quantity of interest in the analysis of pipe flow is the pressure drop ΔP since it is directly related to the power requirements of the fan or pump to maintain flow. We note that dP/dx = constant, and integrating from x = x1 where the pressure is P1 to x = x1 + L where the pressure is P2 gives dP dx = P2 −P1 L (8–19) Substituting Eq. 8–19 into the Vavg expression in Eq. 8–16, the pressure drop is expressed as Laminar flow: ΔP = P1 −P2 = 8𝜇LVavg R2 = 32𝜇LVavg D2 (8–20) The symbol Δ is typically used to indicate the difference between the final and initial values, like Δy = y2 − y1. But in fluid flow, ΔP is used to des ignate pressure drop, and thus it is P1 − P2. A pressure drop due to viscous effects represents an irreversible pressure loss, and it is sometimes called pressure loss ΔPL to emphasize that it is a loss ( just like the head loss hL, which as we shall see is proportional to ΔPL).
Note from Eq. 8–20 that the pressure drop is proportional to the viscosity 𝜇 of the fluid, and ΔP would be zero if there were no friction. Therefore, the drop of pressure from P1 to P2 in this case is due entirely to viscous effects, and Eq. 8–20 represents the pressure loss ΔPL when a fluid of vis cosity 𝜇 flows through a pipe of constant diameter D and length L at aver age velocity Vavg.
In practice, it is convenient to express the pressure loss for all types of fully developed internal flows (laminar or turbulent flows, circular or non circular pipes, smooth or rough surfaces, horizontal or inclined pipes) as (Fig. 8–13) Pressure loss: ΔPL = f L D ρV 2 avg 2 (8–21) where 𝜌V 2 avg/2 is the dynamic pressure and f is the Darcy friction factor, f = 8𝜏w ρV 2 avg (8–22) It is also called the Darcy–Weisbach friction factor, named after the Frenchman Henry Darcy (1803–1858) and the German Julius Weisbach (1806–1871), the two engineers who provided the greatest contribution to Pressure loss: ΔPL = f L D 2 2 1 2g Head loss: hL = = f L ΔPL D ρg D L ΔPL Vavg ρV 2 avg V 2 avg FIGURE 8–13 The relation for pressure loss (and head loss) is one of the most general relations in fluid mechanics, and it is valid for laminar or turbulent flows, circular or noncircular pipes, and pipes with smooth or rough surfaces.
cen96537_ch08_351-442.indd 359 14/01/17 2:58 pm 360 internal flow its development. It should not be confused with the friction coefficient Cf [also called the Fanning friction factor, named after the American engineer John Fanning (1837–1911)], which is defined as Cf = 2𝜏w/(𝜌V 2 avg) = f/4. Setting Eqs. 8–20 and 8–21 equal to each other and solving for f gives the friction factor for fully developed laminar flow in a circular pipe, Circular pipe, laminar: f = 64𝜇 ρDVavg = 64 Re (8–23) This equation shows that in laminar flow, the friction factor is a function of the Reynolds number only and is independent of the roughness of the pipe surface (assuming, of course, that the roughness is not extreme). In the analysis of piping systems, pressure losses are commonly expressed in terms of the equivalent fluid column height, called the head loss hL. Noting from fluid statics that ΔP = 𝜌gh and thus a pressure difference of ΔP cor responds to a fluid height of h = ΔP/𝜌g, the pipe head loss is obtained by dividing ΔPL by 𝜌g to give Head loss: hL = ΔPL ρg = f L D V 2 avg 2g (8–24) The head loss hL represents the additional height that the fluid needs to be raised by a pump in order to overcome the frictional losses in the pipe. The head loss is caused by viscosity, and it is directly related to the wall shear stress. Equations 8–21 and 8–24 are valid for both laminar and turbulent flows in both circular and noncircular pipes, but Eq. 8–23 is valid only for fully developed laminar flow in circular pipes. Once the pressure loss (or head loss) is known, the required pumping power to overcome the pressure loss is determined from W .
pump, L = V · ΔPL = V ·ρghL = m ·ghL (8–25) where V . is the volume flow rate and m . is the mass flow rate. The average velocity for laminar flow in a horizontal pipe is, from Eq. 8–20, Horizontal pipe: Vavg = (P1 −P2)R2 8𝜇L = (P1 −P2)D2 32𝜇L = ΔP D 2 32𝜇L (8–26) Then the volume flow rate for laminar flow through a horizontal pipe of diameter D and length L becomes V · = VavgAc = (P1 −P2)R2 8𝜇L 𝜋R2 = (P1 −P2)𝜋D4 128𝜇L = ΔP 𝜋D4 128𝜇L (8–27) This equation is known as Poiseuille’s law, and this flow is called Hagen– Poiseuille flow in honor of the works of G. Hagen (1797–1884) and J. Poiseuille (1799–1869) on the subject. Note from Eq. 8–27 that for a specified flow rate, the pressure drops and thus the required pumping power is propor tional to the length of the pipe and the viscosity of the fluid, but it is inversely proportional to the fourth power of the radius (or diameter) of the pipe. There fore, the pumping power requirement for a laminar-flow piping system can be reduced by a factor of 16 by doubling the pipe diameter (Fig. 8–14). Of course the benefits of the reduction in the energy costs must be weighed against the increased cost of construction due to using a larger-diameter pipe. The pressure drop ΔP equals the pressure loss ΔPL in the case of a hori zontal pipe, but this is not the case for inclined pipes or pipes with vari able cross-sectional area. This can be demonstrated by writing the energy 2D Wpump = 16 hp ⋅ Wpump = 1 hp /4 ⋅ D Vavg Vavg FIGURE 8–14 The pumping power requirement for a laminar-flow piping system can be reduced by a factor of 16 by doubling the pipe diameter.
cen96537_ch08_351-442.indd 360 14/01/17 2:58 pm 361 CHAPTER 8 equation for steady, incompressible one-dimensional flow in terms of heads as (see Chap. 5) P1 ρg + 𝛼1 V 2 1 2g + z1 + hpump, u = P2 ρg + 𝛼2 V 2 2 2g + z2 + hturbine, e + hL (8–28) where hpump, u is the useful pump head delivered to the fluid, hturbine, e is the turbine head extracted from the fluid, hL is the irreversible head loss between sections 1 and 2, V1 and V2 are the average velocities at sections 1 and 2, respectively, and 𝛼1 and 𝛼2 are the kinetic energy correction factors at sections 1 and 2 (it can be shown that 𝛼 = 2 for fully developed laminar flow and about 1.05 for fully developed turbulent flow). Equation 8–28 can be rearranged as P1 −P2 = ρ(𝛼2V 2 2 −𝛼1V 2 1 )/2 + ρg[(z2 −z1) + hturbine, e −hpump, u + hL] (8–29) Therefore, the pressure drop ΔP = P1 − P2 and pressure loss ΔPL = 𝜌ghL for a given flow section are equivalent if (1) the flow section is horizontal so that there are no hydrostatic or gravity effects (z1 = z2), (2) the flow section does not involve any work devices such as a pump or a turbine since they change the fluid pressure (hpump, u = hturbine, e = 0), (3) the cross- sectional area of the flow section is constant and thus the average flow velocity is constant (V1 = V2), and (4) the velocity profiles at sections 1 and 2 are the same shape (𝛼1 = 𝛼2).
Effect of Gravity on Velocity and Flow Rate in Laminar Flow Gravity has no effect on flow in horizontal pipes, but it has a significant effect on both the velocity and the flow rate in uphill or downhill pipes. Relations for inclined pipes can be obtained in a similar manner from a force balance in the direction of flow. The only additional force in this case is the component of the fluid weight in the flow direction, whose magnitude is dWx = dW sin 𝜃 = ρgdVelement sin 𝜃 = ρg(2𝜋r dr dx) sin 𝜃 (8–30) where 𝜃 is the angle between the horizontal and the flow direction (Fig. 8–15). The force balance in Eq. 8–9 now becomes (2𝜋r dr P)x −(2𝜋r dr P)x+dx + (2𝜋r dx 𝜏)r − (2𝜋r dx 𝜏)r+dr −ρg(2𝜋r dr dx) sin 𝜃 = 0 (8–31) which results in the differential equation 𝜇 r d dr (r du dr) = dP dx + ρg sin 𝜃 (8–32) Following the same solution procedure as previously, the velocity profile is u(r) = −R2 4𝜇 ( dP dx + ρg sin 𝜃 ) (1 −r 2 R2) (8–33) From Eq. 8–33, the average velocity and the volume flow rate relations for laminar flow through inclined pipes are, respectively, Vavg = (ΔP −ρgL sin 𝜃 )D2 32𝜇L and V .
= (ΔP −ρgL sin 𝜃 )𝜋D4 128𝜇L (8–34) which are identical to the corresponding relations for horizontal pipes, except that ΔP is replaced by ΔP − 𝜌gL sin 𝜃. Therefore, the results already θ r + dr τ r τ Px + dx dW Px x r dW sin θ θ dx dr FIGURE 8–15 Free-body diagram of a ring-shaped differential fluid element of radius r, thickness dr, and length dx oriented coaxially with an inclined pipe in fully developed laminar flow.
FIGURE 8–16 The relations developed for fully developed laminar flow through horizontal pipes can also be used for inclined pipes by replacing ΔP with ΔP − 𝜌gL sin 𝜃. Uphill flow: θ > 0 and sin θ > 0 Downhill flow: θ < 0 and sin θ < 0 Laminar Flow in Circular Pipes or turbine in the flow section, and ΔP = P1 – P2) (Fully developed flow with no pump Horizontal pipe: V = ΔP πD4 128μL .
.
Inclined pipe: V = (ΔP – ρgL sin θ)πD4 128μL cen96537_ch08_351-442.indd 361 14/01/17 2:58 pm 362 internal flow obtained for horizontal pipes can also be used for inclined pipes provided that ΔP is replaced by ΔP − 𝜌gL sin 𝜃 (Fig. 8–16). Note that 𝜃 > 0 and thus sin 𝜃 > 0 for uphill flow, and 𝜃 < 0 and thus sin 𝜃 < 0 for downhill flow. In inclined pipes, the combined effect of pressure difference and gravity drives the flow. Gravity helps downhill flow but opposes uphill flow. There fore, much greater pressure differences need to be applied to maintain a specified flow rate in uphill flow although this becomes important only for liquids, because the density of gases is generally low. In the special case of no flow (V . = 0), Eq. 8–34 yields ΔP = 𝜌gL sin 𝜃, which is what we would obtain from fluid statics (Chap. 3).
Laminar Flow in Noncircular Pipes The friction factor f relations are given in Table 8–1 for fully developed laminar flow in pipes of various cross sections. The Reynolds number for flow in these pipes is based on the hydraulic diameter Dh = 4Ac/p, where Ac is the cross-sectional area of the pipe and p is its wetted perimeter.
TABLE 8–1 Friction factor for fully developed laminar flow in pipes of various cross sections (Dh = 4Ac/p and Re = Vavg Dh/𝜈) Tube Geometry a/b or 𝜃° Friction Factor f Circle — 64.00/Re D Rectangle a/b 1 56.92/Re 2 62.20/Re 3 68.36/Re 4 72.92/Re 6 78.80/Re 8 82.32/Re ∞ 96.00/Re Ellipse a/b 1 64.00/Re 2 67.28/Re 4 72.96/Re 8 76.60/Re 16 78.16/Re Isosceles triangle 𝜃 10° 50.80/Re 30° 52.28/Re 60° 53.32/Re 90° 52.60/Re 120° 50.96/Re b a b a θ cen96537_ch08_351-442.indd 362 14/01/17 2:58 pm 363 CHAPTER 8 EXAMPLE 8–1 Laminar Draining from a Pool At the end of the summer, a swimming pool is being drained through a very long, small-diameter hose (Fig. 8–17). The hose is smooth, of inner diameter D = 6.0 cm, and of length L = 65 m. The initial height difference from the pool surface to the outlet of the hose is H = 2.20 m. Calculate the volume flow rate in liters per minute (LPM) at the start of draining.
SOLUTION Water is drained from a pool. The volume flow rate is to be deter mined.
Assumptions 1 The flow is incompressible and quasi-steady (we approxi mate it as steady at the start of the draining since the pool volume is large and drains slowly). 2 Entrance effects are negligible since the hose is so long; the flow is fully developed. 3 Any other losses such as elbows, etc. in the hose are negligible.
Properties The properties of water are 𝜌 = 998 kg/m3 and 𝜇 = 0.001002 kg/m⋅s. Analysis The first step in a problem like this is to wisely choose a control vol ume. The one shown in the sketch cuts just below the surface of the water in the pool (inlet, 1) and slices through the hose discharge (outlet, 2). We assume that the flow is laminar, but we will need to verify this at the end. For fully developed laminar pipe flow at the outlet of the hose, 𝛼2 = 2. Letting V be the average velocity through the hose, the appropriate equations are: Re = ρVD 𝜇 f = 64 Re hL = f L D V 2 2g (1) and the head form of the energy equation from Chap. 5, P1 ρg + 𝛼1 V 2 1 2g + z1 + hpump, u = P2 ρg + 𝛼2 V 2 2 2g + z2 + hturbine, e + hL (2) In this problem, there are no pumps or turbines, so those terms are zero. Both P1 and P2 are equal to atmospheric pressure, so the pressure terms cancel. Water speed V1 at the control volume inlet (the pool surface) is negligibly small since the pool is draining slowly. At the outlet, V2 = V. Thus, the energy equation reduces to 𝛼2 V 2 2g = z1 −z2 −hL = H −hL (3) At this point, we can enter this set of simultaneous equations into an equation solver to obtain the solution. Or, if we are calculating by hand, we must perform algebra on Eqs. 1 and 3 to get one equation and one unknown (V), V 2 + 64𝜇L ρD2𝛼2 V −2gH 𝛼2 = 0 (4) Since Eq. (4) is in standard form for solution by the quadratic rule, and all the constants in the equation are known, we easily solve for V = 0.36969 m/s. Finally, we calculate the volume flow rate, V .
= VAc = V 𝜋D2 4 = (0.36969 m/s) 𝜋(0.0060 m)2 4 ( 1000 L m3 ) ( 60 s min)= 0.6272 L min which we round to our final answer: V . ≈ 0.627 Lpm FIGURE 8–17 Schematic for Example 8–1.
Control volume 2 1 D H cen96537_ch08_351-442.indd 363 14/01/17 2:58 pm 364 internal flow We check the Reynolds number, Re = ρVD 𝜇 = (998 kg/m3)(0.36969 m/s)(0.0060 m) 0.001002 kg/m·s = 2209 Since Re < 2300, we verify that the flow is laminar, although it is close to transitioning to turbulent flow.
Discussion It is important to verify assumptions. It would be wise and useful practice for students to verify the algebra leading to Eq. 4, and to solve it using the quadratic rule, including all units. The hose in the sketch is inclined at some angle. However, the angle was never used in the solution because the only thing that matters is height H, and the result would be the same regardless of whether the pool is shallow with a large inclination of the hose or the pool is deep with a small or even zero inclination of the hose.
EXAMPLE 8–2 Pumping Power Requirement for Oil Flow in a Pipe Consider the flow of oil with 𝜌 = 894 kg/m3 and 𝜇 = 2.33 kg/m·s in a 28-cm-diameter pipeline at an average velocity of 0.5 m/s. A 330-m-long section of the pipeline passes through the icy waters of a lake (Fig. 8–18). Disregarding the entrance effects, determine the pumping power required to overcome the pressure losses and to maintain the flow of oil in the pipe.
SOLUTION Oil flows through a pipeline that passes through icy waters of a lake. The pumping power needed to overcome pressure losses is to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The flow section considered is away from the entrance, and thus the flow is fully developed. 3 The roughness effects are negligible, and thus the inner surfaces are considered to be smooth, 𝜀 ≈ 0.
Properties The properties of oil are given to be 𝜌 = 894 kg/m3 and 𝜇 = 2.33 kg/m⋅s.
Analysis The volume flow rate and the Reynolds number in this case are V .
= VAc = V 𝜋D2 4 = (0.5 m/s)𝜋(0.28 m)2 4 = 0.03079 m3/s Re = ρVD 𝜇 = (894 kg/m3)(0.5 m/s)(0.28 m) 2.33 kg/m·s = 53.72 which is less than 2300. Therefore, the flow is laminar, and the friction factor is f = 64 Re = 64 53.72 = 1.191 Then the pressure drop in the pipe and the required pumping power become ΔP = ΔPL = f L D ρV 2 2 = 1.191 330 m 0.28 m (894 kg/m3)(0.5 m/s)2 2 ( 1 kN 1000 kg · m/s2) ( 1 kPa 1 kN/m2) = 156.9 kPa FIGURE 8–18 Schematic for Example 8–2. Drawing not to scale.
Oil 0.5 m/s L = 330 m Icy lake, 0 °C D = 0.28 m cen96537_ch08_351-442.indd 364 14/01/17 2:58 pm 365 CHAPTER 8 8–5 ■ TURBULENT FLOW IN PIPES Most flows encountered in engineering practice are turbulent, and thus it is important to understand how turbulence affects wall shear stress. However, turbulent flow is a complex mechanism dominated by fluctuations, and despite tremendous amounts of work done in this area by researchers, turbulent flow still is not fully understood. Therefore, we must rely on experiments and the empirical or semi-empirical correlations developed for various situations. Turbulent flow is characterized by disorderly and rapid fluctuations of swirl ing regions of fluid, called eddies, throughout the flow (Fig. 8–19). These fluctuations provide an additional mechanism for momentum and energy transfer. In laminar flow, fluid particles flow in an orderly manner along path lines, and momentum and energy are transferred across streamlines by molec ular diffusion. In turbulent flow, the swirling eddies transport mass, momen tum, and energy to other regions of flow much more rapidly than molecular diffusion, greatly enhancing mass, momentum, and heat transfer. As a result, turbulent flow is associated with much higher values of friction, heat trans fer, and mass transfer coefficients (Fig. 8–20). Even when the average flow is steady, the eddy motion in turbulent flow causes significant fluctuations in the values of velocity, temperature, pres sure, and even density (in compressible flow). Figure 8–21 shows the vari ation of the instantaneous velocity component u with time at a specified location, as can be measured with a hot-wire anemometer probe or other sen sitive device. We observe that the instantaneous values of the velocity fluctu ate about an average value, which suggests that the velocity can be expressed as the sum of an average value u – and a fluctuating component u′, u = u + uʹ (8–35) This is also the case for other properties such as the velocity component υ in the y-direction, and thus υ = υ – + υ′, P = P – + P′, and T = T – + T′. The average value of a property at some location is determined by averaging it over a time interval that is sufficiently large so that the time average levels off to a constant. Therefore, the time average of fluctuating components is zero, e.g., uʹ = 0. The magnitude of u′ is usually just a few percent of u –, but the high frequencies of eddies (on the order of a thousand per second) make them very effective for the transport of momentum, thermal energy, and mass. In time-averaged stationary turbulent flow, the average values of properties (indicated by an overbar) are independent of time. The chaotic fluctuations of fluid particles play a dominant role in pressure drop, and these random motions must be considered in analyses together with the average velocity.
W .
pump = V · ΔP = (0.03079 m3/s)(156.9 kPa)( 1 kW 1 kPa · m3/s) = 4.83 kW Discussion The power input determined is the mechanical power that needs to be imparted to the fluid. The shaft power will be more than this due to pump inefficiency; the electrical power input will be even more due to motor inefficiency.
FIGURE 8–19 Water exiting a tube: (a) laminar flow at low flow rate, (b) turbulent flow at high flow rate, and (c) same as (b) but with a short shutter exposure to capture individual eddies.
Photos by Alex Wouden.
(a) (c) (b) cen96537_ch08_351-442.indd 365 14/01/17 2:58 pm 366 internal flow Perhaps the first thought that comes to mind is to determine the shear stress in an analogous manner to laminar flow from 𝜏 = −𝜇 du –/dr, where u –(r) is the average velocity profile for turbulent flow. But the experimental studies show that this is not the case, and the effective shear stress is much larger due to the turbulent fluctuations. Therefore, it is convenient to think of the turbulent shear stress as consisting of two parts: the laminar component, which accounts for the friction between layers in the flow direction (expressed as 𝜏lam = −𝜇 du –/dr), and the turbulent component, which accounts for the friction between the fluctuating fluid particles and the fluid body (denoted as 𝜏turb and is related to the fluctuation components of velocity). Then the total shear stress in turbulent flow can be expressed as 𝜏total = 𝜏lam + 𝜏turb (8–36) The typical average velocity profile and relative magnitudes of laminar and turbulent components of shear stress for turbulent flow in a pipe are given in Fig. 8–22. Note that although the velocity profile is approximately parabolic in laminar flow, it becomes flatter or “fuller” in turbulent flow, with a sharp drop near the pipe wall. The fullness increases with the Reynolds number, and the velocity profile becomes more nearly uniform, lending support to the commonly utilized uniform velocity profile approximation for fully devel oped turbulent pipe flow. Keep in mind, however, that the flow speed at the wall of a stationary pipe is always zero (no-slip condition).
Turbulent Shear Stress Consider turbulent flow in a horizontal pipe, and the upward eddy motion of a fluid particle from a layer of lower velocity to an adjacent layer of higher velocity through a differential area dA as a result of the velocity fluctuation υ′, as shown in Fig. 8–23. The mass flow rate of the fluid particle rising through dA is 𝜌υ′dA, and its net effect on the layer above dA is a reduction in its aver age flow velocity because of momentum transfer to the fluid particle with lower average flow velocity. This momentum transfer causes the horizontal velocity of the fluid particle to increase by u′, and thus its momentum in the horizontal direction to increase at a rate of (𝜌υ′dA)u′, which must be equal to the decrease in the momentum of the upper fluid layer. Noting that force in a given direction is equal to the rate of change of momentum in that direction, the horizontal force acting on a fluid element above dA due to the passing of fluid particles through dA is 𝛿F = (𝜌υ′dA)(−u′) = −𝜌u′υ′dA. Therefore, the shear force per unit area due to the eddy motion of fluid par ticles 𝛿F/dA = −𝜌u′υ′ can be viewed as the instantaneous turbulent shear stress. Then the turbulent shear stress can be expressed as 𝜏turb = −ρuʹ𝜐ʹ (8–37) where uʹ𝜐ʹ is the time average of the product of the fluctuating velocity com ponents u′ and υ′. Note that uʹ𝜐ʹ ≠0 even though uʹ = 0 and 𝜐ʹ = 0 (and thus uʹ 𝜐ʹ = 0), and experimental results show that uʹ𝜐ʹ is usually a negative quantity. Terms such as −ρuʹ𝜐ʹ or −ρuʹ2 are called Reynolds stresses or turbulent stresses. Many semi-empirical formulations have been developed that model the Reynolds stress in terms of average velocity gradients in order to provide (a) Before turbulence (b) After turbulence FIGURE 8–20 The intense mixing in turbulent flow brings fluid particles at different momentums into close contact and thus enhances momentum transfer.
u u' u – Time, t u = + u' u – FIGURE 8–21 Fluctuations of the velocity component u with time at a specified location in turbulent flow.
FIGURE 8–22 The average or mean velocity profile for turbulent flow in a pipe.
u(r) r 0 υ' ρυ' dA u(y) u u' dA y FIGURE 8–23 Fluid particle moving upward through a differential area dA as a result of the velocity fluctuation 𝜐′.
cen96537_ch08_351-442.indd 366 14/01/17 2:58 pm 367 CHAPTER 8 mathematical closure to the equations of motion. Such models are called turbulence models and are discussed in more detail in Chap. 15. The random eddy motion of groups of particles resembles the random motion of molecules in a gas—colliding with each other after traveling a certain distance and exchanging momentum in the process. Therefore, momentum transport by eddies in turbulent flows is analogous to the molec ular momentum diffusion. In many of the simpler turbulence models, tur bulent shear stress is expressed in an analogous manner as suggested by the French mathematician Joseph Boussinesq (1842–1929) in 1877 as 𝜏turb = −ρuʹ𝜐ʹ = 𝜇t ∂u ∂y (8–38) where 𝜇t is the eddy viscosity or turbulent viscosity, which accounts for momentum transport by turbulent eddies. Then the total shear stress can be expressed conveniently as 𝜏total = (𝜇+ 𝜇t) ∂u ∂y = ρ(𝜈+ 𝜈t) ∂u ∂y (8–39) where 𝜈t = 𝜇t/𝜌 is the kinematic eddy viscosity or kinematic turbulent vis cosity (also called the eddy diffusivity of momentum). The concept of eddy viscosity is very appealing, but it is of no practical use unless its value can be determined. In other words, eddy viscosity must be modeled as a function of the average flow variables; we call this eddy viscosity closure. For example, in the early 1900s, the German engineer L. Prandtl introduced the concept of mixing length lm, which is related to the average size of the eddies that are primarily responsible for mixing, and expressed the turbulent shear stress as 𝜏turb = 𝜇t ∂u ∂y = ρl 2 m( ∂u ∂y) 2 (8–40) But this concept is also of limited use since lm is not a constant for a given flow (in the vicinity of the wall, for example, lm is nearly proportional to the distance from the wall) and its determination is not easy. Final mathemati cal closure is obtained only when lm is written as a function of average flow variables, distance from the wall, etc. Eddy motion and thus eddy diffusivities are much larger than their molec ular counterparts in the core region of a turbulent boundary layer. The eddy motion loses its intensity close to the wall and diminishes at the wall because of the no-slip condition (u′ and υ′ are identically zero at a station ary wall). Therefore, the velocity profile is very slowly changing in the core region of a turbulent boundary layer, but very steep in the thin layer adja cent to the wall, resulting in large velocity gradients at the wall surface. So it is no surprise that the wall shear stress is much larger in turbulent flow than it is in laminar flow (Fig. 8–24). Note that the molecular diffusivity of momentum 𝜈 (as well as 𝜇) is a fluid property, and its value is listed in fluid handbooks. Eddy diffusivity 𝜈t (as well as 𝜇t), however, is not a fluid property, and its value depends on flow conditions. Eddy diffusivity 𝜈t decreases toward the wall, becoming zero at the wall. Its value ranges from zero at the wall to several thousand times the value of the molecular diffusivity in the core region.
y=0 Turbulent flow y ∂u ∂y y=0 Laminar flow y ∂u ∂y ( ) ( ) FIGURE 8–24 The velocity gradients at the wall, and thus the wall shear stress, are much larger for turbulent flow than they are for laminar flow, even though the turbulent boundary layer is thicker than the laminar one for the same value of free-stream velocity.
cen96537_ch08_351-442.indd 367 14/01/17 2:58 pm 368 internal flow Turbulent Velocity Profile Unlike laminar flow, the expressions for the velocity profile in a turbu lent flow are based on both analysis and measurements, and thus they are semi-empirical in nature with constants determined from experimental data. Consider fully developed turbulent flow in a pipe, and let u denote the time-averaged velocity in the axial direction (and thus drop the overbar from u – for simplicity). Typical velocity profiles for fully developed laminar and turbulent flows are given in Fig. 8–25. Note that the velocity profile is parabolic in lami nar flow but is much fuller in turbulent flow, with a sharp drop near the pipe wall. Turbulent flow along a wall can be considered to consist of four regions, characterized by the distance from the wall (Fig. 8–25). The very thin layer next to the wall where viscous effects are dominant is the viscous (or laminar or linear or wall) sublayer. The velocity profile in this layer is very nearly linear, and the flow is streamlined. Next to the viscous sublayer is the buffer layer, in which turbulent effects are becoming significant, but the flow is still dominated by viscous effects. Above the buffer layer is the overlap (or transition) layer, also called the inertial sublayer, in which the turbulent effects are much more significant, but still not dominant. Above that is the outer (or turbulent) layer in the remaining part of the flow in which turbulent effects dominate over molecular diffusion (viscous) effects. Flow characteristics are quite different in different regions, and thus it is difficult to come up with an analytic relation for the velocity profile for the entire flow as we did for laminar flow. The best approach in the turbulent case turns out to be to identify the key variables and functional forms using dimensional analysis, and then to use experimental data to determine the numerical values of any constants. The thickness of the viscous sublayer is very small (typically, much less than 1 percent of the pipe diameter), but this thin layer next to the wall plays a dominant role on flow characteristics because of the large velocity gradi ents it involves. The wall dampens any eddy motion, and thus the flow in this layer is essentially laminar and the shear stress consists of laminar shear stress which is proportional to the fluid viscosity. Considering that velocity changes from zero to nearly the core region value across a layer that is some times no thicker than a hair (almost like a step function), we would expect the velocity profile in this layer to be very nearly linear, and experiments confirm that. Then the velocity gradient in the viscous sublayer remains nearly constant at du/dy = u/y, and the wall shear stress can be expressed as 𝜏w = 𝜇 u y = ρ𝜈 u y or 𝜏w ρ = 𝜈u y (8–41) where y is the distance from the wall (note that y = R − r for a circular pipe). The quantity 𝜏w/𝜌 is frequently encountered in the analysis of turbulent velocity profiles. The square root of 𝜏w/𝜌 has the dimensions of velocity, and thus it is convenient to view it as a fictitious velocity called the friction velocity expressed as u = √𝜏w/ρ. Substituting this into Eq. 8–41, the velocity profile in the viscous sublayer is expressed in dimensionless form as Viscous sublayer: u u = yu 𝜈 (8–42) Laminar flow u(r) r 0 Turbulent flow Turbulent layer Overlap layer Buffer layer Viscous sublayer u(r) r 0 Vavg Vavg FIGURE 8–25 The velocity profile in fully developed pipe flow is parabolic in laminar flow, but much fuller in turbulent flow. Note that u(r) in the turbulent case is the time-averaged velocity component in the axial direction (the overbar on u has been dropped for simplicity).
cen96537_ch08_351-442.indd 368 14/01/17 2:58 pm 369 CHAPTER 8 This equation is known as the law of the wall, and it is found to satisfacto rily correlate with experimental data for smooth surfaces for 0 ≤ yu/𝜈 ≤ 5. Therefore, the thickness of the viscous sublayer is roughly Thickness of viscous sublayer: y = 𝛿sublayer = 5𝜈 u = 25𝜈 u𝛿 (8–43) where u𝛿 is the flow velocity at the edge of the viscous sublayer (where u𝛿 ≈ 5u), which is closely related to the average velocity in a pipe. Thus we conclude that the thickness of the viscous sublayer is proportional to the kinematic viscosity and inversely proportional to the average flow velocity. In other words, the viscous sublayer is suppressed and it gets thinner as the velocity (and thus the Reynolds number) increases. Consequently, the velocity profile becomes nearly flat and thus the velocity distribution becomes more uniform at very high Reynolds numbers. The quantity 𝜈/u has dimensions of length and is called the viscous length; it is used to nondimensionalize the distance y from the surface. In boundary layer analysis, it is convenient to work with nondimensionalized distance and nondimensionalized velocity defined as Nondimensionalized variables: y+ = yu 𝜈 and u+ = u u (8–44) Then the law of the wall (Eq. 8–42) becomes simply Normalized law of the wall: u+ = y+ (8–45) Note that the friction velocity u is used to nondimensionalize both y and u, and y+ resembles the Reynolds number expression. In the overlap layer, the experimental data for velocity are observed to line up on a straight line when plotted against the logarithm of distance from the wall. Dimensional analysis indicates and the experiments confirm that the velocity in the overlap layer is proportional to the logarithm of distance, and the velocity profile can be expressed as The logarithmic law: u u = 1 𝜅 ln yu 𝜈+ B (8–46) where 𝜅 and B are constants whose values are determined experimentally to be about 0.40 and 5.0, respectively. Equation 8–46 is known as the loga rithmic law. Substituting the values of the constants, the velocity profile is determined to be Overlap layer: u u = 2.5 ln yu 𝜈+ 5.0 or u+ = 2.5 ln y+ + 5.0 (8–47) It turns out that the logarithmic law in Eq. 8–47 satisfactorily represents exper imental data for the entire flow region except for the regions very close to the wall and near the pipe center, as shown in Fig. 8–26, and thus it is viewed as a universal velocity profile for turbulent flow in pipes or over surfaces. Note from the figure that the logarithmic-law velocity profile is quite accurate for y+ > 30, but neither velocity profile is accurate in the buffer layer, i.e., the region 5 < y+ < 30. Also, the viscous sublayer appears much larger in the figure than it is since we used a logarithmic scale for distance from the wall.
A good approximation for the outer turbulent layer of pipe flow can be obtained by evaluating the constant B in Eq. 8–46 from the requirement that Viscous sublayer 100 30 25 20 15 10 5 0 101 102 y+ = yu/v u+ = u/u 103 104 Buffer layer Overlap layer Turbulent layer Eq. 8–47 Eq. 8–42 Experimental data FIGURE 8–26 Comparison of the law of the wall and the logarithmic-law velocity profiles with experimental data for fully developed turbulent flow in a pipe.
cen96537_ch08_351-442.indd 369 14/01/17 2:58 pm 370 internal flow maximum velocity in a pipe occurs at the centerline where r = 0. Solving for B from Eq. 8–46 by setting y = R − r = R and u = umax, and substituting it back into Eq. 8–46 together with 𝜅 = 0.4 gives Outer turbulent layer: umax −u u = 2.5 ln R R −r (8–48) The deviation of velocity from the centerline value umax − u is called the velocity defect, and Eq. 8–48 is called the velocity defect law. This relation shows that the normalized velocity profile in the core region of turbulent flow in a pipe depends on the distance from the centerline and is independ ent of the viscosity of the fluid. This is not surprising since the eddy motion is dominant in this region, and the effect of fluid viscosity is negligible. Numerous other empirical velocity profiles exist for turbulent pipe flow. Among those, the simplest and the best known is the power-law velocity profile expressed as Power-law velocity profile: u umax = ( y R) 1/n or u umax = (1 −r R) 1/n (8–49) where the exponent n is a constant whose value depends on the Reynolds number. The value of n increases with increasing Reynolds number. The value n = 7 generally approximates many flows in practice, giving rise to the term one-seventh power-law velocity profile. Various power-law velocity profiles are shown in Fig. 8–27 for n = 6, 8, and 10 together with the velocity profile for fully developed laminar flow for comparison. Note that the turbulent velocity profile is fuller than the laminar one, and it becomes more flat as n (and thus the Reynolds number) increases. Also note that the power-law profile cannot be used to calculate wall shear stress since it gives a velocity gradient of infinity there, and it fails to give zero slope at the centerline. But these regions of discrepancy constitute a small portion of the overall flow, and the power-law profile gives highly accurate results for turbulent flow through a pipe. Despite the small thickness of the viscous sublayer (usually much less than 1 percent of the pipe diameter), the characteristics of the flow in this layer are very important since they set the stage for flow in the rest of the pipe. Any irregularity or roughness on the surface disturbs this layer and affects the flow. Therefore, unlike laminar flow, the friction factor in turbu lent flow is a strong function of surface roughness. It should be kept in mind that roughness is a relative concept, and it has significance when its height 𝜀 is comparable to the thickness of the viscous sublayer (which is a function of the Reynolds number). All materials appear “rough” under a microscope with sufficient magnification. In fluid mechanics, a surface is characterized as being rough when the hills of roughness protrude out of the viscous sublayer. A surface is said to be hydrodynamically smooth when the sublayer submerges the roughness elements. Glass and plastic sur faces are generally considered to be hydrodynamically smooth.
The Moody Chart and Its Associated Equations The friction factor in fully developed turbulent pipe flow depends on the Reynolds number and the relative roughness 𝜀/D, which is the ratio of the mean height of roughness of the pipe to the pipe diameter. The functional 0.2 0 0.4 0.6 0.8 1 0.8 0.6 0.4 0.2 0 u/umax r/R 1 Laminar n = 6 n = 8 n = 10 FIGURE 8–27 Power-law velocity profiles for fully developed turbulent flow in a pipe for different exponents, and its comparison with the laminar velocity profile.
cen96537_ch08_351-442.indd 370 14/01/17 2:58 pm 371 CHAPTER 8 form of this dependence cannot be obtained from a theoretical analysis, and all available results are obtained from painstaking experiments using artificially roughened surfaces (usually by gluing sand grains of a known size on the inner surfaces of the pipes). Most such experiments were conducted by Prandtl’s stu dent J. Nikuradse in 1933, followed by the works of others. The friction factor was calculated from measurements of the flow rate and the pressure drop. The experimental results are presented in tabular, graphical, and func tional forms obtained by curve-fitting experimental data. In 1939, Cyril F. Colebrook (1910–1997) combined the available data for transition and tur bulent flow in smooth as well as rough pipes into the following implicit relation (Fig. 8–28) known as the Colebrook equation: 1 √f = −2.0 log( 𝜀 /D 3.7 + 2.51 Re√f) (turbulent flow) (8–50) We note that the logarithm in Eq. 8–50 is a base 10 rather than a natural logarithm. In 1942, the American engineer Hunter Rouse (1906–1996) veri fied Colebrook’s equation and produced a graphical plot of f as a function of Re and the product Re√f. He also presented the laminar flow relation and a table of commercial pipe roughness. Two years later, Lewis F. Moody (1880–1953) redrew Rouse’s diagram into the form commonly used today. The now famous Moody chart is given in the appendix as Fig. A–12. It presents the Darcy friction factor for pipe flow as a function of Reynolds number and 𝜀/D over a wide range. It is probably one of the most widely accepted and used charts in engineering. Although it is developed for circu lar pipes, it can also be used for noncircular pipes by replacing the diameter with the hydraulic diameter. Commercially available pipes differ from those used in the experiments in that the roughness of pipes in the market is not uniform and it is difficult to give a precise description of it. Equivalent roughness values for some commercial pipes are given in Table 8–2 as well as on the Moody chart. But it should be kept in mind that these values are for new pipes, and the relative roughness of pipes may increase with use as a result of corrosion, scale buildup, and precipitation. As a result, the friction factor may increase by a factor of 5 to 10. Actual operating conditions must be considered in the design of piping systems. Also, the Moody chart and its equivalent Cole brook equation involve several uncertainties (the roughness size, experimen tal error, curve fitting of data, etc.), and thus the results obtained should not be treated as “exact.” They are is usually considered to be accurate to ±15 percent over the entire range in the figure. We make the following observations from the Moody chart: • For laminar flow, the friction factor decreases with increasing Reynolds number, and it is independent of surface roughness.
• The friction factor is a minimum for a smooth pipe (but still not zero be cause of the no-slip condition) and increases with roughness (Fig. 8–29). The Colebrook equation in this case (𝜀 = 0) reduces to the Prandtl equation expressed as 1/√f = 2.0 log(Re√f ) − 0.8.
• The transition region from the laminar to turbulent regime (2300 < Re < 4000) is indicated by the shaded area in the Moody chart (Figs. 8–30 FIGURE 8–28 The Colebrook equation.
The Colebrook equation is implicit in f since f appears on both sides of the equation.
It must be solved iteratively.
+ = –2.0 log 1 f f Re ε/D 3.7 2.51 FIGURE 8–29 The friction factor is minimum for a smooth pipe and increases with roughness.
Relative Roughness, 𝜀/D Friction Factor, f 0.0 0.00001 0.0001 0.0005 0.001 0.005 0.01 0.05 0.0119 0.0119 0.0134 0.0172 0.0199 0.0305 0.0380 0.0716 Smooth surface. All values are for Re = 106 and are calculated from the Colebrook equation.
The uncertainty in these values can be as much as ±60 percent.
TABLE 8–2 Equivalent roughness values for new commercial pipes Roughness, 𝜀 Material ft mm Glass, plastic 0 (smooth) Concrete 0.003–0.03 0.9–9 Wood stave 0.0016 0.5 Rubber, smoothed 0.000033 0.01 Copper or brass tubing 0.000005 0.0015 Cast iron 0.00085 0.26 Galvanized iron 0.0005 0.15 Wrought iron 0.00015 0.046 Stainless steel 0.000007 0.002 Commercial steel 0.00015 0.045 cen96537_ch08_351-442.indd 371 14/01/17 2:58 pm 372 internal flow and A–12). The flow in this region may be laminar or turbulent, depending on flow disturbances, or it may alternate between laminar and turbulent, and thus the friction factor may also alternate between the values for laminar and turbulent flow. The data in this range are the least reliable. At small relative roughnesses, the friction factor increases in the transition region and approaches the value for smooth pipes.
• At very large Reynolds numbers (to the right of the dashed line on the Moody chart) the friction factor curves corresponding to specified relative roughness curves are nearly horizontal, and thus the friction factors are independent of the Reynolds number (Fig. 8–30). The flow in that region is called fully rough turbulent flow or just fully rough flow because the thickness of the viscous sublayer decreases with increasing Reynolds number, and it becomes so thin that it is negligibly small compared to the surface roughness height. The viscous effects in this case are produced in the main flow primarily by the protruding roughness elements, and the contribution of the viscous sublayer is negligible. The Colebrook equation in the fully rough zone (Re →∞) reduces to the von Kármán equation expressed as 1/√f =−2.0 log[(𝜀 /D)/3.7], which is explicit in f. Some authors call this zone completely (or fully) turbulent flow, but this is misleading since the flow to the left of the dashed blue line in Fig. 8–30 is also fully turbulent.
In calculations, we should make sure that we use the actual internal diam eter of the pipe, which may be different than the nominal diameter. For example, the internal diameter of a steel pipe whose nominal diameter is 1 in is 1.049 in (Table 8–3).
Types of Fluid Flow Problems In the design and analysis of piping systems that involve the use of the Moody chart (or the Colebrook equation), we usually encounter three types of problems (the fluid and the roughness of the pipe are assumed to be specified in all cases) (Fig. 8–31): TABLE 8–3 Standard sizes for Schedule 40 steel pipes Nominal Size, in Actual Inside Diameter, in 1 8 1 4 3 8 1 2 3 4 1 11 2 2 21 2 3 5 10 0.269 0.364 0.493 0.622 0.824 1.049 1.610 2.067 2.469 3.068 5.047 10.02 FIGURE 8–30 At very large Reynolds numbers, the friction factor curves on the Moody chart are nearly horizontal, and thus the friction factors are independent of the Reynolds number. See Fig. A–12 for a full-page, more detailed Moody chart.
103 104 105 106 107 108 Re ε/D = 0.001 0.1 0.01 0.001 ƒ Transitional Laminar Fully rough turbulent flow (ƒ levels off) ε/D = 0.01 ε/D = 0.0001 ε/D = 0 Smooth turbulent L, D, V Problem type 1 L, ΔP, V L, D, ΔP ΔP (or hL ) D V 2 3 Given Find ⋅ ⋅ ⋅ FIGURE 8–31 The three types of problems encountered in pipe flow.
cen96537_ch08_351-442.indd 372 14/01/17 2:58 pm 373 CHAPTER 8 1. Determining the pressure drop (or head loss) when the pipe length and diameter are given for a specified flow rate (or velocity) 2. Determining the flow rate when the pipe length and diameter are given for a specified pressure drop (or head loss) 3. Determining the pipe diameter when the pipe length and flow rate are given for a specified pressure drop (or head loss) Problems of the first type are straightforward and can be solved directly by using the Moody chart. Problems of the second type and third type are commonly encountered in engineering design (in the selection of pipe diam eter, for example, that minimizes the sum of the construction and pumping costs), but the use of the Moody chart with such problems requires an itera tive approach—an equation solver is recommended. In problems of the second type, the diameter is given but the flow rate is unknown. A good guess for the friction factor in that case is obtained from the completely turbulent flow region for the given roughness. This is true for large Reynolds numbers, which is often the case in practice. Once the flow rate is obtained, the friction factor is corrected using the Moody chart or the Colebrook equation, and the process is repeated until the solution converges. (Typically only a few iterations are required for convergence to three or four digits of precision.) In problems of the third type, the diameter is not known and thus the Reynolds number and the relative roughness cannot be calculated. Therefore, we start calculations by assuming a pipe diameter. The pressure drop calculated for the assumed diameter is then compared to the specified pressure drop, and calculations are repeated with another pipe diameter in an iterative fashion until convergence. To avoid tedious iterations in head loss, flow rate, and diameter calcula tions, Swamee and Jain (1976) proposed the following explicit relations that are accurate to within 2 percent of the Moody chart: hL = 1.07 V · 2L gD5{ln[ 𝜀 3.7D + 4.62( 𝜈D V · ) 0.9 ] } −2 10−6 < 𝜀 /D < 10−2 3000 < Re < 3 × 108 (8–51) V · = −0.965( gD5hL L ) 0.5 ln[ 𝜀 3.7D + ( 3.17v 2L gD3hL ) 0.5 ] Re > 2000 (8–52) D = 0.66[𝜀 1.25( LV · 2 ghL ) 4.75 + 𝜈V · 9.4( L ghL) 5.2 ] 0.04 10−6 < 𝜀 /D < 10−2 5000 < Re < 3 × 108 (8–53) Note that all quantities are dimensional and the units simplify to the desired unit (for example, to m or ft in the last relation) when consistent units are used. Noting that the Moody chart is accurate to within 15 per cent of experimental data, we should have no reservation in using these approximate relations in the design of piping systems. The Colebrook equation is implicit in f, and thus the determination of the friction factor requires iteration. An approximate explicit relation for f was given by S. E. Haaland in 1983 as 1 √f ≅−1.8 log[ 6.9 Re + ( 𝜀 /D 3.7 ) 1.11 ] (8–54) cen96537_ch08_351-442.indd 373 14/01/17 2:58 pm 374 internal flow The results obtained from this relation are within 2 percent of those obtained from the Colebrook equation. If more accurate results are desired, Eq. 8–54 can be used as a good first guess in a Newton iteration when using a programmable calculator or a spreadsheet to solve for f with Eq. 8–50. We routinely use the Colebrook equation to calculate the friction factor f for fully developed turbulent pipe flow. Indeed, the Moody chart is created using the Colebrook equation. However, in addition to being implicit, the Colebrook equation is valid only for turbulent pipe flow (when the flow is laminar, f = 64/Re). Thus, we need to verify that the Reynolds number is in the turbulent range. An equation was generated by Churchill (1997) that is not only explicit, but is also useful for any Re and any roughness, even for laminar flow, and even in the fuzzy transitional region between laminar and turbulent flow. The Churchill equation is f = 8[( 8 Re) 12 + (A + B)−1.5 ] 1 12 (8–55) where A = {−2.457 ln[( 7 Re) 0.9 + 0.27 𝜀 D] } 16 and B = ( 37,530 Re ) 16 The difference between the Colebrook and Churchill equations is less than one percent. Because it is explicit and valid over the entire range of Reynolds numbers and roughnesses, it is recommended that the Churchill equation be used for determination of friction factor f.
150 m D 0.35 m3/s air FIGURE 8–32 Schematic for Example 8–3.
EXAMPLE 8–3 Determining the Diameter of an Air Duct Heated air at 1 atm and 35°C is to be transported in a 150-m-long circular plastic duct at a rate of 0.35 m3/s (Fig. 8–32). If the head loss in the pipe is not to exceed 20 m, determine the minimum diameter of the duct.
SOLUTION The flow rate and the head loss in an air duct are given. The diam eter of the duct is to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The entrance effects are negligible, and thus the flow is fully developed. 3 The duct involves no components such as bends, valves, and connectors. 4 Air is an ideal gas. 5 The duct is smooth since it is made of plastic. 6 The flow is turbulent (to be verified).
Properties The density, dynamic viscosity, and kinematic viscosity of air at 35°C are 𝜌 = 1.145 kg/m3, 𝜇 = 1.895 × 10−5 kg/m·s, and 𝜈 = 1.655 × 10−5 m2/s.
Analysis This is a problem of the third type since it involves the determination of diameter for specified flow rate and head loss. We can solve this problem using three different approaches: (1) an iterative approach by assuming a pipe diameter, calculating the head loss, comparing the result to the specified head loss, and repeating calculations until the calculated head loss matches the specified value; (2) writing all the relevant equations (leaving the diameter as an unknown) and solving them simultaneously using an equation solver; and (3) using the third Swamee–Jain formula. We will demonstrate the use of the last two approaches.
cen96537_ch08_351-442.indd 374 14/01/17 2:58 pm 375 CHAPTER 8 200 ft 2 in 0.2 ft3/s water FIGURE 8–33 Schematic for Example 8–4. The average velocity, the Reynolds number, the friction factor, and the head loss relations are expressed as (D is in m, V is in m/s, and Re and f are dimen sionless) V = V · Ac = V · 𝜋D2/4 = 0.35 m3/s 𝜋D2/4 Re = VD 𝜈 = VD 1.655 × 10−5 m2/s 1 √f = −2.0 log( 𝜀 /D 3.7 + 2.51 Re√f) = −2.0 log( 2.51 Re√f) hL = f L D V 2 2g → 20 m = f 150 m D V 2 2(9.81 m/s2) The roughness is approximately zero for a plastic pipe (Table 8–2). Therefore, this is a set of four equations and four unknowns, and solving them with an equation solver gives D = 0.267 m, f = 0.0180, V = 6.24 m/s, and Re = 100,800 Therefore, the diameter of the duct should be more than 26.7 cm if the head loss is not to exceed 20 m. Note that Re > 4000, and thus the turbulent flow assumption is verified. The diameter can also be determined directly from the third Swamee–Jain formula to be D = 0.66[𝜀 1.25( LV · 2 ghL ) 4.75 + 𝜈V · 9.4( L ghL) 5.2 ] 0.04 = 0.66[0 + (1.655 × 10−5 m2/s)(0.35 m3/s)9.4( 150 m (9.81 m/s2)(20 m)) 5.2 ] 0.04 = 0.271 m Discussion Note that the difference between the two results is less than 2 percent. Therefore, the simple Swamee–Jain relation can be used with confidence. Finally, the first (iterative) approach requires an initial guess for D. If we use the Swamee–Jain result as our initial guess, the diameter converges to D = 0.267 m in short order.
EXAMPLE 8–4 Determining the Pressure Drop in a Water Pipe Water at 60°F (𝜌 = 62.36 lbm/ft3 and 𝜇 = 7.536 × 10−4 lbm/ft·s) is flowing steadily in a 2-in-diameter horizontal pipe made of stainless steel at a rate of 0.2 ft3/s (Fig. 8–33). Determine the pressure drop, the head loss, and the required pumping power input for flow over a 200-ft-long section of the pipe.
SOLUTION The flow rate through a specified water pipe is given. The pressure drop, the head loss, and the pumping power requirements are to be determined.
cen96537_ch08_351-442.indd 375 14/01/17 2:58 pm 376 internal flow Assumptions 1 The flow is steady and incompressible. 2 The entrance effects are negligible, and thus the flow is fully developed. 3 The pipe involves no compo nents such as bends, valves, and connectors. 4 The piping section involves no work devices such as a pump or a turbine.
Properties The density and dynamic viscosity of water are given to be 𝜌 = 62.36 lbm/ft3 and 𝜇 = 7.536 × 10−4 lbm/ft·s, respectively.
Analysis We recognize this as a problem of the first type, since flow rate, pipe length, and pipe diameter are known. First we calculate the average velocity and the Reynolds number to determine the flow regime: V = V · Ac = V · 𝜋D2/4 = 0.2 ft3/s 𝜋(2/12 ft)2/4 = 9.17 ft/s Re = ρV D 𝜇 = (62.36 lbm/ft3)(9.17 ft/s)(2/12 ft) 7.536 × 10−4 lbm/ft·s = 126,400 Since Re is greater than 4000, the flow is turbulent. The relative roughness of the pipe is estimated using Table 8–3 𝜀 /D = 0.000007 ft 2/12 ft = 0.000042 The friction factor corresponding to this relative roughness and Reynolds number is determined from the Moody chart. To avoid any reading error, we determine f from the Colebrook equation on which the Moody chart is based: 1 √f = −2.0 log( 𝜀 /D 3.7 + 2.51 Re√f)→1 √f = −2.0 log ( 0.000042 3.7 + 2.51 126,400√f) Using an equation solver or an iterative scheme, the friction factor is determined to be f = 0.0174. Then the pressure drop (which is equivalent to pressure loss in this case), head loss, and the required power input become ΔP = ΔPL = f L D ρV 2 2 = 0.0174 200 ft 2/12 ft (62.36 lbm/ft3)(9.17 ft/s)2 2 ( 1 lbf 32.2 lbm·ft/s2) = 1700 lbf/ft2 = 11.8 psi hL = ΔPL ρg = f L D V 2 2g = 0.0174 200 ft 2/12 ft (9.17 ft/s)2 2(32.2 ft/s2) = 27.3 ft W .
pump = V .
ΔP = (0.2 ft3/s)(1700 lbf/ft2)( 1 W 0.737 lbf·ft/s) = 461 W Therefore, power input in the amount of 461 W is needed to overcome the frictional losses in the pipe.
Discussion It is common practice to write our final answers to three significant digits, even though we know that the results are accurate to at most two significant digits because of inherent inaccuracies in the Colebrook equation, as discussed previ ously. The friction factor could also be determined easily from the explicit Haaland relation (Eq. 8–54). It would give f = 0.0172, which is sufficiently close to 0.0174. The Churchill equation (Eq. 8–55) gives f = 0.0173, which is also very close to 0.0174, but does not require iteration. Also, the friction factor corresponding to 𝜀 = 0 in this case is 0.0171, which indicates that this stainless-steel pipe can be approxi mated as smooth with minimal error.
cen96537_ch08_351-442.indd 376 14/01/17 2:58 pm 377 CHAPTER 8 EXAMPLE 8–5 Determining the Flow Rate of Air in a Duct Reconsider Example 8–3. Now the duct length is doubled while its diameter is maintained constant (Fig. 8–34). If the total head loss is to remain constant, determine the drop in the flow rate through the duct.
SOLUTION The diameter and the head loss in an air duct are given. The drop in the flow rate is to be determined.
Analysis This is a problem of the second type since it involves the determina tion of the flow rate for a specified pipe diameter and head loss. The solution involves an iterative approach since the flow rate (and thus the flow velocity) is not known. The average velocity, Reynolds number, friction factor, and the head loss relations are expressed as (D is in m, V is in m/s, and Re and f are dimensionless) V = V · Ac = V · 𝜋D2/4 → V = V · 𝜋(0.267 m)2/4 Re = VD 𝜈 → Re = V(0.267 m) 1.655 × 10−5 m2/s 1 √f = −2.0 log( 𝜀 /D 3.7 + 2.51 Re√f) → 1 √f = −2.0 log( 2.51 Re√f) hL = f L D V 2 2g → 20 m = f 300 m 0.267 m V 2 2(9.81 m/s2) This is a set of four equations in four unknowns and solving them with an equation solver gives V · = 0.24 m3/s, f = 0.0195, V = 4.23 m/s, and Re = 68,300 Then the drop in the flow rate becomes V .
drop = V .
old −V · new = 0.35 −0.24 = 0.11 m3/s (a drop of 31 percent) Therefore, for a specified head loss (or available head or fan pumping power), the flow rate drops by about 31 percent from 0.35 to 0.24 m3/s when the duct length doubles.
Alternative Solution If a computer is not available (as in an exam situation), another option is to set up a manual iteration loop. We have found that the best con vergence is usually realized by first guessing the friction factor f, and then solving for the velocity V. The equation for V as a function of f is Average velocity through the pipe: V = √ 2ghL f L / D Once V is calculated, the Reynolds number can be calculated, from which a cor rected friction factor is obtained from the Moody chart or the Colebrook equation. We repeat the calculations with the corrected value of f until convergence. We guess f = 0.04 for illustration: Iteration f (guess) V, m/s Re Corrected f 1 0.04 2.955 4.724 × 104 0.0212 2 0.0212 4.059 6.489 × 104 0.01973 3 0.01973 4.207 6.727 × 104 0.01957 4 0.01957 4.224 6.754 × 104 0.01956 5 0.01956 4.225 6.756 × 104 0.01956 FIGURE 8–34 Schematic for Example 8–5.
300 m D cen96537_ch08_351-442.indd 377 14/01/17 2:58 pm 378 internal flow Notice that the iteration has converged to three digits in only three iterations and to four digits in only four iterations. The final results are identical to those obtained with the equation solver, yet do not require a computer.
Discussion The new flow rate can also be determined directly from the second Swamee–Jain formula to be V · = −0.965( gD5hL L ) 0.5 ln[ 𝜀 3.7D + ( 3.17𝜈2L gD3hL ) 0.5 ] = −0.965( (9.81 m/s2)(0.267 m)5(20 m) 300 m ) 0.5 × ln[0 + ( 3.17(1.655 × 10−5 m2/s)2(300 m) (9.81 m/s2)(0.267 m)3(20 m) ) 0.5 ] = 0.24 m3/s Note that the result from the Swamee–Jain relation is the same (to two significant digits) as that obtained with the Colebrook equation using the equation solver or using our manual iteration technique. Therefore, the simple Swamee–Jain relation can be used with confidence.
EXAMPLE 8–6 Turbulent Draining from a Pool Reconsider the draining pool example, Example 8–1. The flow rate was very low, so the pool owner uses a larger diameter hose (Fig. 8–35). The new hose has inner diameter D = 2.00 cm and average roughness height 𝜀 = 0.0020 cm. All other parameters remain the same as the previous problem, including the length of the hose. Calculate the volume flow rate in liters per minute (LPM) at the start of draining.
SOLUTION Water is drained from a pool. The volume flow rate is to be determined.
Assumptions 1 The flow is incompressible and quasi-steady. 2 Entrance effects are negligible since the hose is so long; the flow is fully developed. 3 Any other losses such as elbows, etc. in the hose are negligible.
Properties The properties of water are 𝜌 = 998 kg/m3 and 𝜇 = 0.001002 kg/m ⋅ s. Analysis We choose the same (wise) control volume as in Example 8–1. The equations and analysis are in fact identical to the previous problem except that friction factor f is no longer 64/Re since we suspect that the flow will be turbu lent since the hose diameter is larger. We must use instead either the Colebrook or Churchill equation to obtain f. Following the previous example, the energy equation reduces to 𝛼2 V 2 2g = H −hL (1) For fully developed turbulent pipe flow at the outlet of the hose, 𝛼2 = 1.05. For turbulent flow, we cannot obtain a simple quadratic equation for average veloc ity V. Instead, we must solve Eq. (1) simultaneously with either the Colebrook or Churchill equation and the other equations defining Re and hL from Example 8–1. This can be done “by hand” in similar fashion as Example 8–5 or with an equation Control volume 2 1 D H FIGURE 8–35 Schematic for Example 8–6.
cen96537_ch08_351-442.indd 378 14/01/17 2:58 pm 379 CHAPTER 8 8–6 ■ MINOR LOSSES The fluid in a typical piping system passes through various fittings, valves, bends, elbows, tees, inlets, exits, expansions, and contractions in addition to the straight sections of piping. These components interrupt the smooth flow of the fluid and cause additional losses because of the flow separation and mixing they induce. In a typical system with long pipes, these losses are minor com pared to the head loss in the straight sections (the major losses) and are called minor losses. Although this is generally true, in some cases the minor losses may be greater than the major losses. This is the case, for example, in systems with several turns and valves in a short distance. The head loss introduced by a completely open valve, for example, may be negligible. But a partially closed valve may cause the largest head loss in the system, as evidenced by the drop in the flow rate. Flow through valves and fittings is very complex, and a theo retical analysis is generally not plausible. Therefore, minor losses are deter mined experimentally, usually by the manufacturers of the components. Minor losses are usually expressed in terms of the loss coefficient KL (also called the resistance coefficient), defined as (Fig. 8–36) Loss coefficient: KL = hL V 2/(2g) (8–56) where hL is the additional irreversible head loss in the piping system caused by insertion of the component, and is defined as hL = ΔPL/𝜌g. For example, imagine replacing the valve in Fig. 8–36 with a section of constant diameter pipe from location 1 to location 2. ΔPL is defined as the pressure drop from 1 to 2 for the case with the valve, (P1 − P2)valve, minus the pressure drop that would occur in the imaginary straight pipe section from 1 to 2 without the valve, (P1 − P2)pipe at the same flow rate. While the majority of the irreversible head loss occurs locally near the valve, some of it occurs down stream of the valve due to induced swirling turbulent eddies that are pro duced in the valve and continue downstream. These eddies “waste” mechanical energy because they are ultimately dissipated into heat while the flow in the downstream section of pipe eventually returns to fully developed conditions. solver. We choose the latter and calculate V = 0.6536 m/s using the Churchill equa tion. Finally, we calculate the volume flow rate, V .
= VAc = V 𝜋D2 4 = (0.6536 m/s) 𝜋(0.020 m)2 4 ( 1000 L m3 ) ( 60 s min) = 12.32 L min which we round to our final answer: V . ≈ 12.3 LPM.
We check the Reynolds number, Re = ρVD 𝜇 = (998 kg/m3)(0.6536 m/s)(0.020 m) 0.001002 kg/m · s = 13,020 Since Re > 4000, we verify that the flow is indeed turbulent.
Discussion If we use the Colebrook equation, the volume flow rate is 12.38 LPM, only about 0.5 percent different from the Churchill result. Compared to Example 8–1, the flow rate is significantly higher and the pool would drain much faster with the larger diameter hose, as expected.
(P 1 – P2)valve 1 2 ΔP L = (P 1 – P2)valve – (P1 – P2)pipe V 1 2 V (P1 – P2)pipe Pipe section without valve: Pipe section with valve: FIGURE 8–36 For a constant-diameter section of a pipe with a minor loss component, the loss coefficient of the component (such as the gate valve shown) is determined by measuring the additional pressure loss it causes and dividing it by the dynamic pressure in the pipe.
cen96537_ch08_351-442.indd 379 14/01/17 2:58 pm 380 internal flow When measuring minor losses in some minor loss components, such as elbows, for example, location 2 must be considerably far downstream (tens of pipe diameters) in order to fully account for the additional irreversible losses due to these decaying eddies. When the pipe diameter downstream of the component changes, deter mination of the minor loss is even more complicated. In all cases, how ever, it is based on the additional irreversible loss of mechanical energy that would otherwise not exist if the minor loss component were not there. For simplicity, you may think of the minor loss as occurring locally across the minor loss component, but keep in mind that the component influences the flow for several pipe diameters downstream. By the way, this is the reason why most flow meter manufacturers recommend installing their flow meter at least 10 to 20 pipe diameters downstream of any elbows or valves—this allows the swirling turbulent eddies generated by the elbow or valve to largely disappear and the velocity profile to become fully developed before entering the flow meter. (Most flow meters are calibrated with a fully devel oped velocity profile at the flow meter inlet, and yield the best accuracy when such conditions also exist in the actual application.) When the inlet diameter equals the outlet diameter, the loss coefficient of a component can also be determined by measuring the pressure loss across the component and dividing it by the dynamic pressure, KL = ΔPL/(1 2𝜌V2). When the loss coefficient for a component is available, the head loss for that component is determined from Minor loss: hL = KL V 2 2g (8–57) The loss coefficient, in general, depends on the geometry of the component and the Reynolds number, just like the friction factor. However, it is usually assumed to be independent of the Reynolds number. This is a reasonable approximation since most flows in practice have large Reynolds numbers and the loss coefficients (including the friction factor) tend to be independ ent of the Reynolds number at large Reynolds numbers. Minor losses are also expressed in terms of the equivalent length Lequiv, defined as (Fig. 8–37) Equivalent length: hL = KL V 2 2g = f Lequiv D V 2 2g → Lequiv = D f KL (8–58) where f is the friction factor and D is the diameter of the pipe that contains the component. The head loss caused by the component is equivalent to the head loss caused by a section of the pipe whose length is Lequiv. Therefore, the contribution of a component to the head loss is accounted for by simply adding Lequiv to the total pipe length. Both approaches are used in practice, but the use of loss coefficients is more common. Therefore, we also use that approach in this book. Once all the loss coefficients are available, the total head loss in a piping system is determined from Total head loss (general): hL, total = hL, major + hL, minor = ∑ i fi Li Di V 2 i 2g + ∑ j KL, j V 2 j 2g (8–59) FIGURE 8–37 The head loss caused by a component (such as the angle valve shown) is equivalent to the head loss caused by a section of the pipe whose length is the equivalent length.
ΔP = P1 – P2 = P3 – P4 Lequiv D 3 4 1 2 D cen96537_ch08_351-442.indd 380 14/01/17 2:58 pm 381 CHAPTER 8 where i represents each pipe section with constant diameter and j represents each component that causes a minor loss. If the entire piping system being analyzed has a constant diameter, Eq. 8–59 reduces to Total head loss (D = constant): hL, total = (f L D + ∑KL) V 2 2g (8–60) where V is the average flow velocity through the entire system (note that V = constant since D = constant). Representative loss coefficients KL are given in Table 8–4 for inlets, exits, bends, sudden and gradual area changes, and valves. There is considerable uncertainty in these values since the loss coefficients, in general, vary with the pipe diameter, the surface roughness, the Reynolds number, and the details of the design. The loss coefficients of two seemingly identical valves by two different manufacturers, for example, can differ by a factor of 2 or more. Therefore, the particular manufacturer’s data should be consulted in the final design of piping systems rather than relying on the representative values in handbooks. The head loss at the inlet of a pipe is a strong function of geometry. It is almost negligible for well-rounded inlets (KL = 0.03 for r/D > 0.2), but increases to about 0.50 for sharp-edged inlets (Fig. 8–38). That is, a sharp-edged inlet causes half of the velocity head to be lost as the fluid enters the pipe. This is because the fluid cannot make sharp 90° turns easily, especially at high velocities. As a result, the flow separates at the corners, and the flow is constricted into the vena contracta region formed in the midsection of the pipe (Fig. 8–39). Therefore, a sharp-edged inlet acts like a flow constric tion. The velocity increases in the vena contracta region (and the pressure decreases) because of the reduced effective flow area and then decreases as the flow fills the entire cross section of the pipe. There would be negligible loss if the pressure were increased in accordance with Bernoulli’s equation (the velocity head would simply be converted into pressure head). However, this deceleration process is far from ideal and the viscous dissipation caused by intense mixing and the turbulent eddies converts part of the kinetic energy into frictional heating, as evidenced by a slight rise in fluid temperature. The end result is a drop in velocity without much pressure recovery, and the inlet loss is a measure of this irreversible pressure drop. Even slight rounding of the edges can result in significant reduction of KL, as shown in Fig. 8–40. The loss coefficient rises sharply (to about KL = 0.8) when the pipe protrudes into the reservoir since some fluid near the edge in this case is forced to make a 180° turn. The loss coefficient for a submerged pipe exit is often listed in handbooks as KL = 1. More precisely, however, KL is equal to the kinetic energy correction factor 𝛼 at the exit of the pipe. Although 𝛼 is indeed close to 1 for fully developed turbulent pipe flow, it is equal to 2 for fully devel oped laminar pipe flow. To avoid possible errors when analyzing laminar pipe flow, then, it is best to always set KL = 𝛼 at a submerged pipe exit. At any such exit, whether laminar or turbulent, the fluid leaving the pipe loses all of its kinetic energy as it mixes with the reservoir fluid and eventually comes to rest through the irreversible action of viscosity. This is true regard less of the shape of the exit (Table 8–4 and Fig. 8–41). Therefore, there is no advantage to rounding off the sharp edges of pipe exits.
Well-rounded inlet KL = 0.03 D r Sharp-edged inlet KL = 0.50 Recirculating flow Vena contracta FIGURE 8–38 The head loss at the inlet of a pipe is almost negligible for well-rounded inlets (KL = 0.03 for r/D > 0.2) but increases to about 0.50 for sharp-edged inlets.
cen96537_ch08_351-442.indd 381 14/01/17 2:58 pm 382 internal flow TABLE 8–4 Loss coefficients KL of various pipe components for turbulent flow (for use in the relation hL = KLV 2/(2g), where V is the average velocity in the pipe that contains the component) Pipe Inlet Reentrant: KL = 0.80 (t < < D and I ≈ 0.1D) D V l t Sharp-edged: KL = 0.50 D V Well-rounded (r /D > 0.2): KL = 0.03 Slightly rounded (r /D = 0.1): KL = 0.12 (see Fig. 8–40) D V r Pipe Exit Reentrant: KL = 𝛼 V Sharp-edged: KL = 𝛼 V Rounded: KL = 𝛼 V Note: The kinetic energy correction factor is 𝛼 = 2 for fully developed laminar flow, and 𝛼 ≈ 1.05 for fully developed turbulent flow.
Sudden Expansion and Contraction (based on the velocity in the smaller-diameter pipe) Sudden expansion: KL = 𝛼(1 −d 2 D 2) 2 V d D Sudden contraction: See chart.
V d D Gradual Expansion and Contraction (based on the velocity in the smaller-diameter pipe) Expansion (for 𝜃 = 20°): Contraction: KL = 0.30 for d/D = 0.2 KL = 0.02 for 𝜃 = 30° KL = 0.25 for d/D = 0.4 KL = 0.04 for 𝜃 = 45° KL = 0.15 for d/D = 0.6 KL = 0.07 for 𝜃 = 60° KL = 0.10 for d/D = 0.8 0.6 0.4 0.2 00 0.2 0.4 0.6 0.8 1.0 KL d2/D2 KL for sudden contraction V d D θ V D d θ (continues) cen96537_ch08_351-442.indd 382 14/01/17 2:58 pm 383 CHAPTER 8 TABLE 8–4 (Continued) Bends and Branches 90° smooth bend: Flanged: KL = 0.3 Threaded: KL = 0.9 V 90° miter bend (without vanes): KL = 1.1 V 90° miter bend (with vanes): KL = 0.2 V 45° threaded elbow: KL = 0.4 V 45° 180° return bend: Flanged: KL = 0.2 Threaded: KL = 1.5 V Tee (branch flow): Flanged: KL = 1.0 Threaded: KL = 2.0 V Tee (line flow): Flanged: KL = 0.2 Threaded: KL = 0.9 V Threaded union: KL = 0.08 V Valves Globe valve, fully open: KL = 10 Gate valve, fully open: KL = 0.2 Angle valve, fully open: KL = 5 1 4 closed: KL = 0.3 Ball valve, fully open: KL = 0.05 1 2 closed: KL = 2.1 Swing check valve: KL = 2 3 4 closed: KL = 17 These are representative values for loss coefficients. Actual values strongly depend on the design and manufacture of the components and may differ from the given values considerably (especially for valves). Actual manufacturer’s data should be used in the final design.
FIGURE 8–39 Graphical representation of flow contraction and the associated head loss at a sharp-edged pipe inlet.
2 1 Head Pressure head converted to velocity head Remaining pressure head Remaining velocity head Lost velocity head Total head Pressure head P0 ρg P1 ρg P2 ρg 2g /2g KLV2/2g 0 Vena contracta Separated flow 1 2 1 2 2 V 2 1 V 2 cen96537_ch08_351-442.indd 383 14/01/17 2:58 pm 384 internal flow Piping systems often involve sudden or gradual expansion or contraction sections to accommodate changes in flow rates or properties such as density and velocity. The losses are usually much greater in the case of sudden expan sion and contraction (or wide-angle expansion) because of flow separation. By combining the equations of mass, momentum, and energy balance, the loss coefficient for the case of a sudden expansion is approximated as KL = 𝛼(1 − Asmall Alarge) 2 (sudden expansion) (8–61) where Asmall and Alarge are the cross-sectional areas of the small and large pipes, respectively. Note that KL = 0 when there is no area change (Asmall = Alarge) and KL = 𝛼 when a pipe discharges into a reservoir (Alarge ≫ Asmall). No such relation exists for a sudden contraction, and the KL values in that case must be read from a chart or table (e.g., Table 8–4). The losses due to expansions and contractions can be reduced significantly by installing conical gradual area changers (nozzles and diffusers) between the small and large pipes. The KL values for representative cases of gradual expansion and contraction are given in Table 8–4. Note that in head loss calculations, the velocity in the small pipe is to be used as the reference velocity in Eq. 8–57. Losses during expansion are usually much higher than the losses during contraction because of flow separation. Piping systems also involve changes in direction without a change in diameter, and such flow sections are called bends or elbows. The losses in these devices are due to flow separation (just like a car being thrown off the road when it enters a turn too fast) on the inner side and the swirling secondary flows that result. The losses during changes of direction can be minimized by making the turn “easy” on the fluid by using circular arcs (like 90° elbows) instead of sharp turns (like miter bends) (Fig. 8–42). But the use of sharp turns (and thus suffering a penalty in loss coefficient) may be necessary when the turning space is limited. In such cases, the losses can be minimized by utilizing properly placed guide vanes to help the flow turn in an orderly manner without being thrown off the course. The loss coeffi cients for some elbows and miter bends as well as tees are given in Table 8–4. These coefficients do not include the frictional losses along the pipe bend. Such losses should be calculated as in straight pipes (using the length of the centerline as the pipe length) and added to other losses.
0.05 0 0.10 0.15 0.20 0.5 0.4 0.3 0.2 0.1 0 r/D KL 0.25 D r FIGURE 8–40 The effect of rounding of a pipe inlet on the loss coefficient.
Data from ASHRAE Handbook of Fundamentals.
Mixing Entrained ambient fluid Submerged outlet FIGURE 8–41 All the kinetic energy of the flow is “lost” (turned into thermal energy) through friction as the jet decelerates and mixes with ambient fluid downstream of a submerged outlet.
Flanged elbow KL = 0.3 Sharp turn KL = 1.1 FIGURE 8–42 The losses during changes of direction can be minimized by making the turn “easy” on the fluid by using circular arcs instead of sharp turns.
cen96537_ch08_351-442.indd 384 14/01/17 2:59 pm 385 CHAPTER 8 Valves are commonly used in piping systems to control flow rates by sim ply altering the head loss until the desired flow rate is achieved. For valves it is desirable to have a very low loss coefficient when they are fully open, such as with a ball valve, so that they cause minimal head loss during full-load operation (Fig. 8–43b). Several different valve designs, each with its own advantages and disadvantages, are in common use today. The gate valve slides up and down like a gate, the globe valve (Fig. 8–43a) closes a hole placed in the valve, the angle valve is a globe valve with a 90° turn, and the check valve allows the fluid to flow only in one direction like a diode in an electric circuit. Table 8–4 lists the representative loss coefficients of the popular designs. Note that the loss coefficient increases drastically as a valve is closed. Also, the deviation in the loss coefficients for different manufacturers is greatest for valves because of their complex geometries.
FIGURE 8–43 (a) The large head loss in a partially closed globe valve is due to irreversible deceleration, flow separation, and mixing of high-velocity fluid coming from the narrow valve passage. (b) The head loss through a fully-open ball valve, on the other hand, is quite small. Photo by John M. Cimbala. V2 = V1 Vconstriction > V1 V1 V2 Constriction A globe valve (a) (b) 9 cm 6 cm Water 7 m/s 150 kPa 1 2 FIGURE 8–44 Schematic for Example 8–7.
EXAMPLE 8–7 Head Loss and Pressure Rise during Gradual Expansion A 6-cm-diameter horizontal water pipe expands gradually to a 9-cm-diameter pipe (Fig. 8–44). The walls of the expansion section are angled 10° from the axis. The average velocity and pressure of water before the expansion section are 7 m/s and 150 kPa, respectively. Determine the head loss in the expansion section and the pressure in the larger-diameter pipe.
SOLUTION A horizontal water pipe expands gradually into a larger-diameter pipe. The head loss and pressure after the expansion are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The flow at sections 1 and 2 is fully developed and turbulent with 𝛼1 = 𝛼2 ≅ 1.06.
Properties We take the density of water to be 𝜌 = 1000 kg/m3. The loss coefficient for a gradual expansion of total included angle 𝜃 = 20° and diameter ratio d/D = 6/9 is KL = 0.133 (by interpolation using Table 8–4).
Analysis Noting that the density of water remains constant, the downstream velocity of water is determined from conservation of mass to be m · 1 = m · 2 → ρV1A1 = ρV2A2 → V2 = A1 A2 V1 = D 2 1 D 2 2 V1 V2 = (0.06 m)2 (0.09 m)2 (7 m/s) = 3.11 m/s Then the irreversible head loss in the expansion section becomes hL = KL V 2 1 2g = (0.133) (7 m/s)2 2(9.81 m/s2) = 0.333 m Noting that z1 = z2 and there are no pumps or turbines involved, the energy equation for the expansion section is expressed in terms of heads as P1 ρg + 𝛼1 V 2 1 2g + z1 + hpump, u = P2 ρg + 𝛼2 V 2 2 2g + z2 + hturbine, e + hL or P1 ρg + 𝛼1 V 2 1 2g = P2 ρg + 𝛼2 V 2 2 2g + hL → 0 0 ⟶ cen96537_ch08_351-442.indd 385 14/01/17 2:59 pm 386 internal flow 8–7 ■ PIPING NETWORKS AND PUMP SELECTION Series and Parallel Pipes Most piping systems encountered in practice such as the water distribu tion systems in cities or commercial or residential establishments involve numerous parallel and series connections as well as several sources (supply of fluid into the system) and loads (discharges of fluid from the system) (Fig. 8–45). A piping project may involve the design of a new system or the expansion of an existing system. The engineering objective in such projects is to design a piping system that will reliably deliver the specified flow rates at specified pressures at minimum total (initial plus operating and mainte nance) cost. Once the layout of the system is prepared, the determination of the pipe diameters and the pressures throughout the system, while remaining within the budget constraints, typically requires solving the system repeat edly until the optimal solution is reached. Computer modeling and analysis of such systems make this tedious task a simple chore. Piping systems typically involve several pipes connected to each other in series and/or in parallel, as shown in Figs. 8–46 and 8–47. When the pipes are connected in series, the flow rate through the entire system remains con stant regardless of the diameters of the individual pipes in the system. This is a natural consequence of the conservation of mass principle for steady incompressible flow. The total head loss in this case is equal to the sum of the Solving for P2 and substituting, P2 = P1 + ρ{ 𝛼1V 2 1 −𝛼2V 2 2 2 −ghL} = 150 kPa + (1000 kg/m3) × { 1.06(7 m/s)2 −1.06(3.11 m/s)2 2 −(9.81 m/s2)(0.333 m)} × ( 1 kN 1000 kg·m/s2)( 1 kPa 1 kN/m2) = 168 kPa Therefore, despite the head (and pressure) loss, the pressure increases from 150 to 168 kPa after the expansion. This is due to the conversion of dynamic pressure to static pressure when the average flow velocity is decreased in the larger pipe.
Discussion It is common knowledge that higher pressure upstream is neces sary to cause flow, and it may come as a surprise to you that the downstream pressure has increased after the expansion, despite the loss. This is because the flow is driven by the sum of the three heads that comprise the total head (namely, pressure head, velocity head, and elevation head). During flow expansion, the higher velocity head upstream is converted to pressure head downstream, and this increase outweighs the nonrecoverable head loss. Also, you may be tempted to solve this problem using the Bernoulli equation. Such a solution would ignore the head loss (and the associated pressure loss) and result in an incorrect higher pressure for the fluid downstream.
FIGURE 8–45 A piping network in an industrial facility.
© 123RF A fA, LA, DA VA = VB ⋅ ⋅ fB, LB, DB B 1 2 hL, 1-2 = hL, A + hL, B + hL, sudden contraction FIGURE 8–46 For pipes in series, the flow rate is the same in each pipe, and the total head loss is the sum of the head losses in the individual pipes.
cen96537_ch08_351-442.indd 386 14/01/17 2:59 pm 387 CHAPTER 8 head losses in individual pipes in the system, including the minor losses. The expansion or contraction losses at connections are considered to belong to the smaller-diameter pipe since the expansion and contraction loss coefficients are defined on the basis of the average velocity in the smaller-diameter pipe. For a pipe that branches out into two (or more) parallel pipes and then rejoins at a junction downstream, the total flow rate is the sum of the flow rates in the individual pipes. The pressure drop (or head loss) in each indi vidual pipe connected in parallel must be the same since ΔP = PA − PB and the junction pressures PA and PB are the same for all the individual pipes. For a system of two parallel pipes 1 and 2 between junctions A and B with negligible minor losses, this is expressed as hL, 1 = hL, 2 → f1 L1 D1 V 2 1 2g = f2 L 2 D2 V 2 2 2g Then the ratio of the average velocities and the flow rates in the two parallel pipes become V1 V2 = ( f2 f1 L2 L1 D1 D2) 1/2 and V · 1 V · 2 = Ac, 1V1 Ac, 2V2 = D 2 1 D 2 2 ( f2 f1 L2 L1 D1 D2) 1/2 Therefore, the relative flow rates in parallel pipes are established from the requirement that the head loss in each pipe be the same. This result can be extended to any number of pipes connected in parallel. The result is also valid for pipes for which the minor losses are significant if the equivalent lengths for components that contribute to minor losses are added to the pipe length. Note that the flow rate in one of the parallel branches is proportional to its diameter to the power 5/2 and is inversely proportional to the square root of its length and friction factor. The analysis of piping networks, no matter how complex they are, is based on two simple principles: 1. Conservation of mass throughout the system must be satisfied. This is done by requiring the total flow into a junction to be equal to the total flow out of the junction for all junctions in the system. Also, the flow rate must remain constant in pipes connected in series regardless of the changes in diameters.
2. Pressure drop (and thus head loss) between two junctions must be the same for all paths between the two junctions. This is because pressure is a point function and it cannot have two values at a specified point. In practice this rule is used by requiring that the algebraic sum of head losses in a loop (for all loops) be equal to zero. (A head loss is taken to be positive for flow in the clockwise direction and negative for flow in the counterclockwise direction.) Therefore, the analysis of piping networks is very similar to the analysis of electric circuits (Kirchhoff’s laws), with flow rate corresponding to electric current and pressure corresponding to electric potential. However, the situ ation is much more complex here since, unlike the electrical resistance, the “flow resistance” is a highly nonlinear function. Therefore, the analysis of piping networks requires the simultaneous solution of a system of nonlinear equations, which requires software such as EES, Excel, Mathcad, Matlab, etc., or commercially available software designed specifically for such applications.
PA Branch 1 Branch 2 PB < PA A B hL, 1 = hL, 2 VA = V1 + V2 = VB ⋅ ⋅ ⋅ ⋅ f1, L1, D1 f2, L2, D2 FIGURE 8–47 For pipes in parallel, the head loss is the same in each pipe, and the total flow rate is the sum of the flow rates in individual pipes.
cen96537_ch08_351-442.indd 387 14/01/17 2:59 pm 388 internal flow Piping Systems with Pumps and Turbines When a piping system involves a pump and/or turbine, the steady-flow energy equation on a unit-mass basis is expressed as (see Section 5–6) P1 ρ + 𝛼1 V 2 1 2 + gz1 + wpump, u = P2 ρ + 𝛼2 V 2 2 2 + gz2 + wturbine, e + ghL (8–62) or in terms of heads as P1 ρg + 𝛼1 V 2 1 2g + z1 + hpump, u = P2 ρg + 𝛼2 V 2 2 2g + z2 + hturbine, e + hL (8–63) where hpump, u = wpump, u/g is the useful pump head delivered to the fluid, hturbine, e = wturbine, e/g is the turbine head extracted from the fluid, 𝛼 is the kinetic energy correction factor whose value is about 1.05 for most (turbulent) flows encountered in practice, and hL is the total head loss in the piping (including the minor losses if they are significant) between points 1 and 2. The pump head is zero if the piping system does not involve a pump or a fan, the turbine head is zero if the system does not involve a turbine, and both are zero if the system does not involve any mechanical work-producing or work-consuming devices. Many practical piping systems involve a pump to move a fluid from one reservoir to another. Taking points 1 and 2 to be at the free surfaces of the reservoirs (Fig. 8–48), the energy equation is solved for the required useful pump head, yielding hpump, u = (z2 −z1) + hL (8–64) since the velocities at free surfaces are negligible for large reservoirs and the pressures are at atmospheric pressure. Therefore, the useful pump head is equal to the elevation difference between the two reservoirs plus the head loss. If the head loss is negligible compared to z2 − z1, the useful pump head is equal to the elevation difference between the two reservoirs. In the case of z1 > z2 (the first reservoir being at a higher elevation than the second one) with no pump, the flow is driven by gravity at a flow rate that causes a head loss equal to the elevation difference. A similar argument can be given for the turbine head for a hydroelectric power plant by replacing hpump, u in Eq. 8–64 by −hturbine, e. Once the useful pump head is known, the mechanical power that needs to be delivered by the pump to the fluid and the electric power consumed by the motor of the pump for a specified flow rate are determined from W .
pump, shaft = ρV ·ghpump, u 𝜂pump and W .
elect = ρV ·ghpump, u 𝜂pump–motor (8–65) where 𝜂pump–motor is the efficiency of the pump–motor combination, which is the product of the pump and the motor efficiencies (Fig. 8–49). The pump– motor efficiency is defined as the ratio of the net mechanical energy deliv ered to the fluid by the pump to the electric energy consumed by the motor of the pump, and it typically ranges between 50 and 85 percent. The head loss of a piping system increases (usually quadratically) with the flow rate. A plot of required useful pump head hpump, u as a function of flow rate is called the system (or demand) curve. The head produced by a pump z1 z2 Pump Control volume boundary Wpump, u = ρVghpump, u ⋅ ⋅ 1 2 hpump, u = (z2 – z1) + hL FIGURE 8–48 When a pump moves a fluid from one reservoir to another, the useful pump head requirement is equal to the elevation difference between the two reservoirs plus the head loss.
FIGURE 8–49 The efficiency of the pump–motor combination is the product of the pump and the motor efficiencies.
© Alex LMX/Shutterstock RF Motor, ηmotor Pump, ηpump Liquid out Liquid in ηpump–motor = ηpumpηmotor cen96537_ch08_351-442.indd 388 14/01/17 2:59 pm 389 CHAPTER 8 is not a constant either. Both the pump head and the pump efficiency vary with the flow rate, and pump manufacturers supply this variation in tabular or graphical form, as shown in Fig. 8–50. These experimentally determined hpump, u and 𝜂pump, u versus V . curves are called characteristic (or supply or performance) curves. Note that the flow rate of a pump increases as the required head decreases. The intersection point of the pump head curve with the vertical axis typically represents the maximum head (called the shutoff head) the pump can provide, while the intersection point with the horizontal axis indicates the maximum flow rate (called the free delivery) that the pump can supply. The efficiency of a pump is highest at a certain combination of head and flow rate. Therefore, a pump that can supply the required head and flow rate is not necessarily a good choice for a piping system unless the efficiency of the pump at those conditions is sufficiently high. The pump installed in a piping system will operate at the point where the system curve and the characteristic curve intersect. This point of intersection is called the operating point, as shown in Fig. 8–47. The useful head produced by the pump at this point matches the head requirements of the system at that flow rate. Also, the efficiency of the pump during operation is the value corresponding to that flow rate.
40 30 20 10 0 80 60 40 20 0 100 1 2 3 Flow rate, m3/s 4 5 6 0 Head, m Pump efficiency, pump,% Operating point No pipe is attached to the pump (no load to maximize flow rate) System curve Pump exit is closed to produce maximum head (shutoff head) hpump, u Supply curve Free delivery η pump η FIGURE 8–50 Characteristic pump curves for centrifugal pumps, the system curve for a piping system, and the operating point.
EXAMPLE 8–8 Pumping Water through Two Parallel Pipes Water at 20°C is to be pumped from a reservoir (zA = 5 m) to another reservoir at a higher elevation (zB = 13 m) through two 36-m-long pipes connected in parallel, as shown in Fig. 8–51. The pipes are made of commercial steel, and the diameters of the two pipes are 4 and 8 cm. Water is to be pumped by a 70 percent efficient motor–pump combination that draws 8 kW of electric power during operation. The minor losses and the head loss in pipes that connect the parallel pipes to the two reservoirs are considered to be negligible. Determine the total flow rate between the reservoirs and the flow rate through each of the parallel pipes.
cen96537_ch08_351-442.indd 389 14/01/17 2:59 pm 390 internal flow 1 2 zA = 5 m L1 = 36 m D1 = 4 cm Control volume boundary A Pump zB = 13 m B D2 = 8 cm L2 = 36 m FIGURE 8–51 The piping system discussed in Example 8–8.
SOLUTION The pumping power input to a piping system with two parallel pipes is given. The flow rates are to be determined.
Assumptions 1 The flow is steady (since the reservoirs are large) and incom pressible. 2 The entrance effects are negligible, and thus the flow is fully developed. 3 The elevations of the reservoirs remain constant. 4 The minor losses and the head loss in pipes other than the parallel pipes are negligible. 5 Flows through both pipes are turbulent (to be verified).
Properties The density and dynamic viscosity of water at 20°C are 𝜌 = 998 kg/m3 and 𝜇 = 1.002 × 10−3 kg/m·s. The roughness of commercial steel pipe is 𝜀 = 0.000045 m (Table 8–3).
Analysis This problem cannot be solved directly since the velocities (or flow rates) in the pipes are not known. Therefore, we would normally use a trial-and-error approach here. However, equation solvers are widely available, and thus, we simply set up the equations to be solved by an equation solver. The useful head supplied by the pump to the fluid is determined from W .
elect = ρV ·ghpump, u 𝜂pump−motor → 8000 W = (998 kg/m3)V · (9.81 m/s2)hpump, u 0.70 (1) We choose points A and B at the free surfaces of the two reservoirs. Noting that the fluid at both points is open to the atmosphere (and thus PA = PB = Patm) and that the fluid velocities at both points are nearly zero (VA ≈ VB ≈ 0) since the reservoirs are large, the energy equation for a control volume between these two points simplifies to PA ρg + 𝛼A V 2 A 2g + zA + hpump, u = PB ρg + 𝛼B V 2 B 2g + zB + hL or hpump, u = (zB −zA) + hL or hpump, u = (13 m −5 m) + hL (2) where hL = hL, 1 = hL, 2 (3)(4) 0 ↗ 0 ↗ cen96537_ch08_351-442.indd 390 14/01/17 2:59 pm 391 CHAPTER 8 We designate the 4-cm-diameter pipe by 1 and the 8-cm-diameter pipe by 2. Equa tions for the average velocity, the Reynolds number, the friction factor, and the head loss in each pipe are V1 = V · 1 Ac, 1 = V · 1 𝜋D 2 1 /4 → V1 = V · 1 𝜋(0.04 m)2/4 (5) V2 = V · 2 Ac, 2 = V · 2 𝜋D 2 2 /4 → V2 = V · 2 𝜋(0.08 m)2/4 (6) Re1 = ρV1D1 𝜇 → Re1 = (998 kg/m3)V1(0.04 m) 1.002 × 10−3 kg/m·s (7) Re2 = ρV 2D2 𝜇 → Re2 = (998 kg/m3)V2(0.08 m) 1.002 × 10−3 kg/m·s (8) 1 √f1 = −2.0 log( 𝜀 /D1 3.7 + 2.51 Re1√f1) → 1 √f1 = −2.0 log( 0.000045 3.7 × 0.04 + 2.51 Re1√f1) (9) 1 √f2 = −2.0 log( 𝜀 /D2 3.7 + 2.51 Re2√f2) → 1 √f2 = −2.0 log( 0.000045 3.7 × 0.08 + 2.51 Re2√f2) (10) hL, 1 = f1 L1 D1 V 2 1 2g → hL, 1 = f1 36 m 0.04 m V 2 1 2(9.81 m/s2) (11) hL, 2 = f2 L2 D2 V 2 2 2g → hL, 2 = f2 36 m 0.08 m V 2 2 2(9.81 m/s2) (12) V · = V · 1 + V · 2 (13) This is a system of 13 equations in 13 unknowns, and their simultaneous solution by an equation solver gives V . = 0.0300 m3/s, V .
1 = 0.00415 m3/s, V .
2 = 0.0259 m3/s V1 = 3.30 m/s, V 2 = 5.15 m/s, hL = hL, 1 = hL, 2 = 11.1 m, hpump = 19.1 m Re1 = 131,600, Re2 = 410,000, f1 = 0.0221, f2 = 0.0182 Note that Re > 4000 for both pipes, and thus the assumption of turbulent flow is verified.
Discussion The two parallel pipes have the same length and roughness, but the diameter of the first pipe is half the diameter of the second one. Yet only 14 percent of the water flows through the first pipe. This shows the strong dependence of the flow rate on diameter. Also, it can be shown that if the free surfaces of the two res ervoirs were at the same elevation (and thus zA = zB), the flow rate would increase by 20 percent from 0.0300 to 0.0361 m3/s. Alternately, if the reservoirs were as given but the irreversible head losses were negligible, the flow rate would become 0.0715 m3/s (an increase of 138 percent).
cen96537_ch08_351-442.indd 391 14/01/17 2:59 pm 392 internal flow EXAMPLE 8–9 Gravity-Driven Water Flow in a Pipe Water at 10°C flows from a large reservoir to a smaller one through a 5-cm-diame ter cast iron piping system, as shown in Fig. 8–52. Determine the elevation z1 for a flow rate of 6 L/s.
SOLUTION The flow rate through a piping system connecting two reservoirs is given. The elevation of the source is to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The elevations of the reservoirs remain constant. 3 There are no pumps or turbines in the line.
Properties The density and dynamic viscosity of water at 10°C are 𝜌 = 999.7 kg/ m3 and 𝜇 = 1.307 × 10−3 kg/m·s. The roughness of cast iron pipe is 𝜀 = 0.00026 m (Table 8–3).
Analysis The piping system involves 89 m of piping, a sharp-edged entrance (KL = 0.5), two standard flanged elbows (KL = 0.3 each), a fully open gate valve (KL = 0.2), and a submerged exit (KL = 1.06). We choose points 1 and 2 at the free surfaces of the two reservoirs. Noting that the fluid at both points is open to the atmosphere (and thus P1 = P2 = Patm) and that the fluid velocities at both points are nearly zero (V1 ≈ V2 ≈ 0), the energy equation for a control volume between these two points simplifies to P1 ρg + 𝛼1 V 2 1 2g + z1 = P2 ρg + 𝛼2 V 2 2 2g + z2 + hL → z1 = z2 + hL where hL = hL, total = hL, major + hL, minor = (f L D + ∑KL) V 2 2g since the diameter of the piping system is constant. The average velocity in the pipe and the Reynolds number are V = V · Ac = V · 𝜋D2/4 = 0.006 m3/s 𝜋(0.05 m)2/4 = 3.06 m/s Re = ρVD 𝜇 = (999.7 kg/m3)(3.06 m/s)(0.05 m) 1.307 × 10−3 kg/m·s = 117,000 The flow is turbulent since Re > 4000. Noting that 𝜀/D = 0.00026/0.05 = 0.0052, the friction factor is determined from the Colebrook equation (or the Moody chart), 0 0 ↗ ↗ 1 z1 = ?
2 z2 = 4 m D = 5 cm 9 m 80 m Standard elbow, flanged, KL = 0.3 Gate valve, fully open KL = 0.2 Sharp-edged entrance, KL = 0.5 Control volume boundary Exit, KL = 1.06 FIGURE 8–52 The piping system discussed in Example 8–9.
cen96537_ch08_351-442.indd 392 14/01/17 2:59 pm 393 CHAPTER 8 1 √f = −2.0 log( 𝜀 /D 3.7 + 2.51 Re√f) → 1 √f = −2.0 log( 0.0052 3.7 + 2.51 117,000√f) It gives f = 0.0315. The sum of the loss coefficients is ∑KL = KL, entrance + 2KL, elbow + KL, valve + KL, exit = 0.5 + 2 × 0.3 + 0.2 + 1.06 = 2.36 Then the total head loss and the elevation of the source become hL = (f L D + ∑KL) V 2 2g =(0.0315 89 m 0.05 m + 2.36) (3.06 m/s)2 2(9.81 m/s2) = 27.9 m z1 = z2 + hL = 4 + 27.9 = 31.9 m Therefore, the free surface of the first reservoir must be 31.9 m above the ground level to ensure water flow between the two reservoirs at the specified rate.
Discussion Note that fL/D = 56.1 in this case, which is about 24 times the total minor loss coefficient. Therefore, ignoring the sources of minor losses in this case would result in about 4 percent error. It can be shown that at the same flow rate, the total head loss would be 35.9 m (instead of 27.9 m) if the valve were three-fourths closed, and it would drop to 24.8 m if the pipe between the two reservoirs were straight at the ground level (thus eliminating the elbows and the vertical section of the pipe). The head loss could be reduced further (from 24.8 to 24.6 m) by rounding the entrance. The head loss can be reduced significantly (from 27.9 to 16.0 m) by replacing the cast iron pipes by smooth pipes such as those made of plastic.
EXAMPLE 8–10 Effect of Flushing on Flow Rate from a Shower The bathroom plumbing of a building consists of 1.5-cm-diameter copper pipes with threaded connectors, as shown in Fig. 8–53. (a) If the gage pressure at the inlet of the system is 200 kPa during a shower and the toilet reservoir is full (no flow in that branch), determine the flow rate of water through the shower head. (b) Determine the effect of flushing of the toilet on the flow rate through the shower FIGURE 8–53 Schematic for Example 8–10.
5 m 4 m Toilet reservoir with float KL = 14 KL = 0.9 KL = 2 KL = 10 KL = 12 Shower head Globe valve, fully open KL = 10 Cold water 1 m 2 m 3 1 2 cen96537_ch08_351-442.indd 393 14/01/17 2:59 pm 394 internal flow head. Take the loss coefficients of the shower head and the reservoir to be 12 and 14, respectively.
SOLUTION The cold-water plumbing system of a bathroom is given. The flow rate through the shower and the effect of flushing the toilet on the flow rate are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The flow is turbulent and fully developed. 3 The reservoir is open to the atmosphere. 4 The velocity heads are negligible.
Properties The properties of water at 20°C are 𝜌 = 998 kg/m3, 𝜇 = 1.002 × 10−3 kg/m·s, and 𝜈 = 𝜇/𝜌 = 1.004 × 10−6 m2/s. The roughness of copper pipes is 𝜀 = 1.5 × 10−6 m.
Analysis This is a problem of the second type since it involves the determina tion of the flow rate for a specified pipe diameter and pressure drop. The solution involves an iterative approach since the flow rate (and thus the flow velocity) is not known.
(a) The piping system of the shower alone involves 11 m of piping, a tee with line flow (KL = 0.9), two standard elbows (KL = 0.9 each), a fully open globe valve (KL = 10), and a shower head (KL = 12). Therefore, ∑KL = 0.9 + 2 × 0.9 + 10 + 12 = 24.7. Noting that the shower head is open to the atmosphere, and the velocity heads are negligible, the energy equation for a control volume between points 1 and 2 simplifies to P1 ρg + 𝛼1 V 2 1 2g + z1 + hpump, u = P2 ρg + 𝛼2 V 2 2 2g + z2 + hturbine, e + hL → P1, gage ρg = (z2 −z1) + hL Therefore, the head loss is hL = 200,000 N/m2 (998 kg/m3)(9.81 m/s2) −2 m = 18.4 m Also, hL = (f L D + ∑KL) V 2 2g → 18.4 = (f 11 m 0.015 m + 24.7) V 2 2(9.81 m/s2) since the diameter of the piping system is constant. Equations for the average veloc ity in the pipe, the Reynolds number, and the friction factor are V = V · Ac = V · 𝜋D2/4 → V = V · 𝜋(0.015 m)2/4 Re = VD 𝜈 → Re = V(0.015 m) 1.004 × 10−6 m2/s 1 √f = −2.0 log( 𝜀 /D 3.7 + 2.51 Re√f) → 1 √f = −2.0 log( 1.5 × 10−6 m 3.7(0.015 m) + 2.51 Re√f) cen96537_ch08_351-442.indd 394 14/01/17 2:59 pm 395 CHAPTER 8 This is a set of four equations with four unknowns, and solving them with an equa tion solver gives V · = 0.00053 m3/s, f = 0.0218, V = 2.98 m/s, and Re = 44,550 Therefore, the flow rate of water through the shower head is 0.53 L/s.
(b) When the toilet is flushed, the float moves and opens the valve. The discharged water starts to refill the reservoir, resulting in parallel flow after the tee connection. The head loss and minor loss coefficients for the shower branch were determined in (a) to be hL, 2 = 18.4 m and ∑KL, 2 = 24.7, respectively. The corresponding quantities for the reservoir branch can be determined similarly to be hL, 3 = 200,000 N/m2 (998 kg/m3)(9.81 m/s2) −1 m = 19.4 m ∑KL, 3 = 2 + 10 + 0.9 + 14 = 26.9 The relevant equations in this case are V · 1 = V · 2 + V · 3 hL, 2 = f1 5 m 0.015 m V 2 1 2(9.81 m/s2) + (f2 6 m 0.015 m + 24.7) V 2 2 2(9.81 m/s2) = 18.4 hL, 3 = f1 5 m 0.015 m V 2 1 2(9.81 m/s2) + (f3 1 m 0.015 m + 26.9) V 2 3 2(9.81 m/s2) = 19.4 V1 = V · 1 𝜋(0.015 m)2/4 , V2 = V · 2 𝜋(0.015 m)2/4 , V3 = V · 3 𝜋(0.015 m)2/4 Re1 = V1(0.015 m) 1.004 × 10−6m2/s , Re2 = V2(0.015 m) 1.004 × 10−6m2/s , Re3 = V3(0.015 m) 1.004 × 10−6m2/s 1 √f1 = −2.0 log( 1.5 × 10−6 m 3.7(0.015 m) + 2.51 Re1√f1) 1 √f2 = −2.0 log( 1.5 × 10−6 m 3.7(0.015 m) + 2.51 Re2√f2) 1 √f3 = −2.0 log( 1.5 × 10−6 m 3.7(0.015 m) + 2.51 Re3√f3) Solving these 12 equations in 12 unknowns simultaneously using an equation solver, the flow rates are determined to be V · 1 = 0.00090 m3/s, V · 2 = 0.00042 m3/s, and V · 3 = 0.00048 m3/s Therefore, the flushing of the toilet reduces the flow rate of cold water through the shower by 21 percent from 0.53 to 0.42 L/s, causing the shower water to sud denly get very hot (Fig. 8–54).
FIGURE 8–54 Flow rate of cold water through a shower may be affected significantly by the flushing of a nearby toilet.
cen96537_ch08_351-442.indd 395 14/01/17 2:59 pm 396 internal flow 8–8 ■ FLOW RATE AND VELOCITY MEASUREMENT A major application area of fluid mechanics is the determination of the flow rate of fluids, and numerous devices have been developed over the years for the purpose of flow metering. Flowmeters range widely in their level of sophistication, size, cost, accuracy, versatility, capacity, pressure drop, and the operating principle. We give an overview of the meters commonly used to measure the flow rate of liquids and gases flowing through pipes or ducts. We limit our consideration to incompressible flow. Some flowmeters measure the flow rate directly by discharging and recharging a measuring chamber of known volume continuously and keep ing track of the number of discharges per unit time. But most flowmeters measure the flow rate indirectly—they measure the average velocity V or a quantity that is related to average velocity such as pressure and drag, and determine the volume flow rate V . from V · = VAc (8–66) where Ac is the cross-sectional area of flow. Therefore, measuring the flow rate is usually done by measuring flow velocity, and many flowmeters are simply velocimeters used for the purpose of metering flow. The velocity in a pipe varies from zero at the wall to a maximum at the center, and it is important to keep this in mind when taking velocity meas urements. For laminar flow, for example, the average velocity is half the centerline velocity. But this is not the case in turbulent flow, and it may be necessary to take the weighted average or an integral of several local veloc ity measurements to determine the average velocity. The flow rate measurement techniques range from very crude to very elegant. The flow rate of water through a garden hose, for example, can be measured simply by collecting the water in a bucket of known volume and dividing the amount collected by the collection time (Fig. 8–55). A crude way of estimating the flow velocity of a river is to drop a float on the river and measure the drift time between two specified locations. At the other extreme, some flowmeters use the propagation of sound in flowing flu ids while others use the electromotive force generated when a fluid passes through a magnetic field. In this section we discuss devices that are com monly used to measure velocity and flow rate, starting with the Pitot-static probe introduced in Chap. 5.
Pitot and Pitot-Static Probes Pitot probes (also called Pitot tubes) and Pitot-static probes, named after the French engineer Henri de Pitot (1695–1771), are widely used for flow speed measurement. A Pitot probe is just a tube with a pressure tap at the stagnation Discussion If the velocity heads were considered, the flow rate through the shower would be 0.43 instead of 0.42 L/s. Therefore, the assumption of negligible velocity heads is reasonable in this case. Note that a leak in a piping system would cause the same effect, and thus an unexplained drop in flow rate at an end point may signal a leak in the system.
FIGURE 8–55 A primitive (but fairly accurate) way of measuring the flow rate of water through a garden hose involves collecting water in a bucket and recording the collection time.
Nozzle Bucket Garden hose Stopwatch cen96537_ch08_351-442.indd 396 14/01/17 2:59 pm 397 CHAPTER 8 point that measures stagnation pressure, while a Pitot-static probe has both a stagnation pressure tap and several circumferential static pressure taps and it measures both stagnation and static pressures (Figs. 8–56 and 8–57). Pitot was the first person to measure velocity with the upstream pointed tube, while French engineer Henry Darcy (1803–1858) developed most of the features of the instruments we use today, including the use of small openings and the placement of the static tube on the same assembly. Therefore, it is more appropriate to call the Pitot-static probes Pitot–Darcy probes. The Pitot-static probe measures local velocity by measuring the pressure difference in conjunction with the Bernoulli equation. It consists of a slen der double-tube aligned with the flow and connected to a differential pres sure meter. The inner tube is fully open to flow at the nose, and thus it measures the stagnation pressure at that location (point 1). The outer tube is sealed at the nose, but it has holes on the side of the outer wall (point 2) and thus it measures the static pressure. For incompressible flow with suf ficiently high velocities (so that the frictional effects between points 1 and 2 are negligible), the Bernoulli equation is applicable and is expressed as P1 ρg + V 2 1 2g + z1 = P2 ρg + V 2 2 2g + z2 (8–67) Noting that z1 ≅z2 since the static pressure holes of the Pitot-static probe are arranged circumferentially around the tube and V1 = 0 because of the stagnation conditions, the flow velocity V = V2 becomes Pitot formula: V = √ 2(P1 −P2) ρ (8–68) which is known as the Pitot formula. Note that this velocity is theoretical and works best at high Reynolds number; the actual velocity is a bit smaller than this, and some experimentalists multiply by a velocity coefficient CV that ranges between about 0.970 and 0.999, increasing with Reynolds number. If the velocity is measured at a location where the local velocity is equal to the average flow velocity, the volume flow rate can be determined from V . = VAc. The Pitot-static probe is a simple, inexpensive, and highly reliable device since it has no moving parts (Fig. 8–58). It also causes very small pressure drop and usually does not disturb the flow appreciably. However, it is important that it be properly aligned with the flow (typically within ± 10o) to avoid significant errors that may be caused by misalignment. Also, the FIGURE 8–56 (a) A Pitot probe measures stagnation pressure at the nose of the probe, while (b) a Pitot-static probe measures both stagnation pressure and static pressure, from which the flow speed is calculated.
Stagnation pressure To stagnation pressure meter To stagnation pressure meter To static pressure meter Pitot-static probe Pitot probe (a) (b) V V Stagnation pressure Static pressure Wind tunnel wall Flexible tubing Differential pressure transducer or manometer to measure P1 – P2 P1 – P2 Flow Pitot-static probe Stagnation pressure, P 1 Static pressure, P2 FIGURE 8–57 Measuring flow velocity with a Pitot-static probe. (A manometer may be used in place of the differential pressure transducer.) FIGURE 8–58 Close-up of a Pitot-static probe, showing the stagnation pressure hole and two of the five static circumferential pressure holes. Photo by Po-Ya Abel Chuang. Used by permission.
cen96537_ch08_351-442.indd 397 14/01/17 2:59 pm 398 internal flow difference between the static and stagnation pressures (which is the dynamic pressure) is proportional to the density of the fluid and the square of the flow velocity. It is used to measure velocity in both liquids and gases. Not ing that gases have low densities, the flow velocity should be sufficiently high when the Pitot-static probe is used for gas flow such that a measurable dynamic pressure develops.
Obstruction Flowmeters: Orifice, Venturi, and Nozzle Meters Consider incompressible steady flow of a fluid in a horizontal pipe of diam eter D that is constricted to a flow area of diameter d, as shown in Fig. 8–59. The mass balance and the Bernoulli equations between a location before the constriction (point 1) and the location where constriction occurs (point 2) are written as Mass balance: V · = A1V1 = A2V2 → V1 = (A2/A1)V2 = (d/D)2V2 (8–69) Bernoulli equation (z1 = z2): P1 ρg + V 2 1 2g = P2 ρg + V 2 2 2g (8–70) Combining Eqs. 8–69 and 8–70 and solving for velocity V2 gives Obstruction (with no loss): V2 = √ 2(P1 −P2) ρ(1 −𝛽 4) (8–71) where 𝛽 = d/D is the diameter ratio. Once V2 is known, the flow rate can be determined from V . = A2V2 = (𝜋d 2/4)V2. This simple analysis shows that the flow rate through a pipe can be deter mined by constricting the flow and measuring the decrease in pressure due to the increase in velocity at the constriction site. Noting that the pressure drop between two points along the flow is measured easily by a differential pressure transducer or manometer, it appears that a simple flow rate meas urement device can be built by obstructing the flow. Flowmeters based on this principle are called obstruction flowmeters and are widely used to measure flow rates of gases and liquids. The velocity in Eq. 8–71 is obtained by assuming no loss, and thus it is the maximum velocity that can occur at the constriction site. In real ity, some pressure losses due to frictional effects are inevitable, and thus the actual velocity is less. Also, the fluid stream continues to contract past the obstruction, and the vena contracta area is less than the flow area of the obstruction. Both losses can be accounted for by incorporating a correction factor called the discharge coefficient Cd whose value (which is less than 1) is determined experimentally. Then the flow rate for obstruction flowmeters is expressed as Obstruction flowmeters: V · = A0Cd √ 2(P1 −P2) ρ(1 −𝛽 4) (8–72) where A0 = A2 = 𝜋d 2/4 is the cross-sectional area of the throat or orifice and 𝛽 = d/D is the ratio of throat diameter to pipe diameter. The value of Cd depends on both 𝛽 and the Reynolds number Re = V1D/𝜈, and charts and FIGURE 8–59 Flow through a constriction in a pipe.
1 2 D d Obstruction cen96537_ch08_351-442.indd 398 14/01/17 2:59 pm 399 CHAPTER 8 curve-fit correlations for Cd are available for various types of obstruction meters. Note that Cd is actually the product of the velocity coefficient CV and the contraction coefficient CC, i.e., Cd = CVCC, but usually only Cd is listed in manufacturers’ literature. Of the numerous types of obstruction meters available, those most widely used are orifice meters, flow nozzles, and Venturi meters (Fig. 8–60). For standardized geometries, the experimentally determined data for discharge coefficients are expressed as (Miller, 1997) Orifice meters: Cd = 0.5959 + 0.0312𝛽 2.1 −0.184𝛽 8 + 91.71𝛽 2.5 Re0.75 (8–73) Nozzle meters: Cd = 0.9975 −6.53𝛽 0.5 Re0.5 (8–74) These relations are valid for 0.25 < 𝛽 < 0.75 and 104 < Re < 107. Precise values of Cd depend on the particular design of the obstruction, and thus the manufacturer’s data should be consulted when available. Also, the Reynolds number depends on the flow velocity, which is not known a priori. There fore, the solution is iterative in nature when curve-fit correlations are used for Cd. For flows with high Reynolds numbers (Re > 30,000), the value of Cd can be taken to be 0.96 for flow nozzles and 0.61 for orifices. Owing to its streamlined design, the discharge coefficients of Venturi meters are very high, ranging between 0.95 and 0.99 (the higher values are for the higher Reynolds numbers) for most flows. In the absence of specific data, we can take Cd = 0.98 for Venturi meters. The orifice meter has the simplest design and it occupies minimal space as it consists of a plate with a hole in the middle, but there are considerable varia tions in design (Fig. 8–61). Some orifice meters are sharp-edged, while others are beveled or rounded. The sudden change in the flow area in orifice meters causes considerable swirl and thus significant head loss or permanent pressure loss, as shown in Fig. 8–62. In nozzle meters, the plate is replaced by a nozzle, and thus the flow in the nozzle is streamlined. As a result, the vena contracta is practically eliminated and the head loss is smaller. How ever, flow nozzle meters are more expensive than orifice meters. The Venturi meter, invented by the American engineer Clemens Her schel (1842–1930) and named by him after the Italian Giovanni Venturi FIGURE 8–60 Common types of obstruction meters.
D (c) Venturi meter D d d (b) Flow nozzle 21° 15° (a) Orifice meter D d FIGURE 8–61 An orifice meter and schematic showing its built-in pressure transducer and digital readout.
Courtesy of KOBOLD Instruments, Pittsburgh, PA. www.koboldusa.com. Used by permission.
Flow Housing Magnet Bellows Orifice P1 V1 V2 > V1 P1 > P2 P2 V2 Force cen96537_ch08_351-442.indd 399 14/01/17 2:59 pm 400 internal flow (1746–1822) for his pioneering work on conical flow sections, is the most accurate flowmeter in this group, but it is also the most expensive. Its grad ual contraction and expansion prevent flow separation and swirling, and it suffers only frictional losses on the inner wall surfaces. Venturi meters cause very low head losses, and thus, they should be pre ferred for applica tions that cannot allow large pressure drops. When an obstruction flowmeter is placed in a piping system, its net effect on the flow system is like that of a minor loss. The minor loss coefficient of the flowmeter is available from the manufacturer, and should be included when summing minor losses in the system. In general, orifice meters have the highest minor loss coefficients, while Venturi meters have the lowest. Note that the pressure drop P1 − P2 measured to calculate the flow rate is not the same as the total pressure loss caused by the obstruction flowmeter because of the locations of the pressure taps. Note the distinction between pressure drop and pressure loss (Fig. 8–62). Pressure drop is due to conver sion between potential and kinetic energy and is recoverable. Pressure loss, however, is caused by irreversible losses and is not recoverable. Finally, obstruction flowmeters are also used to measure compressible-gas flow rates, but an additional correction factor must be inserted into Eq. 8–72 to account for compressibility effects. In such cases, the equation is written for mass flow rate instead of volume flow rate, and the compressible correc tion factor is typically an empirically curve-fitted equation (like the one for Cd) and is available from the flowmeter manufacturer.
EXAMPLE 8–11 Measuring Flow Rate with an Orifice Meter The flow rate of methanol at 20°C (𝜌 = 788.4 kg/m3 and 𝜇 = 5.857 × 10−4 kg/m·s) through a 4-cm-diameter pipe is to be measured with a 3-cm-diameter orifice meter equipped with a mercury manometer across the orifice plate, as shown in Fig. 8–63. If the differential height of the manometer is 11 cm, determine the flow rate of methanol through the pipe and the average flow velocity.
SOLUTION The flow rate of methanol is to be measured with an orifice meter. For a given pressure drop across the orifice plate, the flow rate and the average flow velocity are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 Our first guess for the discharge coefficient of the orifice meter is Cd = 0.61.
Properties The density and dynamic viscosity of methanol are given to be 𝜌 = 788.4 kg/m3 and 𝜇 = 5.857 × 10−4 kg/m·s, respectively. We take the density of mercury to be 13,600 kg/m3.
Analysis The diameter ratio and the throat area of the orifice are 𝛽 = d D = 3 4 = 0.75 A0 = 𝜋d 2 4 = 𝜋(0.03 m)2 4 = 7.069 × 10−4 m2 The pressure drop across the orifice plate is ΔP = P1 −P2 = (ρHg −ρmet)gh FIGURE 8–63 Schematic for the orifice meter considered in Example 8–11.
1 2 11 cm Mercury manometer FIGURE 8–62 The variation of pressure along a flow section with an orifice meter as measured with piezometer tubes; the lost pressure and the pressure recovery are shown.
Pressure drop across orifice HGL Lost pressure Orifice meter P1 P2 P3 Recovered pressure cen96537_ch08_351-442.indd 400 14/01/17 2:59 pm 401 CHAPTER 8 Then the flow rate relation for obstruction meters becomes V · = A0Cd √ 2(P1 −P2) ρ(1 −𝛽 4) = A0Cd √ 2(ρHg −ρmet)gh ρmet(1 −𝛽 4) = A0Cd √ 2(ρHg/ρmet −1)gh 1 −𝛽 4 Substituting, the flow rate is determined to be V · = (7.069 × 10−4 m2)(0.61)√ 2(13,600/788.4 −1)(9.81 m/s2)(0.11 m) 1 −0.754 = 3.09 × 10−3 m3/s which is equivalent to 3.09 L/s. The average flow velocity in the pipe is determined by dividing the flow rate by the cross-sectional area of the pipe, V = V · Ac = V · 𝜋D2/4 = 3.09 × 10−3 m3/s 𝜋(0.04 m)2/4 = 2.46 m/s The Reynolds number of flow through the pipe is Re = ρVD 𝜇 = (788.4 kg/m3)(2.46 m/s)(0.04 m) 5.857 × 10−4 kg/m·s = 1.32 × 105 Substituting 𝛽 = 0.75 and Re = 1.32 × 105 into the orifice discharge coefficient relation Cd = 0.5959 + 0.0312𝛽 2.1 −0.184𝛽 8 + 91.71𝛽 2.5 Re0.75 gives Cd = 0.601, which differs from the original guessed value of 0.61. Using this refined value of Cd, the flow rate becomes 3.04 L/s, which differs from our original result by 1.6 percent. After a couple iterations, the final converged flow rate is 3.04 L/s, and the average velocity is 2.42 m/s (to three significant digits).
Discussion If the problem is solved using an equation solver it can be formulated using the curve-fit formula for Cd (which depends on the Reynolds number), and all equations can be solved simultaneously by letting the equation solver perform the iterations as necessary.
Positive Displacement Flowmeters When we buy gasoline for our car, we are interested in the total amount of gasoline that flows through the nozzle during the period we fill the tank rather than the flow rate of gasoline. Likewise, we care about the total amount of water or natural gas we use in our homes during a billing period. In these and many other applications, the quantity of interest is the total amount of mass or volume of a fluid that passes through a cross section of a pipe over a certain period of time rather than the instantaneous value of flow rate, and positive displacement flowmeters are well suited for such applications. There are numerous types of displacement meters, and they are based on continuous filling and discharging of the measuring chamber. They operate by trapping a certain amount of incoming fluid, displacing it to the discharge side of the meter, and counting the number of such discharge– recharge cycles to determine the total amount of fluid displaced.
cen96537_ch08_351-442.indd 401 14/01/17 2:59 pm 402 internal flow Figure 8–64 shows a positive displacement flowmeter with two rotating impellers driven by the flowing liquid. Each impeller has three gear lobes, and a pulsed output signal is generated each time a lobe passes by a non intrusive sensor. Each pulse represents a known volume of liquid that is captured in between the lobes of the impellers, and an electronic controller converts the pulses to volume units. The clearance between the impeller and its casing must be controlled carefully to prevent leakage and thus to avoid error. This particular meter has a quoted accuracy of 0.1 percent, has a low pressure drop, and can be used with high- or low-viscosity liquids at tem peratures up to 230°C and pressures up to 7 MPa for flow rates of up to 700 gal/min (or 50 L/s). The most widely used flowmeters to measure liquid volumes are nutating disk flowmeters, shown in Fig. 8–65. They are commonly used as water and gasoline meters. The liquid enters the nutating disk meter through the chamber (A). This causes the disk (B) to nutate or wobble and results in the rotation of a spindle (C) and the excitation of a magnet (D). This signal is transmitted through the casing of the meter to a second magnet (E). The total volume is obtained by counting the number of these signals during a discharge process. Quantities of gas flows, such as the amount of natural gas used in build ings, are commonly metered by using bellows flowmeters that displace a certain amount of gas volume (or mass) during each revolution.
Turbine Flowmeters We all know from experience that a propeller held against the wind rotates, and the rate of rotation increases as the wind velocity increases. You may also have seen that the turbine blades of wind turbines rotate rather slowly at low winds, but quite fast at high winds. These observations suggest that the flow velocity in a pipe can be measured by placing a freely rotating propeller inside a pipe sec tion and doing the necessary calibration. Flow measurement devices that work on this principle are called turbine flowmeters or sometimes propeller flow meters, although the latter is a misnomer since, by definition, propellers add energy to a fluid, while turbines extract energy from a fluid. A turbine flowmeter consists of a cylindrical flow section that houses a turbine (a vaned rotor) that is free to rotate, additional stationary vanes at the inlet to straighten the flow, and a sensor that generates a pulse each time a marked point on the turbine passes by to determine the rate of rotation. The rotational speed of the turbine is nearly proportional to the flow rate of the fluid. Turbine flowmeters give highly accurate results (as accurate as 0.25 percent) over a wide range of flow rates when calibrated properly for the anticipated flow conditions. Turbine flowmeters have very few blades (sometimes just two blades) when used to measure liquid flow, but several blades when used to measure gas flow to ensure adequate torque generation. The head loss caused by the turbine is very small. Turbine flowmeters have been used extensively for flow measurement since the 1940s because of their simplicity, low cost, and accuracy over a wide range of flow conditions. They are commercially available for both liq uids and gases and for pipes of practically all sizes. Turbine flowmeters are also commonly used to measure flow velocities in unconfined flows such as winds, rivers, and ocean currents. The handheld device shown in Fig. 8–66c is used to measure wind velocity.
FIGURE 8–65 A nutating disk flowmeter.
Courtesy of Badger Meter, Inc.
Used by permission.
A B D C E FIGURE 8–64 A positive displacement flowmeter with double helical three-lobe impeller design.
Courtesy of Flow Technology, Inc. Source: www.ftimeters.com cen96537_ch08_351-442.indd 402 14/01/17 2:59 pm 403 CHAPTER 8 Paddlewheel Flowmeters Paddlewheel flowmeters are low-cost alternatives to turbine flowmeters for flows where very high accuracy is not required. In paddlewheel flowmeters, the paddlewheel (the rotor and the blades) is perpendicular to the flow, as shown in Fig. 8–67, rather than parallel as was the case with turbine flow meters. The paddles cover only a portion of the flow cross section (typically less than half), and thus the head loss is smaller compared to that of turbine flowmeters, but the depth of insertion of the paddlewheel into the flow is of critical importance for accuracy. Also, no strainers are required since the pad dlewheels are less susceptible to fouling. A sensor detects the passage of each of the paddlewheel blades and transmits a signal. A microprocessor then con verts this rotational speed information to flow rate or integrated flow quantity.
Variable-Area Flowmeters (Rotameters) A simple, reliable, inexpensive, and easy-to-install flowmeter with reasona bly low pressure drop and no electrical connections that gives a direct read ing of flow rate for a wide range of liquids and gases is the variable-area flowmeter, also called a rotameter or floatmeter. A variable-area flow meter consists of a vertical tapered conical transparent tube made of glass or plastic with a float inside that is free to move, as shown in Fig. 8–68. As fluid flows through the tapered tube, the float rises within the tube to a location where the float weight, drag force, and buoyancy force balance each other and the net force acting on the float is zero. The flow rate is determined by simply matching the position of the float against the gradu ated flow scale outside the tapered transparent tube. The float itself is typi cally either a sphere or a loose-fitting piston-like cylinder (as in Fig. 8–68a). We know from experience that high winds knock down trees, break power lines, and blow away hats or umbrellas. This is because the drag force increases with flow velocity. The weight and the buoyancy force acting on the float are constant, but the drag force changes with flow velocity. Also, the velocity along the tapered tube decreases in the flow direction because of the increase in the cross-sectional area. There is a certain velocity that FIGURE 8–66 (a) An in-line turbine flowmeter to measure liquid flow, with flow from left to right, (b) a cutaway view of the turbine blades inside the flowmeter, and (c) a handheld turbine flowmeter to measure wind speed, measuring no flow at the time the photo was taken so that the turbine blades are visible. The flowmeter in (c) also measures the air temperature for convenience. (a) and (c) Photos by John M. Cimbala; (b) Photo Courtesy of Hoffer Flow Controls, Inc.
(a) (c) (b) cen96537_ch08_351-442.indd 403 14/01/17 2:59 pm 404 internal flow generates enough drag to balance the float weight and the buoyancy force, and the location at which this velocity occurs around the float is the loca tion where the float settles. The degree of tapering of the tube can be made such that the vertical rise changes linearly with flow rate, and thus the tube can be calibrated linearly for flow rates. The transparent tube also allows the fluid to be seen during flow. There are several kinds of variable-area flowmeters. The gravity-based flowmeter, as shown in Fig. 8–68a must be positioned vertically, with fluid entering from the bottom and leaving from the top. In spring-opposed flow meters (Fig. 8–68b), the drag force is balanced by the spring force, and such flowmeters can be installed horizontally. The accuracy of variable-area flowmeters is typically ±5 percent. There fore, these flowmeters are not appropriate for applications that require preci sion measurements. However, some manufacturers quote accuracies of the order of 1 percent. Also, these meters depend on visual checking of the loca tion of the float, and thus they cannot be used to measure the flow rate of fluids that are opaque or dirty, or fluids that coat the float since such fluids block visual access. Finally, glass tubes are prone to breakage and thus they pose a safety hazard if toxic fluids are handled. In such applications, variable-area flowmeters should be installed at locations with minimum traffic.
Ultrasonic Flowmeters It is a common observation that when a stone is dropped into calm water, the waves that are generated spread out as concentric circles uniformly in all directions. But when a stone is thrown into flowing water such as a river, the waves move much faster in the flow direction (the wave and flow veloci ties are added since they are in the same direction) compared to the waves moving in the upstream direction (the wave and flow velocities are subtracted since they are in opposite directions). As a result, the waves appear spread out downstream while they appear tightly packed upstream. The difference between the number of waves in the upstream and downstream parts of the flow per unit length is proportional to the flow velocity, and this suggests that flow velocity can be measured by comparing the propagation of waves in the forward and backward directions with respect to the flow. Ultrasonic flow meters operate on this principle, using sound waves in the ultrasonic range (beyond human hearing ability, typically at a frequency of 1 MHz). Ultrasonic (or acoustic) flowmeters operate by generating sound waves with a transducer and measuring the propagation of those waves through a flowing fluid. There are two basic kinds of ultrasonic flowmeters: tran sit time and Doppler-effect (or frequency shift) flowmeters. The transit time flowmeter transmits sound waves in the upstream and downstream direc tions and measures the difference in travel time. A typical transit time ultra sonic meter is shown schematically in Fig. 8–69. It involves two transducers that alternately transmit and receive ultrasonic waves, one in the direction of flow and the other in the opposite direction. The travel time for each direction can be measured accurately, and the difference in the travel time is calculated. The average flow velocity V in the pipe is proportional to this travel time difference Δt, and is determined from V = KL Δt (8–75) where L is the distance between the transducers and K is a constant.
FIGURE 8–67 Paddlewheel flowmeter to measure liquid flow, with flow from left to right, and a schematic diagram of its operation. Photo by John M. Cimbala.
Paddlewheel sensor Sensor housing Flow Truseal locknut Retainer cap cen96537_ch08_351-442.indd 404 14/01/17 2:59 pm 405 CHAPTER 8 Doppler-Effect Ultrasonic Flowmeters You have probably noticed that when a fast-moving car approaches with its horn blowing, the tone of the high-pitched sound of the horn drops to a lower pitch as the car passes by. This is due to the sonic waves being com pressed in front of the car and being spread out behind it. This shift in fre quency is called the Doppler effect, and it forms the basis for the operation of most ultrasonic flowmeters. Doppler-effect ultrasonic flowmeters measure the average flow velocity along the sonic path. This is done by clamping a piezoelectric transducer on the outside surface of a pipe (or pressing the transducer against the pipe for handheld units). The transducer transmits a sound wave at a fixed frequency through the pipe wall and into the flowing liquid. The waves reflected by impurities, such as suspended solid particles or entrained gas bubbles, are relayed to a receiving transducer. The change in the frequency of the reflected waves is proportional to the flow velocity, and a microprocessor determines the flow velocity by comparing the frequency shift between the transmitted and reflected signals (Figs. 8–70 and 8–71). The flow rate and the total amount of flow can also be determined using the measured velocity by properly configuring the flowmeter for the given pipe and flow conditions. The operation of ultrasonic flowmeters depends on the ultrasound waves being reflected off discontinuities in density. Ordinary ultrasonic flowmeters require the liquid to contain impurities in concentrations greater than 25 parts per million (ppm) in sizes greater than at least 30 μm. But advanced ultrasonic units can also measure the velocity of clean liquids by sensing the waves reflected off turbulent swirls and eddies in the flow stream, provided that they are installed at locations where such disturbances are nonsymmetrical and at a high level, such as a flow section just downstream of a 90° elbow. Ultrasonic flowmeters have the following advantages: • They are easy and quick to install by clamping them on the outside of pipes of 0.6 cm to over 3 m in diameter (Fig. 8–71), and even on open channels.
• They are nonintrusive. Since the meters clamp on, there is no need to stop operation and drill holes into piping, and no production downtime.
• There is no pressure drop since the meters do not interfere with the flow.
• Since there is no direct contact with the fluid, there is no danger of corro sion or clogging.
• They are suitable for a wide range of fluids from toxic chemicals to slur ries to clean liquids, for permanent or temporary flow measurement.
• There are no moving parts, and thus the meters provide reliable and maintenance-free operation.
• They can also measure flow quantities in reverse flow.
• The quoted accuracies are 1 to 2 percent. Ultrasonic flowmeters are noninvasive devices, and the ultrasonic transducers can effectively transmit signals through polyvinyl chloride (PVC), steel, iron, and glass pipe walls. However, coated pipes and concrete pipes are not suit able for this measurement technique since they absorb ultrasonic waves.
FIGURE 8–68 Two types of variable-area flowmeters: (a) an ordinary gravity-based meter and (b) a spring-opposed meter.
(a) Photo by Luke A. Cimbala and (b) Courtesy Insite, Universal Flow Monitors, Inc. Used by permission.
(a) (b) Flow A Reflect-mode configuration B Top view FIGURE 8–69 The operation of a transit time ultrasonic flowmeter equipped with two transducers.
cen96537_ch08_351-442.indd 405 14/01/17 2:59 pm 406 internal flow Electromagnetic Flowmeters It has been known since Faraday’s experiments in the 1830s that when a conductor is moved in a magnetic field, an electromotive force develops across that conductor as a result of magnetic induction. Faraday’s law states that the voltage induced across any conductor as it moves at right angles through a magnetic field is proportional to the velocity of that conductor. This suggests that we may be able to determine flow velocity by replacing the solid conductor by a conducting fluid, and electromagnetic flowme ters do just that. Electromagnetic flowmeters have been in use since the mid-1950s, and they come in various designs such as full-flow and inser tion types. A full-flow electromagnetic flowmeter is a nonintrusive device that con sists of a magnetic coil that encircles the pipe, and two electrodes drilled into the pipe along a diameter flush with the inner surface of the pipe so that the electrodes are in contact with the fluid but do not interfere with the flow and thus do not cause any head loss (Fig. 8–72a). The electrodes are connected to a voltmeter. The coils generate a magnetic field when sub jected to electric current, and the voltmeter measures the electric potential difference between the electrodes. This potential difference is proportional to the flow velocity of the conducting fluid, and thus the flow velocity can be calculated by relating it to the voltage generated. Insertion electromagnetic flowmeters operate similarly, but the magnetic field is confined within a flow channel at the tip of a rod inserted into the flow, as shown in Fig. 8–72b. Electromagnetic flowmeters are well-suited for measuring flow veloci ties of liquid metals such as mercury, sodium, and potassium that are used FIGURE 8–70 The operation of a Doppler-effect ultrasonic flowmeter equipped with a transducer pressed on the outer surface of a pipe.
Transmitting element Receiving element Flow direction Reflectors FIGURE 8–71 Ultrasonic clamp-on flowmeters enable one to measure flow velocity without even contacting (or disturb ing) the fluid by simply pressing a transducer on the outer surface of the pipe.
© J. Matthew Deepe cen96537_ch08_351-442.indd 406 14/01/17 2:59 pm 407 CHAPTER 8 in some nuclear reactors. They can also be used for liquids that are poor conductors, such as water, provided that they contain an adequate amount of charged particles. Blood and seawater, for example, contain sufficient amounts of ions, and thus electromagnetic flowmeters can be used to measure their flow rates. Electromagnetic flowmeters can also be used to measure the flow rates of chemicals, pharmaceuticals, cosmetics, corrosive liquids, beverages, fertilizers, and numerous slurries and sludges, provided that the substances have high enough electrical conductivities. Electromag netic flowmeters are not suitable for use with distilled or deionized water. Electromagnetic flowmeters measure flow velocity indirectly, and thus careful calibration is important during installation. Their use is limited by their relatively high cost, power consumption, and the restrictions on the types of suitable fluids with which they can be used.
Vortex Flowmeters You have probably noticed that when a flow stream such as a river encoun ters an obstruction such as a rock, the fluid separates and moves around the rock. But the presence of the rock is felt for some distance downstream via the swirls generated by it. Most flows encountered in practice are turbulent, and a disk or a short cylinder placed in the flow coaxially sheds vortices (see also Chap. 4). It is observed that these vortices are shed periodically, and the shedding frequency is proportional to the average flow velocity. This suggests that the flow rate can be determined by generating vortices in the flow by placing an obstruc tion in the flow and measuring the shedding frequency. The flow measure ment devices that work on this principle are called vortex flowmeters. The Strouhal number, defined as St = fd/V, where f is the vortex shedding fre quency, d is the characteristic diameter or width of the obstruction, and V is the velocity of the flow impinging on the obstruction, also remains constant in this case, provided that the flow velocity is high enough. A vortex flowmeter consists of a sharp-edged bluff body (strut) placed in the flow that serves as the vortex generator, and a detector (such as a FIGURE 8–72 (a) Full-flow and (b) insertion electromagnetic flowmeters, www.flocat.com.
(a) Full-flow electromagnetic flowmeter (b) Insertion electromagnetic flowmeter Electrodes Flow Flow E E Flow cen96537_ch08_351-442.indd 407 14/01/17 2:59 pm 408 internal flow pressure transducer that records the oscillation in pressure) placed a short distance downstream on the inner surface of the casing to measure the shed ding frequency. The detector can be an ultrasonic, electronic, or fiber-optic sensor that monitors the changes in the vortex pattern and transmits a pul sating output signal (Fig. 8–73). A microprocessor then uses the frequency information to calculate and display the flow velocity or flow rate. The frequency of vortex shedding is proportional to the average velocity over a wide range of Reynolds numbers, and vortex flowmeters operate reliably and accurately at Reynolds numbers from 104 to 107. The vortex flowmeter has the advantage that it has no moving parts and thus is inherently reliable, versatile, and very accurate (usually ±1 percent over a wide range of flow rates), but it obstructs the flow and thus causes considerable head loss.
Thermal (Hot-Wire and Hot-Film) Anemometers Thermal anemometers were introduced in the late 1950s and have been in common use since then in fluid research facilities and labs. As the name implies, thermal anemometers involve an electrically heated sensor, as shown in Fig. 8–74, and utilize a thermal effect to measure flow velocity. Thermal anemometers have extremely small sensors, and thus they can be used to measure the instantaneous velocity at any point in the flow without appreciably disturbing the flow. They can take thousands of velocity meas urements per second with excellent spatial and temporal resolution, and thus they can be used to study the details of fluctuations in turbulent flow. They can measure velocities in liquids and gases accurately over a wide range— from a few centimeters to over a hundred meters per second. A thermal anemometer is called a hot-wire anemometer if the sensing element is a wire, and a hot-film anemometer if the sensor is a thin metallic film (less than 0.1 μm thick) mounted usually on a relatively thick ceramic support having a diameter of about 50 μm. The hot-wire anemometer is characterized by its very small sensor wire—usually a few microns in diameter and a couple of millimeters in length. The sensor is usually made of platinum, tungsten, or platinum–iridium alloys, and it is attached to the probe through needle-like holders. The fine wire sensor of a hot-wire anemometer is very fragile because of its small size and can easily break if the liquid or gas contains excessive amounts of contaminants or particulate matter. This is especially of consequence at high velocities. In such cases, the more rugged hot-film probes should be used. But the sensor of the hot-film probe is larger, has significantly lower frequency response, and interferes more with the flow; thus it is not always suitable for studying the fine details of turbulent flow. The operating principle of a constant-temperature anemometer (CTA), which is the most common type and is shown schematically in Fig. 8–75, is as follows: the sensor is electrically heated to a specified temperature (typi cally about 200°C). The sensor tends to cool as it loses heat to the surround ing flowing fluid, but electronic controls maintain the sensor at a constant temperature by varying the electric current (which is done by varying the voltage) as needed. The higher the flow velocity, the higher the rate of heat FIGURE 8–73 The operation of a vortex flowmeter.
Bluff body (strut) Vortex Transmitting transducer Receiving transducer Flow FIGURE 8–74 The electrically heated sensor and its support, which are the components of a hot-wire probe.
Electric current I Sensor (a thin wire approximately 1 mm long with a diameter of 5 μm) Wire support Flow velocity V cen96537_ch08_351-442.indd 408 14/01/17 2:59 pm 409 CHAPTER 8 transfer from the sensor, and thus the larger the voltage that needs to be applied across the sensor to maintain it at constant temperature. There is a close correlation between the flow velocity and voltage, and the flow veloc ity is determined by measuring the voltage applied by an amplifier or the electric current passing through the sensor. The sensor is maintained at a constant temperature during operation, and thus its thermal energy content remains constant. The conservation of energy principle requires that the electrical Joule heating W .
elect = I 2Rw = E2/Rw of the sensor must be equal to the total rate of heat loss from the sen sor Q .
total, which consists of convection heat transfer since conduction to the wire supports and radiation to the surrounding surfaces are small and can be disregarded. Using proper relations for forced convection, the energy balance is expressed by King’s law as E2 = a + bV n (8–76) where E is the voltage, and the values of the constants a, b, and n are cali brated for a given probe. Once the voltage is measured, this relation gives the flow velocity V directly. Most hot-wire sensors have a diameter of 5 µm and a length of approximately 1 mm and are made of tungsten. The wire is spot-welded to needle-shaped prongs embedded in a probe body, which is connected to the anemometer electronics. Thermal anemometers can be used to measure two- or three-dimensional velocity components simultaneously by using probes with two or three sensors, respectively (Fig. 8–76). When selecting probes, consideration should be given to the type and the contamination level of the fluid, the number of velocity components to be measured, the required spa tial and temporal resolution, and the location of measurement.
FIGURE 8–75 Schematic of a thermal anemometer system.
Bridge Connector box and computer CTA Signal conditioner Sensor Flow Filter Gain Servo loop Probe FIGURE 8–76 Thermal anemometer probes with single, double, and triple sensors to measure (a) one-, (b) two-, and (c) three-dimensional velocity components simultaneously.
(a) (c) (b) cen96537_ch08_351-442.indd 409 14/01/17 2:59 pm 410 internal flow Laser Doppler Velocimetry Laser Doppler velocimetry (LDV), also called laser velocimetry (LV) or laser Doppler anemometry (LDA), is an optical technique to measure flow velocity at any desired point without disturbing the flow. Unlike ther mal anemometry, LDV involves no probes or wires inserted into the flow, and thus it is a nonintrusive method. Like thermal anemometry, it can accu rately measure velocity at a very small volume, and thus it can also be used to study the details of flow at a locality, including turbulent fluctuations, and it can be traversed through the entire flow field without intrusion. The LDV technique was developed in the mid-1960s and has found widespread acceptance because of the high accuracy it provides for both gas and liquid flows; the high spatial resolution it offers; and, in recent years, its ability to measure all three velocity components. Its drawbacks are the relatively high cost; the requirement for sufficient transparency between the laser source, the target location in the flow, and the photodetector; and the requirement for careful alignment of emitted and reflected beams for accuracy. The latter drawback is eliminated for the case of a fiber-optic LDV system, since it is aligned at the factory. The operating principle of LDV is based on sending a highly coherent monochromatic (all waves are in phase and at the same wavelength) light beam toward the target, collecting the light reflected by small particles in the target area, determining the change in frequency of the reflected radia tion due to the Doppler effect, and relating this frequency shift to the flow velocity of the fluid at the target area. LDV systems are available in many different configurations. A basic dual-beam LDV system to measure a single velocity component is shown in Fig. 8–77. In the heart of all LDV systems is a laser power source, which is usually a helium–neon or argon-ion laser with a power output of 10 mW to 20 W. Lasers are preferred over other light sources since laser beams are highly coherent and highly focused. The helium–neon laser, for exam ple, emits radiation at a wavelength of 0.6328 μm, which is in the reddish-orange color range. The laser beam is first split into two parallel beams of equal intensity by a half-silvered mirror called a beam splitter. Both beams then pass through a converging lens that focuses the beams at a point in the flow (the target). The small fluid volume where the two beams inter sect is the region where the velocity is measured and is called the meas urement volume or the focal volume. The measurement volume resembles an ellipsoid, typically of 0.1 mm diameter and 0.5 mm in length. The laser FIGURE 8–77 A dual-beam LDV system in forward scatter mode.
Beam splitter Mirror Laser Bragg cell Measurement volume Receiving lens Photodetector Sending lens V α cen96537_ch08_351-442.indd 410 14/01/17 2:59 pm 411 CHAPTER 8 light is scattered by particles passing through this measurement volume, and the light scattered in a certain direction is collected by a receiving lens and is passed through a photodetector that converts the fluctuations in light intensity into fluctuations in a voltage signal. Finally, a signal pro cessor determines the frequency of the voltage signal and thus the velocity of the flow. The waves of the two laser beams that cross in the measurement volume are shown schematically in Fig. 8–78. The waves of the two beams inter fere in the measurement volume, creating a bright fringe where they are in phase and thus support each other, and creating a dark fringe where they are out of phase and thus cancel each other. The bright and dark fringes form lines parallel to the midplane between the two incident laser beams. Using trigonometry, the spacing s between the fringe lines, which can be viewed as the wavelength of fringes, can be shown to be s = 𝜆/[2 sin(𝛼/2)], where 𝜆 is the wavelength of the laser beam and 𝛼 is the angle between the two laser beams. When a particle traverses these fringe lines at velocity V, the fre quency of the scattered fringe lines is f = V s = 2V sin(𝛼/2) 𝜆 (8–77) This fundamental relation shows the flow velocity to be proportional to the frequency and is known as the LDV equation. As a particle passes through the measurement volume, the reflected light is bright, then dark, then bright, etc., because of the fringe pattern, and the flow velocity is determined by measuring the frequency of the reflected light. The velocity profile at a cross section of a pipe, for example, can be obtained by mapping the flow across the pipe (Fig. 8–79). The LDV method obviously depends on the presence of scattered fringe lines, and thus the flow must contain a sufficient amount of small particles called seeds or seeding particles. These particles must be small enough to fol low the flow closely so that the particle velocity is equal to the flow velocity, but large enough (relative to the wavelength of the laser light) to scatter an ade quate amount of light. Particles with a diameter of 1 μm usually serve the pur pose well. Some fluids such as tap water naturally contain an adequate amount of such particles, and no seeding is necessary. Gases such as air are commonly seeded with smoke or with particles made of latex, oil, or other materials. By using three laser beam pairs at different wavelengths, the LDV system is also used to obtain all three velocity components at any point in the flow.
Particle Image Velocimetry Particle image velocimetry (PIV) is a double-pulsed laser technique used to measure the instantaneous velocity distribution in a plane of flow by photographically determining the displacement of particles in the plane during a very short time interval. Unlike methods like hot-wire anemometry and LDV that measure velocity at a point, PIV provides velocity values simultaneously throughout an entire cross section, and thus it is a whole-field technique. PIV combines the accuracy of LDV with the capability of flow visualization and provides instantaneous flow field mapping. The entire instantaneous velocity profile at a cross section of pipe, for example, can be FIGURE 8–78 Fringes that form as a result of the interference at the intersection of two laser beams of an LDV system (lines represent peaks of waves). The top diagram is a close-up view of two fringes.
Fringe lines V Laser beams Measurement volume Fringe lines s α λ FIGURE 8–79 A time-averaged velocity profile in turbulent pipe flow obtained by an LDV system.
Courtesy Dantec Dynamics, www.dantecdynamics.com. Used by permission.
5 4 3 2 1 –80 V (m/s) –60 –40 x (mm) –20 0 cen96537_ch08_351-442.indd 411 14/01/17 2:59 pm 412 internal flow obtained with a single PIV measurement. A PIV system can be viewed as a camera that can take a snapshot of velocity distribution at any desired plane in a flow. Ordinary flow visualization gives a qualitative picture of the details of flow. PIV also provides an accurate quantitative description of various flow quantities such as the velocity field, and thus the capability to analyze the flow numerically using the velocity data provided. Because of its whole-field capability, PIV is also used to validate computational fluid dynamics (CFD) codes (Chap. 15). The PIV technique has been used since the mid-1980s, and its use and capabilities have grown in recent years with improvements in frame grabber and charge-coupled device (CCD) camera technologies. The accuracy, flex ibility, and versatility of PIV systems with their ability to capture whole-field images with submicrosecond exposure time have made them extremely valuable tools in the study of supersonic flows, explosions, flame propaga tion, bubble growth and collapse, turbulence, and unsteady flow. The PIV technique for velocity measurement consists of two main steps: visualization and image processing. The first step is to speed the flow with suitable particles in order to trace the fluid motion. Then a pulse of laser light sheet illuminates a thin slice of the flow field at the desired plane, and the positions of particles in that plane are determined by detecting the light scattered by particles on a digital video or photographic camera positioned at right angles to the light sheet (Fig. 8–80). After a very short time period Δt (typically in μs), the particles are illuminated again by a second pulse of laser light sheet, and their new positions are recorded. Using the informa tion on these two superimposed camera images, the particle displacements Δs are determined for all particles, and the magnitude of velocity of each particle in the plane of the laser light sheet is determined from Δs/Δt. The direction of motion of the particles is also determined from the two positions, so that two components of velocity in the plane are calculated. The built-in algorithms of PIV systems determine the velocities at hundreds FIGURE 8–80 A PIV system to study flame stabilization.
Computer Pulsed laser Sheet-forming optics Beam dump Synchronizer Pulser Seeded flow Video camera cen96537_ch08_351-442.indd 412 14/01/17 2:59 pm 413 CHAPTER 8 or thousands of area elements called interrogation regions throughout the entire plane and display the velocity field on the computer monitor in any desired form (Fig. 8–81). The PIV technique relies on the laser light scattered by particles, and thus the flow must be seeded if necessary with particles, also called markers, in order to obtain an adequate reflected signal. Seed particles must be able to follow the pathlines in the flow for their motion to be representative of the flow, and this requires the particle density to be equal to the fluid density (so that they are neutrally buoyant) or the particles to be so small (typically μm-sized) that their movement relative to the fluid is insignifi cant. A variety of such particles is available to seed gas or liquid flow. Very small particles must be used in high-speed flows. Silicon carbide particles (mean diameter of 1.5 μm) are suitable for both liquid and gas flow, titanium dioxide particles (mean diameter of 0.2 μm) are usually used for gas flow and are suitable for high-temperature applications, and polystyrene latex particles (nominal diameter of 1.0 μm) are suitable for low-temperature applications. Metallic-coated particles (mean diameter of 9.0 μm) are also used to seed water flows for LDV measurements because of their high reflectivity. Gas bubbles as well as droplets of some liquids such as olive oil or silicon oil are also used as seeding particles after they are atomized to μm-sized spheres. A variety of laser light sources such as argon, copper vapor, and Nd:YAG can be used with PIV systems, depending on the requirements for pulse duration, power, and time between pulses. Nd:YAG lasers are commonly used in PIV systems over a wide range of applications. A beam delivery system such as a light arm or a fiber-optic system is used to generate and deliver a high-energy pulsed laser sheet at a specified thickness.
FIGURE 8–81 Instantaneous PIV velocity vectors superimposed on a hummingbird in hover. Color scale is from low velocity (blue) to high velocity (red).
Photo by Douglas Warrick. Used by permission.
cen96537_ch08_351-442.indd 413 14/01/17 2:59 pm 414 internal flow With PIV, other flow properties such as vorticity and strain rates can also be obtained, and the details of turbulence can be studied. Recent advances in PIV technology have made it possible to obtain three-dimensional veloc ity profiles at a cross section of a flow using two cameras (Fig. 8–82). This is done by recording the images of the target plane simultaneously by both cameras at different angles, processing the information to produce two sepa rate two-dimensional velocity maps, and combining these two maps to gen erate the instantaneous three-dimensional velocity field.
Introduction to Biofluid Mechanics1 Biofluid mechanics can cover a number of physiological systems in the human body but the term also applies to all animal species as there are a number of basic fluid systems that are essentially a series of piping networks to transport a fluid (be it liquid or gas or perhaps both). If we focus on humans, these fluid systems are the cardiovascular, respiratory, lymphatic, ocular, and gastrointestinal to name several. We should keep in mind that all these systems are similar to other mechanical piping networks in that the fundamental constituents for the network include a pump, pipes, valves, and a fluid. For our purposes, we will focus more on the cardiovascular system to demonstrate the basic concepts of a piping network within a human. Figure 8–83 illustrates the cardiovascular system, more specifically, the systemic circulation or the vessels (pipes) that carry the blood (fluid) from the heart, specifically the left ventricle (pump), to the rest of the body. Keep in mind there is a separate network of vessels from the right ventricle to the lungs to oxygenate the blood again. What is unique about the series of pipes in the systemic circulation is that the geometry or cross section is not cir cular but rather elliptical and in fact, unlike the typical mechanical systems for piping networks that have fittings to transition from one size pipe to another size pipe, the cardiovascular system starting with the aorta (the first vessel from the left ventricle) continually tapers from approximately 25 mm in diameter to 5 microns in diameter at the capillary level and then gradu ally increases in diameter to approximately 25 mm at the vena cava, which is the vessel connected to the right ventricle. Another important element of the circulation and specifically the vessels is that they are compliant and can expand to accommodate blood volume as needed to regulate pressure changes to maintain homeostasis. The cardiovascular system is a complex network of pipes that themselves are living and respond to stresses as do blood elements that react when the norm has changed. Even with this network, the system is even more intri cate given that the flow is continually moving based on pulses initiated from the heart to drive blood through the network. This pulsatility propa gates through the blood and the vessel wall creating an interaction of waves and reflections within the system. Because of the discontinuities associated with the branching, bifurcations, and curvature as seen in Fig. 8–83 initial and boundary conditions are not straightforward. Understanding blood flow is a challenging endeavor given the complexities of the vessel network and the components themselves.
FIGURE 8–82 A three-dimensional PIV system set up to study the mixing of an air jet with cross duct flow. Jet flow Flow Jet trajectory Stereoscopic camera setup Light-guide delivery of laser sheet Field of view x y 1 This section was contributed by Professor Keefe Manning of Penn State University.
cen96537_ch08_351-442.indd 414 14/01/17 2:59 pm 415 CHAPTER 8 FIGURE 8–83 The cardiovascular system.
McGraw-Hill Companies, Inc.
Basilar artery Internal carotid artery External carotid artery External jugular vein Internal jugular vein Vertebral arteries Pulmonary arteries Pulmonary veins Heart Celiac trunk Hepatic vein Renal veins Renal artery Gonadal artery Gonadal vein Common iliac vein Common iliac artery Internal iliac artery Internal iliac vein External iliac artery External iliac vein Great saphenous vein Femoral artery Femoral vein Popliteal artery Popliteal vein Small saphenous vein Anterior tibial artery Posterior tibial artery Peroneal artery Anterior/posterior tibial veins Dorsal venous arch Dorsal digital vein Arcuate artery Palmar digital veins Radial artery Ulnar artery Cephalic vein Median cubital vein Basillic vein Brachial artery Descending aorta Inferior vena cava Superior vena cava Aorta Axillary artery Axillary vein Cephalic vein Subclavian vein Subclavian artery Digital artery Dorsal digital arteries Common carotid arteries cen96537_ch08_351-442.indd 415 14/01/17 2:59 pm 416 internal flow Flow measurement techniques like PIV and LDV are extremely useful in characterizing the flow within and about medical devices, particularly those implanted in the cardiovascular system. Much can be ascertained, and design changes can be made, using these techniques with respect to how blood might flow through or about these cardiovascular devices. Furthermore, we can even use these measurements to then estimate levels of blood dam age and the potential for clotting to occur. To ensure we have an accurate representation of the cardiovascular system on the bench, engineers have designed mock circulatory loops or flow loops that allow the experimental ist to simulate cardiac flow and pressure waveforms for bench top studies. For example, Dr. Gus Rosenberg developed the Penn State mock circulatory loop in the early 1970s (Rosenberg et al., 1981). We also need to simu late blood for these particular flow measurement techniques to ensure that the fluid is transparent but also mimics the behavior of blood as a non-Newtonian fluid. We have developed a blood analog that does that and also matches the refractive index of the acrylic models that represent the cardio vascular devices, thus allowing the laser light to pass through the acrylic into the flow field without any refraction. The simulated loop and fluid are critical to ensure that the measurements are acquired under controllable physiological conditions and with sufficient accuracy. The Pennsylvania State University has been developing mechanical cir culatory support devices (blood pumps) since the 1970s, which are devices that help patients stay alive as they await a heart transplant (former Vice President Dick Cheney used such technology while awaiting a heart transplant). Through the years, PIV and LDV have been used quite suc cessfully to measure the flow and make design changes that reduce clot ting. Our recent focus has been the development of a pulsatile pediatric ventricular assist device (PVAD) that helps children stay alive until they can receive a donor heart. The device operates pneumatically with air pulsing into a chamber which then causes a diaphragm to inflate against a polyurethane urea sac (the blood contacting surface within the PVAD). The flow is directed into the device from a tube attached to the left ven tricle, passes through a mechanical heart valve into the PVAD, and then flows through the outlet of the device through another mechanical heart valve and into a tube which is attached to the ascending aorta as shown in Fig. 8–84a. Fig. 8–84b shows the flow path through the PVAD, and it should be noted that it can be placed within the palm of an adult’s hand. One of the first PIV PVAD studies was to determine which type of mechanical heart valve (tilting disc or bileaflet) would be used with the device. Fig. 8–85 illustrates part of the PIV study results (Cooper et al., 2008). Here, we used particle traces as a way to examine how the vortical structure would develop inside the device, which for this technology is a way to ensure adequate wall washing (sufficient wall shear) to prevent clotting on the blood contacting surfaces within the device. The tighter rotation would lead to more momentum over the entire cardiac cycle and create a larger vortical structure. Our research group has also looked at characterizing the flow through mechanical heart valves. In one study (Manning et al., 2008), we focused on the flow within the housing of a Bjork-Shiley Monostrut mechanical heart valve (tilting disc valve) as shown in Fig. 8–86b. We removed part (a) (b) FIGURE 8–84 (a) An artist rendering of the 12-cc pulsatile Penn State pediatric ventricular assist device with the inlet attached to the left atrium and the outlet attached to the ascending aorta (b) The direction of blood through the PVAD.
Photo (b) Permission granted from ASME, from Cooper BT, et al., Journal of Biomechanical Engineering, Volume 130, 2008.
cen96537_ch08_351-442.indd 416 14/01/17 2:59 pm 417 CHAPTER 8 of the housing and inserted an optical window to allow access for the LDV system. Instead of using a flow-through loop for this study, we used a single-shot chamber (Fig. 8–86a) that mimicked the mitral valve posi tion since we are more interested in the closure fluid dynamics. The mitral valve sits between the left atrium and left ventricle. The native heart valves, like the mitral valve, are passive, similar to a check valve, and respond to the pressure changes within the heart’s different structures. In this study, we measured how fast the fluid flowed through the small gap between the tilting disc and valve housing, and also how large was the vortex that is created as the tilting disc closes. Figure 8–87 is a schematic illustration of the flow, and Fig. 8–88 is a time sequence of the flow that was measured using LDV within a couple of milliseconds around impact of the valve housing during closure. The intense vortex can be measured right at impact. These data were collected over hundreds of simulated heart beats. We then used these velocity measurements to estimate the amount of potential blood damage by relating time duration and shear magnitude.
0.8 m/s Vel Mag 1 m/s 0.6 m/s 0.4 m/s 0.2 m/s 0 m/s FIGURE 8–85 Particle traces for the BSM valve configuration at 250 ms (left column) and for the CM valve configuration at 350 ms (right column) for the 7 mm (top row), 8.2 mm (middle row), and 11 mm (bottom row) planes. These images highlight the first time step that the rotational flow pattern is fully developed.
Permission granted from ASME, from Manning, KB, et al., Journal of Biomechanical Engineering, Volume 130, 2008.
cen96537_ch08_351-442.indd 417 14/01/17 2:59 pm 418 internal flow Left Atrium Left Ventricle Bjork-Shiley Monostrut MHV (a) (b) Window FIGURE 8–86 (a) The single shot chamber mimics the closure dynamics of the Bjork-Shiley Monostrut valve. (b) On the lefthand side is a view of the intact Bjork-Shiley Monostrut mechanical heart valve. To the right, the modifi cation to the valve housing is dis played. The window was later filled in with acrylic to maintain similar fluid dynamic patterns and rigidity.
Permission granted from ASME, from Manning, KB, et al., Journal of Biomechanical Engineering, Volume 130, 2008.
1 ms before impact (a) At impact (b) SIDE FRONT 1 ms after impact (c) 2 ms after impact (d) FIGURE 8–87 These schematics depict side and front views of the overall flow structure generated by the closing occluder for four successive times.
Permission granted from ASME, Manning et al. JBME, 2008.
EXAMPLE 8–12 Blood Flow through the Aortic Bifurcation Blood flows from the heart (specifically, the left ventricle) into the aorta to feed the rest of the body oxygen. As blood flow moves from the ascending aorta and downward to the abdominal aorta, some of the volume is directed through a branching network. As the blood reaches the pelvic region, there is a bifurcation (see Fig. 8–89) into the left and right common iliac arteries. This bifurcation is symmetrical but the common iliac vessels are not the same diameter. Given that the kinematic viscosity of blood is 4 cSt (centistokes), the abdominal aorta’s diam eter is 15 mm, the right common iliac artery’s diameter is 10 mm, and the left common iliac artery’s diameter is 8 mm, determine the mean flow rate through the right common iliac artery if the abdominal aorta’s mean velocity is 30 cm/s and the left common iliac artery’s mean velocity is 40 cm/s.
cen96537_ch08_351-442.indd 418 14/01/17 2:59 pm 419 CHAPTER 8 2 1.5 0.5 Z Y 0 0 0 1 1 3 2 1 X (mm) Z (mm) Z (mm) Z (mm) Y (mm) Y (mm) Y (mm) Z (mm) 0 –1 –2 –3 3 2 1 X (mm) 0 –1 –2 –3 3 2 1 X (mm) 0 –1 –2 –3 3 2 1 X (mm) 0 –1 –2 –3 1 2 1.5 0.5 0 0 1 0 1 1 2 7 6 5 4 3 2 1 –1 –2 –3 –4 –5 0 W (m/s) 1.5 0.5 0 1 2 1.5 0.5 1 (a) 1 ms before impact (b) At occluder impact (c) 1 ms after impact (d) 2 ms after impact X 0 FIGURE 8–88 Three-dimensional flow structures are constructed with the vectors indicating direction and the color signifying axial velocity strength. The valve closes right to left, with x = 0 representing the centerline of the leaflet. The four plots show the flow (a) 1 ms before impact, (b) at impact, (c) 1 ms following closure, and (d ) 2 ms after closure.
Permission granted from ASME, Manning et al. JBME, 2008.
SOLUTION The mean velocities for two of the three vessels is provided along with the diameters of all three vessels. Approximate the vessels as rigid pipes.
Assumptions 1 The flow is steady even though the heart contracts and relaxes approximately 75 beats per minute creating a pulsatile flow. 2 The entrance effects are negligible and the flow is considered fully developed. 3 Blood acts as a Newtonian fluid.
Properties The kinematic viscosity at 37°C is 4 cSt.
Analysis Using conservation of mass, we can say the flow rate of the abdominal aorta (V · 1) equals the sum of both common iliac arteries (V · 2 for left and V · 3 for right). Thus, V · 1 = V · 2 + V · 3 Since we are using the mean velocities, we know the diameters, and the density of blood is the same throughout this section of the circulatory system, we can rewrite the equation to be V1Al = V2A2 + V3A3 where V are the average velocities and A are the areas.
By rearranging and solving for V3, the equation becomes, V3 = (V1A1 − V2A2)/A3 Inserting the values we know, V3 = (30 cm/s × (1.5 cm)2 − 40 cm/s × (0.8 cm)2)/(1.0 cm)2 V3 = 41.9 cm/s cen96537_ch08_351-442.indd 419 14/01/17 2:59 pm 420 internal flow Aorta Left com. iliac Right kidney Left renal vessels Transversus abdominis Quadratus lumborum Iliacus Right com. iliac Psoas major Ureter R. Suprarenal gland Diaphragm L. Suprarenal gland Esophagus Hepatic veins Inferior phrenic arteries Internal spermatic vessels Inf. vena cava Left kidney Right renal vessels Discussion Since we assume a steady flow, the mean velocities are appropri ate, but in reality there will be a maximum positive velocity and also some retro grade (or reverse) flow towards the heart as the left ventricle fills during diastole. The velocity profiles through these vessels and many large arteries will vary over a cardiac cycle. It is also assumed that blood will behave as a Newtonian fluid even though it is viscoelastic. Many researchers use this assumption since at this par ticular location, the shear rate is sufficient to reach the asymptotic value for blood viscosity.
Guest Author: Jean Hertzberg1, Brett Fenster2, Jamey Browning1, and Joyce Schroeder2 1Department of Mechanical Engineering, University of Colorado, Boulder, CO.
2National Jewish Health Center, Denver, CO.
MRI (magnetic resonance imaging) can measure the velocity field of blood moving through the human heart, including all three velocity components (u,𝜐,w) with reasonable resolution in 3-D space and time (Bock et al., 2010). Figure 8–90 shows blood moving from the right atrium into the right ventricle at the peak of diastole (the heart-filling phase) of a normal volunteer subject. The black arrow shows the long axis of the ventricle. The smaller arrows show the velocity vector field and are colored by velocity magnitude, with blue at the slow end of the scale, up to red at 0.5 m/s.
APPLICATION SPOTLIGHT ■ PIV Applied to Cardiac Flow FIGURE 8–89 Anatomy of the human body. Note the aorta and left and right common iliac arteries.
cen96537_ch08_351-442.indd 420 14/01/17 2:59 pm 421 CHAPTER 8 The flow patterns change rapidly with time during the approximately one-second long cardiac cycle, and show complex geometry. The flow moves in a subtle helical path from the atrium into the ventricle, as shown by the white stream tube. The tricuspid valve between the atrium and the ventri cle is a set of three thin tissue flaps, which are not visible in this data set. The effect of the valve on flow patterns can be seen as flow curls around one of the flaps, shown by the yellow stream tube. The details of the flow (including vorticity, Chap. 4) are expected to reveal information about the underlying physics of the interaction between the heart and lungs, and lead to improved diagnostics for pathologic conditions like pulmonary hyperten sion (Fenster et al., 2012). After the right ventricle is filled, the tricuspid valve closes, the ventricle con tracts, and blood is ejected into the pulmonary arteries which lead to the lungs, where the blood is oxygenated. After that, the blood goes to the left side of the heart, where the pressure is raised by the contraction of the left ventricle. The oxygenated blood is then ejected into the aorta and is distributed to the body. In this way, the heart functions as two separate positive displacement pumps. Since calibration of these data is difficult, it’s important to check the data for consistency. One useful test, conservation of mass in the ventricle throughout one cardiac cycle, is applied by computing the volume flow of blood entering the ventricle during diastole, and comparing it to the vol ume that leaves during systole. Similarly, the net flow through the right side of the heart must match the net flow through the left side of the heart in each cycle.
References Bock, J., Frydrychowicz, A., Stalder, A.F., Bley, T.A., Burkhardt, H., Hennig, J., and Markl, M., “40 Phase Contrast MRI at 3 T: Effect of Standard and Blood-pool Contrast Agents on SNR, PC-MRA, and Blood Flow Visualization.” Magnetic Resonance in Medicine 63(2):330–338, 2010.
Fenster, B.E., Schroeder, J.D., Hertzberg, J.R., and Chung, J.H., “4-Dimensional Cardiac Magnetic Resonance in a Patient with Bicuspid Pulmonic Valve: Characterization of Post-Stenotic Flow,” J Am Coll Cardiol 59(25):e49, 2012.
Guest Authors: Michael McPhail and Michael Krane, Applied Research Laboratory, Penn State University Particle shadow velocimetry (PSV) is an optical technique to measure fluid velocity without disturbing the flow. Like particle image velocimetry (PIV), PSV provides an instantaneous velocity field in a plane by imaging tracer particles. The velocity fields are estimated by acquiring two subsequent images of the particles in the flow, separated by a short time delay Δt, usually in the µs range. Then, image-processing techniques are used to estimate the vector displacement APPLICATION SPOTLIGHT ■ Multicolor Particle Shadow Velocimetry/Accelerometry SVC Aorta PA RVOT LV RV RA IVC FIGURE 8–90 MRI-PCV (phase contrast velocimetry) measurements of flow through a human heart.
Photo courtesy of Jean Hertzberg.
cen96537_ch08_351-442.indd 421 14/01/17 2:59 pm 422 internal flow Δs⃗ the particles travel in the time between the illumination flashes. As with PIV, the velocity vectors are given by Δs⃗/Δt. The result is a two-dimensional velocity vector field. PSV uses overdriven light-emitting diodes (LEDs) for pulsed-light illumina tion. The LEDs are placed on the opposite side of the flow field from the cam era (Fig. 8–91). The camera images the shadows produced by the particles in the flow field and the LED flash. The camera lens is used to isolate the measure ment region. LEDs have many advantages over lasers as pulsed-light sources for veloc ity measurements. Velocimetry techniques, like PSV and PIV, are often used to measure flows in situations where flow boundaries move, such as with tur bomachinery. These machines may also experience cavitation near their moving surfaces. Laser light scattered from the moving machinery or cavitation bubbles can harm the cameras, or worse, the eyes of the operator. The LED illumination used in PSV is much safer to use in these cases. Another advantage of this technique is that LEDs are much less expensive than lasers. The low cost of LEDs can be further exploited by using multiple LED colors along with a color camera (Goss and Estevadeordal, 2006; Goss et al., 2007). Particles can then be imaged multiple times within the same cam era exposure by firing the different color LEDs with a small time delay between them. This approach is shown in Fig. 8–92 with a bright background from LED illumination and the darker regions that are the particle shadows. Here, the red LEDs were fired first, followed by green, and then blue. The pop out of Fig. 8–92 shows the shadows of a single particle moving from left to right. The left-most shadow in the pop out appears cyan because the particle was in that location during the red flash. That part of the image therefore received only blue and green light, resulting in cyan. This approach effectively triples the acquisition rate by treating the different colors as separate recording channels. We can then acquire velocity fields in rapid succession (Fig. 8–93a,b) and also estimate unsteady acceleration ∂u⃗/∂t (McPhail et al., 2015).
40 0 0 10 20 30 2 1 0 –1 –2 10 20 y (mm) x (mm) 30 (c) Δu1 (m/s) 40 0 0 10 20 30 8 6 4 2 0 10 20 y (mm) x (mm) 30 (b) │u│(m/s) 40 0 0 10 (a) 20 30 8 6 4 2 0 10 20 y (mm) x (mm) 30 │u│(m/s) References Goss, L., and Estevadeordal, J., “Parametric Characterization for Particle-Shadow Velocimetry (PSV),” 25th AIAA Aerodynamics Measurement Technology and Ground Testing Conf. AIAA-2006 2808 (San Francisco, CA), 2006.
Goss. L., Estevadeordal, J., and Crafton. J., “Kilo-hertz Color Particle Shadow Velocimetry (PSV),” 37th AIAA Fluid Dynamics Conf. and Exhibit (Miami, FL), 2007.
McPhail, M.J., Krane, M.H., Fontaine, A.A., Goss, L., and Crafton, J., “Multi color Particle Shadow Accelerometry,” Measurement Science and Technology 26(4): 045301, 2015.
FIGURE 8–91 Hardware layout for particle shadow velocimetry (PSV) technique.
Measurement region Fluid LED lamp Camera FIGURE 8–92 Example image from multicolor particle shadow velocimetry (PSV).
© Michael McPhail and Michael Krane, Applied Research Laboratory, Penn State University.
FIGURE 8–93 Color particle shadow velocimetry measurements of near-wall turbulent pipe flow. The wall is at y = 0 and the flow is from left to right. (a, b) Two subsequent velocity fields, separated in time by 2 ms. Colormaps (shown to the right of each) indicate local veloc ity magnitude. The mean streamwise (u1) velocity has been subtracted from each velocity vector to clearly show the turbulent flow structures. (c) Difference of velocity fields depicted in Figs. 8–93(a) and 8–93(b), used to estimate local acceleration.
© Michael McPhail and Michael Krane, Applied Research Laboratory, Penn State University.
cen96537_ch08_351-442.indd 422 14/01/17 2:59 pm 423 CHAPTER 8 SUMMARY In internal flow, a pipe is completely filled with a fluid. Laminar flow is characterized by smooth streamlines and highly ordered motion, and turbulent flow is characterized by unsteady disorderly velocity fluctuations and highly dis ordered motion. The Reynolds number is defined as Re = Inertial forces Viscous forces = VavgD 𝜈 = ρVavgD 𝜇 Under most practical conditions, the flow in a pipe is lami nar at Re < 2300, turbulent at Re > 4000, and transitional in between. The region of the flow in which the effects of the viscous shearing forces are felt is called the velocity boundary layer. The region from the pipe inlet to the point at which the flow becomes fully developed is called the hydrodynamic entrance region, and the length of this region is called the hydrodynamic entry length Lh. It is given by Lh, laminar D ≅0.05 Re and Lh, turbulent D ≅10 The friction coefficient in the fully developed flow region is constant. The maximum and average velocities in fully developed laminar flow in a circular pipe are umax = 2Vavg and Vavg = ΔPD2 32𝜇L The volume flow rate and the pressure drop for laminar flow in a horizontal pipe are V · = Vavg Ac = ΔP𝜋D4 128𝜇L and ΔP = 32𝜇LVavg D2 The pressure loss and head loss for all types of internal flows (laminar or turbulent, in circular or noncircular pipes, smooth or rough surfaces) are expressed as ΔPL = f L D ρV 2 2 and hL = ΔPL ρg = f L D V 2 2g where 𝜌V 2/2 is the dynamic pressure and the dimensionless quantity f is the friction factor. For fully developed laminar flow in a round pipe, the friction factor is f = 64/Re. For noncircular pipes, the diameter in the previous rela tions is replaced by the hydraulic diameter defined as Dh = 4Ac/p, where Ac is the cross-sectional area of the pipe and p is its wetted perimeter. In fully developed turbulent flow, the friction factor depends on the Reynolds number and the relative roughness 𝜀/D. The friction factor in turbulent flow is given by the Colebrook equation, expressed as 1 √f = −2.0 log( 𝜀 /D 3.7 + 2.51 Re√f ) The plot of this formula is known as the Moody chart. The design and analysis of piping systems involve the determina tion of the head loss, flow rate, or the pipe diameter. Tedious iterations in these calculations can be avoided by the approx imate Swamee–Jain formulas expressed as hL = 1.07 V . 2L gD5{ln[ 𝜀 3.7D + 4.62( 𝜈D V . ) 0.9 ] } −2 10−6 < 𝜀 /D < 10−2 3000 < Re < 3 × 108 V .
= −0.965( gD5hL L ) 0.5 ln[ 𝜀 3.7D + ( 3.17𝜈2L gD3hL ) 0.5 ] Re > 2000 D = 0.66[𝜀 1.25( LV . 2 ghL ) 4.75 + 𝜈V . 9.4( L ghL) 5.2 ] 0.04 10−6 < 𝜀 /D < 10−2 5000 < Re < 3 × 108 The losses that occur in piping components such as fittings, valves, bends, elbows, tees, inlets, exits, expansions, and contractions are called minor losses. The minor losses are usually expressed in terms of the loss coefficient KL. The head loss for a component is determined from hL = KL V 2 2g When all the loss coefficients are available, the total head loss in a piping system is hL, total = hL, major + hL, minor = ∑ i fi Li Di V 2 i 2g + ∑ j KL, j V 2 j 2g If the entire piping system is of constant diameter, the total head loss reduces to hL, total = (f L D + ∑KL) V 2 2g The analysis of a piping system is based on two simple prin ciples: (1) The conservation of mass throughout the system must be satisfied and (2) the pressure drop between two points must be the same for all paths between the two points. When cen96537_ch08_351-442.indd 423 14/01/17 2:59 pm 424 internal flow the pipes are connected in series, the flow rate through the entire system remains constant regardless of the diameters of the individual pipes. For a pipe that branches out into two (or more) parallel pipes and then rejoins at a junction downstream, the total flow rate is the sum of the flow rates in the individual pipes but the head loss in each branch is the same. When a piping system involves a pump and/or turbine, the steady-flow energy equation is expressed as P1 ρg + 𝛼1 V 2 1 2g + z1 + hpump, u = P2 ρg + 𝛼2 V 2 2 2g + z2 + hturbine, e + hL When the useful pump head hpump, u is known, the mechani cal power that needs to be supplied by the pump to the fluid and the electric power consumed by the motor of the pump for a specified flow rate are W · pump, shaft = ρV ·ghpump, u 𝜂pump and W · elect = ρV ·ghpump, u 𝜂pump−motor where 𝜂pump–motor is the efficiency of the pump–motor com bination, which is the product of the pump and the motor efficiencies. The plot of the head loss versus the flow rate V · is called the system curve. The head produced by a pump is not a constant, and the curves of hpump, u and 𝜂pump versus V · are called the characteristic curves. A pump installed in a piping system operates at the operating point, which is the point of intersection of the system curve and the characteristic curve. Flow measurement techniques and devices can be considered in three major categories: (1) volume (or mass) flow rate meas urement techniques and devices such as obstruction flowmeters, turbine meters, positive displacement flowmeters, rotameters, and ultrasonic meters; (2) point velocity measurement tech niques such as the Pitot-static probes, hot-wires, and LDV; and (3) whole-field velocity measurement techniques such as PIV. The emphasis in this chapter has been on flow through pipes, including blood vessels. A detailed treatment of numer ous types of pumps and turbines, including their operation principles and performance parameters, is given in Chap. 14.
REFERENCES AND SUGGESTED READING 1. H. S. Bean (ed.). Fluid Meters: Their Theory and Appli cations, 6th ed. New York: American Society of Mechan ical Engineers, 1971.
2. M. S. Bhatti and R. K. Shah. “Turbulent and Transition Flow Convective Heat Transfer in Ducts.” In Handbook of Single-Phase Convective Heat Transfer, ed. S. Kakaç, R. K. Shah, and W. Aung. New York: Wiley Interscience, 1987.
3. S. W. Churchill. “Friction Factor Equation Spans all Fluid-Flow Regimes,” Chemical Engineering, 7 (1977), pp. 91–92.
4. B. T. Cooper, B. N. Roszelle, T. C. Long, S. Deutsch, and K. B. Manning. “The 12 cc Penn State pulsatile pediatric ventricular assist device: fluid dynamics associated with valve selection.” J. of Biomechonicol Engineering, 130 (2008), pp. 041019. 5. C. F. Colebrook. “Turbulent Flow in Pipes, with Par ticular Reference to the Transition between the Smooth and Rough Pipe Laws,” Journal of the Institute of Civil Engineers London, 11 (1939), pp. 133–156.
6. F. Durst, A. Melling, and J. H. Whitelaw. Principles and Practice of Laser-Doppler Anemometry, 2nd ed. New York: Academic, 1981.
7. Fundamentals of Orifice Meter Measurement. Houston, TX: Daniel Measurement and Control, 1997.
8. S. E. Haaland. “Simple and Explicit Formulas for the Friction Factor in Turbulent Pipe Flow,” Journal of Fluids Engineering, March 1983, pp. 89–90.
9. I. E. Idelchik. Handbook of Hydraulic Resistance, 3rd ed. Boca Raton, FL: CRC Press, 1993.
10. W. M. Kays, M. E. Crawford, and B. Weigand. Convective Heat and Mass Transfer, 4th ed. New York: McGraw-Hill, 2004.
11. K. B. Manning, L. H. Herbertson, A. A. Fontaine, and S. S. Deutsch. “A detailed fluid mechanics study of tilting disk mechanical heart valve closure and the implications to blood damage.” J. Biomech. Eng. 130(4) (2008), pp. 041001-1-4.
12. R. W. Miller. Flow Measurement Engineering Hand book, 3rd ed. New York: McGraw-Hill, 1997.
13. L. F. Moody. “Friction Factors for Pipe Flows,” Transac tions of the ASME 66 (1944), pp. 671–684.
14. G. Rosenberg, W. M. Phillips, D. L. Landis, and W. S. Pierce. “Design and evaluation of the Pennsylvania State University Mock Circulatory System.” ASAIO J. 4 (1981) pp. 41–49.
15. O. Reynolds. “On the Experimental Investigation of the Circumstances Which Determine Whether the Motion cen96537_ch08_351-442.indd 424 14/01/17 3:00 pm 425 CHAPTER 8 of Water Shall Be Direct or Sinuous, and the Law of Resistance in Parallel Channels.” Philosophical Transactions of the Royal Society of London, 174 (1883), pp. 935–982.
16. H. Schlichting. Boundary Layer Theory, 7th ed. New York: Springer, 2000.
17. R. K. Shah and M. S. Bhatti. “Laminar Convective Heat Transfer in Ducts.” In Handbook of Single-Phase Convective Heat Transfer, ed. S. Kakaç, R. K. Shah, and W. Aung. New York: Wiley Interscience, 1987.
18. P. L. Skousen. Valve Handbook. New York: McGraw-Hill, 1998.
19. P. K. Swamee and A. K. Jain. “Explicit Equations for Pipe-Flow Problems,” Journal of the Hydraulics Division. ASCE 102, no. HY5 (May 1976), pp. 657–664. 20. G. Vass. “Ultrasonic Flowmeter Basics,” Sensors, 14, no. 10 (1997).
21. A. J. Wheeler and A. R. Ganji. Introduction to Engineer ing Experimentation. Englewood Cliffs, NJ: Prentice-Hall, 1996.
22. W. Zhi-qing. “Study on Correction Coefficients of Laminar and Turbulent Entrance Region Effects in Round Pipes,” Applied Mathematical Mechanics, 3 (1982), p. 433.
PROBLEMS Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and SI users can ignore them. Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software.
Laminar and Turbulent Flow 8–1C How is the hydrodynamic entry length defined for flow in a pipe? Is the entry length longer in laminar or tur bulent flow?
8–2C Why are liquids usually transported in circular pipes?
8–3C What is the physical significance of the Reynolds number? How is it defined for (a) flow in a circular pipe of inner diameter D and (b) flow in a rectangular duct of cross section a × b?
a b D FIGURE P8–3C 8–4C Consider a person walking first in air and then in water at the same speed. For which motion will the Reynolds number be higher?
8–5C Show that the Reynolds number for flow in a circular pipe of diameter D can be expressed as Re = 4m ./(𝜋D𝜇).
8–6C Which fluid at room temperature requires a larger pump to flow at a specified velocity in a given pipe: water or engine oil? Why?
8–7C What is the generally accepted value of the Reynolds number above which the flow in smooth pipes is turbulent?
8–8C Consider the flow of air and water in pipes of the same diameter, at the same temperature, and at the same mean velocity. Which flow is more likely to be turbulent? Why?
8–9C Consider laminar flow in a circular pipe. Is the wall shear stress 𝜏w higher near the inlet of the pipe or near the exit? Why? What would your response be if the flow were turbulent?
8–10C How does surface roughness affect the pressure drop in a pipe if the flow is turbulent? What would your response be if the flow were laminar?
8–11C What is hydraulic diameter? How is it defined? What is it equal to for a circular pipe of diameter D?
8–12E Shown here is a cool picture of water being released at 300,000 gallons per second in the spring of 2008. This was part of a revitalization effort for the ecosystem of the Grand Canyon and the Colorado River. Estimate the Reynolds num ber of the pipe flow. Is it laminar or turbulent? (Hint: For a length scale, approximate the height of the man in the blue shirt directly above the pipe to be 6 ft.) FIGURE P8–12E Courtesy U.S. Bureau of Reclamation, Public Affairs PN Region.
cen96537_ch08_351-442.indd 425 14/01/17 3:00 pm 426 internal flow Fully Developed Flow in Pipes 8–13C What fluid property is responsible for the develop ment of the velocity boundary layer? For what kinds of fluids will there be no velocity boundary layer in a pipe?
8–14C In the fully developed region of flow in a circular pipe, does the velocity profile change in the flow direction?
8–15C Someone claims that the volume flow rate in a circular pipe with laminar flow can be determined by measuring the velocity at the centerline in the fully developed region, multiplying it by the cross-sectional area, and dividing the result by 2. Do you agree? Explain.
8–16C Someone claims that the average velocity in a circular pipe in fully developed laminar flow can be determined by simply measuring the velocity at R/2 (midway between the wall surface and the centerline). Do you agree? Explain.
8–17C Someone claims that the shear stress at the center of a circular pipe during fully developed laminar flow is zero. Do you agree with this claim? Explain.
8–18C Someone claims that in fully developed turbulent flow in a pipe, the shear stress is a maximum at the pipe wall. Do you agree with this claim? Explain.
8–19C How does the wall shear stress 𝜏w vary along the flow direction in the fully developed region in (a) laminar flow and (b) turbulent flow?
8–20C How is the friction factor for flow in a pipe related to the pressure loss? How is the pressure loss related to the pumping power requirement for a given mass flow rate?
8–21C Discuss whether fully developed pipe flow is one-, two-, or three-dimensional.
8–22C Consider fully developed flow in a circular pipe with negligible entrance effects. If the length of the pipe is doubled, the head loss will (a) double, (b) more than double, (c) less than double, (d ) reduce by half, or (e) remain constant.
8–23C Consider fully developed laminar flow in a circular pipe. If the diameter of the pipe is reduced by half while the flow rate and the pipe length are held constant, the head loss will (a) double, (b) triple, (c) quadruple, (d ) increase by a factor of 8, or (e) increase by a factor of 16.
8–24C Explain why the friction factor is independent of the Reynolds number at very large Reynolds numbers.
8–25C Consider laminar flow of air in a circular pipe with perfectly smooth surfaces. Do you think the friction factor for this flow is zero? Explain.
8–26C Consider fully developed laminar flow in a circular pipe. If the viscosity of the fluid is reduced by half by heat ing while the flow rate is held constant, how does the head loss change?
8–27C How is head loss related to pressure loss? For a given fluid, explain how you would convert head loss to pressure loss.
8–28C What is turbulent viscosity? What causes it? 8–29C What is the physical mechanism that causes the fric tion factor to be higher in turbulent flow?
8–30C The head loss for a certain circular pipe is given by hL = 0.0826fL(V .
2/D5), where f is the friction factor (dimen sionless), L is the pipe length, V . is the volumetric flow rate, and D is the pipe diameter. Determine if the 0.0826 is a dimen sional or dimensionless constant. Is this equation dimen sion ally homogeneous as it stands? 8–31 The velocity profile for the fully developed laminar flow of a Newtonian fluid between two large parallel plates is given by u(y) = 3u0 2 [1 −( y h) 2 ] where 2h is the distance between the two plates, u0 is the velocity at the center plane, and y is the vertical coordinate from the center plane. For a plate width of b, obtain a rela tion for the flow rate through the plates.
8–32 Water at 15°C (𝜌 = 999.1 kg/m3 and 𝜇 = 1.138 × 10−3 kg/m·s) is flowing steadily in a 30-m-long and 6-cm-diameter horizontal pipe made of stainless steel at a rate of 10 L/s. Determine (a) the pressure drop, (b) the head loss, and (c) the pumping power requirement to overcome this pressure drop.
30 m 6 cm 10 L/s FIGURE P8–32 8–33E Water at 70°F passes through 0.75-in-internal- diameter copper tubes at a rate of 0.5 lbm/s. Determine the pumping power per ft of pipe length required to maintain this flow at the specified rate.
8–34E Heated air at 1 atm and 100°F is to be transported in a 400-ft-long circular plastic duct at a rate of 12 ft3/s. If the head loss in the pipe is not to exceed 50 ft, determine the minimum diameter of the duct.
8–35 In fully developed laminar flow in a circular pipe, the velocity at R/2 (midway between the wall surface and the centerline) is measured to be 11 m/s. Determine the velocity at the center of the pipe. Answer: 14.7 m/s 8–36 The velocity profile in fully developed laminar flow in a circular pipe of inner radius R = 2 cm, in m/s, is given cen96537_ch08_351-442.indd 426 14/01/17 3:00 pm 427 CHAPTER 8 by u(r) = 4(1 − r2/R2). Determine the average and maximum velocities in the pipe and the volume flow rate.
R = 2 cm u(r) = 4 1 – r2 –– R2 FIGURE P8–36 8–37 Repeat Prob. 8–36 for a pipe of inner radius 7 cm.
8–38 Water at 10°C (𝜌 = 999.7 kg/m3 and 𝜇 = 1.307 × 10−3 kg/m·s) is flowing steadily in a 0.12-cm-diameter, 15-m-long pipe at an average velocity of 0.9 m/s. Determine (a) the pressure drop, (b) the head loss, and (c) the pumping power requirement to overcome this pressure drop. Answers: (a) 392 kPa, (b) 40.0 m, (c) 0.399 W 8–39 Consider laminar flow of a fluid through a square channel with smooth surfaces. Now the average velocity of the fluid is doubled. Determine the change in the head loss of the fluid. Assume the flow regime remains unchanged.
8–40 Repeat Prob. 8–39 for turbulent flow in smooth pipes for which the friction factor is given as f = 0.184Re−0.2. What would your answer be for fully turbulent flow in a rough pipe?
8–41 Air enters a 10-m-long section of a rectangular duct of cross section 15 cm × 20 cm made of commercial steel at 1 atm and 35°C at an average velocity of 5 m/s. Disregarding the entrance effects, determine the fan power needed to overcome the pressure losses in this section of the duct. Answer: 2.55 W 10 m 15 cm 20 cm Air 5 m/s FIGURE P8–41 8–42 Consider an air solar collector that is 1 m wide and 4 m long and has a constant spacing of 3 cm between the glass cover and the collector plate. Air flows at an aver age temperature of 45°C at a rate of 0.12 m3/s through the 1-m-wide edge of the collector along the 4-m-long passage way. Disregarding the entrance and roughness effects and the 90° bend, determine the pressure drop in the collector. Answer: 17.5 Pa Collector plate Insulation Glass cover 4 m Air 0.12 m3/s FIGURE P8–42 8–43 Oil with 𝜌 = 876 kg/m3 and 𝜇 = 0.24 kg/m·s is flow ing through a 1.5-cm-diameter pipe that discharges into the atmosphere at 88 kPa. The absolute pressure 15 m before the exit is measured to be 135 kPa. Determine the flow rate of oil through the pipe if the pipe is (a) horizontal, (b) inclined 8° upward from the horizontal, and (c) inclined 8° downward from the horizontal.
1.5 cm 15 m 135 kPa Oil FIGURE P8–43 8–44 Glycerin at 40°C with 𝜌 = 1252 kg/m3 and 𝜇 = 0.27 kg/m·s is flowing through a 3-cm-diameter, 25-m-long pipe that discharges into the atmosphere at 100 kPa. The flow rate through the pipe is 0.075 L/s. (a) Determine the absolute pressure 25 m before the pipe exit. (b) At what angle 𝜃 must the pipe be inclined downward from the hori zontal for the pressure in the entire pipe to be atmospheric pressure and the flow rate to be maintained the same?
8–45E Air at 1 atm and 60°F is flowing through a 1 ft × 1 ft square duct made of commercial steel at a rate of 1600 cfm. Determine the pressure drop and head loss per ft of the duct.
1 ft 1 ft Air 1600 ft3/min FIGURE P8–45E cen96537_ch08_351-442.indd 427 14/01/17 3:00 pm 428 internal flow 8–46 Oil with a density of 850 kg/m3 and kinematic viscos ity of 0.00062 m2/s is being discharged by a 8-mm-diameter, 40-m-long horizontal pipe from a storage tank open to the atmosphere. The height of the liquid level above the center of the pipe is 4 m. Disregarding the minor losses, determine the flow rate of oil through the pipe.
Oil tank 4 m 8 mm FIGURE P8–46 8–47 In an air heating system, heated air at 40°C and 105 kPa absolute is distributed through a 0.2 m × 0.3 m rectangular duct made of commercial steel at a rate of 0.5 m3/s. Determine the pressure drop and head loss through a 40-m-long section of the duct. Answers: 124 Pa, 10.8 m 8–48 Glycerin at 40°C with 𝜌 = 1252 kg/m3 and 𝜇 = 0.27 kg/m·s is flowing through a 6-cm-diameter horizontal smooth pipe with an average velocity of 3.5 m/s. Determine the pressure drop per 10 m of the pipe.
8–49 Reconsider Prob. 8–48. Using appropriate soft ware, investigate the effect of the pipe diameter on the pressure drop for the same constant flow rate. Let the pipe diameter vary from 1 to 10 cm in increments of 1 cm. Tabulate and plot the results, and draw conclusions.
8–50 Liquid ammonia at −20°C is flowing through a 20-m-long section of a 5-mm-diameter copper tube at a rate of 0.09 kg/s. Determine the pressure drop, the head loss, and the pumping power required to overcome the frictional losses in the tube. Answers: 1240 kPa, 189 m, 0.167 kW 8–51 Consider the fully developed flow of glycerin at 40°C through a 70-m-long, 4-cm-diameter, horizontal, circular pipe. If the flow velocity at the centerline is measured to be 6 m/s, determine the velocity profile and the pressure differ ence across this 70-m-long section of the pipe, and the useful pumping power required to maintain this flow.
FIGURE P8–51 6 m/s Glycerine 70 m D = 2 cm 8–52 The velocity profile for a steady laminar flow in a circular pipe of radius R is given by u = u0(1 − r 2 R2). If the fluid density varies with radial distance r from the centerline as ρ = ρ0(1 + r R) 1 ╱ 4 where ρ0 is the fluid density at the pipe center, obtain a relation for the bulk fluid density in the tube.
8–53 The generalized Bernoulli equation for unsteady flows can be expressed as P1 ρg + z1 = V 2 2g + 1 g∫ 2 1 ∂V ∂t ds + hL If the valve suddenly opened, the exit velocity will vary with time. Develop an expression for exit velocity V as a function of time. Neglect local losses.
2 L h Valve d 1 FIGURE P8–53 Minor Losses 8–54C What is minor loss in pipe flow? How is the minor loss coefficient KL defined?
8–55C Define equivalent length for minor loss in pipe flow. How is it related to the minor loss coefficient?
8–56C The effect of rounding of a pipe inlet on the loss coefficient is (a) negligible, (b) somewhat significant, or (c) very significant.
8–57C The effect of rounding of a pipe exit on the loss coefficient is (a) negligible, (b) somewhat significant, or (c) very significant.
8–58C Which has a greater minor loss coefficient during pipe flow: gradual expansion or gradual contraction? Why?
8–59C A piping system involves sharp turns, and thus large minor head losses. One way of reducing the head loss is to replace the sharp turns by circular elbows. What is another way?
8–60C During a retrofitting project of a fluid flow system to reduce the pumping power, it is proposed to install vanes into the miter elbows or to replace the sharp turns in 90° miter elbows by smooth curved bends. Which approach will result in a greater reduction in pumping power requirements?
cen96537_ch08_351-442.indd 428 14/01/17 3:00 pm 429 CHAPTER 8 8–61 A horizontal pipe has an abrupt expansion from D1 = 5 cm to D2 = 10 cm. The water velocity in the smaller section is 8 m/s and the flow is turbulent. The pressure in the smaller section is P1 = 410 kPa. Taking the kinetic energy correction factor to be 1.06 at both the inlet and the outlet, determine the downstream pressure P2, and estimate the error that would have occurred if Bernoulli’s equation had been used. Answers: 424 kPa, 16.2 kPa D2 = 10 cm D1 = 5 cm 8 m/s 410 kPa Water FIGURE P8–61 8–62 Consider flow from a water reservoir through a circu lar hole of diameter D at the side wall at a vertical distance H from the free surface. The flow rate through an actual hole with a sharp-edged entrance (KL = 0.5) is considerably less than the flow rate calculated assuming “frictionless” flow and thus zero loss for the hole. Disregarding the effect of the kinetic energy correction factor, obtain a relation for the “equivalent diameter” of the sharp-edged hole for use in fric tionless flow relations.
D Frictionless flow Actual flow Dequiv FIGURE P8–62 8–63 Repeat Prob. 8–62 for a slightly rounded entrance (KL = 0.12).
8–64 Water is to be withdrawn from an 8-m-high water res ervoir by drilling a 2.2-cm-diameter hole at the bottom sur face. Disregarding the effect of the kinetic energy correction factor, determine the flow rate of water through the hole if (a) the entrance of the hole is well-rounded and (b) the entrance is sharp-edged. Piping Systems and Pump Selection 8–65C A piping system equipped with a pump is operating steadily. Explain how the operating point (the flow rate and the head loss) is established.
8–66C Water is pumped from a large lower reservoir to a higher reservoir. Someone claims that if the head loss is negligible, the required pump head is equal to the elevation difference between the free surfaces of the two reservoirs. Do you agree? 8–67C For a piping system, define the system curve, the characteristic curve, and the operating point on a head versus flow rate chart.
8–68C A person filling a bucket with water using a garden hose suddenly remembers that attaching a nozzle to the hose increases the discharge velocity of water and wonders if this increased velocity would decrease the filling time of the bucket. What would happen to the filling time if a nozzle were attached to the hose: increase it, decrease it, or have no effect? Why?
8–69C Consider two identical 2-m-high open tanks filled with water on top of a 1-m-high table. The discharge valve of one of the tanks is connected to a hose whose other end is left open on the ground while the other tank does not have a hose connected to its discharge valve. Now the discharge valves of both tanks are opened. Disregarding any frictional loses in the hose, which tank do you think empties com pletely first? Why?
8–70C A piping system involves two pipes of different diameters (but of identical length, material, and roughness) connected in series. How would you compare the (a) flow rates and (b) pressure drops in these two pipes?
8–71C A piping system involves two pipes of different diameters (but of identical length, material, and roughness) connected in parallel. How would you compare the (a) flow rates and (b) pressure drops in these two pipes?
8–72C A piping system involves two pipes of identical diameters but of different lengths connected in parallel. How would you compare the pressure drops in these two pipes?
8–73 Water at 15°C is drained from a large reservoir using two horizontal plastic pipes connected in series. The first pipe is 13 m long and has a 10-cm diameter, while the second pipe is 35 m long and has a 5-cm diameter. The water level in the reser voir is 18 m above the centerline of the pipe. The pipe entrance is sharp-edged, and the contraction between the two pipes is sudden. Neglecting the effect of the kinetic energy correction factor, determine the discharge rate of water from the reservoir.
Water tank 18 m 13 m 35 m FIGURE P8–73 8–74 A semi-spherical tank of radius R is completely filled with water. Now a hole of cross-sectional area Ah and dis charge coefficient Cd at the bottom of the tank is fully opened and water starts to flow out. Develop an expression for the time needed to empty the tank completely. cen96537_ch08_351-442.indd 429 14/01/17 3:00 pm 430 internal flow h(t) R Water FIGURE P8–74 8–75 The water needs of a small farm are to be met by pumping water from a well that can supply water continuously at a rate of 5 L/s. The water level in the well is 20 m below the ground level, and water is to be pumped to a large tank on a hill, which is 58 m above the ground level of the well, using 6-cm internal diameter plastic pipes. The required length of piping is measured to be 510 m, and the total minor loss coefficient due to the use of elbows, vanes, etc. is estimated to be 12. Taking the efficiency of the pump to be 75 percent, determine the rated power of the pump that needs to be purchased, in kW. The den sity and viscosity of water at anticipated operation conditions are taken to be 1000 kg/m3 and 0.00131 kg/m⋅s, respectively. Is it wise to purchase a suitable pump that meets the total power requirements, or is it necessary to also pay particular attention to the large elevation head in this case? Explain. Answer: 6.89 kW 8–76E Water at 70°F flows by gravity from a large reser voir at a high elevation to a smaller one through a 60-ft-long, 2-in-diameter cast iron piping system that includes four stand ard flanged elbows, a well-rounded entrance, a sharp-edged exit, and a fully open gate valve. Taking the free surface of the lower reservoir as the reference level, determine the ele vation z1 of the higher reservoir for a flow rate of 10 ft3/min.
Answer: 12.6 ft 8–77 A 2.4-m-diameter tank is initially filled with water 4 m above the center of a sharp-edged 10-cm-diameter orifice. The tank water surface is open to the atmosphere, and the orifice drains to the atmosphere. Neglecting the effect of the kinetic energy correction factor, calculate (a) the initial velocity from the tank and (b) the time required to empty the tank. Does the loss coefficient of the orifice cause a signifi cant increase in the draining time of the tank?
Water tank 4 m 2.4 m Sharp-edged orifice FIGURE P8–77 8–78 A 3-m-diameter tank is initially filled with water 2 m above the center of a sharp-edged 10-cm-diameter orifice. The tank water surface is open to the atmosphere, and the ori fice drains to the atmosphere through a 100-m-long pipe. The friction coefficient of the pipe is taken to be 0.015 and the effect of the kinetic energy correction factor can be neglected. Determine (a) the initial velocity from the tank and (b) the time required to empty the tank.
8–79 Reconsider Prob. 8–78. In order to drain the tank faster, a pump is installed near the tank exit as in Fig. P8–79. Determine how much pump power input is nec essary to establish an average water velocity of 4 m/s when the tank is full at z = 2 m. Also, assuming the discharge velocity to remain constant, estimate the time required to drain the tank. Someone suggests that it makes no difference whether the pump is located at the beginning or at the end of the pipe, and that the performance will be the same in either case, but another person argues that placing the pump near the end of the pipe may cause cavitation. The water temperature is 30°C, so the water vapor pressure is Pv = 4.246 kPa = 0.43 m H2O, and the system is located at sea level. Investigate if there is the possibility of cavitation and if we should be concerned about the location of the pump.
Water tank Pump 4 m/s 2 m 3 m FIGURE P8–79 8–80 Gasoline (𝜌 = 680 kg/m3 and 𝜈 = 4.29 × 10−7 m2/s) is transported at a rate of 240 L/s for a distance of 2 km. The surface roughness of the piping is 0.03 mm. If the head loss due to pipe friction is not to exceed 10 m, determine the minimum diameter of the pipe.
8–81E A clothes dryer discharges air at 1 atm and 120°F at a rate of 1.2 ft3/s when its 5-in-diameter, well-rounded vent with negligible loss is not connected to any duct. Determine the flow rate when the vent is connected to a 15-ft-long, 5-in-diameter duct made of galvanized iron, with three 90° flanged smooth bends. Take the friction fac tor of the duct to be 0.019, and assume the fan power input to remain constant.
cen96537_ch08_351-442.indd 430 14/01/17 3:00 pm 431 CHAPTER 8 Hot air Clothes drier 15 ft 5 in FIGURE P8–81E 8–82 Oil at 20°C is flowing through a vertical glass fun nel that consists of a 20-cm-high cylindrical reservoir and a 1-cm-diameter, 40-cm-high pipe. The funnel is always main tained full by the addition of oil from a tank. Assuming the entrance effects to be negligible, determine the flow rate of oil through the funnel and calculate the “funnel effectiveness,” which is defined as the ratio of the actual flow rate through the funnel to the maximum flow rate for the “frictionless” case. Answers: 3.83 × 10−6 m3/s, 1.4 percent 20 cm 40 cm 1 cm Oil Oil FIGURE P8–82 8–83 Repeat Prob. 8–82 assuming (a) the diameter of the pipe is tripled and (b) the length of the pipe is tripled while the diameter is maintained the same.
8–84 A 4-m-high cylindrical tank having a cross-sectional area of AT = 1.5 m2 is filled with equal volumes of water and oil whose specific gravity is SG = 0.75. Now a 1-cm-diameter hole at the bottom of the tank is opened, and water starts to flow out. If the discharge coefficient of the hole is Cd = 0.85, determine how long it will take for the water in the tank, which is open to the atmosphere to empty completely. 8–85E A farmer is to pump water at 70°F from a river to a water storage tank nearby using a 125-ft-long, 5-in-diameter plastic pipe with three flanged 90° smooth bends. The water velocity near the river surface is 6 ft/s, and the pipe inlet is placed in the river normal to the flow direction of water to take advantage of the dynamic pressure. The elevation dif ference between the river and the free surface of the tank is 12 ft. For a flow rate of 1.5 ft3/s and an overall pump effi ciency of 70 percent, determine the required electric power input to the pump.
8–86E Reconsider Prob. 8–85E. Using appropriate software, investigate the effect of the pipe diameter on the required electric power input to the pump. Let the pipe diameter vary from 1 to 10 in, in increments of 1 in. Tabulate and plot the results, and draw conclusions.
8–87 A water tank filled with solar-heated water at 40°C is to be used for showers in a field using gravity-driven flow. The system includes 35 m of 1.5-cm-diameter galvanized iron piping with four miter bends (90°) without vanes and a wide-open globe valve. If water is to flow at a rate of 1.2 L/s through the shower head, determine how high the water level in the tank must be from the exit level of the shower. Disre gard the losses at the entrance and at the shower head, and neglect the effect of the kinetic energy correction factor.
8–88 Two water reservoirs A and B are connected to each other through a 40-m-long, 2-cm-diameter cast iron pipe with a sharp-edged entrance. The pipe also involves a swing check valve and a fully open gate valve. The water level in both reservoirs is the same, but reservoir A is pressurized by compressed air while reservoir B is open to the atmosphere at 95 kPa. If the initial flow rate through the pipe is 1.5 L/s, determine the absolute air pressure on top of reservoir A. Take the water temperature to be 10°C. Answer: 1100 kPa Air 2 cm 40 m FIGURE P8–88 8–89 A vented tanker is to be filled with fuel oil with 𝜌 = 920 kg/m3 and 𝜇 = 0.045 kg/m·s from an under ground reservoir using a 25-m-long, 4-cm-diameter plas tic hose with a slightly rounded entrance and two 90° smooth bends. The elevation difference between the oil level in the reservoir and the top of the tanker where the hose is discharged is 5 m. The capacity of the tanker is 18 m3 and the filling time is 30 min. Taking the kinetic energy correction factor at the hose discharge to be 1.05 and assuming an overall pump efficiency of 82 percent, determine the required power input to the pump.
cen96537_ch08_351-442.indd 431 14/01/17 3:00 pm 432 internal flow 4 cm Pump 5 m 25 m Tanker 18 m3 FIGURE P8–89 8–90 Water at 15°C is to be pumped from a reser voir (zA = 2 m) to another reservoir at a higher eleva tion (zB = 9 m) through two 25-m-long plastic pipes con nected in parallel. The diameters of the two pipes are 3 cm and 5 cm. Water is to be pumped by a 68 percent effi cient motor–pump unit that draws 8 kW of electric power dur ing operation. The minor losses and the head loss in the pipes that connect the parallel pipes to the two reservoirs are consid ered to be negligible. Determine the total flow rate between the reservoirs and the flow rates through each of the parallel pipes.
Pump Reservoir A zA = 2 m 25 m 3 cm 5 cm Reservoir B zB = 9 m FIGURE P8–90 8–91 A certain part of cast iron piping of a water distri bution system involves a parallel section. Both parallel pipes have a diameter of 30 cm, and the flow is fully tur bulent. One of the branches (pipe A) is 1500 m long while the other branch (pipe B) is 2500 m long. If the flow rate through pipe A is 0.4 m3/s, determine the flow rate through pipe B. Disregard minor losses and assume the water temperature to be 15°C. Show that the flow is fully rough, and thus the friction factor is independent of Reynolds number. Answer: 0.310 m3/s 2500 m 30 cm 0.4 m3/s 30 cm 1500 m A B FIGURE P8–91 8–92 Repeat Prob. 8–91 assuming pipe A has a halfway-closed gate valve (KL = 2.1) while pipe B has a fully open globe valve (KL = 10), and the other minor losses are negligible. 8–93 A geothermal district heating system involves the transport of geothermal water at 110°C from a geothermal well to a city at about the same elevation for a distance of 12 km at a rate of 1.5 m3/s in 60-cm-diameter stainless-steel pipes. The fluid pressures at the wellhead and the arrival point in the city are to be the same. The minor losses are negligi ble because of the large length-to-diameter ratio and the rela tively small number of components that cause minor losses. (a) Assuming the pump–motor efficiency to be 80 percent, determine the electric power consumption of the system for pumping. Would you recommend the use of a single large pump or several smaller pumps of the same total pumping power scattered along the pipeline? Explain. (b) Determine the daily cost of power consumption of the system if the unit cost of electricity is $0.06/kWh. (c) The temperature of geo thermal water is estimated to drop 0.5°C during this long flow. Determine if the frictional heating during flow can make up for this drop in temperature.
8–94 Repeat Prob. 8–93 for cast iron pipes of the same diameter.
8–95 Water is transported by gravity through a 10-cm-diameter 550-m-long plastic pipe with an elevation gradi ent of 0.01 (i.e., an elevation drop of 1 m per 100 m of pipe length). Taking 𝜌 = 1000 kg/m3 and 𝜈 = 1 × 10−6 m2/s for water, determine the flow rate of water through the pipe. If the pipe were horizontal, what would the power requirements be to maintain the same flow rate? 8–96 Water to a residential area is transported at a rate of 1.5 m3/s via 70-cm-internal-diameter concrete pipes with a surface roughness of 3 mm and a total length of 1500 m. In order to reduce pumping power requirements, it is proposed to line the interior surfaces of the concrete pipe with 2-cm-thick petroleum-based lining that has a surface roughness thickness of 0.04 mm. There is a concern that the reduction of pipe diameter to 66 cm and the increase in average velocity may offset any gains. Taking 𝜌 = 1000 kg/m3 and 𝜈 = 1 × 10−6 m2/s for water, determine the percent increase or decrease in the pumping power requirements due to pipe frictional losses as a result of lining the concrete pipes. 8–97 In large buildings, hot water in a water tank is circulated through a loop so that the user doesn’t have to wait for all the water in long piping to drain before hot water starts coming out. A certain recirculating loop involves 40-m-long, 1.2-cm-diameter cast iron pipes with six 90° threaded smooth bends and two fully open gate valves. If the average flow velocity through the loop is 2 m/s, determine the required power input for the recir culating pump. Take the average water temperature to be 60°C and the efficiency of the pump to be 70 percent. Answer: 0.111 kW cen96537_ch08_351-442.indd 432 14/01/17 3:00 pm 433 CHAPTER 8 8–98 Reconsider Prob. 8–97. Using appropriate soft ware, investigate the effect of the average flow velocity on the power input to the recirculating pump. Let the velocity vary from 0 to 3 m/s in increments of 0.3 m/s. Tabu late and plot the results.
8–99 Repeat Prob. 8–97 for plastic (smooth) pipes.
8–100 Two pipes of identical length and material are con nected in parallel. The diameter of pipe A is twice the diam eter of pipe B. Assuming the friction factor to be the same in both cases and disregarding minor losses, determine the ratio of the flow rates in the two pipes.
Flow Rate and Velocity Measurements 8–101C What are the primary considerations when select ing a flowmeter to measure the flow rate of a fluid?
8–102C What is the difference between laser Doppler velo cim etry (LDV) and particle image velocimetry (PIV)?
8–103C What is the difference between the operating prin ciples of thermal and laser Doppler anemometers?
8–104C Explain how flow rate is measured with a Pitot-static tube, and discuss its advantages and disadvantages with respect to cost, pressure drop, reliability, and accuracy.
8–105C Explain how flow rate is measured with obstruc tion-type flowmeters. Compare orifice meters, flow nozzles, and Venturi meters with respect to cost, size, head loss, and accuracy.
8–106C How do positive displacement flowmeters operate? Why are they commonly used to meter gasoline, water, and natural gas?
8–107C Explain how flow rate is measured with a turbine flowmeter, and discuss how they compare to other types of flowmeters with respect to cost, head loss, and accuracy.
8–108C What is the operating principle of variable-area flow meters (rotameters)? How do they compare to other types of flowmeters with respect to cost, head loss, and reliability?
8–109 A 15-L kerosene tank (𝜌 = 820 kg/m3) is filled with a 2-cm-diameter hose equipped with a 1.5-cm-diameter noz zle meter. If it takes 18 s to fill the tank, determine the pres sure difference indicated by the nozzle meter.
8–110 A Pitot-static probe is mounted in a 2.5-cm-inner dia meter pipe at a location where the local velocity is approx imately equal to the average velocity. The oil in the pipe has density 𝜌 = 860 kg/m3 and viscosity 𝜇 = 0.0103 kg/m⋅s.
The pressure difference is measured to be 95.8 Pa. Calculate the volume flow rate through the pipe in cubic meters per second.
8–111 Calculate the Reynolds number of the flow of Prob. 8–110. Is it laminar or turbulent?
8–112 A flow nozzle equipped with a differential pres sure gage is used to measure the flow rate of water at 10°C (𝜌 = 999.7 kg/m3 and 𝜇 = 1.307 × 10−3 kg/m·s) through a 3-cm-diameter horizontal pipe. The nozzle exit diameter is 1.5 cm, and the measured pressure drop is 3 kPa. Determine the volume flow rate of water, the average velocity through the pipe, and the head loss.
1.5 cm 3 cm ΔP = 3 kPa Differential pressure gage FIGURE P8–112 8–113 The flow rate of ammonia at 10°C (𝜌 = 624.6 kg/m3 and 𝜇 = 1.697 × 10−4 kg/m·s) through a 2-cm-diameter pipe is to be measured with a 1.5-cm-diameter flow nozzle equipped with a differential pressure gage. If the gage reads a pressure differential of 4 kPa, determine the flow rate of ammonia through the pipe, and the average flow velocity.
8–114E An orifice with a 1.8-in-diameter opening is used to measure the mass flow rate of water at 60°F (𝜌 = 62.36 lbm/ft3 and 𝜇 = 7.536 × 10−4 lbm/ft·s) through a horizontal 4-in-diameter pipe. A mercury manometer is used to measure the pressure difference across the orifice. If the differential height of the manometer is 7 in, determine the volume flow rate of water through the pipe, the average velocity, and the head loss caused by the orifice meter.
1.8 in 4 in 7 in FIGURE P8–114E 8–115E Repeat Prob. 8–114E for a differential height of 10 in.
cen96537_ch08_351-442.indd 433 14/01/17 3:00 pm 434 internal flow 8–116 Air (𝜌 = 1.225 kg/m3 and 𝜇 = 1.789 × 10−5 kg/m⋅s) flows in a wind tunnel, and the wind tunnel speed is measured with a Pitot-static probe. For a certain run, the stagnation pressure is measured to be 560.4 Pa gage and the static pres sure is 12.7 Pa gage. Calculate the wind-tunnel speed.
8–117 A Venturi meter equipped with a differential pres sure gage is used to measure the flow rate of water at 15°C (𝜌 = 999.1 kg/m3) through a 5-cm-diameter horizontal pipe. The diameter of the Venturi neck is 3 cm, and the measured pressure drop is 5 kPa. Taking the discharge coefficient to be 0.98, determine the volume flow rate of water and the aver age velocity through the pipe. Answers: 2.35 L/s and 1.20 m/s 5 cm 3 cm ΔP Differential pressure gage FIGURE P8–117 8–118 Reconsider Prob. 8–117. Letting the pressure drop vary from 1 kPa to 10 kPa, evaluate the flow rate at intervals of 1 kPa, and plot it against the pressure drop.
8–119 The mass flow rate of air at 20°C (𝜌 = 1.204 kg/m3) through a 18-cm-diameter duct is measured with a Venturi meter equipped with a water manometer. The Venturi neck has a diameter of 5 cm, and the manometer has a maximum differential height of 60 cm. Taking the discharge coefficient to be 0.98, determine the maximum mass flow rate of air this Venturi meter/manometer can measure. Answer: 0.230 kg/s 18 cm 5 cm Water manometer h FIGURE P8–119 8–120 Repeat Prob. 8–119 for a Venturi neck diameter of 6 cm.
8–121 A vertical Venturi meter equipped with a differen tial pressure gage shown in Fig. P8–121 is used to measure the flow rate of liquid propane at 10°C (𝜌 = 514.7 kg/m3) through an 10-cm-diameter vertical pipe. For a discharge coefficient of 0.98, determine the volume flow rate of pro pane through the pipe.
10 cm 5 cm 30 cm ΔP = 7 kPa FIGURE P8–121 8–122E The volume flow rate of liquid refrigerant-134a at 10°F (𝜌 = 83.31 lbm/ft3) is to be measured with a horizontal Venturi meter with a diameter of 6 in at the inlet and 2 in at the throat. If a differential pressure meter indicates a pres sure drop of 7.5 psi, determine the flow rate of the refriger ant. Take the discharge coefficient of the Venturi meter to be 0.98.
8–123 The flow rate of water at 20°C (𝜌 = 998 kg/m3 and 𝜇 = 1.002 × 10−3 kg/m·s) through a 60-cm-diameter pipe is measured with an orifice meter with a 30-cm-diameter open ing to be 350 L/s. Determine the pressure difference indi cated by the orifice meter and the head loss.
8–124 The flow rate of water at 20°C (𝜌 = 998 kg/m3 and 𝜇 = 1.002 × 10−3 kg/m·s) through a 4-cm-diameter pipe is measured with a 2-cm-diameter nozzle meter equipped with an inverted air–water manometer. If the manometer indi cates a differential water height of 44 cm, determine the cen96537_ch08_351-442.indd 434 14/01/17 3:00 pm 435 CHAPTER 8 volume flow rate of water and the head loss caused by the nozzle meter.
2 cm 4 cm 44 cm Water FIGURE P8–124 8–125 The flow rate of water through a 10-cm-diameter pipe is to be determined by measuring the water velocity at several locations along a cross section. For the set of meas urements given in the table, determine the flow rate. r, cm V, m/s 0 6.4 1 6.1 2 5.2 3 4.4 4 2.0 5 0.0 Review Problems 8–126 In a laminar flow through a circular tube of radius of R, the velocity and temperature profiles at a cross section are given by u = u0(1 −r2/R2) and T(r) = A + Br2 − Cr4 where A, B and C are positive constants. Obtain a relation for the bulk fluid temperature at that cross section. 8–127 Water enters into a cone of height H and base radius R through a small hole of cross-sectional area Ah and the dis charge coefficient is Cd at the base with a constant uniform velocity of V. Obtain a relation for the variation of water height h from the cone base with time. Air escapes the cone through the tip at the top as water enters the cone from the bottom.
8–128 The conical container with a thin horizontal tube attached at the bottom, shown in Fig. P8–128, is to be used to measure the viscosity of an oil. The flow through the tube is laminar. The discharge time needed for the oil level to drop from h1 to h2 is to be measured by a stopwatch. Develop an expression for the viscosity of oil in the container as a func tion of the discharge time t.
H R h2 d h1 FIGURE P8–128 8–129 In a geothermal district heating system, 10,000 kg/s of hot water must be delivered a distance of 10 km in a horizontal pipe. The minor losses are negligible, and the only significant energy loss arises from pipe friction. The friction factor is taken to be 0.015. Specifying a larger-diameter pipe would reduce water velocity, velocity head, pipe friction, and thus power consumption. But a larger pipe would also cost more money initially to purchase and install. Otherwise stated, there is an optimum pipe diameter that will minimize the sum of pipe cost and future electric power cost. Assume the system will run 24 h/day, every day, for 30 years. During this time the cost of electricity remains constant at $0.06/kWh. Assume system performance stays constant over the decades (this may not be true, especially if highly miner alized water is passed through the pipeline—scale may form). The pump has an overall efficiency of 80 percent. The cost to purchase, install, and insulate a 10-km pipe depends on the diameter D and is given by Cost = $106 D2, where D is in m. Assuming zero inflation and interest rate for simplicity and zero salvage value and zero maintenance cost, determine the optimum pipe diameter.
8–130 The compressed air requirements of a manufactur ing facility are met by a 120-hp compressor that draws in air from the outside through an 9-m-long, 22-cm-diameter duct made of thin galvanized iron sheets. The compressor takes in air at a rate of 0.27 m3/s at the outdoor conditions of 15°C and 95 kPa. Disregarding any minor losses, determine the useful power used by the compressor to overcome the fric tional losses in this duct. Answer: 6.74 W cen96537_ch08_351-442.indd 435 14/01/17 3:00 pm 436 internal flow 9 m 22 cm Air, 0.27 m3/s 15°C, 95 kPa Air compressor 120 hp FIGURE P8–130 8–131 A house built on a riverside is to be cooled in sum mer by utilizing the cool water of the river. A 20-m-long section of a circular stainless-steel duct of 20-cm diameter passes through the water. Air flows through the underwa ter section of the duct at 4 m/s at an average temperature of 15°C. For an overall fan efficiency of 62 percent, determine the fan power needed to overcome the flow resistance in this section of the duct.
River Air Air, 4 m/s FIGURE P8–131 8–132 The velocity profile in fully developed laminar flow in a circular pipe, in m/s, is given by u(r) = 6(1 − 100r2), where r is the radial distance from the centerline of the pipe in m. Determine (a) the radius of the pipe, (b) the average velocity through the pipe, and (c) the maximum velocity in the pipe.
8–133 Oil at 20°C is flowing steadily through a 6-cm-diameter 33-m-long pipe. The pressures at the pipe inlet and outlet are measured to be 745 and 97.0 kPa, respectively, and the flow is expected to be laminar. Determine the flow rate of oil through the pipe, assuming fully developed flow and that the pipe is (a) horizontal, (b) inclined 15° upward, and (c) inclined 15° downward. Also, verify that the flow through the pipe is laminar.
8–134 Two pipes of identical diameter and material are connected in parallel. The length of pipe A is five times the length of pipe B. Assuming the flow is fully turbulent in both pipes and thus the friction factor is independent of the Reynolds number and disregarding minor losses, determine the ratio of the flow rates in the two pipes. Answer: 0.447 8–135 Repeat Prob. 8–134 except let the length of pipe A be three times that of pipe B. Compare this result to that of Prob. 8–134. Does the difference agree with your intuition? Explain.
8–136 Shell-and-tube heat exchangers with hundreds of tubes housed in a shell are commonly used in practice for heat transfer between two fluids. Such a heat exchanger used in an active solar hot-water system transfers heat from a water-antifreeze solution flowing through the shell and the solar collector to fresh water flowing through the tubes at an average temperature of 60°C at a rate of 15 L/s. The heat exchanger contains 80 brass tubes 1 cm in inner diam eter and 1.5 m in length. Disregarding inlet, exit, and header losses, determine the pressure drop across a single tube and the pumping power required by the tube-side fluid of the heat exchanger. After operating a long time, 1-mm-thick scale builds up on the inner surfaces with an equivalent roughness of 0.4 mm. For the same pumping power input, determine the percent reduction in the flow rate of water through the tubes.
1.5 m 80 tubes 1 cm Water FIGURE P8–136 8–137 Water at 15°C is to be discharged from a reservoir at a rate of 18 L/s using two horizontal cast iron pipes con nected in series and a pump between them. The first pipe is 20 m long and has a 6-cm diameter, while the second pipe is 35 m long and has a 3-cm diameter. The water level in the reservoir is 30 m above the centerline of the pipe. The pipe entrance is sharp-edged, and losses associated with the con nection of the pump are negligible. Neglecting the effect of the kinetic energy correction factor, determine the required pumping head and the minimum pumping power to maintain the indicated flow rate.
cen96537_ch08_351-442.indd 436 14/01/17 3:00 pm 437 CHAPTER 8 Water tank Pump 35 m 20 m 30 m 6 cm 3 cm FIGURE P8–137 8–138 Reconsider Prob. 8–137. Using appropriate software, investigate the effect of the second pipe diameter on the required pumping head to maintain the indicated flow rate. Let the diameter vary from 1 to 10 cm in increments of 1 cm. Tabulate and plot the results.
8–139 Consider flow from a reservoir through a horizon tal pipe of length L and diameter D that penetrates into the side wall at a vertical distance H from the free surface. The flow rate through an actual pipe with a reentrant section (KL = 0.8) is considerably less than the flow rate through the hole calculated assuming “frictionless” flow and thus zero loss. Obtain a relation for the “equivalent diameter” of the reentrant pipe for use in relations for frictionless flow through a hole and determine its value for a pipe friction factor, length, and diameter of 0.018, 10 m, and 0.04 m, respectively. Assume the friction factor of the pipe to remain constant and the effect of the kinetic energy correction factor to be negligible.
8–140 A pipeline that transports oil at 40°C at a rate of 3 m3/s branches out into two parallel pipes made of commer cial steel that reconnect downstream. Pipe A is 500 m long and has a diameter of 30 cm while pipe B is 800 m long and has a diameter of 45 cm. The minor losses are considered to be negligible. Determine the flow rate through each of the parallel pipes.
500 m 30 cm 800 m 3 m3/s Oil A B 45 cm FIGURE P8–140 8–141 Repeat Prob. 8–140 for hot-water flow of a district heating system at 100°C.
8–142 Water is to be withdrawn from a 7-m-high water reservoir by drilling a well-rounded 5-cm-diameter hole with negligible loss near the bottom and attaching a horizontal 90° bend of negligible length. Taking the kinetic energy cor rection factor to be 1.05, determine the flow rate of water through the bend if (a) the bend is a flanged smooth bend and (b) the bend is a miter bend without vanes. Answers: (a) 19.8 L/s, (b) 15.7 L/s 7 m FIGURE P8–142 8–143 The compressed air requirements of a textile fac tory are met by a large compressor that draws in 0.6 m3/s air at atmospheric conditions of 20°C and 1 bar (100 kPa) and consumes 300 kW electric power when operating. Air is compressed to a gage pressure of 8 bar (absolute pressure of 900 kPa), and compressed air is transported to the produc tion area through a 15-cm-internal-diameter, 83-m-long, gal vanized steel pipe with a surface roughness of 0.15 mm. The average temperature of compressed air in the pipe is 60°C. The compressed air line has 8 elbows with a loss coefficient of 0.6 each. If the compressor efficiency is 85 percent, deter mine the pressure drop and the power wasted in the transpor tation line. Answers: 1.40 kPa, 0.125 kW 8–144 Reconsider Prob. 8–143. In order to reduce the head losses in the piping and thus the power wasted, someone sug gests doubling the diameter of the 83-m-long compressed air pipes. Calculating the reduction in wasted power, and deter mine if this is a worthwhile idea. Considering the cost of replacement, does this proposal make sense to you?
8–145E A water fountain is to be installed at a remote loca tion by attaching a cast iron pipe directly to a water main through which water is flowing at 70°F and 60 psig. The entrance to the pipe is sharp-edged, and the 70-ft-long piping system involves three 90° miter bends without vanes, a fully open gate valve, and an angle valve with a loss coefficient of 5 when fully open. If the system is to provide water at a rate of 15 gal/min and the elevation difference between the pipe and the fountain is negligible, determine the minimum diam eter of the piping system. Answer: 0.713 in cen96537_ch08_351-442.indd 437 14/01/17 3:00 pm 438 internal flow 60 psig 70 ft 15 gpm Water main FIGURE P8–145E 8–146E Repeat Prob. 8–145E for plastic (smooth) pipes.
8–147 In a hydroelectric power plant, water at 20°C is supplied to the turbine at a rate of 0.55 m3/s through a 200-m-long, 0.35-m-diameter cast iron pipe. The elevation difference between the free surface of the reservoir and the tur bine discharge is 140 m, and the combined turbine–generator efficiency is 85 percent. Disregarding the minor losses because of the large length-to-diameter ratio, determine the electric power output of this plant.
8–148 In Prob. 8–147, the pipe diameter is tripled in order to reduce the pipe losses. Determine the percent increase in the net power output as a result of this modification.
8–149E The drinking water needs of an office are met by large water bottles. One end of a 0.35-in-diameter, 6-ft-long plastic hose is inserted into the bottle placed on a high stand, while the other end with an on/off valve is maintained 3 ft below the bottom of the bottle. If the water level in the bot tle is 1 ft when it is full, determine how long it would take to fill an 8-oz glass (= 0.00835 ft3) (a) when the bottle is first opened and (b) when the bottle is almost empty. Take the total minor loss coefficient, including the on/off valve, to be 2.8 when it is fully open. Assume the water temperature to be the same as the room temperature of 70°F. Answers: (a) 2.4 s, (b) 2.8 s 3 ft 1 ft 6 ft 0.35 in FIGURE P8–149E 8–150E Reconsider Prob. 8–149E. Using appropriate software, investigate the effect of the hose diameter on the time required to fill a glass when the bottle is full. Let the diameter vary from 0.2 to 2 in, in increments of 0.2 in. Tabulate and plot the results.
8–151E Reconsider Prob. 8–149E. The office worker who set up the siphoning system purchased a 12-ft-long reel of the plastic tube and wanted to use the whole thing to avoid cutting it in pieces, thinking that it is the elevation difference that makes siphoning work, and the length of the tube is not important. So he used the entire 12-ft-long tube. Assum ing the turns or constrictions in the tube are not significant (being very optimistic) and the same elevation is maintained, determine the time it takes to fill a glass of water for both cases (bottle nearly full and bottle nearly empty).
8–152 A system that consists of two interconnected cylin drical tanks with D1 = 30 cm and D2 = 12 cm is to be used to determine the discharge coefficient of a short D0 = 5 mm diameter orifice. At the beginning (t = 0 s), the fluid heights in the tanks are h1 = 45 cm and h2 = 15 cm, as shown in Fig. P8–152. If it takes 200 s for the fluid levels in the two tanks to equalize and the flow to stop, determine the discharge coefficient of the orifice. Disregard any other losses associ ated with this flow.
Orifice h1 h h2 Tank 2 Tank 1 FIGURE P8–152 8–153 The water at 20°C in a 10-m-diameter, 2-m-high aboveground swimming pool is to be emptied by unplug ging a 5-cm-diameter, 25-m-long horizontal plastic pipe attached to the bottom of the pool. Determine the initial rate of discharge of water through the pipe and the time (hours) it would take to empty the swimming pool completely assum ing the entrance to the pipe is well-rounded with negligible loss. Take the friction factor of the pipe to be 0.022. Using the initial discharge velocity, check if this is a reasonable value for the friction factor. Answers: 3.55 L/s, 24.6 h cen96537_ch08_351-442.indd 438 14/01/17 3:00 pm 439 CHAPTER 8 2 m Swimming pool 10 m 25 m 5 cm FIGURE P8–153 8–154 Reconsider Prob. 8–153. Using appropriate software, investigate the effect of the discharge pipe diameter on the time required to empty the pool com pletely. Let the diameter vary from 1 to 10 cm, in increments of 1 cm. Tabulate and plot the results.
8–155 Repeat Prob. 8–153 for a sharp-edged entrance to the pipe with KL = 0.5. Is this “minor loss” truly “minor” or not?
8–156 A combined tank, shown in the figure, contains water. Neglecting the viscous effects, determine the total time needed to empty the combined tank through a small opening with 1 cm diameter at point A where the discharge coefficient is Cd = 0.67.
D = 0.6 m A 0.3 m 0.6 m FIGURE P8–156 8–157 Find the total volume flow rate leaving a tank through a slot having rectangular cross section a × b as a function of the given parameters.
FIGURE P8–157 h Water a b Front view 8–158 Water enters a large rigid tank steadily at a rate of V . = 0.5 m3/s, and exits through a large rectangular opening through which the velocity changes linearly in the vertical direction from 4 m/s to 5 m/s. Knowing that the tank is full, determine the magnitude of the velocity through the circular opening at the top whose diameter is D = 15 cm.
FIGURE P8–158 V = 4 m/s V = 5 m/s D = 15 cm 40 cm 20 cm Front view x y V = 0.5 m3/s ⋅ 8–159 Water is siphoned from a reservoir open to the atmosphere by a pipe with diameter D, total length L, and friction factor f as shown in Fig. P8–159. Atmos pheric pressure is 99.27 kPa. Minor losses may be ignored.
(a) Verify the following equation for the ratio of exit velocity for the siphon with nozzle to the exit velocity for the siphon without the nozzle, VD/VC VD/VC = √ (f L D + 1)╱(f L D d 4 D4 + 1) Calculate its value for L = 28 m, h1 = 1.8 m, h2 = 12 m, D = 12 cm, d = 3 cm, and f = 0.02.
(b) Show that static pressure at B in the siphon with the nozzle is greater and the velocity is smaller with respect to the siphon without the nozzle. Explain how this situation can be useful with respect to flow physics.
(c) Determine the maximum value of h2 in the piping system to avoid cavitation. The density and vapor pressure of water are ρ = 1000 kg/m3 and Pv = 4.25 kPa, and L1 = 4 m.
FIGURE P8–159 A Water B d D C D C Nozzle L1 h1 h2 cen96537_ch08_351-442.indd 439 14/01/17 3:00 pm 440 internal flow 8–160 It is a well–known fact that Roman aqueduct custom ers obtained extra water by attaching a diffuser to their pipe exits. The figure shows a simulation with a smooth inlet pipe, with and without a 20° diffuser expanding to a 6-cm-diameter exit. The pipe entrance is sharp-edged. Calculate the flow rate (a) without, and (b) with the diffuser and interpret the results. (c) For the case of a well-rounded entrance, Kent = 0.05, between the reservoir and pipe inlet for the system with a diffuser, how much would the flow rate increase? For water at 20°C, take ρ = 998 kg/m3 and μ = 0.001 kg/m·s.
FIGURE P8–160 D1 = 4 cm L = 2 m D2 = 6 cm Water 3 m 2 V1 V2 20º Diffuser 8–161 In an industrial application, water (ρ = 1000 kg/ m3, μ = 1.00 × 10–3 kg/m·s) is pumped from one large tank to another large tank that is at higher elevation, and is located in a different section of the plant. The free surfaces of both reservoirs are exposed to atmospheric pressure, and the elevation difference between the two surfaces is 5.00 m. The pipe is 1.27 cm I.D. PVC plastic pipe. The total pipe length is 47.48 m. The entrance and exit are sharp, and both the entrance and exit are submerged. There are five regular flanged smooth 90-degree elbows, and two fully open globe valves. The pump’s performance (supply curve) is approximated by the expression Havailable = hpump, u supply = H0 − aV ⋅ 2 where shutoff head H0 = 18.0 m of water col umn, coefficient a = 0.045 m/LPM2, available pump head Havailable is in units of meters of water column, and volume flow rate V ⋅ is in units of liters per minute (LPM). For con sistency, use g = 9.807 m/s2, and use 𝛼 = 1.05 for fully developed turbulent pipe flow.
(a) Predict the volume flow rate of water through this pip ing system. Note: simultaneous equation solver software is recommended here, especially for parts (b) and (c) below, where you need to make some simple changes and re-run your computer program. Your answer should lie between 9 and 10 LPM.
(b) Jennifer tells her boss that the flow rate can be increased very inexpensively if they simply replace the globe valves with ball valves. Repeat the calculations of part (a). By what percentage does the volume flow rate increase com pared to the base case of part (a)?
(c) Brad recommends that they also replace the entire piping system with PVC pipe that is about twice the diameter (2.54 cm I.D.). Predict the volume flow rate for this case, assum ing the same pump, the same total pipe length (47.48 m), the same number and type of elbows, but also with two fully open ball valves in the new system as in part (b) to maximize the flow rate. By what percentage does the volume flow rate increase compared to the base case of part (a)?
Fundamentals of Engineering (FE) Exam Problems 8–162 In a piping system, what is used to control the flow rate?
(a) Pipe (b) Valve (c) Fitting (d) Pump (e) Elbow 8–163 The Reynolds number can be viewed as the ratio of (a) Drag force/Dynamic force (b) Buoyancy force/Viscous force (c) Wall friction force/Viscous force (d) Inertial force/Gravitational force (e) Inertial force/Viscous force 8–164 Water at 10°C flows in a 3-cm-diameter pipe at a velocity of 1.25 m/s. The Reynolds number for this flow is (a) 19,770 (b) 23,520 (c) 28,680 (d) 32,940 (e) 36,050 8–165 Air at 1 atm and 20°C flows in a 3-cm-diameter tube. The maximum velocity of air to keep the flow laminar is (a) 0.87 m/s (b) 0.95 m/s (c) 1.16 m/s (d) 1.32 m/s (e) 1.44 m/s 8–166 Consider laminar flow of water in a 0.8-cm-diameter pipe at a rate of 1.15 L/min. The velocity of water halfway between the surface and the center of the pipe is (a) 0.381 m/s (b) 0.762 m/s (c) 1.15 m/s (d) 0.874 m/s (e) 0.572 m/s 8–167 Water at 10°C flows in a 1.2-cm-diameter pipe at a rate of 1.33 L/min. The hydrodynamic entry length is (a) 0.60 m (b) 0.94 m (c) 1.08 m (d) 1.20 m (e) 1.33 m 8–168 Engine oil at 20°C flows in a 15-cm-diameter pipe at a rate of 800 L/min. The friction factor for this flow is (a) 0.746 (b) 0.533 (c) 0.115 (d) 0.0826 (e) 0.0553 8–169 Engine oil at 40°C (𝜌 = 876 kg/m3, 𝜇 = 0.2177 kg/m⋅s) flows in a 20-cm-diameter pipe at a velocity of 0.9 m/s. The pressure drop of oil for a pipe length of 20 m is (a) 3135 Pa (b) 4180 Pa (c) 5207 Pa (d) 6413 Pa (e) 7620 Pa 8–170 Water flows in a 15-cm-diameter pipe at a velocity of 1.8 m/s. If the head loss along the pipe is estimated to be 16 m, the required pumping power to overcome this head loss is (a) 3.22 kW (b) 3.77 kW (c) 4.45 kW (d) 4.99 kW (e) 5.54 kW cen96537_ch08_351-442.indd 440 14/01/17 3:00 pm 441 CHAPTER 8 8–171 The pressure drop for a given flow is determined to be 100 Pa. For the same flow rate, if we reduce the diameter of the pipe by half, the pressure drop will be (a) 25 Pa (b) 50 Pa (c) 200 Pa (d) 400 Pa (e) 1600 Pa 8–172 Engine oil at 40°C (ρ = 876 kg/m3, μ = 0.2177 kg/m⋅s) flows in a 20-cm-diameter pipe at a velocity of 1.6 m/s. The head loss of oil for a pipe length of 130 m is (a) 0.86 m (b) 1.30 m (c) 2.27 m (d) 3.65 m (e) 4.22 m 8–173 Air at 1 atm and 25°C flows in a 4-cm-diameter glass pipe at a velocity of 7 m/s. The friction factor for this flow is (a) 0.0266 (b) 0.0293 (c) 0.0313 (d) 0.0176 (e) 0.0157 8–174 Hot combustion gases approximated as air at 1 atm and 350°C flow in a 16-cm-diameter steel pipe at a veloc ity of 3.5 m/s. The roughness of the pipe is 0.045 mm. The required power input to overcome the pressure drop for a pipe length of 60 m is (a) 0.55 W (b) 1.33 W (c) 2.85 W (d) 4.82 W (e) 6.35 W 8–175 Air at 1 atm and 40°C flows in a 8-cm-diameter pipe at a rate of 2500 L/min. The friction factor is deter mined from the Moody chart to be 0.027. The required power input to overcome the pressure drop for a pipe length of 150 m is (a) 310 W (b) 188 W (c) 132 W (d) 81.7 W (e) 35.9 W 8–176 Air at 1 atm and 20°C is to be transported in a 60-m-long circular steel duct at a rate of 2200 L/min. The roughness of the duct is 0.11 mm. If the head loss in the pipe is not to exceed 8 m, the minimum diameter of the duct is (a) 5.9 cm (b) 11.7 cm (c) 13.5 cm (d) 16.1 cm (e) 20.7 cm 8–177 Air at 1 atm and 20°C is to be transported in a 60-m-long circular steel duct at a rate of 5100 L/min. The roughness of the duct is 0.25 mm. If the pressure drop in the pipe is not to exceed 90 Pa, the maximum velocity of the air is (a) 3.99 m/s (b) 4.32 m/s (c) 6.68 m/s (d) 7.32 m/s (e) 8.90 m/s 8–178 The valve in a piping system causes a 3.1 m head loss. If the velocity of the flow is 4 m/s, the loss coefficient of this valve is (a) 1.7 (b) 2.2 (c) 2.9 (d) 3.3 (e) 3.8 8–179 A water flow system involves a 180° return bend (threaded) and a 90° miter bend (without vanes). The velocity of water is 1.2 m/s. The minor losses due to these bends are equivalent to a pressure loss of (a) 648 Pa (b) 933 Pa (c) 1255 Pa (d) 1872 Pa (e) 2600 Pa 8–180 Air flows in an 8-cm-diameter, 33-m-long pipe at a velocity of 5.5 m/s. The piping system involves multiple flow restrictions with a total minor loss coefficient of 2.6. The friction factor of pipe is obtained from the Moody chart to be 0.025. The total head loss of this piping system is (a) 13.5 m (b) 7.6 m (c) 19.9 m (d) 24.5 m (e) 4.2 m 8–181 Consider a pipe that branches out into two parallel pipes and then rejoins at a junction downstream. The two parallel pipes have the same lengths and friction factors. The diameters of the pipes are 2 cm and 4 cm. If the flow rate in one pipe is 10 L/min, the flow rate in the other pipe is (a) 10 L/min (b) 3.3 L/min (c) 100 L/min (d) 40 L/min (e) 56.6 L/min 8–182 Consider a pipe that branches out into two parallel pipes and then rejoins at a junction downstream. The two parallel pipes have the same lengths and friction factors. The diameters of the pipes are 2 cm and 4 cm. If the head loss in one pipe is 0.5 m, the head loss in the other pipe is (a) 0.5 m (b) 1 m (c) 0.25 m (d) 2 m (e) 0.125 m 8–183 Consider a pipe that branches out into three paral lel pipes and then rejoins at a junction downstream. All three pipes have the same diameters (D = 3 cm) and friction fac tors ( f = 0.018). The lengths of pipe 1 and pipe 2 are 5 m and 8 m, respectively, while the velocities of the fluid in pipe 2 and pipe 3 are 2 m/s and 4 m/s, respectively. The velocity of the fluid in pipe 1 is (a) 1.75 m/s (b) 2.12 m/s (c) 2.53 m/s (d) 3.91 m/s (e) 7.68 m/s 8–184 A pump moves water from a reservoir to another reservoir through a piping system at a rate of 0.1 m3/min. Both reservoirs are open to the atmosphere. The elevation difference between the two reservoirs is 35 m and the total head loss is estimated to be 4 m. If the efficiency of the motor-pump unit is 65 percent, the electrical power input to the motor of the pump is (a) 1660 W (b) 1472 W (c) 1292 W (d) 981 W (e) 808 W Design and Essay Problems 8–185 Electronic boxes such as computers are commonly cooled by a fan. Write an essay on forced air cooling of elec tronic boxes and on the selection of the fan for electronic devices.
8–186 Discuss equations used to calculate unknown flow rates or diameters for each pipe section in piping systems such as pipe networks and branching pipes, and how their solution is carried out. Define the analogy between electric current in electric circuits and fluid flow in pipe networks.
cen96537_ch08_351-442.indd 441 14/01/17 3:00 pm 442 internal flow 8–187 Design an experiment to measure the viscosity of liquids using a vertical funnel with a cylindrical reservoir of height h and a narrow flow section of diameter D and length L. Making appropriate assumptions, obtain a relation for viscosity in terms of easily measurable quantities such as density and volume flow rate. Is there a need for the use of a correction factor?
8–188 A pump is to be selected for a waterfall in a garden. The water collects in a pond at the bottom, and the eleva tion difference between the free surface of the pond and the location where the water is discharged is 3 m. The flow rate of water is to be at least 8 L/s. Select an appropriate motor– pump unit for this job and identify three manufactur ers with product model numbers and prices. Make a selec tion and explain why you selected that particular product. Also estimate the cost of annual power consumption of this unit assuming continuous operation.
8–189 During a camping trip you notice that water is discharged from a high reservoir to a stream in the valley through a 30-cm-diameter plastic pipe. The elevation differ ence between the free surface of the reservoir and the stream is 70 m. You conceive the idea of generating power from this water. Design a power plant that will produce the most power from this resource. Also, investigate the effect of power gen eration on the discharge rate of water. What discharge rate maximizes the power production?
cen96537_ch08_351-442.indd 442 14/01/17 3:00 pm 9 CHAPTER 443 OBJECTIVES When you finish reading this chapter, you should be able to ■ ■ Understand how the differen tial equation of conservation of mass and the differential linear momentum equation are derived and applied ■ ■ Calculate the stream function and pressure field, and plot streamlines for a known velocity field ■ ■ Obtain analytical solutions of the equations of motion for simple flow fields D I F F E R E N T I AL AN ALYSI S O F F LUI D F LOW I n this chapter we derive the differential equations of fluid motion, namely, conservation of mass (the continuity equation) and Newton’s second law (the Navier–Stokes equation). These equations apply to every point in the flow field and thus enable us to solve for all details of the flow everywhere in the flow domain. Unfortunately, most differential equations encountered in fluid mechanics are very difficult to solve and often require the aid of a computer. Also, these equations must be combined when necessary with additional equations, such as an equation of state and an equation for energy and/or species transport. We provide a step-by-step procedure for solving this set of differential equations of fluid motion and obtain analytical solutions for several simple examples. We also introduce the concept of the stream function; curves of constant stream function turn out to be streamlines in two-dimensional flow fields.
The fundamental differential equations of fluid motion are derived in this chapter, and we show how to solve them analytically for some simple flows. More complicated flows, such as the air flow induced by a tornado shown here, cannot be solved exactly although they can often be solved approximately with reasonable accuracy.
© Corbis RF cen96537_ch09_443-518.indd 443 14/01/17 3:08 pm 444 Differential Analysis of Fluid Flow 9–1 ■ INTRODUCTION In Chap. 5, we derived control volume versions of the laws of conservation of mass and energy, and in Chap. 6 we did the same for momentum. The control volume technique is useful when we are interested in the overall features of a flow, such as mass flow rate into and out of the control volume or net forces applied to bodies. An example is sketched in Fig. 9–1a for the case of wind flowing around a satellite dish. A rectangular control volume is taken around the vicinity of the satellite dish, as sketched. If we know the air velocity along the entire control surface, we can calculate the net reaction force on the stand without ever knowing any details about the geometry of the satellite dish. The interior of the control volume is in fact treated like a “black box” in control volume analysis—we cannot obtain detailed knowledge about flow properties such as velocity or pressure at points inside the control volume. Differential analysis, on the other hand, involves application of differen tial equations of fluid motion to any and every point in the flow field over a region called the flow domain. You can think of the differential technique as the analysis of millions of tiny control volumes stacked end to end and on top of each other all throughout the flow field. In the limit as the number of tiny control volumes goes to infinity, and the size of each control volume shrinks to a point, the conservation equations simplify to a set of partial differential equations that are valid at any point in the flow. When solved, these differen tial equations yield details about the velocity, density, pressure, etc., at every point throughout the entire flow domain. In Fig. 9–1b, for example, differential analysis of airflow around the satellite dish yields streamline shapes, a detailed pressure distribution around the dish, etc. From these details, we can integrate to find gross features of the flow such as the net force on the satellite dish. In a fluid flow problem such as the one illustrated in Fig. 9–1 in which air density and temperature changes are insignificant, it is sufficient to solve two differential equations of motion—conservation of mass and Newton’s second law (the linear momentum equation). For three-dimensional incom pressible flow, there are four unknowns (velocity components u, 𝜐, w, and pressure P) and four equations (one from conservation of mass, which is a scalar equation, and three from Newton’s second law, which is a vector equation). As we shall see, the equations are coupled, meaning that some of the variables appear in all four equations; the set of differential equations must therefore be solved simultaneously for all four unknowns. In addition, boundary conditions for the variables must be specified at all boundaries of the flow domain, including inlets, outlets, and walls. Finally, if the flow is unsteady, we must march our solution along in time as the flow field changes. You can see how differential analysis of fluid flow can become quite complicated and difficult. Computers are a tremendous help here, as discussed in Chap. 15. Nevertheless, there is much we can do analytically, and we start by deriving the differential equation for conservation of mass.
9–2 ■ CONSERVATION OF MASS— THE CONTINUITY EQUATION Through application of the Reynolds transport theorem (Chap. 4), we have the following general expression for conservation of mass as applied to a control volume: Control volume Flow out Flow out Flow in F F (a) Flow domain Flow out Flow out Flow in (b) FIGURE 9–1 (a) In control volume analysis, the interior of the control volume is treated like a black box, but (b) in differential analysis, all the details of the flow are solved at every point within the flow domain.
cen96537_ch09_443-518.indd 444 14/01/17 3:08 pm 445 CHAPTER 9 Conservation of mass for a CV: 0 = ∫CV ∂ρ ∂t dV + ∫CS ρV › ·n › dA (9–1) Recall that Eq. 9–1 is valid for both fixed and moving control volumes, provided that the velocity vector is the absolute velocity (as seen by a fixed observer). When there are well-defined inlets and outlets, Eq. 9–1 is rewritten as ∫CV ∂ρ ∂t dV = ∑ in m · −∑ out m · (9–2) In words, the net rate of change of mass within the control volume is equal to the rate at which mass flows into the control volume minus the rate at which mass flows out of the control volume. Equation 9–2 applies to any control volume, regardless of its size. To generate a differential equation for conservation of mass, we imagine the control volume shrinking to infinitesi mal size, with dimensions dx, dy, and dz (Fig. 9–2). In the limit, the entire control volume shrinks to a point in the flow.
Derivation Using the Divergence Theorem The quickest and most straightforward way to derive the differential form of conservation of mass is to apply the divergence theorem to Eq. 9–1. The divergence theorem is also called Gauss’s theorem, named after the German mathematician Johann Carl Friedrich Gauss (1777–1855). The divergence theorem allows us to transform a volume integral of the diver gence of a vector into an area integral over the surface that defines the vol ume. For any vector G ›, the divergence of G › is defined as ∇ ›·G ›, and the divergence theorem is written as Divergence theorem: ∫V ∇ › ·G › dV = ∮ A G › ·n › dA (9–3) The circle on the area integral is used to emphasize that the integral must be evaluated around the entire closed area A that surrounds volume V . Note that the control surface of Eq. 9–1 is a closed area, even though we do not always add the circle to the integral symbol. Equation 9–3 applies to any vol ume, so we choose the control volume of Eq. 9–1. We also let G › = 𝜌V › since G › can be any vector. Substitution of Eq. 9–3 into Eq. 9–1 converts the area integral into a volume integral, 0 = ∫CV ∂ρ ∂t dV + ∫CV ∇ › ·(ρV ›) dV We now combine the two volume integrals into one, ∫CV [ ∂ρ ∂t + ∇ › ·(ρV ›)] dV = 0 (9–4) Finally, we argue that Eq. 9–4 must hold for any control volume regardless of its size or shape. This is possible only if the integrand (the terms within dx dz dy CV x1 y1 z1 y z x FIGURE 9–2 To derive a differential conservation equation, we imagine shrinking a control volume to infinitesimal size.
cen96537_ch09_443-518.indd 445 14/01/17 3:08 pm 446 Differential Analysis of Fluid Flow square brackets) is identically zero. Hence, we have a general differential equation for conservation of mass, better known as the continuity equation: Continuity equation: ∂ρ ∂t + ∇ › ·(ρV › ) = 0 (9–5) Equation 9–5 is the compressible form of the continuity equation since we have not assumed incompressible flow. It is valid at any point in the flow domain.
Derivation Using an Infinitesimal Control Volume We derive the continuity equation in a different way, by starting with a con trol volume on which we apply conservation of mass. Consider an infini tesimal box-shaped control volume aligned with the axes in Cartesian coor dinates (Fig. 9–3). The dimensions of the box are dx, dy, and dz, and the center of the box is shown at some arbitrary point P from the origin (the box can be located anywhere in the flow field). At the center of the box we define the density as 𝜌 and the velocity components as u, 𝜐, and w, as shown. At locations away from the center of the box, we use a Taylor series expansion about the center of the box (point P). [The series expansion is named in honor of its creator, the English mathematician Brook Taylor (1685–1731).] For example, the center of the right-most face of the box is located a distance dx/2 from the middle of the box in the x-direction; the value of 𝜌u at that point is (ρu)center of right face = ρu + ∂(ρu) ∂x dx 2 + 1 2! ∂2(ρu) ∂x2 ( dx 2 ) 2 + ⋯ (9–6) As the box representing the control volume shrinks to a point, however, sec ond-order and higher terms become negligible. For example, suppose dx/L = 10−3, where L is some characteristic length scale of the flow domain. Then (dx/L)2 = 10−6, a factor of a thousand less than dx/L. In fact, the smaller dx, the better the assumption that second-order terms are negligible. Applying this truncated Taylor series expansion to the density times the normal veloc ity component at the center point of each of the six faces of the box, we have Center of right face: (ρu)center of right face ≅ρu + ∂(ρu) ∂x dx 2 Center of left face: (ρu)center of left face ≅ρu −∂( ρu) ∂x dx 2 Center of front face: (ρw)center of front face ≅ρw + ∂(ρw) ∂z dz 2 Center of rear face: ( ρw)center of rear face ≅ρw −∂(ρw) ∂z dz 2 Center of top face: ( ρ𝜐)center of top face ≅ρ𝜐+ ∂(ρ𝜐) ∂y dy 2 Center of bottom face: ( ρ𝜐)center of bottom face ≅ρ𝜐−∂(ρ𝜐) ∂y dy 2 y z x dx dz dy u υ w P ρ FIGURE 9–3 A small box-shaped control volume centered at point P is used for derivation of the differential equation for conservation of mass in Cartesian coordinates; the red dots indicate the center of each face.
cen96537_ch09_443-518.indd 446 14/01/17 3:08 pm 447 CHAPTER 9 The mass flow rate into or out of one of the faces is equal to the density times the normal velocity component at the center point of the face times the surface area of the face. In other words, m . = 𝜌VnA at each face, where Vn is the magnitude of the normal velocity through the face and A is the surface area of the face (Fig. 9–4). The mass flow rate through each face of our infinitesimal control volume is illustrated in Fig. 9–5. We could con struct truncated Taylor series expansions at the center of each face for the remaining (nonnormal) velocity components as well, but this is unnecessary since these components are tangential to the face under consideration. For example, the value of 𝜌𝜐 at the center of the right face can be estimated by a similar expansion, but since 𝜐 is tangential to the right face of the box, it contributes nothing to the mass flow rate into or out of that face. As the control volume shrinks to a point, the value of the volume integral on the left-hand side of Eq. 9–2 becomes Rate of change of mass within CV: ∫CV ∂ρ ∂t dV ≅∂ρ ∂t dx dy dz (9–7) since the volume of the box is dx dy dz. We now apply the approximations of Fig. 9–5 to the right-hand side of Eq. 9–2. We add up all the mass flow rates into and out of the control volume through the faces. The left, bottom, and back faces contribute to mass inflow, and the first term on the right-hand side of Eq. 9–2 becomes Net mass flow rate into CV: ∑ in m · ≅(ρu −∂(ρu) ∂x dx 2 ) dy dz+ (ρ𝜐−∂(ρ𝜐) ∂y dy 2 ) dx dz+ (ρw −∂(ρw) ∂z dz 2 ) dx dy left face bottom face rear face y z x A = surface area Vn = average normal velocity component FIGURE 9–4 The mass flow rate through a surface is equal to 𝜌VnA.
y z x dx dz dy dx dz dy ∂(ρυ) ( ) 2 ∂y ρυ + dx dy dz ∂(ρw) ( ) 2 ∂z ρw – dy dz dx ∂(ρu) ( ) 2 ∂x ρu + dx dz dy ∂(ρυ) ( ) 2 ∂y ρυ – dx dy dz ∂(ρw) ( ) 2 ∂z ρw + dy dz dz ∂(ρu) ( ) 2 ∂x ρu – FIGURE 9–5 The inflow or outflow of mass through each face of the differential control volume; the red dots indicate the center of each face.
cen96537_ch09_443-518.indd 447 14/01/17 3:08 pm 448 Differential Analysis of Fluid Flow Similarly, the right, top, and front faces contribute to mass outflow, and the second term on the right-hand side of Eq. 9–2 becomes Net mass flow rate out of CV: ∑ out m · ≅(ρu+ ∂(ρu) ∂x dx 2 ) dy dz+ (ρ𝜐+ ∂(ρ𝜐) ∂y dy 2 ) dx dz+ (ρw+ ∂(ρw) ∂z dz 2 ) dx dy right face top face front face We substitute Eq. 9–7 and these two equations for mass flow rate into Eq. 9–2. Many of the terms cancel each other out; after combining and sim plifying the remaining terms, we are left with ∂ρ ∂t dx dy dz = − ∂(ρu) ∂x dx dy dz −∂( ρ𝜐) ∂y dx dy dz −∂(ρw) ∂z dx dy dz The volume of the box, dx dy dz, appears in each term and can be elimi nated. After rearrangement we end up with the following differential equation for conservation of mass in Cartesian coordinates: Continuity equation in Cartesian coordinates: ∂ρ ∂t + ∂(ρu) ∂x + ∂( ρ𝜐) ∂y + ∂(ρw) ∂z = 0 (9–8) Equation 9–8 is the compressible form of the continuity equation in Cartesian coordinates. It is written in more compact form by recognizing the divergence operation (Fig. 9–6), yielding the same equation as Eq. 9–5.
Cylinder L(t) ρ(t) y VP υ Lbottom Piston Time t Time t = 0 FIGURE 9–7 Fuel and air being compressed by a piston in a cylinder of an internal combustion engine.
The Divergence Operation Cartesian coordinates: + ∂r Cylindrical coordinates: 1 r • (ρV) = Δ Δ ∂(rρur) + ∂θ 1 r ∂(ρuθ) ∂z ∂(ρuz) ∂ ∂x ∂ ∂y ∂ ∂z • (ρV ) = (ρu) + (ρυ) + (ρw) FIGURE 9–6 The divergence operation in Cartesian and cylindrical coordinates.
EXAMPLE 9–1 Compression of an Air–Fuel Mixture An air–fuel mixture is compressed by a piston in a cylinder of an internal combus tion engine (Fig. 9–7). The origin of coordinate y is at the top of the cylinder, and y points straight down as shown. The piston is assumed to move up at constant speed VP. The distance L between the top of the cylinder and the piston decreases with time according to the linear approximation L = Lbottom − VPt, where Lbottom is the location of the piston when it is at the bottom of its cycle at time t = 0, as sketched in Fig. 9–7. At t = 0, the density of the air–fuel mixture in the cylinder is every where equal to 𝜌(0). Estimate the density of the air–fuel mixture as a function of time and the given parameters during the piston’s up stroke.
SOLUTION The density of the air–fuel mixture is to be estimated as a function of time and the given parameters in the problem statement.
Assumptions 1 Density varies with time, but not space; in other words, the den sity is uniform throughout the cylinder at any given time, but changes with time: 𝜌 = 𝜌(t). 2 Velocity component 𝜐 varies with y and t, but not with x or z; in other words 𝜐 = 𝜐(y, t) only. 3 u = w = 0. 4 No mass escapes from the cylinder during the compression.
Analysis First we need to establish an expression for velocity component 𝜐 as a function of y and t. Clearly 𝜐 = 0 at y = 0 (the top of the cylinder), and 𝜐 = −VP at y = L. For simplicity, we approximate that 𝜐 varies linearly between these two boundary conditions, Vertical velocity component: 𝜐= −VP y L (1) cen96537_ch09_443-518.indd 448 14/01/17 3:08 pm 449 CHAPTER 9 Alternative Form of the Continuity Equation We expand Eq. 9–5 by using the product rule on the divergence term, ∂ρ ∂t + ∇ › ·(ρV › ) = ∂ρ ∂t + V › ·∇ › ρ + ρ∇ › ·V ›= 0 (9–9) Material derivative of 𝜌 Recognizing the material derivative in Eq. 9–9 (see Chap. 4), and dividing by 𝜌, we write the compressible continuity equation in an alternative form, Alternative form of the continuity equation: 1 ρ Dρ Dt + ∇ › ·V ›= 0 (9–10) Equation 9–10 shows that as we follow a fluid element through the flow field (we call this a material element), its density changes as ∇ ›·V › changes (Fig. 9–9). 5 4 3 2 ρ 1 0 0.2 0.4 0.6 t 0.8 1 FIGURE 9–8 Nondimensional density as a function of nondimensional time for Example 9–1.
Streamline FIGURE 9–9 As a material element moves through a flow field, its density changes according to Eq. 9–10.
where L is a function of time, as given. The compressible continuity equation in Cartesian coordinates (Eq. 9–8) is appropriate for solution of this problem.
∂ρ ∂t + ∂(ρu) ∂x + ∂(ρ𝜐) ∂y + ∂(ρw) ∂z = 0 → ∂ρ ∂t + ∂(ρ𝜐) ∂y = 0 0 since u = 0 0 since w = 0 By assumption 1, however, density is not a function of y and can therefore come out of the y-derivative. Substituting Eq. 1 for 𝜐 and the given expression for L, differen tiating, and simplifying, we obtain ∂ρ ∂t = −ρ ∂𝜐 ∂y = −ρ ∂ ∂y (−VP y L) = ρ VP L = ρ VP Lbottom −VP t (2) By assumption 1 again, we replace ∂𝜌/∂t by d𝜌/dt in Eq. 2. After separating vari ables we obtain an expression that can be integrated analytically, ∫ ρ ρ=ρ(0) dρ ρ = ∫ t t=0 VP Lbottom −VP t dt → ln ρ ρ(0) = ln Lbottom Lbottom −VP t (3) Finally then, we have the desired expression for 𝜌 as a function of time, ρ = ρ(0) Lbottom Lbottom −VP t (4) In keeping with the convention of nondimensionalizing results, Eq. 4 is rewritten as ρ ρ(0) = 1 1 −VP t/Lbottom → 𝞺 = 1 1 −t (5) where 𝜌 = 𝜌/𝜌(0) and t = VPt/Lbottom. Equation 5 is plotted in Fig. 9–8.
Discussion At t = 1, the piston hits the top of the cylinder and 𝜌 goes to infin ity. In an actual internal combustion engine, the piston stops before reaching the top of the cylinder, forming what is called the clearance volume, which typically consti tutes 4 to 12 percent of the maximum cylinder volume. The assumption of uniform density within the cylinder is the weakest link in this simplified analysis. In reality, 𝜌 may be a function of both space and time.
cen96537_ch09_443-518.indd 449 14/01/17 3:08 pm 450 Differential Analysis of Fluid Flow On the other hand, if changes in the density of the material element are negligibly small compared to the magnitude of the density itself as the element moves around, then both terms in Eq. 9–10 are negligibly small; ∇ ›·V › ≅0 and 𝜌−1 D𝜌 / Dt ≅0, and the flow is approximated as incompressible.
Continuity Equation in Cylindrical Coordinates Many problems in fluid mechanics are more conveniently solved in cylin drical coordinates (r, 𝜃, z) (often called cylindrical polar coordinates), rather than in Cartesian coordinates. For simplicity, we introduce cylindri cal coordinates in two dimensions first (Fig. 9–10a). By convention, r is the radial distance from the origin to some point (P), and 𝜃 is the angle measured from the x-axis (𝜃 is always defined as mathematically positive in the counterclockwise direction). Velocity components, ur and u𝜃, and unit vectors, e → r and e → 𝜃, are also shown in Fig. 9–10a. In three dimensions, imag ine sliding everything in Fig. 9–10a out of the page along the z-axis (nor mal to the xy-plane) by some distance z. We have attempted to draw this in Fig. 9–10b. In three dimensions, we have a third velocity component, uz, and a third unit vector, e → z, also sketched in Fig. 9–10b. The following coordinate transformations are obtained from Fig. 9–10: Coordinate transformations: r = √x2 + y2 x = r cos 𝜃 y = r sin 𝜃 𝜃= tan−1 y x (9–11) Coordinate z is the same in cylindrical and Cartesian coordinates. To obtain an expression for the continuity equation in cylindrical coordi nates, we have two choices. First, we can use Eq. 9–5 directly, since it was derived without regard to our choice of coordinate system. We simply look up the expression for the divergence operator in cylindrical coordinates in a vector calculus book (e.g., Spiegel, 1968; see also Fig. 9–6). Second, we can draw a three-dimensional infinitesimal fluid element in cylindrical coor dinates and analyze mass flow rates into and out of the element, similar to what we did before in Cartesian coordinates. Either way, we end up with Continuity equation in cylindrical coordinates: ∂ρ ∂t + 1 r ∂(rρur) ∂r + 1 r ∂(ρu𝜃) ∂𝜃 + ∂(ρuz) ∂z = 0 (9–12) Details of the second method can be found in Fox and McDonald (1998).
Special Cases of the Continuity Equation We now look at two special cases, or simplifications, of the continuity equa tion. In particular, we first consider steady compressible flow, and then incompressible flow.
Special Case 1: Steady Compressible Flow If the flow is compressible but steady, ∂/∂t of any variable is equal to zero. Thus, Eq. 9–5 reduces to Steady continuity equation: ∇ › ·(ρV ›) = 0 (9–13) y y x x x y z eθ ur uθ P r r θ er eθ ur uθ P er ez uz z θ (b) (a) FIGURE 9–10 Velocity components and unit vectors in cylindrical coordinates: (a) two-dimensional flow in the xy- or r𝜃-plane, (b) three-dimensional flow.
cen96537_ch09_443-518.indd 450 14/01/17 3:08 pm 451 CHAPTER 9 In Cartesian coordinates, Eq. 9–13 reduces to ∂(ρu) ∂x + ∂(ρ𝜐) ∂y + ∂(ρw) ∂z = 0 (9–14) In cylindrical coordinates, Eq. 9–13 reduces to 1 r ∂(rρur) ∂r + 1 r ∂(ρu𝜃) ∂𝜃 + ∂(ρuz) ∂z = 0 (9–15) Special Case 2: Incompressible Flow If the flow is approximated as incompressible, density is not a function of time or space. Thus the unsteady term in Eq. 9–5 disappears and 𝜌 can be taken outside of the divergence operator. Equation 9–5 therefore reduces to Incompressible continuity equation: ∇ › ·V ›= 0 (9–16) The same result is obtained if we start with Eq. 9–10 and recognize that for an incompressible flow, density does not change appreciably following a fluid particle, as pointed out previously. Thus the material derivative of 𝜌 is approximately zero, and Eq. 9–10 reduces immediately to Eq. 9–16. You may have noticed that no time derivatives remain in Eq. 9–16. We conclude from this that even if the flow is unsteady, Eq. 9–16 applies at any instant in time. Physically, this means that as the velocity field changes in one part of an incompressible flow field, the entire rest of the flow field immediately adjusts to the change such that Eq. 9–16 is satisfied at all times. For compressible flow this is not the case. In fact, a disturbance in one part of the flow is not even felt by fluid particles some distance away until the sound wave from the disturbance reaches that distance. Very loud noises, such as that from a gun or explosion, generate a shock wave that actually travels faster than the speed of sound. (The shock wave produced by an explosion is illustrated in Fig. 9–11.) Shock waves and other manifes tations of compressible flow are discussed in Chap. 12. In Cartesian coordinates, Eq. 9–16 is Incompressible continuity equation in Cartesian coordinates: ∂u ∂x + ∂𝜐 ∂y + ∂w ∂z = 0 (9–17) Equation 9–17 is the form of the continuity equation you will probably encounter most often. It applies to steady or unsteady, incompressible, three-dimensional flow, and you would do well to memorize it. In cylindrical coordinates, Eq. 9–16 is Incompressible continuity equation in cylindrical coordinates: 1 r ∂(rur) ∂r + 1 r ∂(u𝜃) ∂𝜃 + ∂(uz) ∂z = 0 (9–18) EXAMPLE 9–2 Design of a Compressible Converging Duct A two-dimensional converging duct is being designed for a high-speed wind tun nel. The bottom wall of the duct is to be flat and horizontal, and the top wall is to be curved in such a way that the axial wind speed u increases approxi mately linearly from u1 = 100 m/s at section (1) to u2 = 300 m/s at section Shock wave Observer Pow!
FIGURE 9–11 The disturbance from an explosion is not felt until the shock wave reaches the observer.
cen96537_ch09_443-518.indd 451 14/01/17 3:08 pm 452 Differential Analysis of Fluid Flow (2) (Fig. 9–12). Meanwhile, the air density 𝜌 is to decrease approximately lin early from 𝜌1 = 1.2 kg/m3 at section (1) to 𝜌2 = 0.85 kg/m3 at section (2). The converging duct is 2.0 m long and is 2.0 m high at section (1). (a) Predict the y-component of velocity, 𝜐(x, y), in the duct. (b) Plot the approximate shape of the duct, ignoring friction on the walls. (c) How high should the duct be at sec tion (2), the exit of the duct?
SOLUTION For given velocity component u and density 𝜌, we are to predict velocity component 𝜐, plot an approximate shape of the duct, and predict its height at the duct exit.
Assumptions 1 The flow is steady and two-dimensional in the xy-plane. 2 Friction on the walls is ignored. 3 Axial velocity u increases linearly with x, and density 𝜌 decreases linearly with x.
Properties The fluid is air at room temperature (25°C). The speed of sound is about 346 m/s, so the flow is subsonic, but compressible.
Analysis (a) We write expressions for u and 𝜌, forcing them to be linear in x, u = u1 + Cux where Cu = u2 −u1 Δx = (300 −100) m/s 2.0 m = 100 s−1 (1) and ρ = ρ1 + Cρ x where Cρ = ρ2 −ρ1 Δx = (0.85 −1.2) kg/m3 2.0 m (2) = −0.175 kg/m4 The steady continuity equation (Eq. 9–14) for this two-dimensional compressible flow simplifies to ∂(ρu) ∂x + ∂(ρ𝜐) ∂y + ∂(ρw) ∂z = 0 → ∂(ρ𝜐) ∂y = − ∂(ρu) ∂x (3) 0 (2-D) Substituting Eqs. 1 and 2 into Eq. 3 and noting that Cu and C𝜌 are constants, ∂(ρ𝜐) ∂y = − ∂[(ρ1 + Cρx)(u1 + Cux)] ∂x = −(ρ1Cu + u1Cρ) −2CuCρx Integration with respect to y gives ρ𝜐= −(ρ1Cu + u1Cρ)y −2CuCρ xy + f (x) (4) Note that since the integration is a partial integration, we have added an arbi trary function of x instead of simply a constant of integration. Next, we apply boundary conditions. We argue that since the bottom wall is flat and horizontal, 𝜐 must equal zero at y = 0 for any x. This is possible only if f (x) = 0. Solving Eq. 4 for 𝜐 gives 𝜐= −(ρ1Cu + u1Cρ)y −2CuCρxy ρ → 𝝊= −(𝞺1Cu + u1C𝞺)y −2CuC𝞺xy 𝞺1 + C𝞺 x (5) (b) Using Eqs. 1 and 5 and the technique described in Chap. 4, we plot several streamlines between x = 0 and x = 2.0 m in Fig. 9–13. The streamline starting at x = 0, y = 2.0 m approximates the top wall of the duct.
2.0 m (1) (2) x y ∆x = 2.0 m FIGURE 9–12 Converging duct, designed for a high-speed wind tunnel (not to scale).
y 0 0.5 1 1.5 2 0 0.5 1 1.5 2 Top wall x Bottom wall FIGURE 9–13 Streamlines for the converging duct of Example 9–2.
cen96537_ch09_443-518.indd 452 14/01/17 3:08 pm 453 CHAPTER 9 (c) At section (2), the top streamline crosses y = 0.941 m at x = 2.0 m. Thus, the predicted height of the duct at section (2) is 0.941 m.
Discussion You can verify that the combination of Eqs. 1, 2, and 5 satisfies the continuity equation. However, this alone does not guarantee that the density and velocity components will actually follow these equations if the duct were to be built as designed here. The actual flow depends on the pressure drop between sections (1) and (2); only one unique pressure drop can yield the desired flow acceleration. Temperature may also change considerably in this kind of compressible flow in which the air accelerates toward sonic speeds.
EXAMPLE 9–3 Incompressibility of an Unsteady Two-Dimensional Flow Consider the velocity field of Example 4–5—an unsteady, two-dimensional velocity field given by V › = (u, 𝜐) = (0.5 + 0.8x)i → + [1.5 + 2.5 sin (𝜔t ) − 0.8y] j → , where angular frequency 𝜔 is equal to 2𝜋 rad/s (a physical frequency of 1 Hz). Verify that this flow field can be approximated as incompressible.
SOLUTION We are to verify that a given velocity field is incompressible.
Assumptions 1 The flow is two-dimensional, implying no z-component of velocity and no variation of u or 𝜐 with z.
Analysis The components of velocity in the x- and y-directions, respectively, are u = 0.5 + 0.8x and 𝜐= 1.5 + 2.5 sin (𝜔t) −0.8y If the flow is incompressible, Eq. 9–16 must apply. More specifically, in Cartesian coordinates Eq. 9–17 must apply. Let’s check: ∂u ∂x + ∂𝜐 ∂y + ∂w ∂z = 0 → 0.8 −0.8 = 0 0.8 −0.8 0 since 2-D So we see that the incompressible continuity equation is indeed satisfied at any instant in time, and this flow field may be approximated as incompressible.
Discussion Although there is an unsteady term in 𝜐, it has no y-derivative and drops out of the continuity equation.
EXAMPLE 9–4 Finding a Missing Velocity Component The u velocity component of a steady, two-dimensional, incompressible flow field is u = ax + by, where a and b are constants. Velocity component 𝜐 is missing (Fig. 9–14). Generate an expression for 𝜐 as a function of x and y.
SOLUTION We are to find the y component of velocity 𝜐, using a given expression for u. Assumptions 1 The flow is steady. 2 The flow is incompressible. 3 The flow is two-dimensional in the xy-plane, implying that w = 0 and neither u nor 𝜐 depends on z.
Analysis We plug the velocity components into the steady incompressible conti nuity equation, For Sale: 6-mo. old computer $300 OBO 862-2720 Need a pl to Lewis D This Friday 234-228 Missing: y velocity component If found, call 1-800-CON-UITY FIGURE 9–14 The continuity equation can be used to find a missing velocity component.
cen96537_ch09_443-518.indd 453 14/01/17 3:08 pm 454 Differential Analysis of Fluid Flow Condition for incompressibility: ∂𝜐 ∂y = − ∂u ∂x −∂w ∂z → ∂𝜐 ∂y = −a a 0 Next we integrate with respect to y. Note that since the integration is a partial inte gration, we must add some arbitrary function of x instead of simply a constant of integration. Solution: 𝝊 = −ay + f(x) If the flow were three-dimensional, we would add a function of x and z instead.
Discussion To satisfy the incompressible continuity equation, any function of x will work since there are no derivatives of 𝜐 with respect to x in the continu ity equation. Not all functions of x are necessarily physically possible, however, since the flow may not be able to satisfy the steady conservation of momentum equation.
EXAMPLE 9–5 Two-Dimensional, Incompressible, Vortical Flow Consider a two-dimensional, incompressible flow in cylindrical coordinates; the tangential velocity component is u𝜃 = K/r, where K is a constant. This represents a class of vortical flows. Generate an expression for the other velocity compo nent, ur.
SOLUTION For a given tangential velocity component, we are to generate an expression for the radial velocity component.
Assumptions 1 The flow is two-dimensional in the xy- (r𝜃-) plane (velocity is not a function of z, and uz = 0 everywhere). 2 The flow is incompressible.
Analysis The incompressible continuity equation (Eq. 9–18) for this two- dimensional case simplifies to 1 r ∂(rur) ∂r + 1 r ∂u𝜃 ∂𝜃+ ∂uz ∂z = 0 → ∂(rur) ∂r = − ∂u𝜃 ∂𝜃 (1) 0 (2-D) The given expression for u𝜃 is not a function of 𝜃, and therefore Eq. 1 reduces to ∂(rur) ∂r = 0 → rur = f(𝜃, t) (2) where we have introduced an arbitrary function of 𝜃 and t instead of a con stant of integration, since we performed a partial integration with respect to r. Solving for ur, ur = f(𝜃, t) r (3) Thus, any radial velocity component of the form given by Eq. 3 yields a two-dimensional, incompressible velocity field that satisfies the continuity equation. We discuss some specific cases. The simplest case is when f(𝜃, t) = 0 (ur = 0, u𝜃 = K/r). This yields the line vortex discussed in Chap. 4, as sketched in Fig. 9–15a. Another simple case is when f(𝜃, t) = C, where C is a constant. This yields a radial velocity whose magnitude decays as 1/r. For negative C, imagine a spiraling line vortex/sink flow, in which fluid elements not only revolve around the origin, but get sucked into a sink at the origin (actually a line sink along the z-axis). This is illustrated in Fig. 9–15b.
uθ r uθ = K r ur = 0 uθ r uθ = K r ur = C r (b) (a) FIGURE 9–15 Streamlines and velocity profiles for (a) a line vortex flow and (b) a spiraling line vortex/sink flow.
cen96537_ch09_443-518.indd 454 14/01/17 3:08 pm 455 CHAPTER 9 Discussion Other more complicated flows can be obtained by setting f (𝜃, t) to some other function. For any function f(𝜃, t), the flow satisfies the two-dimen sional, incompressible continuity equation at a given instant in time.
EXAMPLE 9–6 Comparison of Continuity and Volumetric Strain Rate Recall the volumetric strain rate, defined in Chap. 4. In Cartesian coordinates, 1 V DV Dt = 𝜀xx + 𝜀yy + 𝜀zz = ∂u ∂x + ∂𝜐 ∂y + ∂w ∂z (1) Show that volumetric strain rate is zero for incompressible flow. Discuss the physi cal interpretation of volumetric strain rate for incompressible and compressible flows.
SOLUTION We are to show that volumetric strain rate is zero in an incompress ible flow, and discuss its physical significance in incompressible and compressible flow.
Analysis If the flow is incompressible, Eq. 9–16 applies. More specifically, Eq. 9–17, in Cartesian coordinates, applies. Comparing Eq. 9–17 to Eq. 1, 1 V DV Dt = 0 for incompressible flow Thus, volumetric strain rate is zero in an incompressible flow field. In fact, you can define incompressibility by DV/Dt = 0. Physically, as we follow a fluid element, parts of it may stretch while other parts shrink, and the element may translate, distort, and rotate, but its volume remains constant along its entire path through the flow field (Fig. 9–16a). This is true whether the flow is steady or unsteady, as long as it is incompressible. If the flow were compressible, the volumetric strain rate would not be zero, implying that fluid elements may expand in volume (dilate) or shrink in volume as they move around in the flow field (Fig. 9–16b). Specifically, consider Eq. 9–10, an alternative form of the continuity equation for compressible flow. By definition, 𝜌 = m/V, where m is the mass of a fluid element. For a material element (following the fluid element as it moves through the flow field), m must be constant. Applying some algebra to Eq. 9–10 yields 1 ρ Dρ Dt = V m D(m/V ) Dt = − V m m V 2 DV Dt = − 1 V DV Dt = −∇ › ·V › → 1 V DV Dt = ∇ › ·V › Discussion The final result is general—not limited to Cartesian coordinates. It applies to unsteady as well as steady flows.
EXAMPLE 9–7 Conditions for Incompressible Flow Consider a steady velocity field given by V › = (u, 𝜐, w) = a(x 2y + y 2)i → + bxy 2 j → + cxk →, where a, b, and c are constants. Under what conditions is this flow field incompressible?
Time = t1 Time = t2 Time = t2 Volume = V2 = V1 Volume = V1 Volume = V1 Volume = V2 Time = t1 (a) (b) FIGURE 9–16 (a) In an incompressible flow field, fluid elements may translate, distort, and rotate, but they do not grow or shrink in volume; (b) in a compressible flow field, fluid elements may grow or shrink in volume as they translate, distort, and rotate.
cen96537_ch09_443-518.indd 455 14/01/17 3:08 pm 456 Differential Analysis of Fluid Flow 9–3 ■ THE STREAM FUNCTION The Stream Function in Cartesian Coordinates Consider the simple case of incompressible, two-dimensional flow in the xy-plane. The continuity equation (Eq. 9–17) in Cartesian coordinates reduces to ∂u ∂x + ∂𝜐 ∂y = 0 (9–19) A clever variable transformation enables us to rewrite Eq. 9–19 in terms of one dependent variable (𝜓) instead of two dependent variables (u and 𝜐). We define the stream function 𝜓 as (Fig. 9–17) Incompressible, two-dimensional stream function in Cartesian coordinates: u = ∂𝜓 ∂y and 𝜐= − ∂𝜓 ∂x (9–20) The stream function and the corresponding velocity potential function (Chap. 10) were first introduced by the Italian mathematician Joseph Louis Lagrange (1736–1813). Substitution of Eq. 9–20 into Eq. 9–19 yields ∂ ∂x ( ∂𝜓 ∂y ) + ∂ ∂y (− ∂𝜓 ∂x ) = ∂2𝜓 ∂x ∂y −∂2𝜓 ∂y ∂x = 0 which is identically satisfied for any smooth function 𝜓(x, y), because the order of differentiation (y then x versus x then y) is irrelevant. You may ask why we chose to put the negative sign on 𝜐 rather than on u. (We could have defined the stream function with the signs reversed, and continuity would still have been identically satisfied.) The answer is that although the sign is arbitrary, the definition of Eq. 9–20 leads to flow from left to right as 𝜓 increases in the y-direction, which is usually preferred. Most fluid mechanics books define 𝜓 in this way, although sometimes 𝜓 is SOLUTION We are to determine a relationship between constants a, b, and c that ensures incompressibility.
Assumptions 1 The flow is steady. 2 The flow is incompressible (under certain constraints to be determined).
Analysis We apply Eq. 9–17 to the given velocity field, ∂u ∂x + ∂𝜐 ∂y + ∂w ∂z = 0 → 2axy + 2bxy = 0 2axy 2bxy 0 Thus to guarantee incompressibility, constants a and b must be equal in magnitude but opposite in sign. Condition for incompressibility: a = −b Discussion If a were not equal to −b, this might still be a valid flow field, but density would have to vary with location in the flow field. In other words, the flow would be compressible, and Eq. 9–14 would need to be satisfied in place of Eq. 9–17.
Stream Function 2-D, incompressible, Cartesian coordinates: and ∂ψ ∂y = ∂ψ ∂x = 2-D, incompressible, cylindrical coordinates: and ∂ψ ∂θ 1 = ∂ψ ∂r r r = Axisymmetric, incompressible, cylindrical coordinates: and ∂ψ ∂z 1 = ∂ψ ∂r 1 r = 2-D, compressible, Cartesian coordinates: and ρu = ur uz ur u uθ ∂ψρ ∂y ρυ = ∂ψρ ∂x υ FIGURE 9–17 There are several definitions of the stream function, depending on the type of flow under consideration as well as the coordinate system being used.
cen96537_ch09_443-518.indd 456 14/01/17 3:08 pm 457 CHAPTER 9 defined with the opposite signs (e.g., in some British textbooks and in the indoor air quality field, Heinsohn and Cimbala, 2003). What have we gained by this transformation? First, as already mentioned, a single variable (𝜓) replaces two variables (u and 𝜐)—once 𝜓 is known, we can generate both u and 𝜐 via Eq. 9–20, and we are guaranteed that the solution satisfies continuity, Eq. 9–19. Second, it turns out that the stream function has useful physical significance (Fig. 9–18). Namely, Curves of constant 𝜓 are streamlines of the flow.
This is easily proven by considering a streamline in the xy-plane, as sketched in Fig. 9–19. Recall from Chap. 4 that along such a streamline, Along a streamline: dy dx = 𝜐 u → −𝜐 dx + u dy = 0 ∂𝜓/∂x ∂𝜓/∂y where we have applied Eq. 9–20, the definition of 𝜓. Thus, Along a streamline: ∂𝜓 ∂x dx + ∂𝜓 ∂y dy = 0 (9–21) But for any smooth function 𝜓 of two variables x and y, we know by the chain rule of mathematics that the total change of 𝜓 from point (x, y) to another point (x + dx, y + dy) some infinitesimal distance away is Total change of 𝜓: d𝜓= ∂𝜓 ∂x dx + ∂𝜓 ∂y dy (9–22) By comparing Eq. 9–21 to Eq. 9–22 we see that d𝜓 = 0 along a streamline; thus we have proven the statement that 𝜓 is constant along streamlines.
EXAMPLE 9–8 Calculation of the Velocity Field from the Stream Function A steady, two-dimensional, incompressible flow field in the xy-plane has a stream function given by 𝜓 = ax3 + by + cx, where a, b, and c are constants: a = 0.50 (m·s)−1, b = −2.0 m/s, and c = −1.5 m/s. (a) Obtain expressions for velocity components u and 𝜐. (b) Verify that the flow field satisfies the incom pressible continuity equation. (c) Plot several streamlines of the flow in the upper-right quadrant.
SOLUTION For a given stream function, we are to calculate the velocity compo nents, verify incompressibility, and plot flow streamlines.
Assumptions 1 The flow is steady. 2 The flow is incompressible (this assump tion is to be verified). 3 The flow is two-dimensional in the xy-plane, implying that w = 0 and neither u nor 𝜐 depend on z.
Analysis (a) We use Eq. 9–20 to obtain expressions for u and 𝜐 by differentiating the stream function, u = ∂𝜓 ∂y = b and 𝜐= −∂𝜓 ∂x = −3ax2 −c (b) Since u is not a function of x, and 𝜐 is not a function of y, we see immediately that the two-dimensional, incompressible continuity equation (Eq. 9–19) is satis fied. In fact, since 𝜓 is smooth in x and y, the two-dimensional, incompressible V dr y x Streamline Point (x, y) Point (x + dx, y + dy) dx dy u υ FIGURE 9–19 Arc length dr› = (dx, dy) and local velocity vector V ›= (u, 𝜐) along a two-dimensional streamline in the xy-plane.
y x Streamlines ψ = ψ1 ψ = ψ2 ψ = ψ3 ψ = ψ4 FIGURE 9–18 Curves of constant stream function represent streamlines of the flow.
cen96537_ch09_443-518.indd 457 14/01/17 3:08 pm 458 Differential Analysis of Fluid Flow continuity equation in the xy-plane is automatically satisfied by the very defini tion of 𝜓. We conclude that the flow is indeed incompressible.
(c) To plot streamlines, we solve the given equation for either y as a function of x and 𝜓, or x as a function of y and 𝜓. In this case, the former is easier, and we have Equation for a streamline: y = 𝜓−ax3 −cx b This equation is plotted in Fig. 9–20 for several values of 𝜓, and for the provided values of a, b, and c. The flow is nearly straight down at large values of x, but veers upward for x < 1 m.
Discussion You can verify that 𝜐 = 0 at x = 1 m. In fact, 𝜐 is nega tive for x > 1 m and positive for x < 1 m. The direction of the flow can also be determined by picking an arbitrary point in the flow, say (x = 3 m, y = 4 m), and calculating the velocity there. We get u = −2.0 m/s and 𝜐 = −12.0 m/s at this point, either of which shows that fluid flows to the lower left in this region of the flow field. For clarity, the velocity vector at this point is also plotted in Fig. 9–20; it is clearly parallel to the streamline near that point. Velocity vectors at three other locations are also plotted.
EXAMPLE 9–9 Calculation of Stream Function for a Known Velocity Field Consider a steady, two-dimensional, incompressible velocity field with u = ax + b and 𝜐 = −ay + cx, where a, b, and c are constants: a = 0.50 s−1, b = 1.5 m/s, and c = 0.35 s−1. Generate an expression for the stream function and plot some streamlines of the flow in the upper-right quadrant.
SOLUTION For a given velocity field we are to generate an expression for 𝜓 and plot several streamlines for given values of constants a, b, and c.
Assumptions 1 The flow is steady. 2 The flow is incompressible. 3 The flow is two-dimensional in the xy-plane, implying that w = 0 and neither u nor 𝜐 depend on z.
Analysis We start by picking one of the two parts of Eq. 9–20 that define the stream function (it doesn’t matter which part we choose—the solution will be iden tical).
∂𝜓 ∂y = u = ax + b Next we integrate with respect to y, noting that this is a partial integration, so we add an arbitrary function of the other variable, x, rather than a constant of integra tion, 𝜓= axy + by + g(x) (1) Now we choose the other part of Eq. 9–20, differentiate Eq. 1, and rearrange as follows: 𝜐= − ∂𝜓 ∂x = −ay −gʹ(x) (2) where g′(x) denotes dg/dx since g is a function of only one variable, x. We now have two expressions for velocity component 𝜐, the equation given in the 5 4 3 2 1 0 –1 0 1 2 3 Scale for velocity vectors: 10 m/s ψ = –5 m2/s x, m 4 5 y, m ψ = 0 –10 –7.5 60 50 40 30 20 10 5 FIGURE 9–20 Streamlines for the velocity field of Example 9–8; the value of constant 𝜓 is indicated for each streamline, and velocity vectors are shown at four locations.
cen96537_ch09_443-518.indd 458 14/01/17 3:08 pm 459 CHAPTER 9 There is another physically significant fact about the stream function: The difference in the value of 𝜓 from one streamline to another is equal to the volume flow rate per unit width between the two streamlines.
This statement is illustrated in Fig. 9–22. Consider two streamlines, 𝜓1 and 𝜓2, and imagine two-dimensional flow in the xy-plane, of unit width into the page (1 m in the −z-direction). By definition, no flow can cross a streamline. Thus, the fluid that happens to occupy the space between these two stream lines remains confined between the same two streamlines. It follows that the mass flow rate through any cross-sectional slice between the streamlines is the same at any instant in time. The cross-sectional slice can be any shape, provided that it starts at streamline 1 and ends at streamline 2. In Fig. 9–22, for example, slice A is a smooth arc from one streamline to the other while slice B is wavy. For steady, incompressible, two-dimensional flow in the xy-plane, the volume flow rate V . between the two streamlines (per unit width) must therefore be a constant. If the two streamlines spread apart, as they do from cross-sectional slice A to cross-sectional slice B, the average velocity between the two streamlines decreases accordingly, such that the volume flow rate remains the same (V .
A = V .
B). In Fig. 9–20 of Example 9–8, velocity vec tors at four locations in the flow field between streamlines 𝜓 = 0 m2/s and 𝜓 = 5 m2/s are plotted. You can clearly see that as the streamlines diverge from each other, the velocity vector decays in magnitude. Likewise, when streamlines converge, the average velocity between them must increase.
problem statement and Eq. 2. We equate these and integrate with respect to x to find g(x), 𝜐= −ay + cx = −ay −gʹ(x) → gʹ(x) = −cx → g(x) = −c x2 2 + C (3) Note that here we have added an arbitrary constant of integration C since g is a function of x only. Finally, substituting Eq. 3 into Eq. 1 yields the final expression for 𝜓, Solution: 𝟁= axy + by −c x2 2 + C (4) To plot the streamlines, we note that Eq. 4 represents a family of curves, one unique curve for each value of the constant (𝜓 − C). Since C is arbitrary, it is com mon to set it equal to zero, although it can be set to any desired value. For simplicity we set C = 0 and solve Eq. 4 for y as a function of x, yielding Equation for streamlines: y = 𝜓+ cx2/ 2 ax + b (5) For the given values of constants a, b, and c, we plot Eq. 5 for several values of 𝜓 in Fig. 9–21; these curves of constant 𝜓 are streamlines of the flow. From Fig. 9–21 we see that this is a smoothly converging flow in the upper-right quadrant.
Discussion It is always good to check your algebra. In this example, you should substitute Eq. 4 into Eq. 9–20 to verify that the correct velocity components are obtained.
–6 –4 –2 0 2 6 8 10 12 14 16 5 4 3 2 1 0 –1 0 1 2 3 x, m 4 5 y, m ψ = 4 m2/s FIGURE 9–21 Streamlines for the velocity field of Example 9–9; the value of constant 𝜓 is indicated for each streamline.
y x Streamline 1 Streamline 2 .
.
.
A B VB = VA VA ψ = ψ1 ψ = ψ2 FIGURE 9–22 For two-dimensional streamlines in the xy-plane, the volume flow rate V . per unit width between two streamlines is the same through any cross-sectional slice.
cen96537_ch09_443-518.indd 459 14/01/17 3:08 pm 460 Differential Analysis of Fluid Flow We prove the given statement mathematically by considering a control vol ume bounded by the two streamlines of Fig. 9–22 and by cross-sectional slice A and cross-sectional slice B (Fig. 9–23). An infinitesimal length ds along slice B is illustrated in Fig. 9–23a, along with its unit normal vector n →. A mag nified view of this region is sketched in Fig. 9–23b for clarity. As shown, the two components of ds are dx and dy; thus the unit normal vector is n ›= dy ds i ›−dx ds j › The volume flow rate per unit width through segment ds of the control surface is d V · = V › ·n › dA = (u i ›+ 𝜐 j › ) · ( dy ds i ›−dx ds j › ) ds (9–23) ds where dA = ds times 1 = ds, where the 1 indicates a unit width into the page, regardless of the unit system. When we expand the dot product of Eq. 9–23 and apply Eq. 9–20, we get d V · = u dy −𝜐 dx = ∂𝜓 ∂y dy + ∂𝜓 ∂x dx = d𝜓 (9–24) We find the total volume flow rate through cross-sectional slice B by inte grating Eq. 9–24 from streamline 1 to streamline 2, V · B = ∫B V › · n › dA = ∫B d V · = ∫ 𝜓=𝜓2 𝜓=𝜓1 d𝜓= 𝜓2 −𝜓1 (9–25) Thus, the volume flow rate per unit width through slice B is equal to the difference between the values of the two stream functions that bound slice B. Now consider the entire control volume of Fig. 9–23a. Since we know that no flow crosses the streamlines, conservation of mass demands that the volume flow rate into the control volume through slice A be iden tical to the volume flow rate out of the control volume through slice B. Finally, since we may choose a cross-sectional slice of any shape or location between the two streamlines, the statement is proven. When dealing with stream functions, the direction of flow is obtained by what we might call the “left-side convention.” Namely, if you are looking down the z-axis at the xy-plane (Fig. 9–24) and are moving in the direction of the flow, the stream function increases to your left.
The value of 𝜓 increases to the left of the direction of flow in the xy-plane.
In Fig. 9–24, for example, the stream function increases to the left of the flow direction, regardless of how much the flow twists and turns. Notice also that when the streamlines are far apart (lower right of Fig. 9–24), the mag nitude of velocity (the fluid speed) in that vicinity is small relative to the speed in locations where the streamlines are close together (middle region of Fig. 9–24). This is easily explained by conservation of mass. As the stream lines converge, the cross-sectional area between them decreases, and the velocity must increase to maintain the flow rate between the streamlines.
y x ds dy/ds ds dy dx ds dx u υ y x Streamline 1 Streamline 2 A B CV CV Control surface ψ = ψ1 ψ = ψ2 V V n n (b) (a) FIGURE 9–23 (a) Control volume bounded by streamlines 𝜓1 and 𝜓2 and slices A and B in the xy-plane; (b) magnified view of the region around infinitesimal length ds.
y x ψ = 7 ψ = 6 ψ = 5 FIGURE 9–24 Illustration of the “left-side convention.” In the xy-plane, the value of the stream function always increases to the left of the flow direction.
cen96537_ch09_443-518.indd 460 14/01/17 3:08 pm 461 CHAPTER 9 FIGURE 9–25 Streaklines produced by Hele–Shaw flow over an inclined plate. The streaklines model streamlines of potential flow (Chap. 10) over a two-dimensional inclined plate of the same cross-sectional shape.
Original © D.H. Peregrine, School of Mathematics, University of Bristol. Courtesy of Onno Bokhove and Valerie Zwart.
EXAMPLE 9–10 Relative Velocity Deduced from Streamlines Hele–Shaw flow is produced by forcing a liquid through a thin gap between paral lel plates. An example of Hele–Shaw flow is provided in Fig. 9–25 for flow over an inclined plate. Streaklines are generated by introducing dye at evenly spaced points upstream of the field of view. Since the flow is steady, the streaklines are coincident with streamlines. The fluid is water and the glass plates are 1.0 mm apart. Discuss how you can tell from the streamline pattern whether the flow speed in a particular region of the flow field is (relatively) large or small.
SOLUTION For the given set of streamlines, we are to discuss how we can tell the relative speed of the fluid.
Assumptions 1 The flow is steady. 2 The flow is incompressible. 3 The flow models two-dimensional potential flow in the xy-plane.
Analysis When equally spaced streamlines of a stream function spread away from each other, it indicates that the flow speed has decreased in that region. Likewise, if the streamlines come closer together, the flow speed has increased in that region. In Fig. 9–25 we infer that the flow far upstream of the plate is straight and uniform, since the streamlines are equally spaced. The fluid decelerates as it approaches the underside of the plate, especially near the stagnation point, as indicated by the wide gap between streamlines. The flow accelerates rapidly to very high speeds around the sharp corners of the plate, as indicated by the tightly spaced streamlines.
Discussion The streaklines of Hele–Shaw flow turn out to be similar to those of potential flow, which is discussed in Chap. 10.
EXAMPLE 9–11 Volume Flow Rate Deduced from Streamlines Water is sucked through a narrow slot on the bottom wall of a water channel. The water in the channel flows from left to right at uniform velocity V = 1.0 m/s. The slot is perpendicular to the xy-plane, and runs along the z-axis across the entire channel, which is w = 2.0 m wide. The flow is thus approximately two-dimen sional in the xy-plane. Several streamlines of the flow are plotted and labeled in Fig. 9–26.
cen96537_ch09_443-518.indd 461 14/01/17 3:08 pm 462 Differential Analysis of Fluid Flow The thick streamline in Fig. 9–26 is called the dividing streamline because it divides the flow into two parts. Namely, all the water below this dividing streamline gets sucked into the slot, while all the water above the dividing streamline continues on its way downstream. What is the volume flow rate of water being sucked through the slot? Estimate the magnitude of the velocity at point A.
SOLUTION For the given set of streamlines, we are to determine the volume flow rate through the slot and estimate the fluid speed at a point.
Assumptions 1 The flow is steady. 2 The flow is incompressible. 3 The flow is two-dimensional in the xy-plane. 4 Friction along the bottom wall is neglected.
Analysis By Eq. 9–25, the volume flow rate per unit width between the bottom wall (𝜓wall = 0) and the dividing streamline (𝜓dividing = 1.0 m2/s) is V · w = 𝜓dividing −𝜓wall = (1.0 −0) m2/s = 1.0 m2/s All of this flow must go through the slot. Since the channel is 2.0 m wide, the total volume flow rate through the slot is V · = V · w w = (1.0 m2/s)(2.0 m) = 2.0 m3/s To estimate the speed at point A, we measure the distance 𝛿 between the two streamlines that enclose point A. We find that streamline 1.8 is about 0.21 m away from streamline 1.6 in the vicinity of point A. The volume flow rate per unit width (into the page) between these two streamlines is equal to the difference in value of the stream function. We thus estimate the speed at point A, VA ≅V · w𝛿 = 1 𝛿 V · w = 1 𝛿 (𝜓1.8 −𝜓1.6) = 1 0.21 m (1.8 −1.6) m2/s = 0.95 m/s Our estimate is close to the known free-stream speed (1.0 m/s), indicating that the fluid in the vicinity of point A flows at nearly the same speed as the free-stream flow, but points slightly downward.
Discussion The streamlines of Fig. 9–26 were generated by superposition of a uniform stream and a line sink, assuming irrotational (potential) flow. We discuss such superposition in Chap. 10.
2 1.5 1 0.5 0 –3 –2 –1 1 x, m 3 2 V w .
A 2.0 1.8 0.8 1.6 0.6 1.4 1.2 0.2 1.0 0.4 y, m FIGURE 9–26 Streamlines for free-stream flow along a wall with a narrow suction slot; streamline values are shown in units of m2/s; the thick streamline is the dividing streamline. The direction of the velocity vector at point A is determined by the left-side convention.
cen96537_ch09_443-518.indd 462 14/01/17 3:08 pm 463 CHAPTER 9 The Stream Function in Cylindrical Coordinates For two-dimensional flow in the xy-plane, we can also define the stream func tion in cylindrical coordinates, which is more convenient for many problems. Note that by two-dimensional we mean that there are only two relevant inde pendent spatial coordinates—with no dependence on the third component. There are two possibilities. The first is planar flow, just like that of Eqs. 9–19 and 9–20, but in terms of (r, 𝜃) and (ur, u𝜃) instead of (x, y) and (u, 𝜐) (see Fig. 9–10a). In this case, there is no dependence on coordinate z. We simplify the incompressible continuity equation, Eq. 9–18, for two-dimensional planar flow in the r𝜃-plane, ∂(rur) ∂r + ∂(u𝜃) ∂𝜃 = 0 (9–26) We define the stream function as follows: Incompressible, planar stream function in cylindrical coordinates: ur = 1 r ∂𝜓 ∂𝜃 and u𝜃= − ∂𝜓 ∂r (9–27) We note again that the signs are reversed in some textbooks. You can substi tute Eq. 9–27 into Eq. 9–26 to convince yourself that Eq. 9–26 is identically satisfied for any smooth function 𝜓(r, 𝜃), since the order of differentiation (r then 𝜃 versus 𝜃 then r) is irrelevant for a smooth function. The second type of two-dimensional flow in cylindrical coordinates is axisymmetric flow, in which r and z are the relevant spatial variables, ur and uz are the nonzero velocity components, and there is no dependence on 𝜃 (Fig. 9–27). Examples of axisymmetric flow include flow around spheres, bullets, and the fronts of many objects like torpedoes and missiles, which would be axisymmetric everywhere if not for their fins. For incompressible axisymmetric flow, the continuity equation is 1 r ∂(rur) ∂r + ∂(uz) ∂z = 0 (9–28) The stream function 𝜓 is defined such that it satisfies Eq. 9–28 exactly, pro vided of course that 𝜓 is a smooth function of r and z, Incompressible, axisymmetric stream function in cylindrical coordinates: ur = − 1 r ∂𝜓 ∂z and uz = 1 r ∂𝜓 ∂r (9–29) We also note that there is another way to describe axisymmetric flows, namely, by using Cartesian coordinates (x, y) and (u, 𝜐), but forcing coor dinate x to be the axis of symmetry. This can lead to confusion because the equations of motion must be modified accordingly to account for the axi symmetry. Nevertheless, this is often the approach used in CFD codes. The advantage is that after one sets up a grid in the xy-plane, the same grid can be used for both planar flow (flow in the xy-plane with no z-dependence) and axisymmetric flow (flow in the xy-plane with rotational symmetry about the x-axis). We do not discuss the equations for this alternative description of axisymmetric flows.
z y r r z ur uz Rotational symmetry Axisymmetric body x θ FIGURE 9–27 Flow over an axisymmetric body in cylindrical coordinates with rotational symmetry about the z-axis; neither the geometry nor the velocity field depend on 𝜃, and u𝜃 = 0.
cen96537_ch09_443-518.indd 463 14/01/17 3:08 pm 464 Differential Analysis of Fluid Flow The Compressible Stream Function We extend the stream function concept to steady, compressible, two-dimensional flow in the xy-plane. The compressible continuity equation (Eq. 9–14) in Cartesian coordinates reduces to the following for steady two-dimensional flow: ∂(ρu) ∂x + ∂(ρ𝜐) ∂y = 0 (9–30) EXAMPLE 9–12 Stream Function in Cylindrical Coordinates Consider a line vortex, defined as steady, planar, incompressible flow in which the velocity components are ur = 0 and u𝜃 = K/r, where K is a constant. This flow is represented in Fig. 9–15a. Derive an expression for the stream function 𝜓(r, 𝜃), and prove that the streamlines are circles.
SOLUTION For a given velocity field in cylindrical coordinates, we are to derive an expression for the stream function and show that the streamlines are cir cular.
Assumptions 1 The flow is steady. 2 The flow is incompressible. 3 The flow is planar in the r𝜃-plane.
Analysis We use the definition of stream function given by Eq. 9–27. We can choose either component to start with; we choose the tangential component, ∂𝜓 ∂r = −u𝜃= − K r → 𝜓= −K ln r + f (𝜃) (1) Now we use the other component of Eq. 9–27, ur = 1 r ∂𝜓 ∂𝜃= 1 r f ʹ(𝜃) (2) where the prime denotes a derivative with respect to 𝜃. By equating ur from the given information to Eq. 2, we see that f ʹ(𝜃) = 0 → f (𝜃) = C where C is an arbitrary constant of integration. Equation 1 is thus Solution: 𝟁= −K ln r + C (3) Finally, we see from Eq. 3 that curves of constant 𝜓 are produced by setting r to a constant value. Since curves of constant r are circles by definition, streamlines (curves of constant 𝜓) must therefore be circles about the origin, as in Fig. 9–15a. For given values of C and 𝜓, we solve Eq. 3 for r to plot the streamlines, Equation for streamlines: r = e−(𝜓−C )/K (4) For K = 10 m2/s and C = 0, streamlines from 𝜓 = 0 to 22 are plotted in Fig. 9–28.
Discussion Notice that for a uniform increment in the value of 𝜓, the streamlines get closer and closer together near the origin as the tangential velocity increases. This is a direct result of the statement that the difference in the value of 𝜓 from one streamline to another is equal to the volume flow rate per unit width between the two streamlines.
–1 –0.5 x 0 0.5 1 1 0.5 0 –0.5 –1 y 14 8 ψ = 0 m2/s 22 12 10 6 4 2 FIGURE 9–28 Streamlines for the velocity field of Example 9–12, with K = 10 m2/s and C = 0; the value of constant 𝜓 is indicated for several streamlines.
This section can be skipped without loss of continuity (no pun intended).
cen96537_ch09_443-518.indd 464 14/01/17 3:08 pm 465 CHAPTER 9 We introduce a compressible stream function, which we denote as 𝜓𝜌, Steady, compressible, two-dimensional stream function in Cartesian coordinates: ρu = ∂𝜓ρ ∂y and ρ𝜐= − ∂𝜓ρ ∂x (9–31) By definition, 𝜓𝜌 of Eq. 9–31 satisfies Eq. 9–30 exactly, provided that 𝜓𝜌 is a smooth function of x and y. Many of the features of the compressible stream function are the same as those of the incompressible 𝜓 as discussed previously. For example, curves of constant 𝜓𝜌 are still streamlines. How ever, the difference in 𝜓𝜌 from one streamline to another is mass flow rate per unit width rather than volume flow rate per unit width. Although not as popular as its incompressible counterpart, the compressible stream function finds use in some commercial CFD codes.
9–4 ■ THE DIFFERENTIAL LINEAR MOMENTUM EQUATION—CAUCHY’S EQUATION Through application of the Reynolds transport theorem (Chap. 4), we have the general expression for the linear momentum equation as applied to a control volume, ∑F ›= ∫CV ρg › dV + ∫CS 𝜎ij ·n › dA = ∫CV ∂ ∂t (ρV ›) dV + ∫CS (ρV ›)V › · n › dA (9–32) where 𝜎ij is the stress tensor introduced in Chap. 6. Components of 𝜎ij on the positive faces of an infinitesimal rectangular control volume are shown in Fig. 9–29. Equation 9–32 applies to both fixed and moving con trol volumes, provided that V › is the absolute velocity (as seen from a fixed observer). For the special case of flow with well defined inlets and outlets, Eq. 9–32 is simplified as follows: ∑F ›= ∑F › body + ∑F › surface = ∫CV ∂ ∂t (ρV › ) dV + ∑ out 𝛽 m · V ›−∑ in 𝛽 m · V › (9–33) where V › in the last two terms is taken as the average velocity at an inlet or outlet, and 𝛽 is the momentum flux correction factor (Chap. 6). In words, the total force acting on the control volume is equal to the rate at which momen tum changes within the control volume plus the rate at which momentum flows out of the control volume minus the rate at which momentum flows into the control volume. Equation 9–33 applies to any control volume, regardless of its size. To generate a differential linear momentum equa tion, we imagine the control volume shrinking to infinitesimal size. In the limit, the entire control volume shrinks to a point in the flow (Fig. 9–2). We take the same approach here as we did for conservation of mass; namely, we show more than one way to derive the differential form of the linear momen tum equation. Derivation Using the Divergence Theorem The most straightforward (and most elegant) way to derive the differen tial form of the momentum equation is to apply the divergence theorem of Eq. 9–3. A more general form of the divergence theorem applies not only to vectors, but to other quantities as well, such as tensors, as illustrated in dx dy dz σxx σxz σxy σyy σyx σyz σzz σzy σzx FIGURE 9–29 Positive components of the stress tensor in Cartesian coordinates on the positive (right, top, and front) faces of an infinitesimal rectangular control volume. The red dots indicate the center of each face. Positive components on the negative (left, bottom, and back) faces are in the opposite direction of those shown here.
cen96537_ch09_443-518.indd 465 14/01/17 3:08 pm 466 Differential Analysis of Fluid Flow Fig. 9–30. Specifically, if we replace Gij in the extended divergence theorem of Fig. 9–30 with the quantity (𝜌V ›)V ›, a second-order tensor, the last term in Eq. 9–32 becomes ∫CS (ρV › )V › ·n › dA = ∫CV ∇ › ·(ρV › V ›) dV (9–34) where V › V › is a vector product called the outer product of the velocity vector with itself. (The outer product of two vectors is not the same as the inner or dot product, nor is it the same as the cross product of the two vectors.) Similarly, if we replace Gij in Fig. 9–30 by the stress tensor 𝜎ij, the second term on the left-hand side of Eq. 9–32 becomes ∫CS 𝜎ij·n › dA = ∫CV ∇ › ·𝜎ij dV (9–35) Thus, the two surface integrals of Eq. 9–32 become volume integrals by applying Eqs. 9–34 and 9–35. We combine and rearrange the terms, and rewrite Eq. 9–32 as ∫CV [ ∂ ∂t (ρV › ) + ∇ › ·(ρV › V ›) −ρg ›−∇ › ·𝜎ij] dV = 0 (9–36) Finally, we argue that Eq. 9–36 must hold for any control volume regardless of its size or shape. This is possible only if the integrand (enclosed by square brackets) is identically zero. Hence, we have a general differential equation for linear momentum, known as Cauchy’s equation, Cauchy’s equation: ∂ ∂t (ρV ›) + ∇ › ·(ρV › V ›) = ρg ›+ ∇ › ·𝜎ij (9–37) Equation 9–37 is named in honor of the French engineer and mathemati cian Augustin Louis de Cauchy (1789–1857). It is valid for compressible as well as incompressible flow since we have not made any assumptions about incompressibility. It is valid at any point in the flow domain (Fig. 9–31). Note that Eq. 9–37 is a vector equation, and thus represents three scalar equations, one for each coordinate axis in three-dimensional problems.
Derivation Using an Infinitesimal Control Volume We derive Cauchy’s equation a second way, using an infinitesimal control volume on which we apply the linear momentum equation (Eq. 9–33). We consider the same box-shaped control volume we used to derive the con tinuity equation (Fig. 9–3). At the center of the box, as previously, we define the density as 𝜌 and the velocity components as u, 𝜐, and w. We also define the stress tensor as 𝜎ij at the center of the box. For simplicity, we consider the x-component of Eq. 9–33, obtained by setting Σ F › equal to its x-component, ΣFx, and V › equal to its x-component, u. This not only simplifies the diagrams, but enables us to work with a scalar equation, namely, ∑Fx = ∑Fx, body + ∑Fx, surface = ∫CV ∂ ∂t (ρu) dV + ∑ out 𝛽 m · u −∑ in 𝛽 m · u (9–38) As the control volume shrinks to a point, the first term on the right-hand side of Eq. 9–38 becomes Rate of change of x-momentum within the control volume: ∫CV ∂ ∂t (ρu) dV = ∂ ∂t (ρu) dx dy dz (9–39) V • Gij d V = A Gij • n dA Δ The Extended Divergence Theorem ∫ ∮ V • Gd V = A G • n dA or Δ ∫ ∮ FIGURE 9–30 An extended form of the divergence theorem is useful not only for vectors, but also for tensors. In the equation, Gij (or G → → ) is a second-order tensor, V is a volume, and A is the surface area that encloses and defines the volume.
Equation of the Day ∂ ∂t (ρV) + • (ρVV ) = ρg + • σij Δ Δ Cauchy’s Equation FIGURE 9–31 Cauchy’s equation is a differential form of the linear momentum equation. It applies to any type of fluid.
cen96537_ch09_443-518.indd 466 14/01/17 3:08 pm 467 CHAPTER 9 since the volume of the differential element is dx dy dz. We apply first-order truncated Taylor series expansions at locations away from the center of the control volume to approximate the inflow and outflow of momentum in the x-direction. Figure 9–32 shows these momentum fluxes at the center point of each of the six faces of the infinitesimal control volume. Only the normal velocity component at each face needs to be considered, since the tangential velocity components contribute no mass flow out of (or into) the face, and hence no momentum flow through the face either. By summing all the outflows and subtracting all the inflows shown in Fig. 9–32, we obtain an approximation for the last two terms of Eq. 9–38, Net outflow of x-momentum through the control surface: ∑ out 𝛽 m · u −∑ in 𝛽 m · u ≅( ∂ ∂x (ρuu) + ∂ ∂y (ρ𝜐u) + ∂ ∂z (ρwu)) dx dy dz (9–40) where 𝛽 is set equal to one at all faces, consistent with our first-order approximation. Next, we sum all the forces acting on our infinitesimal control volume in the x-direction. As was done in Chap. 6, we need to consider both body forces and surface forces. Gravity force (weight) is the only body force we take into account. For the general case in which the coordinate system may not be aligned with the z-axis (or with any coordinate axis for that matter), as sketched in Fig. 9–33, the gravity vector is written as g ›= gx i ›+ gy j ›+ gz k › Thus, in the x-direction, the body force on the control volume is ∑dFx, body = ∑dFx, gravity = ρgx dx dy dz (9–41) Next we consider the net surface force in the x-direction. Recall that stress tensor 𝜎ij has dimensions of force per unit area. Thus, to obtain a force, we must multiply each stress component by the surface area of the face on y z x dx dz dy dx dz ∂y ( ) ρυu + ∂(ρυu) dy 2 dx dy ∂z ( ) ρwu + ∂(ρwu) dz 2 dy dz ∂x ( ) ρuu +∂(ρuu) dx 2 dx dy ∂z ( ) ρwu – ∂(ρwu) dz 2 dx dz ∂y ( ) ρυu – ∂(ρυu) dy 2 dy dz ∂x ( ) ρuu – ∂(ρuu) dx 2 FIGURE 9–32 Inflow and outflow of the x-component of linear momentum through each face of an infinitesimal control volume; the red dots indicate the center of each face.
y g z x dy dx dz dFgravity FIGURE 9–33 The gravity vector is not neces sarily aligned with any particular axis, in general, and there are three components of the body force acting on an infinitesimal fluid element.
cen96537_ch09_443-518.indd 467 14/01/17 3:08 pm 468 Differential Analysis of Fluid Flow which it acts. We need to consider only those components that point in the x- (or −x-) direction. (The other components of the stress tensor, although they may be nonzero, do not contribute to a net force in the x-direction.) Using truncated Taylor series expansions, we sketch all the surface forces that contribute to a net x-component of surface force acting on our differential fluid element (Fig. 9–34). Summing all the surface forces illustrated in Fig. 9–34, we obtain an approximation for the net surface force acting on the differential fluid ele ment in the x-direction, ∑dFx, surface = ( ∂ ∂x 𝜎xx + ∂ ∂y 𝜎yx + ∂ ∂z 𝜎zx) dx dy dz (9–42) We now substitute Eqs. 9–39 through 9–42 into Eq. 9–38, noting that the volume of the differential element of fluid, dx dy dz, appears in all terms and can be eliminated. After some rearrangement we obtain the differential form of the x-momentum equation, ∂(ρu) ∂t + ∂(ρuu) ∂x + ∂(ρ𝜐u) ∂y + ∂(ρwu) ∂z = ρgx + ∂ ∂x 𝜎xx + ∂ ∂y 𝜎yx + ∂ ∂z 𝜎zx (9–43) In similar fashion, we generate differential forms of the y- and z-momentum equations, ∂(ρ𝜐) ∂t + ∂(ρu𝜐) ∂x + ∂(ρ𝜐𝜐) ∂y + ∂(ρw𝜐) ∂z = ρgy + ∂ ∂x 𝜎xy + ∂ ∂y 𝜎yy + ∂ ∂z 𝜎zy (9–44) and ∂(ρw) ∂t + ∂(ρuw) ∂x + ∂(ρ𝜐w) ∂y + ∂(ρww) ∂z = ρgz + ∂ ∂x 𝜎xz + ∂ ∂y 𝜎yz + ∂ ∂z 𝜎zz (9–45) respectively. Finally, we combine Eqs. 9–43 through 9–45 into one vector equation, Cauchy’s equation: ∂ ∂t (ρV ›) + ∇ › ·(ρV › V ›) = ρg ›+ ∇ › ·𝜎ij This equation is identical to Cauchy’s equation (Eq. 9–37); thus we confirm that our derivation using the differential fluid element yields the same result y z x dx dz dy dx dy ∂z ( ) σzx – ∂σzx dz 2 dx dz ∂y ( ) σyx + ∂σyx dy 2 dy dz ∂x ( ) σxx + ∂σxx dx 2 dx dy ∂z ( ) σzx + ∂σzx dz 2 dy dz ∂x ( ) σxx – ∂σxx dx 2 dx dz ∂y ( ) σyx – ∂σyx dy 2 FIGURE 9–34 Sketch illustrating the surface forces acting in the x-direction due to the appropriate stress tensor component on each face of the differential control volume; the red dots indicate the center of each face.
cen96537_ch09_443-518.indd 468 14/01/17 3:08 pm 469 CHAPTER 9 as our derivation using the divergence theorem. Note that the product V › V › is a second-order tensor (Fig. 9–35).
Alternative Form of Cauchy’s Equation Applying the product rule to the first term on the left side of Eq. 9–37, we get ∂ ∂t (ρV › ) = ρ ∂V › ∂t + V › ∂ρ ∂t (9–46) The second term of Eq. 9–37 is written as ∇ › ·(ρV › V ›) = V › ∇ › ·(ρV ›) + ρ(V › ·∇ ›)V › (9–47) Thus we have eliminated the second-order tensor represented by V › V ›. After some rearrangement, substitution of Eqs. 9–46 and 9–47 into Eq. 9–37 yields ρ ∂V › ∂t + V › [ ∂ρ ∂t + ∇ › ·(ρV ›)] + ρ(V › ·∇ ›)V ›= ρg ›+ ∇ › ·𝜎ij But the expression in square brackets in this equation is identically zero by the continuity equation, Eq. 9–5. By combining the remaining two terms on the left side, we write Alternative form of Cauchy’s equation: ρ[ ∂V › ∂t + (V › ·∇ ›)V › ] = ρ DV › Dt = ρg ›+ ∇ › ·𝜎ij (9–48) where we have recognized the expression in square brackets as the material acceleration—the acceleration following a fluid particle (see Chap. 4).
Derivation Using Newton’s Second Law We derive Cauchy’s equation by yet a third method. Namely, we take the differential fluid element as a material element instead of a control volume. In other words, we think of the fluid within the differential element as a tiny system of fixed identity, moving with the flow (Fig. 9–36). The acceleration of this fluid element is a › = DV ›/Dt by definition of the material accelera tion. By Newton’s second law applied to a material element of fluid, ∑dF ›= dma ›= dm DV › Dt = ρ dx dy dz DV › Dt (9–49) At the instant in time represented in Fig. 9–36, the net force on the differen tial fluid element is found in the same way as that calculated earlier on the differential control volume. Thus the total force acting on the fluid element is the sum of Eqs. 9–41 and 9–42, extended to vector form. Substituting these into Eq. 9–49 and dividing by dx dy dz, we once again generate the alternative form of Cauchy’s equation, ρ DV › Dt = ρg ›+ ∇ › ·𝜎ij (9–50) Equation 9–50 is identical to Eq. 9–48. In hindsight, we could have started with Newton’s second law from the beginning, avoiding some algebra. Nev ertheless, derivation of Cauchy’s equation by three methods certainly boosts our confidence in the validity of the equation!
VV = uu uυ uw υu υυ υw wu wυ ww FIGURE 9–35 The outer product of vector V › = (u, 𝜐, w) with itself is a second-order tensor. The product shown is in Cartesian coordinates and is illustrated as a nine-component matrix.
y z dz a dx dy x Streamline dF Σ FIGURE 9–36 If the differential fluid element is a material element, it moves with the flow and Newton’s second law applies directly.
cen96537_ch09_443-518.indd 469 14/01/17 3:08 pm 470 Differential Analysis of Fluid Flow We must be very careful when expanding the last term of Eq. 9–50, which is the divergence of a second-order tensor. In Cartesian coordinates, the three components of Cauchy’s equation are x-component: ρ Du Dt = ρgx + ∂𝜎xx ∂x + ∂𝜎yx ∂y + ∂𝜎zx ∂z (9–51a) y-component: ρ D𝜐 Dt = ρgy + ∂𝜎xy ∂x + ∂𝜎yy ∂y + ∂𝜎zy ∂z (9–51b) z-component: ρ Dw Dt = ρgz + ∂𝜎xz ∂x + ∂𝜎yz ∂y + ∂𝜎zz ∂z (9–51c) We conclude this section by noting that we cannot solve any fluid mechan ics problems using Cauchy’s equation by itself (even when combined with continuity). The problem is that the stress tensor 𝜎ij needs to be expressed in terms of the primary unknowns in the problem, namely, density, pressure, and velocity. This is done for the most common type of fluid in Section 9–5.
9–5 ■ THE NAVIER–STOKES EQUATION Introduction Cauchy’s equation (Eq. 9–37 or its alternative form Eq. 9–48) is not very useful to us as is, because the stress tensor 𝜎ij contains nine components, six of which are independent (because of symmetry). Thus, in addition to den sity and the three velocity components, there are six additional unknowns, for a total of 10 unknowns. (In Cartesian coordinates the unknowns are 𝜌, u, 𝜐, w, 𝜎xx, 𝜎xy, 𝜎xz, 𝜎yy, 𝜎yz, and 𝜎zz.) Meanwhile, we have discussed only four equations so far—continuity (one equation) and Cauchy’s equation (three equations). Of course, to be mathematically solvable, the number of equa tions must equal the number of unknowns, and thus we need six more equa tions. These equations are called constitutive equations, and they enable us to write the components of the stress tensor in terms of the velocity field and pressure field. The first thing we do is separate the pressure stresses and the viscous stresses. When a fluid is at rest, the only stress acting at any surface of any fluid element is pressure P, which always acts inward and normal to the surface (Fig. 9–37). Thus, regardless of the orientation of the coordinate axes, for a fluid at rest the stress tensor reduces to Fluid at rest: 𝜎ij = ( 𝜎xx 𝜎xy 𝜎xz 𝜎yx 𝜎yy 𝜎yz 𝜎zx 𝜎zy 𝜎zz ) = ( −P 0 0 0 −P 0 0 0 −P ) (9–52) Pressure P in Eq. 9–52 is the same as the thermodynamic pressure with which we are familiar from our study of thermodynamics. P is related to temperature and density through some type of equation of state (e.g., the ideal gas law). As a side note, this further complicates a compressible fluid flow analysis because we introduce yet another unknown, namely, tempera ture T. This new unknown requires another equation—the differential form of the energy equation—which is not discussed in this text.
y z x dx dz dy P P P P P P FIGURE 9–37 For fluids at rest, the only stress on a fluid element is the hydrostatic pressure, which always acts inward and normal to any surface. Note that we are ignoring gravity in this case; otherwise pressure would increase in the direction of the gravitational acceleration.
cen96537_ch09_443-518.indd 470 14/01/17 3:08 pm 471 CHAPTER 9 When a fluid is moving, pressure still acts inwardly normal, but viscous stresses may also exist. We generalize Eq. 9–52 for moving fluids as Moving fluids: 𝜎ij = ( 𝜎xx 𝜎xy 𝜎xz 𝜎yx 𝜎yy 𝜎yz 𝜎zx 𝜎zy 𝜎zz ) = ( −P 0 0 0 −P 0 0 0 −P ) + ( 𝜏xx 𝜏xy 𝜏xz 𝜏yx 𝜏yy 𝜏yz 𝜏zx 𝜏zy 𝜏zz) (9–53) where we have introduced a new tensor, 𝜏ij, called the viscous stress tensor or the deviatoric stress tensor. Mathematically, we have not helped the sit uation because we have replaced the six unknown components of 𝜎ij with six unknown components of 𝜏ij, and have added another unknown, pressure P. Fortunately, however, there are constitutive equations that express 𝜏ij in terms of the velocity field and measurable fluid properties such as viscosity. The actual form of the constitutive relations depends on the type of fluid, as discussed shortly. As a side note, there are some subtleties associated with the pressure in Eq. 9–53. If the fluid is incompressible, we have no equation of state (it is replaced by the equation 𝜌 = constant), and we can no longer define P as the thermodynamic pressure. Instead, we define P in Eq. 9–53 as the mechanical pressure, Mechanical pressure: Pm = −1 3 (𝜎xx + 𝜎yy + 𝜎zz) (9–54) We see from Eq. 9–54 that mechanical pressure is the mean normal stress acting inwardly on a fluid element. It is therefore also called mean pressure by some authors. Thus, when dealing with incompressible fluid flows, pres sure variable P is always interpreted as the mechanical pressure Pm. For compressible flow fields however, pressure P in Eq. 9–53 is the thermody namic pressure, but the mean normal stress felt on the surfaces of a fluid element is not necessarily the same as P (pressure variable P and mechani cal pressure Pm are not necessarily equivalent). You are referred to Panton (1996) or Kundu et al. (2011) for a more detailed discussion of mechanical pressure.
Newtonian versus Non-Newtonian Fluids The study of the deformation of flowing fluids is called rheology; the rhe ological behavior of various fluids is sketched in Fig. 9–38. In this text, we concentrate on Newtonian fluids, defined as fluids for which the stress tensor is linearly proportional to the strain rate tensor. Newtonian fluids (stress proportional to strain rate) are analogous to elastic solids (Hooke’s law: stress proportional to strain). Many common fluids, such as air and other gases, water, kerosene, gasoline, and other oil-based liquids, are New tonian fluids. Fluids for which the stress tensor is not linearly related to the strain rate tensor are called non-Newtonian fluids. Examples include slurries and colloidal suspensions, polymer solutions, blood, paste, and cake batter. Some non-Newtonian fluids exhibit a “memory”—the shear stress depends not only on the local strain rate, but also on its history. A fluid that returns (partially) to its original shape after the applied stress is released is called viscoelastic.
Stress Strain rate Yield stress Shear thinning Bingham plastic Newtonian Shear thickening FIGURE 9–38 Rheological behavior of fluids—stress as a function of strain rate.
cen96537_ch09_443-518.indd 471 14/01/17 3:08 pm 472 Differential Analysis of Fluid Flow Some non-Newtonian fluids are called shear thinning fluids or pseudoplastic fluids, because the more the fluid is sheared, the less viscous it becomes. A good example is paint. Paint is very viscous when poured from the can or when picked up by a paintbrush, since the shear rate is small. How ever, as we apply the paint to the wall, the thin layer of paint between the paintbrush and the wall is subjected to a large shear rate, and it becomes much less viscous. Plastic fluids are those in which the shear thinning effect is extreme. In some fluids a finite stress called the yield stress is required before the fluid begins to flow at all; such fluids are called Bingham plastic fluids. Certain pastes such as acne cream and toothpaste are examples of Bingham plastic fluids. If you hold the tube upside down, the paste does not flow, even though there is a nonzero stress due to gravity. However, if you squeeze the tube (greatly increasing the stress), the paste flows like a very viscous fluid. Other fluids show the opposite effect and are called shear thickening fluids or dilatant fluids; the more the fluid is sheared, the more viscous it becomes. The best example is quicksand, a thick mixture of sand and water. As we all know from Hollywood movies, it is easy to move slowly through quicksand, since the viscosity is low; but if you panic and try to move quickly, the vis cous resistance increases considerably and you get “stuck” (Fig. 9–39). You can create your own quicksand by mixing two parts cornstarch with one part water—try it! Shear thickening fluids are used in some exercise equipment— the faster you pull, the more resistance you encounter.
Derivation of the Navier–Stokes Equation for Incompressible, Isothermal Flow From this point on, we limit our discussion to Newtonian fluids, where by definition the stress tensor is linearly proportional to the strain rate ten sor. The general result (for compressible flow) is rather involved and is not included here. Instead, we assume incompressible flow (𝜌 = constant). We also assume nearly isothermal flow—namely, that local changes in tem perature are small or nonexistent; this eliminates the need for a differential energy equation. A further consequence of the latter assumption is that fluid properties, such as dynamic viscosity 𝜇 and kinematic viscosity 𝜈, are con stant as well (Fig. 9–40). With these assumptions, it can be shown (Kundu et al., 2011) that the viscous stress tensor reduces to Viscous stress tensor for an incompressible Newtonian fluid with constant properties: 𝜏ij = 2𝜇𝜀ij (9–55) where 𝜀ij is the strain rate tensor defined in Chap. 4. Equation 9–55 shows that stress is linearly proportional to strain. In Cartesian coordinates, the nine components of the viscous stress tensor are listed, only six of which are independent due to symmetry: 𝜏ij = ( 𝜏xx 𝜏xy 𝜏xz 𝜏yx 𝜏yy 𝜏yz 𝜏zx 𝜏zy 𝜏zz) = ( 2𝜇 ∂u ∂x 𝜇( ∂u ∂y + ∂𝜐 ∂x) 𝜇( ∂u ∂z + ∂w ∂x ) 𝜇( ∂𝜐 ∂x + ∂u ∂y) 2𝜇 ∂𝜐 ∂y 𝜇( ∂𝜐 ∂z + ∂w ∂y ) 𝜇( ∂w ∂x + ∂u ∂z) 𝜇( ∂w ∂y + ∂𝜐 ∂z) 2𝜇 ∂w ∂z ) (9–56) I think he means quicksand.
?
Help!
I fell into a dilatant fluid!
FIGURE 9–39 When an engineer falls into quicksand (a dilatant fluid), the faster he tries to move, the more viscous the fluid becomes.
For a fluid flow that is both incompressible and isothermal: • ρ = constant • μ = constant And therefore: • ν = constant FIGURE 9–40 The incompressible flow approxima tion implies constant density, and the isothermal approximation implies constant viscosity.
cen96537_ch09_443-518.indd 472 14/01/17 3:08 pm 473 CHAPTER 9 In Cartesian coordinates the stress tensor of Eq. 9–53 thus becomes 𝜎 ij = ( −P 0 0 0 −P 0 0 0 −P ) + ( 2𝜇 ∂u ∂x 𝜇( ∂u ∂y + ∂𝜐 ∂x) 𝜇( ∂u ∂z + ∂w ∂x ) 𝜇( ∂𝜐 ∂x + ∂u ∂y) 2𝜇 ∂𝜐 ∂y 𝜇( ∂𝜐 ∂z + ∂w ∂y ) 𝜇( ∂w ∂x + ∂u ∂z) 𝜇( ∂w ∂y + ∂𝜐 ∂z) 2𝜇 ∂w ∂z ) (9–57) Now we substitute Eq. 9–57 into the three Cartesian components of Cauchy’s equation. Let’s consider the x-component first. Equation 9–51a becomes ρ Du Dt = −∂P ∂x + ρgx + 2𝜇 ∂2u ∂x2 + 𝜇 ∂ ∂y ( ∂𝜐 ∂x + ∂u ∂y) + 𝜇 ∂ ∂z ( ∂w ∂x + ∂u ∂z) (9–58) Notice that since pressure consists of a normal stress only, it contributes only one term to Eq. 9–58. However, since the viscous stress tensor con sists of both normal and shear stresses, it contributes three terms. (This is a direct result of taking the divergence of a second-order tensor, by the way.) We note that as long as the velocity components are smooth functions of x, y, and z, the order of differentiation is irrelevant. For example, the first part of the last term in Eq. 9–58 can be rewritten as 𝜇 ∂ ∂z ( ∂w ∂x ) = 𝜇 ∂ ∂x ( ∂w ∂z ) After some clever rearrangement of the viscous terms in Eq. 9–58, ρ Du Dt = − ∂P ∂x + ρgx + 𝜇[ ∂2u ∂x2 + ∂ ∂x ∂u ∂x + ∂ ∂x ∂𝜐 ∂y + ∂2u ∂y2 + ∂ ∂x ∂w ∂z + ∂2u ∂z2] = − ∂P ∂x + ρgx + 𝜇[ ∂ ∂x ( ∂u ∂x + ∂𝜐 ∂y + ∂w ∂z ) + ∂2u ∂x2 + ∂2u ∂y2 + ∂2u ∂z2] The term in parentheses is zero because of the continuity equation for incompressible flow (Eq. 9–17). We also recognize the last three terms as the Laplacian of velocity component u in Cartesian coordinates (Fig. 9–41). Thus, we write the x-component of the momentum equation as ρ Du Dt = − ∂P ∂x + ρgx + 𝜇∇2u (9–59a) Similarly, the y- and z-components of the momentum equation reduce to ρ D𝜐 Dt = − ∂P ∂y + ρgy + 𝜇∇2𝜐 (9–59b) and ρ Dw Dt = − ∂P ∂z + ρgz + 𝜇∇2w (9–59c) respectively. Finally, we combine the three components into one vector equation; the result is the Navier–Stokes equation for incompressible flow with constant viscosity.
Δ = + + ∂2 ∂x2 r r ∂y2 ∂z2 ∂2 ∂2 Cartesian coordinates: Cylindrical coordinates: 2 Δ = + + ∂ 1 ∂r r2 ∂2 1 ∂θ2 ∂ ∂r ∂z2 ∂2 2 The Laplacian Operator ( ) FIGURE 9–41 The Laplacian operator, shown here in both Cartesian and cylindrical coordinates, appears in the viscous term of the incompressible Navier–Stokes equation.
cen96537_ch09_443-518.indd 473 14/01/17 3:08 pm 474 Differential Analysis of Fluid Flow Incompressible Navier–Stokes equation: ρ DV › Dt = −∇ › P + ρg ›+ 𝜇∇2V › (9–60) Although we derived the components of Eq. 9–60 in Cartesian coordi nates, the vector form of Eq. 9–60 is valid in any orthogonal coordinate system. This famous equation is named in honor of the French engineer Louis Marie Henri Navier (1785–1836) and the English mathematician Sir George Gabriel Stokes (1819–1903), who both developed the viscous terms, although independently of each other. The Navier–Stokes equation is the cornerstone of fluid mechanics (Fig. 9–42). It may look harmless enough, but it is an unsteady, nonlinear, second-order, partial differential equation. If we were able to solve this equation for flows of any geometry, this book would be about half as thick. Unfortunately, analytical solutions are unobtainable except for very simple flow fields. It is not too far from the truth to say that the rest of this book is devoted to solving Eq. 9–60! In fact, many researchers have spent their entire careers trying to solve the Navier–Stokes equation. Equation 9–60 has four unknowns (three velocity components and pres sure), yet it represents only three equations (three components since it is a vector equation). Obviously we need another equation to make the prob lem solvable. The fourth equation is the incompressible continuity equation (Eq. 9–16). Before we attempt to solve this set of differential equations, we need to choose a coordinate system and expand the equations in that coordi nate system.
Continuity and Navier–Stokes Equations in Cartesian Coordinates The continuity equation (Eq. 9–16) and the Navier–Stokes equation (Eq. 9–60) are expanded in Cartesian coordinates (x, y, z) and (u, 𝜐, w): Incompressible continuity equation: ∂u ∂x + ∂𝜐 ∂y + ∂w ∂z = 0 (9–61a) x-component of the incompressible Navier–Stokes equation: ρ( ∂u ∂t + u ∂u ∂x + 𝜐 ∂u ∂y + w ∂u ∂z) = − ∂P ∂x + ρgx + 𝜇( ∂2u ∂x2 + ∂2u ∂y2 + ∂2u ∂z2) (9–61b) y-component of the incompressible Navier–Stokes equation: ρ( ∂𝜐 ∂t + u ∂𝜐 ∂x + 𝜐 ∂𝜐 ∂y + w ∂𝜐 ∂z) = − ∂P ∂y + ρgy + 𝜇( ∂2𝜐 ∂x2 + ∂2𝜐 ∂y2 + ∂2𝜐 ∂z2) (9–61c) z-component of the incompressible Navier–Stokes equation: ρ( ∂w ∂t + u ∂w ∂x + 𝜐 ∂w ∂y + w ∂w ∂z ) = − ∂P ∂z + ρgz + 𝜇( ∂2w ∂x2 + ∂2w ∂y2 + ∂2w ∂z2) (9–61d) FIGURE 9–42 The Navier–Stokes equation is the cornerstone of fluid mechanics.
cen96537_ch09_443-518.indd 474 14/01/17 3:08 pm 475 CHAPTER 9 Continuity and Navier–Stokes Equations in Cylindrical Coordinates The continuity equation (Eq. 9–16) and the Navier–Stokes equation (Eq. 9–60) are expanded in cylindrical coordinates (r, 𝜃, z) and (ur, u𝜃, uz): Incompressible continuity equation: 1 r ∂(rur) ∂r + 1 r ∂(u𝜃) ∂𝜃 + ∂(uz) ∂z = 0 (9–62a) r-component of the incompressible Navier–Stokes equation: ρ( ∂ur ∂t + ur ∂ur ∂r + u𝜃 r ∂ur ∂𝜃−u𝜃 2 r + uz ∂ur ∂z ) = − ∂P ∂r + ρgr + 𝜇 [ 1 r ∂ ∂r (r ∂ur ∂r ) −ur r2 + 1 r2 ∂2ur ∂𝜃2 −2 r2 ∂u𝜃 ∂𝜃+ ∂2ur ∂z2 ] (9–62b) 𝜃-component of the incompressible Navier–Stokes equation: ρ( ∂u𝜃 ∂t + ur ∂u𝜃 ∂r + u𝜃 r ∂u𝜃 ∂𝜃+ uru𝜃 r + uz ∂u𝜃 ∂z ) = − 1 r ∂P ∂𝜃+ ρg𝜃+ 𝜇[ 1 r ∂ ∂r (r ∂u𝜃 ∂r ) −u𝜃 r2 + 1 r2 ∂2u𝜃 ∂𝜃2 + 2 r2 ∂ur ∂𝜃+ ∂2u𝜃 ∂z2 ] (9–62c) z-component of the incompressible Navier–Stokes equation: ρ( ∂uz ∂t + ur ∂uz ∂r + u𝜃 r ∂uz ∂𝜃+ uz ∂uz ∂z ) = − ∂P ∂z + ρgz + 𝜇[ 1 r ∂ ∂r(r ∂uz ∂r ) + 1 r2 ∂2uz ∂𝜃2 + ∂2uz ∂z2 ] (9–62d) The first two viscous terms in Eqs. 9–62b and 9–62c can be manipulated to a different form that is often more useful when solving these equations (Fig. 9–43). The derivation is left as an exercise. The “extra” terms on both sides of the r- and 𝜃-components of the Navier–Stokes equation (Eqs. 9–62b and 9–62c) arise because of the special nature of cylindrical coordinates. Namely, as we move in the 𝜃-direction, the unit vector e› r also changes direction; thus the r- and 𝜃-components are coupled (Fig. 9–44). (This coupling effect is not present in Cartesian coordinates, and thus there are no “extra” terms in Eqs. 9–61.) For completeness, the six independent components of the viscous stress tensor are listed here in cylindrical coordinates, 𝜏ij = ( 𝜏rr 𝜏r𝜃 𝜏rz 𝜏𝜃r 𝜏𝜃𝜃 𝜏𝜃z 𝜏zr 𝜏z𝜃 𝜏zz) = ( 2𝜇 ∂ur ∂r 𝜇[r ∂ ∂r ( u𝜃 r ) + 1 r ∂ur ∂𝜃] 𝜇( ∂ur ∂z + ∂uz ∂r ) 𝜇[r ∂ ∂r( u𝜃 r ) + 1 r ∂ur ∂𝜃] 2𝜇( 1 r ∂u𝜃 ∂𝜃+ ur r ) 𝜇( ∂u𝜃 ∂z + 1 r ∂uz ∂𝜃) 𝜇( ∂ur ∂z + ∂uz ∂r ) 𝜇( ∂u𝜃 ∂z + 1 r ∂uz ∂𝜃) 2𝜇 ∂uz ∂z ) (9–63) y eθ x eθ er r2 r1 θ2 θ1 er FIGURE 9–44 Unit vectors e› r and e› 𝜃 in cylindrical coordinates are coupled: movement in the 𝜃-direction causes e› r to change direction, and leads to extra terms in the r- and 𝜃-components of the Navier–Stokes equation.
FIGURE 9–43 An alternative form for the first two viscous terms in the r- and 𝜃-components of the Navier–Stokes equation.
Alternative Form of the Viscous Terms It can be shown that and ∂ ∂r ∂ur ∂r ur (rur) r2 1 r 1 r r ∂ ∂r ∂ ∂r ∂ ∂r ∂u ∂r u (ru ) r2 1 r 1 r r ∂ ∂r ∂ ∂r θ θ θ ( ( ( ( ) ) ) ) cen96537_ch09_443-518.indd 475 14/01/17 3:09 pm 476 Differential Analysis of Fluid Flow 9–6 ■ DIFFERENTIAL ANALYSIS OF FLUID FLOW PROBLEMS In this section we show how to apply the differential equations of motion in both Cartesian and cylindrical coordinates. There are two types of problems for which the differential equations (continuity and Navier–Stokes) are useful: • Calculating the pressure field for a known velocity field • Calculating both the velocity and pressure fields for a flow of known geometry and known boundary conditions For simplicity, we consider only incompressible flow, eliminating calcula tion of 𝜌 as a variable. In addition, the form of the Navier–Stokes equa tion derived in Section 9–5 is valid only for Newtonian fluids with constant properties (viscosity, thermal conductivity, etc.). Finally, we assume negligi ble temperature variations, so that T is not a variable. We are left with four variables or unknowns (pressure plus three components of velocity), and we have four differential equations (Fig. 9–45).
Calculation of the Pressure Field for a Known Velocity Field The first set of examples involves calculation of the pressure field for a known velocity field. Since pressure does not appear in the continuity equa tion, we can theoretically generate a velocity field based solely on con servation of mass. However, since velocity appears in both the continuity equation and the Navier–Stokes equation, these two equations are coupled. In addition, pressure appears in all three components of the Navier–Stokes equation, and thus the velocity and pressure fields are also coupled. This intimate coupling between velocity and pressure enables us to calculate the pressure field for a known velocity field.
EXAMPLE 9–13 Calculating the Pressure Field in Cartesian Coordinates Consider the steady, two-dimensional, incompressible velocity field of Example 9–9, namely, V ›= (u, 𝜐) = (ax + b) i ›+ (−ay + cx) j ›. Calculate the pressure as a func tion of x and y.
SOLUTION For a given velocity field, we are to calculate the pressure field.
Assumptions 1 The flow is steady and incompressible. 2 The fluid has constant properties. 3 The flow is two-dimensional in the xy-plane. 4 Gravity does not act in either the x- or y-direction.
Analysis First we check whether the given velocity field satisfies the two-dimensional, incompressible continuity equation: ∂u ∂x + ∂𝜐 ∂y + ∂w ∂z = a −a = 0 (1) a −a 0 (2-D) Three-Dimensional Incompressible Flow Four variables or unknowns: Four equations of motion: • Pressure P • Three components of velocity V • Continuity, •V = 0 • Three components of Navier–Stokes, DV Dt Δ Δ Δ ρ = – P + ρg + μ 2V FIGURE 9–45 A general three-dimensional but incompressible flow field with constant properties requires four equations to solve for four unknowns.
cen96537_ch09_443-518.indd 476 14/01/17 3:09 pm 477 CHAPTER 9 = ∂2P ∂x ∂y ∂2P ∂y ∂x P(x, y) is a smooth function of x and y only if the order of differentiation does not matter: Cross-Differentiation, xy-Plane FIGURE 9–46 For a two-dimensional flow field in the xy-plane, cross-differentiation reveals whether pressure P is a smooth function.
Thus, continuity is indeed satisfied by the given velocity field. If continuity were not satisfied, we would stop our analysis—the given velocity field would not be physically possible, and we could not calculate a pressure field. Next, we consider the y-component of the Navier–Stokes equation: ρ( ∂𝜐 ∂t + u ∂𝜐 ∂x + 𝜐 ∂𝜐 ∂y + w ∂𝜐 ∂z) = − ∂P ∂y + ρgy + 𝜇( ∂2𝜐 ∂x2 + ∂2𝜐 ∂y2 + ∂2𝜐 ∂z2) 0 (steady) (ax + b)c (−ay + cx)(−a) 0 (2-D) 0 0 0 0 (2-D) The y-momentum equation reduces to ∂P ∂y = ρ(−acx −bc −a2y + acx) = ρ(−bc −a2y) (2) The y-momentum equation is satisfied if we can generate a pressure field that satis fies Eq. 2. In similar fashion, the x-momentum equation reduces to ∂P ∂x = ρ(−a2x −ab) (3) The x-momentum equation is satisfied if we can generate a pressure field that satis fies Eq. 3. In order for a steady flow solution to exist, P cannot be a function of time. Furthermore, a physically realistic steady, incompressible flow field requires a pressure field P(x, y) that is a smooth function of x and y (there can be no sudden discontinuities in either P or a derivative of P ). Mathematically, this requires that the order of differentiation (x then y versus y then x) should not matter (Fig. 9–46). We check whether this is so by cross-differentiating Eqs. 2 and 3, respectively, ∂2P ∂x ∂y = ∂ ∂x ( ∂P ∂y) = 0 and ∂2P ∂y ∂x = ∂ ∂y ( ∂P ∂x) = 0 (4) Equation 4 shows that P is indeed a smooth function of x and y. Thus, the given velocity field satisfies the steady, two-dimensional, incompressible Navier–Stokes equation. If at this point in the analysis, the cross-differentiation of pressure were to yield two incompatible relationships (in other words if the equation in Fig. 9–46 were not satisfied) we would conclude that the given velocity field could not satisfy the steady, two-dimensional, incompressible Navier–Stokes equation, and we would abandon our attempt to calculate a steady pressure field. To calculate P(x, y), we partially integrate Eq. 2 (with respect to y ) Pressure field from y-momentum: P(x, y) = ρ(−bcy −a2y2 2 ) + g(x) (5) Note that we add an arbitrary function of the other variable x rather than a constant of integration since this is a partial integration. We then take the partial derivative of Eq. 5 with respect to x to obtain ∂P ∂x = gʹ(x) = ρ(−a2x −ab) (6) cen96537_ch09_443-518.indd 477 14/01/17 3:09 pm 478 Differential Analysis of Fluid Flow Notice that the final equation (Eq. 8) for pressure in Example 9–13 con tains an arbitrary constant C1. This illustrates an important point about the pressure field in an incompressible flow; namely, The velocity field in an incompressible flow is not affected by the absolute magnitude of pressure, but only by pressure differences.
This should not be surprising if we look at the Navier–Stokes equation, where P appears only as a gradient, never by itself. Another way to explain this statement is that it is not the absolute magnitude of pressure that matters, but only pressure differences (Fig. 9–47). A direct result of the statement is that we can calculate the pressure field to within an arbitrary constant, but in order to determine that constant (C1 in Example 9–13), we must measure (or otherwise obtain) P somewhere in the flow field. In other words, we require a pressure boundary condition. We illustrate this point with an example generated using computational fluid dynamics (CFD), where the continuity and Navier– Stokes equations are solved numerically (Chap. 15). Consider downward flow of air through a channel in which there is a nonsymmetrical blockage (Fig. 9–48). (Note that the computational flow domain extends much fur ther upstream and downstream than shown in Fig. 9–48.) We calculate two cases that are identical except for the pressure condition. In case 1 we set the gage pressure far downstream of the blockage to zero. In case 2 we set the pressure at the same location to 500 Pa gage pressure. The gage pressure at the top center of the field of view and at the bottom center of the field of view are shown in Fig. 9–48 for both cases, as generated by the two CFD solutions. You can see that the pressure field for case 2 is identical to that of case 1 except that the pressure is everywhere increased by 500 Pa. Also shown in Fig. 9–48 are a velocity vector plot and a streamline plot for each case. The results are identical, confirming our statement that the velocity field is not affected by the absolute magnitude of the pressure, but only by pressure differences. Subtracting the pressure at the bottom from that at the top, we see that ∆P = 12.784 Pa for both cases. The statement about pressure differences is not true for compressible flow fields, where P is the thermodynamic pressure rather than the mechanical where we have equated our result to Eq. 3 for consistency. We now integrate Eq. 6 to obtain the function g(x): g(x) = ρ(−a2x2 2 −abx) + C1 (7) where C 1 is an arbitrary constant of integration. Finally, we substitute Eq. 7 into Eq. 5 to obtain our final expression for P(x, y). The result is P(x, y) = 𝞺(−a2x2 2 −a2y2 2 −abx −bcy) + C1 (8) Discussion For practice, and as a check of our algebra, you should differentiate Eq. 8 with respect to both y and x, and compare to Eqs. 2 and 3. In addition, try to obtain Eq. 8 by starting with Eq. 3 rather than Eq. 2; you should get the same answer.
DV Dt Δ ρ = + ρg + μ 2V FIGURE 9–47 Since pressure appears only as a gradient in the incompressible Navier–Stokes equation, the absolute magnitude of pressure is not relevant—only pressure differences matter.
cen96537_ch09_443-518.indd 478 14/01/17 3:09 pm 479 CHAPTER 9 pressure. In such cases, P is coupled with density and temperature through an equation of state, and the absolute magnitude of pressure is important. A compressible flow solution requires not only mass and momentum equa tions, but also an energy equation and an equation of state. We take this opportunity to comment further about the CFD results shown in Fig. 9–48. You can learn a lot about the physics of fluid flow by study ing relatively simple flows like this. Notice that most of the pressure drop occurs across the throat of the channel where the flow is rapidly acceler ated. There is also flow separation downstream of the blockage; rapidly moving air cannot turn around a sharp corner, and the flow separates off the walls as it exits the opening. The streamlines indicate large recirculating regions on both sides of the channel downstream of the blockage. Pressure is low in these recirculating regions. The velocity vectors indicate an inverse bell-shaped velocity profile exiting the opening—much like an exhaust jet. Because of the nonsymmetric nature of the geometry, the jet turns to the right, and the flow reattaches to the right wall much sooner than to the left wall. The pressure increases somewhat in the region where the jet impinges on the right wall, as you might expect. Finally, notice that as the air acceler ates to squeeze through the orifice, the streamlines converge (as discussed in Section 9–3). As the jet of air fans out downstream, the streamlines diverge somewhat. Notice also that the streamlines in the recirculating zones are very far apart, indicating that the velocities are relatively small there; this is veri fied by the velocity vector plots. Finally, we note that most CFD codes do not calculate pressure by inte gration of the Navier–Stokes equation as we have done in Example 9–13. Instead, some kind of pressure correction algorithm is used. Most of the commonly used algorithms work by combining the continuity and Navier– Stokes equations in such a way that pressure appears in the continuity equa tion. The most popular pressure correction algorithms result in a form of Poisson’s equation for the change in pressure ∆P from one iteration (n) to the next (n + 1), Poisson’s equation for ∆P: ∇2(ΔP) = RHS(n) (9–64) Then, as the computer iterates toward a solution, the modified continuity equation is used to “correct” the pressure field at iteration (n + 1) from its values at iteration (n), Correction for P: P(n+ 1) = P(n) + ΔP Details associated with the development of pressure correction algorithms is beyond the scope of the present text. An example for two-dimensional flows is developed in Gerhart, Gross, and Hochstein (1992).
P = –3.562 Pa gage (a) P = 9.222 Pa gage P = 496.438 Pa gage (b) P = 509.222 Pa gage FIGURE 9–48 Filled pressure contour plot, velocity vector plot, and streamlines for downward flow of air through a channel with blockage: (a) case 1; (b) case 2—identical to case 1, except P is everywhere increased by 500 Pa. On the contour plots, blue is low pressure and red is high pressure.
EXAMPLE 9–14 Calculating the Pressure Field in Cylindrical Coordinates Consider the steady, two-dimensional, incompressible velocity field of Example 9–5 with function f(𝜃, t) equal to 0. This represents a line vortex whose axis lies along the z-coordinate (Fig. 9–49). The velocity components are ur = 0 and u𝜃 = K/r, where K is a constant. Calculate the pressure as a function of r and 𝜃.
cen96537_ch09_443-518.indd 479 14/01/17 3:09 pm 480 Differential Analysis of Fluid Flow SOLUTION For a given velocity field, we are to calculate the pressure field.
Assumptions 1 The flow is steady. 2 The fluid is incompressible with constant properties. 3 The flow is two-dimensional in the r𝜃-plane. 4 Gravity does not act in either the r- or the 𝜃-direction.
Analysis The flow field must satisfy both the continuity and the momentum equations, Eqs. 9–62. For steady, two-dimensional, incompressible flow, Incompressible continuity: 1 r ∂(rur) ∂r + 1 r ∂(u𝜃) ∂𝜃 + ∂(uz) ∂z = 0 0 0 0 Thus, the incompressible continuity equation is satisfied. Now we look at the 𝜃 component of the Navier–Stokes equation (Eq. 9–62c): ρ ( ∂u𝜃 ∂t + ur ∂u𝜃 ∂r + u𝜃 r ∂u𝜃 ∂𝜃+ uru𝜃 r + uz ∂u𝜃 ∂z ) 0 (steady) (0)(−K r2) ( K r2)(0) 0 0 (2-D) = −1 r ∂P ∂𝜃+ ρg𝜃+ 𝜇 ( 1 r ∂ ∂r (r ∂u𝜃 ∂r ) −u𝜃 r2 + 1 r2 ∂2u𝜃 ∂𝜃2 + 2 r2 ∂ur ∂𝜃+ ∂2u𝜃 ∂z2 ) 0 K r3 K r3 0 0 0 (2-D) The 𝜃-momentum equation therefore reduces to 𝜃-momentum: ∂P ∂𝜃= 0 (1) Thus, the 𝜃-momentum equation is satisfied if we can generate an appropriate pressure field that satisfies Eq. 1. In similar fashion, the r-momentum equation (Eq. 9–62b) reduces to r-momentum: ∂P ∂r = ρ K2 r3 (2) Thus, the r-momentum equation is satisfied if we can generate a pressure field that satisfies Eq. 2. In order for a steady flow solution to exist, P cannot be a function of time. Furthermore, a physically realistic steady, incompressible flow field requires a pressure field P(r, 𝜃) that is a smooth function of r and 𝜃. Mathematically, this requires that the order of differentiation (r then 𝜃 versus 𝜃 then r ) should not matter (Fig. 9–50). We check whether this is so by cross-differentiating the pressure: ∂2P ∂r ∂𝜃= ∂ ∂r ( ∂P ∂𝜃) = 0 and ∂2P ∂𝜃 ∂r = ∂ ∂𝜃 ( ∂P ∂r ) = 0 (3) Equation 3 shows that P is indeed a smooth function of r and 𝜃. Thus, the given velocity field satisfies the steady, two-dimensional, incompressible Navier– Stokes equation. We integrate Eq. 1 with respect to 𝜃 to obtain an expression for P (r, 𝜃), Pressure field from 𝜃-momentum: P(r, 𝜃) = 0 + g(r) (4) = ∂2P ∂r ∂θ ∂2P ∂θ ∂r P(r, θ) is a smooth function of r and θ only if the order of differentiation does not matter: Cross-Differentiation, rθ-Plane FIGURE 9–50 For a two-dimensional flow field in the r𝜃-plane, cross-differentiation reveals whether pressure P is a smooth function.
uθ r uθ = K r ur = 0 FIGURE 9–49 Streamlines and velocity profiles for a line vortex.
cen96537_ch09_443-518.indd 480 14/01/17 3:09 pm 481 CHAPTER 9 Exact Solutions of the Continuity and Navier–Stokes Equations The remaining example problems in this section are exact solutions of the differential equation set consisting of the incompressible continuity and Navier–Stokes equations. As you will see, these problems are by necessity simple, so that they are solvable. Most of them assume infinite boundaries and fully developed conditions so that the advective terms on the left side of the Navier–Stokes equation disappear. In addition, they are laminar, two-dimensional, and either steady or dependent on time in a predefined manner. There are six basic steps in the procedure used to solve these problems, as listed in Fig. 9–52. Step 2 is especially critical, since the boundary conditions determine the uniqueness of the solution. Step 4 is not possible analytically except for simple problems. In step 5, enough boundary conditions must be available to solve for all the constants of integration produced in step 4. Step 6 involves verifying that all the differential equations and boundary conditions Note that we added an arbitrary function of the other variable r, rather than a con stant of integration, since this is a partial integration. We take the partial derivative of Eq. 4 with respect to r to obtain ∂P ∂r = gʹ(r) = ρ K 2 r 3 (5) where we have equated our result to Eq. 2 for consistency. We integrate Eq. 5 to obtain the function g(r): g(r) = −1 2 ρ K2 r 2 + C (6) where C is an arbitrary constant of integration. Finally, we substitute Eq. 6 into Eq. 4 to obtain our final expression for P (r, 𝜃). The result is P(r, 𝞱) = −1 2 𝞺 K2 r 2 + C (7) Thus the pressure field for a line vortex decreases like 1/r2 as we approach the origin. (The origin itself is a singular point.) This flow field is a simplistic model of a tornado or hurricane, and the low pressure at the center is the “eye of the storm” (Fig. 9–51). We note that this flow field is irrotational, and thus Bernoulli’s equation can be used instead to calculate the pressure. If we call the pressure P∞ far away from the origin (r → ∞), where the local velocity approaches zero, Bernoulli’s equation shows that at any distance r from the origin, Bernoulli equation: P + 1 2 ρV 2 = P∞ → P = P∞−1 2 ρ K2 r2 (8) Equation 8 agrees with our solution (Eq. 7) from the Navier–Stokes equation if we set constant C equal to P∞. A region of rotational flow near the origin would avoid the singularity there and would yield a more physically realistic model of a tornado.
Discussion For practice, try to obtain Eq. 7 by starting with Eq. 2 rather than Eq. 1; you should get the same answer.
r P ∞ P FIGURE 9–51 The two-dimensional line vortex is a simple approximation of a tornado; the lowest pressure is at the center of the vortex.
Step 1: Set up the problem and geometry (sketches are helpful), identifying all relevant dimensions and parameters.
Step 2: List all appropriate assumptions, approximations, simplifications, and boundary conditions.
Step 5: Apply boundary conditions to solve for the constants of integration.
Step 6: Verify your results.
Step 4: Integrate the equations, leading to one or more constants of integration.
Step 3: Simplify the differential equations of motion (continuity and Navier–Stokes) as much as possible.
FIGURE 9–52 Procedure for solving the incompressible continuity and Navier–Stokes equations.
cen96537_ch09_443-518.indd 481 14/01/17 3:09 pm 482 Differential Analysis of Fluid Flow are satisfied. We advise you to follow these steps, even in cases where some of the steps seem trivial, in order to learn the procedure. While the examples shown here are simple, they adequately illustrate the procedure used to solve these differential equations. In Chap. 15 we discuss how computers have enabled us to solve the Navier–Stokes equations numeri cally for much more complicated flows using computational fluid dynamics (CFD). You will see that the same procedure is used there—specification of geometry, application of boundary conditions, integration of the differential equations, etc., although the steps are not always followed in the same order.
Boundary Conditions Since boundary conditions are so critical to a proper solution, we discuss the types of boundary conditions that are commonly encountered in fluid flow analyses. The most-used boundary condition is the no-slip condition, which states that for a fluid in contact with a solid wall, the velocity of the fluid must equal that of the wall, No-slip boundary condition: V › fluid = V › wall (9–65) In other words, as its name implies, there is no “slip” between the fluid and the wall. Fluid particles adjacent to the wall adhere to the surface of the wall and move at the same velocity as the wall. A special case of Eq. 9–65 is for a stationary wall with V › wall = 0; the fluid adjacent to a stationary wall has zero velocity. For cases in which temperature effects are also considered, the tem perature of the fluid must equal that of the wall, i.e., Tfluid = Twall. You must be careful to assign the no-slip condition according to your chosen frame of reference. Consider, for example, the thin film of oil between a piston and its cylinder wall (Fig. 9–53). From a stationary frame of reference, the fluid adja cent to the cylinder is at rest, and the fluid adjacent to the moving piston has velocity V › fluid = V › wall = VP j ›. From a frame of reference moving with the pis ton, however, the fluid adjacent to the piston has zero velocity, but the fluid adjacent to the cylinder has velocity V › fluid = V › wall = −VP j ›. An exception to the no-slip condition occurs in rarefied gas flows, such as during reentry of a spaceship or in the study of motion of extremely small (submicron) particles. In such flows the air can actually slip along the wall, but these flows are beyond the scope of the present text. When two fluids (fluid A and fluid B) meet at an interface, the interface boundary conditions are Interface boundary conditions: V › A = V › B and 𝜏s, A = 𝜏s, B (9–66) where, in addition to the condition that the velocities of the two fluids must be equal, the shear stress 𝜏s acting on a fluid particle adjacent to the interface in the direction parallel to the interface must also match between the two flu ids (Fig. 9–54). Note that in the figure, 𝜏s, A is drawn on the top of the fluid particle in fluid A, while 𝜏s, B is drawn on the bottom of the fluid particle in fluid B, and we have considered the direction of shear stress carefully. Because of the sign convention on shear stress, the direction of the arrows in Fig. 9–54 is opposite (a consequence of Newton’s third law). We note that although velocity is continuous across the interface, its slope is not. Also, if temperature effects are considered, TA = TB at the interface, but there may be a discontinuity in the slope of temperature at the interface as well.
VP Cylinder Oil film Piston Magnifying glass y x FIGURE 9–53 A piston moving at speed VP in a cylinder. A thin film of oil is sheared between the piston and the cylinder; a magnified view of the oil film is shown. The no-slip boundary condition requires that the velocity of fluid adjacent to a wall equal that of the wall.
τs, B τs, A Fluid B Fluid A VA VB s n FIGURE 9–54 At an interface between two fluids, the velocity of the two fluids must be equal. In addition, the shear stress parallel to the interface must be the same in both fluids.
cen96537_ch09_443-518.indd 482 14/01/17 3:09 pm 483 CHAPTER 9 What about pressure at an interface? If surface tension effects are neg ligible or if the interface is nearly flat, PA = PB. If the interface is sharply curved, however, as in the meniscus of liquid rising in a capillary tube, the pressure on one side of the interface can be substantially different than that on the other side. You should recall from Chap. 2 that the pressure jump across an interface is inversely proportional to the radius of curvature of the interface, as a result of surface tension effects. A degenerate form of the interface boundary condition occurs at the free surface of a liquid, meaning that fluid A is a liquid and fluid B is a gas (usu ally air). We illustrate a simple case in Fig. 9–55 where fluid A is liquid water and fluid B is air. The interface is flat and surface tension effects are negligi ble, but the water is moving horizontally (like water flowing in a calm river). In this case, the air and water velocities must match at the surface and the shear stress acting on a water particle on the surface of the water must equal that acting on an air particle just above the surface. According to Eq. 9–66, Boundary conditions at water–air interface: uwater = uair and 𝜏s, water = 𝜇water ∂u ∂y) water = 𝜏s, air = 𝜇air ∂u ∂y) air (9–67) A quick glance at the fluid property tables reveals that 𝜇water is over 50 times greater than 𝜇air . In order for the shear stresses to be equal, Eq. 9–67 requires that slope (∂u/∂y)air be more than 50 times greater than (∂u/∂y)water . Thus, it is reasonable to approximate the shear stress acting at the surface of the water as negligibly small compared to shear stresses elsewhere in the water. Another way to say this is that the moving water drags air along with it with little resistance from the air; in contrast, the air doesn’t slow down the water by any significant amount. In summary, for the case of a liquid in contact with a gas, and with negligible surface tension effects, the free- surface boundary conditions are Free-surface boundary conditions: Pliquid = Pgas and 𝜏s, liquid ≅0 (9–68) Other boundary conditions arise depending on the problem setup. For example, we often need to define inlet boundary conditions at a boundary of a flow domain where fluid enters the domain. Likewise, we define out let boundary conditions at an outflow. Symmetry boundary conditions are useful along an axis or plane of symmetry. For example, the appropri ate symmetry boundary conditions along a horizontal plane of symmetry are illustrated in Fig. 9–56. For unsteady flow problems we also need to define initial conditions (at the starting time, usually t = 0). In Examples 9–15 through 9–19, we apply boundary conditions from Eqs. 9–65 through 9–68 where appropriate. These and other boundary con ditions are discussed in much greater detail in Chap. 15 where we apply them to CFD solutions.
EXAMPLE 9–15 Fully Developed Couette Flow Consider steady, incompressible, laminar flow of a Newtonian fluid in the narrow gap between two infinite parallel plates (Fig. 9–57). The top plate is moving at speed V, and the bottom plate is stationary. The distance between these two plates is h, and gravity acts in the negative z-direction (into the page in Fig. 9–57). There is no applied pressure other than hydrostatic pressure due to gravity. This Fluid B—air Fluid A—water y uair ∂u ∂y uwater u x air ∂u ∂y water ) ) FIGURE 9–55 Along a horizontal free surface of water and air, the water and air velocities must be equal and the shear stresses must match. However, since 𝜇air < < 𝜇water, a good approximation is that the shear stress at the water surface is negligibly small.
P = continuous u Symmetry plane y x ∂u ∂y= 0 υ = 0 FIGURE 9–56 Boundary conditions along a plane of symmetry are defined so as to ensure that the flow field on one side of the symmetry plane is a mirror image of that on the other side, as shown here for a horizontal symmetry plane.
h y V x Fluid: ρ, μ Moving plate Fixed plate FIGURE 9–57 Geometry of Example 9–15: viscous flow between two infinite plates; upper plate moving and lower plate stationary.
cen96537_ch09_443-518.indd 483 14/01/17 3:09 pm 484 Differential Analysis of Fluid Flow flow is called Couette flow. Calculate the velocity and pressure fields, and estimate the shear force per unit area acting on the bottom plate.
SOLUTION For a given geometry and set of boundary conditions, we are to calculate the velocity and pressure fields, and then estimate the shear force per unit area acting on the bottom plate.
Assumptions 1 The plates are infinite in x and z. 2 The flow is steady, i.e., ∂/∂t of anything is zero. 3 This is a parallel flow (we assume that the y-component of velocity, v, is zero). 4 The fluid is incompressible and Newtonian with constant properties, and the flow is laminar. 5 Pressure P = constant with respect to x. In other words, there is no applied pressure gradient pushing the flow in the x-direction; the flow establishes itself due to viscous stresses caused by the mov ing upper plate. 6 The velocity field is purely two-dimensional, meaning here that w = 0 and ∂/∂z of any velocity component is zero. 7 Gravity acts in the negative z-direction (into the page in Fig. 9–57). We express this mathematically as g›= −gk ›, or gx = gy = 0 and gz =−g.
Analysis To obtain the velocity and pressure fields, we follow the step-by-step procedure outlined in Fig. 9–52.
Step 1 Set up the problem and the geometry. See Fig. 9–57.
Step 2 List assumptions and boundary conditions. We have numbered and listed seven assumptions (above). The boundary conditions come from imposing the no-slip condition: (1) At the bottom plate (y = 0), u = 𝜐 = w = 0. (2) At the top plate (y = h), u = V, 𝜐 = 0, and w = 0.
Step 3 Simplify the differential equations. We start with the incompressible continuity equation in Cartesian coordinates, Eq. 9–61a, ∂u ∂x + ∂𝜐 ∂y + ∂w ∂z = 0 → ∂u ∂x = 0 (1) assumption 3 assumption 6 Equation 1 tells us that u is not a function of x. In other words, it doesn’t matter where we place our origin—the flow is the same at any x-location. The phrase fully developed is often used to describe this situation (Fig. 9–58). This can also be obtained directly from assumption 1, which tells us that there is nothing special about any x-location since the plates are infinite in length. Furthermore, since u is not a function of time (assumption 2) or z (assumption 6), we conclude that u is at most a function of y, Result of continuity: u = u(y) only (2) We now simplify the x-momentum equation (Eq. 9–61b) as far as possible. It is good practice to list the reason for crossing out a term, as we do here: ρ( ∂u ∂t + u ∂u ∂x + 𝜐 ∂u ∂y + w ∂u ∂z) = −∂P ∂x + ρgx assumption 2 continuity assumption 3 assumption 6 assumption 5 assumption 7 + 𝜇( ∂2u ∂x2 + ∂2u ∂y2 + ∂2u ∂z2) → d 2u dy2 = 0 (3) continuity assumption 6 y x h V V x = x1 x = x2 FIGURE 9–58 A fully developed region of a flow field is a region where the velocity profile does not change with down stream distance. Fully developed flows are encountered in long, straight channels and pipes. Fully developed Couette flow is shown here—the velocity profile at x2 is identical to that at x1.
cen96537_ch09_443-518.indd 484 14/01/17 3:09 pm 485 CHAPTER 9 Notice that the material acceleration (left-hand side of Eq. 3) is zero, implying that fluid particles are not accelerating in this flow field, neither by local (unsteady) acceleration, nor by advective acceleration. Since the advective acceleration terms make the Navier–Stokes equation nonlinear, this greatly simplifies the problem. In fact, all other terms in Eq. 3 have disappeared except for a lone viscous term, which must then itself equal zero. Also notice that we have changed from a partial derivative (∂/∂y) to a total derivative (d/dy) in Eq. 3 as a direct result of Eq. 2. We do not show the details here, but you can show in similar fashion that every term except the pressure term in the y-momentum equation (Eq. 9–61c) goes to zero, forcing that lone term to also be zero, ∂P ∂y = 0 (4) In other words, P is not a function of y. Since P is also not a function of time (assumption 2) or x (assumption 5), P is at most a function of z, Result of y-momentum: P = P(z) only (5) Finally, by assumption 6 the z-component of the Navier–Stokes equation (Eq. 9–61d) simplifies to ∂P ∂z = −ρg → dP dz = −ρg (6) where we used Eq. 5 to convert from a partial derivative to a total derivative.
Step 4 Solve the differential equations. Continuity and y-momentum have already been “solved,” resulting in Eqs. 2 and 5, respectively. Equation 3 (x-momentum) is integrated twice to get u = C1y + C2 (7) where C1 and C2 are constants of integration. Equation 6 (z-momentum) is integrated once, resulting in P = −ρgz + C3 (8) Step 5 Apply boundary conditions. We begin with Eq. 8. Since we have not specified boundary conditions for pressure, C3 remains an arbitrary constant. (Recall that for incompressible flow, the absolute pressure can be specified only if P is known somewhere in the flow.) For example, if we let P = P0 at z = 0, then C3 = P0 and Eq. 8 becomes Final solution for pressure field: P = P0 −𝞺gz (9) Alert readers will notice that Eq. 9 represents a simple hydrostatic pressure distribution (pressure decreasing linearly as z increases). We conclude that, at least for this problem, hydrostatic pressure acts independently of the flow. More generally, we make the following statement (see also Fig. 9–59): For incompressible flow fields without free surfaces, hydrostatic pressure does not contribute to the dynamics of the flow field.
In fact, in Chap. 10 we show how hydrostatic pressure can actually be removed from the equations of motion through use of a modified pressure.
z x or y Phydrostatic g FIGURE 9–59 For incompressible flow fields without free surfaces, hydrostatic pressure does not contribute to the dynamics of the flow field.
cen96537_ch09_443-518.indd 485 14/01/17 3:09 pm 486 Differential Analysis of Fluid Flow We next apply boundary conditions (1) and (2) from step 2 to obtain constants C1 and C2. Boundary condition (1): u = C1 × 0 + C2 = 0 → C2 = 0 and Boundary condition (2): u = C1 × h + 0 = V → C1 = V/h Finally, Eq. 7 becomes Final result for velocity field: u = V y h (10) The velocity field reveals a simple linear velocity profile from u = 0 at the bottom plate to u = V at the top plate, as sketched in Fig. 9–60.
Step 6 Verify the results. Using Eqs. 9 and 10, you can verify that all the differential equations and boundary conditions are satisfied. To calculate the shear force per unit area acting on the bottom plate, we consider a rectangular fluid element whose bottom face is in contact with the bottom plate (Fig. 9–61). Mathematically positive viscous stresses are shown. In this case, these stresses are in the proper direction since fluid above the differential element pulls it to the right while the wall below the element pulls it to the left. From Eq. 9–56, we write out the components of the viscous stress tensor, 𝜏 ij = ( 2𝜇 ∂u ∂x 𝜇( ∂u ∂y + ∂𝜐 ∂x) 𝜇( ∂u ∂z + ∂w ∂x ) 𝜇( ∂𝜐 ∂x + ∂u ∂y) 2𝜇 ∂𝜐 ∂y 𝜇( ∂𝜐 ∂z + ∂w ∂y ) 𝜇( ∂w ∂x + ∂u ∂z) 𝜇( ∂w ∂y + ∂𝜐 ∂z) 2𝜇 ∂w ∂z ) = ( 0 𝜇V h 0 𝜇 V h 0 0 0 0 0 ) (11) Since the dimensions of stress are force per unit area by definition, the force per unit area acting on the bottom face of the fluid element is equal to 𝜏yx = 𝜇V/h and acts in the negative x-direction, as sketched. The shear force per unit area on the wall is equal and opposite to this (Newton’s third law); hence, Shear force per unit area acting on the wall: F › A = 𝞵 V h i › (12) The direction of this force agrees with our intuition; namely, the fluid tries to pull the bottom wall to the right, due to viscous effects (friction).
Discussion The z-component of the linear momentum equation is uncoupled from the rest of the equations; this explains why we get a hydrostatic pressure dis tribution in the z-direction, even though the fluid is not static, but moving. Equation 11 reveals that the viscous stress tensor is constant everywhere in the flow field, not just at the bottom wall (notice that none of the components of 𝜏ij is a function of location). You may be questioning the usefulness of the final results of Example 9–15. After all, when do we encounter two infinite parallel plates, one of which is moving? Actually there are several practical flows for which the Couette flow solution is a very good approximation. One such flow occurs inside a rotational viscometer (Fig. 9–62), an instrument used FIGURE 9–61 Stresses acting on a (greatly magni fied) differential two-dimensional rect angular fluid element whose bottom face is in contact with the bottom plate of Example 9–15. Note that we are ignoring gravity in this case; otherwise pressure would increase in the direc tion of the gravitational acceleration.
FIGURE 9–60 The linear velocity profile of Example 9–15: Couette flow between parallel plates.
h y x u = V y Vh y x u(y) P P P dy τxy τyx τxy τyx dx P cen96537_ch09_443-518.indd 486 14/01/17 3:09 pm 487 CHAPTER 9 to measure viscosity. It is constructed of two concentric circular cylinders of length L—a solid, rotating inner cylinder of radius Ri and a hollow, station ary outer cylinder of radius Ro. (L is into the page in Fig. 9–62; the z-axis is out of the page.) The gap between the two cylinders is very small and contains the fluid whose viscosity is to be measured. The magnified region of Fig. 9–62 is a nearly identical setup as that of Fig. 9–57 since the gap is small, i.e., (Ro − Ri) ≪ Ro. In a viscosity measurement, the angular veloc ity of the inner cylinder, 𝜔, is measured, as is the applied torque, Tapplied, required to rotate the cylinder. From Example 9–15, we know that the vis cous shear stress acting on a fluid element adjacent to the inner cylinder is approximately equal to 𝜏= 𝜏yx ≅𝜇 V Ro −Ri = 𝜇 𝜔Ri Ro −Ri (9–69) where the speed V of the moving upper plate in Fig. 9–57 is replaced by the counterclockwise speed 𝜔Ri of the rotating wall of the inner cylinder. In the magnified region at the bottom of Fig. 9–62, 𝜏 acts to the right on the fluid element adjacent to the inner cylinder wall; hence, the force per unit area acting on the inner cylinder at this location acts to the left with magnitude given by Eq. 9–69. The total clockwise torque acting on the inner cylinder wall due to fluid viscosity is thus equal to this shear stress times the wall area times the moment arm, Tviscous = 𝜏ARi ≅𝜇 𝜔Ri Ro −Ri (2𝜋 RiL)Ri (9–70) Under steady conditions, the clockwise torque Tviscous is balanced by the applied counterclockwise torque Tapplied. Equating these and solving Eq. 9–70 for the fluid viscosity yields Viscosity of the fluid: 𝜇= Tapplied (Ro −Ri) 2𝜋 𝜔Ri 3L A similar analysis can be performed on an unloaded journal bearing in which a viscous oil flows in the small gap between the inner rotating shaft and the stationary outer housing. (When the bearing is loaded, the inner and outer cylinders cease to be concentric and a more involved analysis is required.) Fluid: ρ, μ Magnifying glass Rotating inner cylinder Stationary outer cylinder ω τ Ro Ri FIGURE 9–62 A rotational viscometer; the inner cylinder rotates at angular velocity 𝜔, and a torque Tapplied is applied, from which the viscosity of the fluid is calculated.
h x1 y V x Fluid: ρ, μ Moving plate Fixed plate x2 P2 P1 ∂P ∂x = P2 – P1 x2 – x1 FIGURE 9–63 Geometry of Example 9–16: viscous flow between two infinite plates with a constant applied pressure gradient ∂P/∂x; the upper plate is moving and the lower plate is stationary.
EXAMPLE 9–16 Couette Flow with an Applied Pressure Gradient Consider the same geometry as in Example 9–15, but instead of pressure being con stant with respect to x, let there be an applied pressure gradient in the x-direction (Fig. 9–63). Specifically, let the pressure gradient in the x-direction, ∂P/∂x, be some constant value given by Applied pressure gradient: ∂P ∂x = P2 −P1 x2 −x1 = constant (1) where x1 and x2 are two arbitrary locations along the x-axis, and P1 and P2 are the pressures at those two locations. Everything else is the same as for Example 9–15. (a) Calculate the velocity and pressure field. (b) Plot a family of velocity profiles in dimensionless form.
cen96537_ch09_443-518.indd 487 14/01/17 3:09 pm 488 Differential Analysis of Fluid Flow SOLUTION We are to calculate the velocity and pressure field for the flow sketched in Fig. 9–63 and plot a family of velocity profiles in dimensionless form.
Assumptions The assumptions are identical to those of Example 9–15, except assumption 5 is replaced by the following: A constant pressure gradient is applied in the x-direction such that pressure changes linearly with respect to x according to Eq. 1.
Analysis (a) We follow the same procedure as in Example 9–15. Much of the algebra is identical, so to save space we discuss only the differences.
Step 1 See Fig. 9–63.
Step 2 Same as Example 9–15 except for assumption 5.
Step 3 The continuity equation is simplified in the same way as in Example 9–15, Result of continuity: u = u(y) only (2) The x-momentum equation is simplified in the same manner as in Example 9–15 except that the pressure gradient term remains. The result is Result of x-momentum: d2u dy2 = 1 𝜇 ∂P ∂x (3) Likewise, the y-momentum and z-momentum equations simplify to Result of y-momentum: ∂P ∂y = 0 (4) and Result of z-momentum: ∂P ∂z = −ρg (5) We cannot convert from a partial derivative to a total derivative in Eq. 5, because P is a function of both x and z in this problem, unlike in Example 9–15 where P was a function of z only.
Step 4 We integrate Eq. 3 (x-momentum) twice, noting that ∂P/∂x is a constant, Integration of x-momentum: u = 1 2𝜇 ∂P ∂x y2 + C1y + C2 (6) where C1 and C2 are constants of integration. Equation 5 (z-momentum) is integrated once, resulting in Integration of z-momentum: P = −ρgz + f (x) (7) Note that since P is now a function of both x and z, we add a function of x instead of a constant of integration in Eq. 7. This is a partial integration with respect to z, and we must be careful when performing partial integrations (Fig. 9–64).
Step 5 From Eq. 7, we see that the pressure varies hydrostatically in the z-direction, and we have specified a linear change in pressure in the x-direction. Thus the function f (x) must equal a constant plus ∂P/∂x times x. If we set P = P0 along the line x = 0, z = 0 (the y-axis), Eq. 7 becomes Final result for pressure field: P = P0 + 𝝏P 𝝏x x −𝞺gz (8) CAUTION! WHEN PERFORMING A PARTIAL INTEGRATION, ADD A FUNCTION OF THE OTHER VARIABLE(S) FIGURE 9–64 A caution about partial integration.
cen96537_ch09_443-518.indd 488 14/01/17 3:09 pm 489 CHAPTER 9 We next apply the velocity boundary conditions (1) and (2) from step 2 of Example 9–15 to obtain constants C1 and C2. Boundary condition (1): u = 1 2𝜇 ∂P ∂x × 0 + C1 × 0 + C2 = 0 → C2 = 0 and Boundary condition (2): u = 1 2𝜇 ∂P ∂x h2 + C1 × h + 0 = V → C1 = V h −1 2𝜇 ∂P ∂x h Finally, Eq. 6 becomes u = Vy h + 1 2𝞵 𝝏P 𝝏x (y2 −hy) (9) Equation 9 indicates that the velocity field consists of the superposition of two parts: a linear velocity profile from u = 0 at the bottom plate to u = V at the top plate, and a parabolic distribution that depends on the magnitude of the applied pressure gradient. If the pressure gradient is zero, the parabolic portion of Eq. 9 disappears and the profile is linear, just as in Example 9–15; this is sketched as the dashed red line in Fig. 9–65. If the pressure gradient is negative (pressure decreasing in the x-direction, causing flow to be pushed from left to right), ∂P/∂x < 0 and the velocity profile looks like the one sketched in Fig. 9–65. A special case is when V = 0 (top plate stationary); the linear portion of Eq. 9 vanishes, and the velocity profile is parabolic and symmetric about the center of the channel (y = h/2); this is sketched as the dotted line in Fig. 9–65.
Step 6 You can use Eqs. 8 and 9 to verify that all the differential equations and boundary conditions are satisfied.
(b) We use dimensional analysis to generate the dimensionless groups (Π groups). We set up the problem in terms of velocity component u as a function of y, h, V, 𝜇, and ∂P/∂x. There are six variables (including the dependent variable u), and since there are three primary dimensions represented in the problem (mass, length, and time), we expect 6 − 3 = 3 dimensionless groups. When we pick h, V, and 𝜇 as our repeating variables, we get the following result using the method of repeating vari ables (details are left for you to do on your own—this is a good review of Chap. 7 material): Result of dimensional analysis: u V = f ( y h, h2 𝜇V ∂P ∂x ) (10) Using these three dimensionless groups, we rewrite Eq. 9 as Dimensionless form of velocity field: u = y + 1 2 P y (y −1) (11) where the dimensionless parameters are u = u V y = y h P = h2 𝜇V ∂P ∂x In Fig. 9–66, u is plotted as a function of y for several values of P, using Eq. 11.
y h V u(y) x FIGURE 9–65 The velocity profile of Example 9–16: Couette flow between parallel plates with an applied negative pressure gradient; the dashed red line indicates the profile for a zero pressure gradient, and the dotted line indicates the profile for a negative pressure gradient with the upper plate stationary (V = 0).
cen96537_ch09_443-518.indd 489 14/01/17 3:09 pm 490 Differential Analysis of Fluid Flow 1 0.8 0.6 0.4 y = y/h 0.2 0 –1.5 –1 –0.5 0 u = u/V 0.5 1 2.5 1.5 2 P = 15 10 5 0 –5 –10 –15 FIGURE 9–66 Nondimensional velocity profiles for Couette flow with an applied pressure gradient; profiles are shown for several values of nondimensional pressure gradient.
h y u(y) x FIGURE 9–67 The velocity profile for fully developed two-dimensional channel flow (planar Poiseuille flow).
h z x g Oil film: ρ, μ Fixed wall Air P = Patm FIGURE 9–68 Geometry of Example 9–17: a viscous film of oil falling by gravity along a vertical wall.
Discussion When the result is nondimensionalized, we see that Eq. 11 repre sents a family of velocity profiles. We also see that when the pressure gradient is positive (flow being pushed from right to left) and of sufficient magnitude, we can have reverse flow in the bottom portion of the channel. For all cases, the boundary conditions reduce to u = 0 at y = 0 and u = 1 at y = 1. If there is a pressure gradient but both walls are stationary, the flow is called two-dimensional channel flow, or planar Poiseuille flow (Fig. 9–67). We note, however, that most authors reserve the name Poiseuille flow for fully developed pipe flow—the axisymmetric analog of two-dimensional channel flow (see Example 9–18).
EXAMPLE 9–17 Oil Film Flowing Down a Vertical Wall by Gravity Consider steady, incompressible, parallel, laminar flow of a film of oil falling slowly down an infinite vertical wall (Fig. 9–68). The oil film thickness is h, and gravity acts in the negative z-direction (downward in Fig. 9–68). There is no applied (forced) pressure driving the flow—the oil falls by gravity alone. Cal culate the velocity and pressure fields in the oil film and sketch the normal ized velocity profile. You may neglect changes in the hydrostatic pressure of the surrounding air.
SOLUTION For a given geometry and set of boundary conditions, we are to calculate the velocity and pressure fields and plot the velocity profile.
Assumptions 1 The wall is infinite in the yz-plane (y is into the page for a right-handed coordinate system). 2 The flow is steady (all partial derivatives with respect to time are zero). 3 The flow is parallel (the x-component of velocity, u, is zero everywhere). 4 The fluid is incompressible and Newtonian with constant proper ties, and the flow is laminar. 5 Pressure P = Patm = constant at the free surface. In other words, there is no applied pressure gradient pushing the flow; the flow estab lishes itself due to a balance between gravitational forces and viscous forces. In addition, since there is no gravity force in the horizontal direction, P = Patm every where. 6 The velocity field is purely two-dimensional, which implies that velocity cen96537_ch09_443-518.indd 490 14/01/17 3:09 pm 491 CHAPTER 9 NOTICE If u = u(x) only, change from PDE to ODE: ∂u ∂x du dx FIGURE 9–69 In Examples 9–15 through 9–18, the equations of motion are reduced from partial differential equations to ordinary differential equations, making them much easier to solve.
component 𝜈 = 0 and all partial derivatives with respect to y are zero. 7 Gravity acts in the negative z-direction. We express this mathematically as g›= −gk ›, or gx = gy = 0 and gz = −g.
Analysis We obtain the velocity and pressure fields by following the step-by-step procedure for differential fluid flow solutions (Fig. 9–52).
Step 1 Set up the problem and the geometry. See Fig. 9–68.
Step 2 List assumptions and boundary conditions. We have listed seven assumptions. The boundary conditions are: (1) There is no slip at the wall; at x = 0, u = 𝜐 = w = 0. (2) At the free surface (x = h), there is negligible shear (Eq. 9–68), which for a vertical free surface in this coordinate system means ∂w/∂x = 0 at x = h.
Step 3 Write out and simplify the differential equations. We start with the incompressible continuity equation in Cartesian coordinates, ∂u ∂x + ∂𝜐 ∂y + ∂w ∂z = 0 → ∂w ∂z = 0 (1) assumption 3 assumption 6 Equation 1 tells us that w is not a function of z; i.e., it doesn’t matter where we place our origin—the flow is the same at any z-location. In other words, the flow is fully developed. Since w is not a function of time (assumption 2), z (Eq. 1), or y (assumption 6), we conclude that w is at most a function of x, Result of continuity: w = w(x) only (2) We now simplify each component of the Navier–Stokes equation as far as possible. Since u = 𝜐 = 0 everywhere, and gravity does not act in the x- or y-directions, the x- and y-momentum equations are satisfied exactly (in fact all terms are zero in both equations). The z-momentum equation reduces to ρ( ∂w ∂t + u ∂w ∂x + 𝜐 ∂w ∂y + w ∂w ∂z ) = −∂P ∂z + ρgz assumption 2 assumption 3 assumption 6 continuity assumption 5 −𝜌g + 𝜇( ∂2w ∂x2 + ∂2w ∂y2 + ∂2w ∂z2 ) → d 2w dx2 = ρg 𝜇 (3) assumption 6 continuity The material acceleration (left side of Eq. 3) is zero, implying that fluid particles are not accelerating in this flow field, neither by local nor advective acceleration. Since the advective acceleration terms make the Navier–Stokes equation nonlinear, this greatly simplifies the problem. We have changed from a partial derivative (∂/∂x) to a total derivative (d/dx) in Eq. 3 as a direct result of Eq. 2, reducing the partial differential equation (PDE) to an ordinary differential equation (ODE). ODEs are of course much easier than PDEs to solve (Fig. 9–69).
Step 4 Solve the differential equations. The continuity and x- and y-momentum equations have already been “solved.” Equation 3 (z-momentum) is integrated twice to get w = ρg 2𝜇 x2 + C1x + C2 (4) cen96537_ch09_443-518.indd 491 14/01/17 3:09 pm 492 Differential Analysis of Fluid Flow The solution procedure used in Examples 9–15 through 9–17 in Car tesian coordinates can also be used in any other coordinate system. In Example 9–18 we present the classic problem of fully developed flow in a round pipe, for which we use cylindrical coordinates.
Step 5 Apply boundary conditions. We apply boundary conditions (1) and (2) from step 2 to obtain constants C1 and C2, Boundary condition (1): w = 0 + 0 + C2 = 0 C2 = 0 and Boundary condition (2): dw dx ) x=h = ρg 𝜇 h + C1 = 0 → C1 = −ρgh 𝜇 Finally, Eq. 4 becomes Velocity field: w = 𝞺g 2𝞵 x2 −𝞺g 𝞵 hx = 𝞺gx 2𝞵 (x −2h) (5) Since x < h in the film, w is negative everywhere, as expected (flow is downward). The pressure field is trivial; namely, P = Patm everywhere.
Step 6 Verify the results. You can verify that all the differential equations and boundary conditions are satisfied. We normalize Eq. 5 by inspection: we let x = x/h and w = w𝜇/(𝜌gh2). Equation 5 becomes Normalized velocity profile: w = x 2 (x −2) (6) We plot the normalized velocity field in Fig. 9–70.
Discussion The velocity profile has a large slope near the wall due to the no-slip condition there (w = 0 at x = 0), but zero slope at the free surface, where the bound ary condition is zero shear stress (∂w/∂x = 0 at x = h). We could have introduced a factor of −2 in the definition of w so that w would equal 1 instead of −1 2 at the free surface.
EXAMPLE 9–18 Fully Developed Flow in a Round Pipe—Poiseuille Flow Consider steady, incompressible, laminar flow of a Newtonian fluid in an infinitely long round pipe of diameter D or radius R = D/2 (Fig. 9–71). We ignore the effects of gravity. A constant pressure gradient ∂P/∂x is applied in the x-direction, Applied pressure gradient: ∂P ∂x = P2 −P1 x2 −x1 = constant (1) where x1 and x2 are two arbitrary locations along the x-axis, and P1 and P2 are the pressures at those two locations. Note that we adopt a modified cylindri cal coordinate system here with x instead of z for the axial component, namely, (r, 𝜃, x) and (ur, u𝜃, u). Derive an expression for the velocity field inside the pipe and estimate the viscous shear force per unit surface area acting on the pipe wall.
Free surface Wall 0 –0.1 –0.2 –0.3 –0.4 –0.5 –0.6 w 0 0.2 0.4 0.6 0.8 1 x FIGURE 9–70 The normalized velocity profile of Example 9–17: an oil film falling down a vertical wall.
r R P2 P1 x1 x2 Fluid: ρ, μ Pipe wall x D ∂x x2 – x1 P2 – P1 ∂P = V FIGURE 9–71 Geometry of Example 9–18: steady laminar flow in a long round pipe with an applied pressure gradient ∂P/∂x pushing fluid through the pipe. The pressure gradient is usually produced by a pump and/or gravity.
cen96537_ch09_443-518.indd 492 14/01/17 3:09 pm 493 CHAPTER 9 FIGURE 9–72 Exact analytical solutions of the Navier– Stokes equations, as in the examples provided here, are not possible if the flow is turbulent.
SOLUTION For flow inside a round pipe we are to calculate the velocity field, and then estimate the viscous shear stress acting on the pipe wall.
Assumptions 1 The pipe is infinitely long in the x-direction. 2 The flow is steady (all partial time derivatives are zero). 3 This is a parallel flow (the r-component of velocity, ur, is zero). 4 The fluid is incompressible and New tonian with constant properties, and the flow is laminar (Fig. 9–72). 5 A constant pressure gradient is applied in the x-direction such that pressure changes linearly with respect to x according to Eq. 1. 6 The velocity field is axisymmetric with no swirl, imply ing that u𝜃 = 0 and all partial derivatives with respect to 𝜃 are zero. 7 We ignore the effects of gravity.
Analysis To obtain the velocity field, we follow the step-by-step procedure out lined in Fig. 9–52.
Step 1 Lay out the problem and the geometry. See Fig. 9–71.
Step 2 List assumptions and boundary conditions. We have listed seven assumptions. The first boundary condition comes from imposing the no-slip condition at the pipe wall: (1) at r = R, V ›= 0. The second boundary condition comes from the fact that the centerline of the pipe is an axis of symmetry: (2) at r = 0, ∂u/∂r = 0.
Step 3 Write out and simplify the differential equations. We start with the incompressible continuity equation in cylindrical coordinates, a modified version of Eq. 9–62a, 1 r ∂(rur) ∂r + 1 r ∂(u𝜃) ∂𝜃 + ∂u ∂x = 0 → ∂u ∂x = 0 (2) assumption 3 assumption 6 Equation 2 tells us that u is not a function of x. In other words, it doesn’t matter where we place our origin—the flow is the same at any x-location. This can also be inferred directly from assumption 1, which tells us that there is nothing special about any x-location since the pipe is infinite in length—the flow is fully developed. Furthermore, since u is not a function of time (assumption 2) or 𝜃 (assumption 6), we conclude that u is at most a function of r, Result of continuity: u = u(r) only (3) We now simplify the axial momentum equation (a modified version of Eq. 9–62d) as far as possible: ρ( ∂u ∂t + ur ∂u ∂r + u𝜃 r ∂u ∂𝜃 + u ∂u ∂x) assumption 2 assumption 3 assumption 6 continuity = −∂P ∂x + ρgx + 𝜇( 1 r ∂ ∂r(r ∂u ∂r) + 1 r2 ∂2u ∂𝜃2 + ∂2u ∂x2) assumption 7 assumption 6 continuity or 1 r d dr (r du dr) = 1 𝜇 ∂P ∂x (4) As in Examples 9–15 through 9–17, the material acceleration (entire left side of the x-momentum equation) is zero, implying that fluid particles are not cen96537_ch09_443-518.indd 493 14/01/17 3:09 pm 494 Differential Analysis of Fluid Flow accelerating at all in this flow field, and linearizing the Navier–Stokes equation (Fig. 9–73). We have replaced the partial derivative operators for the u-derivatives with total derivative operators because of Eq. 3. In similar fashion, every term in the r-momentum equation (Eq. 9–62b) except the pressure gradient term is zero, forcing that lone term to also be zero, r-momentum: ∂P ∂r = 0 (5) In other words, P is not a function of r. Since P is also not a function of time (assumption 2) or 𝜃 (assumption 6), P can be at most a function of x, Result of r-momentum: P = P(x) only (6) Therefore, we replace the partial derivative operator for the pressure gradient in Eq. 4 by the total derivative operator since P varies only with x. Finally, all terms of the 𝜃-component of the Navier–Stokes equation (Eq. 9–62c) go to zero.
Step 4 Solve the differential equations. Continuity and r-momentum have already been “solved,” resulting in Eqs. 3 and 6, respectively. The 𝜃-momentum equation has vanished, and thus we are left with Eq. 4 (x-momentum). After multiplying both sides by r, we integrate once to obtain r du dr = r2 2𝜇 dP dx + C1 (7) where C1 is a constant of integration. Note that the pressure gradient dP/dx is a constant here. Dividing both sides of Eq. 7 by r, we integrate a second time to get u = r2 4𝜇 dP dx + C1 ln r + C2 (8) where C2 is a second constant of integration.
Step 5 Apply boundary conditions. First, we apply boundary condition (2) to Eq. 7, Boundary condition (2): 0 = 0 + C1 → C1 = 0 An alternative way to interpret this boundary condition is that u must remain finite at the centerline of the pipe. This is possible only if constant C1 is equal to 0, since ln(0) is undefined in Eq. 8. Now we apply boundary condition (1), Boundary condition (1): u = R2 4𝜇 dP dx + 0 + C2 = 0 → C2 = −R2 4𝜇 dP dx Finally, Eq. 8 becomes Axial velocity: u = 1 4𝞵 dP dx (r2 −R2) (9) The axial velocity profile is thus in the shape of a paraboloid, as sketched in Fig. 9–74.
Step 6 Verify the results. You can verify that all the differential equations and boundary conditions are satisfied.
∂V ∂t + (V • )V = – P + ρg + μ 2V ρ Δ Δ Δ The Navier–Stokes Equation Nonlinear term ( ) FIGURE 9–73 For incompressible flow solutions in which the advective terms in the Navier–Stokes equation are zero, the equation becomes linear since the advective term is the only nonlinear term in the equation.
r R u(r) V = uavg = umax/2 umax θ x D FIGURE 9–74 Axial velocity profile of Example 9–18: steady laminar flow in a long round pipe with an applied constant-pressure gradient dP/dx pushing fluid through the pipe.
cen96537_ch09_443-518.indd 494 14/01/17 3:09 pm 495 CHAPTER 9 We calculate some other properties of fully developed laminar pipe flow as well. For example, the maximum axial velocity obviously occurs at the centerline of the pipe (Fig. 9–74). Setting r = 0 in Eq. 9 yields Maximum axial velocity: umax = −R2 4𝜇 dP dx (10) The volume flow rate through the pipe is found by integrating Eq. 9 through a cross section of the pipe, V · = ∫ 2𝜋 𝜃=0 ∫ R r=0 ur dr d𝜃= 2𝜋 4𝜇 dP dx ∫ R r=0 (r2 −R2)r dr = −𝜋 R4 8𝜇 dP dx (11) Since volume flow rate is also equal to the average axial velocity times cross-sectional area, we easily determine the average axial velocity V: Average axial velocity: V = V · A = (−𝜋 R4/8𝜇) (dP/dx) 𝜋 R2 = −R2 8𝜇 dP dx (12) Comparing Eqs. 10 and 12 we see that for fully developed laminar pipe flow, the average axial velocity is equal to exactly half of the maximum axial velocity. To calculate the viscous shear force per unit surface area acting on the pipe wall, we consider a differential fluid element adjacent to the bottom portion of the pipe wall (Fig. 9–75). Pressure stresses and mathematically positive viscous stresses are shown. From Eq. 9–63 (modified for our coordinate system), we write the viscous stress tensor as 𝜏ij = ( 𝜏rr 𝜏r𝜃 𝜏rx 𝜏𝜃r 𝜏𝜃𝜃 𝜏𝜃x 𝜏xr 𝜏x𝜃 𝜏xx) = ( 0 0 𝜇 ∂u ∂r 0 0 0 𝜇 ∂u ∂r 0 0 ) (13) We use Eq. 9 for u, and set r = R at the pipe wall; component 𝜏rx of Eq. 13 reduces to Viscous shear stress at the pipe wall: 𝜏rx = 𝜇 du dr = R 2 dP dx (14) For flow from left to right, dP/dx is negative, so the viscous shear stress on the bottom of the fluid element at the wall is in the direction opposite to that indicated in Fig. 9–75. (This agrees with our intuition since the pipe wall exerts a retarding force on the fluid.) The shear force per unit area on the wall is equal and opposite to this; hence, Viscous shear force per unit area acting on the wall: F › A = −R 2 dP dx i › (15) The direction of this force again agrees with our intuition; namely, the fluid tries to pull the bottom wall to the right, due to friction, when dP/dx is negative.
Discussion Since du/dr = 0 at the centerline of the pipe, 𝜏rx = 0 there. You are encouraged to try to obtain Eq. 15 by using a control volume approach instead, taking your control volume as the fluid in the pipe between any two x-locations, Centerline Pipe wall dr x dx P r P P + dP dx dx 2 P – dP dx dx 2 ∂r τrx – ∂τrx dr 2 ∂r τxr – ∂τxr dx 2 ∂r τxr + ∂τxr dx 2 ∂r τrx + ∂τrx dr 2 FIGURE 9–75 Pressure and viscous shear stresses acting on a differential fluid element whose bottom face is in contact with the pipe wall. Note that we are ignoring gravity in this case; otherwise pressure would increase in the direc tion of the gravitational acceleration.
cen96537_ch09_443-518.indd 495 14/01/17 3:09 pm 496 Differential Analysis of Fluid Flow So far, all our Navier–Stokes solutions have been for steady flow. You can imagine how much more complicated the solutions must get if the flow is allowed to be unsteady, and the time derivative term in the Navier–Stokes equation does not disappear. Nevertheless, there are some unsteady flow prob lems that can be solved analytically. We present one of these in Example 9–19.
x1 and x2 (Fig. 9–76). You should get the same answer. (Hint: Since the flow is fully developed, the axial velocity profile at location 1 is identical to that at loca tion 2.) Note that when the volume flow rate through the pipe exceeds a critical value, instabilities in the flow occur, and the solution presented here is no longer valid. Specifically, flow in the pipe becomes turbulent rather than laminar; turbu lent pipe flow is discussed in more detail in Chap. 8. This problem is also solved in Chap. 8 using an alternative approach.
EXAMPLE 9–19 Sudden Motion of an Infinite Flat Plate Consider a viscous Newtonian fluid on top of an infinite flat plate lying in the xy-plane at z = 0 (Fig. 9–77). The fluid is at rest until time t = 0, when the plate suddenly starts moving at speed V in the x-direction. Gravity acts in the −z- direction. Determine the pressure and velocity fields.
SOLUTION The velocity and pressure fields are to be calculated for the case of fluid on top of an infinite flat plate that suddenly starts moving.
Assumptions 1 The wall is infinite in the x- and y-directions; thus, nothing is special about any particular x- or y-location. 2 The flow is parallel everywhere (w = 0). 3 Pressure P = constant with respect to x. In other words, there is no applied pressure gradient pushing the flow in the x-direction; flow occurs due to viscous stresses caused by the moving plate. 4 The fluid is incompressible and Newtonian with constant properties, and the flow is laminar. 5 The velocity field is two-dimensional in the xz-plane; therefore, 𝜐 = 0, and all partial derivatives with respect to y are zero. 6 Gravity acts in the −z-direction.
Analysis To obtain the velocity and pressure fields, we follow the step-by-step procedure outlined in Fig. 9–52.
Step 1 Lay out the problem and the geometry. (See Fig. 9–77.) Step 2 List assumptions and boundary conditions. We have listed six assumptions. The boundary conditions are: (1) At t = 0, u = 0 everywhere (no flow until the plate starts moving); (2) at z = 0, u = V for all values of x and y (no-slip condition at the plate); (3) as z → ∞, u = 0 (far from the plate, the effect of the moving plate is not felt); and (4) at z = 0, P = Pwall (the pressure at the wall is constant at any x- or y-location along the plate).
Step 3 Write out and simplify the differential equations. We start with the incompressible continuity equation in Cartesian coordinates (Eq. 9–61a), ∂u ∂x + ∂𝜐 ∂y + ∂w ∂z = 0 → ∂u ∂x = 0 (1) assumption 5 assumption 2 Equation 1 tells us that u is not a function of x. Furthermore, since u is not a function of y (assumption 5), we conclude that u is at most a function of z and t, r R P2 P1 x1 x2 Fluid: ρ, μ Pipe wall x CV dx x2 – x1 P2 – P1 dP = FIGURE 9–76 Control volume used to obtain Eq. 15 of Example 9–18 by an alternative method.
z V g = –gk Fluid: ρ, μ x Infinite flat plate FIGURE 9–77 Geometry and setup for Example 9–19; the y-coordinate is into the page.
cen96537_ch09_443-518.indd 496 14/01/17 3:09 pm 497 CHAPTER 9 Equation of the Day The 1-D Diffusion Equation ∂t ∂u ∂z2 ∂2u = v FIGURE 9–78 The one-dimensional diffusion equation is linear, but it is a partial differential equation (PDE). It occurs in many fields of science and engineering. Result of continuity: u = u (z, t) only (2) The y-momentum equation reduces to ∂P ∂y = 0 (3) by assumptions 5 and 6 (all terms with 𝜐, the y-component of velocity, vanish, and gravity does not act in the y -direction). Equation 3 simply tells us that pressure is not a function of y; hence, Result of y-momentum: P = P(z, t) only (4) Similarly the z-momentum equation reduces to ∂P ∂z = −ρg (5) We now simplify the x-momentum equation (Eq. 9–61b) as far as possible. ρ( ∂u ∂t + u ∂u ∂x + 𝜐 ∂u ∂y + w ∂u ∂z) = −∂P ∂x + ρgx continuity assumption 5 assumption 2 assumption 3 assumption 6 + 𝜇( ∂2u ∂x2 + ∂2u ∂y2 + ∂2u ∂z2) → ρ ∂u ∂t = 𝜇 ∂2u ∂z (6) continuity assumption 5 It is convenient to combine the viscosity and density into the kinematic viscosity, defined as 𝜈 = 𝜇/𝜌. Equation 6 reduces to the well-known one-dimensional diffusion equation (Fig. 9–78), Result of x-momentum: ∂u ∂t = 𝜈 ∂2u ∂z2 (7) Step 4 Solve the differential equations. Continuity and y-momentum have already been “solved,” resulting in Eqs. 2 and 4, respectively. Equation 5 (z-momentum) is integrated once, resulting in P = −ρgz + f (t) (8) where we have added a function of time instead of a constant of integration since P is a function of two variables, z and t (see Eq. 4). Equation 7 (x-momentum) is a linear partial differential equation whose solution is obtained by combining the two independent variables z and t into one independent variable. The result is called a similarity solution, the details of which are beyond the scope of this text. Note that the one-dimensional diffusion equation occurs in many other fields of engineering, such as diffusion of species (mass diffusion) and diffusion of heat (conduction); details about the solution can be found in books on these subjects. The solution of Eq. 7 is intimately tied to the boundary condition that the plate is impulsively started, and the result is Integration of x-momentum: u = C1[1 −erf( z 2√𝜈t)] (9) where erf in Eq. 9 is the error function, defined as Error function: erf(𝜉 ) = 2 √𝜋 ∫ 𝜉 0 e−𝜂2d𝜂 (10) cen96537_ch09_443-518.indd 497 14/01/17 3:09 pm 498 Differential Analysis of Fluid Flow The error function is commonly used in probability theory and is plotted in Fig. 9–79. Tables of the error function can be found in many reference books, and some calculators and spreadsheets can calculate the error function directly.
Step 5 Apply boundary conditions. We begin with Eq. 8 for pressure. Boundary condition (4) requires that P = Pwall at z = 0 for all times, and Eq. 8 becomes Boundary condition (4): P = 0 + f (t) = Pwall → f (t) = Pwall In other words, the arbitrary function of time, f(t), turns out not to be a function of time at all, but merely a constant. Thus, Final result for pressure field: P = Pwall −𝞺gz (11) which is simply hydrostatic pressure. We conclude that hydrostatic pressure acts independently of the flow. Boundary conditions (1) and (3) from step 2 have already been applied in order to obtain the solution of the x-momentum equation in step 4. Since erf(0) = 0, the second boundary condition yields Boundary condition (2): u = C1(1 −0) = V → C1 = V and Eq. 9 becomes Final result for velocity field: u = V[1 −erf( z 2√𝝼t)] (12) Several velocity profiles are plotted in Fig. 9–80 for the specific case of water at room temperature (𝜈 = 1.004 × 10−6 m2/s) with V = 1.0 m/s. At t = 0, there is no flow. As time goes on, the motion of the plate is felt farther and farther into the fluid, as expected. Notice how long it takes for viscous diffusion to penetrate into the fluid—after 15 min of flow, the effect of the moving plate is not felt beyond about 10 cm above the plate! We define normalized variables u and z as Normalized variables: u = u V and z = z 2√𝜈t Then we rewrite Eq. 12 in terms of nondimensional parameters: Normalized velocity field: u = 1 −erf(z ) (13) The combination of unity minus the error function occurs often in engineering and is given the special name complementary error function and symbol erfc. Thus Eq. 13 can also be written as Alternative form of the velocity field: u = erfc(z ) (14) The beauty of the normalization is that this one equation for u as a function of z is valid for any fluid (with any kinematic viscosity 𝜈) above a plate moving at any speed V and at any location z in the fluid at any time t! The normalized velocity profile of Eq. 13 is sketched in Fig. 9–81. All the profiles of Fig. 9–80 collapse into the single profile of Fig. 9–81; such a profile is called a similarity profile.
Step 6 Verify the results. You can verify that all the differential equations and boundary conditions are satisfied.
1 0.8 0.6 0.4 erf(ξ) 0.2 0 0 0.5 1 1.5 ξ 2 2.5 3 FIGURE 9–79 The error function ranges from 0 at 𝜉 = 0 to 1 as 𝜉 → ∞.
0.2 0.15 0.1 0.05 z, m 0 0 0.2 0.4 0.6 u, m/s 0.8 1 24 h 8 h 3 h 1 h 15 min 5 min 30 s FIGURE 9–80 Velocity profiles of Example 9–19: flow of water above an impulsively started infinite plate; 𝜈 = 1.004 × 10−6 m2/s and V = 1.0 m/s.
cen96537_ch09_443-518.indd 498 14/01/17 3:09 pm 499 CHAPTER 9 Examples 9–15 through 9–19 are for incompressible laminar flow. The same set of differential equations (incompressible continuity and Navier– Stokes) is valid for incompressible turbulent flow. However, turbulent flow solutions are much more complicated because the flow contains disordered, unsteady, three-dimensional eddies that mix the fluid. Furthermore, these eddies may range in size over several orders of magnitude. In a turbu lent flow field, none of the terms in the equations can be ignored (with the exception of the gravity term in some cases), and thus solutions can be obtained only through numerical computations. Computational fluid dynamics (CFD) is discussed in Chap. 15.
Differential Analysis of Biofluid Mechanics Flows In Example 9–18 we derived fully developed flow in a round pipe, or what is commonly referred to as Poiseuille flow. The solution to the Navier– Stokes equation for this particular example is quite straightforward but is based on a number of assumptions and approximations. These approxima tions hold true for standard pipe flow with most water systems, for example. However, when applied to blood flow in the human body, the approxima tions must be closely monitored and evaluated for their applicability. Tra ditionally as a first-order attempt, cardiovascular fluid dynamists have used the Poiseuille flow derivation to understand blood flow in arteries. This can provide the engineer with a first-order approximation for the velocity and flow rate, but if the engineer were interested in a more sophisticated and, frankly realistic, understanding of blood flow, it is important to examine the main approximations used to arrive at Poiseuille flow. Before delving in, let’s retain the basic approximations about the fluid, or blood in this case. The fluid will remain incompressible, the flow will continue to be laminar, and gravity remains negligible. The approximation of fully developed flow will also remain, though in reality this is not appli cable in the cardiovascular system. Based on only these approximations, this leaves the other main approximations of steady, parallel, axisymmetric Newtonian flow, and the pipe approximated as a rigid circular tube. Recall that the heart pumps blood continuously at an average rate of 75 beats per minute for a healthy adult human at rest. As an example of the flow waveform generated by the ventricular contraction simulated in a mock circulatory system (Fig. 9–82), the flow rate changes temporally for this 800 ms cycle. Therefore, fundamentally to model blood flow through Discussion The time required for momentum to diffuse into the fluid seems much longer than we would expect based on our intuition. This is because the solution presented here is valid only for laminar flow. It turns out that if the plate’s speed is large enough, or if there are significant vibrations in the plate or disturbances in the fluid, the flow will become turbulent. In a turbulent flow, large eddies mix rap idly moving fluid near the wall with slowly moving fluid away from the wall. This mixing process occurs rather quickly, so that turbulent diffusion is usually orders of magnitude faster than laminar diffusion.
3 2.5 2 1.5 z 2 νt 0 0 0.2 0.4 0.6 u/V 0.8 1 1 0.5 √ FIGURE 9–81 Normalized velocity profile of Example 9–19: laminar flow of a viscous fluid above an impulsively started infinite plate.
This section was contributed by Professor Keefe Manning of Penn State University.
FIGURE 9–82 The flow waveform created during ejection from a ventricular assist device in a mock circulatory loop. This is similar to the waveform created during left ventricular ejection.
–5 0 100 200 300 400 Cycle time (ms) Flow rate (LPM) 500 600 700 800 0 5 10 15 20 cen96537_ch09_443-518.indd 499 14/01/17 3:09 pm 500 Differential Analysis of Fluid Flow the arteries, the steady flow approximation is inappropriate, making model ing blood flow as Poiseuille flow unsuitable for just this one approximation alone. There is a rapid acceleration and deceleration of flow within a short time period (~300 ms). However, the wave propagation that is initiated at the heart diminishes with distance from it, and as the arteries become progressively smaller to the capillary level, the magnitude of pulsatility decreases. When focused on the venous side as blood returns to the heart, the steady flow approximation can be applied with more confidence, but it should be noted that there remains flow disruption, in particular, from the lower limbs as venous valves (similar to heart valves) help bring blood back to the heart. The rigid, circular tube approximation is equally as inappropriate when applied to cardiovascular blood flow. As mentioned in Chap. 8, the blood vessels continually taper from the main vessel (the aorta) to smaller vessels (arteries, arterioles, and capillaries). There are no abrupt changes in diam eter as might be seen in a commercial piping network. Therefore, one geo metric consideration is the fact that a segment of blood vessel from one end to the other end will have a continual change in diameter. With respect to a circular tube cross section, the vessels are not perfectly circular but rather more elliptical in their cross section, so there is a major axis and minor axis. The most important approximation here that applies to Poiseuille flow is the fact that pipes are typically considered rigid. However, healthy vessels are not rigid; these structures are compliant and flexible. For example, the aorta emanating from the left ventricle can double in diameter to accommodate the sharp increase in blood volume during left ventricular ejection over a brief time period. One of the major exceptions to using this approximation is when studying pathologic states like atherosclerosis or studying blood flow in the elderly. The basic result of both is that the vessels will harden. In doing so, the rigidity approximation can be applied. There is also a secondary effect as the vessels harden, namely, the pulsatility of blood dampens more quickly, which can influence the steady flow approximation in the arterioles in these particular patient populations. With respect to parallel flow and axisymmetric flow, these both can be invalidated as inappropriate approximations applied to blood flow, by focusing on one location of the cardiovascular system. Considering the aorta in Fig. 9–83 (ascending from the left ventricle, the aortic arch, and descending from the arch), there are significant changes in geometry that influence the flow field. What is commonly not displayed in two-dimen sional pictures of the cardiovascular system (like Fig. 8–83) is the fact that the aorta does not remain in one plane as typically depicted. Actually, the aorta (as one looks at another person) will start from the left ventricle and move towards the spinal column (towards the back of the person) moving the flow into other planes due to pure anatomy. What this geometry does is create Dean flow in this region. As a result, the flow that is created moving around this bend and backwards, is a double helical swirling pattern (think about the DNA helix but the helixes are streamlines). With all this swirling, the approximations of parallel and axisymmetric flow are inappropriate. This is the most extreme case of flow in the human body (except for cases of pathology or with medical device intervention). The parallel and axisymmet ric flow approximations can be used with more confidence in the rest of the circulatory system.
FIGURE 9–83 An anatomical figure illustrating the ascending aorta, aortic arch, and descending aorta coming from the left ventricle (on the backside of the heart in this view). The illustration demonstrates how the aorta moves toward the spinal cord.
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Ascending aorta Descending aorta Aortic arch cen96537_ch09_443-518.indd 500 14/01/17 3:09 pm 501 CHAPTER 9 It should be mentioned that flow within the capillaries is not Poisueille flow since the red blood cells have to squeeze into these vessels and what results is a two-phase flow where a red blood cell is followed by plasma, which is in turn followed by a red blood cell; this continues, creating a unique flow field to facilitate oxygen and nutrient exchange. Finally, blood is not Newtonian, as illustrated in Example 9–20.
EXAMPLE 9–20 Fully Developed Flow in a Round Pipe with a Simple Blood Viscosity Model Consider Example 9–18 and all the approximations to arrive at Poiseuille flow and the axial velocity profile shown in Fig. 9–74. In this example, we will change the basic assumption of a Newtonian fluid and instead use a non-Newtonian fluid viscosity model. Blood behaves as a viscoelastic fluid but for our purposes, we assume a shear thinning or pseudoplastic model and apply a generalized power law viscosity model. The power law model effectively comes from the viscous stress tensor and is 𝜏rz = −𝜇 ( du dr) n where we introduce a negative sign for direction, and where 0 < n < 1.
SOLUTION We take Example 9–18 up to Equation 4 in that example: 1 r d dr (r du dr) = 1 𝜇 dP dx. Through rearrangement and one integration with respect to r, we arrive at r 2 dP dx = 𝜇 dP dx, which is also r 2 dP dx = 𝜇 dP dx = 𝜏rz. Then we can equate the power law model to this as well, and arrive at a new relationship, r 2 dP dx = −𝜇 ( du dr) n . When we move the negative sign to the other side, multiple by 1/n on both sides, and solve for du dr, we arrive at du dr = (−r 2𝜇 dP dx ) 1 n. We integrate and then apply the second boundary condition from Example 9–18 (centerline of the pipe is an axis of symmetry). Our velocity then becomes u = R( n+ 1 n ) −r( n+ 1 n ) ( n + 1 n ) ( 1 2𝜇 dP dx ) 1 n We now have a generalized velocity profile for a power law fluid or a type of non-Newtonian fluid, which might be a rudimentary model for blood. As mentioned, we approximate blood as a pseudoplastic fluid; as such, we arbitrarily set n = 0.5. The actual velocity then becomes u = R3 −r3 3 ( 1 2𝜇 dP dx ) 2 Note that if we were to use n = 1 instead, we would get the following, u = (R2 −r2) ( 1 4𝜇 dP dx ), which is the axial velocity for a Newtonian fluid. We plot both the Newtonian and pseudoplastic velocity profiles in Fig. 9–84. Note how the viscosity alters the flow profile making it more blunt. To calculate FIGURE 9–84 Assuming all values are the same in the velocity equations and the pipe is the same diameter, the pseudoplastic fluid causes the velocity profile to be more blunt compared to the parabolic profile generated for a Newtonian fluid.
Newtonian fluid Pseudoplastic fluid u r cen96537_ch09_443-518.indd 501 14/01/17 3:09 pm 502 Differential Analysis of Fluid Flow the volume flow rate, we integrate over the cross section of the pipe using the equation V · = ∫R 0 2𝜋 ru dr and using the generalized form for u. Once we integrate and do some algebraic manipulation, our flow rate becomes V · = n𝜋 R3 3n + 1 ( R 2𝜇 dP dx ) 1 n For our example pseudoplastic fluid (n = 0.5), the flow rate simplifies to V · = 𝜋 R5 5 ( 1 2𝜇 dP dx ) 2 Discussion When n = 1, the general equation for volume flow rate reduces to that for Poiseuille flow, as it must.
SUMMARY In this chapter we derive the differential forms of conser vation of mass (the continuity equation) and the linear momentum equation (the Navier–Stokes equation). For incompressible flow of a Newtonian fluid with constant properties, the continuity equation is ∇ ›·V ›= 0 and the Navier–Stokes equation is ρ DV › Dt = −∇ › P + ρg ›+ 𝜇∇2V › For incompressible two-dimensional flow, we also define the stream function 𝜓. In Cartesian coordinates, u = ∂𝜓 ∂y 𝜐= −∂𝜓 ∂x We show that the difference in the value of 𝜓 from one streamline to another is equal to the volume flow rate per unit width between the two streamlines and that curves of constant 𝜓 are streamlines of the flow. We provide several examples showing how the differential equations of fluid motion are used to generate an expression for the pressure field for a given velocity field and to gener ate expressions for both velocity and pressure fields for a flow with specified geometry and boundary conditions. The solution procedure learned here can be extended to much more compli cated flows whose solutions require the aid of a computer. The Navier–Stokes equation is the cornerstone of fluid mechanics. Although we know the necessary differential equations that describe fluid flow (continuity and Navier– Stokes), it is another matter to solve them. For some simple (usually infinite) geometries, the equations reduce to equa tions that we can solve analytically. For more complicated geometries, the equations are nonlinear, coupled, second-order, partial differential equations that cannot be solved with pencil and paper. We must then resort to either approxi mate solutions (Chap. 10) or numerical solutions (Chap. 15).
REFERENCES AND SUGGESTED READING 1. R. W. Fox and A. T. McDonald. Introduction to Fluid Mechanics, 8th ed. New York: Wiley, 2011.
2. P. M. Gerhart, R. J. Gross, and J. I. Hochstein. Fun damentals of Fluid Mechanics, 2nd ed. Reading, MA: Addison-Wesley, 1992.
3. R. J. Heinsohn and J. M. Cimbala. Indoor Air Quality Engineering. New York: Marcel-Dekker, 2003.
4. P. K. Kundu, I. M. Cohen., and D. R. Dowling. Fluid Mechanics, ed. 5. San Diego, CA: Academic Press, 2011.
5. R. L. Panton. Incompressible Flow, 2nd ed. New York: Wiley, 2005.
6. M. R. Spiegel. Vector Analysis, Schaum’s Outline Series, Theory and Problems. New York: McGraw-Hill Trade, 1968.
7. M. Van Dyke. An Album of Fluid Motion. Stanford, CA: The Parabolic Press, 1982.
cen96537_ch09_443-518.indd 502 14/01/17 3:09 pm 503 CHAPTER 9 Guest Author: Minami Yoda, Georgia Institute of Technology The boundary conditions for a fluid in contact with a solid states that there is no “slip” between the fluid and the solid. The boundary condi tion for a fluid in contact with a different fluid also states that there is no slip between the two fluids. Yet why would different substances—fluid and solid molecules, or molecules of different fluids—have the same behavior? The no-slip boundary condition is widely accepted because it has been ver ified by observation, and because measurements of quantities derived from the velocity field, such as the shear stress, are in agreement with a veloc ity profile that assumes that the tangential velocity component is zero at a stationary wall. Interestingly, Navier (of the Navier–Stokes equations) did not propose a no-slip boundary condition. He instead proposed the partial-slip boundary condi tion (Fig. 9–85) for a fluid in contact with a solid boundary: the fluid velocity component parallel to the wall at the wall, uf, is proportional to the fluid shear stress at the wall, 𝜏s: uf = b𝜏s = b𝜇f ∂u ∂y) f (1) where the constant of proportionality b, which has dimensions of length, is called the slip length. The no-slip condition is the special case of Eq. 1 where b = 0. Although some recent studies in very small (< 0.1 mm diameter) chan nels suggest that the no-slip condition may not hold within a few nanometers of the wall (recall that 1 nm = 10−9 m = 10 Ångstroms), the no-slip condition appears to be the correct boundary condition for a fluid in contact with a wall for a fluid that is a continuum. Nevertheless, engineers also exploit the no-slip boundary condition to reduce friction (or viscous) drag. As discussed in this chapter, the no-slip boundary condition at a free surface, or a water-air interface, makes the viscous stress 𝜏s, and thus the friction drag, very small in the liquid (Eq. 9–68). One way to cre ate a free surface over a solid surface, like the hull of a ship, is to inject air to create a film of air that (at least partially) covers the hull surface (Fig. 9–86). In theory, the drag on the ship, and hence its fuel consumption, can be greatly reduced by creating a free-surface boundary condition over the ship hull. Maintaining a stable air film remains a major engineering challenge, however.
References Lauga, E., Brenner, M., and Stone, H., “Microfluidics: The No-Slip Boundary Condition,” Springer Handbook of Experimental Fluid Mechanics (eds. C. Tropea, A. Yarin, J. F. Foss), Ch. 19, pp. 1219-1240, 2007.
APPLICATION SPOTLIGHT ■ The No-Slip Boundary Condition FIGURE 9–85 Navier’s partial-slip boundary condition.
u(y) uf y b Slope = Fluid Wall ∂u ∂y f ⎛ ⎜ ⎝ FIGURE 9–86 Proposed injection of air bubbles to form an air film over the bottom hull of a cargo ship [based on a picture courtesy of Y. Murai and Y. Oishi, Hokkaido University and the Monohakobi Technology Institute (MTI), Nippon Yusen Kaisha (NYK) and NYK-Hinode Lines].
cen96537_ch09_443-518.indd 503 14/01/17 3:09 pm 504 Differential Analysis of Fluid Flow PROBLEMS General and Mathematical Background Problems 9–1C Explain the fundamental differences between a flow domain and a control volume.
9–2C What does it mean when we say that two or more differential equations are coupled?
9–3C The divergence theorem is ∫V ∇ ›·G › dV = ∮ A G ›·n › dA where G › is a vector, V is a volume, and A is the surface area that encloses and defines the volume. Express the divergence theorem in words.
9–4C For a three-dimensional, unsteady, incompressible flow field in which temperature variations are insignificant, how many unknowns are there? List the equations required to solve for these unknowns.
9–5C For an unsteady, compressible flow field that is two-dimensional in the xy-plane and in which temperature and density variations are significant, how many unknowns are there? List the equations required to solve for these unknowns. (Note: Assume other flow properties like viscos ity, thermal conductivity, etc., can be treated as constants.) 9–6C For a three-dimensional, unsteady, compressible flow field in which temperature and density variations are sig nificant, how many unknowns are there? List the equations required to solve for these unknowns. (Note: Assume other flow properties like viscosity and thermal conductivity can be treated as constants.) 9–7 Transform the position x › = (2, 4, −1) from Cartesian (x, y, z) coordinates to cylindrical (r, 𝜃, z) coordinates, includ ing units. The values of x › are in units of meters.
9–8 Transform the position x = (3 m, 𝜋/4 radians, 0.96 m) from cylindrical (r, 𝜃, z) coordinates to Cartesian (x, y, z) coordinates, including units. Write all three components of x › in units of meters.
9–9 Let vector G › be given by G ›= 2xzi ›−1 2 x2 j ›−z2 k ›.Cal culate the divergence of G ›, and simplify as much as possible. Is there anything special about your result? Answer: 0 9–10 On many occasions we need to transform a veloc ity from Cartesian (x, y, z) coordinates to cylindrical (r, 𝜃, z) coordinates (or vice versa). Using Fig. P9–10 as a guide, transform cylindrical velocity components (ur, u𝜃, uz) into Carte sian velocity components (u, 𝜐, w). (Hint: Since the z-component of velocity remains the same in such a trans formation, we need only to consider the xy-plane, as in Fig. P9–10.) FIGURE P9–10 x y r ur uθ u υ θ V 9–11 Using Fig. P9–10 as a guide, transform Cartesian velocity components (u, 𝜐, w) into cylindrical velocity com ponents (ur, u𝜃, uz). (Hint: Since the z-component of velocity remains the same in such a transformation, we need only to consider the xy-plane.) 9–12 Beth is studying a rotating flow in a wind tunnel. She measures the u and 𝜐 components of velocity using a hot-wire anemometer. At x = 0.40 m and y = 0.20 m, u = 10.3 m/s and 𝜐 = −5.6 m/s. Unfortunately, the data analysis program requires input in cylindrical coordinates (r, 𝜃) and (ur, u𝜃). Help Beth transform her data into cylindrical coordinates. Specifically, calculate r, 𝜃, ur, and u𝜃 at the given data point.
9–13 A steady, two-dimensional, incompressible veloc ity field has Cartesian velocity components u = Cy/(x2 + y2) and 𝜐 = −Cx/(x2 + y2), where C is a constant. Transform these Cartesian velocity components into cylindrical veloc ity components ur and u𝜃, simplifying as much as possible. You should recognize this flow. What kind of flow is this? Answer: 0, −C/r, line vortex 9–14 A Taylor series expansion of function f (x) about some x-location x0 is given as f(x0 + Δx) = f(x0) + ( df dx) x=x0 Δx + 1 2! ( d 2f dx2) x=x0 (Δx)2 + 1 3! ( d 3f dx3) x=x0 (Δx)3 + ⋯ Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and SI users can ignore them. Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software.
cen96537_ch09_443-518.indd 504 14/01/17 3:09 pm 505 CHAPTER 9 Consider the function f (x) = exp(x) = ex. Suppose we know the value of f (x) at x = x0, i.e., we know the value of f (x0), and we want to estimate the value of this function at some x location near x0. Generate the first four terms of the Taylor series expansion for the given function (up to order (Δx)3 as in the above equation). For x0 = 0 and Δx = −0.1, use your truncated Taylor series expansion to estimate f (x0 + Δx). Compare your result with the exact value of e−0.1. How many digits of accuracy do you achieve with your truncated Taylor series?
9–15E Alex is measuring the time-averaged velocity components in a pump using a laser Doppler velocimeter (LDV). Since the laser beams are aligned with the radial and tangential directions of the pump, he measures the ur and u𝜃 com ponents of velocity. At r = 5.20 in and 𝜃 = 30.0°, ur = 2.06 ft/s and u𝜃 = 4.66 ft/s. Unfortunately, the data analysis program requires input in Cartesian coordinates (x, y) in feet and (u, 𝜐) in ft/s. Help Alex transform his data into Cartesian coordinates. Specifically, calculate x, y, u, and 𝜐 at the given data point.
9–16 Let vector G › be given by G ›= 4xz i›−y2 j ›+ yz k › and let V be the volume of a cube of unit length with its cor ner at the origin, bounded by x = 0 to 1, y = 0 to 1, and z = 0 to 1 (Fig. P9–16). Area A is the surface area of the cube. Perform both integrals of the divergence theorem and verify that they are equal. Show all your work.
FIGURE P9–16 1 1 1 x y z A V 9–17 The product rule can be applied to the divergence of scalar f times vector G › as: ∇ ›·( fG ›) = G ›·∇ ›f + f ∇ ›· G ›. Expand both sides of this equation in Cartesian coordinates and verify that it is correct.
9–18 The outer product of two vectors is a second-order tensor with nine components. In Cartesian coordinates, it is F › G ›= [ FxGx FxGy FxGz FyGx FyGy FyGz FzGx FzGy FzGz] The product rule applied to the divergence of the product of two vectors F › and G › is written as ∇ ›·(F ›G ›) = G ›(∇ ›·F ›) + (F ›·∇ ›)G ›. Expand both sides of this equation in Cartesian coordinates and verify that it is correct.
9–19 Use the product rule of Prob. 9–18 to show that ∇ ›·(𝜌V › V ›) = V › ∇ ›·(𝜌V ›) + 𝜌(V ›·∇ ›)V ›.
Continuity Equation 9–20C If a flow field is compressible, what can we say about the material derivative of density? What about if the flow field is incompressible?
9–21C In this chapter we derive the continuity equation in two ways: by using the divergence theorem and by summing mass flow rates through each face of an infinitesimal control volume. Explain why the former is so much less involved than the latter.
9–22 Repeat Example 9–1 (gas compressed in a cylinder by a piston), but without using the continuity equation. Instead, consider the fundamental definition of density as mass divided by volume. Verify that Eq. 5 of Example 9–1 is correct.
9–23 Consider the steady, two-dimensional velocity field given by V › = (u, 𝜐) = (1.6 + 2.8x) i › + (1.5 − 2.8y) j ›. Verify that this flow field is incompressible.
9–24 The compressible form of the continuity equation is (∂𝜌/∂t) + ∇ ›·(𝜌V ›) = 0. Expand this equation as far as possible in Cartesian coordinates (x, y, z) and (u, 𝜐, w).
9–25 In Example 9–6 we derive the equation for volumetric strain rate, (1/V)(DV/Dt) = ∇ ›·V ›. Write this as a word equa tion and discuss what happens to the volume of a fluid ele ment as it moves around in a compressible fluid flow field (Fig. P9–25).
FIGURE P9–25 Time = t1 Time = t2 Time = t3 9–26 Consider a spiraling line vortex/sink flow in the xy- or r𝜃-plane as sketched in Fig. P9–26. The two-dimensional cylindrical velocity components (ur, u𝜃) for this flow field are ur = C/2𝜋r and u𝜃 = Γ/2𝜋r, where C and Γ are constants (m is negative and Γ is positive). Verify that this spiraling line vortex/sink flow in the r𝜃-plane satisfies the two-dimen sional incompressible continuity equation. What happens to conservation of mass at the origin? Discuss.
cen96537_ch09_443-518.indd 505 14/01/17 3:09 pm 506 Differential Analysis of Fluid Flow FIGURE P9–26 y x 9–27 Verify that the steady, two-dimensional, incompress ible velocity field of Prob. 9–13 satisfies the continuity equa tion. Stay in Cartesian coordinates and show all your work.
9–28 Consider steady flow of water through an axisym metric garden hose nozzle (Fig. P9–28). The axial compo nent of velocity increases linearly from uz, entrance to uz, exit as sketched. Between z = 0 and z = L, the axial velocity com ponent is given by uz = uz,entrance + [(uz,exit − uz,entrance)/L]z. Generate an expression for the radial velocity component ur between z = 0 and z = L. You may ignore frictional effects on the walls.
FIGURE P9–28 Dexit Dentrance uz, entrance z uz, exit z = L z = 0 r 9–29 Consider the following steady, three-dimensional veloc ity field in Cartesian coordinates: V › = (u, 𝜐, w) = (axy2 − b) i › −2cy3 j › + dxyk →, where a, b, c, and d are con stants. Under what conditions is this flow field incompress ible? Answer: a = 6c 9–30 Consider the following steady, three-dimensional velocity field in Cartesian coordinates: V › = (u, v, w) = (ax2y + b) i › + cxy2 j › + dx2y k → where a, b, c, and d are constants. Under what conditions is this flow field incompressible?
9–31 Two velocity components of a steady, incompressible flow field are known: u = 2ax + bxy + cy2 and 𝜐 = axz − byz2, where a, b, and c are constants. Velocity component w is miss ing. Generate an expression for w as a function of x, y, and z.
9–32 Imagine a steady, two-dimensional, incompressible flow that is purely circular in the xy- or r𝜃-plane. In other words, velocity component u𝜃 is nonzero, but ur is zero every where (Fig. P9–32). What is the most general form of velocity component u𝜃 that does not violate conservation of mass?
FIGURE P9–32 y x r uθ θ 9–33 The u velocity component of a steady, two-dimensional, incompressible flow field is u = 3ax2 − 2bxy, where a and b are constants. Velocity component 𝜐 is unknown. Generate an expression for 𝜐 as a function of x and y.
9–34 Imagine a steady, two-dimensional, incompressible flow that is purely radial in the xy- or r𝜃-plane. In other words, velocity component ur is nonzero, but u𝜃 is zero every where (Fig. P9–34). What is the most general form of velocity component ur that does not violate conservation of mass?
FIGURE P9–34 y x r ur θ 9–35 The u velocity component of a steady, two-dimensional, incompressible flow field is u = ax + b, where a and b are constants. Velocity component 𝜐 is unknown. Generate an expression for 𝜐 as a function of x and y.
9–36 A two-dimensional diverging duct is being designed to diffuse the high-speed air exiting a wind tunnel. The x-axis is the centerline of the duct (it is symmetric about the x-axis), and the top and bottom walls are to be curved in such a way that the axial wind speed u decreases approximately linearly from u1 = 300 m/s at sec tion 1 to u2 = 100 m/s at section 2 (Fig. P9–36). Meanwhile, the air density 𝜌 is to increase approximately linearly from cen96537_ch09_443-518.indd 506 14/01/17 3:09 pm 507 CHAPTER 9 𝜌1 = 0.85 kg/m3 at section 1 to 𝜌2 = 1.2 kg/m3 at section 2. The diverging duct is 2.0 m long and is 1.60 m high at sec tion 1 (only the upper half is sketched in Fig. P9–36; the half-height at section 1 is 0.80 m). (a) Predict the y-component of velocity, 𝜐(x, y), in the duct. (b) Plot the approximate shape of the duct, ignoring friction on the walls. (c) What should be the half-height of the duct at section 2?
FIGURE P9–36 y x Δx = 2.0 m 0.8 m (1) (2) Stream Function 9–37C What is significant about curves of constant stream function? Explain why the stream function is useful in fluid mechanics.
9–38C In CFD lingo, the stream function is often called a non-primitive variable, while velocity and pressure are called primitive variables. Why do you suppose this is the case?
9–39C What restrictions or conditions are imposed on stream function 𝜓 so that it exactly satisfies the two-dimensional incompressible continuity equation by definition? Why are these restrictions necessary?
9–40C Consider two-dimensional flow in the xy-plane. What is the significance of the difference in value of stream function 𝜓 from one streamline to another?
9–41 Consider a steady, two-dimensional, incompress ible flow field called a uniform stream. The fluid speed is V everywhere, and the flow is aligned with the x-axis (Fig. P9–41). The Cartesian velocity components are u = V and 𝜐 = 0. Generate an expression for the stream function for this flow. Suppose V = 5.08 m/s. If 𝜓2 is a horizontal line at y = 0.5 m and the value of 𝜓 along the x-axis is zero, cal culate the volume flow rate per unit width (into the page of Fig. P9–41) between these two streamlines.
FIGURE P9–41 ψ0 = 0 ψ1 –ψ1 –ψ2 ψ2 y V x 9–42 A common flow encountered in practice is the cross-flow of a fluid approaching a long cylinder of radius R at a free stream speed of U∞. For incompressible inviscid flow, the velocity field of the flow is given as ur = U∞(1 −R2 r2 )cos 𝜃 u𝜃= −U∞(1 + R2 r2 )sin 𝜃 Show that the velocity field satisfies the continuity equa tion, and determine the stream function corresponding to this velocity field.
9–43 The stream function of an unsteady two-dimensional flow field is given by 𝜓= 4x y2 t Sketch a few streamlines for the given flow on the xy-plane, and derive expressions for the velocity components u(x, y, t) and v(x, y, t). Also determine the pathlines at t = 0.
9–44 Consider steady, incompressible, axisymmetric flow (r, z) and (ur, uz) for which the stream function is defined as ur = −(1/r)(∂𝜓/∂z) and uz = (1/r)(∂𝜓/∂r). Verify that 𝜓 so defined satisfies the continuity equation. What conditions or restrictions are required on 𝜓?
9–45 Consider fully developed Couette flow—flow between two infinite parallel plates separated by distance h, with the top plate moving and the bottom plate stationary as illus trated in Fig. P9–45. The flow is steady, incompressible, and two-dimensional in the xy-plane. The velocity field is given by V › = (u, 𝜐) = (Vy/h) i › + 0 j ›. Generate an expression for stream function 𝜓 along the vertical dashed line in Fig. P9–45. For convenience, let 𝜓 = 0 along the bottom wall of the channel. What is the value of 𝜓 along the top wall? Answers: Vy 2/2h, Vh/2 FIGURE P9–45 h y x u = V y Vh 9–46 As a follow-up to Prob. 9–45, calculate the volume flow rate per unit width into the page of Fig. P9–45 from first principles (integration of the velocity field). Compare your result to that obtained directly from the stream function. Discuss.
cen96537_ch09_443-518.indd 507 14/01/17 3:09 pm 508 Differential Analysis of Fluid Flow 9–47E Consider the Couette flow of Fig. P9–45. For the case in which V = 10.0 ft/s and h = 1.20 in, plot several streamlines using evenly spaced values of stream function. Are the stream lines themselves equally spaced? Discuss why or why not.
9–48 Consider fully developed, two-dimensional channel flow—flow between two infinite parallel plates separated by distance h, with both the top plate and bottom plate station ary, and a forced pressure gradient dP/dx driving the flow as illustrated in Fig. P9–48. (dP/dx is constant and negative.) The flow is steady, incompressible, and two-dimensional in the xy-plane. The velocity components are given by u = (1/2𝜇)(dP/dx) (y2 − hy) and 𝜐 = 0, where 𝜇 is the fluid’s viscosity. Generate an expression for stream function 𝜓 along the vertical dashed line in Fig. P9–48. For convenience, let 𝜓 = 0 along the bot tom wall of the channel. What is the value of 𝜓 along the top wall?
FIGURE P9–48 h y u(y) x 9–49 As a follow-up to Prob. 9–48, calculate the volume flow rate per unit width into the page of Fig. P9–48 from first principles (integration of the velocity field). Compare your result to that obtained directly from the stream function. Discuss.
9–50 Consider the channel flow of Fig. P9–48. The fluid is water at 20°C. For the case in which dP/dx = −20,000 N/m3 and h = 1.20 mm, plot several streamlines using evenly spaced values of stream function. Are the streamlines themselves equally spaced? Discuss why or why not.
9–51 In the field of air pollution control, one often needs to sample the quality of a moving airstream. In such measure ments a sampling probe is aligned with the flow as sketched in Fig. P9–51. A suction pump draws air through the probe at volume flow rate V · as sketched. For accurate sampling, the air speed through the probe should be the same as that of the airstream (isokinetic sampling). However, if the applied suction is too large, as sketched in Fig. P9–51, the air speed through the probe is greater than that of the air stream (super iso kinetic sampling). For simplicity consider a two-dimensional case in which the sampling probe height is h = 4.58 mm and its width (into the page of Fig. P9–51) is W = 39.5 mm. The values of the stream function corre sponding to the lower and upper dividing streamlines are 𝜓l = 0.093 m2/s and 𝜓u = 0.150 m2/s, respectively. Calcu late the volume flow rate through the probe (in units of m3/s) and the average speed of the air sucked through the probe. Answers: 0.00225 m3/s, 12.4 m/s FIGURE P9–51 Vfree stream Sampling probe Dividing streamlines V .
ψ = ψu ψ = ψl h V avg 9–52 Suppose the suction applied to the sampling probe of Prob. 9–51 were too weak instead of too strong. Sketch what the streamlines would look like in that case. What would you call this kind of sampling? Label the lower and upper divid ing streamlines.
9–53 Consider the air sampling probe of Prob. 9–51. If the upper and lower streamlines are 7.85 mm apart in the air stream far upstream of the probe, estimate the free stream speed Vfree stream.
9–54 Flow separates at a sharp corner along a wall and forms a recirculating separation bubble as sketched in Fig. P9–54 (streamlines are shown). The value of the stream function at the wall is zero, and that of the uppermost streamline shown is some positive value 𝜓upper. Discuss the value of the stream function inside the separation bubble. In particular, is it positive or negative? Why? Where in the flow is 𝜓 a minimum?
FIGURE P9–54 ψ = ψupper ψ = 0 Separation bubble 9–55 A uniform stream of speed V is inclined at angle 𝛼 from the x-axis (Fig. P9–55). The flow is steady, two-dimen cen96537_ch09_443-518.indd 508 14/01/17 3:09 pm 509 CHAPTER 9 sional, and incompressible. The Cartesian velocity compo nents are u = V cos 𝛼 and 𝜐 = V sin 𝛼. Generate an expres sion for the stream function for this flow.
α ψ0 ψ1 –ψ1 –ψ2 ψ2 y V x FIGURE P9–55 9–56 A steady, two-dimensional, incompressible flow field in the xy-plane has the following stream function: 𝜓 = ax2 + bxy + cy2, where a, b, and c are constants. (a) Obtain expressions for velocity components u and 𝜐. (b) Verify that the flow field satisfies the incompressible continuity equation.
9–57 For the velocity field of Prob. 9–56, plot streamlines 𝜓 = 0, 1, 2, 3, 4, 5, and 6 m2/s. Let constants a, b, and c have the following values: a = 0.50 s−1, b = −1.3 s−1, and c = 0.50 s−1. For consistency, plot streamlines between x = −2 and 2 m, and y = −4 and 4 m. Indicate the direction of flow with arrows.
9–58 A steady, two-dimensional, incompressible flow field in the xy-plane has a stream function given by 𝜓 = ax2 − by2 + cx + dxy, where a, b, c, and d are constants. (a) Obtain expressions for velocity components u and 𝜐. (b) Verify that the flow field satisfies the incompressible con tinuity equation.
9–59 Repeat Prob. 9–58, except make up your own stream function. You may create any function 𝜓(x, y) that you desire, as long as it contains at least three terms and is not the same as an example or problem in this text. Discuss.
9–60 We briefly mention the compressible stream function 𝜓𝜌 in this chapter, defined in Cartesian coordinates as 𝜌u = (∂𝜓𝜌/∂y) and 𝜌𝜐 = −(∂𝜓𝜌/∂x). What are the primary dimen sions of 𝜓𝜌? Write the units of 𝜓𝜌 in primary SI units and in primary English units.
9–61 A steady, incompressible, two-dimensional CFD cal culation of flow through an asymmetric two-dimensional branching duct reveals the streamline pattern sketched in Fig. P9–61, where the values of 𝜓 are in units of m2/s, and W is the width of the duct into the page. The values of stream function 𝜓 on the duct walls are shown. What percentage of the flow goes through the upper branch of the duct? Answer: 53.9% FIGURE P9–61 ψ = 4.35 ψ = 2.03 ψ = 3.10 h 9–62 If the average velocity in the main branch of the duct of Prob. 9–61 is 11.5 m/s, calculate duct height h in units of cm. Obtain your result in two ways, showing all your work. You may use the results of Prob. 9–61 in only one of the methods.
9–63E Consider a steady, two-dimensional, incompress ible flow field for which the u velocity component is u = ax2 − bxy, where a = 0.45 (ft·s)−1, and b = 0.75 (ft·s)−1. Let 𝜐 = 0 for all values of x when y = 0 (that is, 𝜐 = 0 along the x-axis). Generate an expression for the stream function and plot some streamlines of the flow. For consistency, set 𝜓 = 0 along the x-axis, and plot in the range 0 < x < 3 ft and 0 < y < 4 ft.
9–64 Consider the garden hose nozzle of Prob. 9–28. Gen erate an expression for the stream function corresponding to this flow field.
9–65E Consider the garden hose nozzle of Probs. 9–28 and 9–64. Let the entrance and exit nozzle diameters be 0.50 and 0.14 in, respectively, and let the nozzle length be 2.0 in. The volume flow rate through the nozzle is 2.0 gal/min. (a) Cal culate the axial speeds (ft/s) at the nozzle entrance and at the nozzle exit. (b) Plot several streamlines in the rz-plane inside the nozzle, and design the appropriate nozzle shape.
9–66 There are numerous occasions in which a fairly uni form free-stream flow of speed V in the x-direction encoun ters a long circular cylinder of radius a aligned normal to the flow (Fig. P9–66). Examples include air flowing around a car antenna, wind blowing against a flag pole or telephone pole, wind hitting electric wires, and ocean currents impinging on the submerged round beams that support oil platforms. In all these cases, the flow at the rear of the cylinder is separated and unsteady and usually turbulent. However, the flow in the front half of the cylinder is much more steady and pre dictable. In fact, except for a very thin boundary layer near the cylinder surface, the flow field can be approximated by the following steady, two-dimensional stream function in the xy- or r𝜃-plane, with the cylinder centered at the origin: 𝜓 = V sin 𝜃(r − a2/r). Generate expressions for the radial and tan gential velocity components.
cen96537_ch09_443-518.indd 509 14/01/17 3:09 pm 510 Differential Analysis of Fluid Flow FIGURE P9–66 y r = a V r θ x y r = a V r θ x 9–67 A graduate student is running a CFD code for his MS research project and generates a plot of flow streamlines (contours of constant stream function). The contours are of equally spaced values of stream function. Professor I. C. Flows looks at the plot and immediately points to a region of the flow and says, “Look how fast the flow is moving here!” What did Professor Flows notice about the streamlines in that region and how did she know that the flow was fast in that region?
9–68E A sketch of flow streamlines (contours of con stant stream function) is shown in Fig. P9–68E for steady, incompressible, two-dimensional flow of air in a curved duct. (a) Draw arrows on the streamlines to indicate the direction of flow. (b) If h = 1.14 in, what is the approxi mate speed of the air at point P? (c) Repeat part (b) if the fluid were water instead of air. Discuss. Answers: (b) 1.37 ft/s, (c) 1.37 ft/s P h ψ = 0.32 ft2/s ψ = 0.45 ft2/s FIGURE P9–68E 9–69 In Example 9–2, we provide expressions for u, 𝜐, and 𝜌 for flow through a compressible converging duct. Generate an expression for the compressible stream function 𝜓𝜌 that describes this flow field. For consistency, set 𝜓𝜌 = 0 along the x-axis.
9–70 In Prob. 9–36 we developed expressions for u, 𝜐, and 𝜌 for flow through the compressible, two-dimensional, diverging duct of a high-speed wind tunnel. Generate an expression for the compressible stream function 𝜓𝜌 that describes this flow field. For consistency, set 𝜓𝜌 = 0 along the x-axis. Plot several streamlines and verify that they agree with those you plotted in Prob. 9–36. What is the value of 𝜓𝜌 at the top wall of the diverging duct?
9–71 Steady, incompressible, two-dimensional flow over a newly designed small hydrofoil of chord length c = 9.0 mm is modeled with a commercial computational fluid dynamics (CFD) code. A close-up view of flow streamlines (contours of constant stream function) is shown in Fig. P9–71. Values of the stream function are in units of m2/s. The fluid is water at room temperature. (a) Draw an arrow on the plot to indi cate the direction and relative magnitude of the velocity at point A. Repeat for point B. Discuss how your results can be used to explain how such a body creates lift. (b) What is the approximate speed of the air at point A? (Point A is between streamlines 1.65 and 1.66 in Fig. P9–71.) c A B 1.60 1.62 1.64 1.66 1.68 1.70 1.61 1.63 1.65 1.67 1.69 1.71 FIGURE P9–71 9–72 Time-averaged, turbulent, incompressible, two-dimensional flow over a square block of dimension h = 1 m sitting on the ground is modeled with a computational fluid dynamics (CFD) code. A close-up view of flow streamlines (contours of constant stream function) is shown in Fig. P9–72. The fluid is air at room temperature. Note that contours of constant compressible stream function are plotted in Fig. P9–72, even though the flow itself is approximated as incompressible. Values of 𝜓𝜌 are in units of kg/m·s. (a) Draw an arrow on the plot to indicate the direction and relative magnitude of the velocity at point A. Repeat for point B. (b) What is the approximate speed of the air at point B? (Point B is between streamlines 5 and 6 in Fig. P9–72.) A B 3 4 6 10 2 1 h h FIGURE P9–72 cen96537_ch09_443-518.indd 510 14/01/17 3:09 pm 511 CHAPTER 9 9–73 Consider steady, incompressible, two-dimensional flow due to a line source at the origin (Fig. P9–73). Fluid is created at the origin and spreads out radially in all direc tions in the xy-plane. The net volume flow rate of created fluid per unit width is V · /L (into the page of Fig. P9–73), where L is the width of the line source into the page in Fig. P9–73. Since mass must be conserved everywhere except at the origin (a singular point), the volume flow rate per unit width through a circle of any radius r must also be V · /L. If we (arbitrarily) specify stream function 𝜓 to be zero along the positive x-axis (𝜃 = 0), what is the value of 𝜓 along the positive y-axis (𝜃 = 90°)? What is the value of 𝜓 along the negative x-axis (𝜃 = 180°)?
y x V L r ur θ ⋅ FIGURE P9–73 9–74 Repeat Prob. 9–73 for the case of a line sink instead of a line source. Let V · /L be a positive value, but the flow is everywhere in the opposite direction.
9–75 Streaklines are shown in Fig. P9–75 for flow of water over the front portion of a blunt, axisymmetric cylinder aligned with the flow. Streaklines are generated by introducing air bubbles at evenly spaced points upstream of the field of view. Only the top half is shown since the flow is symmetric about the horizontal axis. Since the flow is steady, the streaklines are coincident with streamlines. Discuss how you can tell from the streamline pattern whether the flow speed in a particular region of the flow field is (relatively) large or small.
FIGURE P9–75 Courtesy ONERA. Photograph by Werlé.
Linear Momentum Equation, Boundary Conditions, and Applications 9–76C What is the main distinction between a Newtonian fluid and a non-Newtonian fluid? Name at least three Newto nian fluids and three non-Newtonian fluids.
9–77C What is mechanical pressure Pm, and how is it used in an incompressible flow solution?
9–78C What are constitutive equations, and to which fluid mechanics equation are they applied?
9–79C An airplane flies at constant velocity V › airplane (Fig. P9–79C). Discuss the velocity boundary conditions on the air adjacent to the surface of the airplane from two frames of reference: (a) standing on the ground, and (b) moving with the airplane. Likewise, what are the far-field velocity bound ary conditions of the air (far away from the airplane) in both frames of reference?
Vairplane FIGURE P9–79C 9–80C Define or describe each type of fluid: (a) viscoelas tic fluid, (b) pseudoplastic fluid, (c) dilatant fluid, (d) Bing ham plastic fluid.
9–81C The general control volume form of the linear momen tum equation is ∫CV ρg › dV + ∫CS 𝜎ij·n › dA I II = ∫CV ∂ ∂t (ρV › ) dV + ∫CS (ρV › ) V › ·n › dA III IV Discuss the meaning of each term in this equation. The terms are labeled for convenience. Write the equation as a word equa tion.
9–82 Consider the steady, two-dimensional, incompressible velocity field, V › = (u, 𝜐) = (ax + b) i › + (−ay + c) j ›, where a, b, and c are constants. Calculate the pressure as a function of x and y.
9–83 Consider the following steady, two-dimensional, incom pressible velocity field: V › = (u, 𝜐) = (−ax2) i › + (2axy) j ›, where a is a constant. Calculate the pressure as a function of x and y.
9–84 Consider the following steady, two-dimensional, incom pressible velocity field: V › = (u, 𝜐) = (ax + b) i › + (−ay + cx2) j ›, where a, b, and c are constants. Calculate the pressure as a function of x and y. Answer: cannot be found cen96537_ch09_443-518.indd 511 14/01/17 3:09 pm 512 Differential Analysis of Fluid Flow 9–85 Consider liquid in a cylindrical tank. Both the tank and the liquid rotate as a rigid body (Fig. P9–85). The free surface of the liquid is exposed to room air. Surface ten sion effects are negligible. Discuss the boundary conditions required to solve this problem. Specifically, what are the velocity boundary conditions in terms of cylindrical coordi nates (r, 𝜃, z) and velocity components (ur, u𝜃, uz) at all sur faces, including the tank walls and the free surface? What pressure boundary conditions are appropriate for this flow field? Write mathematical equations for each boundary con dition and discuss.
Liquid Free surface r z ω ρ g R P = Patm FIGURE P9–85 9–86 Engine oil at T = 60°C is forced to flow between two very large, stationary, parallel flat plates separated by a thin gap height h = 3.60 mm (Fig. P9–86). The plate dimensions are L = 1.25 m and W = 0.550 m. The outlet pressure is atmo spheric, and the inlet pressure is 1 atm gage pressure. Esti mate the volume flow rate of oil. Also calculate the Reynolds number of the oil flow, based on gap height h and average velocity V. Is the flow laminar or turbulent? Answers: 2.39 × 10−3 m3/s, 51.8, laminar y x h L W V Pout Pin FIGURE P9–86 9–87 Consider steady, two-dimensional, incompressible flow due to a spiraling line vortex/sink flow centered on the z-axis. Streamlines and velocity components are shown in Fig. P9–87. The velocity field is ur = C/r and u𝜃 = K/r, where C and K are constants. Calculate the pressure as a function of r and 𝜃.
uθ r uθ = K r ur = C r FIGURE P9–87 9–88 Consider steady, incompressible, parallel, laminar flow of a viscous fluid falling between two infinite vertical walls (Fig. P9–88). The distance between the walls is h, and gravity acts in the negative z-direction (downward in the figure). There is no applied (forced) pressure driving the flow—the fluid falls by gravity alone. The pressure is constant everywhere in the flow field. Calculate the velocity field and sketch the velocity profile using appropriate nondimensionalized variables.
h z x g Fluid: ρ, μ Fixed wall Fixed wall FIGURE P9–88 9–89 For the fluid falling between two parallel vertical walls (Prob. 9–88), generate an expression for the volume flow rate per unit width (V · /L) as a function of 𝜌, 𝜇, h, and g. Compare your result to that of the same fluid falling along one vertical wall with a free surface replacing the second wall (Example 9–17), all else being equal. Discuss the differences and provide a physical explanation. Answer: 𝜌gh3/12𝜇 downward 9–90 Repeat Example 9–17, except for the case in which the wall is inclined at angle 𝛼 (Fig. P9–90). Generate expres sions for both the pressure and velocity fields. As a check, make sure that your result agrees with that of Example 9–17 when 𝛼 = 90°. [Hint: It is most convenient to use the (s, y, n) coordinate system with velocity components (us, 𝜐, un), where y is into the page in Fig. P9–90. Plot the dimensionless veloc ity profile us versus n for the case in which 𝛼 = 60°.] cen96537_ch09_443-518.indd 512 14/01/17 3:09 pm 513 CHAPTER 9 z x g s n Fixed wall Air P = Patm Oil film: ρ, μ h α FIGURE P9–90 9–91 For the falling oil film of Prob. 9–90, generate an expression for the volume flow rate per unit width of oil falling down the wall (V · /L) as a function of 𝜌, 𝜇, h, and g. Calculate (V · /L) for an oil film of thickness 10 mm with 𝜌 = 888 kg/m3 and 𝜇 = 0.80 kg/m·s.
9–92 The first two viscous terms in the 𝜃-component of the Navier–Stokes equation (Eq. 9–62c) are 𝜇[ 1 r ∂ ∂r (r ∂u𝜃 ∂r )− u𝜃 r2] .
Expand this expression as far as possible using the product rule, yielding three terms. Now combine all three terms into one term. (Hint: Use the product rule in reverse—some trial and error may be required.) 9–93 An incompressible Newtonian liquid is confined between two concentric circular cylinders of infinite length— a solid inner cylinder of radius Ri and a hollow, stationary outer cylinder of radius Ro (Fig. P9–93; the z-axis is out of the page). The inner cylinder rotates at angular velocity 𝜔i. The flow is steady, laminar, and two-dimensional in the r𝜃-plane. The flow is also rotationally symmetric, meaning that nothing is a function of coordinate 𝜃 (u𝜃 and P are func tions of radius r only). The flow is also circular, meaning that velocity component ur = 0 everywhere. Generate an exact expression for velocity component u𝜃 as a function of radius r and the other parameters in the problem. You may ignore gravity. (Hint: The result of Prob. 9–92 is useful.) 9–94 Repeat Prob. 9–93, but let the inner cylinder be sta tionary and the outer cylinder rotate at angular velocity 𝜔o. Generate an exact solution for u𝜃(r) using the step-by-step procedure discussed in this chapter.
9–95 Analyze and discuss two limiting cases of Prob. 9–93: (a) The gap is very small. Show that the velocity profile approaches linear from the outer cylinder wall to the inner cylinder wall. In other words, for a very tiny gap the velocity profile reduces to that of simple two-dimensional Couette flow. (Hint: Define y = Ro − r, h = gap thickness = Ro − Ri, and V = speed of the “upper plate” = Ri𝜔i.) (b) The outer cylinder radius approaches infinity, while the inner cylinder radius is very small. What kind of flow does this approach?
9–96 Repeat Prob. 9–93 for the more general case. Namely, let the inner cylinder rotate at angular velocity 𝜔i and let the outer cylinder rotate at angular velocity 𝜔o. All else is the same as Prob. 9–93. Generate an exact expression for veloc ity component u𝜃 as a function of radius r and the other parameters in the problem. Verify that when 𝜔o = 0 your result simplifies to that of Prob. 9–93.
9–97 Analyze and discuss a limiting case of Prob. 9–96 in which there is no inner cylinder (Ri = 𝜔i = 0). Gen erate an expression for u𝜃 as a function of r. What kind of flow is this? Describe how this flow could be set up experimentally. Answer: 𝜔or 9–98 Consider steady, incompressible, laminar flow of a Newtonian fluid in an infinitely long round pipe annulus of inner radius Ri and outer radius Ro (Fig. P9–98). Ignore the effects of gravity. A constant negative pressure gradi ent ∂P/∂x is applied in the x-direction, (∂P/dx) = (P2 − P1)/ (x2 − x1), where x1 and x2 are two arbitrary locations along Liquid: ρ, μ Rotating inner cylinder Stationary outer cylinder Ro Ri ωi FIGURE P9–93 r Ro P2 P1 Ri x1 x2 Fluid: ρ, μ Outer pipe wall x ∂x x2 – x1 P2 – P1 ∂P = FIGURE P9–98 cen96537_ch09_443-518.indd 513 14/01/17 3:09 pm 514 Differential Analysis of Fluid Flow the x-axis, and P1 and P2 are the pressures at those two loca tions. The pressure gradient may be caused by a pump and/ or gravity. Note that we adopt a modified cylindrical coordi nate system here with x instead of z for the axial component, namely, (r, 𝜃, x) and (ur, u𝜃, u). Derive an expression for the velocity field in the annular space in the pipe.
9–99 Consider again the pipe annulus sketched in Fig. P9–98. Assume that the pressure is constant everywhere (there is no forced pressure gradient driving the flow). How ever, let the inner cylinder be moving at steady velocity V to the right. The outer cylinder is stationary. (This is a kind of axisymmetric Couette flow.) Generate an expression for the x-component of velocity u as a function of r and the other parameters in the problem.
9–100 Repeat Prob. 9–99 except swap the stationary and moving cylinder. In particular, let the inner cylinder be stationary, and let the outer cylinder be moving at steady velocity V to the right, all else being equal. Generate an expression for the x-component of velocity u as a function of r and the other parameters in the problem.
9–101 Consider a modified form of Couette flow in which there are two immiscible fluids sandwiched between two infi nitely long and wide, parallel flat plates (Fig. P9–101). The flow is steady, incompressible, parallel, and laminar. The top plate moves at velocity V to the right, and the bottom plate is stationary. Gravity acts in the −z-direction (downward in the figure). There is no forced pressure gradient pushing the fluids through the channel—the flow is set up solely by viscous effects created by the moving upper plate. You may ignore surface tension effects and assume that the interface is horizontal. The pressure at the bottom of the flow (z = 0) is equal to P0. (a) List all the appropriate boundary conditions on both velocity and pressure. (Hint: There are six required boundary conditions.) (b) Solve for the velocity field. (Hint: Split up the solution into two portions, one for each fluid. Generate expressions for u1 as a function of z and u2 as a function of z.) (c) Solve for the pressure field. (Hint: Again split up the solution. Solve for P1 and P2.) (d) Let fluid 1 be water and let fluid 2 be unused engine oil, both at 80°C. Also let h1 = 5.0 mm, h2 = 8.0 mm, and V = 10.0 m/s. Plot u as a function of z across the entire channel. Discuss the results.
h2 h1 Interface Moving wall Fluid 2 ρ2, μ2 Fluid 1 ρ1, μ1 z x V FIGURE P9–101 9–102 Consider dimensionless velocity distribution in Cou ette flow (which is also called generalized Couette flow) with an applied pressure gradient which is obtained in the follow ing form in Example 9–16 as u = y + 1 2P y (y −1) u = u V y = y h P = h2 ∂P 𝜇V ∂x where u, V, ∂P/∂x, and h represent fluid velocity, upper plate velocity, pressure gradient, and distance between parallel plates, respectively. Also, u, y, and P represent dimen sionless velocity, dimensionless distance between the plates, and dimensionless pressure gradient, respectively. (a) Explain why the velocity distribution is a superposition of Couette flow with a linear velocity distribution and Poiseuille flow with a parabolic velocity distribution. (b) Show that if P > 2, backflow begins at the lower wall and it never occurs at the upper wall. Plot u versus y for this situation. (c) Find the position and magnitude of maximum dimensionless velocity.
u = 0 y x 0 h u = V FIGURE P9–102 9–103 Consider steady, incompressible, laminar flow of a Newtonian fluid in an infinitely long round pipe of diameter D or radius R = D/2 inclined at angle 𝛼 (Fig. P9–103). There is no applied pressure gradient (∂P/∂x = 0). Instead, the fluid flows down the pipe due to gravity alone. We adopt the coor dinate system shown, with x down the axis of the pipe. Derive an expression for the x-component of velocity u as a function of radius r and the other parameters of the problem. Calculate the volume flow rate and average axial velocity through the pipe. Answers: 𝜌g (sin 𝛼)(R2 − r2)/4𝜇, 𝜌g (sin 𝛼)𝜋R4/8𝜇, 𝜌g (sin 𝛼)R2/8𝜇 R α Fluid: ρ, μ Pipe wall D g r x FIGURE P9–103 cen96537_ch09_443-518.indd 514 14/01/17 3:09 pm 515 CHAPTER 9 9–104 A stirrer mixes liquid chemicals in a large tank (Fig. P9–104). The free surface of the liquid is exposed to room air. Surface tension effects are negligible. Discuss the boundary conditions required to solve this problem. Specifi cally, what are the velocity boundary conditions in terms of cylindrical coordinates (r, 𝜃, z) and velocity components (ur, u𝜃, uz) at all surfaces, including the blades and the free surface? What pressure boundary conditions are appropriate for this flow field? Write mathematical equations for each boundary condition and discuss.
Free surface z r ω ρ, μ Rtank P = Patm D FIGURE P9–104 9–105 Repeat Prob. 9–104, but from a frame of reference rotating with the stirrer blades at angular velocity 𝜔.
9–106 The r𝜃-component of the viscous stress tensor in cylindrical coordinates is 𝜏r𝜃= 𝜏𝜃r = 𝜇[r ∂ ∂r ( u𝜃 r ) + 1 r ∂ur ∂𝜃] (1) Some authors write this component instead as 𝜏r𝜃= 𝜏𝜃r = 𝜇[ 1 r ( ∂ur ∂𝜃−u𝜃) + ∂u𝜃 ∂r ] (2) Are these the same? In other words, is Eq. 2 equivalent to Eq. 1, or do these other authors define their viscous stress tensor differently? Show all your work.
Review Problems 9–107C List the six steps used to solve the Navier–Stokes and continuity equations for incompressible flow with con stant fluid properties. (You should be able to do this without peeking at the chapter.) 9–108C Explain why the incompressible flow approxima tion and the constant temperature approximation usually go hand in hand.
9–109C Discuss the relationship between volumetric strain rate and the continuity equation. Base your discussion on fundamental definitions.
9–110C For each part, write the official name for the dif ferential equation, discuss its restrictions, and describe what the equation represents physically.
(a) ∂ρ ∂t + ∇ › ·(ρV › ) = 0 (b) ∂ ∂t (ρV › ) + ∇ › ·(ρV › V ›) = ρg ›+ ∇ › ·𝜎 ij (c) ρ DV › Dt = −∇ › P + ρg ›+ 𝜇∇2V › 9–111C For each statement, choose whether the statement is true or false and discuss your answer briefly. For each statement it is assumed that the proper boundary conditions and fluid properties are known.
(a) A general incompressible flow problem with constant fluid properties has four unknowns.
(b) A general compressible flow problem has five unknowns.
(c) For an incompressible fluid mechanics problem, the continuity equation and Cauchy’s equation provide enough equations to match the number of unknowns.
(d ) For an incompressible fluid mechanics problem involv ing a Newtonian fluid with constant properties, the continu ity equation and the Navier–Stokes equation provide enough equations to match the number of unknowns.
9–112 Repeat Example 9–17, except for the case in which the wall is moving upward at speed V. As a check, make sure that your result agrees with that of Example 9–17 when V = 0. Nondimensionalize your velocity profile equa tion using the same normalization as in Example 9–17, and show that a Froude number and a Reynolds number emerge. Plot the profile w versus x for cases in which Fr = 0.5 and Re = 0.5, 1.0, and 5.0. Discuss.
9–113 For the falling oil film of Prob. 9–112, calculate the volume flow rate per unit width of oil falling down the wall (V · /L) as a function of wall speed V and the other parameters in the problem. Calculate the wall speed required such that there is no net volume flow of oil either up or down. Give your answer for V in terms of the other parameters in the problem, namely, 𝜌, 𝜇, h, and g. Calculate V for zero vol ume flow rate for an oil film of thickness 4.12 mm with 𝜌 = 888 kg/m3 and 𝜇 = 0.801 kg/m·s. Answer: 0.0615 m/s 9–114 Look up the definition of Poisson’s equation in one of your math textbooks or on the Internet. Write Poisson’s equation in standard form. How is Poisson’s equation similar to Laplace’s equation? How do these two equations differ?
9–115 Consider the following steady, three-dimensional velocity field in Cartesian coordinates: V › = (u, 𝜐, w) = (axz2 − by) i › + cxyz j › + (dz3 + exz2)k →, where a, b, c, d, and e are constants. Under what conditions is this flow field incompressible? What are the primary dimensions of con stants a, b, c, d, and e?
9–116 Simplify the Navier–Stokes equation as much as possible for the case of an incompressible liquid being accel erated as a rigid body in an arbitrary direction (Fig. P9–116). Gravity acts in the −z-direction. Begin with the incompressible cen96537_ch09_443-518.indd 515 14/01/17 3:09 pm 516 Differential Analysis of Fluid Flow vector form of the Navier–Stokes equation, explain how and why some terms can be simplified, and give your final result as a vector equation.
Free surface Fluid particle Liquid g a a FIGURE P9–116 9–117 Simplify the Navier–Stokes equation as much as pos sible for the case of incompressible hydrostatics, with gravity acting in the negative z-direction. Begin with the incompress ible vector form of the Navier–Stokes equation, explain how and why some terms can be simplified, and give your final result as a vector equation. Answer: ∇ →P = −𝜌gk → 9–118 For each of the listed equations, write down the equation in vector form and decide if it is linear or nonlinear. If it is nonlinear, which term(s) make it so? (a) incompress ible continuity equation, (b) compressible continuity equation, and (c) incompressible Navier–Stokes equation.
9–119 A boundary layer is a thin region near a wall in which viscous (frictional) forces are very important due to the no-slip boundary condition. The steady, incompressible, two-dimensional, boundary layer developing along a flat plate aligned with the free-stream flow is sketched in Fig. P9–119. The flow upstream of the plate is uniform, but boundary layer thickness 𝛿 grows with x along the plate due to viscous effects. Sketch some streamlines, both within the boundary layer and above the boundary layer. Is 𝛿(x) a streamline? (Hint: Pay particular attention to the fact that for steady, incompressible, two-dimensional flow the volume flow rate per unit width between any two streamlines is constant.) Boundary layer y x V∞ δ(x) δ(x) FIGURE P9–119 9–120 Consider steady, two-dimensional, incompressible flow in the xz-plane rather than in the xy-plane. Curves of constant stream function are shown in Fig. P9–120. The nonzero velocity components are (u, w). Define a stream function such that flow is from right to left in the xz-plane when 𝜓 increases in the z-direction.
z x Streamlines ψ = ψ3 ψ = ψ2 ψ = ψ1 FIGURE P9–120 9–121 A block slides down a long, straight, inclined wall at speed V, riding on a thin film of oil of thickness h (Fig. P9–121). The weight of the block is W, and its surface area in contact with the oil film is A. Suppose V is measured, and W, A, angle 𝛼, and viscosity 𝜇 are also known. Oil film thickness h is not known. (a) Generate an exact analytical expression for h as a function of the known parameters V, A, W, 𝛼, and 𝜇. (b) Use dimensional analysis to generate a dimensionless expression for h as a function of the given parameters. Construct a relationship between your Π’s that matches the exact analytical expression of part (a).
A h ρ, μ α g V FIGURE P9–121 9–122E A group of students is designing a small, round (axisymmetric), low-speed wind tunnel for their senior design project (Fig. P9–122E). Their design calls for the axial component of velocity to increase linearly in the contraction section from uz, 0 to uz, L. The air speed through the test section is to be uz, L = 120 ft/s. The length of the contraction is L = 3.0 ft, and the entrance and exit diam eters of the contraction are D0 = 5.0 ft and DL = 1.5 ft, respectively. The air is at standard temperature and pressure. (a) Verify that the flow can be approximated as incompress ible. (b) Generate an expression for the radial velocity com ponent ur between z = 0 and z = L, staying in variable form. You may ignore frictional effects (boundary layers) on the walls. (c) Generate an expression for the stream function 𝜓 as a function of r and z. (d) Plot some streamlines and design the shape of the contraction, assuming that frictional effects cen96537_ch09_443-518.indd 516 14/01/17 3:09 pm 517 CHAPTER 9 along the walls of the wind tunnel contraction are negligible.
D0 DL Test section Contraction uz, 0 uz, L z = L z = 0 r z FIGURE P9–122E 9–123 Water flows down a long, straight, inclined pipe of diameter D and length L (Fig. P9–123). There is no forced pressure gradient between points 1 and 2; in other words, the water flows through the pipe by gravity alone, and P1 = P2 = Patm. The flow is steady, fully developed, and laminar. We adopt a coordinate system in which x follows the axis of the pipe. (a) Use the control volume technique of Chap. 8 to gen erate an expression for average velocity V as a function of the given parameters 𝜌, g, D, ∆z, 𝜇, and L. (b) Use differential analysis to generate an expression for V as a function of the given parameters. Compare with your result of part (a) and discuss. (c) Use dimensional analysis to generate a dimen sionless expression for V as a function of the given param eters. Construct a relationship between your Π’s that matches the exact analytical expression.
Δz D P1 P2 V ρ, μ α x L g FIGURE P9–123 9–124 Taking all the Poiseuille flow approximations except that the fluid is Newtonian, determine the velocity profile and flow rate assuming blood is a Bingham plastic fluid based on the shear stress relationship below. Plot the velocity profile of a Newtonian fluid, a pseudoplastic fluid, and a Bingham plastic fluid. How do they differ? Determine the flow rate assuming a Bingham plastic fluid.
𝜏rz = −𝜇 du dr + 𝜏y 9–125 Bob uses a computational fluid dynamics code to model steady flow of an incompressible fluid through a two-dimensional sudden contraction as sketched in Fig. P9–125. Channel height changes from H1 = 12.0 cm to H2 = 4.6 cm. Uniform velocity V › 1 = 18.5i i › m/s is to be specified on the left boundary of the computational domain. The CFD code uses a numerical scheme in which the stream function must be specified along all boundaries of the computational domain. As shown in Fig. P9–125, 𝜓 is specified as zero along the entire bottom wall of the channel. (a) What value of 𝜓 should Bob specify on the top wall of the channel? (b) How should Bob specify 𝜓 on the left side of the computational domain? (c) Discuss how Bob might specify 𝜓 on the right side of the computational domain.
H1 V1 H2 y x ψ = 0 FIGURE P9–125 9–126 Consider the steady, two-dimensional, incompress ible velocity field, V › = (u, 𝜐) = (−ax + b)i › + (ay + c) j ›, where a, b, and c are constants. (a) Verify that continuity is satisfied. (b) Calculate the pressure as a function of x and y.
9–127 Consider the steady, two-dimensional, incompress ible velocity field, V › = (u, 𝜐) = (−ax + b)i › + (ay + c)j ›, where a, b, and c are constants. In the previous problem, we calculated the pressure as a function of x and y. During that solution you may have noticed that the viscous terms in both the x- and y-components of the Navier–Stokes equa tion reduced to zero. Since viscous effects are apparently not important in this flow field, we suspect that the flow field may be irrotational. (a) Calculate the vorticity and check whether this flow field is indeed irrotational. (b) Regardless of your answer to part (a), assume that the flow field is irro tational, and thus the Bernoulli equation should apply every where, not just along streamlines. Use the Bernoulli equation to generate an expression for P(x,y), and compare with the answer we had obtained in the previous problem. Are they the same? Explain.
Fundamentals of Engineering (FE) Exam Problems 9–128 The continuity equation is also known as (a) Conservation of mass (b) Conservation of energy (c) Conservation of momentum (d ) Newton’s second law (e) Cauchy’s equation cen96537_ch09_443-518.indd 517 14/01/17 3:09 pm 518 Differential Analysis of Fluid Flow 9–129 The Navier–Stokes equation is also known as (a) Newton’s first law (b) Newton’s second law (c) Newton’s third law (d ) Continuity equation (e) Energy equation 9–130 Which choice is not correct regarding the Navier– Stokes equation? (a) Nonlinear equation (b) Unsteady equation (c) Second-order equation (d ) Partial differential equation (e) None of these 9–131 In fluid flow analyses, which boundary condition can be expressed as V › fluid = V › wall?
(a) No-slip (b) Interface (c) Free-surface (d ) Symmetry (e) Inlet 9–132 Which choice is the general differential equation form of the continuity equation for a control volume?
(a) ∫CS ρV › ·n › dA = 0 (b) ∫CV ∂ρ ∂t dV + ∫CS ρV › ·n › dA = 0 (c) ∇ › · (ρV › ) = 0 (d ) ∂ρ ∂t + ∇ › ·(ρV ›) = 0 (e) None of these 9–133 Which choice is the differential, incompressible, two-dimensional continuity equation in Cartesian coordinates?
(a) ∫CS ρV › ·n › dA = 0 (b) 1 r ∂(rur) ∂r + 1 r ∂(u𝜃) ∂𝜃 = 0 (c) ∇ › ·(ρV ›) = 0 (d ) ∇ › ·V ›= 0 (e) ∂u ∂x + ∂𝜐 ∂y = 0 9–134 A steady, two-dimensional, incompressible flow field in the xy-plane has a stream function given by 𝜓 = ax2 + by2 + cy, where a, b, and c are constants. The expression for the velocity component v is (a) 2ax (b) 2by + c (c) −2ax (d ) −2by − c (e) 2ax + 2by + c 9–135 A steady, two-dimensional, incompressible flow field in the xy-plane has a stream function given by 𝜓 = ax2 + by2 + cy, where a, b, and c are constants. The expression for the velocity component u is (a) 2ax (b) 2by + c (c) −2ax (d ) −2by − c (e) 2ax + 2by + c 9–136 A steady velocity field is given by V › = (u, 𝜐, w) = 2ax2y i ›+ 3bxy2 j › + cyk →, where a, b, and c are constants. Under what conditions is this flow field incompressible? (a) a = b (b) a = −b (c) 2a = −3b (d ) 3a = 2b (e) a = 2b 9–137 Which choice is the incompressible Navier–Stokes equation with constant viscosity?
(a) ρ DV › Dt + ∇ ›P −ρg ›= 0 (b) −∇ › P + ρg ›+ 𝜇∇ › 2V ›= 0 (c) ρ DV › Dt = −∇ ›P −𝜇∇ › 2V › (d ) ρ DV › Dt = −∇ ›P + ρg › + 𝜇∇ › 2V › (e) ρ DV › Dt = −∇ ›P + ρg › + 𝜇∇ › 2V › + ∇ › ·V ›= 0 cen96537_ch09_443-518.indd 518 14/01/17 3:09 pm 10 CHAPTER 519 OBJECTIVES When you finish reading this chapter, you should be able to ■ ■ Appreciate why approximations are necessary to solve many fluid flow problems, and know when and where such approximations are appropriate ■ ■ Understand the effects of the lack of inertial terms in the creeping flow approximation, including the disappearance of density from the equations ■ ■ Understand superposition as a method of solving potential flow problems ■ ■ Predict boundary layer thickness and other boundary layer properties APPROXIMATE SOLUTIONS OF THE NAVIER–STOKES EQUATION I n this chapter we look at several approximations that eliminate term(s), reducing the Navier–Stokes equation to a simplified form that is more easily solvable. Sometimes these approximations are appropriate in a whole flow field, but in most cases, they are appropriate only in certain regions of the flow field. We first consider creeping flow, where the Reynolds number is so low that the viscous terms dominate (and eliminate) the inertial terms. Following that, we look at two approximations that are appropriate in regions of flow away from walls and wakes: inviscid flow and irrotational flow (also called potential flow). In these regions, the opposite holds; i.e., inertial terms dominate viscous terms. Finally, we discuss the boundary layer approximation, in which both inertial and viscous terms remain, but some of the viscous terms are negligible. This last approximation is appropriate at very high Reynolds numbers (the opposite of creeping flow) and near walls, the opposite of potential flow.
In this chapter, we discuss several approximations that simplify the Navier– Stokes equation, including creeping flow, where viscous terms dominate inertial terms. The flow of lava from a volcano is an example of creeping flow—the viscosity of molten rock is so large that the Reynolds number is small even though the length scales are large.
© Getty Images RF cen96537_ch10_519-568.indd 519 29/12/16 3:44 pm 520 APPROXIMATE SOLUTIONS OF THE N–S EQ 10–1 ■ INTRODUCTION In Chap. 9, we derived the differential equation of linear momentum for an incompressible Newtonian fluid with constant properties—the Navier– Stokes equation. We showed some examples of analytical solutions to the continuity and Navier–Stockes equations for simple (usually infinite) geom etries, in which most of the terms in the component equations are eliminated and the resulting differential equations are analytically solvable. Unfortu nately, there aren’t very many known analytical solutions available in the literature; in fact, we can count the number of such solutions on the fingers of a few students. The vast majority of practical fluid mechanics problems cannot be solved analytically and require either (1) further approximations or (2) computer assistance. We consider option 1 here; option 2 is discussed in Chap. 15. For simplicity, we consider only incompressible flow of Newtonian fluids in this chapter. We emphasize first that the Navier–Stokes equation itself is not exact, but rather is a model of fluid flow that involves several inherent approxima tions (Newtonian fluid, constant thermodynamic and transport properties, etc.). Nevertheless, it is an excellent model and is the foundation of modern fluid mechanics. In this chapter we distinguish between “exact” solutions and approximate solutions (Fig. 10–1). The term exact is used when the solution starts with the full Navier–Stokes equation. The solutions discussed in Chap. 9 are exact solutions because we begin each of them with the full form of the equation. Some terms are eliminated in a specific problem due to the specified geometry or other simplifying assumptions in the prob lem. In a different solution, the terms that get eliminated may not be the same ones, but depend on the geometry and assumptions of that particular problem. We define an approximate solution, on the other hand, as one in which the Navier–Stokes equation is simplified in some region of the flow before we even start the solution. In other words, term(s) are eliminated a priori depending on the class of problem, which may differ from one region of the flow to another. For example, we have already discussed one approximation, namely, fluid statics (Chap. 3). This can be considered to be an approximation of the Navier–Stokes equation in a region of the flow field where the fluid veloc ity is not necessarily zero, but the fluid is nearly stagnant, and we neglect all terms involving velocity. In this approximation, the Navier–Stokes equa tion reduces to just two terms, pressure and gravity, i.e., ∇ ›P = 𝜌g›. The approximation is that the inertial and viscous terms in the Navier–Stokes equation are negligibly small compared to the pressure and gravity terms. Although approximations render the problem more tractable, there is a danger associated with any approximate solution. Namely, if the approxima tion is not appropriate to begin with, the solution will be incorrect—even if we perform all the mathematics correctly. Why? Because we start with equations that do not apply to the problem at hand. For example, we may solve a problem using the creeping flow approximation and obtain a solu tion that satisfies all assumptions and boundary conditions. However, if the Reynolds number of the flow is too high, the creeping flow approximation is inappropriate from the start, and our solution (regardless of how proud of it we may be) is not physically correct. Another common mistake is to “Exact” solution Full Navier–Stokes equation Analysis Solution Approximate solution Analysis Solution Simplified Navier–Stokes equation FIGURE 10–1 “Exact” solutions begin with the full Navier–Stokes equation, while approximate solutions begin with a simplified form of the Navier–Stokes equation right from the start.
cen96537_ch10_519-568.indd 520 29/12/16 3:44 pm 521 CHAPTER 10 assume irrotational flow in regions of the flow where the assumption of irrotationality is not appropriate. The bottom line is that we must be very careful of the approximations we apply, and we should always verify and justify our approximations wherever possible. Finally, we stress that in most practical fluid flow problems, a particular approximation may be appropriate in a certain region of the flow field, but not in other regions, where a different approximation may perhaps be more appropriate. Figure 10–2 illustrates this point qualitatively for flow of a liq uid from one tank to another. The fluid statics approximation is appropriate in a region of the supply tank far away from the connecting pipe, and to a lesser extent in the receiving tank. The irrotational flow approximation is appropriate near the inlet to the connecting pipe and through the middle portion of the pipe where strong viscous effects are absent. Near the walls, the boundary layer approximation is appropriate. The flow in some regions does not meet the criteria for any approximations, and the full Navier– Stokes equation must be solved there (e.g., downstream of the pipe outlet in the receiving tank). How do we determine if an approximation is appropri ate? We do this by comparing the orders of magnitude of the various terms in the equations of motion to see if any terms are negligibly small compared to other terms.
10–2 ■ NONDIMENSIONALIZED EQUATIONS OF MOTION Our goal in this section is to nondimensionalize the equations of motion so that we can properly compare the orders of magnitude of the various terms in the equations. We begin with the incompressible continuity equation, ∇ › ·V › = 0 (10–1) and the vector form of the Navier–Stokes equation, valid for incompressible flow of a Newtonian fluid with constant properties, ρ DV › Dt = ρ [ ∂V › ∂t + (V ›·∇ ›)V › ]= −∇ › P + ρg ›+ 𝜇∇2V › (10–2) We introduce in Table 10–1 some characteristic (reference) scaling param eters that are used to nondimensionalize the equations of motion.
FIGURE 10–2 A particular approximation of the Navier–Stokes equation is appropriate only in certain regions of the flow field; other approximations may be appropriate in other regions of the flow field.
Supply tank Receiving tank Boundary layer region Fluid statics region Fluid statics region Full Navier– Stokes region Irrotational flow region cen96537_ch10_519-568.indd 521 29/12/16 3:44 pm 522 APPROXIMATE SOLUTIONS OF THE N–S EQ We then define several nondimensional variables and one nondimensional operator based on the scaling parameters in Table 10–1, t = ft x › = x › L V › = V › V P = P −P∞ P0 −P∞ g › = g › g ∇ › = L∇ › (10–3) Notice that we define the nondimensional pressure variable in terms of a pressure difference, based on our discussion about pressure versus pressure differences in Chap. 9. Each of the starred quantities in Eq. 10–3 is nondi mensional. For example, although each component of the gradient operator ∇ › has dimensions of {L−1}, each component of ∇ › has dimensions of {1} (Fig. 10–3). We substitute Eq. 10–3 into Eqs. 10–1 and 10–2, treating each term carefully. For example, ∇ › = ∇ ›/L and V › = V V › , so the advective accelera tion term in Eq. 10–2 becomes ρ(V › ·∇ › )V ›= ρ( VV › · ∇ › L ) VV › = ρV 2 L (V › ·∇ › )V › We perform similar algebra on each term in Eqs. 10–1 and 10–2. Equa tion 10–1 is rewritten in terms of nondimensional variables as V L ∇ › ·V › = 0 After dividing both sides by V/L to make the equation dimensionless, we get Nondimensionalized continuity: ∇ › ·V › = 0 (10–4) Similarly, Eq. 10–2 is rewritten as ρVf ∂V › ∂t + ρV 2 L (V › ·∇ › )V › = −P0 −P∞ L ∇ › P + ρgg › + 𝜇V L2 ∇2V › which, after multiplication by the collection of constants L/(𝜌V 2) to make all the terms dimensionless, becomes [ fL V ] ∂V › ∂t + ( V › ·∇ › )V › = −[ P0 −P∞ ρV 2 ]∇ ›P + [ gL V 2]g › + [ 𝜇 ρVL]∇2V › (10–5) Each of the terms in square brackets in Eq. 10–5 is a nondimensional grouping of parameters—a Pi group (Chap. 7). With the help of Table 7–5, we name each of these dimensionless parameters: The one on the left is the TABLE 10–1 Scaling parameters used to nondimensionalize the continuity and momentum equa tions, along with their primary dimensions Scaling Parameter Description Primary Dimensions L Characteristic length {L} V Characteristic speed {Lt−1} f Characteristic frequency {t−1} P0 − P∞ Reference pressure difference {mL−1t−2} g Gravitational acceleration {Lt−2} FIGURE 10–3 The gradient operator is nondimensionalized by Eq. 10–3, regardless of our choice of coordinate system. = , , ∂ ∂x ∂ L∂ , ∂ ∂y ∂ ∂z = = , , ∂ ∂x 1 L 1 L 1 ∂y ∂ ∂z , ∂ L∂ ∂ L∂ x L y L z L Δ Cartesian coordinates , , ∂ ∂r , ∂ ∂θ ∂ L∂ , ∂ ∂θ ∂ ∂z , ∂ ∂z = = , ∂ ∂r 1 L 1 L 1 r ∂ ∂θ 1 L ∂ L∂ = Cylindrical coordinates 1 r r L r L z L = = Δ Δ Δ cen96537_ch10_519-568.indd 522 29/12/16 3:44 pm 523 CHAPTER 10 Strouhal number, St = fL/V; the first one on the right is the Euler number, Eu = (P0 − P∞)/𝜌V 2; the second one on the right is the reciprocal of the square of the Froude number, Fr2 = V 2/gL; and the last one is the reciprocal of the Reynolds number, Re = 𝜌VL/𝜇. Equation 10–5 thus becomes Nondimensionalized Navier–Stokes: [St] ∂V › ∂t + (V › ·∇ ›)V › =−[Eu]∇ ›P + [ 1 Fr2]g › + [ 1 Re]∇2V › (10–6) Before we discuss specific approximations in detail, there is much to comment about the nondimensionalized equation set consisting of Eqs. 10–4 and 10–6: • The nondimensionalized continuity equation contains no additional dimensionless parameters. Hence, Eq. 10–4 must be satisfied as is—we cannot simplify continuity further, because all the terms are of the same order of magnitude.
• The order of magnitude of the nondimensional variables is unity if they are nondimensionalized using a length, speed, frequency, etc., that are characteristic of the flow field. Thus, t ∼ 1, ∣x ›∣ ∼ 1, ∣V › ∣ ∼ 1, etc., where we use the notation ∼ to denote order of magnitude. It follows that the magnitudes of terms like (V ›⋅∇ ›)V › and ∇ ›P in Eq. 10–6 are also of order of magnitude unity and are the same order of magnitude as each other. Thus, the relative importance of the terms in Eq. 10–6 depends only on the relative magnitudes of the dimensionless parameters St, Eu, Fr, and Re. For example, if St and Eu are of order 1, but Fr and Re are very large, we may consider ignoring the gravitational and viscous terms in the Navier–Stokes equation.
• Since there are four dimensionless parameters in Eq. 10–6, dynamic similarity between a model and a prototype requires all four of these to be the same for the model and the prototype (Stmodel = Stprototype, Eumodel = Euprototype, Frmodel = Frprototype, and Remodel = Reprototype), as illustrated in Fig. 10–4.
• If the flow is steady, then f = 0 and the Strouhal number drops out of the list of dimensionless parameters (St = 0). The first term on the left side of Eq. 10–6 then disappears, as does its corresponding unsteady term 𝜕V ›/𝜕t in Eq. 10–2. If the characteristic frequency f is very small such that St ≪ 1, the flow is called quasi-steady. This means that at any instant in time (or at any phase of a slow periodic cycle), we can solve the problem as if the flow were steady, and the unsteady term in Eq. 10–6 again drops out.
• The effect of gravity is usually important only in flows with free-surface effects (e.g., waves, ship motion, spillways from hydroelectric dams, flow of rivers). For many engineering problems there is no free surface (pipe flow, fully submerged flow around a submarine or torpedo, automobile motion, flight of airplanes, birds, insects, etc.). In such cases, the only effect of gravity on the flow dynamics is a hydrostatic pressure distribution in the vertical direction superposed on the pressure field due to the fluid flow. In other words, For flows without free-surface effects, gravity does not affect the dynamics of the flow—its only effect is to superpose a hydrostatic pressure on the dynamic pressure field.
FIGURE 10–4 For complete dynamic similarity between prototype (subscript p) and model (subscript m), the model must be geometrically similar to the prototype, and (in general) all four dimensionless parameters, St, Eu, Fr, and Re, must match. As discussed in Chap. 7, however, this may not always be possible in a model test.
(Top) © James Gritz/Getty Images RF.
gp P ∞, p P0, p Lp fp Vp Stprototype, Euprototype, Frprototype, Reprototype Prototype gm P ∞, m P0, m fm Stmodel, Eumodel, Frmodel, Remodel Model Vm Lm cen96537_ch10_519-568.indd 523 29/12/16 3:44 pm 524 APPROXIMATE SOLUTIONS OF THE N–S EQ • We define a modified pressure P′ that absorbs the effect of hydrostatic pressure. For the case in which z is defined vertically upward (opposite to the direction of the gravity vector), and in which we define some arbitrary reference datum plane at z = 0, Modified pressure: Pʹ = P + ρgz (10–7) The idea is to replace the two terms −∇ ›P + 𝜌g› in Eq. 10–2 with one term −∇ ›P′ using the modified pressure of Eq. 10–7. The Navier–Stokes equation (Eq. 10–2) is written in modified form as ρ DV › Dt = ρ [ ∂V › ∂t + (V › ·∇ › )V › ] = −∇ › Pʹ + 𝜇∇2V › (10–8) With P replaced by P′, and with the gravity term removed from Eq. 10–2, the Froude number drops out of the list of dimensionless parameters. The advantage is that we can solve a form of the Navier–Stokes equation that has no gravity term. After solving the Navier–Stokes equation in terms of modified pressure P′, it is a simple matter to add back the hydrostatic pressure distribution using Eq. 10–7. An example is shown in Fig. 10–5 for the case of two-dimensional Couette flow. Modified pressure is often used in computational fluid dynamics (CFD) codes to separate gravitational effects (hydrostatic pressure in the vertical direction) from fluid flow (dynamic) effects. Note that modified pressure should not be used in flows with free-surface effects. Now we are ready to make some approximations, in which we eliminate one or more of the terms in Eq. 10–2 by comparing the relative magnitudes of the dimensionless parameters associated with the corresponding terms in Eq. 10–6.
10–3 ■ THE CREEPING FLOW APPROXIMATION Our first approximation is the class of fluid flow called creeping flow. Other names for this class of flow include Stokes flow and low Reynolds number flow. As the latter name implies, these are flows in which the Reynolds number is very small (Re ≪ 1). By inspection of the definition of the Reynolds number, Re = 𝜌VL/𝜇, we see that creeping flow is encoun tered when either 𝜌, V, or L is very small or viscosity is very large (or some combination of these). You encounter creeping flow when you pour syrup (a very viscous liquid) on your pancakes or when you dip a spoon into a jar of honey (also very viscous) to add to your tea (Fig. 10–6). Another example of creeping flow is all around us and inside us, although we can’t see it, namely, flow around microscopic organisms. Microorgan isms live their entire lives in the creeping flow regime since they are very small, their size being of order a few microns (1 μm = 10−6 m), and they move very slowly, even though they may move in air or swim in water with a viscosity that can hardly be classified as “large” (𝜇air ≅ 1.8 × 10−5 N·s/m2 and 𝜇water ≅ 1.0 × 10−3 N·s/m2 at room temperature). Figure 10–7 shows a Salmonella bacterium swimming through water. The bacterium’s body is only about 1 μm long; its flagella (hairlike tails) extend several microns behind the body and serve as its propulsion mechanism. The Reynolds number associated with its motion is much smaller than 1.
P P' V z Hydrostatic pressure g x (a) P P' V g z x (b) FIGURE 10–5 Pressure and modified pressure distribution on the right face of a fluid element in Couette flow between two infinite, parallel, horizontal plates: (a) z = 0 at the bottom plate, and (b) z = 0 at the top plate. The modified pressure P′ is constant, but the actual pressure P is not constant in either case. The shaded area in (b) represents the hydrostatic pressure component.
FIGURE 10–6 The slow flow of a very viscous liquid like honey is classified as creeping flow.
cen96537_ch10_519-568.indd 524 29/12/16 3:44 pm 525 CHAPTER 10 Creeping flow also occurs in the flow of lubricating oil in the very small gaps and channels of a lubricated bearing. In this case, the speeds may not be small, but the gap size is very small (on the order of tens of microns), and the viscosity is relatively large (𝜇oil ∼ 1 N·s/m2 at room temperature). For simplicity, we assume that gravitational effects are negligible, or that they contribute only to a hydrostatic pressure component, as discussed previ ously. We also assume either steady flow or oscillating flow, with a Strouhal number of order unity (St ∼ 1) or smaller, so that the unsteady acceleration term [St] 𝜕V ›/𝜕t is orders of magnitude smaller than the viscous term [1/Re]∇ ›2V › (the Reynolds number is very small). The advective term in Eq. 10–6 is of order 1, (V ›⋅∇ ›)V › ∼ 1, so this term drops out as well. Thus, we ignore the entire left side of Eq. 10–6, which reduces to Creeping flow approximation: [Eu]∇ ›P ≅[ 1 Re]∇2V › (10–9) In words, pressure forces in the flow (left side) must be large enough to bal ance the (relatively) large viscous forces on the right side. However, since the nondimensional variables in Eq. 10–9 are of order 1, the only way for the two sides to balance is if Eu is of the same order of magnitude as 1/Re. Equating these, [Eu] = P0 −P∞ ρV 2 ∼[ 1 Re] = 𝜇 ρVL After some algebra, Pressure scale for creeping flow: P0 −P∞∼𝜇V L (10–10) Equation 10–10 reveals two interesting properties of creeping flow. First, we are used to inertially dominated flows, in which pressure differences scale like 𝜌V 2 (e.g., the Bernoulli equation). Here, however, pressure differ ences scale like 𝜇V/L instead, since creeping flow is a viscously dominated flow. In fact, all the inertial terms of the Navier–Stokes equation disappear in creeping flow. Second, density has completely dropped out as a param eter in the Navier–Stokes equation (Fig. 10–8). We see this more clearly by writing the dimensional form of Eq. 10–9, Approximate Navier–Stokes equation for creeping flow: ∇ › P ≅𝜇∇2V › (10–11) Alert readers may point out that density still has a minor role in creeping flow. Namely, it is needed in the calculation of the Reynolds number. How ever, once we have determined that Re is very small, density is no longer needed since it does not appear in Eq. 10–11. Density also pops up in the hydrostatic pressure term, but this effect is usually negligible in creeping flow, since the vertical distances involved are often measured in millime ters or micrometers. Besides, if there are no free-surface effects, we can use modified pressure instead of physical pressure in Eq. 10–11. Let’s discuss the lack of inertia terms in Eq. 10–11 in somewhat more detail. You rely on inertia when you swim (Fig. 10–9). For example, you take a stroke, and then you are able to glide for some distance before you need to take another stroke. When you swim, the inertial terms in the Navier–Stokes equation are much larger than the viscous terms, since the FIGURE 10–8 In the creeping flow approximation, density does not appear in the momentum equation.
Density?
What is density?
∇P ≅ 휇∇2V FIGURE 10–7 (a) Salmonella typhimurium invading cultured human cells. (b) The bacterium Salmonella swimming through water.
(a) NIAID, NIH, Rocky Mantain Laboratories. (b) © MedicalRF.com/Getty Images RF.
(a) (b) 1 μm cen96537_ch10_519-568.indd 525 29/12/16 3:44 pm 526 APPROXIMATE SOLUTIONS OF THE N–S EQ Reynolds number is very large. (Believe it or not, even extremely slow swimmers move at very large Reynolds numbers!) For microorganisms swimming in the creeping flow regime, however, there is negligible inertia, and thus no gliding is possible. In fact, the lack of inertial terms in Eq. 10–11 has a substantial impact on how microorganisms are designed to swim. A flapping tail like that of a dolphin would get them nowhere. Instead, their long, narrow tails ( flagella) undulate in a sinusoidal motion to propel them forward, as illustrated in Fig. 10–10 for the case of a sperm. Without any inertia, the sperm does not move unless his tail is mov ing. The instant his tail stops, the sperm stops moving. If you have ever seen a video clip of swimming sperm or other microorganisms, you may have noticed how hard they have to work just to move a short distance. That is the nature of creeping flow, and it is due to the lack of inertia. Careful study of Fig. 10–10 reveals that the sperm’s tail has completed approximately two complete undulation cycles, yet the sperm’s head has moved to the left by only about two head lengths. It is very difficult for us humans to imagine moving in creeping flow conditions, since we are so used to the effects of inertia. Some authors have suggested that you imagine trying to swim in a vat of honey. We suggest instead that you go to a fast-food restaurant where they have a children’s play area and watch a child play in a pool of plastic spheres (Fig. 10–11). When the child tries to “swim” among the balls (without touching the walls or the bottom), he or she can move forward only by certain snakelike wrig gling body motions. The instant the child stops wriggling, all motion stops, since there is negligible inertia. The child must work very hard to move for ward a short distance. There is a weak analogy between a child “swimming” in this kind of situation and a microorganism swimming in creeping flow conditions. We next discuss the lack of density in Eq. 10–11. At high Reynolds numbers, the aerodynamic drag on an object increases proportionally with 𝜌. (Denser fluids exert more pressure force on the body as the fluid impacts the body.) However, this is actually an inertial effect, and inertia is negligible in creeping flow. In fact, aerodynamic drag cannot even be a function of density in a creeping flow, since density has disappeared from the Navier–Stokes equation. Example 10–1 illustrates this situation through the use of dimensional analysis.
FIGURE 10–9 A person swims at a very high Reynolds number, and inertial terms are large; thus the person is able to glide long distances without moving.
FIGURE 10–10 A sperm of the sea squirt Ciona swimming in seawater; flash photographs at 200 frames per second, with each image positioned directly below the one before it.
Courtesy of Professor Charlotte Omoto, Washington State University, School of Biological Sciences.
10 μm EXAMPLE 10–1 Drag on an Object in Creeping Flow Since density has vanished from the Navier–Stokes equation, aerodynamic drag on an object in creeping flow is a function only of its speed V, some characteristic length scale L of the object, and fluid viscosity 𝜇 (Fig. 10–12). Use dimensional analysis to generate a relationship for FD as a function of these independent variables.
SOLUTION We are to use dimensional analysis to generate a functional rela tionship between FD and variables V, L, and 𝜇.
Assumptions 1 We assume Re ≪ 1 so that the creeping flow approximation applies. 2 Gravitational effects are irrelevant. 3 No parameters other than those listed in the problem statement are relevant to the problem.
cen96537_ch10_519-568.indd 526 29/12/16 3:44 pm 527 CHAPTER 10 Drag on a Sphere in Creeping Flow As shown in Example 10–1, the drag force FD on a three-dimensional object of characteristic dimension L moving under creeping flow conditions at speed V through a fluid with viscosity 𝜇 is FD = constant⋅𝜇VL. Dimensional analysis cannot predict the value of the constant, since it depends on the shape and orientation of the body in the flow field. For the particular case of a sphere, Eq. 10–11 can be solved analytically. The details are beyond the scope of this text, but can be found in graduate-level fluid mechanics books (White, 2005; Panton, 2005). It turns out that the constant in the drag equation is equal to 3𝜋 if L is taken as the sphere’s diameter D (Fig. 10–13).
Drag force on a sphere in creeping flow: FD = 3𝜋 𝜇VD (10–12) As a side note, two-thirds of this drag is due to viscous forces and the other one-third is due to pressure forces. This confirms that the viscous terms and the pressure terms in Eq. 10–11 are of the same order of magnitude, as mentioned previously.
Analysis We follow the step-by-step method of repeating variables discussed in Chap. 7; the details are left as an exercise. There are four parameters in this prob lem (n = 4). There are three primary dimensions: mass, length, and time, so we set j = 3 and use independent variables V, L, and 𝜇 as our repeating variables. We expect only one Pi since k = n − j = 4 − 3 = 1, and that Pi must equal a constant. The result is F D = constant · 𝞵VL Thus, we have shown that for creeping flow around any three-dimensional object, the aerodynamic drag force is simply a constant multiplied by 𝜇VL.
Discussion This result is significant, because all that is left to do is find the con stant, which is a function only of the shape of the object.
EXAMPLE 10–2 Terminal Velocity of a Particle from a Volcano A volcano has erupted, spewing stones, steam, and ash several thousand meters into the atmosphere (Fig. 10–14). After some time, the particles begin to settle to the ground. Consider a nearly spherical ash particle of diameter 50 µm, falling in air whose temperature is −50°C and whose pressure is 55 kPa. The density of the particle is 1240 kg/m3. Estimate the terminal velocity of this particle at this altitude.
SOLUTION We are to estimate the terminal velocity of a falling ash particle.
Assumptions 1 The Reynolds number is very small (we will need to verify this assumption after we obtain the solution). 2 The particle is spherical.
Properties At the given temperature and pressure, the ideal gas law gives 𝜌 = 0.8588 kg/m3. Since viscosity is a very weak function of pres sure, we use the value at −50°C and atmospheric pressure, 𝜇 = 1.474 × 10−5 kg/m·s.
Analysis We treat the problem as quasi-steady. Once the falling particle has reached its terminal settling velocity, the net downward force (weight) FIGURE 10–11 A child trying to move in a pool of plastic balls is analogous to a microorganism trying to propel itself without the benefit of inertia.
Photo by Laura Pauley.
FIGURE 10–12 For creeping flow over a three-dimensional object, the aerodynamic drag on the object does not depend on density, but only on speed V, some characteristic size of the object L, and fluid viscosity 𝜇.
V L 휇 FD FIGURE 10–13 The aerodynamic drag on a sphere of diameter D in creeping flow is equal to 3𝜋𝜇VD.
V 휇 FD D cen96537_ch10_519-568.indd 527 29/12/16 3:44 pm 528 APPROXIMATE SOLUTIONS OF THE N–S EQ A consequence of the disappearance of density from the equations of motion for creeping flow is clearly seen in Example 10–2. Namely, air den sity is not important in any calculations except to verify that the Reynolds number is small. (Note that since 𝜌air is so small compared to 𝜌particle, the buoyancy force could have been ignored with negligible loss of accuracy.) Suppose instead that the air density were one-half of the actual density in Example 10–2, but all other properties were unchanged. The terminal veloc ity would be the same (to three significant digits), except that the Reynolds number would be smaller by a factor of 2. Thus, The terminal velocity of a dense, small particle in creeping flow conditions is nearly independent of fluid density, but highly dependent on fluid viscosity.
Since the viscosity of air varies with altitude by only about 25 percent, a small particle settles at nearly constant speed regardless of elevation, even though the air density increases by more than a factor of 10 as the particle falls from an altitude of 50,000 ft (15,000 m) to sea level.
balances the net upward force (aerodynamic drag + buoyancy), as illustrated in Fig. 10–15.
Downward force: Fdown = W = 𝜋 D3 6 ρparticle g (1) The aerodynamic drag force acting on the particle is obtained from Eq. 10–12, and the buoyancy force is the weight of the displaced air. Thus, Upward force: Fup = FD + Fbuoyancy = 3𝜋 𝜇VD + 𝜋 D3 6 ρair g (2) We equate Eqs. 1 and 2, and solve for terminal velocity V, V = D2 18𝜇= (ρparticle −ρair)g = (50 × 10−6 m)2 18(1.474 × 10−5 kg/m·s) (1240 −0.8588) kg/m3 = 0.115 m/s Finally, we verify that the Reynolds number is small enough that creeping flow is an appropriate approximation, Re = ρairVD 𝜇 = (0.8588 kg/m3)(0.115 m/s)(50 × 10−6 m) 1.474 × 10−5 kg/m·s = 0.335 Thus the Reynolds number is less than 1, but certainly not much less than 1.
Discussion Although the equation for creeping flow drag on a sphere (Eq. 10–12) was derived for a case with Re ≪ 1, it turns out that the approximation is reasonable up to Re ≅ 1. A more involved calculation, including a Reynolds number correction and a correction based on the mean free path of air molecules, yields a terminal velocity of 0.110 m/s (Heinsohn and Cimbala, 2003); the error of the creeping flow approxima tion is less than 5 percent.
FIGURE 10–14 Small ash particles spewed from a volcanic eruption settle slowly to the ground; the creeping flow approximation is reasonable for this type of flow field.
Terminal velocity V FIGURE 10–15 A particle falling at a steady terminal velocity has no acceleration; therefore, its weight is balanced by aerodynamic drag and the buoyancy force acting on the particle.
V ρparticle ρair, μair Fbuoyancy D FD W cen96537_ch10_519-568.indd 528 29/12/16 3:44 pm 529 CHAPTER 10 For nonspherical three-dimensional objects, the creeping flow aerodynamic drag is still given by FD = constant⋅𝜇VL; however, the constant is not 3𝜋, but depends on both the shape and orientation of the body. The constant can be thought of as a kind of drag coefficient for creeping flow.
10–4 ■ APPROXIMATION FOR INVISCID REGIONS OF FLOW There is much confusion in the fluid mechanics literature about the word inviscid and the phrase inviscid flow. The apparent meaning of inviscid is not viscous. Inviscid flow would then seem to refer to flow of a fluid with no viscosity. However, that is not what is meant by the phrase inviscid flow! All fluids of engineering relevance have viscosity, regardless of the flow field. Authors who use the phrase inviscid flow actually mean flow of a viscous fluid in a region of the flow in which net viscous forces are negligible compared to pressure and/or inertial forces (Fig. 10–16). Some authors use the phrase “frictionless flow” as a synonym of inviscid flow. This causes more confusion, because even in regions of the flow where net viscous forces are negligible, friction still acts on fluid elements, and there may still be significant viscous stresses. It’s just that these stresses cancel each other out, leaving no significant net viscous force on fluid elements. It can be shown that significant viscous dissipation may also be present in such regions. As is discussed in Section 10–5, fluid elements in an irrota tional region of the flow also have negligible net viscous forces—not because there is no friction, but because the frictional (viscous) stresses cancel each other out. Because of the confusion caused by the terminology, the pres ent authors discourage use of the phrases “inviscid flow” and “frictionless flow.” Instead, we advocate use of the phrases inviscid regions of flow or regions of flow with negligible net viscous forces. Regardless of the terminology used, if net viscous forces are very small compared to inertial and/or pressure forces, the last term on the right side of Eq. 10–6 is negligible. This is true only if 1/Re is small. Thus, inviscid regions of flow are regions of high Reynolds number—the opposite of creep ing flow regions. In such regions, the Navier–Stokes equation (Eq. 10–2) loses its viscous term and reduces to the Euler equation, Euler equation: ρ[ ∂V › ∂t + (V › · ∇ › )V › ] =−∇ › P + ρg› (10–13) The Euler equation is simply the Navier–Stokes equation with the viscous term neglected; it is an approximation of the Navier–Stokes equation. Because of the no-slip condition at solid walls, frictional forces are not negligible in a region of flow very near a solid wall. In such a region, called a boundary layer, the velocity gradients normal to the wall are large enough to offset the small value of 1/Re. An alternate explanation is that the characteristic length scale of the body (L) is no longer the most appropriate length scale inside a boundary layer and must be replaced by a much smaller length scale associated with the distance from the wall. When we define the Reynolds number with this smaller length scale, Re is no longer large, and the viscous term in the Navier–Stokes equation cannot be neglected.
FIGURE 10–16 An inviscid region of flow is a region where net viscous forces are negligible compared to inertial and/or pressure forces because the Reynolds number is large; the fluid itself is still a viscous fluid.
ρ, μ Streamlines ∂V ∂t ρ[ + (V . ) V] = – Δ Δ Δ P + ρg + μ 2V negligible cen96537_ch10_519-568.indd 529 29/12/16 3:44 pm 530 APPROXIMATE SOLUTIONS OF THE N–S EQ A similar argument can be made in the wake of a body, where velocity gradients are relatively large and the viscous terms are not negligible com pared to inertial terms (Fig. 10–17). In practice, therefore, it turns out that The Euler equation approximation is appropriate in high Reynolds number regions of the flow, where net viscous forces are negligible, away from walls and wakes. The term that is neglected in the Euler approximation of the Navier–Stokes equation (𝜇∇2V ›) is the term that contains the highest-order derivatives of velocity. Mathematically, loss of this term reduces the number of boundary conditions that we can specify. It turns out that when we use the Euler equa tion approximation, we cannot specify the no-slip boundary condition at solid walls, although we still specify that fluid cannot flow through the wall (the wall is impermeable). Solutions of the Euler equation are therefore not physi cally meaningful near solid walls, since flow is allowed to slip there. Never theless, as we show in Section 10–6, the Euler equation is often used as the first step in a boundary layer approximation. Namely, the Euler equation is applied over the whole flow field, including regions close to walls and wakes, where we know the approximation is not appropriate. Then, a thin boundary layer is inserted in these regions as a correction to account for viscous effects. Finally, we point out that the Euler equation (Eq. 10–13) is sometimes used as a first approximation in CFD calculations in order to reduce CPU time (and cost).
Derivation of the Bernoulli Equation in Inviscid Regions of Flow In Chap. 5, we derived the Bernoulli equation along a streamline. Here we show an alternative derivation based on the Euler equation. For simplicity, we assume steady incompressible flow. The advective term in Eq. 10–13 can be rewritten through use of a vector identity, Vector identity: (V › · ∇ › )V ›= ∇ › ( V 2 2 ) −V ›× (∇ ›× V › ) (10–14) where V is the magnitude of vector V ›. We recognize the second term in parentheses on the right side as the vorticity vector 𝜁 → (see Chap. 4); thus, (V › · ∇ ›)V ›= ∇ › ( V 2 2 ) −V ›× 𝜁 › and an alternate form of the steady Euler equation is written as ∇ › ( V 2 2 ) −V ›× 𝜁 ›= −∇ › P ρ + g›= ∇ › (−P ρ) + g› (10–15) where we have divided each term by the density and moved 𝜌 within the gradient operator, since density is constant in an incompressible flow. We make the further assumption that gravity acts only in the −z-direction (Fig. 10–18), so that g › = −gk › = −g∇ › z = ∇ › (−gz) (10–16) where we have used the fact that the gradient of coordinate z is unit vector k → in the z-direction. Note also that g is a constant, which allows us to move it FIGURE 10–17 The Euler equation is an approximation of the Navier–Stokes equation, appropriate only in regions of the flow where the Reynolds number is large and where net viscous forces are negligible compared to inertial and/or pressure forces.
Euler equation valid Euler equation not valid FIGURE 10–18 When gravity acts in the −z-direction, gravity vector g› can be written as ∇ ›(−gz).
i ∂z ∂x 0 z = vertical distance k = unit vector in z-direction Thus, g = –gk = –g z = (–gz) (z) = j ∂z ∂y 0 k = k + + ∂z ∂z 1 g Δ Δ Δ cen96537_ch10_519-568.indd 530 29/12/16 3:44 pm 531 CHAPTER 10 (and the negative sign) within the gradient operator. We substitute Eq. 10–16 into Eq. 10–15, and rearrange by combining three terms within one gradient operator, ∇ › ( P ρ + V 2 2 + gz) = V ›× 𝜁 › (10–17) From the definition of the cross product of two vectors, C › = A › × B ›, the vector C › is perpendicular to both A → and B →. The left side of Eq. 10–17 must therefore be a vector everywhere perpendicular to the local velocity vector V ›, since V › appears in the cross product on the right side of Eq. 10–17. Now consider flow along a three-dimensional streamline (Fig. 10–19), which by definition is everywhere parallel to the local velocity vector. At every point along the streamline, ∇ ›(P/𝜌 + V 2/2 + gz) must be perpendicular to the streamline. Now dust off your vector algebra book and recall that the gradi ent of a scalar points in the direction of maximum increase of the scalar. Furthermore, the gradient of a scalar is a vector that points perpendicular to an imaginary surface on which the scalar is constant. Thus, we argue that the scalar (P/𝜌 + V 2/2 + gz) must be constant along a streamline. This is true even if the flow is rotational (𝜁 › ≠ 0). Thus, we have derived a version of the steady incompressible Bernoulli equation, appropriate in regions of flow with negligible net viscous forces, i.e., in so-called inviscid regions of flow.
Steady incompressible Bernoulli equation in inviscid regions of flow: P ρ + V 2 2 + gz = C = constant along streamlines (10–18) Note that the Bernoulli “constant” C in Eq. 10–18 is constant only along a streamline; the constant may change from streamline to streamline. You may be wondering if it is physically possible to have a rotational region of flow that is also inviscid, since rotationality is usually caused by viscosity. Yes, it is possible, and we give one simple example—solid body rotation (Fig. 10–20). Although the rotation may have been generated by viscous forces, a region of flow in solid body rotation has no shear and no net viscous force; it is an inviscid region of flow, even though it is also rotational. As a consequence of the rotational nature of this flow field, Eq. 10–18 applies to every streamline in the flow, but the Bernoulli constant C differs from streamline to streamline, as illustrated in Fig. 10–20.
FIGURE 10–19 Along a streamline, ∇ ›(P/𝜌 + V 2/2 + gz) is a vector everywhere perpendicular to the streamline; hence, P/𝜌 + V 2/2 + gz is constant along the streamline.
Streamline ∇ ρ 2 V2 V 휁 P + + gz x, i z, k y, j FIGURE 10–20 Solid body rotation is an example of an inviscid region of flow that is also rotational. The Bernoulli constant C differs from streamline to streamline but is constant along any particular streamline.
u휃 u휃 = 휔r r P ρ + + gz = C V2 2 C = C1 C = C2 C = C3 EXAMPLE 10–3 Pressure Field in Solid Body Rotation A fluid is rotating as a rigid body (solid body rotation) around the z-axis as illus trated in Fig. 10–20. The steady incompressible velocity field is given by ur = 0, u𝜃 = 𝜔r, and uz = 0. The pressure at the origin is equal to P0. Calculate the pres sure field everywhere in the flow, and determine the Bernoulli constant along each streamline.
SOLUTION For a given velocity field, we are to calculate the pressure field and the Bernoulli constant along each streamline.
cen96537_ch10_519-568.indd 531 29/12/16 3:44 pm 532 APPROXIMATE SOLUTIONS OF THE N–S EQ Assumptions 1 The flow is steady and incompressible. 2 Since there is no flow in the z- (vertical) direction, a hydrostatic pressure distribution exists in the verti cal direction. 3 The entire flow field is approximated as an inviscid region of flow since viscous forces are zero. 4 There is no variation of any flow variable in the 𝜃-direction.
Analysis Equation 10–18 can be applied directly because of assumption 3, Bernoulli equation: P = ρC −1 2 ρV 2 −ρgz (1) where C is the Bernoulli constant that changes radially across streamlines as illustrated in Fig. 10–20. At any radial location r, V2 = 𝜔2r2, and Eq. 1 becomes P = ρC −ρ 𝜔2r2 2 −ρgz (2) At the origin (r = 0, z = 0), the pressure is equal to P0 (from the given boundary condition). Thus we calculate C = C0 at the origin (r = 0), Boundary condition at the origin: P0 = ρC0 → C0 = P0 ρ But how can we find C at an arbitrary radial location r? Equation 2 alone is insufficient since both C and P are unknowns. The answer is that we must use the Euler equation. Since there is no free surface, we employ the modified pressure of Eq. 10–7. The r-component of the Euler equation in cylindrical coordinates (see Eq. 9–62b without the viscous terms) reduces to r-component of Euler equation: ∂Pʹ ∂r = ρ u𝜃 2 r = ρ𝜔2r (3) where we have substituted the given value of u𝜃. Since hydrostatic pressure is already included in the modified pressure, P′ is not a function of z. By assumptions 1 and 4, respectively, P′ is also not a function of t or 𝜃. Thus P′ is a function of r only, and we replace the partial derivative in Eq. 3 with a total derivative. Integra tion yields Modified pressure field: Pʹ = ρ 𝜔2r 2 2 + B1 (4) where B1 is a constant of integration. At the origin, modified pressure P′ is equal to actual pressure P, since z = 0 there. Thus, constant B1 is found by applying the known pressure boundary condition at the origin. It turns out therefore that B1 is equal to P0. We now convert Eq. 4 back to actual pressure using Eq. 10–7, P = P′ − 𝜌gz, Actual pressure field: P = 𝞺 ω2r2 2 + P0 −𝞺gz (5) At the reference datum plane (z = 0), we plot nondimensional pressure as a function of nondimensional radius, where some arbitrary radial location r = R is chosen as a characteristic length scale in the flow (Fig. 10–21). The pressure distribution is parabolic with respect to r. Finally, we equate Eqs. 2 and 5 to solve for C, Bernoulli constant as a function of r : C = P0 𝞺+ ω2r2 (6) At the origin, C = C0 = P0/𝜌, which agrees with our previous calculation.
5 3.5 1.5 1 P – P0 ρ휔2R2 0.5 0 0 0.5 1 1.5 r/R 2 2.5 3 2 4 4.5 2.5 3 FIGURE 10–21 Nondimensional pressure as a function of nondimensional radial location at zero elevation for a fluid in solid body rotation.
cen96537_ch10_519-568.indd 532 29/12/16 3:44 pm 533 CHAPTER 10 10–5 ■ THE IRROTATIONAL FLOW APPROXIMATION As was pointed out in Chap. 4, there are regions of flow in which fluid par ticles have no net rotation; these regions are called irrotational. You must keep in mind that the assumption of irrotationality is an approximation, which may be appropriate in some regions of a flow field, but not in other regions (Fig. 10–22). In general, inviscid regions of flow far away from solid walls and wakes of bodies are also irrotational, although as pointed out previously, there are situations in which an inviscid region of flow may not be irrotational (e.g., solid body rotation). Solutions obtained for the class of flow defined by irrotationality are thus approximations of full Navier– Stokes solutions. Mathematically, the approximation is that vorticity is negligibly small, Irrotational approximation: 𝜁 › = ∇ ›× V › ≅ 0 (10–19) We now examine the effect of this approximation on both the continuity and momentum equations.
Continuity Equation If you shake some more dust off your vector algebra book, you will find a vector identity concerning the curl of the gradient of any scalar function 𝜙, and hence the curl of any vector V ›, Vector identity: ∇ ›× ∇ › 𝜙= 0 Thus, if ∇ ›× V ›= 0, then V ›= ∇ › 𝜙. (10–20) This can easily be proven in Cartesian coordinates (Fig. 10–23), but applies to any orthogonal coordinate system as long as 𝜙 is a smooth function. In words, if the curl of a vector is zero, the vector can be expressed as the gra dient of a scalar function 𝜙, called the potential function. In fluid mechan ics, vector V › is the velocity vector, the curl of which is the vorticity vector 𝜁 ›, and thus we call 𝜙 the velocity potential function. We write For irrotational regions of flow: V ›= ∇ › 𝜙 (10–21) We should point out that the sign convention in Eq. 10–21 is not universal— in some fluid mechanics textbooks, a negative sign is inserted in the definition of the velocity potential function. We state Eq. 10–21 in words as follows: In an irrotational region of flow, the velocity vector can be expressed as the gradient of a scalar function called the velocity potential function.
Regions of irrotational flow are therefore also called regions of potential flow. Note that we have not restricted ourselves to two-dimensional flows; Discussion For a fluid in solid body rotation, the Bernoulli constant increases as r2. This is not surprising, since fluid particles move faster at larger values of r, and thus they possess more energy. In fact, Eq. 5 reveals that pressure itself increases as r2. Physically, the pressure gradient in the (inward) radial direction provides the centripetal force necessary to keep fluid particles revolving about the origin.
FIGURE 10–22 The irrotational flow approximation is appropriate only in certain regions of the flow where the vorticity is negligible.
Irrotational flow region Rotational flow region FIGURE 10–23 The vector identity of Eq. 10–20 is easily proven by expanding the terms in Cartesian coordinates.
Proof of the vector identity: × ϕ = 0 Expand in Cartesian coordinates, The identity is proven if ϕ is a smooth function of x, y, and z.
∇ ∇ × ϕ = ∇ ∇ i ∂2ϕ ∂y ∂z ∂2ϕ ∂z ∂y – j ∂2ϕ ∂z ∂x ∂2ϕ ∂x ∂z – + k = 0 ∂2ϕ ∂x ∂y ∂2ϕ ∂y ∂x – + cen96537_ch10_519-568.indd 533 29/12/16 3:44 pm 534 APPROXIMATE SOLUTIONS OF THE N–S EQ Eq. 10–21 is valid for three-dimensional flow fields, as long as the approxi mation of irrotationality is appropriate in the region of flow under study. In Cartesian coordinates, u = ∂𝜙 ∂x 𝜐= ∂𝜙 ∂y w = ∂𝜙 ∂z (10–22) and in cylindrical coordinates, ur = ∂𝜙 ∂r u𝜃= 1 r ∂𝜙 ∂𝜃 uz = ∂𝜙 ∂z (10–23) The usefulness of Eq. 10–21 becomes apparent when it is substituted into Eq. 10–1, the incompressible continuity equation: ∇ ›⋅V › = 0 → ∇ ›⋅∇ ›𝜙 = 0, or For irrotational regions of flow: ∇2𝜙= 0 (10–24) where the Laplacian operator ∇2 is a scalar operator defined as ∇ ›⋅∇ ›, and Eq. 10–24 is called the Laplace equation. We stress that Eq. 10–24 is valid only in regions where the irrotational flow approximation is reasonable (Fig. 10–24). In Cartesian coordinates, ∇2𝜙= ∂2𝜙 ∂x2 + ∂2𝜙 ∂y2 + ∂2𝜙 ∂z2 = 0 and in cylindrical coordinates, ∇2𝜙= 1 r ∂ ∂r(r ∂𝜙 ∂r ) + 1 r2 ∂2𝜙 ∂𝜃2 + ∂2𝜙 ∂z2 = 0 The beauty of this approximation is that we have combined three unknown velocity components (u, 𝜐, and w, or ur, u𝜃, and uz, depending on our choice of coordinate system) into one unknown scalar variable 𝜙, eliminating two of the equations required for a solution (Fig. 10–25). Once we obtain a solu tion of Eq. 10–24 for 𝜙, we can calculate all three components of the velocity field using Eq. 10–22 or 10–23. The Laplace equation is well known since it shows up in several fields of physics, applied mathematics, and engineering. Various solution techniques, both analytical and numerical, are available in the literature. Solutions of the Laplace equation are dominated by the geometry (i.e., boundary conditions). Although Eq. 10–24 comes from conservation of mass, mass itself (or density, which is mass per unit volume) has dropped out of the equation altogether. With a given set of boundary conditions surrounding the entire irrotational region of the flow field, we can thus solve Eq. 10–24 for 𝜙, regardless of the fluid properties. Once we have calculated 𝜙, we can then calculate V › everywhere in that region of the flow field (using Eq. 10–21), without ever having to solve the Navier–Stokes equation. The solution is valid for any incompressible fluid, regardless of its density or its viscosity, in regions of the flow in which the irrotational approximation is appropriate. The solution is even valid instantaneously for an unsteady flow, since time does not appear in the incompressible continuity equation. In other words, at any instant in time, the incompressible flow field instantly adjusts itself so as to satisfy the Laplace equation and the boundary conditions that exist at that instant in time.
FIGURE 10–24 The Laplace equation for velocity potential function 𝜙 is valid in both two and three dimensions and in any coordinate system, but only in irrotational regions of flow (generally away from walls and wakes).
∇2휙 = 0 ∇2휙 ≠ 0 FIGURE 10–25 In irrotational regions of flow, three unknown scalar components of the velocity vector are combined into one unknown scalar function—the velocity potential function.
General 3-D incompressible flow: • Unknowns = 푢, 푣, 푤, and P • Four equations required Approximation Irrotational region of flow: • Unknowns = 휙 and P • Two equations required cen96537_ch10_519-568.indd 534 29/12/16 3:44 pm 535 CHAPTER 10 Momentum Equation We now turn our attention to the differential linear momentum equation— the Navier–Stokes equation (Eq. 10–2). We have just shown that in an irro tational region of flow, we can obtain the velocity field without application of the Navier–Stokes equation. Why then do we need it at all? The answer is that once we have established the velocity field through use of the velocity potential function, we use the Navier–Stokes equation to solve for the pres sure field. A simplified form of the Navier–Stokes equation is the second required equation mentioned in Fig. 10–25 for solution of two unknowns, 𝜙 and P, in an irrotational region of flow. We begin our analysis by applying the irrotational flow approximation, (Eq. 10–21), to the viscous term of the Navier–Stokes equation (Eq. 10–2). Provided that 𝜙 is a smooth function, that term becomes 𝜇∇2V ›= 𝜇∇2( ∇ › 𝜙) = 𝜇∇ ›(∇2𝜙) = 0 0 where we have applied Eq. 10–24. Thus, the Navier–Stokes equation reduces to the Euler equation in irrotational regions of the flow, For irrotational regions of flow: ρ[ ∂V › ∂t + (V › · ∇ › )V › ] =−∇ › P + ρg› (10–25) We emphasize that although we get the same Euler equation as we did for an inviscid region of flow (Eq. 10–13), the viscous term vanishes here for a different reason, namely, that the flow in this region is assumed to be irrota tional rather than inviscid (Fig. 10–26).
Derivation of the Bernoulli Equation in Irrotational Regions of Flow In Section 10–4 we derived the Bernoulli equation along a streamline for inviscid regions of flow, based on the Euler equation. We now do a simi lar derivation beginning with Eq. 10–25 for irrotational regions of flow. For simplicity, we again assume steady incompressible flow. We use the same vector identity used previously (Eq. 10–14), leading to the alterna tive form of the Euler equation of Eq. 10–15. Here, however, the vorticity vector 𝜁 › is negligibly small since we are considering an irrotational region of flow (Eq. 10–19). Thus, for gravity acting in the negative z-direction, Eq. 10–17 reduces to ∇ › ( P ρ + V 2 2 + gz) = 0 (10–26) We now argue that if the gradient of some scalar quantity (the quantity in parentheses in Eq. 10–26) is zero everywhere, the scalar quantity itself must be a constant. Thus, we generate the Bernoulli equation for irrotational regions of flow, Steady incompressible Bernoulli equation in irrotational regions of flow: P ρ + V 2 2 + gz = C = constant everywhere (10–27) It is useful to compare Eqs. 10–18 and 10–27. In an inviscid region of flow, the Bernoulli equation holds along streamlines, and the Bernoulli constant } FIGURE 10–26 An irrotational region of flow is a region where net viscous forces are negligible compared to inertial and/or pressure forces because of the irrotational approximation. All irrotational regions of flow are therefore also inviscid, but not all inviscid regions of flow are irrotational. The fluid itself is still a viscous fluid in either case.
ρ + (V . ∇)V = –∇P + ρg + 휇∇2V [ ] ρ, 휇 ∂V ∂t Streamlines 0 cen96537_ch10_519-568.indd 535 29/12/16 3:44 pm 536 APPROXIMATE SOLUTIONS OF THE N–S EQ may change from streamline to streamline. In an irrotational region of flow, the Bernoulli constant is the same everywhere, so the Bernoulli equation holds everywhere in the irrotational region of flow, even across streamlines. Thus, the irrotational approximation is more restrictive than the inviscid approximation. A summary of the equations and solution procedure relevant to irrotational regions of flow is provided in Fig. 10–27. In a region of irrotational flow, the velocity field is obtained first by solution of the Laplace equation for veloc ity potential function 𝜙 (Eq. 10–24), followed by application of Eq. 10–21 to obtain the velocity field. To solve the Laplace equation, we must provide boundary conditions for 𝜙 everywhere along the boundary of the flow field of interest. Once the velocity field is known, we use the Bernoulli equation (Eq. 10–27) to obtain the pressure field, where the Bernoulli constant C is obtained from a boundary condition on P somewhere in the flow. Example 10–4 illustrates a situation in which the flow field consists of two separate regions—an inviscid, rotational region and an inviscid, irrota tional region.
FIGURE 10–27 Flowchart for obtaining solutions in an irrotational region of flow. Calculate 휙 from continuity: ∇2휙 = 0 Calculate V from irrotationality: V = ∇휙 Calculate P from Bernoulli: P ρ + + gz = C V2 2 FIGURE 10–28 (a) A horizontal slice through a torna do can be modeled by two regions—an inviscid but rotational inner region of flow (r < R) and an irrotational outer region of flow (r > R). (b) A “bathtub vortex” with air in the inner region and water in the outer region.
Photo (b) © Girish Kumar Rajan. Used by permission.
Inner region Outer region r = R x r P = P∞ 휃 y (a) EXAMPLE 10–4 A Two-Region Model of a Tornado A horizontal slice through a tornado (Fig. 10–28a) is modeled by two dis tinct regions. The inner or core region (0 < r < R) is modeled by solid body rotation—a rotational but inviscid region of flow as discussed earlier. The outer region (r > R) is modeled as an irrotational region of flow. The flow is two-dimensional in the r𝜃-plane, and the components of the velocity field V › = (ur, u𝜃) are given by Velocity components: ur = 0 u𝜃= { 𝜔r 𝜔R2 r 0 < r < R r > R (1) where 𝜔 is the magnitude of the angular velocity in the inner region. The ambient pressure (far away from the tornado) is equal to P∞. Calculate the pressure field in a horizontal slice of the tornado for 0 < r < ∞. What is the pressure at r = 0? Plot the pressure and velocity fields.
SOLUTION We are to calculate the pressure field P(r) in a horizontal radial slice through a tornado for which the velocity components are approximated by Eq. 1. We are also to calculate the pressure in this horizontal slice at r = 0.
Assumptions 1 The flow is steady and incompressible. 2 Although R increases and 𝜔 decreases with increasing elevation z, R and 𝜔 are assumed to be constants when considering a particular horizontal slice. 3 The flow in the horizontal slice is two-dimensional in the r𝜃-plane (no dependence on z and no w-component of velocity). 4 The effects of gravity are negligible within a particular horizontal slice (an additional hydrostatic pressure field exists in the z-direction, of course, but this does not affect the dynamics of the flow, as discussed previously).
Analysis In the inner region, the Euler equation is an appropriate approximation of the Navier–Stokes equation, and the pressure field is found by integration. In Example 10–3 we showed that for solid body rotation, Pressure field in inner region (r < R): P = ρ 𝜔2r 2 2 + P0 (2) cen96537_ch10_519-568.indd 536 29/12/16 3:44 pm 537 CHAPTER 10 where P0 is the (unknown) pressure at r = 0 and we have neglected the gravity term. Since the outer region is a region of irrotational flow, the Bernoulli equa tion is appropriate and the Bernoulli constant is the same everywhere from r = R outward to r → ∞. The Bernoulli constant is found by applying the boundary con dition far from the tornado, namely, as r → ∞, u𝜃 → 0 and P → P∞ (Fig. 10–29). Equation 10–27 yields As r → ∞: P ρ + V 2 2 + gz = C → C = P∞ ρ (3) P∞/𝜌 V→0 as r→∞ assumption 4 The pressure field anywhere in the outer region is obtained by substituting the value of constant C from Eq. 3 into the Bernoulli equation (Eq. 10–27). Neglecting gravity, In outer region (r > R): P = ρC −1 2 ρV 2 = P∞−1 2 ρV 2 (4) We note that V 2 = u𝜃 2. After substitution of Eq. 1 for u𝜃, Eq. 4 reduces to Pressure field in outer region (r > R): P = P∞−𝞺 2 ω2R4 r2 (5) At r = R, the interface between the inner and outer regions, the pressure must be continuous (no sudden jumps in P ), as illustrated in Fig. 10–30. Equating Eqs. 2 and 5 at this interface yields Pressure at r = R: Pr=R = ρ 𝜔2R2 2 + P0 = P∞−ρ 2 𝜔2R 4 R2 (6) from which the pressure P0 at r = 0 is found, Pressure at r = 0: P0 = P∞−𝞺ω2R2 (7) Equation 7 provides the value of pressure in the middle of the tornado—the eye of the storm. This is the lowest pressure in the flow field. Substitution of Eq. 7 into Eq. 2 enables us to rewrite Eq. 2 in terms of the given far-field ambient pressure P∞, In inner region (r < R): P = P∞−𝞺ω2(R2 −r2 2 ) (8) Instead of plotting P as a function of r in this horizontal slice, we plot a nondimen sional pressure distribution instead, so that the plot is valid for any horizontal slice. In terms of nondimensional variables, Inner region (r < R): u𝜃 𝜔R = r R P −P∞ ρ𝜔2R2 = 1 2( r R) 2 −1 Outer region (r > R): u𝜃 𝜔R = R r P −P∞ ρ𝜔2R2 = −1 2( R r ) 2 (9) Figure 10–31 shows both nondimensional tangential velocity and nondimensional pressure as functions of nondimensional radial location.
Discussion In the outer region, pressure increases as speed decreases— a direct result of the Bernoulli equation, which applies with the same Bernoulli constant everywhere in the outer region. You are encouraged to calculate P } } } FIGURE 10–29 A good place to obtain boundary conditions for this problem is the far field; this is true for many problems in fluid mechanics.
Hint of the Day Look to the far field. There you may find what you seek.
FIGURE 10–30 For our model of the tornado to be valid, the pressure can have a discontinuity in slope at r = R, but cannot have a sudden jump of value there; (a) is valid, but (b) is not.
P r r = R P r r = R (a) (b) cen96537_ch10_519-568.indd 537 29/12/16 3:44 pm 538 APPROXIMATE SOLUTIONS OF THE N–S EQ FIGURE 10–31 Nondimensional tangential velocity distribution (blue curve) and nondimensional pressure distribution (black curve) along a horizontal radial slice through a tornado. The inner and outer regions of flow are marked.
1 0.4 0 0 –0.2 –0.4 –0.6 –0.8 –1 0 1 2 3 r/R 4 5 0.6 0.8 0.2 P – P∞ 휌휔2R2 Nondimensional pressure u휃 휔R Outer region Inner region Nondimensional tangential velocity FIGURE 10–32 The lowest pressure occurs at the center of the tornado, and the flow in that region can be approximated by solid body rotation.
Auntie Em!
in the outer region by an alternate method—direct integration of the Euler equa tion without use of the Bernoulli equation; you should get the same result. In the inner region, P increases parabolically with r even though speed also increases; this is because the Bernoulli constant changes from streamline to streamline (as also pointed out in Example 10–3). Notice that even though there is a discontinu ity in the slope of tangential velocity at r /R = 1, the pressure has a fairly smooth transition between the inner and outer regions. The pressure is lowest in the center of the tornado and rises to atmospheric pressure in the far field (Fig. 10–32). The flow in the inner region is rotational but inviscid, since viscosity plays no role in that region of the flow. In fact, the inner region can even consist of a different fluid than the outer region. For example, a tank of water draining through a hole in the bottom of a tank often forms a “bathtub vortex” (Fig. 10–28b) that can be approximated as an inner region of air and an outer region of water. The flow in the outer region is irrotational but viscous. Note, however, that viscosity still acts on fluid particles in the outer region. (Viscosity causes the fluid particles to shear and distort, even though the net viscous force on any fluid particle in the outer region is zero.) Two-Dimensional Irrotational Regions of Flow In irrotational regions of flow, Eqs. 10–24 and 10–21 apply for both two- and three-dimensional flow fields, and we solve for the velocity field in these regions by solving the Laplace equation for velocity potential func tion 𝜙. If the flow is also two-dimensional, we are able to make use of the stream function as well (Fig. 10–33). The two-dimensional approximation is not limited to flow in the xy-plane, nor is it limited to Cartesian coordinates. In fact, we can assume two-dimensionality in any region of the flow where only two directions of motion are important and where there is no signifi cant variation in the third direction. The two most common examples are planar flow (flow in a plane with negligible variation in the direction nor mal to the plane) and axisymmetric flow (flow in which there is rotational symmetry about some axis). We may also choose to work in Cartesian coor dinates, cylindrical coordinates, or spherical polar coordinates, depending on the geometry of the problem at hand.
cen96537_ch10_519-568.indd 538 29/12/16 3:44 pm 539 CHAPTER 10 Planar Irrotational Regions of Flow We consider planar flow first, since it is the simplest. For a steady, incom pressible, planar, irrotational region of flow in the xy-plane in Cartesian coordinates (Fig. 10–34), the Laplace equation for 𝜙 is ∇2𝜙= ∂2𝜙 ∂x2 + ∂2𝜙 ∂y2 = 0 (10–28) For incompressible planar flow in the xy-plane, the stream function 𝜓 is defined as (Chap. 9) Stream function: u = ∂𝜓 ∂y 𝜐= −∂𝜓 ∂x (10–29) Note that Eq. 10–29 holds whether the region of flow is rotational or irrota tional. In fact, the stream function is defined such that it always satisfies the continuity equation, regardless of rotationality. If we restrict our approxima tion to irrotational regions of flow, Eq. 10–19 must also hold; namely, the vorticity is zero or negligibly small. For general two-dimensional flow in the xy-plane, the z-component of vorticity is the only nonzero component. Thus, in an irrotational region of flow, 𝜁z = ∂𝜐 ∂x −∂u ∂y = 0 Substitution of Eq. 10–29 into this equation yields ∂ ∂x (−∂𝜓 ∂x ) −∂ ∂y ( ∂𝜓 ∂y ) = −∂2𝜓 ∂x2 −∂2𝜓 ∂y2 = 0 We recognize the Laplacian operator in this latter equation. Thus, ∇2𝜓 = ∂2𝜓 ∂x2 + ∂2𝜓 ∂y2 = 0 (10–30) We conclude that the Laplace equation is applicable, not only for 𝜙 (Eq. 10–28), but also for 𝜓 (Eq. 10–30) in steady, incompressible, irrotational, planar regions of flow. Curves of constant values of 𝜓 define streamlines of the flow, while curves of constant values of 𝜙 define equipotential lines. (Note that some authors use the phrase equipotential lines to refer to both streamlines and lines of constant 𝜙 rather than exclusively for lines of constant 𝜙.) In pla nar irrotational regions of flow, it turns out that streamlines intersect equi potential lines at right angles, a condition known as mutual orthogonality (Fig. 10–35). In addition, the potential functions 𝜓 and 𝜙 are intimately related to each other—both satisfy the Laplace equation, and from either 𝜓 or 𝜙 we can determine the velocity field. Mathematicians call solutions of 𝜓 and 𝜙 harmonic functions, and 𝜓 and 𝜙 are called harmonic conjugates of each other. Although 𝜓 and 𝜙 are related, their origins are somewhat opposite; it is perhaps best to say that 𝜓 and 𝜙 are complementary to each other: • The stream function is defined by continuity; the Laplace equation for 𝜓 results from irrotationality.
• The velocity potential is defined by irrotationality; the Laplace equation for 𝜙 results from continuity.
FIGURE 10–33 Two-dimensional flow is a subset of three-dimensional flow; in two-dimensional regions of flow we can define a stream function, but we cannot do so in three-dimensional flow. The velocity potential function, however, can be defined for any irrotational region of flow.
3-D irrotational region of flow: • V = ∇휙 • ∇2휙 = 0 • Cannot define 휓 2-D irrotational region of flow: • V = ∇휙 • ∇2휙 = 0 • Can also define 휓 • ∇2휓 = 0 FIGURE 10–34 Velocity components and unit vectors in Cartesian coordinates for planar two-dimensional flow in the xy-plane. There is no variation normal to this plane.
y x y u ʋ x V j i cen96537_ch10_519-568.indd 539 29/12/16 3:44 pm 540 APPROXIMATE SOLUTIONS OF THE N–S EQ In practice, we may perform a potential flow analysis using either 𝜓 or 𝜙, and we should achieve the same results either way. However, it is often more convenient to use 𝜓, since boundary conditions on 𝜓 are usually easier to specify. Planar flow in the xy-plane can also be described in cylindrical coordi nates (r, 𝜃) and (ur, u𝜃), as shown in Fig. 10–36. Again, there is no z-compo nent of velocity, and velocity does not vary in the z-direction. In cylindrical coordinates, Laplace equation, planar flow in (r, 𝜃): 1 r ∂ ∂r(r ∂𝜙 ∂r ) + 1 r2 ∂2𝜙 ∂𝜃2 = 0 (10–31) The stream function 𝜓 for planar flow in Cartesian coordinates is defined by Eq. 10–29, and the irrotationality condition causes 𝜓 to also satisfy the Laplace equation. In cylindrical coordinates we perform a similar analysis. Recall from Chap. 9, Stream function: ur = 1 r ∂𝜓 ∂𝜃 u𝜃= −∂𝜓 ∂r (10–32) It is left as an exercise for you to show that the stream function defined by Eq. 10–32 also satisfies the Laplace equation in cylindrical coordinates for regions of two-dimensional planar irrotational flow. (Verify your results by replacing 𝜙 by 𝜓 in Eq. 10–31 to obtain the Laplace equation for the stream function.) Axisymmetric Irrotational Regions of Flow Axisymmetric flow is a special case of two-dimensional flow that can be described in either cylindrical coordinates or spherical polar coordinates. In cylindrical coordinates, r and z are the relevant spatial variables, and ur and uz are the nonzero velocity components (Fig. 10–37). There is no dependence on angle 𝜃 since rotational symmetry is defined about the z-axis. This is a type of two-dimensional flow because there are only two independent spatial variables, r and z. (Imagine rotating the radial component r in Fig. 10–37 in the 𝜃-direction about the z-axis without changing the magnitude of r.) Because of rotational symmetry about the z-axis, the magnitudes of velocity compo nents ur and uz remain unchanged after such a rotation. The Laplace equa tion for velocity potential 𝜙 for the case of axisymmetric irrotational regions of flow in cylindrical coordinates is 1 r ∂ ∂r (r ∂𝜙 ∂r ) + ∂2𝜙 ∂z2 = 0 In order to obtain expressions for the stream function for axisymmetric flow, we begin with the incompressible continuity equation in r- and z-coordinates, 1 r ∂ ∂r (rur) + ∂uz ∂z = 0 (10–33) After some algebra, we define a stream function that identically satisfies Eq. 10–33, Stream function: ur = −1 r ∂𝜓 ∂z uz = 1 r ∂𝜓 ∂r FIGURE 10–35 In planar irrotational regions of flow, curves of constant 𝜙 (equipotential lines) and curves of constant 𝜓 (streamlines) are mutually orthogonal, meaning that they intersect at 90° angles everywhere.
Streamlines 90° Equipotential lines FIGURE 10–36 Velocity components and unit vectors in cylindrical coordinates for planar flow in the r𝜃-plane. There is no variation normal to this plane.
y x y r x V er eθ θ uθ ur cen96537_ch10_519-568.indd 540 29/12/16 3:45 pm 541 CHAPTER 10 Following the same procedure as for planar flow, we generate an equation for 𝜓 for axisymmetric irrotational regions of flow by forcing the vorticity to be zero. In this case, only the 𝜃-component of vorticity is relevant since the velocity vector always lies in the rz-plane. Thus, in an irrotational region of flow, ∂ur ∂z −∂uz ∂r = ∂ ∂z (−1 r ∂𝜓 ∂z ) −∂ ∂r ( 1 r ∂𝜓 ∂r ) = 0 After taking r outside the z-derivative (since r is not a function of z), we get r ∂ ∂r ( 1 r ∂𝜓 ∂r ) + ∂2𝜓 ∂z2 = 0 (10–34) Note that Eq. 10–34 is not the same as the Laplace equation for 𝜓. You cannot use the Laplace equation for the stream function in axisymmetric irrotational regions of flow (Fig. 10–38).
For planar irrotational regions of flow, the Laplace equation is valid for both 𝜙 and 𝜓; but for axisymmetric irrotational regions of flow, the Laplace equation is valid for 𝜙 but not for 𝜓.
A direct consequence of this statement is that curves of constant 𝜓 and curves of constant 𝜙 in axisymmetric irrotational regions of flow are not mutually orthogonal. This is a fundamental difference between planar and axisymmetric flows. Finally, even though Eq. 10–34 is not the same as the Laplace equation, it is still a linear partial differential equation. This allows us to use the technique of superposition with either 𝜓 or 𝜙 when solving for the flow field in axisymmetric irrotational regions of flow. Superposition is discussed shortly.
Summary of Two-Dimensional Irrotational Regions of Flow Equations for the velocity components for both planar and axisymmetric irrotational regions of flow are summarized in Table 10–2.
FIGURE 10–38 The equation for the stream function in axisymmetric irrotational flow (Eq. 10–34) is not the Laplace equation.
TABLE 10–2 Velocity components for steady, incompressible, irrotational, two-dimensional regions of flow in terms of velocity potential function and stream function in various coordi nate systems Description and Coordinate System Velocity Component 1 Velocity Component 2 Planar; Cartesian coordinates u = ∂𝜙 ∂x = ∂𝜓 ∂y 𝜐= ∂𝜙 ∂y = −∂𝜓 ∂x Planar; cylindrical coordinates ur = ∂𝜙 ∂r = 1 r ∂𝜓 ∂𝜃 u𝜃= 1 r ∂𝜙 ∂𝜃= −∂𝜓 ∂r Axisymmetric; cylindrical coordinates ur = ∂𝜙 ∂r = −1 r ∂𝜓 ∂z uz = ∂𝜙 ∂z = 1 r ∂𝜓 ∂r FIGURE 10–37 Flow over an axisymmetric body in cylindrical coordinates with rotational symmetry about the z-axis. Neither the geometry nor the velocity field depends on 𝜃; and u𝜃 = 0.
z y r r z ur uz Rotational symmetry Axisymmetric body x θ cen96537_ch10_519-568.indd 541 29/12/16 3:45 pm 542 APPROXIMATE SOLUTIONS OF THE N–S EQ Superposition in Irrotational Regions of Flow Since the Laplace equation is a linear homogeneous differential equation, the linear combination of two or more solutions of the equation must also be a solution. For example, if ϕ1 and ϕ2 are each solutions of the Laplace equation, then Aϕ1 + Bϕ2 + C is also a solution, where A, B, and C are arbitrary constants. By extension, you may combine several solutions of the Laplace equation, and the combination is guaranteed to also be a solution. If a region of irrotational flow is modeled by the sum of two or more sep arate irrotational flow fields, e.g., a source located in a free-stream flow, one can simply add the velocity potential functions for each individual flow to describe the combined flow field. This process of adding two or more known solutions to create a third, more complicated solution is known as superposition (Fig. 10–39). For the case of two-dimensional irrotational flow regions, a similar anal ysis can be performed using the stream function rather than the velocity potential function. We stress that the concept of superposition is useful, but is valid only for irrotational flow fields for which the equations for 𝜙 and 𝜓 are linear. You must be careful to ensure that the two flow fields you wish to add vectorially are both irrotational. For example, the flow field for a jet should never be added to the flow field for an inlet or for free-stream flow, because the velocity field associated with a jet is strongly affected by vis cosity, is not irrotational, and cannot be described by potential functions. It also turns out that since the potential function of the composite field is the sum of the potential functions of the individual flow fields, the velocity at any point in the composite field is the vector sum of the velocities of the individual flow fields. We prove this in Cartesian coordinates by consider ing a planar irrotational flow field that is the superposition of two inde pendent planar irrotational flow fields denoted by subscripts 1 and 2. The composite velocity potential function is given by Superposition of two irrotational flow fields: 𝜙= 𝜙1 + 𝜙2 Using the equations for planar irrotational flow in Cartesian coordinates in Table 10–2, the x-component of velocity of the composite flow is u = ∂𝜙 ∂x = ∂(𝜙1 + 𝜙2) ∂x = ∂𝜙1 ∂x + ∂𝜙2 ∂x = u1 + u2 You can generate an analogous expression for 𝜐. Thus, superposition enables us to simply add the individual velocities vectorially at any location in the flow region to obtain the velocity of the composite flow field at that loca tion (Fig. 10–40).
Composite velocity field from superposition: V ›= V › 1 + V › 2 (10–35) Elementary Planar Irrotational Flows Superposition enables us to add two or more simple irrotational flow solu tions to create a more complex (and hopefully more physically significant) flow field. It is therefore useful to establish a collection of elementary building block irrotational flows, with which we can construct a variety of more practical flows (Fig. 10–41). Elementary planar irrotational flows are FIGURE 10–39 Superposition is the process of adding two or more irrotational flow solu tions together to generate a third (more complicated) solution.
ϕ1 + ϕ2 = ϕ + = FIGURE 10–40 In the superposition of two irrotational flow solutions, the two velocity vectors at any point in the flow region add vectorially to produce the composite velocity at that point.
+ = + = V1 V2 V cen96537_ch10_519-568.indd 542 29/12/16 3:45 pm 543 CHAPTER 10 described in xy- and/or r𝜃-coordinates, depending on which pair is more useful in a particular problem.
Building Block 1—Uniform Stream The simplest building block flow we can think of is a uniform stream of flow moving at constant velocity V in the x-direction (left to right). In terms of the velocity potential and stream function (Table 10–2), Uniform stream: u = ∂𝜙 ∂x = ∂𝜓 ∂y = V 𝜐= ∂𝜙 ∂y = −∂𝜓 ∂x = 0 By integrating the first of these with respect to x, and then differentiat ing the result with respect to y, we generate an expression for the velocity potential function for a uniform stream, 𝜙= Vx + f(y) → 𝜐= ∂𝜙 ∂y = f ʹ(y) = 0 → f(y) = constant The constant is arbitrary since velocity components are always derivatives of 𝜙. We set the constant equal to zero, knowing that we can always add an arbitrary constant later on if desired. Thus, Velocity potential function for a uniform stream: 𝜙= Vx (10–36) In a similar manner we generate an expression for the stream function for this elementary planar irrotational flow, Stream function for a uniform stream: 𝜓 = Vy (10–37) Shown in Fig. 10–42 are several streamlines and equipotential lines for a uniform stream. Notice the mutual orthogonality. It is often convenient to express the stream function and velocity potential function in cylindrical coordinates rather than rectangular coordinates, par ticularly when superposing a uniform stream with some other planar irrota tional flow(s). The conversion relations are obtained from the geometry of Fig. 10–36, x = r cos 𝜃 y = r sin 𝜃 r = √x2 + y2 (10–38) From Eq. 10–38 and a bit of trigonometry, we derive relationships for u and 𝜐 in terms of cylindrical coordinates, Transformation: u = ur cos 𝜃−u𝜃 sin 𝜃 𝜐= ur sin 𝜃+ u𝜃 cos 𝜃 (10–39) In cylindrical coordinates, Eqs. 10–36 and 10–37 for 𝜙 and 𝜓 become Uniform stream: 𝜙= Vr cos 𝜃 𝜓 = Vr sin 𝜃 (10–40) We may modify the uniform stream so that the fluid flows uniformly at speed V at an angle of inclination 𝛼 from the x-axis. For this situation, u = V cos 𝛼 and 𝜐 = V sin 𝛼 as shown in Fig. 10–43. It is left as an exercise to show that the velocity potential function and stream function for a uniform stream inclined at angle 𝛼 are Uniform stream inclined at angle 𝛼: 𝜙= V(x cos 𝛼 + y sin 𝛼 ) 𝜓 = V(y cos 𝛼 −x sin 𝛼 ) (10–41) When necessary, Eq. 10–41 can easily be converted to cylindrical coordi nates through use of Eq. 10–38.
FIGURE 10–41 With superposition we build up a complicated irrotational flow field by adding together elementary “building block” irrotational flow fields.
ø4 ø4 ø3 ø3 ø 1 ø 1 ø1 ø1 ø 2 ø 2 ø 4 ø 4 ø 1 ø 1 ø 2 ø 2 ø 2 ø 2 ø 3 ø 3 ø 4 ø 4 ø2 ø2 ø2 ø2 ø 2 ø 2 ø 2 ø 2 ø 3 ø 3 ø 4 ø 4 ø1 ø1 ø2 ø2 ø2 ø2 ø3 ø3 ø4 ø4 ø 1 ø 1 ø 4 ø 4 FIGURE 10–42 Streamlines (solid) and equipotential lines (dashed) for a uniform stream in the x-direction.
ϕ1 ϕ2 y V x ϕ = 0 –ϕ2 –ϕ1 ψ = 0 ψ3 ψ2 ψ1 –ψ1 –ψ2 FIGURE 10–43 Streamlines (solid) and equipotential lines (dashed) for a uniform stream inclined at angle 𝛼.
y x V ψ = 0 ϕ = 0 –ψ2 –ϕ2 –ϕ1 ϕ1 ϕ2 α –ψ1 ψ1 ψ2 cen96537_ch10_519-568.indd 543 29/12/16 3:45 pm 544 APPROXIMATE SOLUTIONS OF THE N–S EQ Building Block 2—Line Source or Line Sink Our second building block flow is a line source. Imagine a line segment of length L parallel to the z-axis, along which fluid emerges and flows uni formly outward in all directions normal to the line segment (Fig. 10–44). The total volume flow rate is equal to V .
. As length L approaches infinity, the flow becomes two-dimensional in planes perpendicular to the line, and the line from which the fluid escapes is called a line source. For an infinite line, V . also approaches infinity; thus, it is more convenient to consider the volume flow rate per unit depth, V .
/L, called the line source strength (often given the symbol m). A line sink is the opposite of a line source; fluid flows into the line from all directions in planes normal to the axis of the line sink. By convention, positive V .
/L signifies a line source and negative V .
/L signifies a line sink. The simplest case occurs when the line source is located at the origin of the xy-plane, with the line itself lying along the z-axis. In the xy-plane, the line source looks like a point at the origin from which fluid is spewed outward in all directions in the plane (Fig. 10–45). At any radial distance r from the line source, the radial velocity component ur is found by applying conservation of mass. Namely, the entire volume flow rate per unit depth from the line source must pass through the circle defined by radius r. Thus, V · L = 2𝜋 rur ur = V ·/L 2𝜋 r (10–42) Clearly, ur decreases with increasing r as we would expect. Notice also that ur is infinite at the origin since r is zero in the denominator of Eq. 10–42. We call this a singular point or a singularity—it is certainly unphysical, but keep in mind that planar irrotational flow is merely an approximation, and the line source is still useful as a building block for superposition in irro tational flow. As long as we stay away from the immediate vicinity of the center of the line source, the rest of the flow field produced by superposi tion of a line source and other building block(s) may still be a good repre sentation of a region of irrotational flow in a physically realistic flow field. We now generate expressions for the velocity potential function and the stream function for a line source of strength V .
/L. We use cylindrical coor dinates, beginning with Eq. 10–42 for ur and also recognize that u𝜃 is zero everywhere. Using Table 10–2, the velocity components are Line source: ur = ∂𝜙 ∂r = 1 r ∂𝜓 ∂𝜃= V ·/L 2𝜋 r u𝜃= 1 r ∂𝜙 ∂𝜃= −∂𝜓 ∂r = 0 To generate the stream function, we (arbitrarily) choose one of these equa tions (we choose the second one), integrate with respect to r, and then dif ferentiate with respect to the other variable 𝜃, ∂𝜓 ∂r = −u𝜃= 0 → 𝜓 = f(𝜃) → ∂𝜓 ∂𝜃= f ʹ(𝜃) = rur = V ·/L 2𝜋 from which we integrate to obtain f(𝜃) = V ·/L 2𝜋 𝜃+ constant Again we set the arbitrary constant of integration equal to zero, since we can add back a constant as desired at any time without changing the flow. FIGURE 10–44 Fluid emerging uniformly from a finite line segment of length L. As L approaches infinity, the flow becomes a line source, and the xy-plane is taken as normal to the axis of the source.
L y x xy-plane z FIGURE 10–45 Line source of strength V .
/L located at the origin in the xy-plane; the total volume flow rate per unit depth through a circle of radius r must equal V .
/L regardless of the value of r.
y x V/L r ur θ ⋅ cen96537_ch10_519-568.indd 544 29/12/16 3:45 pm 545 CHAPTER 10 After a similar analysis for 𝜙, we obtain the following expressions for a line source at the origin: Line source at the origin: 𝜙= V ·/L 2𝜋 ln r and 𝜓 = V ·/L 2𝜋 𝜃 (10–43) Several streamlines and equipotential lines are sketched for a line source in Fig. 10–46. As expected, the streamlines are rays (lines of constant 𝜃), and the equipotential lines are circles (lines of constant r). The streamlines and equipotential lines are mutually orthogonal everywhere except at the origin, which is a singular point. In situations where we would like to place a line source somewhere other than the origin, we must transform Eq. 10–43 carefully. Sketched in Fig. 10–47 is a source located at an arbitrary point (a, b) in the xy-plane. We define r1 as the distance from the source to some point P in the flow, where P is located at (x, y) or (r, 𝜃). Similarly, we define 𝜃1 as the angle from the source to point P, as measured from a line parallel to the x-axis. We analyze the flow as if the source were at a new origin at absolute location (a, b). Equations 10–43 for 𝜙 and 𝜓 are thus still usable, but r and 𝜃 must be replaced by r1 and 𝜃1. Some trigonometry is required to convert r1 and 𝜃1 back to (x, y) or (r, 𝜃). In Cartesian coordinates, for example, Line source at point (a, b): 𝜙= V ·/L 2𝜋 ln r1 = V ·/L 2𝜋 ln √(x −a)2 + (y −b)2 𝜓 = V ·/L 2𝜋 𝜃1 = V ·/L 2𝜋 arctan y −b x −a (10–44) EXAMPLE 10–5 Superposition of a Source and Sink of Equal Strength Consider an irrotational region of flow composed of a line source of strength V .
/L at location (−a, 0) and a line sink of the same strength (but opposite sign) at (a, 0), as sketched in Fig. 10–48. Generate an expression for the stream function in both Cartesian and cylindrical coordinates.
SOLUTION We are to superpose a source and a sink, and generate an expres sion for 𝜓 in both Cartesian and cylindrical coordinates.
Assumptions The region of flow under consideration is incompressible and irrotational.
Analysis We use Eq. 10–44 to obtain 𝜓 for the source, Line source at (−a, 0): 𝜓 1 = V ·/L 2𝜋 𝜃1 where 𝜃1 = arctan y x + a (1) Similarly for the sink, Line sink at (a, 0): 𝜓 2 = −V ·/L 2𝜋 𝜃2 where 𝜃2 = arctan y x −a (2) Superposition enables us to simply add the two stream functions, Eqs. 1 and 2, to obtain the composite stream function, Composite stream function: 𝜓 = 𝜓 1 + 𝜓 2 = V ·/L 2𝜋 (𝜃1 −𝜃2) (3) FIGURE 10–46 Streamlines (solid) and equipotential lines (dashed) for a line source of strength V .
/L located at the origin in the xy-plane. y x r θ ψ1 ϕ1 ϕ2 ϕ3 ψ3 ψ4 ψ5 ψ6 ψ7 ψ8 ψ2 V/L ⋅ FIGURE 10–47 Line source of strength V .
/L located at some arbitrary point (a, b) in the xy-plane.
r a b P r1 θ1 θ y x V/L ⋅ FIGURE 10–48 Superposition of a line source of strength V .
/L at (−a, 0) and a line sink (source of strength −V .
/L) at (a, 0).
– r r2 r1 x y P θ2 θ1 θ a a V/L ⋅ V/L ⋅ cen96537_ch10_519-568.indd 545 29/12/16 3:45 pm 546 APPROXIMATE SOLUTIONS OF THE N–S EQ Building Block 3—Line Vortex Our third building block flow is a line vortex parallel to the z-axis. As with the previous building block, we start with the simple case in which the line vortex is located at the origin (Fig. 10–50). Again we use cylindrical coordinates for convenience. The velocity components are Line vortex: ur = ∂𝜙 ∂r = 1 r ∂𝜓 ∂𝜃= 0 u𝜃= 1 r ∂𝜙 ∂𝜃= −∂𝜓 ∂r = Γ 2𝜋 r (10–45) where Г is called the circulation or the vortex strength. Following the stan dard convention in mathematics, positive Г represents a counterclockwise vortex, while negative Г represents a clockwise vortex. It is left as an exer cise to integrate Eq. 10–45 to obtain expressions for the stream function and the velocity potential function, Line vortex at the origin: 𝜙= Γ 2𝜋 𝜃 𝜓 = −Γ 2𝜋 ln r (10–46) Comparing Eqs. 10–43 and 10–46, we see that a line source and line vortex are somewhat complementary in the sense that the expressions for 𝜙 and 𝜓 are reversed. For situations in which we would like to place the vortex somewhere other than the origin, we must transform Eq. 10–46 as we did for a line source. Sketched in Fig. 10–51 is a line vortex located at an arbitrary point (a, b) We rearrange Eq. 3 and take the tangent of both sides to get tan 2𝜋 𝜓 V ·/L = tan (𝜃1 −𝜃2) = tan 𝜃1 −tan 𝜃2 1 + tan 𝜃1 tan 𝜃2 (4) where we have used a trigonometric identity (Fig. 10–49). We substitute Eqs. 1 and 2 for 𝜃1 and 𝜃2 and perform some algebra to obtain an expression for the stream function, tan 2𝜋 𝜓 V ·/L = y x + a − y x −a 1 + y x + a y x −a = −2ay x2 + y2 −a2 or, taking the arctangent of both sides, Final result, Cartesian coordinates: 𝟁= −V ·/L 2𝜋 arctan 2ay x2 + y2 −a2 (5) We translate to cylindrical coordinates by using Eqs. 10–38, Final result, cylindrical coordinates: 𝟁= −V ·/L 2𝜋 arctan 2ar sin 𝞱 r 2 −a2 (6) Discussion If the source and sink were to switch places, the result would be the same, except that the negative sign on source strength V .
/L would disappear.
FIGURE 10–49 Some useful trigonometric identities.
Useful Trigonometric Identities sin(α + β) = sin α cos β + cos α sin β cos(α + β) = cos α cos β – sin α sin β tan(α + β) = tan α + tan β cot(α + β) = cot β cot α – 1 1– tan α tan β cot β + cot α FIGURE 10–50 Line vortex of strength Г located at the origin in the xy-plane.
θ uθ y r x L FIGURE 10–51 Line vortex of strength Г located at some arbitrary point (a, b) in the xy-plane.
y x P θ1 θ a b r r1 L cen96537_ch10_519-568.indd 546 29/12/16 3:45 pm 547 CHAPTER 10 in the xy-plane. We define r1 and 𝜃1 as previously (Fig. 10–47). To obtain expressions for 𝜙 and 𝜓, we replace r and 𝜃 by r1 and 𝜃1 in Eqs. 10–46 and then transform to regular coordinates, either Cartesian or cylindrical. In Cartesian coordinates, Line vortex at point (a, b): 𝜙= Γ 2𝜋 𝜃1 = Γ 2𝜋 arctan y −b x −a 𝜓 = −Γ 2𝜋 ln r1 = −Γ 2𝜋 ln √(x −a)2 + (y −b)2 (10–47) EXAMPLE 10–6 Velocity in a Flow Composed of Three Components An irrotational region of flow is formed by superposing a line source of strength (V .
/L)1 = 2.00 m2/s at (x, y ) = (0, −1), a line source of strength (V .
/L)2 = −1.00 m2/s at (x, y ) = (1, −1), and a line vortex of strength Г = 1.50 m2/s at (x, y ) = (1, 1), where all spatial coordinates are in meters. [Source number 2 is actually a sink, since (V .
/L)2 is negative.] The locations of the three building blocks are shown in Fig. 10–52. Calculate the fluid velocity at the point (x, y ) = (1, 0).
SOLUTION For the given superposition of two line sources and a vortex, we are to calculate the velocity at the point (x, y ) = (1, 0).
Assumptions 1 The region of flow being modeled is steady, incompressible, and irrotational. 2 The velocity at the location of each component is infinite (they are singularities), and the flow in the vicinity of each of these singularities is unphysi cal; however, these regions are ignored in the present analysis.
Analysis There are several ways to solve this problem. We could sum the three stream functions using Eqs. 10–44 and 10–47, and then take derivatives of the composite stream function to calculate the velocity components. Alternatively, we could do the same for velocity potential function. An easier approach is to recognize that velocity itself can be superposed; we simply add the velocity vectors induced by each of the three individual singularities to form the composite velocity at the given point. This is illustrated in Fig. 10–53. Since the vortex is located 1 m above the point (1, 0), the velocity induced by the vortex is to the right and has a magnitude of Vvortex = Γ 2𝜋 rvortex = 1.50 m2/s 2𝜋 (1.00 m) = 0.239 m/s (1) Similarly, the first source induces a velocity at point (1, 0) at a 45° angle from the x-axis as shown in Fig. 10–53. Its magnitude is Vsource 1 = ∣(V ·/L)1∣ 2𝜋 rsource 1 = 2.00 m2/s 2𝜋 (√2 m) = 0.225 m/s (2) Finally, the second source (the sink) induces a velocity straight down with magni tude Vsource 2 = ∣(V ·/L)2∣ 2𝜋 rsource 2 = ∣−1.00 m2/s∣ 2𝜋 (1.00 m) = 0.159 m/s (3) FIGURE 10–52 Superposition of two line sources and a line vortex in the xy-plane (Example 10–6).
Point of interest (V/L)2 1 0 –1 0 1 y, m x, m rvortex rsource 1 rsource 2 L ⋅ (V/L)1 ⋅ cen96537_ch10_519-568.indd 547 29/12/16 3:45 pm 548 APPROXIMATE SOLUTIONS OF THE N–S EQ Building Block 4—Doublet Our fourth and final building block flow is called a doublet. Although we treat it as a building block for use with superposition, the doublet itself is generated by superposition of two earlier building blocks, namely, a line source and a line sink of equal magnitude, as discussed in Example 10–5. The composite stream function was obtained in that example problem and the result is repeated here: Composite stream function: 𝜓 = −V ·/L 2𝜋 arctan 2ar sin 𝜃 r2 −a2 (10–48) Now imagine that the distance a from the origin to the source and from the origin to the sink approaches zero (Fig. 10–55). You should recall that arctan 𝛽 approaches 𝛽 for very small values of angle 𝛽 in radians. Thus, as distance a approaches zero, Eq. 10–48 reduces to Stream function as a → 0: 𝜓 →−a(V ·/L)r sin 𝜃 𝜋 (r2 −a2) (10–49) If we shrink a while maintaining the same source and sink strengths (V .
/L and −V .
/L), the source and sink cancel each other out when a = 0, leav ing us with no flow at all. However, imagine that as the source and sink approach each other, their strength V .
/L increases inversely with distance a such that the product a(V .
/L) remains constant. In that case, r ≫ a at any point P except very close to the origin, and Eq. 10–49 reduces to Doublet along the x-axis: 𝜓 = −a(V ·/L) 𝜋 sin 𝜃 r = −K sin 𝜃 r (10–50) where we have defined doublet strength K = a(V .
/L)/𝜋 for convenience. The velocity potential function is obtained in similar fashion, Doublet along the x-axis: 𝜙= K cos 𝜃 r (10–51) Several streamlines and equipotential lines for a doublet are plotted in Fig. 10–56. It turns out that the streamlines are circles tangent to the x-axis, and the equipotential lines are circles tangent to the y-axis. The circles inter sect at 90° angles everywhere except at the origin, which is a singular point.
We sum these velocities vectorially by completing the parallelograms, as illustrated in Fig. 10–54. Using Eq. 10–35, the resultant velocity is V ›= V › vortex + V › source 1 + V › source 2 = (0.398i ›+ 0 j ›) m/s (4) 0.239i → m/s ( 0.225 √2 i ›+ 0.225 √2 j › ) m/s −0.159j → m/s The superposed velocity at point (1, 0) is 0.398 m/s to the right.
Discussion This example demonstrates that velocity can be superposed just as stream function or velocity potential function can be superposed. Superposition of velocity is valid in irrotational regions of flow because the differential equations for 𝜙 and 𝜓 are linear; the linearity extends to their derivatives as well.
⏟ ⏟ ⏟ FIGURE 10–53 Induced velocity due to (a) the vortex, (b) source 1, and (c) source 2 (noting that source 2 is negative) (Example 10–6).
y, m x, m Vvortex 0 1 L 0 1 y, m x, m Vsource 1 0 1 –1 0 y, m x, m Vsource 2 0 1 –1 0 (<0) rsource 2 rsource 1 rvortex (c) (b) (a) (V/L)2 ⋅ (V/L)1 ⋅ FIGURE 10–54 Vector summation of the three induced velocities of Example 10–6.
x Vsource 2 Vsource 1 Vvortex Point (1, 0) Resultant velocity cen96537_ch10_519-568.indd 548 29/12/16 3:45 pm 549 CHAPTER 10 If K is negative, the doublet is “backwards,” with the sink located at x = 0− (infinitesimally to the left of the origin) and the source located at x = 0+ (infinitesimally to the right of the origin). In that case all the streamlines in Fig. 10–56 would be identical in shape, but the flow would be in the oppo site direction. It is left as an exercise to construct expressions for a doublet that is aligned at some angle 𝛼 from the x-axis.
Irrotational Flows Formed by Superposition Now that we have a set of building block irrotational flows, we are ready to construct some more interesting irrotational flow fields by the super position technique. We limit our examples to planar flows in the xy-plane; examples of superposition with axisymmetric flows can be found in more advanced textbooks (e.g., Kundu et al., 2011; Panton, 2005; Heinsohn and Cimbala, 2003). Note that even though 𝜓 for axisymmetric irrotational flow does not satisfy the Laplace equation, the differential equation for 𝜓 (Eq. 10–34) is still linear, and thus superposition is still valid.
Superposition of a Line Sink and a Line Vortex Our first example is superposition of a line source of strength V .
/L (V .
/L is a negative quantity in this example) and a line vortex of strength Г, both located at the origin (Fig. 10–57). This represents a region of flow above a drain in a sink or bathtub where fluid spirals in toward the drain. We can superpose either 𝜓 or 𝜙. We choose 𝜓 and generate the composite stream function by adding 𝜓 for a source (Eq. 10–43) and 𝜓 for a line vortex (Eq. 10–46), Superposition: 𝜓 = V ·/L 2𝜋 𝜃−Г 2𝜋 ln r (10–52) To plot streamlines of the flow, we pick a value of 𝜓 and then solve for either r as a function of 𝜃 or 𝜃 as a function of r. We choose the former; after some algebra we get Streamlines: r = exp( (V ·/L)𝜃−2𝜋 𝜓 Г ) (10–53) We pick some arbitrary values for V .
/L and Г so that we can generate a plot; namely, we set V .
/L = −1.00 m2/s and Г = 1.50 m2/s. Note that V .
/L is nega tive for a sink. Also note that the units for V .
/L and Г are obtained easily since we know that the dimensions of stream function in planar flow are {length2/time}. Streamlines are calculated for several values of 𝜓 using Eq. 10–53 and are plotted in Fig. 10–58. The velocity components at any point in this irrotational flow are obtained by differentiating Eq. 10–52, Velocity components: ur = 1 r ∂𝜓 ∂𝜃= V ·/L 2𝜋 r u𝜃= −∂𝜓 ∂r = Γ 2𝜋 r We notice that in this simple example, the radial velocity component is due entirely to the sink, since there is no contribution to radial velocity from the vortex. Similarly, the tangential velocity component is due entirely to the vortex. The composite velocity at any point in the flow is the vector sum of these two components, as sketched in Fig. 10–57.
FIGURE 10–55 A doublet is formed by superposition of a line source at (−a, 0) and a line sink at (a, 0); a decreases to zero while V . /L increases to infinity such that the product aV . /L remains constant.
r r2 r1 x y P θ2 θ1 θ a 0 ∞ a 0 – –∞ V/L ⋅ V/L ⋅ FIGURE 10–56 Streamlines (solid) and equipotential lines (dashed) for a doublet of strength K located at the origin in the xy-plane and aligned with the x-axis.
r x K y θ –ϕ1 ϕ1 –ψ1 ψ1 –ψ2 ψ2 –ψ3 ψ3 –ϕ2 ϕ2 –ϕ3 ϕ3 cen96537_ch10_519-568.indd 549 29/12/16 3:45 pm 550 APPROXIMATE SOLUTIONS OF THE N–S EQ Superposition of a Uniform Stream and a Doublet— Flow over a Circular Cylinder Our next example is a classic in the field of fluid mechanics, namely, the superposition of a uniform stream of speed V∞ and a doublet of strength K located at the origin (Fig. 10–59). We superpose the stream function by adding Eq. 10–40 for a uniform stream and Eq. 10–50 for a doublet at the origin. The composite stream function is thus Superposition: 𝜓 = V∞r sin 𝜃−K sin 𝜃 r (10–54) For convenience we set 𝜓 = 0 when r = a (the reason for this will soon become apparent). Equation 10–54 if then solved for doublet strength K, Doublet strength: K = V∞a2 and Eq. 10–54 becomes Alternate form of stream function: 𝜓 = V∞ sin 𝜃(r −a2 r ) (10–55) It is clear from Eq. 10–55 that one of the streamlines (𝜓 = 0) is a circle of radius a (Fig. 10–60). We can plot this and other streamlines by solving Eq. 10–55 for r as a function of 𝜃 or vice versa. However, as you should be aware by now, it is usually better to present results in terms of nondimen sional parameters. By inspection, we define three nondimensional parameters, 𝜓 = 𝜓 V∞a r = r a 𝜃 where angle 𝜃 is already dimensionless. In terms of these parameters, Eq. 10–55 is written as 𝜓 = sin 𝜃(r −1 r) (10–56) We solve Eq. 10–56 for r as a function of 𝜃 through use of the quadratic rule, Nondimensional streamlines: r = 𝜓 ± √(𝜓 )2 + 4 sin2 𝜃 2 sin 𝜃 (10–57) Using Eq. 10–57, we plot several nondimensional streamlines in Fig. 10–61. Now you see why we chose the circle r = a (or r = 1) as the zero streamline— this streamline can be thought of as a solid wall, and this flow represents potential flow over a circular cylinder. Not shown are streamlines inside the circle—they exist, but are of no concern to us. There are two stagnation points in this flow field, one at the nose of the cylinder and one at the tail. Streamlines near the stagnation points are far apart since the flow is very slow there. By contrast, streamlines near the top and bottom of the cylinder are close together, indicating regions of fast flow. Physically, fluid must accelerate around the cylinder since it is acting as an obstruction to the flow. Notice also that the flow is symmetric about both the x- and y-axes. While top-to-bottom symmetry is not surprising, fore-to-aft symmetry is perhaps unexpected, since we know that real flow around a cylinder generates a FIGURE 10–57 Superposition of a line source of strength V .
/L and a line vortex of strength Г located at the origin. Vector velocity addition is shown at some arbitrary location in the xy-plane.
x y V Vvortex Vsink L V/L (V/L is negative here.) ⋅ ⋅ FIGURE 10–58 Streamlines created by superposition of a line sink and a line vortex at the origin. Values of 𝜓 are in units of m2/s.
2 1 y, m 0 –1 –2 –2 x, m –1 0 1 2 0.8 0.9 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 FIGURE 10–59 Superposition of a uniform stream and a doublet; vector velocity addition is shown at some arbitrary location in the xy-plane.
y K x V V∞ Vuniform stream Vdoublet cen96537_ch10_519-568.indd 550 29/12/16 3:45 pm 551 CHAPTER 10 wake region behind the cylinder, and the streamlines are not symmetric. However, we must keep in mind that the results here are only approxima tions of a real flow. We have assumed irrotationality everywhere in the flow field, and we know that this approximation is not true near walls and in wake regions. We calculate the velocity components everywhere in the flow field by dif ferentiating Eq. 10–55, ur = 1 r ∂𝜓 ∂𝜃= V ∞ cos 𝜃(1 −a2 r2) u𝜃= −∂𝜓 ∂r = −V ∞ sin 𝜃(1 + a2 r2) (10–58) A special case is on the surface of the cylinder itself (r = a), where Eqs. 10–58 reduce to On the surface of the cylinder: ur = 0 u𝜃= −2V ∞ sin 𝜃 (10–59) Since the no-slip condition at solid walls cannot be satisfied when making the irrotational approximation, there is slip at the cylinder wall. In fact, at the top of the cylinder (𝜃 = 90°), the fluid speed at the wall is twice that of the free stream.
EXAMPLE 10–7 Pressure Distribution on a Circular Cylinder Using the irrotational flow approximation, calculate and plot the nondimensional static pressure distribution on the surface of a circular cylinder of radius a in a uni form stream of speed V∞ (Fig. 10–62). Discuss the results. The pressure far away from the cylinder is P∞.
SOLUTION We are to calculate and plot the nondimensional static pressure dis tribution along the surface of a circular cylinder in a free-stream flow.
Assumptions 1 The region of flow being modeled is steady, incompressible, and irrotational. 2 The flow field is two-dimensional in the xy-plane.
Analysis First of all, static pressure is the pressure that would be measured by a pressure probe moving with the fluid. Experimentally, we measure this pressure on a surface through use of a static pressure tap, which is basically a tiny hole drilled normal to the surface (Fig. 10–63). At the other end of the tap is a tube leading to a pressure measuring device. Experimental data of the static pressure distribution along the surface of a cylinder are available in the literature, and we compare our results to some of those experimental data. From Chap. 7 we recognize that the appropriate nondimensional pressure is the pressure coefficient, Pressure coefficient: Cp = P −P ∞ 1 2 ρV∞ 2 (1) Since the flow in the region of interest is irrotational, we use the Bernoulli equa tion (Eq. 10–27) to calculate the pressure anywhere in the flow field. Ignoring the effects of gravity, Bernoulli equation: P ρ + V 2 2 = constant = P∞ ρ + V∞ 2 2 (2) FIGURE 10–60 Superposition of a uniform stream and a doublet yields a streamline that is a circle.
y K r = a x V∞ ψ = 0 FIGURE 10–61 Nondimensional streamlines created by superposition of a uniform stream and a doublet at the origin; 𝜓 = 𝜓/(V∞a), Δ𝜓 = 0.2, x = x/a, and y = y/a, where a is the cylinder radius.
2 1 0 –1 –2 –2 –1 0 1 2 y x ψ = 0 1 FIGURE 10–62 Planar flow over a circular cylinder of radius a immersed in a uniform stream of speed V∞ in the xy-plane. Angle 𝛽 is defined from the front of the cylinder by convention.
y a V∞ β = π – θ θ β x cen96537_ch10_519-568.indd 551 29/12/16 3:45 pm 552 APPROXIMATE SOLUTIONS OF THE N–S EQ Rearranging Eq. 2 into the form of Eq. 1, we get Cp = P −P ∞ 1 2 ρV ∞ 2 = 1 −V 2 V ∞ 2 (3) We substitute our expression for tangential velocity on the cylinder surface, Eq. 10–59, since along the surface V 2 = u𝜃 2; Eq. 3 becomes Surface pressure coefficient: Cp = 1 −(−2V ∞ sin 𝜃)2 V ∞ 2 = 1 −4 sin2 𝜃 In terms of angle 𝛽, defined from the front of the body (Fig. 10–62), we use the transformation 𝛽 = 𝜋 − 𝜃 to obtain Cp in terms of angle 𝛽: Cp = 1 −4 sin2 𝞫 (4) We plot the pressure coefficient on the top half of the cylinder as a function of angle 𝛽 in Fig. 10–64, solid blue curve. (Because of top–bottom symmetry, there is no need to also plot the pressure distribution on the bottom half of the cylin der.) The first thing we notice is that the pressure distribution is symmetric fore and aft. This is not surprising since we already know that the streamlines are also symmetric fore and aft (Fig. 10–61). The front and rear stagnation points (at 𝛽 = 0° and 180°, respectively) are labeled SP on Fig. 10–64. The pressure coefficient is unity there, and these two points have the highest pressure in the entire flow field. In physical vari ables, static pressure P at the stagnation points is equal to P∞ + 𝜌V ∞ 2 /2. In other words, the full dynamic pressure (also called impact pressure) of the oncoming fluid is felt as a static pressure on the nose of the body as the fluid is decelerated to zero speed at the stagnation point. At the very top of the cylinder (𝛽 = 90°), the speed along the surface is twice the free-stream velocity (V = 2V∞), and the pressure coefficient is lowest there (Cp = −3). Also marked on Fig. 10–64 are the two locations where Cp = 0, namely at 𝛽 = 30° and 150°. At these locations, the static pressure along the surface is equal to that of the free stream (P = P∞).
Discussion Typical experimental data for laminar and turbulent flow over the surface of a circular cylinder are indicated by the green circles and red circles, respectively, in Fig. 10–64. It is clear that near the front of the cylinder, the irrota tional flow approximation is excellent. However, for 𝛽 greater than about 60°, and especially near the rear portion of the cylinder (right side of the plot), the irrota tional flow results do not match well at all with experimental data. In fact, it turns out that for flow over bluff body shapes like this, the irrotational flow approxima tion usually does a fairly good job on the front half of the body, but a very poor job on the rear half of the body. The irrotational flow approximation agrees better with experimental turbulent data than with experimental laminar data; this is because flow separation occurs farther downstream for the case with a turbulent boundary layer, as discussed in more detail in Section 10–6.
FIGURE 10–63 Static pressure on a surface is measured through use of a static pressure tap connected to a pressure manometer or electronic pressure transducer.
Body surface To pressure transducer Flexible tubing P Pressure tap FIGURE 10–64 Pressure coefficient as a function of angle 𝛽 along the surface of a circular cylinder; the solid blue curve is the irrotational flow approximation, green circles are from experimental data at Re = 2 × 105 − laminar boundary layer separation, and red circles are from typical experimental data at Re = 7 × 105 − turbulent boundary layer separation.
Data from Kundu et al. (2011).
90 1 0 –1 –2 –3 0 30 60 120 150 180 Free-stream pressure Cp β, degrees Front SP Rear SP Top One immediate consequence of the symmetry of the pressure distribution in Fig. 10–64 is that there is no net pressure drag on the cylinder (pressure forces in the front half of the body are exactly balanced by those on the rear half of the body). In this irrotational flow approximation, the pressure fully recovers at the rear stagnation point, so that the pressure there is the same as that at the front stagnation point. We also predict that there is no net viscous cen96537_ch10_519-568.indd 552 29/12/16 3:45 pm 553 CHAPTER 10 drag on the body, since we cannot specify the no-slip condition on the body surface when we make the irrotational approximation. Hence, the net aero dynamic drag on the cylinder in irrotational flow is identically zero. This is one example of a more general statement that applies to bodies of any shape (even unsymmetrical shapes) when the irrotational flow approximation is made, namely, the famous paradox first stated by Jean-le-Rond d’Alembert (1717–1783) in the year 1752: D’Alembert’s paradox: With the irrotational flow approximation, the aerodynamic drag force on any nonlifting body of any shape immersed in a uniform stream is zero.
D’Alembert recognized the paradox of his statement, of course, knowing that there is aerodynamic drag on real bodies immersed in real fluids. In a real flow, the pressure on the back surface of the body is significantly less than that on the front surface, leading to a nonzero pressure drag on the body. This pressure difference is enhanced if the body is bluff and there is flow separation, as sketched in Fig. 10–65. Even for streamlined bodies, however (such as airplane wings at low angles of attack), the pressure near the back of the body never fully recovers. In addition, the no-slip condition on the body surface leads to a nonzero viscous drag as well. Thus, the irro tational flow approximation falls short in its prediction of aerodynamic drag for two reasons: it predicts no pressure drag and it predicts no viscous drag. The pressure distribution at the front end of any rounded body shape is qualitatively similar to that plotted in Fig. 10–64. Namely, the pressure at the front stagnation point (SP) is the highest pressure on the body: PSP = P∞ + 𝜌V 2/2, where V is the free-stream velocity (we have dropped the sub script ∞), and Cp = 1 there. Moving downstream along the body surface, pressure drops to some minimum value for which P is less than P∞ (Cp < 0). This point, where the velocity just above the body surface is largest and the pressure is smallest, is often called the aerodynamic shoulder of the body. Beyond the shoulder, the pressure slowly rises. With the irrotational flow approximation, the pressure always rises back to the dynamic pressure at the rear stagnation point, where Cp = 1. However, in a real flow, the pressure never fully recovers, leading to pressure drag as discussed previously. Somewhere between the front stagnation point and the aerodynamic shoul der is a point on the body surface where the speed just above the body is equal to V, the pressure P is equal to P∞, and Cp = 0. This point is called the zero pressure point, where the phrase is obviously based on gage pressure, not absolute pressure. At this point, the pressure acting normal to the body surface is the same (P = P∞), regardless of how fast the body moves through the fluid. This fact is a factor in the location of fish eyes (Fig. 10–66). If a fish’s eye were located closer to its nose, the eye would experience an increase in water pressure as the fish swims—the faster it would swim, the higher the water pressure on its eye would be. This would cause the soft eyeball to dis tort, affecting the fish’s vision. Likewise, if the eye were located farther back, near the aerodynamic shoulder, the eye would experience a relative suction pressure when the fish would swim, again distorting its eyeball and blurring its vision. Experiments have revealed that the fish’s eye is instead located very close to the zero-pressure point where P = P∞, and the fish can swim at any speed without distorting its vision. Incidentally, the back of the gills FIGURE 10–65 (a) D’Alembert’s paradox is that the aerodynamic drag on any nonlifting body of any shape is predicted to be zero when the irrotational flow approximation is invoked; (b) in real flows there is a nonzero drag on bodies immersed in a uniform stream.
Irrotational flow approximation (a) (b) Real (rotational) flow field Aerodynamic drag = 0 Aerodynamic drag ≠ 0 V FD V FIGURE 10–66 A fish’s body is designed such that its eye is located near the zero-pressure point so that its vision is not distorted while it swims. Data shown are along the side of a bluefish. 1.0 Cp 0.5 0.0 –0.5 cen96537_ch10_519-568.indd 553 29/12/16 3:45 pm 554 APPROXIMATE SOLUTIONS OF THE N–S EQ is located near the aerodynamic shoulder so that the suction pressure there helps the fish to “exhale.” The heart is also located near this lowest-pressure point to increase the heart’s stroke volume during rapid swimming. If we think about the irrotational flow approximation a little more closely, we realize that the circle we modeled as a solid cylinder in Example 10–7 is not really a solid wall at all—it is just a streamline in the flow field that we are modeling as a solid wall. The particular streamline we model as a solid wall just happens to be a circle. We could have just as easily picked some other streamline in the flow to model as a solid wall. Since flow cannot cross a streamline by definition, and since we cannot satisfy the no-slip condition at a wall, we state the following: With the irrotational flow approximation, any streamline can be thought of as a solid wall.
For example, we can model any streamline in Fig. 10–61 as a solid wall. Let’s take the first streamline above the circle, and model it as a wall. (This streamline has a nondimensional value of 𝜓 = 0.2.) Several streamlines are plotted in Fig. 10–67; we have not shown any stream lines below the streamline 𝜓 = 0.2—they are still there, it’s just that we are no longer concerned with them. What kind of flow does this repre sent? Well, imagine wind flowing over a hill; the irrotational approxima tion shown in Fig. 10–67 is representative of this flow. We might expect inconsistencies very close to the ground, and perhaps on the downstream side of the hill, but the approximation is probably very good on the front side of the hill. You may have noticed a problem with this kind of superposition. Namely, we perform the superposition first, and then try to define some physical problems that might be modeled by the flow we generate. While useful as a learning tool, this technique is not always practical in real-life engineer ing. For example, it is unlikely that we will encounter a hill shaped exactly like the one modeled in Fig. 10–67. Instead, we usually already have a geometry and wish to model flow over or through this geometry. There are more sophisticated superposition techniques available that are better suited to engineering design and analysis. Namely, there are techniques in which numerous sources and sinks are placed at appropriate locations so as to model flow over a predetermined geometry. These techniques can even be extended to fully three-dimensional irrotational flow fields, but require a com puter because of the amount of calculations involved (Kundu et al., 2011). We do not discuss these techniques here.
FIGURE 10–67 The same nondimensionalized streamlines as in Fig. 10–61, except streamline 𝜓 = 0.2 is modeled as a solid wall. This flow represents flow of air over a symmetric hill.
4 3 2 y 1 0 –2 –1 0 x 1 2 1 ψ = 0.2 Floor Sink w (a) (b) y x b Floor y L z x b V ⋅ FIGURE 10–68 Vacuum cleaner hose with floor attachment; (a) three-dimensional view with floor in the xz-plane, and (b) view of a slice in the xy-plane with suction modeled by a line sink.
EXAMPLE 10–8 Flow into a Vacuum Cleaner Attachment Consider the flow of air into the floor attachment nozzle of a typical household vacuum cleaner (Fig. 10–68a). The width of the nozzle inlet slot is w = 2.0 mm, and its length is L = 35.0 cm. The slot is held a distance b = 2.0 cm above the floor, as shown. The total volume flow rate through the vacuum hose is V . = 0.110 m3/s. Predict the flow field in the center plane of the attachment (the xy-plane in Fig. 10–68a). Specifically, plot several streamlines and calculate the velocity and pressure distribution along the x-axis. What is the maximum speed along the floor, and where does it occur? Where along the floor is the vacuum cleaner most effective?
cen96537_ch10_519-568.indd 554 29/12/16 3:45 pm 555 CHAPTER 10 SOLUTION We are to predict the flow field in the center plane of a vacuum cleaner attachment, plot velocity and pressure along the floor (x-axis), predict the location and value of the maximum velocity along the floor, and predict where along the floor the vacuum cleaner is most effective.
Assumptions 1 The flow is steady and incompressible. 2 The flow in the xy-plane is two-dimensional (planar). 3 The majority of the flow field is irrota tional. 4 The room is infinitely large and free of air currents that might influence the flow.
Analysis We approximate the slot on the vacuum cleaner attachment as a line sink (a line source with negative source strength), located at distance b above the x-axis, as sketched in Fig. 10–68b. With this approximation, we are ignor ing the finite width of the slot (w); instead we model flow into the slot as flow into the line sink, which is simply a point in the xy-plane at (0, b). We are also ignoring any effects of the hose or the body of the attachment. The strength of the line source is obtained by dividing total volume flow rate by the length L of the slot, Strength of line source: V · L = −0.110 m3/s 0.35 m = −0.314 m2/s (1) where we include a negative sign since this is a sink instead of a source. Clearly this line sink by itself (Fig. 10–68b) is not sufficient to model the flow, since air would flow into the sink from all directions, including up through the floor. To avoid this problem, we add another elementary irrotational flow (building block) to model the effect of the floor. A clever way to do this is through the method of images. With this technique, we place a second identi cal sink below the floor at point (0, −b). We call this second sink the image sink. Since the x-axis is now a line of symmetry, the x-axis is itself a streamline of the flow, and hence can be thought of as the floor. The irrotational flow field to be analyzed is sketched in Fig. 10–69. Two sources of strength V .
/L are shown. The top one is called the flow source, and represents suction into the vacuum cleaner attachment. The bottom one is the image source. Keep in mind that source strength V .
/L is negative in this problem (Eq. 1), so that both sources are actually sinks. We use superposition to generate the stream function for the irrotational approximation of this flow field. The algebra here is similar to that of Example 10–5; in that case we had a source and a sink on the x-axis, while here we have two sources on the y-axis. We use Eq. 10–44 to obtain 𝜓 for the flow source, Line source at (0, b): 𝜓 1 = V ·/L 2𝜋 𝜃1 where 𝜃1 = arctan y −b x (2) Similarly for the image source, Line source at (0, −b): 𝜓 2 = V ·/L 2𝜋 𝜃2 where 𝜃2 = arctan y + b x (3) Superposition enables us to simply add the two stream functions, Eqs. 2 and 3, to obtain the composite stream function, Composite stream function: 𝜓 = 𝜓 1 + 𝜓 2 = V ·/L 2𝜋 (𝜃1 + 𝜃2) (4) FIGURE 10–69 Superposition of a line source of strength V .
/L at (0, b) and a line source of the same strength at (0, −b). The bottom source is a mirror image of the top source, making the x-axis a streamline.
θ θ1 y x Flow source Floor Image source P b b θ2 r2 r1 r V/L ⋅ V/L ⋅ cen96537_ch10_519-568.indd 555 29/12/16 3:45 pm 556 APPROXIMATE SOLUTIONS OF THE N–S EQ We rearrange Eq. 4 and take the tangent of both sides to get tan 2𝜋 𝜓 V ·/L = tan(𝜃1 + 𝜃2) = tan 𝜃1 + tan 𝜃2 1 −tan 𝜃1tan 𝜃2 (5) where we have again used a trigonometric identity (Fig. 10–49). We substitute Eqs. 2 and 3 for 𝜃1 and 𝜃2 and perform some algebra to obtain our final expression for the stream function in Cartesian coordinates, 𝜓 = V ·/L 2𝜋 arctan 2xy x2 −y2 + b2 (6) We translate to cylindrical coordinates using Eq. 10–38 and nondimensionalize. After some algebra, Nondimensional stream function: 𝜓 = arctan sin 2 𝜃 cos 2 𝜃+ 1/r2 (7) where 𝜓 = 2𝜋𝜓/(V .
/L), r = r /b, and we used trigonometric identities from Fig. 10–49. Because of symmetry about the x-axis, all the air that is produced by the upper line source must remain above the x-axis. Likewise, all the image air that is pro duced at the lower line source must remain below the x-axis. If we were to color air from the upper (north) source blue, and air from the lower (south) source gray (Fig. 10–70), all the blue air would stay above the x-axis, and all the gray air would stay below the x-axis. Thus, the x-axis acts as a dividing streamline, separating the blue from the gray. Furthermore, recall from Chap. 9 that the dif ference in value of 𝜓 from one streamline to the next in planar flow is equal to the volume flow rate per unit width flowing between the two streamlines. We set 𝜓 equal to zero along the positive x-axis. Following the left-side convention, introduced in Chap. 9, we know that 𝜓 on the negative x-axis must equal the total volume flow rate per unit width produced by the upper line source, i.e., V .
/L. Namely, 𝜓 −x-axis −𝜓 +x-axis = V ·/L → 𝜓−x-axis = 2𝜋 (8) These streamlines are labeled in Fig. 10–70. In addition, the nondimensional streamline 𝜓 = 𝜋 is also labeled. It coincides with the y-axis since there is sym metry about that axis as well. The origin (0, 0) is a stagnation point, since the velocity induced by the lower source exactly cancels out that induced by the upper source. For the case of the vacuum cleaner being modeled here, the source strengths are negative (they are sinks). Thus, the direction of flow is reversed, and the values of 𝜓 are of opposite sign to those in Fig. 10–70. Using the left-side convention again, we plot the nondimensional stream function for −2𝜋 < 𝜓 < 0 (Fig. 10–71). To do so, we solve Eq. 7 for r as a function of 𝜃 for various values of 𝜓, Nondimensional streamlines: r = ±√ tan 𝜓 sin 2𝜃−cos 2 𝜃 tan 𝜓 (9) Only the upper half is plotted, since the lower half is symmetric and is merely the mirror image of the upper half. For the case of negative V .
/L, air gets sucked into the vacuum cleaner from all directions as indicated by the arrows on the streamlines. To calculate the velocity distribution on the floor (the x-axis), we can either differentiate Eq. 6 and apply the definition of stream function for planar flow 0 FIGURE 10–70 The x-axis is the dividing streamline that separates air produced by the top source (blue) from air produced by the bottom source (gray).
y x ψ = 0 ψ = π ψ = 2π V/L ⋅ V/L ⋅ FIGURE 10–71 Nondimensional streamlines for the two sources of Fig. 10–69 for the case in which the source strengths are negative (they are sinks). 𝜓 is incremented uniformly from −2𝜋 (negative x-axis) to 0 (positive x-axis), and only the upper half of the flow is shown. The flow is toward the sink at location (0, 1).
3 2 1 0 –2 –1 0 1 x ψ = –2π ψ = –π ψ = 0 y 2 4 cen96537_ch10_519-568.indd 556 29/12/16 3:45 pm 557 CHAPTER 10 (Eq. 10–29), or we can do a vector summation. The latter is simpler and is illustrated in Fig. 10–72 for an arbitrary location along the x-axis. The induced velocity from the upper source (or sink) has magnitude (V .
/L)/(2𝜋r1), and its direction is in line with r1 as shown. Because of symmetry, the induced velocity from the image source has identical magnitude, but its direction is in line with r2. The vector sum of these two induced velocities lies along the x-axis since the two horizontal components add together, but the two vertical components cancel each other out. After a bit of trigonometry, we conclude that Axial velocity along the x-axis: u = V = (V ·/L)x 𝜋 (x2 + b2) (10) where V is the magnitude of the resultant velocity vector along the floor as sketched in Fig. 10–72. Since we have made the irrotational flow approximation, the Bernoulli equation can be used to generate the pressure field. Ignoring gravity, Bernoulli equation: P ρ + V 2 2 = constant = P∞ ρ + V∞ 2 2 (11) To generate a pressure coefficient, we need a reference velocity for the denominator. Having none, we generate one from the known parameters, namely Vref = −(V .
/L)/b, where we insert the negative sign to make Vref positive (since V .
/L is negative for our model of the vacuum cleaner). Then we define Cp as Pressure coefficient: Cp = P −P∞ 1 2 ρVref 2 = −V 2 Vref 2 = −b2V 2 (V ·/L)2 (12) where we have also applied Eq. 11. Substituting Eq. 10 for V, we get Cp = − b2x2 𝜋 2(x2 + b2)2 (13) We introduce nondimensional variables for axial velocity and distance, Nondimensional variables: u = u Vref = −ub V ·/L x = x b (14) We note that Cp is already nondimensional. In dimensionless form, Eqs. 10 and 13 become Along the floor: u = −1 𝜋 x 1 + x2 Cp = −( 1 𝜋 x 1 + x2) 2 = −u2 (15) Curves showing u and Cp as functions of x are plotted in Fig. 10–73. We see from Fig. 10–73 that u increases slowly from 0 at x = −∞ to a maximum value of about 0.159 at x = −1. The velocity is positive (to the right) for negative values of x as expected since air is being sucked into the vacuum cleaner. As speed increases, pressure decreases; Cp is 0 at x = −∞ and decreases to its minimum value of about −0.0253 at x = −1. Between x = −1 and x = 0 the speed decreases to zero while the pressure increases to zero at the stag nation point directly below the vacuum cleaner nozzle. To the right of the nozzle (positive values of x), the velocity is antisymmetric, while the pressure is sym metric. The maximum speed (minimum pressure) along the floor occurs at x = ±1, which is the same distance as the height of the nozzle above the floor ⏟ 0 FIGURE 10–72 Vector sum of the velocities induced by the two sources; the resultant velocity is horizontal at any location on the x-axis due to symmetry.
y b b 2πr1 2πr2 V r1 r2 Image source Suction source Floor x V/L ⋅ V/L ⋅ V/L ⋅ V/L ⋅ cen96537_ch10_519-568.indd 557 29/12/16 3:45 pm 558 APPROXIMATE SOLUTIONS OF THE N–S EQ (Fig. 10–74). In dimensional terms, the maximum speed along the floor occurs at x = ±b, and the speed there is Maximum speed along the floor: ∣u∣max = −∣u∣max V ·/L b = −0.159( −0.314 m2/s 0.020 m ) = 2.50 m/s (16) We expect that the vacuum cleaner is most effective at sucking up dirt from the floor when the speed along the floor is greatest and the pressure along the floor is lowest. Thus, contrary to what you may have thought, the best perfor mance is not directly below the suction inlet, but rather at x = ±b, as illustrated in Fig. 10–74.
Discussion Notice that we never used the width w of the vacuum nozzle in our analysis, since a line sink has no length scale. You can convince yourself that a vacuum cleaner works best at x ≅ ±b by performing a simple experiment with a vacuum cleaner and some small granular material (like sugar or salt) on a hard floor. It turns out that the irrotational approximation is quite realistic for flow into the inlet of a vacuum cleaner everywhere except very close to the floor, because the flow is rotational there.
FIGURE 10–73 Nondimensional axial velocity (blue curve) and pressure coefficient (green curve) along the floor below a vacuum cleaner modeled as an irrotational region of flow.
0.2 0.05 –0.15 –0.2 0.005 0 –0.005 –0.01 –0.015 –0.02 –0.025 –0.03 –5 –4 –3 –2 Normalized distance along the floor, x –1 0 5 –0.1 0.1 0.15 –0.05 0 1 2 3 4 u Cp Pressure coefficient Axial velocity FIGURE 10–74 Based on an irrotational flow approximation, the maximum speed along the floor beneath a vacuum cleaner nozzle occurs at x = ±b. A stagnation point occurs directly below the nozzle.
Vacuum nozzle Stagnation point Maximum speed w b b b y x V ⋅ We conclude this section by emphasizing that although the irrotational flow approximation is mathematically simple, and velocity and pressure fields are easy to obtain, we must be very careful where we apply it. The irrotational flow approximation breaks down in regions of non-negligible vorticity, especially near solid walls, where fluid particles rotate because of viscous stresses caused by the no-slip condition at the wall. This leads us to the final section in this chapter (Section 10–6) in which we discuss the boundary layer approximation.
10–6 ■ THE BOUNDARY LAYER APPROXIMATION As discussed in Sections 10–4 and 10–5, there are at least two flow situ ations in which the viscous term in the Navier–Stokes equation can be neglected. The first occurs in high Reynolds number regions of flow where net viscous forces are known to be negligible compared to inertial and/or cen96537_ch10_519-568.indd 558 29/12/16 3:45 pm 559 CHAPTER 10 pressure forces; we call these inviscid regions of flow. The second situation occurs when the vorticity is negligibly small; we call these irrotational or potential regions of flow. In either case, removal of the viscous terms from the Navier–Stokes equation yields the Euler equation (Eq. 10–13 and also Eq. 10–25). While the math is greatly simplified by dropping the viscous terms, there are some serious deficiencies associated with application of the Euler equation to practical engineering flow problems. High on the list of deficiencies is the inability to specify the no-slip condition at solid walls. This leads to unphysical results such as zero viscous shear forces on solid walls and zero aerodynamic drag on bodies immersed in a free stream. We can therefore think of the Euler equation and the Navier–Stokes equation as two mountains separated by a huge chasm (Fig. 10–75a). We make the fol lowing statement about the boundary layer approximation: The boundary layer approximation bridges the gap between the Euler equation and the Navier–Stokes equation, and between the slip condition and the no-slip condition at solid walls (Fig. 10–75b). From a historical perspective, by the mid-1800s, the Navier–Stokes equation was known, but couldn’t be solved except for flows of very sim ple geometries. Meanwhile, mathematicians were able to obtain beautiful analytical solutions of the Euler equation and of the potential flow equa tions for flows of complex geometry, but their results were often physi cally meaningless. Hence, the only reliable way to study fluid flows was empirically, i.e., with experiments. A major breakthrough in fluid mechan ics occurred in 1904 when Ludwig Prandtl (1875–1953) introduced the boundary layer approximation. Prandtl’s idea was to divide the flow into two regions: an outer flow region that is inviscid and/or irrotational, and an inner flow region called a boundary layer—a very thin region of flow near a solid wall where viscous forces and rotationality cannot be ignored (Fig. 10–76). In the outer flow region, we use the continuity and Euler equations to obtain the outer flow velocity field, and the Bernoulli equa tion to obtain the pressure field. Alternatively, if the outer flow region is irrotational, we may use the potential flow techniques discussed in Section 10–5 (e.g., superposition) to obtain the outer flow velocity field. In either case, we solve for the outer flow region first, and then fit in a thin boundary layer in regions where rotationality and viscous forces cannot be neglected. Within the boundary layer we solve the boundary layer equations, to be discussed shortly. (Note that the boundary layer equations are themselves approximations of the full Navier–Stokes equation, as we will see.) The boundary layer approximation corrects some of the major deficien cies of the Euler equation by providing a way to enforce the no-slip con dition at solid walls. Hence, viscous shear forces can exist along walls, bodies immersed in a free stream can experience aerodynamic drag, and flow separation in regions of adverse pressure gradient can be predicted more accurately. The boundary layer concept therefore became the workhorse of engineering fluid mechanics throughout most of the 1900s. However, the advent of fast, inexpensive computers and computational fluid dynamics (CFD) software in the latter part of the twentieth century enabled numeri cal solution of the Navier–Stokes equation for flows of complex geome try. Today, therefore, it is no longer necessary to split the flow into outer flow regions and boundary layer regions—we can use CFD to solve the FIGURE 10–75 (a) A huge gap exists between the Euler equation (which allows slip at walls) and the Navier–Stokes equation (which supports the no-slip condition); (b) the boundary layer approximation bridges that gap.
Euler equation Slip (a) No slip Navier– Stokes equation Euler equation Slip Boundary layer approximation (b) No slip Navier– Stokes equation FIGURE 10–76 Prandtl’s boundary layer concept splits the flow into an outer flow region and a thin boundary layer region (not to scale).
y δ(x) Outer flow (inviscid and/or irrotational region of flow) Boundary layer (rotational with non-negligible viscous forces) V x cen96537_ch10_519-568.indd 559 29/12/16 3:45 pm 560 APPROXIMATE SOLUTIONS OF THE N–S EQ full set of equations of motion (continuity plus Navier–Stokes) throughout the whole flow field. Nevertheless, boundary layer theory is still useful in some engineering applications, since it takes much less time to arrive at a solution. In addition, there is a lot we can learn about the behavior of flow ing fluids by studying boundary layers. We stress again that boundary layer solutions are only approximations of full Navier–Stokes solutions, and we must be careful where we apply this or any approximation. The key to successful application of the boundary layer approximation is the assumption that the boundary layer is very thin. The classic example is a uniform stream flowing parallel to a long flat plate aligned with the x-axis. Boundary layer thickness 𝛿 at some location x along the plate is sketched in Fig. 10–77. By convention, 𝛿 is usually defined as the distance away from the wall at which the velocity component parallel to the wall is 99 percent of the fluid speed outside the boundary layer. It turns out that for a given fluid and plate, the higher the free-stream speed V, the thinner the boundary layer (Fig. 10–77). In nondimensional terms, we define the Reynolds number based on distance x along the wall, Reynolds number along a flat plate: Rex = ρVx 𝜇 = Vx 𝜈 (10–60) Hence, At a given x-location, the higher the Reynolds number, the thinner the boundary layer.
In other words, the higher the Reynolds number, all else being equal, the more reliable the boundary layer approximation. We are confident that the bound ary layer is thin when 𝛿 ≪ x (or, expressed nondimensionally, 𝛿/x ≪ 1). The shape of the boundary layer profile can be obtained experimentally by flow visualization. An example is shown in Fig. 10–78 for a laminar boundary layer on a flat plate. Taken over 60 years ago by F. X. Wortmann, this is now considered a classic photograph of a laminar flat plate boundary layer profile. FIGURE 10–78 Flow visualization of a laminar flat plate boundary layer profile. Photograph taken by F. X. Wortmann in 1953 as visualized with the tellurium method. Flow is from left to right, and the leading edge of the flat plate is far to the left of the field of view.
Wortmann, F. X., 1977 AGARD Conf. Proc. no. 224, paper 12.
FIGURE 10–77 Flow of a uniform stream parallel to a flat plate (drawings not to scale): (a) Rex ∼ 102, (b) Rex ∼ 104. The larger the Reynolds number, the thinner the boundary layer along the plate at a given x-location.
y δ(x) Rex ~ 102 V x (a) y δ(x) Rex ~ 104 V x (b) cen96537_ch10_519-568.indd 560 29/12/16 3:45 pm 561 CHAPTER 10 The no-slip condition is clearly verified at the wall, and the smooth increase in flow speed away from the wall verifies that the flow is indeed laminar. Note that although we are discussing boundary layers in connection with the thin region near a solid wall, the boundary layer approximation is not limited to wall-bounded flow regions. The same equations may be applied to free shear layers such as jets, wakes, and mixing layers (Fig. 10–79), provided that the Reynolds number is sufficiently high that these regions are thin. The regions of these flow fields with non-negligible viscous forces and finite vorticity can also be considered to be boundary layers, even though a solid wall boundary may not even be present. Boundary layer thickness 𝛿(x) is labeled in each of the sketches in Fig. 10–79. As you can see, by con vention 𝛿 is usually defined based on half of the total thickness of the free shear layer. We define 𝛿 as the distance from the centerline to the edge of the boundary layer where the change in speed is 99 percent of the maximum change in speed from the centerline to the outer flow. Boundary layer thick ness is not a constant, but varies with downstream distance x. In the exam ples discussed here (flat plate, jet, wake, and mixing layer), 𝛿(x) increases with x. There are flow situations however, such as rapidly accelerating outer flow along a wall, in which 𝛿(x) decreases with x. A common misunderstanding among beginning students of fluid mechan ics is that the curve representing 𝛿 as a function of x is a streamline of the flow—it is not! In Fig. 10–80 we sketch both streamlines and 𝛿(x) for the boundary layer growing on a flat plate. As the boundary layer thick ness grows downstream, streamlines passing through the boundary layer must diverge slightly upward in order to satisfy conservation of mass. The amount of this upward displacement is smaller than the growth of 𝛿(x). Since streamlines cross the curve 𝛿(x), 𝛿(x) is clearly not a streamline (streamlines cannot cross each other or else mass would not be conserved). For a laminar boundary layer growing on a flat plate, as in Fig. 10–80, boundary layer thickness 𝛿 is at most a function of V, x, and fluid properties 𝜌 and 𝜇. It is a simple exercise in dimensional analysis to show that 𝛿/x is a function of Rex. In fact, it turns out that 𝛿 is proportional to the square root of Rex. You must note, however, that this result is valid only for a laminar boundary layer on a flat plate. As we move down the plate to larger and larger values of x, Rex increases linearly with x. At some point, infinitesi mal disturbances in the flow begin to grow, and the boundary layer can not remain laminar—it begins a transition process toward turbulent flow. For a smooth flat plate with a uniform free stream, the transition process begins at a critical Reynolds number, Rex, critical ≅ 1 × 105, and contin ues until the boundary layer is fully turbulent at the transition Reynolds number, Rex, transition ≅ 3 × 106 (Fig. 10–81). The transition process is quite complicated, and details are beyond the scope of this text.
FIGURE 10–79 Three additional flow regions where the boundary layer approximation may be appropriate: (a) jets, (b) wakes, and (c) mixing layers.
(a) (b) (c) δ(x) x δ(x) x V V δ(x) x V2 V1 V y Streamlines d(x) δ(x) Boundary layer x δ(x) FIGURE 10–80 Comparison of streamlines and the curve representing 𝛿 as a function of x for a flat plate boundary layer. Since streamlines cross the curve 𝛿(x), 𝛿(x) cannot itself be a streamline of the flow.
cen96537_ch10_519-568.indd 561 29/12/16 3:45 pm 562 APPROXIMATE SOLUTIONS OF THE N–S EQ Note that in Fig. 10–81 the vertical scale has been greatly exaggerated, and the horizontal scale has been shortened (in reality, since Rex, transition ≅ 30 times Rex, critical, the transitional region is much longer than indicated in the figure). To give you a better feel for how thin a boundary layer actually is, we have plotted 𝛿 as a function of x to scale in Fig. 10–82. To gener ate the plot, we carefully selected the parameters such that Rex = 100,000x regardless of the units of x. Thus, Rex, critical occurs at x ≅ 1 and Rex, transition occurs at x ≅ 30 in the plot. Notice how thin the boundary layer is and how long the transitional region is when plotted to scale. In real-life engineering flows, transition to turbulent flow usually occurs more abruptly and much earlier (at a lower value of Rex) than the values given for a smooth flat plate with a calm free stream. Factors such as roughness along the surface, free-stream disturbances, acoustic noise, flow unsteadi ness, vibrations, and curvature of the wall contribute to an earlier transition location. Because of this, an engineering critical Reynolds number of Rex, cr = 5 × 105 is often used to determine whether a boundary layer is most likely laminar (Rex < Rex, cr) or most likely turbulent (Rex > Rex, cr). It is also com mon in heat transfer to use this value as the critical Re; in fact, relations for average friction and heat transfer coefficients are derived by assuming the flow to be laminar for Rex lower than Rex, cr, and turbulent otherwise. The logic here is to ignore transition by treating the first part of transition as laminar and the remaining part as turbulent. We follow this convention throughout the rest of the book unless noted otherwise. The transition process is unsteady as well and is difficult to predict, even with modern CFD codes. In some cases, engineers install rough sandpaper or wires called trip wires along the surface, in order to force transition at a desired location (Fig. 10–83). The eddies from the trip wire cause enhanced local mixing and create disturbances that very quickly lead to a turbulent boundary layer. Again, the vertical scale in Fig. 10–83 is greatly exaggerated for illustrative purposes.
FIGURE 10–81 Transition of the laminar boundary layer on a flat plate into a fully turbulent boundary layer (not to scale).
Laminar Rex ≅ 105 Transitional Turbulent Rex ≅ 3 × 106 y x V δ(x) FIGURE 10–82 Thickness of the boundary layer on a flat plate, drawn to scale. Laminar, transitional, and turbulent regions are indicated for the case of a smooth wall with calm free-stream conditions.
5 0 0 5 10 15 20 25 30 35 40 Laminar Transitional Turbulent δ(x) V y x FIGURE 10–83 A trip wire is often used to initiate early transition to turbulence in a boundary layer (not to scale).
Trip wire Transitional Turbulent y x V Laminar cen96537_ch10_519-568.indd 562 29/12/16 3:45 pm 563 CHAPTER 10 EXAMPLE 10–9 Laminar or Turbulent Boundary Layer?
An aluminum canoe moves horizontally along the surface of a lake at 3.5 mi/h (Fig. 10–84). The temperature of the lake water is 50°F. The bottom of the canoe is 20 ft long and is flat. Is the boundary layer on the canoe bottom laminar or turbulent?
SOLUTION We are to assess whether the boundary layer on the bottom of a canoe is laminar or turbulent. Assumptions 1 The flow is steady and incompressible. 2 Ridges, dings, and other nonuniformities in the bottom of the canoe are ignored—the bottom is assumed to be a smooth flat plate aligned exactly with the direction of flow. 3 From the frame of reference of the canoe, the water below the boundary layer under the canoe moves at uniform speed V = 3.5 mph.
Properties The kinematic viscosity of water at T = 50oF is 𝜈 = 1.407 × 10−5 ft2/s.
Analysis First, we calculate the Reynolds number at the stern of the canoe, Rex = Vx v = (3.5 mi/h)(20 ft) 1.407 × 10−5 ft2/s( 5280 ft 1 mi ) ( 1 h 3600 s) = 7.30 × 106 Since Rex is much greater than Rex,cr (5 × 105), and is even greater than Rex,transition (50 × 105), the boundary layer is definitely turbulent by the back of the canoe.
Discussion Since the canoe bottom is neither perfectly smooth nor perfectly flat, and since we expect some disturbances in the lake water due to waves, the paddles, swimming fish, etc., transition to turbulence is expected to occur much earlier and more rapidly than illustrated for the ideal case in Fig. 10–81. Hence we are even more confident that this boundary layer is turbulent.
The Boundary Layer Equations Now that we have a physical feel for boundary layers, we need the equations of motion to be used in boundary layer calculations—the boundary layer equations. For simplicity we consider only steady, two-dimensional flow in the xy-plane in Cartesian coordinates. The methodology used here can be extended, however, to axisymmetric boundary layers or to three-dimensional boundary layers in any coordinate system. We neglect gravity since we are not dealing with free surfaces or with buoyancy-driven flows (free convection flows), where gravitational effects dominate. We consider only laminar boundary layers; turbulent boundary layer equations are beyond the scope of this text. For the case of a boundary layer along a solid wall, we adopt a coordinate system in which x is everywhere paral lel to the wall and y is everywhere normal to the wall (Fig. 10–85). This coordinate system is called a boundary layer coordinate system. When we solve the boundary layer equations, we do so at one x-location at a time, using this coordinate system locally, and it is locally orthogonal. It is not critical where we define x = 0, but for flow over a body, as in Fig. 10–85, we typically set x = 0 at the front stagnation point.
FIGURE 10–85 The boundary layer coordinate system for flow over a body; x follows the surface and is typically set to zero at the front stagnation point of the body, and y is everywhere normal to the surface locally.
Boundary layer V x = 0 y y x x x L y x Boundary layer δ(x) V FIGURE 10–84 Schematic for Example 10–9.
cen96537_ch10_519-568.indd 563 29/12/16 3:45 pm 564 APPROXIMATE SOLUTIONS OF THE N–S EQ We begin with the nondimensionalized Navier–Stokes equation derived at the beginning of this chapter. With the unsteady term and the gravity term neglected, Eq. 10–6 becomes (V › ·∇ ›)V › = −[Eu]∇ › P + [ 1 Re]∇2V › (10–61) The Euler number is of order unity, since pressure differences outside the boundary layer are determined by the Bernoulli equation and ΔP = P − P∞ ∼ 𝜌V 2. We note that V is a characteristic velocity scale of the outer flow, typically equal to the free-stream velocity for bodies immersed in a uniform flow. The characteristic length scale used in this nondimensionalization is L, some characteristic size of the body. For boundary layers, x is of order of magnitude L, and the Reynolds number in Eq. 10–61 can be thought of as Rex (Eq. 10–60). Rex is very large in typical applications of the boundary layer approximation. It would seem then that we could neglect the last term in Eq. 10–61 in boundary layers. However, doing so would result in the Euler equation, along with all its deficiencies discussed previously. So, we must keep at least some of the viscous terms in Eq. 10–61. How do we decide which terms to keep and which to neglect? To answer this question, we redo the nondimensionalization of the equations of motion based on appropriate length and velocity scales within the boundary layer. A magnified view of a portion of the boundary layer of Fig. 10–85 is sketched in Fig. 10–86. Since the order of magnitude of x is L, we use L as an appro priate length scale for distances in the streamwise direction and for deriva tives of velocity and pressure with respect to x. However, this length scale is much too large for derivatives with respect to y. It makes more sense to use 𝛿 as the length scale for distances in the direction normal to the streamwise direction and for derivatives with respect to y. Similarly, while the charac teristic velocity scale is V for the whole flow field, it is more appropriate to use U as the characteristic velocity scale for boundary layers, where U is the magnitude of the velocity component parallel to the wall at a location just above the boundary layer (Fig. 10–86). U is in general a function of x. Thus, within the boundary layer at some value of x, the orders of magnitude are u ∼U P −P∞∼ρU2 ∂ ∂x ∼1 L ∂ ∂y ∼1 𝛿 (10–62) The order of magnitude of velocity component 𝜐 is not specified in Eq. 10–62, but is instead obtained from the continuity equation. Applying the orders of magnitude in Eq. 10–62 to the incompressible continuity equation in two dimensions, ∂u ∂x + ∂𝜐 ∂y = 0 → U L ∼𝜐 𝛿 ∼U/L ∼𝜐/𝛿 Since the two terms have to balance each other, they must be of the same order of magnitude. Thus we obtain the order of magnitude of velocity component 𝜐, 𝜐∼U𝛿 L (10–63) Since 𝛿/L ≪ 1 in a boundary layer (the boundary layer is very thin), we conclude that 𝜐 ≪ u in a boundary layer (Fig. 10–87). From Eqs. 10–62 } } FIGURE 10–86 Magnified view of the boundary layer along the surface of a body, showing length scales x and 𝛿 and velocity scale U.
Wall Boundary layer x U = U(x) y δ FIGURE 10–87 Highly magnified view of the boundary layer along the surface of a body, showing that velocity component υ is much smaller than u.
Wall Boundary layer x u ʋ U y δ cen96537_ch10_519-568.indd 564 29/12/16 3:45 pm 565 CHAPTER 10 and 10–63, we define the following nondimensional variables within the boundary layer: x = x L y = y 𝛿 u = u U 𝜐 = 𝜐L U𝛿 P = P −P∞ ρU2 Since we used appropriate scales, all these nondimensional variables are of order unity—i.e., they are normalized variables (Chap. 7). We now consider the x- and y-components of the Navier–Stokes equation. We substitute these nondimensional variables into the y-momentum equa tion, giving u ∂𝜐 ∂x + 𝜐 ∂𝜐 ∂y = −1 ρ ∂P ∂y + 𝜈 ∂2𝜐 ∂x2 + 𝜈 ∂2𝜐 ∂y2 ∂ ∂x 𝜐U𝛿 L2 ∂ ∂y 𝜐U𝛿 L𝛿 1 ρ ∂ ∂y PρU2 𝛿 𝜈 ∂2 ∂x2 𝜐U𝛿 L3 𝜈 ∂2 ∂y2 𝜐U𝛿 L𝛿 2 After some algebra and after multiplying each term by L2/(U2𝛿), we get u ∂𝜐 ∂x + 𝜐 ∂𝜐 ∂y = −( L 𝛿 ) 2 ∂P ∂y + ( 𝜈 UL) ∂2𝜐 ∂x2 + ( 𝜈 UL) ( L 𝛿 ) 2 ∂2𝜐 ∂y2 (10–64) Comparing terms in Eq. 10–64, the middle term on the right side is clearly orders of magnitude smaller than any other term since ReL = UL/𝜈 ≫ 1. For the same reason, the last term on the right is much smaller than the first term on the right. Neglecting these two terms leaves the two terms on the left and the first term on the right. However, since L ≫ 𝛿, the pressure gradient term is orders of magnitude greater than the advective terms on the left side of the equation. Thus, the only term left in Eq. 10–64 is the pressure term. Since no other term in the equation can balance that term, we have no choice but to set it equal to zero. Thus, the nondimensional y-momentum equation reduces to ∂P ∂y ≅0 or, in terms of the physical variables, Normal pressure gradient through a boundary layer: ∂P ∂y ≅0 (10–65) In words, although pressure may vary along the wall (in the x-direction), there is negligible change in pressure in the direction normal to the wall. This is illustrated in Fig. 10–88. At x = x1, P = P1 at all values of y across the boundary layer from the wall to the outer flow. At some other x-location, x = x2, the pressure may have changed, but P = P2 at all values of y across that portion of the boundary layer.
The pressure across a boundary layer ( y-direction) is nearly constant. Physically, because the boundary layer is so thin, streamlines within the boundary layer have negligible curvature when observed at the scale of the boundary layer thickness. Curved streamlines require a centripetal accelera tion, which comes from a pressure gradient along the radius of curvature. Since the streamlines are not significantly curved in a thin boundary layer, there is no significant pressure gradient across the boundary layer.
} } } } } } } uU 𝜐 U𝛿 L FIGURE 10–88 Pressure may change along a boundary layer (x-direction), but the change in pressure across a boundary layer (y-direction) is negligible.
Wall Boundary layer Outer flow x x2 x1 P2 P2 P1 P1 P1 P1 P2 P2 y δ cen96537_ch10_519-568.indd 565 29/12/16 3:45 pm 566 APPROXIMATE SOLUTIONS OF THE N–S EQ One immediate consequence of Eq. 10–65 and the statement just presented is that at any x-location along the wall, the pressure at the outer edge of the boundary layer (y ≅ 𝛿) is the same as that at the wall (y = 0). This leads to a tremendous practical application; namely, the pressure at the outer edge of a boundary layer can be measured experimentally by a static pressure tap at the wall directly beneath the boundary layer (Fig. 10–89). Experimental ists routinely take advantage of this fortunate situation, and countless airfoil shapes for airplane wings and turbomachinery blades were tested with such pressure taps over the past century. The experimental pressure data shown in Fig. 10–64 for flow over a cir cular cylinder were measured with pressure taps at the cylinder’s surface, yet they are used to compare with the pressure calculated by the irrotational outer flow approximation. Such a comparison is valid, because the pressure obtained outside of the boundary layer (from the Euler equation or poten tial flow analysis coupled with the Bernoulli equation) applies all the way through the boundary layer to the wall. Returning to the development of the boundary layer equations, we use Eq. 10–65 to greatly simplify the x-component of the momentum equation. Specifically, since P is not a function of y, we replace 𝜕P/𝜕x by dP/dx, where P is the value of pressure calculated from our outer flow approxima tion (using either continuity plus Euler, or the potential flow equations plus Bernoulli). The x-component of the Navier–Stokes equation becomes u ∂u ∂x + 𝜐 ∂u ∂y = −1 ρ dP dx + 𝜈 ∂2u ∂x2 + 𝜈 ∂2u ∂y2 ∂ ∂x uU L ∂ ∂y uU 𝛿 1 ρ ∂ ∂x PρU 2 L 𝜐 ∂2 ∂x2 uU L2 𝜐 ∂2 ∂y2 uU 𝛿 2 After some algebra, and after multiplying each term by L/U2, we get u ∂u ∂x + 𝜐 ∂u ∂y = −dP dx + ( 𝜈 UL) ∂2u ∂x2 + ( 𝜈 UL) ( L 𝛿 ) 2 ∂2u ∂y2 (10–66) Comparing terms in Eq. 10–66, the middle term on the right side is clearly orders of magnitude smaller than the terms on the left side, since ReL = UL/𝜈 ≫ 1. What about the last term on the right? If we neglect this term, we throw out all the viscous terms and are back to the Euler equation. Clearly this term must remain. Furthermore, since all the remaining terms in Eq. 10–66 are of order unity, the combination of parameters in parentheses in the last term on the right side of Eq. 10–66 must also be of order unity, ( 𝜈 UL) ( L 𝛿 ) 2 ∼1 Again recognizing that ReL = UL/𝜈, we see immediately that 𝛿 L ∼ 1 √ReL (10–67) This confirms our previous statement that at a given streamwise location along the wall, the larger the Reynolds number, the thinner the boundary layer. If we substitute x for L in Eq. 10–67, we also conclude that for a lami nar boundary layer on a flat plate, where U(x) = V = constant, 𝛿 grows like the square root of x (Fig. 10–90).
} } } } ⏟ ⏟ ⏟ uU 𝜐 U𝛿 L FIGURE 10–89 The pressure in the irrotational region of flow outside of a boundary layer can be measured by static pressure taps in the surface of the wall. Two such pressure taps are sketched.
Wall Pressure taps Boundary layer Outer flow x x2 P1 P2 x1 P2 P1 P1 P2 y δ FIGURE 10–90 An order-of-magnitude analysis of the laminar boundary layer equations along a flat plate reveals that 𝛿 grows like √x (not to scale).
y δ(x) δ(x) ~ √x U(x) = V V x cen96537_ch10_519-568.indd 566 29/12/16 3:45 pm 567 CHAPTER 10 In terms of the original (physical) variables, Eq. 10–66 is written as x-momentum boundary layer equation: u ∂u ∂x + 𝜐 ∂u ∂y = −1 ρ dP dx + 𝜈 ∂2u ∂y2 (10–68) Note that the last term in Eq. 10–68 is not negligible in the boundary layer, since the y-derivative of velocity gradient 𝜕u/𝜕y is sufficiently large to offset the (typically small) value of kinematic viscosity 𝜈. Finally, since we know from our y-momentum equation analysis that the pressure across the boundary layer is the same as that outside the boundary layer (Eq. 10–65), we apply the Bernoulli equation to the outer flow region. Differentiating with respect to x we get P ρ + 1 2 U 2 = constant → 1 ρ dP dx = −U dU dx (10–69) where we note that both P and U are functions of x only, as illustrated in Fig. 10–91. Substitution of Eq. 10–69 into Eq. 10–68 yields u ∂u ∂x + 𝜐 ∂u ∂y = U dU dx + 𝜈 ∂2u ∂y2 (10–70) and we have eliminated pressure from the boundary layer equations. We summarize the set of equations of motion for a steady, incompress ible, laminar boundary layer in the xy-plane without significant gravitational effects, Boundary layer equations: ∂u ∂x + ∂𝜐 ∂y = 0 u ∂u ∂x + 𝜐 ∂u ∂y = U dU dx + 𝜈 ∂2u ∂y2 (10–71) Mathematically, the full Navier–Stokes equation is elliptic in space, which means that boundary conditions are required over the entire boundary of the flow domain. Physically, flow information is passed in all directions, both upstream and downstream. On the other hand, the x-momentum boundary layer equation (the second equation of Eq. 10–71) is parabolic. This means that we need to specify boundary conditions on only three sides of the (two-dimensional) flow domain. Physically, flow information is not passed in the direction opposite to the flow (from downstream). This fact greatly reduces the level of difficulty in solving the boundary layer equations. Specifically, we don’t need to specify boundary conditions downstream, only upstream and on the top and bottom of the flow domain (Fig. 10–92). For a typical boundary layer problem along a wall, we specify the no-slip condition at the wall (u = 𝜐 = 0 at y = 0), the outer flow condition at the edge of the bound ary layer and beyond [u = U(x) as y → ∞], and a starting profile at some upstream location [u = ustarting(y) at x = xstarting, where xstarting may or may not be zero]. With these boundary conditions, we simply march downstream in the x-direction, solving the boundary layer equations as we go. This is particularly attractive for numerical boundary layer computations, because once we know the profile at one x-location (xi), we can march to the next x-location (xi+1), and then use this newly calculated profile as the starting profile to march to the next x-location (xi+2), etc.
FIGURE 10–91 Outer flow speed parallel to the wall is U(x) and is obtained from the outer flow pressure, P(x). This speed appears in the x-component of the boundary layer momentum equation, Eq. 10–70.
Boundary layer Wall P1 P2 x2 x U1 x1 U2 y P = P(x), U = U(x) δ(x) FIGURE 10–92 The boundary layer equation set is par abolic, so boundary conditions need to be specified on only three sides of the flow domain.
No boundary conditions on downstream edge of flow domain Flow domain u = U(x) u = ustarting(y) u = ʋ = 0 xstarting y x cen96537_ch10_519-568.indd 567 29/12/16 3:45 pm 568 APPROXIMATE SOLUTIONS OF THE N–S EQ The Boundary Layer Procedure When the boundary layer approximation is employed, we use a general step-by-step procedure. We outline the procedure here and in condensed form in Fig. 10–93.
Step 1 Solve for the outer flow, ignoring the boundary layer (assuming that the region of flow outside the boundary layer is approximately inviscid and/or irrotational). Transform coordinates as necessary to obtain U(x).
Step 2 Assume a thin boundary layer—so thin, in fact, that it does not affect the outer flow solution of step 1.
Step 3 Solve the boundary layer equations (Eqs. 10–71), using appropriate boundary conditions: the no-slip boundary condition at the wall, u = 𝜐 = 0 at y = 0; the known outer flow condition at the edge of the boundary layer, u → U(x) as y → ∞; and some known starting profile, u = ustarting(y) at x = xstarting.
Step 4 Calculate quantities of interest in the flow field. For example, once the boundary layer equations have been solved (step 3), we calculate 𝛿(x), shear stress along the wall, total skin friction drag, etc.
Step 5 Verify that the boundary layer approximations are appropriate. In other words, verify that the boundary layer is thin—otherwise the approximation is not justified. Before we do any examples, we list here some of the limitations of the boundary layer approximation. These are red flags to look for when performing boundary layer calculations: • The boundary layer approximation breaks down if the Reynolds number is not large enough. How large is large enough? It depends on the desired accuracy of the approximation. Using Eq. 10–67 as a guideline, 𝛿/L ∼ 0.03 (3 percent) for ReL = 1000, and 𝛿/L ∼ 0.01 (1 percent) for ReL = 10,000.
• The assumption of zero pressure gradient in the y-direction (Eq. 10–65) breaks down if the wall curvature is of similar magnitude as 𝛿 (Fig. 10–94). In such cases, centripetal acceleration effects due to streamline curvature cannot be ignored. Physically, the boundary layer is not thin enough for the approximation to be appropriate when 𝛿 is not ≪ R.
• When the Reynolds number is too high, the boundary layer does not remain laminar, as discussed previously. The boundary layer approxima tion itself may still be appropriate, but Eqs. 10–71 are not valid if the flow is transitional or fully turbulent. As noted before, the laminar boundary layer on a smooth flat plate under clean flow conditions begins to transition toward turbulence at Rex ≅ 1 × 105. In practical engineering applications, walls may not be smooth and there may be vibrations, noise, and fluctuations in the free-stream flow above the wall, all of which contribute to an even earlier start of the transition process.
• If flow separation occurs, the boundary layer approximation is no longer appropriate in the separated flow region. The main reason for this is that a separated flow region contains reverse flow, and the parabolic nature of the boundary layer equations is lost.
Step 1: Calculate U(x) (outer flow).
Step 2: Assume a thin boundary layer.
Step 3: Solve boundary layer equations.
Step 4: Calculate quantities of interest.
Step 5: Verify that boundary layer is thin.
FIGURE 10–93 Summary of the boundary layer procedure for steady, incompressible, two-dimensional boundary layers in the xy-plane.
Boundary layer Wall U(x) R y x δ FIGURE 10–94 When the local radius of curvature of the wall (R) is small enough to be of the same magnitude as 𝛿, centripetal acceleration effects cannot be ignored and 𝜕P/𝜕y ≠ 0. The thin boundary layer approximation is not appropriate in such regions.
cen96537_ch10_519-568.indd 568 29/12/16 3:45 pm 569 CHAPTER 10 EXAMPLE 10–10 Laminar Boundary Layer on a Flat Plate A uniform free stream of speed V flows parallel to an infinitesimally thin semi-infinite flat plate as sketched in Fig. 10–95. The coordinate system is defined such that the plate begins at the origin. Since the flow is symmetric about the x-axis, only the upper half of the flow is considered. Calculate the boundary layer velocity profile along the plate and discuss.
SOLUTION We are to calculate the boundary layer velocity profile (u as a function of x and y ) as the laminar boundary layer grows along the flat plate.
Assumptions 1 The flow is steady, incompressible, and two-dimensional in the xy-plane. 2 The Reynolds number is high enough that the boundary layer approximation is reasonable. 3 The boundary layer remains laminar over the range of interest.
Analysis We follow the step-by-step procedure outlined in Fig. 10–93.
Step 1 The outer flow is obtained by ignoring the boundary layer altogether, since it is assumed to be very, very thin. Recall that any streamline in an irrotational flow can be thought of as a wall since there is no flow through a streamline. In this case, the x-axis can be thought of as a streamline of uniform free-stream flow, one of our building block flows in Section 10–5; this streamline can also be thought of as an infinitesimally thin plate (Fig. 10–96). Thus, Outer flow: U(x) = V = constant (1) For convenience, we use U instead of U(x) from here on, since it is a constant.
Step 2 We assume a very thin boundary layer along the wall (Fig. 10–97). The key here is that the boundary layer is so thin that it has negligible effect on the outer flow calculated in step 1.
Step 3 We must now solve the boundary layer equations. We see from Eq. 1 that dU/dx = 0; in other words, no pressure gradient term remains in the x-momentum boundary layer equation. This is why the boundary layer on a flat plate is often called a zero pressure gradient boundary layer. The continuity and x-momentum equations for the boundary layer (Eqs. 10–71) become ∂u ∂x + ∂𝜐 ∂y = 0 u ∂u ∂x + 𝜐 ∂u ∂y = 𝜈∂2u ∂y2 (2) There are four required boundary conditions, u = 0 at y = 0 u = U as y →∞ 𝜐 = 0 at y = 0 u = U for all y at x = 0 (3) The last of the boundary conditions in Eq. 3 is the starting profile; we assume that the plate has not yet influenced the flow at the starting location of the plate (x = 0). These equations and boundary conditions seem simple enough, but unfortu nately no convenient analytical solution is available. However, a series solu tion of Eqs. 2 was obtained in 1908 by P. R. Heinrich Blasius (1883–1970). As a side note, Blasius was a Ph.D. student of Prandtl. In those days, of course, computers were not yet available, and all the calculations were performed by hand. Today we can solve these equations on a computer in a few seconds. The key to the solution is the assumption of similarity. In simple terms, Infinitesimally thin flat plate ρ, μ, v y x V FIGURE 10–95 Setup for Example 10–10; flow of a uniform stream parallel to a semi- infinite flat plate along the x-axis.
y x U(x) = V V FIGURE 10–96 The outer flow of Example 10–10 is trivial since the x-axis is a streamline of the flow, and U(x) = V = constant.
Boundary layer y x U(x) = V V FIGURE 10–97 The boundary layer is so thin that it does not affect the outer flow; boundary layer thickness is exaggerated here for clarity.
cen96537_ch10_569-610.indd 569 29/12/16 4:06 pm 570 APPROXIMATE SOLUTIONS OF THE N–S EQ similarity can be assumed here because there is no characteristic length scale in the geometry of the problem. Physically, since the plate is infinitely long in the x-direction, we always see the same flow pattern no matter how much we zoom in or zoom out (Fig. 10–98). Blasius introduced a similarity variable 𝜂 that combines independent variables x and y into one nondimensional independent variable, 𝜂 = y√ U 𝜈x (4) and he solved for a nondimensionalized form of the x-component of velocity, f ʹ = u U = function of 𝜂 (5) When we substitute Eqs. 4 and 5 into Eqs. 2, subjected to the boundary conditions of Eq. 3, we get an ordinary differential equation for non dimen sional speed f′(𝜂) = u/U as a function of similarity variable 𝜂. We use the popular Runge–Kutta numerical technique to obtain the results shown in Table 10–3 and in Fig. 10–99. Details of the numerical technique are beyond the scope of this text (see Heinsohn and Cimbala, 2003). There is also a small y-component of velocity 𝜐 away from the wall, but 𝜐 ≪ u, and is not discussed here. The beauty of the similarity solution is that this one unique velocity profile shape applies to any x-location when plotted in similarity variables, as in Fig. 10–99. The agreement of the calculated profile shape in Fig. 10–99 to experimentally obtained data (circles in Fig. 10–99) and to the visualized profile shape of Fig. 10–78 is remarkable. The Blasius solution is a stunning success.
TABLE 10–3 Solution of the Blasius laminar flat plate boundary layer in similarity variables 𝜂 f″ f′ f 𝜂 f″ f′ f 0.0 0.33206 0.00000 0.00000 0.1 0.33205 0.03321 0.00166 0.2 0.33198 0.06641 0.00664 0.3 0.33181 0.09960 0.01494 0.4 0.33147 0.13276 0.02656 0.5 0.33091 0.16589 0.04149 0.6 0.33008 0.19894 0.05973 0.8 0.32739 0.26471 0.10611 1.0 0.32301 0.32978 0.16557 1.2 0.31659 0.39378 0.23795 1.4 0.30787 0.45626 0.32298 1.6 0.29666 0.51676 0.42032 1.8 0.28293 0.57476 0.52952 2.0 0.26675 0.62977 0.65002 2.2 0.24835 0.68131 0.78119 2.4 0.22809 0.72898 0.92229 2.6 0.20645 0.77245 1.07250 2.8 0.18401 0.81151 1.23098 3.0 0.16136 0.84604 1.39681 3.5 0.10777 0.91304 1.83770 4.0 0.06423 0.95552 2.30574 4.5 0.03398 0.97951 2.79013 5.0 0.01591 0.99154 3.28327 5.5 0.00658 0.99688 3.78057 6.0 0.00240 0.99897 4.27962 6.5 0.00077 0.99970 4.77932 7.0 0.00022 0.99992 5.27923 8.0 0.00001 1.00000 6.27921 9.0 0.00000 1.00000 7.27921 10.0 0.00000 1.00000 8.27921 𝜂 is the similarity variable defined in Eq. 4 above, and function f(𝜂) is solved using the Runge–Kutta numerical technique. Note that f″ is proportional to the shear stress 𝜏, f′ is proportional to the x-component of velocity in the boundary layer (f′ = u/U), and f itself is proportional to the stream function. f′ is plotted as a function of 𝜂 in Fig. 10–99.
Magnifying glass or zoom tool + (a) y U(x) = V V y x (b) δ(x) u x δ(x) u V U(x) = V FIGURE 10–98 A useful result of the similarity assumption is that the flow looks the same (is similar) regardless of how far we zoom in or out; (a) view from a distance, as a person might see, (b) close-up view, as an ant might see.
cen96537_ch10_569-610.indd 570 29/12/16 4:06 pm 571 CHAPTER 10 Step 4 We next calculate several quantities of interest in this boundary layer. First, based on a numerical solution with finer resolution than that shown in Table 10–3, we find that u/U = 0.990 at 𝜂 ≅ 4.91. This 99 percent boundary layer thick ness is sketched in Fig. 10–99. Using Eq. 4 and the definition of 𝛿, we conclude that y = 𝛿 when 𝜂 = 4.91 = √ U 𝜈x 𝛿 → 𝜹 x = 4.91 √Rex (6) This result agrees qualitatively with Eq. 10–67, obtained from a simple order-of-magnitude analysis. The constant 4.91 in Eq. 6 is rounded to 5.0 by many authors, but we prefer to express the result to three significant digits for consistency with other quantities obtained from the Blasius profile. Another quantity of interest is the shear stress at the wall 𝜏w, 𝜏w = 𝜇 ∂u ∂y) y=0 (7) Sketched in Fig. 10–99 is the slope of the nondimensional velocity profile at the wall ( y = 0 and 𝜂 = 0). From our similarity results (Table 10–3), the nondimen sional slope at the wall is d (u/U) d𝜂 ) 𝜂 =0 = f ʺ(0) = 0.332 (8) After substitution of Eq. 8 into Eq. 7 and some algebra (transformation of similar ity variables back to physical variables), we obtain Shear stress in physical variables: 𝜏w = 0.332 ρU 2 √Rex (9) Thus, we see that the wall shear stress decays with x like x −1/2, as sketched in Fig. 10–100. At x = 0, Eq. 9 predicts that 𝜏w is infinite, which is unphysical. The boundary layer approximation is ‑not appropriate at the leading edge (x = 0), because the boundary layer thickness is not small compared to x. Furthermore, any real flat plate has finite thickness, and there is a stagnation point at the front of the plate, with the outer flow accelerating quickly to U(x) = V. We may ignore the region very close to x = 0 without loss of accuracy in the rest of the flow. Equation 9 is nondimensionalized by defining a skin friction coefficient (also called a local friction coefficient), Local friction coefficient, laminar flat plate: Cf, x = 𝝉w 1 2ힺU2 = 0.664 √Rex (10) Notice that Eq. 10 for Cf, x has the same form as Eq. 6 for 𝛿/x, but with a different constant—both decay like the inverse of the square root of Reynolds number. In Chap. 11, we integrate Eq. 10 to obtain the total friction drag on a flat plate of length L.
6 5 4 3 η 2 1 0 0 0.2 0.4 0.6 f' = u/U 0.8 Slope at the wall 99% boundary layer thickness 1 FIGURE 10–99 The Blasius profile in similarity variables for the boundary layer growing on a semi-infinite flat plate. Experimental data (circles) are at Rex = 3.64 × 105.
Boundary layer y x τw τw τw τw V δ(x) U(x) = V u (∂u/∂y)y = 0 FIGURE 10–100 For a laminar flat plate boundary layer, wall shear stress decays like x−1/2 as the slope 𝜕u/𝜕y at the wall decreases downstream. The front portion of the plate contributes more skin friction drag than does the rear portion.
cen96537_ch10_569-610.indd 571 29/12/16 4:06 pm 572 APPROXIMATE SOLUTIONS OF THE N–S EQ Step 5 We need to verify that the boundary layer is thin. Consider the practical example of flow over the hood of your car (Fig. 10–101) while you are driving downtown at 20 mi/h on a hot day. The kinematic viscosity of the air is ν = 1.8 × 10−4 ft2/s. We approximate the hood as a flat plate of length 3.5 ft moving hori zontally at a speed of V = 20 mi/h. First, we approximate the Reynolds number at the end of the hood using Eq. 10–60, Rex = Vx 𝜈= (20 mi/h) (3.5 ft) 1.8 × 10−4 ft2/s ( 5280 ft mi ) ( h 3600 s) = 5.7 × 105 Since Rex is very close to the ballpark critical Reynolds number, Rex, cr = 5 × 105, the assumption of laminar flow may or may not be appropriate. Nevertheless, we use Eq. 6 to estimate the thickness of the boundary layer, assuming that the flow remains laminar, 𝛿= 4.91x √Rex = 4.91(3.5 ft) √5.7 × 105 ( 12 in ft ) = 0.27 in (11) By the end of the hood the boundary layer is only about a quarter of an inch thick, and our assumption of a very thin boundary layer is verified.
Discussion The Blasius boundary layer solution is valid only for flow over a flat plate perfectly aligned with the flow. However, it is often used as a quick approxi mation for the boundary layer developing along solid walls that are not necessarily flat nor exactly parallel to the flow, as in the car hood. As illustrated in step 5, it is not difficult in practical engineering problems to achieve Reynolds numbers greater than the critical value for transition to turbulence. You must be careful not to apply the laminar boundary layer solution presented here when the boundary layer becomes turbulent.
Boundary layer Hood U(x) V x y FIGURE 10–101 The boundary layer growing on the hood of a car. Boundary layer thickness is exaggerated for clarity.
V x U(x) = V Outer flow streamline Boundary layer δ(x) y δ(x) FIGURE 10–102 Displacement thickness defined by a streamline outside of the boundary layer. Boundary layer thickness is exaggerated.
Displacement Thickness As was shown in Fig. 10–80, streamlines within and outside a boundary layer must bend slightly outward away from the wall in order to satisfy con servation of mass as the boundary layer thickness grows downstream. This is because the y-component of velocity, 𝜐, is small but finite and positive. Outside of the boundary layer, the outer flow is affected by this deflection of the streamlines. We define displacement thickness 𝛿 as the distance that a streamline just outside of the boundary layer is deflected, as sketched in Fig. 10–102.
Displacement thickness is the distance that a streamline just outside of the boundary layer is deflected away from the wall due to the effect of the boundary layer. We generate an expression for 𝛿 for the boundary layer along a flat plate by performing a control volume analysis using conservation of mass. The details are left as an exercise for the reader; the result at any x-location along the plate is Displacement thickness: 𝛿 = ∫ ∞ 0 (1 −u U) dy (10–72) Note that the upper limit of the integral in Eq. 10–72 is shown as ∞, but since u = U everywhere above the boundary layer, it is necessary to integrate cen96537_ch10_569-610.indd 572 29/12/16 4:06 pm 573 CHAPTER 10 only out to some finite distance above 𝛿. Obviously 𝛿 grows with x as the boundary layer grows (Fig. 10–103). For a laminar flat plate, we integrate the numerical (Blasius) solution of Example 10–10 to obtain Displacement thickness, laminar flat plate: 𝛿 x = 1.72 √Rex (10–73) The equation for 𝛿 is the same as that for 𝛿, but with a different constant. In fact, for laminar flow over a flat plate, 𝛿 at any x-location turns out to be approximately three times smaller than 𝛿 at that same x-location (Fig. 10–103). There is an alternative way to explain the physical meaning of 𝛿 that turns out to be more useful for practical engineering applications. Namely, we can think of displacement thickness as an imaginary or apparent increase in thick ness of the wall from the point of view of the inviscid and/or irrotational outer flow region. For our flat plate example, the outer flow no longer “sees” an infinitesimally thin flat plate; rather it sees a finite-thickness plate shaped like the displacement thickness of Eq. 10–73, as illustrated in Fig. 10–104.
Displacement thickness is the imaginary increase in thickness of the wall, as seen by the outer flow, due to the effect of the growing boundary layer. If we were to solve the Euler equation for the flow around this imaginary thicker plate, the outer flow velocity component U(x) would differ from the original calculation. We could then use this apparent U(x) to improve our boundary layer analysis. You can imagine a modification to the boundary layer procedure of Fig. 10–93 in which we go through the first four steps, calculate 𝛿(x), and then go back to step 1, this time using the imaginary (thicker) body shape to calculate an apparent U(x). Following this, we re-solve the boundary layer equations. We could repeat the loop as many times as necessary until convergence. In this way, the outer flow and the boundary layer would be more consistent with each other. The usefulness of this interpretation of displacement thickness becomes obvious if we consider uniform flow entering a channel bounded by two parallel walls (Fig. 10–105). As the boundary layers grow on the upper and lower walls, the irrotational core flow must accelerate to satisfy con servation of mass (Fig. 10–105a). From the point of view of the core flow between the boundary layers, the boundary layers cause the channel walls to appear to converge—the apparent distance between the walls decreases as x increases. This imaginary increase in thickness of one of the walls is equal to 𝛿(x), and the apparent U(x) of the core flow must increase accordingly, as sketched, to satisfy conservation of mass.
V Boundary layer x U(x) = V y δ(x) δ(x) FIGURE 10–103 For a laminar flat plate boundary layer, the displacement thickness is roughly one-third of the 99 percent boundary layer thickness.
V Boundary layer x Apparent U(x) y δ(x) δ(x) Actual wall Apparent wall FIGURE 10–104 The boundary layer affects the irrotational outer flow in such a way that the wall appears to take the shape of the displacement thickness. The apparent U(x) differs from the original approximation because of the “thicker” wall.
x Apparent U(x) y δ(x) δ(x) x Core flow y δ(x) (b) (a) Boundary layer Boundary layer FIGURE 10–105 The effect of boundary layer growth on flow entering a two-dimensional channel: the irrotational flow between the top and bottom boundary layers accelerates as indicated by (a) actual velocity profiles, and (b) change in apparent core flow due to the displacement thickness of the boundary layer (boundary layers greatly exaggerated for clarity).
cen96537_ch10_569-610.indd 573 29/12/16 4:06 pm 574 APPROXIMATE SOLUTIONS OF THE N–S EQ EXAMPLE 10–11 Displacement Thickness in the Design of a Wind Tunnel A small low-speed wind tunnel (Fig. 10–106) is being designed for calibration of hot wires. The air is at 19°C. The test section of the wind tunnel is 30 cm in diam eter and 30 cm in length. The flow through the test section must be as uniform as possible. The wind tunnel speed ranges from 1 to 8 m/s, and the design is to be optimized for an air speed of V = 4.0 m/s through the test section. (a) For the case of nearly uniform flow at 4.0 m/s at the test section inlet, by how much will the centerline air speed accelerate by the end of the test section? (b) Recommend a design that will lead to a more uniform test section flow.
SOLUTION The acceleration of air through the round test section of a wind tunnel is to be calculated, and a redesign of the test section is to be recommended.
Assumptions 1 The flow is steady and incompressible. 2 The walls are smooth, and disturbances and vibrations are kept to a minimum. 3 The boundary layer is laminar.
Properties The kinematic viscosity of air at 19°C is ν = 1.507 × 10−5 m2/s.
Analysis (a) The Reynolds number at the end of the test section is approximately Rex = Vx 𝜈= (4.0 m/s)(0.30 m) 1.507 × 10−5 m2/s = 7.96 × 104 Since Rex is lower than the engineering critical Reynolds number, Rex, cr = 5 × 105, and is even lower than Rex, critical = 1 × 105, and since the walls are smooth and the flow is clean, we may assume that the boundary layer on the wall remains laminar throughout the length of the test section. As the boundary layer grows along the wall of the wind tunnel test section, air in the region of irrotational flow in the central portion of the test section accelerates as in Fig. 10–105 in order to satisfy conservation of mass. We use Eq. 10–73 to estimate the displacement thickness at the end of the test section, 𝛿 ≅1.72x √Rex = 1.72(0.30 m) √7.96 × 104 = 1.83 × 10−3 m = 1.83 mm (1) Two cross-sectional views of the test section are sketched in Fig. 10–107, one at the beginning and one at the end of the test section. The effective radius at the end of the test section is reduced by 𝛿 as calculated by Eq. 1. We apply conser vation of mass to calculate the average air speed at the end of the test section, Vend Aend = Vbeginning Abeginning → Vend = Vbeginning 𝜋R2 𝜋(R −𝛿)2 (2) which yields Vend = (4.0 m/s) (0.15 m)2 (0.15 m −1.83 × 10−3 m)2 = 4.10 m/s (3) Thus the air speed increases by approximately 2.5 percent through the test section, due to the effect of displacement thickness.
(b) What recommendation can we make for a better design? One possibility is to design the test section as a slowly diverging duct, rather than as a straight-walled Flow straighteners V Diffuser Silencer Test section Fan FIGURE 10–106 Schematic diagram of the wind tunnel of Example 10–11.
(a) (b) R R – δ δ FIGURE 10–107 Cross-sectional views of the test section of the wind tunnel of Example 10–11: (a) beginning of test section and (b) end of test section.
cen96537_ch10_569-610.indd 574 29/12/16 4:06 pm 575 CHAPTER 10 cylinder (Fig. 10–108). If the radius were designed so as to increase like 𝛿(x) along the length of the test section, the displacement effect of the boundary layer would be eliminated, and the test section air speed would remain fairly constant. Note that there is still a boundary layer growing on the wall, as illustrated in Fig. 10–108. However, the core flow speed outside the boundary layer remains con stant, unlike the situation of Fig. 10–105. The diverging wall recommendation would work well at the design operating condition of 4.0 m/s and would help somewhat at other flow speeds. Another option is to apply suction along the wall of the test section in order to remove some of the air along the wall. The advan tage of this design is that the suction can be carefully adjusted as wind tunnel speed is varied so as to ensure constant air speed through the test section at any operating condition. This recommendation is the more complicated, and probably more expensive, option.
Discussion Wind tunnels have been constructed that use either the diverging wall option or the wall suction option to carefully control the uniformity of the air speed through the wind tunnel test section. The same displacement thickness technique is applied to larger wind tunnels, where the boundary layer is turbulent; however, a different equation for 𝛿(x) is required.
x V Original test section wall Core flow δ(x) δ(x) V x Original test section wall Modified test section wall Apparent core flow δ(x) δ(x) (a) (b) Modified test section wall FIGURE 10–108 A diverging test section would eliminate flow acceleration due to the displacement effect of the boundary layer: (a) actual flow and (b) apparent irrotational core flow.
Momentum Thickness Another measure of boundary layer thickness is momentum thickness, commonly given the symbol 𝜃. Momentum thickness is best explained by analyzing the control volume of Fig. 10–109 for a flat plate boundary layer. Since the bottom of the control volume is the plate itself, no mass or momentum can cross that surface. The top of the control volume is taken as a streamline of the outer flow. Since no flow can cross a streamline, there can be no mass or momentum flux across the upper surface of the control volume. When we apply conservation of mass to this control volume, we find that the mass flow entering the control volume from the left (at x = 0) must equal the mass flow exiting from the right (at some arbitrary location x along the plate), 0 = ∫CS ρV ›·n › dA = wρ ∫ Y+𝛿 0 u dy −wρ ∫ Y 0 U dy (10–74) at location x at x = 0 where w is the width into the page in Fig. 10–109, which we take arbitrarily as unit width, and Y is the distance from the plate to the outer streamline at x = 0, as indicated in Fig. 10–109. Since u = U = constant everywhere along the left surface of the control volume, and since u = U between y = Y and y = Y + 𝛿 along the right surface of the control volume, Eq. 10–74 reduces to ∫ Y 0 (U −u) dy = U𝛿 (10–75) Physically, the mass flow deficit within the boundary layer (the lower blue-shaded region in Fig. 10–109) is replaced by a chunk of free-stream flow of thickness 𝛿 (the upper blue-shaded region in Fig. 10–109). Equation 10–75 verifies that these two shaded regions have the same area. We zoom in to show these areas more clearly in Fig. 10–110.
V x Outer flow streamline δ(x) y Boundary layer δ(x) U(x) = V F D, x u Y FIGURE 10–109 A control volume is defined by the thick dashed line, bounded above by a streamline outside of the boundary layer, and bounded below by the flat plate; FD, x is the viscous force of the plate acting on the control volume.
cen96537_ch10_569-610.indd 575 29/12/16 4:06 pm 576 APPROXIMATE SOLUTIONS OF THE N–S EQ Now consider the x-component of the control volume momentum equa tion. Since no momentum crosses the upper or lower control surfaces, the net force acting on the control volume must equal the momentum flux exit ing the control volume minus that entering the control volume, Conservation of x-momentum for the control volume: ∑Fx = −FD, x = ∫CS ρuV ›·n › dA = ρw ∫ Y+𝛿 0 u2 dy −ρw∫ Y 0 U2 dy (10–76) at location x at x = 0 where FD, x is the drag force due to friction on the plate from x = 0 to loca tion x. After some algebra, including substitution of Eq. 10–75, Eq. 10–76 reduces to FD, x = ρw ∫ Y 0 u(U −u) dy (10–77) Finally, we define momentum thickness 𝜃 such that the viscous drag force on the plate per unit width into the page is equal to 𝜌U 2 times 𝜃, i.e., FD, x w = ρ ∫ Y 0 u (U −u) dy ≡ρU2𝜃 (10–78) In words, Momentum thickness is defined as the loss of momentum flux per unit width divided by ρU2 due to the presence of the growing boundary layer.
Equation 10–78 reduces to 𝜃= ∫ Y 0 u U (1 −u U) dy (10–79) Streamline height Y can be any value, as long as the streamline taken as the upper surface of the control volume is above the boundary layer. Since u = U for any y greater than Y, we may replace Y by infinity in Eq. 10–79 with no change in the value of 𝜃, Momentum thickness: 𝜃= ∫ ∞ 0 u U (1 −u U) dy (10–80) For the specific case of the Blasius solution for a laminar flat plate bound ary layer (Example 10–10), we integrate Eq. 10–80 numerically to obtain Momentum thickness, laminar flat plate: 𝜃 x = 0.664 √Rex (10–81) We note that the equation for 𝜃 is the same as that for 𝛿 or for 𝛿 but with a different constant. In fact, for laminar flow over a flat plate, 𝜃 turns out to be approximately 13.5 percent of 𝛿 at any x-location, as indicated in Fig. 10–111.
It is no coincidence that 𝜃/x (Eq. 10–81) is identical to Cf, x (Eq. 10 of Example 10–10)—both are derived from skin friction drag on the plate.
Turbulent Flat Plate Boundary Layer It is beyond the scope of this text to derive or attempt to solve the turbulent flow boundary layer equations. Expressions for the boundary layer profile Free-stream mass flow Mass flow deficit due to boundary layer Wall x δ(x) δ(x) u U(x) y x FIGURE 10–110 Comparison of the area under the boundary layer profile, representing the mass flow deficit, and the area generated by a chunk of free-stream fluid of thickness 𝛿. To satisfy conservation of mass, these two areas must be identical.
V x U(x) = V Boundary layer δ(x) δ(x) y θ(x) FIGURE 10–111 For a laminar flat plate boundary layer, displacement thickness is 35.0 percent of 𝛿, and momentum thickness is 13.5 percent of 𝛿.
cen96537_ch10_569-610.indd 576 29/12/16 4:06 pm 577 CHAPTER 10 shape and other properties of the turbulent boundary layer are obtained empirically (or at best semi-empirically), since we cannot solve the bound ary layer equations for turbulent flow. Note also that turbulent flows are inherently unsteady, and the instantaneous velocity profile shape varies with time (Fig. 10–112). Thus, all turbulent expressions discussed here represent time-averaged values. One common empirical approximation for the time-averaged velocity profile of a turbulent flat plate boundary layer is the one-seventh-power law, u U ≅( y 𝛿) 1/ 7 for y ≤𝛿, → u U ≅1 for y > 𝛿 (10–82) Note that in the approximation of Eq. 10–82, 𝛿 is not the 99 percent boundary layer thickness, but rather the actual edge of the boundary layer, unlike the definition of 𝛿 for laminar flow. Equation 10–82 is plotted in Fig. 10–113. For comparison, the laminar flat plate boundary layer profile (a numerical solution of the Blasius equations Fig. 10–99) is also plotted in Fig. 10–113, using y/𝛿 for the vertical axis in place of similarity variable 𝜂. You can see that if the laminar and turbulent boundary layers were the same thickness, the turbulent one would be much fuller than the laminar one. In other words, the turbulent boundary layer would “hug” the wall more closely, filling the boundary layer with higher-speed flow close to the wall. This is due to the large turbulent eddies that transport high-speed fluid from the outer part of the boundary layer down to the lower parts of the boundary layer (and vice versa). In other words, a turbulent boundary layer has a much greater degree of mixing when compared to a laminar bound ary layer. In the laminar case, fluid mixes slowly due to viscous diffusion. However, the large eddies in a turbulent flow promote much more rapid and thorough mixing. The approximate turbulent boundary layer velocity profile shape of Eq. 10–82 is not physically meaningful very close to the wall (y → 0) since it predicts that the slope (𝜕u/𝜕y) is infinite at y = 0. While the slope at the wall is very large for a turbulent boundary layer, it is nevertheless finite. This large slope at the wall leads to a very high wall shear stress, 𝜏w = 𝜇(𝜕u/𝜕y)y=0, and, therefore, correspondingly high skin friction along the surface of the plate (as compared to a laminar boundary layer of the same thickness). The skin friction drag produced by both laminar and turbulent boundary layers is discussed in greater detail in Chap. 11. A nondimensionalized plot such as that of Fig. 10–113 is somewhat mis leading, since the turbulent boundary layer would actually be much thicker than the corresponding laminar boundary layer at the same Reynolds number. This fact is illustrated in physical variables in Example 10–12. We compare in Table 10–4 expressions for 𝛿, 𝛿, 𝜃, and Cf, x for lami nar and turbulent boundary layers on a smooth flat plate. The turbulent expressions are based on the one-seventh-power law of Eq. 10–82. Note that the expressions in Table 10–4 for the turbulent flat plate boundary layer are valid only for a very smooth surface. Even a small amount of surface roughness greatly affects properties of the turbulent boundary layer, such as momentum thickness and local skin friction coefficient. The effect of surface roughness on a turbulent flat plate boundary layer is discussed in greater detail in Chap. 11.
y U 0 u δ FIGURE 10–112 Illustration of the unsteadiness of a turbulent boundary layer; the thin, wavy black lines are instantaneous profiles, and the thick blue line is a long time-averaged profile.
1.2 0.8 1 0.6 0.4 — δ 0.2 0 0 0.2 0.4 0.6 u/U 0.8 1 y Laminar Turbulent FIGURE 10–113 Comparison of laminar and turbulent flat plate boundary layer profiles, nondimensionalized by boundary layer thickness.
cen96537_ch10_569-610.indd 577 29/12/16 4:06 pm 578 APPROXIMATE SOLUTIONS OF THE N–S EQ TABLE 10–4 Summary of expressions for laminar and turbulent boundary layers on a smooth flat plate aligned parallel to a uniform stream Property Laminar (a) Turbulent(†) (b) Turbulent(‡) Boundary layer thickness Displacement thickness Momentum thickness Local skin friction coefficient 𝛿 x = 4.91 √Rex 𝛿 x = 1.72 √Rex 𝜃 x = 0.664 √Rex Cf, x = 0.664 √Rex 𝛿 x ≅ 0.16 (Rex)1/7 𝛿 x ≅0.020 (Rex)1/7 𝜃 x ≅0.016 (Rex)1/7 Cf, x ≅0.027 (Rex)1/7 𝛿 x ≅ 0.38 (Rex)1/5 𝛿 x ≅0.048 (Rex)1/5 𝜃 x ≅0.037 (Rex)1/5 Cf, x ≅0.059 (Rex)1/5 Laminar values are exact and are listed to three significant digits, but turbulent values are listed to only two significant digits due to the large uncertainty affiliated with all turbulent flow fields.
† Obtained from one-seventh-power law.
‡ Obtained from one-seventh-power law combined with empirical data for turbulent flow through smooth pipes.
y x L V δlaminar δturbulent U(x) = V FIGURE 10–114 Comparison of laminar and turbulent boundary layers for flow of air over a flat plate for Example 10–12 (bound ary layer thickness exaggerated).
EXAMPLE 10–12 Comparison of Laminar and Turbulent Boundary Layers Air at 20°C flows at V = 10.0 m/s over a smooth flat plate of length L = 1.52 m (Fig. 10–114). (a) Plot and compare the laminar and turbulent boundary layer profiles in physical variables (u as a function of y) at x = L. (b) Compare the values of local skin friction coefficient for the two cases at x = L. (c) Plot and com pare the growth of the laminar and turbulent boundary layers.
SOLUTION We are to compare laminar versus turbulent boundary layer pro files, local skin friction coefficient, and boundary layer thickness at the end of a flat plate.
Assumptions 1 The plate is smooth, and the free stream is calm and uniform. 2 The flow is steady in the mean. 3 The plate is infinitesimally thin and is aligned parallel to the free stream.
Properties The kinematic viscosity of air at 20°C is ν = 1.516 × 10−5 m2/s.
Analysis (a) First we calculate the Reynolds number at x = L, Rex = Vx 𝜈= (10.0 m/s)(1.52 m) 1.516 × 10−5 m2/s = 1.00 × 106 This value of Rex is in the transitional region between laminar and turbulent, according to Fig. 10–81. Thus, a comparison between the laminar and turbulent velocity profiles is appropriate. For the laminar case, we multiply the y/𝛿 values of Fig. 10–113 by 𝛿laminar, where 𝛿laminar = 4.91x √Rex = 4.91(1520 mm) √1.00 × 106 = 7.46 mm (1) cen96537_ch10_569-610.indd 578 29/12/16 4:06 pm 579 CHAPTER 10 This gives us y-values in units of mm. Similarly, we multiply the u/U values of Fig. 10–113 by U (U = V = 10.0 m/s) to obtain u in units of m/s. We plot the laminar boundary layer profile in physical variables in Fig. 10–115. We calculate the turbulent boundary layer thickness at this same x-location using the equation provided in Table 10–4, column (a), 𝛿turbulent ≅0.16x (Rex)1/7 = 0.16(1520 mm) (1.00 × 106)1/7 = 34 mm (2) [The value of 𝛿turbulent based on column (b) of Table 10–4 is somewhat higher, namely 36 mm.] Comparing Eqs. 1 and 2, we see that the turbulent bound ary layer is about 4.5 times thicker than the laminar boundary layer at a Reynolds number of 1.0 × 106. The turbulent boundary layer velocity profile of Eq. 10–82 is converted to physical variables and plotted in Fig. 10–115 for comparison with the laminar profile. The two most striking features of Fig. 10–115 are (1) the turbulent boundary layer is much thicker than the lami nar one, and (2) the slope of u versus y near the wall is much steeper for the turbulent case. (Keep in mind, of course, that very close to the wall the one-seventh-power law does not adequately represent the actual turbulent boundary layer profile.) (b) We use the expressions in Table 10–4 to compare the local skin friction coef ficient for the two cases. For the laminar boundary layer, Cf, x, laminar = 0.664 √Rex = 0.664 √1.00 × 106 = 6.64 × 10−4 (3) and for the turbulent boundary layer, column (a), Cf, x, turbulent ≅0.027 (Rex)1/7 = 0.027 (1.00 × 106)1/7 = 3.8 × 10−3 (4) Comparing Eqs. 3 and 4, the turbulent skin friction value is more than five times larger than the laminar value. If we had used the other expression for turbulent skin friction coefficient, column (b) of Table 10–4, we would have obtained Cf, x, turbulent = 3.7 × 10−3, very close to the value calculated in Eq. 4.
(c) The turbulent calculation assumes that the boundary layer is turbulent from the beginning of the plate. In reality, there is a region of laminar flow, followed by a transition region, and then finally a turbulent region, as illustrated in Fig. 10–81. Nevertheless, it is interesting to compare how 𝛿laminar and 𝛿turbulent grow as functions of x for this flow, assuming either all laminar flow or all turbulent flow. Using the expressions in Table 10–4, both of these are plotted in Fig. 10–116 for comparison.
40 30 20 10 0 0 2 4 6 u, m/s 8 Laminar Turbulent 10 y, mm FIGURE 10–115 Comparison of laminar and turbulent flat plate boundary layer profiles in physical variables at the same x-location. The Reynolds number is Rex = 1.0 × 106.
40 30 20 10 0 0 0.5 1 x, m Laminar Turbulent (a) 1.5 δ, mm Turbulent (b) FIGURE 10–116 Comparison of the growth of a laminar boundary layer and a turbulent boundary layer for the flat plate of Example 10–12.
cen96537_ch10_569-610.indd 579 29/12/16 4:06 pm 580 APPROXIMATE SOLUTIONS OF THE N–S EQ The one-seventh-power law is not the only turbulent boundary layer approximation used by fluid mechanicians. Another common approximation is the log law, a semi-empirical expression that turns out to be valid not only for flat plate boundary layers but also for fully developed turbulent pipe flow velocity profiles (Chap. 8). In fact, the log law turns out to be applicable for nearly all wall-bounded turbulent boundary layers, not just flow over a flat plate. (This fortunate situation enables us to employ the log law approximation close to solid walls in computational fluid dynamics codes, as discussed in Chap. 15.) The log law is commonly expressed in variables nondimensional ized by a characteristic velocity called the friction velocity u. (Note that most authors use u instead of u. We use a subscript to distinguish u, a dimen sional quantity, from u, which we use to indicate a nondimensional velocity.) The log law: u u = 1 𝜅 ln yu 𝜈+ B (10–83) where Friction velocity: u = √ 𝜏w ρ (10–84) and 𝜅 and B are constants; their usual values are 𝜅 = 0.40 to 0.41 and B = 5.0 to 5.5. Unfortunately, the log law suffers from the fact that it does not work very close to the wall (ln 0 is undefined). It also deviates from experimental values close to the boundary layer edge. Nevertheless, Eq. 10–83 applies across a significant portion of the turbulent flat plate boundary layer and is useful because it relates the velocity profile shape to the local value of wall shear stress through Eq. 10–84. A clever expression that is valid all the way to the wall was created by D. B. Spalding in 1961 and is called Spalding’s law of the wall, yu 𝜈= u u + e−𝜅 B [e𝜅 (u/u) −1 −𝜅 (u/u) − [𝜅 (u/u)]2 2 −[𝜅 (u/u)]3 6 ] (10–85) While Eq. 10–85 does a better job than Eq. 10–83 very close to the wall, neither equation is valid in the outer portion of the boundary layer, often called the outer layer or turbulent layer. Coles (1956) introduced an empirical formula called the wake function or the law of the wake that fits the data nicely in this region. Coles’ equation is added to the log law, yielding what some call the wall-wake law, u u = 1 𝜅 ln yu 𝜈+ B + 2Π 𝜅 W( y 𝛿 ) (10–86) Discussion The ordinate in Fig. 10–116 is in mm, while the abscissa is in m for clarity—the boundary layer is incredibly thin, even for the turbulent case. The difference between the turbulent (a) and (b) cases (see Table 10–4) is explained by discrepancies between empirical curve fits and semi-empirical approximations used to obtain the expressions in Table 10–4. This reinforces our decision to report turbulent boundary layer values to at most two significant digits. The real value of 𝛿 will most likely lie somewhere between the laminar and turbulent values plotted in Fig. 10–116 since the Reynolds number by the end of the plate is within the transitional region.
cen96537_ch10_569-610.indd 580 29/12/16 4:06 pm 581 CHAPTER 10 Where П = 0.44 for a flat plate boundary layer, and several expressions for W have been suggested, all of which smoothly change from 0 at the wall (y/𝛿 = 0) to 1 at the outer edge of the boundary layer (y/𝛿 = 1). One popular expression is W( y 𝛿) = sin2( 𝜋 2( y 𝛿)) for y 𝛿< 1 (10–87) EXAMPLE 10–13 Comparison of Turbulent Boundary Layer Profile Equations Air at 20°C flows at V = 10.0 m/s over a smooth flat plate of length L = 15.2 m (Fig. 10–117). Plot the turbulent boundary layer profile in physical variables (u as a function of y) at x = L. Compare the profile generated by the one-seventh-power law, the log law, and Spalding’s law of the wall, assuming that the boundary layer is fully turbulent from the beginning of the plate.
SOLUTION We are to plot the mean boundary layer profile u( y) at the end of a flat plate using three different approximations.
Assumptions 1 The plate is smooth, but there are free-stream fluctuations that tend to cause the boundary layer to transition to turbulence sooner than usual—the boundary layer is turbulent from the beginning of the plate. 2 The flow is steady in the mean. 3 The plate is infinitesimally thin and is aligned parallel to the free stream.
Properties The kinematic viscosity of air at 20°C is ν = 1.516 × 10−5 m2/s.
Analysis First we calculate the Reynolds number at x = L, Rex = Vx 𝜈= (10.0 m/s)(15.2 m) 1.516 × 10−5 m2/s = 1.00 × 107 This value of Rex is well above the transitional Reynolds number for a flat plate boundary layer (Fig. 10–81), so the assumption of turbulent flow from the begin ning of the plate is reasonable. Using the column (a) values of Table 10–4, we estimate the boundary layer thick ness and the local skin friction coefficient at the end of the plate, 𝛿≅0.16x (Rex)1/7 = 0.240 m Cf, x ≅0.027 (Rex)1/7 = 2.70 × 10−3 (1) We calculate the friction velocity by using its definition (Eq. 10–84) and the defini tion of Cf, x (left part of Eq. 10 of Example 10–10), u = √ 𝜏w ρ = U√ Cf, x 2 = (10.0 m/s) √ 2.70 × 10−3 2 = 0.367 m/s (2) where U = constant = V everywhere for a flat plate. It is trivial to generate a plot of the one-seventh-power law (Eq. 10–82), but the log law (Eq. 10–83) is implicit for u as a function of y. Instead, we solve Eq. 10–83 for y as a function of u, y = 𝜈 u e𝜅 (u/u−B) (3) Since we know that u varies from 0 at the wall to U at the boundary layer edge, we are able to plot the log law velocity profile in physical variables using Eq. 3. Finally, Spalding’s law of the wall (Eq. 10–85) is also written in terms of y as a function of u. We plot all three profiles on the same plot for comparison (Fig. 10–118). All three are close, and we cannot distinguish the log law from Spalding’s law on this scale.
y x L V U(x) = V δ(x) FIGURE 10–117 The turbulent boundary layer generated by flow of air over a flat plate for Example 10–13 (boundary layer thickness exaggerated).
250 150 200 100 50 0 0 2 4 6 u, m/s 8 10 y, mm 1/7th power Log law Spalding FIGURE 10–118 Comparison of turbulent flat plate boundary layer profile expressions in physical variables at Rex = 1.0 × 107: one-seventh-power approximation, log law, and Spalding’s law of the wall.
cen96537_ch10_569-610.indd 581 29/12/16 4:06 pm 582 APPROXIMATE SOLUTIONS OF THE N–S EQ Instead of a physical variable plot with linear axes as in Fig. 10–118, a semi-log plot of nondimensional variables is often drawn to magnify the near-wall region. The most common notation in the boundary layer literature for the nondimensional variables is y+ and u+ (inner variables or law of the wall variables), where Law of the wall variables: y+ = yu 𝝼 u+ = u u (4) As you can see, y+ is a type of Reynolds number, and friction velocity u is used to nondimensionalize both y and u. Figure 10–118 is redrawn in Fig. 10–119 using law of the wall variables. The differences between the three approximations, especially near the wall, are much clearer when plotted in this fashion. Typical experimental data are also plotted in Fig. 10–119 for comparison. Spalding’s for mula does the best job overall and is the only expression that follows experimental data near the wall. In the outer part of the boundary layer, the experimental values of u+ level off beyond some value of y+, as does the one-seventh-power law. How ever, both the log law and Spalding’s formula continue indefinitely as a straight line on this semi-log plot.
Discussion Also plotted in Fig. 10–119 is the linear equation u+ = y+. The region very close to the wall (0 < y+ < 5 or 6) is called the viscous sublayer. In this region, turbulent fluctuations are suppressed due to the close proximity of the wall, and the velocity profile is nearly linear. Other names for this region are linear sublayer and laminar sublayer. We see that Spalding’s equation captures the viscous sublayer and blends smoothly into the log law by about y+ = 100. Neither the one-seventh-power law nor the log law is valid as we approach the wall.
30 20 10 0 y+ Log law Spalding 104 103 102 10 1 u+ u+ = y+ Experimental data Wall-wake law 1/7th power FIGURE 10–119 Comparison of turbulent flat plate boundary layer profile expressions in law of the wall variables at Rex = 1.0 × 107: one-seventh-power approximation, log law, Spalding’s law of the wall, and wall-wake law. Typical experimental data and the viscous sublayer equation (u+ = y+) are also shown for comparison.
Boundary layer Boundary layer (a) (b) FIGURE 10–120 Boundary layers with nonzero pressure gradients occur in both external flows and internal flows: (a) boundary layer developing along the fuselage of an airplane and into the wake, and (b) boundary layer growing on the wall of a diffuser (boundary layer thickness exaggerated in both cases).
Boundary Layers with Pressure Gradients So far we have spent most of our discussion on flat plate boundary lay ers. Of more practical concern for engineers are boundary layers on walls of arbitrary shape. These include external flows over bodies immersed in a free stream (Fig. 10–120a), as well as some internal flows like the walls of wind tunnels and other large ducts in which boundary layers develop along the walls (Fig. 10–120b). Just as with the zero pressure gradient flat plate boundary layer discussed earlier, boundary layers with nonzero pressure gradients may be laminar or turbulent. We often use the flat plate boundary layer results as ballpark estimates for such things as location of transition to turbulence, cen96537_ch10_569-610.indd 582 29/12/16 4:06 pm 583 CHAPTER 10 boundary layer thickness, skin friction, etc. However, when more accuracy is needed we must solve the boundary layer equations (Eqs. 10–71 for the steady, laminar, two-dimensional case) using the procedure outlined in Fig. 10–93. The analysis is harder than that for a flat plate since the pressure gradient term (U dU/dx) in the x-momentum equation is nonzero. Such an analysis can quickly get quite involved, especially for the case of three-dimensional flows. Therefore, we discuss only some qualitative features of boundary layers with pressure gradients, leaving detailed solutions of the boundary layer equations to higher-level fluid mechanics textbooks (e.g., Panton, 2005; White, 2005). First some terminology. When the flow in the inviscid and/or irrotational outer flow region (outside of the boundary layer) accelerates, U(x) increases and P(x) decreases. We refer to this as a favorable pressure gradient. It is favorable or desirable because the boundary layer in such an accelerating flow is usually thin, hugs closely to the wall, and therefore is not likely to separate from the wall. When the outer flow decelerates, U(x) decreases, P(x) increases, and we have an unfavorable or adverse pressure gradient. As its name implies, this condition is not desirable because the boundary layer is usually thicker, does not hug closely to the wall, and is much more likely to separate from the wall. In a typical external flow, such as flow over an airplane wing (Fig. 10–121), the boundary layer in the front portion of the body is subjected to a favorable pressure gradient, while that in the rear portion is subjected to an adverse pressure gradient. If the adverse pressure gradient is strong enough (dP/dx = −U dU/dx is large), the boundary layer is likely to separate off the wall. Examples of flow separation are shown in Fig. 10–122 for both external and internal flows. In Fig. 10–122a is sketched an airfoil at a moderate angle of attack. The boundary layer remains attached over the entire lower surface of the airfoil, but it separates somewhere near the rear of the upper sur face as sketched. The closed streamline indicates a region of recirculating flow called a separation bubble. As pointed out previously, the boundary layer equations are parabolic, meaning that no information can be passed upstream from the downstream boundary. However, separation leads to reverse flow near the wall, destroying the parabolic nature of the flow field, and rendering the boundary layer equations inapplicable.
The boundary layer equations are not valid downstream of a separation point because of reverse flow in the separation bubble.
In such cases, the full Navier–Stokes equations must be used in place of the boundary layer approximation. From the point of view of the boundary layer procedure of Fig. 10–93, the procedure breaks down because the outer flow calculated in step 1 is no longer valid when separation occurs, especially beyond the separation point (compare Fig. 10–121 to Fig. 10–122a). Figure 10–122b shows the classic case of an airfoil at too high of an angle of attack, in which the separation point moves near the front of the airfoil; Favorable Adverse FIGURE 10–121 The boundary layer along a body immersed in a free stream is typically exposed to a favorable pressure gradient in the front portion of the body and an adverse pressure gradient in the rear portion of the body.
Separation point Separation point Separation point (a) (b) (c) FIGURE 10–122 Examples of boundary layer separation in regions of adverse pressure gradient: (a) an airplane wing at a moderate angle of attack, (b) the same wing at a high angle of attack (a stalled wing), and (c) a wide-angle diffuser in which the boundary layer cannot remain attached and separates on one side.
cen96537_ch10_569-610.indd 583 29/12/16 4:06 pm 584 APPROXIMATE SOLUTIONS OF THE N–S EQ the separation bubble covers nearly the entire upper surface of the airfoil—a condition known as stall. Stall is accompanied by a loss of lift and a marked increase in aerodynamic drag, as discussed in more detail in Chap. 11. Flow separation may also occur in internal flows, such as in the adverse pressure gradient region of a diffuser (Fig. 10–122c). As sketched, separation often occurs asymmetrically on one side of the diffuser only. As with an airfoil with flow separation, the outer flow calculation in the diffuser is no longer meaningful, and the boundary layer equations are not valid. Flow separation in a diffuser leads to a significant decrease of pressure recovery, and such conditions in a diffuser are also referred to as stall conditions. We can learn a lot about the velocity profile shape under various pressure gradient conditions by examining the boundary layer momentum equation right at the wall. Since the velocity is zero at the wall (no-slip condition), the entire left side of Eq. 10–71b disappears, leaving only the pressure gra dient term and the viscous term, which must balance, At the wall: 𝜈 ( ∂2u ∂y2) y=0 = −U dU dx = 1 ρ dP dx (10–88) Under favorable pressure gradient conditions (accelerating outer flow), dU/dx is positive, and by Eq. 10–88, the second derivative of u at the wall is negative, i.e., (𝜕2u/𝜕y2)y=0 < 0. We know that 𝜕2u/𝜕y2 must remain negative as u approaches U(x) at the edge of the boundary layer. Thus, we expect the velocity profile across the boundary layer to be rounded, without any inflection point, as sketched in Fig. 10–123a. Under zero pressure gradient conditions, (𝜕2u/𝜕y2)y=0 is zero, implying a linear growth of u with respect to y near the wall, as sketched in Fig. 10–123b. (This is verified by the Bla sius boundary layer profile for the zero pressure gradient boundary layer on a flat plate, as shown in Fig. 10–99.) For adverse pressure gradients, dU/dx is negative and Eq. 10–86 demands that (𝜕2u/𝜕y2)y=0 be positive. However, since 𝜕2u/𝜕y2 must be negative as u approaches U(x) at the edge of the boundary layer, there has to be an inflection point (𝜕2u/𝜕y2 = 0) somewhere in the boundary layer, as illustrated in Fig. 10–123c. The first derivative of u with respect to y at the wall is directly proportional to 𝜏w, the wall shear stress [𝜏w = 𝜇 (𝜕u/𝜕y)y=0]. Comparison of (𝜕u/𝜕y)y=0 in Fig. 10–123a through c reveals that 𝜏w is largest for favorable pressure gradients and smallest for adverse pressure gradients. Boundary layer thickness increases as the pressure gradient changes sign, as also illustrated in Fig. 10–123.
If the adverse pressure gradient is large enough, (𝜕u/𝜕y)y=0 becomes zero (Fig. 10–123d ); this location along a wall is the separation point, beyond which there is reverse flow and a separation bubble (Fig. 10–123e). Notice that beyond the separation point 𝜏w is negative due to the negative value of (𝜕u/𝜕y)y=0. As mentioned previously, the boundary layer equations break down in regions of reverse flow. Thus, the boundary layer approximation may be appropriate up to the separation point, but not beyond. We use computational fluid dynamics (CFD) to illustrate flow separation for the case of flow over a bump along a wall. The flow is steady and two-dimensional, and Fig. 10–124a shows outer flow streamlines generated by a solution of the Euler equation. Without the viscous terms there is no separation, and the streamlines are symmetric fore and aft. As indicated on the figure, the front portion of the bump experiences an accelerating flow and hence a x δ(x) u U(x) y τw x δ(x) u U(x) y τw x δ(x) u U(x) y τw x δ(x) u U(x) y τw = 0 x δ(x) u U(x) y τw (a) (b) (c) (e) (d) Reverse flow FIGURE 10–123 Comparison of boundary layer profile shape as a function of pressure gradient (dP/dx = −U dU/dx): (a) favorable, (b) zero, (c) mild adverse, (d) critical adverse (separation point), and (e) large adverse; inflection points are indicated by red circles, and wall shear stress 𝜏w = 𝜇 (𝜕u/𝜕y)y=0 is sketched for each case.
cen96537_ch10_569-610.indd 584 29/12/16 4:06 pm 585 CHAPTER 10 Flow direction (a) (d) Favorable Bump surface Adverse Approximate location of separation point Reverse flow Flow direction (b) Separation bubble Reverse flow Approximate location of separation point Bump surface (c) Approximate location of separation point Reverse flow Dividing streamline FIGURE 10–124 CFD calculations of flow over a two-dimensional bump: (a) solution of the Euler equation with outer flow streamlines plotted (no flow separation), (b) laminar flow solution showing flow separation on the downstream side of the bump, (c) close-up view of streamlines near the separation point, and (d ) close-up view of velocity vectors, same view as (c). The dashed red line is a dividing streamline—fluid below this streamline is “trapped” in the recirculating separation bubble.
cen96537_ch10_569-610.indd 585 29/12/16 4:06 pm 586 APPROXIMATE SOLUTIONS OF THE N–S EQ favorable pressure gradient. The rear portion experiences a decelerating flow and an adverse pressure gradient. When the full (laminar) Navier–Stokes equation is solved, the viscous terms lead to flow separation off the rear end of the bump, as seen in Fig. 10–124b. Keep in mind that this is a Navier– Stokes solution, not a boundary layer solution; nevertheless it illustrates the process of flow separation in the boundary layer. The approximate location of the separation point is indicated in Fig. 10–124b, and the dashed red line is a type of dividing streamline. Fluid below this streamline is caught in the separation bubble, while fluid above this streamline continues downstream. A close-up view of streamlines is shown in Fig. 10–124c, and velocity vectors are plotted in Fig. 10–124d using the same close-up view. Reverse flow in the lower portion of the separation bubble is clearly visible. Also, there is a strong y-component of velocity beyond the separation point, and the outer flow is no longer nearly parallel to the wall. In fact, the separated outer flow is nothing like the original outer flow of Fig. 10–124a. This is typical and represents a serious deficiency in the boundary layer approach. Namely, the boundary layer equations may be able to predict the location of the separation point fairly well, but cannot predict anything beyond the separation point. In some cases the outer flow changes significantly upstream of the separation point as well, and the boundary layer approximation gives erroneous results.
The boundary layer approximation is only as good as the outer flow solution; if the outer flow is significantly altered by flow separation, the boundary layer approximation is erroneous. The boundary layers sketched in Fig. 10–123 and the flow separation velocity vectors plotted in Fig. 10–124 are for laminar flow. Turbulent boundary layers have qualitatively similar behavior, although as discussed previously, the mean velocity profile of a turbulent boundary layer is much fuller than a laminar boundary layer under similar conditions. Thus a stron ger adverse pressure gradient is required to separate a turbulent boundary layer. We make the following general statement: Turbulent boundary layers are more resistant to flow separation than are laminar boundary layers exposed to the same adverse pressure gradient.
Experimental evidence for this statement is shown in Fig. 10–125, in which the outer flow is attempting a sharp turn through a 20° angle. The laminar FIGURE 10–125 Flow visualization comparison of laminar and turbulent boundary layers in an adverse pressure gradient; flow is from left to right. (a) The laminar boundary layer separates at the corner, but (b) the turbulent one does not. Photographs taken by M. R. Head in 1982 as visualized with titanium tetrachloride.
Head, M. R. 1982 in Flow Visualization II, W. Merzkirch, ed., pp. 399–403. Washington: Hemisphere.
(a) (b) cen96537_ch10_569-610.indd 586 29/12/16 4:06 pm 587 CHAPTER 10 Outer flow Bump surface FIGURE 10–126 CFD calculation of turbulent flow over the same bump as that of Fig. 10–124. Compared to the laminar result of Fig. 10–124b, the turbulent boundary layer is more resistant to flow separation and does not separate in the adverse pressure gradient region in the rear portion of the bump.
boundary layer (Fig. 10–125a) cannot negotiate the sharp turn, and separates at the corner. The turbulent boundary layer on the other hand (Fig. 10–125b) manages to remain attached around the sharp corner. As another example, flow over the same bump as that of Fig. 10–124 is recalculated, but with turbulence modeled in the simulation. Stream lines generated by the turbulent CFD calculation are shown in Fig. 10–126. Notice that the turbulent boundary layer remains attached (no flow separa tion), in contrast to the laminar boundary layer that separates off the rear portion of the bump. In the turbulent case, the outer flow Euler solution (Fig. 10–124a) is a reasonable approximation over the entire bump since there is no flow separation and since the boundary layer remains very thin. A similar situation occurs for flow over bluff objects like spheres. A smooth golf ball, for example, would maintain a laminar boundary layer on its surface, and the boundary layer would separate fairly easily, leading to large aerodynamic drag. Golf balls have dimples (a type of surface rough ness) in order to create an early transition to a turbulent boundary layer. Flow still separates from the golf ball surface, but much farther downstream in the boundary layer, resulting in significantly reduced aerodynamic drag. This is discussed in more detail in Chap. 11.
The Momentum Integral Technique for Boundary Layers In many practical engineering applications, we do not need to know all the details inside the boundary layer; rather we seek reasonable estimates of gross features of the boundary layer such as boundary layer thickness and skin friction coefficient. The momentum integral technique utilizes a con trol volume approach to obtain such quantitative approximations of bound ary layer properties along surfaces with zero or nonzero pressure gradients. The momentum integral technique is straightforward, and in some applications does not require use of a computer. It is valid for both laminar and turbulent boundary layers. We begin with the control volume sketched in Fig. 10–127. The bottom of the control volume is the wall at y = 0, and the top is at y = Y, high enough to enclose the entire height of the boundary layer. The control volume is an infinitesimally thin slice of width dx in the x-direction. In accordance with τw U(x) δ(x) Y u δ(x + dx) y CV BL x dx Pleft face Pright face x + dx FIGURE 10–127 Control volume (dashed red line) used in derivation of the momentum integral equation.
cen96537_ch10_569-610.indd 587 29/12/16 4:06 pm 588 APPROXIMATE SOLUTIONS OF THE N–S EQ the boundary layer approximation, 𝜕P/𝜕y = 0, so we assume that pressure P acts along the entire left face of the control volume, Pleft face = P In the general case with nonzero pressure gradient, the pressure on the right face of the control volume differs from that on the left face. Using a first-order truncated Taylor series approximation (Chap. 9), we set Pright face = P + dP dx dx In a similar manner we write the incoming mass flow rate through the left face as m · left face = ρw ∫ Y 0 u dy (10–89) and the outgoing mass through the right face as m · right face = ρw[∫ Y 0 u dy + d dx (∫ Y 0 u dy) dx] (10–90) where w is the width of the control volume into the page in Fig. 10–127. If you prefer, you can set w to unit width; it will cancel out later anyway. Since Eq. 10–90 differs from Eq. 10–89, and since no flow crosses the bottom of the control volume (the wall), mass must flow into or out of the top face of the control volume. We illustrate this in Fig. 10–128 for the case of a growing boundary layer in which m . right face < m . left face, and m . top is positive (mass flows out). Conservation of mass over the control volume yields m · top = −ρw d dx (∫ Y 0 u dy) dx (10–91) We now apply conservation of x-momentum for the chosen control vol ume. The x-momentum is brought in through the left face and is removed through the right and top faces of the control volume. The net momentum flux out of the control volume must be balanced by the force due to the shear stress acting on the control volume by the wall and the net pressure force on the control surface, as shown in Fig. 10–127. The steady control volume x-momentum equation is thus ∑Fx, body + ∑Fx, surface ignore gravity YwP −Yw(P + dP dx dx) −w dx 𝜏w = ∫left face ρuV › ·n › dA + ∫right face ρuV ›·n › dA + ∫top ρuV › ·n › dA where the momentum flux through the top surface of the control volume is taken as the mass flow rate through that surface times U. Some of the terms cancel, and we rewrite the equation as −Y dP dx −𝜏w = ρ d dx (∫ Y 0 u2 dy) −ρU d dx (∫ Y 0 u dy) (10–92) −ρw∫ Y 0 u2 dy ρw[∫ Y 0 u2 dy + d dx (∫ Y 0 u2 dy)dx] m · topU dmtop BL y x dx x + dx ⋅ mright face ⋅ mleft face ⋅ FIGURE 10–128 Mass flow balance on the control volume of Fig. 10–127.
cen96537_ch10_569-610.indd 588 29/12/16 4:06 pm 589 CHAPTER 10 where we have used Eq. 10–89 for m · top, and w and dx cancel from each remaining term. For convenience we note that Y = ∫Y 0 dy. From the outer flow (Euler equation), dP/dx = −𝜌U dU/dx. After dividing each term in Eq. 10–90 by density 𝜌, we get U dU dx ∫ Y 0 dy − 𝜏w ρ = d dx (∫ Y 0 u2 dy) −U d dx (∫ Y 0 u dy) (10–93) We simplify Eq. 10–93 by utilizing the product rule of differentiation in reverse (Fig. 10–129). After some rearrangement, Eq. 10–91 becomes d dx (∫ Y 0 u(U −u) dy) + dU dx ∫ Y 0 (U −u) dy = 𝜏w ρ where we are able to put U inside the integrals since at any given x-location, U is constant with respect to y (U is a function of x only). We multiply and divide the first term by U2 and the second term by U to get d dx (U2 ∫ ∞ 0 u U (1 −u U) dy) + U dU dx ∫ ∞ 0 (1 −u U) dy = 𝜏w ρ (10–94) where we have also substituted infinity in place of Y in the upper limit of each integral since u = U for all y greater than Y, and thus the value of the integral does not change by this substitution. We previously defined displacement thickness 𝛿 (Eq. 10–72) and momentum thickness 𝜃 (Eq. 10–80) for a flat plate boundary layer. In the general case with nonzero pressure gradient, we define 𝛿 and 𝜃 in the same way, except we use the local value of outer flow velocity, U = U(x), at a given x-location in place of the constant U since U now varies with x. Equation 10–94 is thus written in more compact form as Kármán integral equation: d dx (U2 𝜃) + U dU dx 𝛿 = 𝜏w ρ (10–95) Equation 10–95 is called the Kármán integral equation in honor of Theodor von Kármán (1881–1963), a student of Prandtl, who was the first to derive the equation in 1921. An alternate form of Eq. 10–95 is obtained by performing the product rule on the first term, dividing by U2, and rearranging, Kármán integral equation, alternative form: Cf, x 2 = d𝜃 dx + (2 + H) 𝜃 U dU dx (10–96) where we define shape factor H as Shape factor: H = 𝛿 𝜃 (10–97) and local skin friction coefficient Cf, x as Local skin friction coefficient: Cf, x = 𝜏w 1 2 ρU2 (10–98) Note that both H and Cf, x are functions of x for the general case of a bound ary layer with a nonzero pressure gradient developing along a surface.
Product rule: dx d U u dy = u dy U U dx dU Product rule in reverse: Y 0 – Y 0 dx d dx d + = ∫ ∫ u dy Y 0 ∫ dx dU u dy Y 0 ∫ ( ( ) dx d U u dy Y 0 ∫ ( ) ) u dy Y 0 ∫ ( ) FIGURE 10–129 The product rule is utilized in reverse in the derivation of the momentum integral equation.
cen96537_ch10_569-610.indd 589 29/12/16 4:06 pm 590 APPROXIMATE SOLUTIONS OF THE N–S EQ We emphasize again that the derivation of the Kármán integral equation and Eqs. 10–95 through 10–98 are valid for any steady incompressible boundary layer along a wall, regardless of whether the boundary layer is laminar, turbu lent, or somewhere in between. For the special case of the boundary layer on a flat plate, U(x) = U = constant, and Eq. 10–96 reduces to Kármán integral equation, flat plat boundary layer: Cf, x = 2 d𝜃 dx (10–99) V x U(x) = V δ y u Cf, x δ(x) FIGURE 10–130 The turbulent boundary layer generated by flow over a flat plate for Example 10–14 (boundary layer thickness exaggerated).
EXAMPLE 10–14 Flat Plate Boundary Layer Analysis Using the Kármán Integral Equation Suppose we know only two things about the turbulent boundary layer over a flat plate, namely, the local skin friction coefficient (Fig. 10–130), Cf, x ≅0.027 (Rex)1/7 (1) and the one-seventh-power law approximation for the boundary layer profile shape, u U ≅( y 𝛿) 1/7 for y ≤𝛿 u U ≅1 for y > 𝛿 (2) Using the definitions of displacement thickness and momentum thickness and employing the Kármán integral equation, estimate how 𝛿, 𝛿, and 𝜃 vary with x.
SOLUTION We are to estimate 𝛿, 𝛿, and 𝜃 based on Eqs. 1 and 2.
Assumptions 1 The flow is turbulent, but steady in the mean. 2 The plate is thin and is aligned parallel to the free stream, so that U(x) = V = constant.
Analysis First we substitute Eq. 2 into Eq. 10–80 and integrate to find momen tum thickness, 𝜃= ∫ ∞ 0 u U(1 −u U) dy = ∫ 𝛿 0( y 𝛿) 1/7 (1 −( y 𝛿) 1/7 ) dy = 7 72 𝛿 (3) Similarly, we find displacement thickness by integrating Eq. 10–72, 𝛿 = ∫ ∞ 0 (1 −u U) dy = ∫ 𝛿 0 (1 −( y 𝛿) 1/7 ) dy = 1 8 𝛿 (4) The Kármán integral equation reduces to Eq. 10–97 for a flat plate boundary layer. We substitute Eq. 3 into Eq. 10–97 and rearrange to get Cf, x = 2 d𝜃 dx = 14 72 d𝛿 dx from which d𝛿 dx = 72 14 Cf, x = 72 14 0.027(Rex)−1/7 (5) where we have substituted Eq. 1 for the local skin friction coefficient. Equation 5 can be integrated directly, yielding Boundary layer thickness: 𝜹 x ≅ 0.16 (Rex)1/7 (6) cen96537_ch10_569-610.indd 590 29/12/16 4:06 pm 591 CHAPTER 10 Finally, substitution of Eqs. 3 and 4 into Eq. 6 gives approximations for 𝛿 and 𝜃, Displacement thickness: 𝜹 x ≅0.020 (Rex)1/7 (7) and Momentum thickness: 𝞱 x ≅0.016 (Rex)1/7 (8) Discussion The results agree with the expressions given in column (a) of Table 10–4 to two significant digits. Indeed, many of the expressions in Table 10–4 were generated with the help of the Kármán integral equation.
CAUTION INTEGRATION REQUIRED FIGURE 10–131 Integration of a known (or assumed) velocity profile is required when using the Kármán integral equation. While fairly simple to use, the momentum integral technique suffers from a serious deficiency. Namely, we must know (or guess) the boundary layer profile shape in order to apply the Kármán integral equation (Fig. 10–131). For the case of boundary layers with pressure gradients, boundary layer shape changes with x (as illustrated in Fig. 10–123), further complicating the analysis. Fortunately, the shape of the velocity profile does not need to be known precisely, since integration is very forgiving. Several techniques have been developed that utilize the Kármán integral equation to predict gross features of the boundary layer. Some of these techniques, such as Thwaite’s method, do a very good job for laminar boundary layers. Unfortu nately, the techniques that have been proposed for turbulent boundary layers have not been as successful. Many of the techniques require the assistance of a computer and are beyond the scope of the present textbook.
EXAMPLE 10–15 Drag on the Wall of a Wind Tunnel Test Section A boundary layer develops along the walls of a rectangular wind tunnel. The air is at 20°C and atmospheric pressure. The boundary layer starts upstream of the con traction and grows into the test section (Fig. 10–132). By the time it reaches the test section, the boundary layer is fully turbulent. The boundary layer profile and its thickness are measured at both the beginning (x = x1) and the end (x = x2) of the bottom wall of the wind tunnel test section. The test section is 1.8 m long and 0.50 m wide (into the page in Fig. 10–132). The following measurements are made: 𝛿1 = 4.2 cm 𝛿2 = 7.7 cm V = 10.0 m/s (1) At both locations the boundary layer profile fits better to a one-eighth-power law approximation than to the standard one-seventh-power law approximation, u U ≅( y 𝛿) 1/8 for y ≤𝛿 u U ≅1 for y > 𝛿 (2) Estimate the total skin friction drag force FD acting on the bottom wall of the wind tunnel test section.
SOLUTION We are to estimate the skin friction drag force on the bottom wall of the test section of the wind tunnel (between x = x1 and x = x2).
Properties For air at 20°C, ν = 1.516 × 10−5 m2/s and 𝜌 = 1.204 kg/m3.
δ(x) δ1 δ2 BL y Contraction Test section Boundary layer u FD x1 x x2 x1 V x2 U(x) = V Diffuser (a) (b) FIGURE 10–132 Boundary layer developing along the wind tunnel walls of Example 10–15: (a) overall view, and (b) close-up view of the bottom wall of the test section (boundary layer thickness exaggerated).
cen96537_ch10_569-610.indd 591 29/12/16 4:06 pm 592 APPROXIMATE SOLUTIONS OF THE N–S EQ We end this chapter with some illuminating results from CFD calculations of flow over a two-dimensional, infinitesimally thin flat plate aligned with the free stream (Fig. 10–133). In all cases the plate is 1 m long (L = 1 m), and the fluid is air with constant properties 𝜌 = 1.23 kg/m3 and 𝜇 = 1.79 × 10−5 kg/m·s. We vary free-stream velocity V so that the Reynolds number at the end of the plate (ReL = 𝜌VL/𝜇) ranges from 10−1 (creeping flow) to 105 (laminar but ready to start transitioning to turbulent). All cases are incompressible, steady, laminar Navier–Stokes solutions generated by a commercial CFD code. In Fig. 10–134, we plot velocity vectors for four Reynolds number cases at three x-locations: x = 0 (beginning of the plate), x = 0.5 m (middle of the plate), and x = 1 m (end of the plate). We also plot streamlines in the vicinity of the plate for each case. In Fig. 10–134a, ReL = 0.1, and the creeping flow approximation is rea sonable. The flow field is nearly symmetric fore and aft—typical of creep ing flow over symmetric bodies. Notice how the flow diverges around the plate as if it were of finite thickness. This is due to the large displacement effect caused by viscosity and the no-slip condition. In essence, the flow Assumptions 1 The flow is steady in the mean. 2 The wind tunnel walls diverge slightly to ensure that U(x) = V = constant.
Analysis First we substitute Eq. 2 into Eq. 10–80 and integrate to find momen tum thickness 𝜃, 𝜃= ∫ ∞ 0 u U (1 −u U) dy = ∫ 𝛿 0 ( y 𝛿) 1/8 [1 −( y 𝛿) 1/8 ] dy = 4 45 𝛿 (3) The Kármán integral equation reduces to Eq. 10–97 for a flat plate boundary layer. In terms of the shear stress along the wall, Eq. 10–97 is 𝜏w = 1 2 ρU2Cf, x = ρU2 d𝜃 dx (4) We integrate Eq. 4 from x = x1 to x = x2 to find the skin friction drag force, FD = w ∫ x2 x1 𝜏w dx = wρU2 ∫ x2 x1 d𝜃 dx dx = wρU2(𝜃2 −𝜃1) (5) where w is the width of the wall into the page in Fig. 10–132. After substitution of Eq. 3 into Eq. 5 we obtain FD = wρU2 4 45 (𝛿2 −𝛿1) (6) Finally, substitution of the given numerical values into Eq. 6 yields the drag force, FD = (0.50 m)(1.204 kg/m3)(10.0 m/s)2 4 45 (0.077 −0.042) m ( s2·N kg·m)= 0.19 N Discussion This is a very small force since the newton is itself a small unit of force. The Kármán integral equation would be more difficult to apply if the outer flow velocity U (x) were not constant.
Flat plate Fluid properties V y x L ρ, μ FIGURE 10–133 Flow over an infinitesimally thin flat plate of length L. CFD calculations are reported for ReL ranging from 10−1 to 105.
cen96537_ch10_569-610.indd 592 29/12/16 4:06 pm 593 CHAPTER 10 Plate (a) ReL = 1 × 10–1 (b) ReL = 1 × 101 (c) ReL = 1 × 103 (d) ReL = 1 × 105 y x L FIGURE 10–134 CFD calculations of steady, incompressible, two-dimensional lami nar flow from left to right over a 1-m-long flat plate of infinitesimal thickness; velocity vectors are shown in the left column at three locations along the plate, and streamlines near the plate are shown in the right column. ReL = (a) 0.1, (b) 10, (c) 1000, and (d ) 100,000; only the upper half of the flow field is solved— the lower half is a mirror image. The computational domain extends hundreds of plate lengths beyond what is shown here in order to approximate “infinite” far-field conditions at the edges of the computational domain.
cen96537_ch10_569-610.indd 593 29/12/16 4:06 pm 594 APPROXIMATE SOLUTIONS OF THE N–S EQ velocity near the plate is so small that the rest of the flow “sees” it as a blockage around which the flow must be diverted. The y-component of velocity is significant near both the front and rear of the plate. Finally, the influence of the plate extends tens of plate lengths in all directions into the rest of the flow, which is also typical of creeping flows. The Reynolds number is increased by two orders of magnitude to ReL = 10 in the results shown in Fig. 10–134b. This Reynolds number is too high to be considered creeping flow, but too low for the boundary layer approxima tion to be appropriate. We notice some of the same features as those of the lower Reynolds number case, such as a large displacement of the stream lines and a significant y-component of velocity near the front and rear of the plate. The displacement effect is not as strong, however, and the flow is no longer symmetric fore and aft. We are seeing the effects of inertia as fluid leaves the end of the flat plate; inertia sweeps fluid into the developing wake behind the plate. The influence of the plate on the rest of the flow is still large, but much less so than for the flow at ReL = 0.1. In Fig. 10–134c are shown results of the CFD calculations at ReL = 1000, another increase of two orders of magnitude. At this Reynolds number, inertial effects are starting to dominate over viscous effects throughout the majority of the flow field, and we can start calling this a boundary layer (albeit a fairly thick one). In Fig. 10–135 we calculate the boundary layer thickness using the laminar expression given in Table 10–4. The pre dicted value of 𝛿(L) is about 15 percent of the plate length at ReL = 1000, which is in reasonable agreement with the velocity vector plot at x = L in Fig. 10–134c. Compared to the lower Reynolds number cases of Fig. 10–134a and b, the displacement effect is greatly reduced and any trace of fore–aft symmetry is gone. Finally, the Reynolds number is once again increased by two orders of magnitude to ReL = 100,000 in the results shown in Fig. 10–134d. There is no question about the appropriateness of the boundary layer approxima tion at this large Reynolds number. The CFD results show an extremely thin boundary layer with negligible effect on the outer flow. The streamlines of Fig. 10–134d are nearly parallel everywhere, and you must look closely to see the thin wake region behind the plate. The streamlines in the wake are slightly farther apart there than in the rest of the flow because in the wake region, the velocity is significantly less than the free-stream velocity.
The y-component of velocity is negligible, as is expected in a very thin boundary layer, since the displacement thickness is so small. Profiles of the x-component of velocity are plotted in Fig. 10–136 for each of the four Reynolds numbers of Fig. 10–134, plus some additional cases at other values of ReL. We use a log scale for the vertical axis (y in units of m), since y spans several orders of magnitude. We nondi mensionalize the abscissa as u/U so that the velocity profile shapes can be compared. All the profiles have a somewhat similar shape when plotted in this fashion. However, we notice that some of the profiles have a sig nificant velocity overshoot (u > U) near the outer portion of the velocity profile. This is a direct result of the displacement effect and the effect of inertia as discussed before. At very low values of ReL (ReL ≤ 100), where the displacement effect is most prominent, the velocity overshoot is almost nonexistent. This is explained by the lack of inertia at these low δ(L) = 4.91(1 m) = 0.155 m 1000 x L V U(x) = V FIGURE 10–135 Calculation of boundary layer thickness for a laminar boundary layer on a flat plate at ReL = 1000. This result is compared to the CFD-generated velocity profile at x = L shown in Fig. 10–134c at this same Reynolds number.
cen96537_ch10_569-610.indd 594 29/12/16 4:06 pm 595 CHAPTER 10 1000 100 10 1 y, m 0.1 0.01 0.001 0 0.2 0.4 0.6 u/U 0.8 1 1.2 ReL = 10–1 100 101 102 106 103 105 104 FIGURE 10–136 CFD calculations of steady, incompressible, two-dimensional laminar flow over a flat plate of infinitesimal thickness: nondimensional x velocity component u/U at the end of the plate (x = L) is plotted against vertical distance from the plate, y. Prominent velocity overshoot is observed at moderate Reynolds numbers, but disappears at very low and very high values of ReL.
Reynolds numbers. Without inertia, there is no mechanism to accelerate the flow around the plate; rather, viscosity retards the flow everywhere in the vicinity of the plate, and the influence of the plate extends tens of plate lengths beyond the plate in all directions. For example, at ReL = 10−1, u does not reach 99 percent of U until y ≅ 320 m—more than 300 plate lengths above the plate! At moderate values of the Reynolds number (ReL between about 101 and 104), the displacement effect is significant, and inertial terms are no longer negligible. Hence, fluid is able to acceler ate around the plate and the velocity overshoot is significant. For example, the maximum velocity overshoot is about 5 percent at ReL = 102. At very high values of the Reynolds number (ReL ≥ 105), inertial terms dominate viscous terms, and the boundary layer is so thin that the displacement effect is almost negligible. The small displacement effect leads to very small velocity overshoot. For example, at ReL = 106 the maximum velocity overshoot is only about 0.4 percent. Beyond ReL = 106, laminar flow is no longer physically realistic, and the CFD calculations would need to include the effects of turbulence.
SUMMARY Since the Navier–Stokes equation is difficult to solve, approx imations are often used for practical engineering analyses. As with any approximation, however, we must be sure that the approximation is appropriate in the region of flow being analyzed. In this chapter we examine sev eral approximations and show examples of flow situations in which they are useful. First we nondimensionalize the Navier–Stokes equation, yielding several nondimensional parameters: the Strouhal number (St), Froude number (Fr), Euler number (Eu), and Reynolds number (Re). Further more, for flows without free-surface effects, the hydrostatic pressure component due to gravity can be incorporated into a modified pressure P′, effectively eliminating the grav ity term (and the Froude number) from the Navier–Stokes equation. The nondimensionalized Navier–Stokes equation with modified pressure is [St] ∂V › ∂t + (V › · ∇ › )V › = −[Eu]∇ › Pʹ + [ 1 Re]∇ › 2V › cen96537_ch10_569-610.indd 595 29/12/16 4:06 pm 596 APPROXIMATE SOLUTIONS OF THE N–S EQ When the nondimensional variables (indicated by ) are of order of magnitude unity, the relative importance of each term in the equation depends on the relative magnitude of the nondimensional parameters. For regions of flow in which the Reynolds number is very small, the last term in the equation dominates the terms on the left side, and hence pressure forces must balance viscous forces. If we ignore inertial forces completely, we make the creeping flow approximation, and the Navier–Stokes equation reduces to ∇ › Pʹ ≅𝜇∇2V › Creeping flow is foreign to our everyday observations since our bodies, our automobiles, etc., move about at relatively high Reynolds numbers. The lack of inertia in the creeping flow approximation leads to some very interesting peculiari ties, as discussed in this chapter. We define inviscid regions of flow as regions where the viscous terms are negligible compared to the inertial terms (opposite of creeping flow). In such regions of flow the Navier–Stokes equation reduces to the Euler equation, ρ( ∂V › ∂t + (V › · ∇ › )V › ) = −∇ › Pʹ In inviscid regions of flow, the Euler equation can be manip ulated to derive the Bernoulli equation, valid along stream lines of the flow. Regions of flow in which individual fluid particles do not rotate are called irrotational regions of flow. In such regions, the vorticity of fluid particles is negligibly small, and the viscous term in the Navier–Stokes equation can be neglected, leaving us again with the Euler equation. In addi tion, the Bernoulli equation becomes less restrictive, since the Bernoulli constant is the same everywhere, not just along streamlines. A nice feature of irrotational flow is that ele mentary flow solutions (building block flows) can be added together to generate more complicated flow solutions, a pro cess known as superposition. Since the Euler equation cannot support the no-slip bound ary condition at solid walls, the boundary layer approximation is useful as a bridge between an Euler equation approximation and a full Navier–Stokes solution. We assume that an inviscid and/or irrotational outer flow exists everywhere except in very thin regions close to solid walls or within wakes, jets, and mixing layers. The boundary layer approximation is appropri ate for high Reynolds number flows. However, we recognize that no matter how large the Reynolds number, viscous terms in the Navier–Stokes equation are still important within the thin boundary layer, where the flow is rotational and viscous. The boundary layer equation for steady, incompressible, two-dimensional, laminar flow are ∂u ∂x + ∂𝜐 ∂y = 0 and u ∂u ∂x + 𝜐 ∂u ∂y = U dU dx + 𝜈 ∂2u ∂y2 We define several measures of boundary layer thickness, including the 99 percent thickness 𝛿, the displacement thick ness 𝛿, and the momentum thickness 𝜃. These quantities can be calculated exactly for a laminar boundary layer growing along a flat plate, under conditions of zero pressure gradient. As the Reynolds number increases down the plate, the boundary layer transitions to turbulence; semi-empirical expressions are given in this chapter for a turbulent flat plate boundary layer. The Kármán integral equation is valid for both laminar and turbulent boundary layers exposed to arbitrary nonzero pressure gradients, d dx (U 2𝜃) + U dU dx 𝛿 = 𝜏w ρ This equation is useful for “back of the envelope” estimations of gross boundary layer properties such as boundary layer thickness and skin friction. The approximations presented in this chapter are applied to many practical problems in engineering. Potential flow analysis is useful for calculation of airfoil lift (Chap. 11). We utilize the inviscid approximation in the analysis of compressible flow (Chap. 12), open-channel flow (Chap. 13), and turbo machinery (Chap. 14). In cases where these approximations are not justified, or where more precise calculations are required, the continuity and Navier–Stokes equations are solved numerically using CFD (Chap. 15).
REFERENCES AND SUGGESTED READING 1. D. E. Coles. “The Law of the Wake in the Turbulent Boundary Layer,” J. Fluid Mechanics, 1, pp. 191–226.
2. R. J. Heinsohn and J. M. Cimbala. Indoor Air Quality Engineering. New York: Marcel-Dekker, 2003.
3. P. K. Kundu, I. M. Cohen., and D. R. Dowling. Fluid Mechanics, ed. 5. San Diego, CA: Academic Press, 2011.
4. R. L. Panton. Incompressible Flow, 3rd ed. New York: Wiley, 2005.
5. M. Van Dyke. An Album of Fluid Motion. Stanford, CA: The Parabolic Press, 1982.
6. F. M. White. Viscous Fluid Flow, 3rd ed. New York: McGraw-Hill, 2005.
7. G. T. Yates. “How Microorganisms Move through Water,” American Scientist, 74, pp. 358–365, July–August, 1986.
cen96537_ch10_569-610.indd 596 29/12/16 4:06 pm 597 CHAPTER 10 Guest Author: James A. Liburdy and Brian Daniels, Oregon State University Droplet formation is a complex interaction of inertial, surface tension, and viscous forces. The actual break-off of a drop from a stream of liquid, although studied for almost 200 years, has still not been fully explained. Droplet-on Demand (DoD) is used for such diverse applications as ink-jet printing and DNA analysis in microscale “lab-on-a-chip” devices. DoD requires very uniform droplet sizes, controlled velocities and trajectories, and a high rate of sequential droplet formation. For example, in ink-jet print ing, the typical size of a droplet is 25 to 50 microns (barely visible with the naked eye), the velocities are on the order of 10 m/s, and the droplet forma tion rate can be higher than 20,000 per second. The most common method for forming droplets involves accelerating a stream of liquid, and then allowing surface tension to induce an instability in the stream, which breaks up into individual droplets. In 1879, Lord Rayleigh developed a classical theory for the instability associated with this break-up; his theory is still widely used today to define droplet break-up conditions. A small perturbation to the surface of the liquid stream sets up an undulating pattern along the length of the stream, which causes the stream to break up into droplets whose size is determined by the radius of the stream and the sur face tension of the liquid. However, most DoD systems rely on acceleration of the stream with time-dependent forcing functions in the form of a pressure wave exerted at the inlet of a nozzle. If the pressure wave is very rapid, vis cous effects at the walls are negligible, and the potential flow approximation can be used to predict the flow. Two important nondimensional parameters in DoD are the Ohnesorge number Oh = 𝜇/(𝜌𝜎sa)1/2 and the Weber number We = 𝜌Va/𝜎s, where a is the radius of the nozzle, 𝜎s is the surface tension, and V is the velocity. The Ohnesorge num ber determines when viscous forces are important relative to surface tension forces. In addition, the nondimensional pressure required to form an unstable fluid stream, Pc = Pa/𝜎s, is called the capillary pressure, and the associated capillary time scale for droplets to form is tc = (𝜌a/𝜎s)1/2. When Oh is small, the potential flow approximation is applicable, and the surface shape is con trolled by a balance between surface tension and fluid acceleration. Example surfaces of flow emerging from a nozzle are shown in Fig. 10–137a and b. Surface shape depends on the pressure amplitude and the time scale of the perturbation, and is predicted well using the potential flow approximation. When the pressure is large enough and the pulse is fast enough, the surface ripples, and the center forms a jet stream that eventually breaks off into a drop let (Fig. 10–137c). An area of active research is how to control the size and velocity of these droplets, while producing thousands per second.
References Rayleigh, Lord, “On the Instability of Jets,” Proc. London Math. Soc., 10, pp. 4–13, 1879.
Daniels, B. J., and Liburdy, J. A., “Oscillating Free-Surface Displacement in an Orifice Leading to Droplet Formation,” J. Fluids Engr., 10, pp. 7–8, 2004.
APPLICATION SPOTLIGHT ■ Droplet Formation FIGURE 10–137 Droplet formation starts when a surface becomes unstable to a pressure pulse. Shown here are water surfaces in (a) an 800-micron orifice disturbed by a 5000-Hz pulse and (b) a 1200-micron orifice disturbed by an 8100-Hz pulse. Reflection from the surface causes the image to appear as if the surface wave is both up and down. The wave is axisymmetric, at least for small-amplitude pressure pulses. The higher the frequency, the shorter the wavelength and the smaller the central node. The size of the central node defines the diameter of the liquid jet, which then breaks up into a droplet. (c) Droplet formation from a high-frequency pressure pulse ejected from a 50-micron-diameter orifice. The center liquid stream produces the droplet and is only about 25 percent of the orifice diameter. Ideally, a single droplet forms, but unwanted, “satellite” droplets are often generated along with the main droplet.
Courtesy of James A. Liburdy and Brian Daniels, Oregon State University. Used by permission.
(a) (b) (c) cen96537_ch10_569-610.indd 597 29/12/16 4:06 pm 598 APPROXIMATE SOLUTIONS OF THE N–S EQ PROBLEMS Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and SI users can ignore them. Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software.
Introductory Problems and Modified Pressure 10–1C Discuss how nondimensionalization of the Navier– Stokes equation is helpful in obtaining approximate solutions. Give an example.
10–2C What criteria can you use to determine whether an approximation of the Navier–Stokes equation is appropriate or not? Explain.
10–3C Explain the difference between an “exact” solution of the Navier–Stokes equation (as discussed in Chap. 9) and an approximate solution (as discussed in this chapter).
10–4C Which nondimensional parameter in the nondimen sionalized Navier–Stokes equation is eliminated by use of modified pressure instead of actual pressure? Explain.
10–5C In the nondimensionalized incompressible Navier– Stokes equation (Eq. 10–6), there are four nondimensional parameters. Name each one, explain its physical significance (e.g., the ratio of pressure forces to viscous forces), and dis cuss what it means physically when the parameter is very small or very large.
10–6C What is the most important criterion for use of the modified pressure P′ rather than the thermodynamic pressure P in a solution of the Navier–Stokes equation?
10–7C What is the most significant danger associated with an approximate solution of the Navier–Stokes equation? Give an example that is different than the ones given in this chapter.
10–8C A box fan sits on the floor of a very large room (Fig. P10–8C). Label regions of the flow field that may be approximated as static. Label regions in which the irrota tional approximation is likely to be appropriate. Label regions where the boundary layer approximation may be appropriate. Finally, label regions in which the full Navier–Stokes equa tion most likely needs to be solved (i.e., regions where no approximation is appropriate).
Box fan FIGURE P10–8C 10–9 Write out the three components of the Navier–Stokes equation in Cartesian coordinates in terms of modified pressure. Insert the definition of modified pressure and show that the x-, y-, and z-components are identical to those in terms of regu lar pressure. What is the advantage of using modified pressure?
10–10 Consider steady, incompressible, laminar, fully devel oped, planar Poiseuille flow between two parallel, horizontal plates (velocity and pressure profiles are shown in Fig. P10–10).
At some horizontal location x = x1, the pressure varies linearly with vertical distance z, as sketched. Choose an appropriate datum plane (z = 0), sketch the profile of modi fied pressure all along the vertical slice, and shade in the region representing the hydrostatic pressure component. Discuss.
x x1 u P g FIGURE P10–10 10–11 Consider the planar Poiseuille flow of Prob. 10–10. Discuss how modified pressure varies with downstream dis tance x. In other words, does modified pressure increase, stay the same, or decrease with x? If P′ increases or decreases with x, how does it do so (e.g., linearly, quadratically, expo nentially)? Use a sketch to illustrate your answer.
10–12 In Example 9–18 we solved the Navier–Stokes equa tion for steady, fully developed, laminar flow in a round pipe (Poiseuille flow), neglecting gravity. Now, add back the effect of gravity by re-solving that same problem, but use modified pressure P′ instead of actual pressure P. Specifically, calcu late the actual pressure field and the velocity field. Assume the pipe is horizontal, and let the datum plane z = 0 be at some arbitrary distance under the pipe. Is the actual pressure at the top of the pipe greater than, equal to, or less than that at the bottom of the pipe? Discuss.
10–13 Consider flow of water through a small hole in the bottom of a large cylindrical tank (Fig. P10–13). The flow is laminar everywhere. Jet diameter d is much smaller than tank diameter D, but D is of the same order of magnitude as tank height H. Carrie reasons that she can use the fluid statics approximation everywhere in the tank except near the hole, but wants to validate this approximation mathematically. She lets the characteristic velocity scale in the tank be V = Vtank. The characteristic length scale is tank height H, the characteristic time is the time required to drain the tank tdrain, and the refer ence pressure difference is 𝜌gH (pressure difference from the water surface to the bottom of the tank, assuming fluid statics). cen96537_ch10_569-610.indd 598 29/12/16 4:06 pm 599 CHAPTER 10 Substitute all these scales into the nondimensionalized incom pressible Navier–Stokes equation (Eq. 10–6) and verify by order-of-magnitude analysis that for d ≪ D, only the pressure and gravity terms remain. In particular, compare the order of magnitude of each term and each of the four nondimensional parameters St, Eu, Fr, and Re. (Hint: Vjet ∼√gH.) Under what criteria is Carrie’s approximation appropriate?
FIGURE P10–13 D d H Vtank Vjet ρ, μ g → 10–14 A flow field is simulated by a computational fluid dynamics code that uses the modified pressure in its calcu lations. A profile of modified pressure along a vertical slice through the flow is sketched in Fig. P10–14. The actual pres sure at a point midway through the slice is known, as indi cated on Fig. P10–14. Sketch the profile of actual pressure all along the vertical slice. Discuss.
FIGURE P10–14 z x P' P at a point g → 10–15 In Chap. 9 (Example 9–15), we generated an “exact” solution of the Navier–Stokes equation for fully developed Couette flow between two horizontal flat plates (Fig. P10–15), with gravity acting in the negative z-direction (into the page of Fig. P10–15). We used the actual pressure in that example. Repeat the solution for the x-component of velocity u and pressure P, but use the modified pres sure in your equations. The pressure is P0 at z = 0. Show that you get the same result as previously. Discuss.
Answers: u = Vy/h, P = P0 − 𝜌gz h y, ʋ x, u V Fluid: ρ, μ Moving plate Fixed plate FIGURE P10–15 Creeping Flow 10–16C Discuss why fluid density has negligible influence on the aerodynamic drag on a particle moving in the creeping flow regime.
10–17C Write a one-word description of each of the five terms in the incompressible Navier–Stokes equation, ρ ∂V › ∂t + ρ(V › ·∇ ›)V ›= −∇ ›P + ρg ›+ 𝜇∇2V › I II III IV V When the creeping flow approximation is made, only two of the five terms remain. Which two terms remain, and why is this significant?
10–18 A person drops 3 aluminum balls of diameters 2 mm, 4 mm, and 10 mm into a tank filled with glycerin at 22°C (𝜇 = 1 kg·m/s), and measured the terminal velocities to be 3.2 mm/s, 12.8 mm/s, and 60.4 mm/s, respectively. The measurements are to be compared with theory using Stokes law for drag force acting on a spherical object of diameter D expressed as FD = 3𝜋𝜇 DV for Re ≪ 1. Compare experimental velocities values with those predicted theoretically.
10–19 Repeat Prob. 10–18 by considering the general form of the Stokes law expressed as FD = 3𝜋𝜇 DV + (9𝜋/16)𝜌V 2D2.
10–20 The viscosity of clover honey is listed as a function of temperature in Table P10–20. The specific gravity of the honey is about 1.42 and is not a strong function of tempera ture. The honey is squeezed through a small hole of diameter D = 6.0 mm in the lid of an inverted honey jar. The room and the honey are at T = 20°C. Estimate the maximum speed of the honey through the hole such that the flow can be approx imated as creeping flow. (Assume that Re must be less than 0.1 for the creeping flow approximation to be appropriate.) Repeat your calculation if the temperature is 50°C. Discuss. Answers: 0.22 m/s, 0.012 m/s TABLE P10–20 Viscosity of clover honey at 16 percent moisture content T, °C 𝜇, poise 14 20 30 40 50 70 600 190 65 20 10 3 Poise = g/cm·s Data from Airborne Honey, Ltd., www.airborne.co.nz.
cen96537_ch10_569-610.indd 599 29/12/16 4:06 pm 600 APPROXIMATE SOLUTIONS OF THE N–S EQ 10–21 Estimate the speed at which you would need to swim in room temperature water to be in the creeping flow regime. (An order-of-magnitude estimate will suffice.) Discuss.
10–22 Estimate the speed and Reynolds number of the sperm shown in Fig. 10–10. Is this microorganism swimming under creeping flow conditions? Assume it is swimming in room-temperature water.
10–23 A good swimmer can swim 100 m in about a min ute. If a swimmer’s body is 1.85 m long, how many body lengths does he swim per second? Repeat the calculation for the sperm of Fig. 10–10. In other words, how many body lengths does the sperm swim per second? Use the sperm’s whole body length, not just that of his head, for the calcula tion. Compare the two results and discuss.
10–24 A drop of water in a rain cloud has diameter D = 57.5 μm (Fig. P10–24). The air temperature is 25°C, and its pressure is standard atmospheric pressure. How fast does the air have to move vertically so that the drop will remain suspended in the air? Answer: 0.0971 m/s V D ρ, μ FIGURE P10–24 10–25 For each case, calculate an appropriate Reynolds number and indicate whether the flow can be approximated by the creeping flow equations. (a) A microorganism of diameter 5.0 μm swims in room temperature water at a speed of 0.75 mm/s. (b) Engine oil at 140°C flows in the small gap of a lubricated automobile bearing. The gap is 0.0016 mm thick, and the characteristic velocity is 15 m/s. (c) A fog droplet of diameter 10 μm falls through 30°C air at a speed of 4.0 mm/s.
10–26 A slipper-pad bearing (Fig. P10–26) is often encoun tered in lubrication problems. Oil flows between two blocks; the upper one is stationary, and the lower one is moving in this case. The drawing is not to scale; in actuality, h ≪ L. The thin gap between the blocks converges with increasing x. Specifically, gap height h decreases linearly from h0 at x = 0 to hL at x = L. Typically, the gap height length scale h0 is much smaller than the axial length scale L. This problem is more complicated than simple Couette flow between parallel plates because of the changing gap height. In particular, axial velocity component u is a function of both x and y, and pres sure P varies nonlinearly from P = P0 at x = 0 to P = PL at x = L. (𝜕P/𝜕x is not constant). Gravity forces are negligible in this flow field, which we approximate as two-dimensional, steady, and laminar. In fact, since h is so small and oil is so vis cous, the creeping flow approximations are used in the analy sis of such lubrication problems. Let the characteristic length scale associated with x be L, and let that associated with y be h0 (x ∼ L and y ∼ h0). Let u ∼ V. Assuming creeping flow, gener ate a characteristic scale for pressure difference ΔP = P − P0 in terms of L, h0, 𝜇, and V. Answer: 𝜇VL/h0 2 y x L h0 h(x) u(x, y) μ hL V FIGURE P10–26 10–27 Consider the slipper-pad bearing of Prob. 10–26. (a) Generate a characteristic scale for 𝜐, the y-component of velocity. (b) Perform an order-of-magnitude analysis to com pare the inertial terms to the pressure and viscous terms in the x-momentum equation. Show that when the gap is small (h0 ≪ L) and the Reynolds number is small (Re = 𝜌Vh0/𝜇 ≪ 1), the creeping flow approximation is appropriate. (c) Show that when h0 ≪ L, the creeping flow equations may still be appropriate even if the Reynolds number (Re = 𝜌Vh0/𝜇) is not less than 1. Explain. Answer: (a) Vh0/L 10–28 Consider again the slipper-pad bearing of Prob. 10–26. Perform an order-of-magnitude analysis on the y-momentum equation, and write the final form of the y-momentum equa tion. (Hint: You will need the results of Probs. 10–26 and 10–27.) What can you say about pressure gradient 𝜕P/𝜕y?
10–29 Consider again the slipper-pad bearing of Prob. 10–26. (a) List appropriate boundary conditions on u. (b) Solve the creeping flow approximation of the x-momentum equation to obtain an expression for u as a function of y (and indirectly as a function of x through h and dP/dx, which are functions of x). You may assume that P is not a function of y. Your final expres sion should be written as u(x, y) = f(y, h, dP/dx, V, and 𝜇).
Name the two distinct components of the velocity profile in your result. (c) Nondimensionalize your expression for u using these appropriate scales: x = x/L, y = y/h0, h = h/h0, u = u/V, and P = (P − P0)h0 2/𝜇VL.
10–30 Consider the slipper-pad bearing of Fig. P10–30. The drawing is not to scale; in actuality, h ≪ L. This case differs from that of Prob. 10–26 in that h(x) is not linear; rather h is some known, arbitrary function of x. Write an expression for axial velocity component u as a function of y, h, dP/dx, V, and 𝜇. Discuss any differences between this result and that of Prob. 10–29.
cen96537_ch10_569-610.indd 600 29/12/16 4:06 pm 601 CHAPTER 10 y x L h0 h(x) u(x, y) μ hL V FIGURE P10–30 10–31 For the slipper-pad bearing of Prob. 10–26, use the continuity equation, appropriate boundary conditions, and the one-dimensional Leibniz theorem (see Chap. 4) to show that d dx ∫ h 0 u dy = 0.
10–32 Combine the results of Probs. 10–29 and 10–31 to show that for a two-dimensional slipper-pad bearing, pressure gradient dP/dx is related to gap height h by d dx (h3 dP dx )= 6𝜇U dh dx. This is the steady, two-dimensional form of the more general Reynolds equation for lubrication (Panton, 2005).
10–33 Consider flow through a two-dimensional slipper-pad bearing with linearly decreasing gap height from h0 to hL (Fig. P10–26), namely, h = h0 + 𝛼x, where 𝛼 is the nondimen sional convergence of the gap, 𝛼 = (hL − h0)/L. We note that tan 𝛼 ≅ 𝛼 for very small values of 𝛼. Thus, 𝛼 is approximately the angle of convergence of the upper plate in Fig. P10–26 (𝛼 is negative for this case). Assume that the oil is exposed to atmospheric pressure at both ends of the slipper-pad, so that P = P0 = Patm at x = 0 and P = PL = Patm at x = L. Integrate the Reynolds equation (Prob. 10–32) for this slipper-pad bear ing to generate an expression for P as a function of x.
10–34E A slipper-pad bearing with linearly decreas ing gap height (Fig. P10–26) is being designed for an amusement park ride. Its dimensions are h0 = 1/1000 in (2.54 × 10−5 m), hL = 1/2000 in (1.27 × 10−5 m), and L = 1.0 in (0.0254 m). The lower plate moves at speed V = 10.0 ft/s (3.048 m/s) relative to the upper plate. The oil is engine oil at 40°C. Both ends of the slipper-pad are exposed to atmospheric pressure, as in Prob. 10–33. (a) Calculate the convergence 𝛼, and verify that tan 𝛼 ≅ 𝛼 for this case. (b) Calculate the gage pressure halfway along the slipper-pad (at x = 0.5 in). Comment on the magnitude of the gage pressure. (c) Plot P as a function of x, where x = x/L and P = (P − Patm)h0 2/𝜇VL. (d) Approximately how many pounds (lbf) of weight (load) can this slipper-pad bearing support if it is b = 6.0 in deep (into the page of Fig. P10–26)?
10–35 Discuss what happens when the oil temperature increases significantly as the slipper-pad bearing of Prob. 10–34E is subjected to constant use at the amusement park. In particular, would the load-carrying capacity increase or decrease? Why?
10–36 Is the slipper-pad flow of Prob. 10–34E in the creep ing flow regime? Discuss. Are the results reasonable?
10–37 We saw in Prob. 10–34E that a slipper-pad bear ing can support a large load. If the load were to increase, the gap height would decrease, thereby increasing the pressure in the gap. In this sense, the slipper-pad bearing is “self-adjusting” to varying loads. If the load increases by a fac tor of 2, calculate how much the gap height decreases. Specifi cally, calculate the new value of h0 and the percentage change. Assume that the slope of the upper plate and all other parame ters and dimensions stay the same as those in Prob. 10–34E.
10–38 Consider creeping flow of a sphere of diameter D moving through a fluid at speed V. We gave an expression for drag force, FD = 3𝜋𝜇VD. As we will see in Chapter 11, the drag coefficient CD over three-dimensional bodies is typically defined as CD = FD 1 2 ρV 2 A where A is the frontal area of the body (the area you “see” when looking at the body from upstream). Generate an expression for CD in terms of Reynolds number for this flow.
Inviscid Flow 10–39C What is the main difference between the steady, incompressible Bernoulli equation for irrotational regions of flow, and the steady incompressible Bernoulli equation for rotational but inviscid regions of flow?
10–40C In what way is the Euler equation an approxima tion of the Navier–Stokes equation? Where in a flow field is the Euler equation an appropriate approximation?
10–41 Write out the components of the Euler equation as far as possible in Cartesian coordinates (x, y, z) and (u, 𝜐, w). Assume gravity acts in some arbitrary direction.
10–42 Write out the components of the Euler equation as far as possible in cylindrical coordinates (r, 𝜃, z) and (ur, u𝜃, uz). Assume gravity acts in some arbitrary direction.
10–43 In a certain region of steady, two-dimensional, incom pressible flow, the velocity field is given by V › = (u, 𝜐) = (ax + b) i › + (−ay + cx) j ›. Show that this region of flow can be considered inviscid.
10–44 In the derivation of the Bernoulli equation for regions of inviscid flow, we rewrite the steady, incompress ible Euler equation into a form showing that the gradient of three scalar terms is equal to the velocity vector crossed with the vorticity vector, noting that z is vertically upward, ∇ › ( P ρ + V 2 2 + gz) = V ›× 𝜁 › We then employ some arguments about the direction of the gradient vector and the direction of the cross product of two cen96537_ch10_569-610.indd 601 29/12/16 4:06 pm 602 APPROXIMATE SOLUTIONS OF THE N–S EQ vectors to show that the sum of the three scalar terms must be constant along a streamline. In this problem you will use a different approach to achieve the same result. Namely, take the dot product of both sides of the Euler equation with velocity vector V › and apply some fundamental rules about the dot product of two vectors. Sketches may be helpful.
10–45 In the derivation of the Bernoulli equation for regions of inviscid flow, we use the vector identity (V ›·∇ › )V ›= ∇ › ( V 2 2 ) −V ›× (∇ ›× V ›) Show that this vector identity is satisfied for the case of velocity vector V › in Cartesian coordinates, i.e., V › = u i › + 𝜐 j › + w k ›. For full credit, expand each term as far as possible and show all your work.
10–46 Water at T = 20°C rotates as a rigid body about the z-axis in a spinning cylindrical container (Fig. P10–46).
There are no viscous stresses since the water moves as a solid body; thus the Euler equation is appropriate. (We neglect viscous stresses caused by air acting on the water surface.) Integrate the Euler equation to generate an expression for pressure as a function of r and z everywhere in the water. Write an equation for the shape of the free surface (zsurface as a function of r). (Hint: P = Patm everywhere on the free surface. The flow is rotationally symmetric about the z-axis.) Answer: zsurface = 𝜔2r 2/2g Water Free surface r z ω R P = Patm uθ = ωr ur = uz = 0 FIGURE P10–46 10–47 Repeat Prob. 10–46, except let the rotating fluid be engine oil at 60°C. Discuss.
10–48 Using the results of Prob. 10–46, calculate the Bernoulli constant as a function of radial coordinate r.
Answer: P atm ρ + 𝜔2r2 10–49 Consider steady, incompressible, two-dimensional flow of fluid into a converging duct with straight walls (Fig. P10–49). The volume flow rate is V ·, and the velocity is in the radial direction only, with ur a function of r only. Let b be the width into the page. At the inlet into the converging duct (r = R), ur is known; ur = ur(R). Assuming inviscid flow everywhere, generate an expression for ur as a function of r, R, and ur(R) only. Sketch what the velocity profile at radius r would look like if friction were not neglected (i.e., a real flow) at the same volume flow rate.
r = R r ur(r) Δθ FIGURE P10–49 Irrotational (Potential) Flow 10–50C What flow property determines whether a region of flow is rotational or irrotational? Discuss.
10–51C Consider the flow field produced by a hair dryer (Fig. P10–51C). Identify regions of this flow field that can be approximated as irrotational, and those for which the irrota tional flow approximation would not be appropriate (rotational flow regions).
FIGURE P10–51C 10–52C In an irrotational region of flow, the velocity field can be calculated without need of the momentum equation by solving the Laplace equation for velocity potential function 𝜙, and then solving for the components of V › from the definition of 𝜙, namely, V › = ∇ ›𝜙. Discuss the role of the momentum equation in an irrotational region of flow.
10–53C A subtle point, often missed by students of fluid mechanics (and even their professors!), is that an inviscid region of flow is not the same as an irrotational (poten tial) region of flow (Fig. P10–53C). Discuss the differences and similarities between these two approximations. Give an example of each.
cen96537_ch10_569-610.indd 602 29/12/16 4:06 pm 603 CHAPTER 10 Inviscid?
Irrotational?
?
FIGURE P10–53C 10–54C What is D’Alembert’s paradox? Why is it a paradox?
10–55 Write the Bernoulli equation, and discuss how it dif fers between an inviscid, rotational region of flow and a viscous, irrotational region of flow. Which case is more restrictive (in regards to the Bernoulli equation)?
10–56 Consider a planar irrotational region of flow in the r𝜃-plane. Show that stream function 𝜓 satisfies the Laplace equation in cylindrical coordinates.
10–57 Consider the following steady, two-dimensional, incom pressible velocity field: V › = (u, 𝜐) = (ax + b) i › + (−ay + c) j ›.
Is this flow field irrotational? If so, generate an expression for the velocity potential function. Answers: Yes, a(x2 − y2)/2 + bx + cy + constant 10–58 Consider the following steady, two-dimensional, incom pressible velocity field: V › = (u, 𝜐) = (1 2ay2 + b) i › + (axy + c) j ›. Is this flow field irrotational? If so, generate an expression for the velocity potential function.
10–59 Consider the following steady, two-dimensional, incompressible velocity field: V › = (u, 𝜐) = (1 2ay2 + b) i › + (axy2 + c) j ›. Is this flow field irrotational? If so, generate an expression for the velocity potential function.
10–60 Consider an irrotational line source of strength V . / L in the xy- or r𝜃-plane. The velocity components are ur = ∂𝜙 ∂r = 1 r ∂𝜓 ∂𝜃= V ·/L 2𝜋r and u𝜃= 1 r ∂𝜙 ∂𝜃= −∂𝜓 ∂r = 0. In this chapter, we started with the equation for u𝜃 to generate expressions for the velocity potential function and the stream function for the line source. Repeat the analysis, except start with the equation for ur, showing all your work.
10–61 Consider a steady, two-dimensional, incompress ible, irrotational velocity field specified by its velocity potential function, 𝜙 = 3(x2 − y2) + 3xy − 2x − 5y + 2. (a) Calculate velocity components u and 𝜐. (b) Verify that the velocity field is irrotational in the region in which 𝜙 applies. (c) Generate an expression for the stream function in this region.
10–62 In this chapter, we describe axisymmetric irrotational flow in terms of cylindrical coordinates r and z and velocity components ur and uz. An alternative description of axisymmetric flow arises if we use spherical polar coordinates and set the x-axis as the axis of symmetry. The two relevant directional components are now r and 𝜃, and their corresponding velocity components are ur and u𝜃. In this coordinate system, radial location r is the distance from the origin, and polar angle 𝜃 is the angle of inclination between the radial vector and the axis of rotational symmetry (the x-axis), as sketched in Fig. P10–62; a slice defining the r𝜃-plane is shown. This is a type of two-dimensional flow because there are only two independent spatial vari ables, r and 𝜃. In other words, a solution of the velocity and pressure fields in any r𝜃-plane is sufficient to characterize the entire region of axisymmetric irrotational flow. Write the Laplace equation for 𝜙 in spherical polar coordinates, valid in regions of axisymmetric irrotational flow. (Hint: You may consult a textbook on vector analysis.) Rotational symmetry y or z r ur uθ Axisymmetric body θ x ϕ FIGURE P10–62 10–63 Show that the incompressible continuity equation for axisymmetric flow in spherical polar coordinates, 1 r ∂ ∂r (r2ur) + 1 sin 𝜃 ∂ ∂𝜃 (u𝜃 sin 𝜃) = 0, is identically satisfied by a stream function defined as ur = − 1 r2 sin 𝜃 ∂𝜓 ∂𝜃 and u𝜃= 1 r sin 𝜃 ∂𝜓 ∂r , so long as 𝜓 is a smooth function of r and 𝜃.
10–64 Streamlines in a steady, two-dimensional, incom pressible flow field are sketched in Fig. P10–64. The flow in the region shown is also approximated as irrotational. Sketch what a few equipotential curves (curves of constant potential function) might look like in this flow field. Explain how you arrive at the curves you sketch.
cen96537_ch10_569-610.indd 603 29/12/16 4:06 pm 604 APPROXIMATE SOLUTIONS OF THE N–S EQ Streamlines FIGURE P10–64 10–65 Consider a uniform stream of magnitude V inclined at angle 𝛼 (Fig. P10–65). Assuming incompressible planar irrotational flow, find the velocity potential function and the stream function. Show all your work. Answers: 𝜙 = Vx cos𝛼 + Vy sin𝛼, 𝜓 = Vy cos𝛼 − Vx sin𝛼 y x V α FIGURE P10–65 10–66 In an irrotational region of flow, we write the velocity vector as the gradient of the scalar velocity potential func tion, V › = ∇ ›𝜙. The components of V › in cylindrical coordi nates, (r, 𝜃, z) and (ur, u𝜃, uz), are ur = ∂𝜙 ∂r u𝜃= 1 r ∂𝜙 ∂𝜃 uz = ∂𝜙 ∂z From Chap. 9, we also write the components of the vorticity vector in cylindrical coordinates as 𝜁r = 1 r ∂uz ∂𝜃−∂u𝜃 ∂z , 𝜁𝜃= ∂ur ∂z −∂uz ∂r , and 𝜁z = 1 r ∂ ∂r (ru𝜃) −1 r ∂ur ∂𝜃. Substitute the velocity components into the vorticity components to show that all three components of the vorticity vector are indeed zero in an irrotational region of flow.
10–67 Substitute the components of the velocity vector given in Prob. 10–66 into the Laplace equation in cylindrical coordinates. Showing all your algebra, verify that the Laplace equation is valid in an irrotational region of flow.
10–68 Consider an irrotational line vortex of strength Г in the xy- or r𝜃-plane. The velocity components are ur = ∂𝜙 ∂r = 1 r ∂𝜓 ∂𝜃= 0 and u𝜃= 1 r ∂𝜙 ∂𝜃= −∂𝜓 ∂r = Γ 2𝜋r. Generate expres-sions for the velocity potential function and the stream func tion for the line vortex, showing all your work.
10–69 Water at atmospheric pressure and temperature (𝜌 = 998.2 kg/m3, and 𝜇 = 1.003 × 10−3 kg/m⋅s) at free stream velocity V = 0.100481 m/s flows over a two-dimensional circular cylinder of diameter d = 1.00 m. Approximate the flow as potential flow. (a) Calculate the Reynolds number, based on cylinder diameter. Is Re large enough that potential flow should be a reasonable approximation? (b) Estimate the minimum and maximum speeds |V|min and |V|max (speed is the magnitude of velocity) and the maximum and minimum pressure difference P − P∞ in the flow, along with their respective locations.
10–70 The stream function for steady, incompressible, two-dimensional flow over a circular cylinder of radius a and free-stream velocity V∞ is 𝜓 = V∞ sin𝜃(r − a2/r) for the case in which the flow field is approximated as irrotational (Fig. P10–70).
Generate an expression for the velocity potential function 𝜙 for this flow as a function of r and 𝜃, and parameters V∞ and a.
y a V∞ r θ x FIGURE P10–70 10–71 Superpose a uniform stream of velocity V∞ and a line source of strength V .
/L at the origin. This generates poten tial flow over a two-dimensional half-body called the Rankine half-body (Fig. P10–71). One unique streamline is the dividing streamline that forms a dividing line between free-stream fluid coming from the left and fluid coming from the source. (a) Generate an equation for the dividing stream function 𝜓dividing as a function of V .
/L. (Hint: The dividing streamline intersects the stagnation point at the nose of the body.) (b) Generate an expression for half-height b as a function of V∞ and V .
/L. (Hint: Consider the flow far downstream.) (c) Generate an equa tion for the dividing stream function in the form of r as a func tion of 𝜃, V∞, and V .
/L. (d) Generate an expression for stagnation point distance a as a function of V∞ and V .
/L. (e) Generate an expression for (V/V∞)2 (the squared nondimensional velocity magnitude) anywhere in the flow as a function of a, r, and 𝜃.
cen96537_ch10_569-610.indd 604 29/12/16 4:06 pm 605 CHAPTER 10 V∞ y a b x FIGURE P10–71 Boundary Layers 10–72C We usually think of boundary layers as occurring along solid walls. However, there are other flow situations in which the boundary layer approximation is also appropriate. Name three such flows, and explain why the boundary layer approximation is appropriate.
10–73C In this chapter, we make a statement that the boundary layer approximation “bridges the gap” between the Euler equation and the Navier–Stokes equation. Explain.
10–74C A laminar boundary layer growing along a flat plate is sketched in Fig. P10–74C. Several velocity profiles and the boundary layer thickness 𝛿(x) are also shown. Sketch several streamlines in this flow field. Is the curve represent ing 𝛿(x) a streamline?
10–81C Explain the difference between a favorable and an adverse pressure gradient in a boundary layer. In which case does the pressure increase downstream? Why?
10–82C For each statement, choose whether the statement is true or false and discuss your answer briefly. These statements concern a laminar boundary layer on a flat plate (Fig. P10–82C).
(a) At a given x-location, if the Reynolds number were to increase, the boundary layer thickness would also increase. (b) As outer flow velocity increases, so does the boundary layer thickness.
(c) As the fluid viscosity increases, so does the boundary layer thickness.
(d) As the fluid density increases, so does the boundary layer thickness.
y U(x) = V δ(x) Outer flow Boundary layer V x FIGURE P10–82C V y U(x) = V x δ(x) Boundary layer FIGURE P10–74C 10–75C What is a trip wire, and what is its purpose?
10–76C Discuss the implication of an inflection point in a boundary layer profile. Specifically, does the existence of an inflection point infer a favorable or adverse pressure gradient? Explain.
10–77C Compare flow separation for a laminar versus tur bulent boundary layer. Specifically, which case is more resis tant to flow separation? Why? Based on your answer, explain why golf balls have dimples.
10–78C In your own words, summarize the five steps of the boundary layer procedure.
10–79C In your own words, list at least three “red flags” to look out for when performing laminar boundary layer calculations.
10–80C Two definitions of displacement thickness are given in this chapter. Write both definitions in your own words. For the laminar boundary layer growing on a flat plate, which is larger—boundary layer thickness 𝛿 or dis placement thickness 𝛿? Discuss.
10–83 On a hot day (T = 30°C), a truck moves along the high way at 29.1 m/s. The flat side of the truck is treated as a sim ple, smooth flat–plate boundary layer, to first approximation. Estimate the x-location along the plate where the boundary layer begins to transition to turbulence. How far downstream from the beginning of the plate do you expect the boundary layer to become fully turbulent? Give both answers to one significant digit.
10–84E A boat moves through water (T = 40°F), at 18.0 mi/h.
A flat portion of the boat hull is 3.3 ft long, and is treated as a simple smooth flat plate boundary layer, to first approx imation. Is the boundary layer on this flat part of the hull laminar, transitional, or turbulent? Discuss.
10–85 Air flows parallel to a speed limit sign along the highway at speed V = 8.5 m/s. The temperature of the air is 25°C, and the width W of the sign parallel to the flow direc tion (i.e., its length) is 0.45 m. Is the boundary layer on the sign laminar or turbulent or transitional?
10–86E Air flows through the test section of a small wind tunnel at speed V = 7.5 ft/s. The temperature of the air is cen96537_ch10_569-610.indd 605 29/12/16 4:06 pm 606 APPROXIMATE SOLUTIONS OF THE N–S EQ 80°F, and the length of the wind tunnel test section is 1.5 ft.
Assume that the boundary layer thickness is negligible prior to the start of the test section. Is the boundary layer along the test section wall laminar or turbulent or transi tional? Answer: laminar 10–87E For the small wind tunnel of Prob. 10–86E, assume the flow remains laminar, and estimate the boundary layer thickness, the displacement thickness, and the momentum thickness of the boundary layer at the end of the test section. Give your answers in inches, compare the three results, and discuss.
10–88 Consider the Blasius solution for a laminar flat plate boundary layer. The nondimensional slope at the wall is given by Eq. 8 of Example 10–10. Transform this result to physical variables, and show that Eq. 9 of Example 10–10 is correct.
10–89 Calculate the value of shape factor H for the limiting case of a boundary layer that is infinitesimally thin (Fig. P10–89). This value of H is the minimum possible value.
U(x) x FIGURE P10–89 10–90 A laminar flow wind tunnel has a test section that is 30 cm in diameter and 80 cm in length. The air is at 20°C. At a uniform air speed of 2.0 m/s at the test section inlet, by how much will the centerline air speed accelerate by the end of the test section? Answer: Approx. 6% 10–91 Repeat the calculation of Prob. 10–90, except for a test section of square rather than round cross section, with a 30 cm × 30 cm cross section and a length of 80 cm. Compare the result to that of Prob. 10–90 and discuss.
10–92 Air at 20°C flows at V = 8.5 m/s parallel to a flat plate (Fig. P10–92). The front of the plate is well rounded, and the plate is 40 cm long. The plate thickness is h = 0.75 cm, but because of boundary layer displacement effects, the flow outside the boundary layer “sees” a plate that has larger apparent thickness. Calculate the apparent thickness of the plate (include both sides) at downstream distance x = 10 cm. Answer: 0.895 cm V h x FIGURE P10–92 10–93E A small, axisymmetric, low-speed wind tunnel is built to calibrate hot wires. The diameter of the test section is 6.68 in, and its length is 12.5 in. The air is at 60°F. At a uniform air speed of 5.0 ft/s at the test section inlet, by how much will the centerline air speed accelerate by the end of the test section? What should the engineers do to eliminate this acceleration?
10–94E Air at 70°F flows parallel to a smooth, thin, flat plate at 15.5 ft/s. The plate is 10.6 ft long. Determine whether the boundary layer on the plate is most likely laminar, tur bulent, or somewhere in between (transitional). Compare the boundary layer thickness at the end of the plate for two cases: (a) the boundary layer is laminar everywhere, and (b) the boundary layer is turbulent everywhere. Discuss.
10–95 In order to avoid boundary layer interference, engi neers design a “boundary layer scoop” to skim off the bound ary layer in a large wind tunnel (Fig. P10–95). The scoop is constructed of thin sheet metal. The air is at 20°C, and flows at V = 45.0 m/s. How high (dimension h) should the scoop be at downstream distance x = 1.45 m?
V h x FIGURE P10–95 10–96 Air at 20°C flows at V = 80.0 m/s over a smooth flat plate of length L = 17.5 m. Plot the turbu lent boundary layer profile in physical variables (u as a function of y) at x = L. Compare the profile generated by the one-seventh-power law, the log law, and Spalding’s law of the wall, assuming that the boundary layer is fully turbulent from the beginning of the plate.
10–97 The streamwise velocity component of a steady, incompressible, laminar, flat plate boundary layer of boundary layer thickness 𝛿 is approximated by the simple linear expres sion, u = Uy/𝛿 for y < 𝛿, and u = U for y > 𝛿 (Fig. P10–97). Generate expressions for displacement thickness and momen tum thickness as functions of 𝛿, based on this linear approxi mation. Compare the approximate values of 𝛿/𝛿 and 𝜃/𝛿 to the values of 𝛿/𝛿 and 𝜃/𝛿 obtained from the Blasius solution. Answers: 0.500, 0.167 U(x) = V δ(x) x V FIGURE P10–97 cen96537_ch10_569-610.indd 606 29/12/16 4:06 pm 607 CHAPTER 10 10–98 For the linear approximation of Prob. 10–97, use the definition of local skin friction coefficient and the Kármán integral equation to generate an expression for 𝛿/x. Compare your result to the Blasius expression for 𝛿/x. (Note: You will need the results of Prob. 10–97 to do this problem.) 10–99 Compare shape factor H (defined in Eq. 10–95) for a laminar versus a turbulent boundary layer on a flat plate, assuming that the turbulent boundary layer is turbulent from the beginning of the plate. Discuss. Specifically, why do you suppose H is called a “shape factor”? Answers: 2.59, 1.25 to 1.30 10–100 One dimension of a rectangular flat plate is twice the other. Air at uniform speed flows parallel to the plate, and a laminar boundary layer forms on both sides of the plate. Which orientation—long dimension parallel to the wind (Fig. P10–100a) or short dimension parallel to the wind (Fig. P10–100b)—has the higher drag? Explain.
(a) (b) V V FIGURE P10–100 10–101 Integrate Eq. 5 to obtain Eq. 6 of Example 10–14, showing all your work.
10–102 Consider a turbulent boundary layer on a flat plate. Suppose only two things are known: Cf, x ≅ 0.059 · (Rex)−1/5 and 𝜃 ≅ 0.097𝛿. Use the Kármán integral equation to gener ate an expression for 𝛿/x, and compare your result to column (b) of Table 10–4.
10–103 Air at 30°C flows at a uniform speed of 30.0 m/s along a smooth flat plate. Calculate the approximate x-location along the plate where the boundary layer begins the transition process toward turbulence. At approximately what x-location along the plate is the boundary layer likely to be fully turbulent? Answers: 5 to 6 cm, 1 to 2 m 10–104 Static pressure P is measured at two locations along the wall of a laminar boundary layer (Fig. P10–104).
The measured pressures are P1 and P2, and the distance between the taps is small compared to the characteristic body dimension (Δx = x2 − x1 ≪ L). The outer flow velocity above the boundary layer at point 1 is U1. The fluid density and viscosity are 𝜌 and 𝜇, respectively. Generate an approximate expression for U2, the outer flow velocity above the boundary layer at point 2, in terms of P1, P2, Δx, U1, 𝜌, and 𝜇.
Wall Pressure taps Boundary layer Outer flow x x2 P1 P2 x1 P2 P1 U1 U2 δ FIGURE P10–104 10–105 Consider two pressure taps along the wall of a lami nar boundary layer as in Fig. P10–104. The fluid is air at 25°C, U1 = 13.7 m/s, and the static pressure P1 is 2.96 Pa greater than static pressure P2, as measured by a very sensitive dif ferential pressure transducer. Is outer flow velocity U2 greater than, equal to, or less than outer flow velocity U1? Explain. Estimate U2. Answers: Less than, 13.5 m/s Review Problems 10–106C For each statement, choose whether the statement is true or false, and discuss your answer briefly.
(a) The velocity potential function can be defined for three-dimensional flows.
(b) The vorticity must be zero in order for the stream function to be defined.
(c) The vorticity must be zero in order for the velocity poten tial function to be defined.
(d) The stream function can be defined only for two-dimensional flow fields.
10–107 In this chapter, we discuss solid body rotation (Fig. P10–107) as an example of an inviscid flow that is also rota tional. The velocity components are ur = 0, u𝜃 = 𝜔r, and uz = 0. Compute the viscous term of the 𝜃-component of the Navier–Stokes equation, and discuss. Verify that this velocity field is indeed rotational by computing the z-component of vorticity. Answer: 𝜉z = 2𝜔 cen96537_ch10_569-610.indd 607 29/12/16 4:06 pm 608 APPROXIMATE SOLUTIONS OF THE N–S EQ uθ uθ = ωr r FIGURE P10–107 10–108 Calculate the nine components of the viscous stress tensor in cylindrical coordinates (see Chap. 9) for the veloc ity field of Prob. 10–107. Discuss your results.
10–109 In this chapter, we discuss the line vortex (Fig. P10–109) as an example of an irrotational flow field. The velocity components are ur = 0, u𝜃 = Г/(2𝜋r), and uz = 0.
Compute the viscous term of the 𝜃-component of the Navier–Stokes equation, and discuss. Verify that this veloc ity field is indeed irrotational by computing the z-component of vorticity.
uθ r 2πr L uθ = FIGURE P10–109 10–110 Calculate the nine components of the viscous stress tensor in cylindrical coordinates (see Chap. 9) for the veloc ity field of Prob. 10–109. Discuss.
10–111 Consider a steady, two-dimensional, incompressible, irrotational velocity field specified by its velocity potential function, 𝜙= 6(x2 −y2) + 3x −5y. (a) Calculate velocity components u and 𝜐.
(b) Verify that the velocity field is irrotational in the region in which ϕ applies.
(c) Generate an expression for the stream function in this region.
10–112 The Blasius boundary layer profile is an exact solution of the boundary layer equations for flow over a flat plate. However, the results are somewhat cumbersome to use, since the data appear in tabular form (the solution is numerical). Thus, a simple sine wave approxima tion (Fig. P10–112) is often used in place of the Blasius solu-tion, namely, u(y) ≅U sin ( 𝜋 2 y 𝛿) for y < 𝛿, and u = U for y ≪ 𝛿, where 𝛿 is the boundary layer thickness. Plot the Bla sius profile and the sine wave approximation on the same plot, in nondimensional form (u/U versus y/𝛿), and compare. Is the sine wave profile a reasonable approximation?
U(x) = V δ(x) x V FIGURE P10–112 10–113 The streamwise velocity component of a steady, incompressible, laminar, flat plate boundary layer of bound ary layer thickness 𝛿 is approximated by the sine wave pro file of Prob. 10–112. Generate expressions for displacement thickness and momentum thickness as functions of 𝛿, based on this sine wave approximation. Compare the approximate values of 𝛿/𝛿 and 𝜃/𝛿 to the values of 𝛿/𝛿 and 𝜃/𝛿 obtained from the Blasius solution.
10–114 For the sine wave approximation of Prob. 10–112, use the definition of local skin friction coefficient and the Kármán integral equation to generate an expression for 𝛿/x. Compare your result to the Blasius expression for 𝛿/x. (Note: You will also need the results of Prob. 10–113 to do this problem.) 10–115 Water falls down a vertical pipe by gravity alone. The flow between vertical locations z1 and z2 is fully developed, and velocity profiles at these two locations are sketched in Fig. P10–115. Since there is no forced pressure gradient, pressure P is constant everywhere in the flow (P = Patm). Calculate the modified pressure at locations z1 and z2. Sketch profiles of modified pressure at locations z1 and z2. Discuss.
z z = z1 z = z2 g FIGURE P10–115 cen96537_ch10_569-610.indd 608 29/12/16 4:06 pm 609 CHAPTER 10 10–116 Suppose the vertical pipe of Prob. 10–115 is now horizontal instead. In order to achieve the same volume flow rate as that of Prob. 10–115, we must supply a forced pres sure gradient. Calculate the required pressure drop between two axial locations in the pipe that are the same distance apart as z2 and z1 of Fig. P10–115. How does modified pres sure P′ change between the vertical and horizontal cases?
Fundamentals of Engineering (FE) Exam Problems 10–117 Which choice is not a scaling parameter used to non dimensionalize the equations of motion? (a) Characteristic length, L (b) Characteristic speed, V (c) Characteristic viscosity, 𝜇 (d ) Characteristic frequency, f (e) Gravitational acceleration, g 10–118 Which choice is not a nondimensional variable defined to nondimensionalize the equations of motion?
(a) t = ft (b) x › = x › L (c) V › = V › V (d ) g › = g › g (e) P = P P0 10–119 Which dimensionless parameter does not appear in the nondimensionalized Navier–Stokes equation? (a) Reynolds number (b) Prandtl number (c) Strouhal number (d ) Euler number (e) Froude number 10–120 If pressure P is replaced by modified pressure P′ = P + 𝜌gz in the nondimensionalized Navier–Stokes equation, which dimensionless parameter drops out? (a) Froude number (b) Reynolds number (c) Strouhal number (d ) Euler number (e) Prandtl number 10–121 In creeping flow, the value of Reynolds number is typically (a) Re < 1 (b) Re ≪ 1 (c) Re > 1 (d ) Re ≫ 1 (e) Re = 0 10–122 Which equation is the proper approximate Navier– Stokes equation in dimensional form for creeping flow?
(a) ∇ › P −ρg›= 0 (b) −∇ › P + 𝜇∇ ›2V ›= 0 (c) −∇ › P + ρg›+ 𝜇∇ ›2V ›= 0 (d ) ρ DV › Dt = −∇ › P + ρg›+ 𝜇∇ ›2V › (e) ρ DV › Dt + ∇ › P −ρg›= 0 10–123 If the fluid velocity is zero in a flow field, the Navier–Stokes equation becomes (a) ∇ › P −ρg›= 0 (b) −∇ › P + ρg›+ 𝜇∇ › 2V ›= 0 (c) ρ DV › Dt = −∇ › P + 𝜇∇ ›2V › (d ) ρ DV › Dt = −∇ › P + ρg›+ 𝜇∇ ›2V › (e) ρ DV › Dt + ∇ › P −ρg›= 0 10–124 For creeping flow over a three-dimensional object, the aerodynamic drag on the object does not depend on (a) Velocity, V (b) Fluid viscosity, 𝜇 (c) Characteristic length, L (d ) Fluid density, 𝜌 (e) None of these 10–125 Consider a spherical ash particle of diameter 65 µm, falling from a volcano at a high elevation in air whose tem perature is −50°C and whose pressure is 55 kPa. The density of air is 0.8588 kg/m3 and its viscosity is 1.474 × 10−5 kg/m⋅s.
The density of the particle is 1240 kg/m3. The drag force on a sphere in creeping flow is given by FD = 3𝜋𝜇VD. The terminal velocity of this particle at this altitude is (a) 0.096 m/s (b) 0.123 m/s (c) 0.194 m/s (d ) 0.225 m/s (e) 0.276 m/s 10–126 Which statement is not correct regarding inviscid regions of flow? (a) Inertial forces are not negligible.
(b) Pressure forces are not negligible.
(c) Reynolds number is large.
(d ) Not valid in boundary layers and wakes.
(e) Solid body rotation of a fluid is an example.
10–127 For which regions of flow is the Laplace equation ∇ ›2𝜙= 0 applicable? (a) Irrotational (b) Inviscid (c) Boundary layer (d ) Wake (e) Creeping 10–128 A very thin region of flow near a solid wall where viscous forces and rotationality cannot be ignored is called (a) Inviscid region of flow (b) Irrotational flow (c) Boundary layer (d ) Outer flow region (e) Creeping flow 10–129 Which one of the following is not a flow region where the boundary layer approximation may be appropriate?
(a) Jet (b) Inviscid region (c) Wake (d ) Mixing layer (e) Thin region near a solid wall 10–130 Which statement is not correct regarding the bound ary layer approximation? (a) The higher the Reynolds number, the thinner the bound ary layer.
(b) The boundary layer approximation may be appropriate for free shear layers.
(c) The boundary layer equations are approximations of the Navier–Stokes equation.
(d ) The curve representing boundary layer thickness 𝛿 as a function of x is a streamline.
(e) The boundary layer approximation bridges the gap between the Euler equation and the Navier–Stokes equation.
cen96537_ch10_569-610.indd 609 29/12/16 4:06 pm 610 APPROXIMATE SOLUTIONS OF THE N–S EQ 10–131 For a laminar boundary layer growing on a horizontal flat plate, the boundary layer thickness 𝛿 is not a function of (a) Velocity, V (b) Distance from the leading edge, x (c) Fluid density, 𝜌 (d ) Fluid viscosity, 𝜇 (e) Gravitational acceleration, g 10–132 For flow along a flat plate with x being the distance from the leading edge, the boundary layer thickness grows like (a) x (b) √x (c) x2 (d ) 1/x (e) 1/x2 10–133 Air flows at 25°C with a velocity of 3 m/s in a wind tunnel whose test section is 25 cm long. The displace ment thickness at the end of the test section is (the kinematic viscosity of air is 1.562 × 10−5 m2/s).
(a) 0.955 mm (b) 1.18 mm (c) 1.33 mm (d ) 1.70 mm (e) 1.96 mm 10–134 Air at 15°C flows at 10 m/s over a flat plate of length 3 m. Using one-seventh power law of the turbulent flow, what is the ratio of local skin friction coefficient for the turbulent and laminar flow cases? (The kinematic viscosity of air is 1.470 × 10−5 m2/s.) (a) 4.25 (b) 5.72 (c) 6.31 (d ) 7.29 (e) 8.54 10–135 Water flows at 20°C with a velocity of 1.1 m/s over a flat plate whose length is 15 cm. The boundary layer thickness at the end of the plate is (the density and viscosity of water are 998 kg/m3 and 1.002 × 103 kg/m⋅s, respectively).
(a) 1.14 mm (b) 1.35 mm (c) 1.56 mm (d ) 1.82 mm (e) 2.09 mm 10–136 Air flows at 15°C with a velocity of 12 m/s over a flat plate whose length is 80 cm. Using one-seventh power law of the turbulent flow, what is the boundary layer thick ness at the end of the plate? (The kinematic viscosity of air is 1.470 × 10−5 m2/s.) (a) 1.54 cm (b) 1.89 cm (c) 2.16 cm (d ) 2.45 cm (e) 2.82 cm 10–137 Air flows at 25°C with a velocity of 9 m/s over a flat plate whose length is 40 cm. The momentum thickness at the center of the plate is (the kinematic viscosity of air is 1.562 × 10−5 m2/s).
(a) 0.391 mm (b) 0.443 mm (c) 0.479 mm (d ) 0.758 mm (e) 1.06 mm Design and Essay Problem 10–138 Explain why there is a significant velocity over shoot for the midrange values of the Reynolds number in the velocity profiles of Fig. 10–136, but not for the very small values of Re or for the very large values of Re.
cen96537_ch10_569-610.indd 610 29/12/16 4:06 pm 11 CHAPTER 611 OBJECTIVES When you finish reading this chapter, you should be able to ■ ■ Have an intuitive understand ing of the various physical phenomena associated with external flow such as drag, friction and pressure drag, drag reduction, and lift ■ ■ Calculate the drag force associated with flow over common geometries ■ ■ Understand the effects of flow regime on the drag coefficients associated with flow over cylinders and spheres ■ ■ Understand the fundamentals of flow over airfoils, and cal culate the drag and lift forces acting on airfoils E X T E R N A L F LOW: D RAG A N D L I FT I n this chapter we consider external flow—flow over bodies that are im mersed in a fluid, with emphasis on the resulting lift and drag forces. In external flow, the viscous effects are confined to a portion of the flow field such as the boundary layers and wakes, which are surrounded by an outer flow region that involves small velocity and temperature gradients. When a fluid moves over a solid body, it exerts pressure forces normal to the surface and shear forces parallel to the surface of the body. We are usually interested in the resultant of the pressure and shear forces acting on the body rather than the details of the distributions of these forces along the entire surface of the body. The component of the resultant pressure and shear forces that acts in the flow direction is called the drag force (or just drag), and the component that acts normal to the flow direction is called the lift force (or just lift). We start this chapter with a discussion of drag and lift, and explore the concepts of pressure drag, friction drag, and flow separation. We continue with the drag coefficients of various two- and three-dimensional geometries encountered in practice and determine the drag force using experimentally determined drag coefficients. We then examine the development of the velocity boundary layer during parallel flow over a flat surface, and develop relations for the skin friction and drag coefficients for flow over flat plates, cylinders, and spheres. Finally, we dis cuss the lift developed by airfoils and the factors that affect the lift characteristics of bodies.
The wake of a Boeing 767 disrupts the top of a cumulus cloud and clearly shows the counter-rotating trailing vortices. Photo by Steve Morris. Used by permission.
cen96537_ch11_611-666.indd 611 14/01/17 3:15 pm 612 EXTERNAL FLOW: DRAG AND LIFT 11–1 ■ INTRODUCTION Fluid flow over solid bodies frequently occurs in practice, and it is responsi ble for numerous physical phenomena such as the drag force acting on auto mobiles, power lines, trees, and underwater pipelines; the lift developed by bird or airplane wings; upward draft of rain, snow, hail, and dust particles in high winds; the transportation of red blood cells by blood flow; the entrain ment and disbursement of liquid droplets by sprays; the vibration and noise generated by bodies moving in a fluid; and the power generated by wind turbines (Fig. 11–1). Therefore, developing a good understanding of exter nal flow is important in the design of many engineering systems such as aircraft, automobiles, buildings, ships, submarines, and all kinds of turbines. Late-model cars, for example, have been designed with particular empha sis on aerodynamics. This has resulted in significant reductions in fuel consumption and noise, and considerable improvement in handling. Sometimes a fluid moves over a stationary body (such as the wind blow ing over a building), and other times a body moves through a quiescent fluid (such as a car moving through air). These two seemingly different processes are equivalent to each other; what matters is the relative motion between the fluid and the body. Such motions are conveniently analyzed by fixing the coordinate system on the body and are referred to as flow over bodies or external flow. The aerodynamic aspects of different airplane wing designs, for example, are studied conveniently in a lab by placing the wings in a wind tunnel and blowing air over them by large fans. Also, a flow can be classi fied as being steady or unsteady, depending on the reference frame selected. Flow around an airplane, for example, is always unsteady with respect to the ground, but it is steady with respect to a frame of reference moving with the airplane at cruise conditions. The flow fields and geometries for most external flow problems are too complicated to be solved analytically, and thus we have to rely on correla tions based on experimental data. The availability of high-speed computers has made it possible to conduct a series of “numerical experiments” quickly by solving the governing equations numerically (Chap. 15), and to resort to the expensive and time-consuming testing and experimentation only in the final stages of design. Such testing is done in wind tunnels. H. F. Phillips (1845–1912) built the first wind tunnel in 1894 and measured lift and drag. In this chapter we mostly rely on relations developed experimentally. The velocity of the fluid approaching a body is called the free-stream velocity and is denoted by V. It is also denoted by u∞ or U∞ when the flow is aligned with the x-axis since u is used to denote the x-component of velocity. The fluid velocity ranges from zero at the body surface (the no-slip condition) to the free-stream value away from the body surface, and the subscript “infinity” serves as a reminder that this is the value at a distance where the presence of the body is not felt. The free-stream velocity may vary with location and time (e.g., the wind blowing past a building). But in the design and analysis, the free-stream velocity is usually assumed to be uniform and steady for convenience, and this is what we do in this chapter. The shape of a body has a profound influence on the flow over the body and the velocity field. The flow over a body is said to be two-dimensional when the body is very long and of constant cross section and the flow is cen96537_ch11_611-666.indd 612 14/01/17 3:15 pm 613 CHAPTER 11 (a) (b) (c) (d) (e) (f) (g) FIGURE 11–1 Flow over bodies is commonly encountered in practice.
(a) Corbis RF; (b) © Imagestate Media/John Foxx RF; (c) © IT Stock/age fotostock RF; (d) © Corbis RF; (e) © StockTrek/Superstock RF; ( f ) © Corbis RF; (g) © Roy H. Photography/Getty Images RF cen96537_ch11_611-666.indd 613 14/01/17 3:15 pm 614 EXTERNAL FLOW: DRAG AND LIFT normal to the body. The wind blowing over a long pipe perpendicular to its axis is an example of two-dimensional flow. Note that the velocity component in the axial direction is zero in this case, and thus the velocity is two-dimensional. The two-dimensional idealization is appropriate when the body is suffi ciently long so that the end effects are negligible and the approach flow is uniform. Another simplification occurs when the body possesses rotational symmetry about an axis in the flow direction. The flow in this case is also two-dimensional and is said to be axisymmetric. A bullet piercing through air is an example of axisymmetric flow. The velocity in this case varies with the axial distance x and the radial distance r. Flow over a body that cannot be modeled as two-dimensional or axisymmetric, such as flow over a car, is three-dimensional (Fig. 11–2). Flow over bodies can also be classified as incompressible flows (e.g., flows over automobiles, submarines, and buildings) and compressible flows (e.g., flows over high-speed aircraft, rockets, and missiles). Compressibility effects are negligible at low velocities (flows with Ma ≲ 0.3), and such flows can be treated as incompressible with little loss in accuracy. Compressible flow is discussed in Chap. 12, and flows that involve partially immersed bodies with a free surface (such as a ship cruising in water) are beyond the scope of this introductory text. Bodies subjected to fluid flow are classified as being streamlined or bluff, depending on their overall shape. A body is said to be streamlined if a conscious effort is made to align its shape with the anticipated streamlines in the flow. Streamlined bodies such as race cars and airplanes appear to be contoured and sleek. Otherwise, a body (such as a building) tends to block the flow and is said to be bluff or blunt. Usually it is much easier to force a streamlined body through a fluid, and thus streamlining has been of great importance in the design of vehicles and airplanes (Fig. 11–3).
11–2 ■ DRAG AND LIFT It is a common experience that a body meets some resistance when it is forced to move through a fluid, especially a liquid. As you may have noticed, it is very difficult to walk in water because of the much greater resistance it offers to motion compared to air. Also, you may have seen high winds knocking down trees, power lines, and even trailers and felt the strong “push” the wind exerts on your body (Fig. 11–4). You experience the same feeling when you extend your arm out of the window of a moving car. A fluid may exert forces and moments on a body in and about various directions. The force a flowing fluid exerts on a body in the flow direction is called drag. The drag force can be measured directly by simply attaching the body subjected to fluid flow to a calibrated spring and measuring the displacement in the flow direction (just like measuring weight with a spring scale). More sophisticated drag-measuring devices, called drag balances, use flexible beams fitted with strain gages to measure the drag electronically. Drag is usually an undesirable effect, like friction, and we do our best to minimize it. Reduction of drag is closely associated with the reduction of fuel consumption in automobiles, submarines, and aircraft; improved safety and durability of structures subjected to high winds; and reduction of noise Wind FIGURE 11–2 Two-dimensional, axisymmetric, and three-dimensional flows.
(a) Photo by John M. Cimbala; (b) © Corbis RF (c) © Hannu Liivaar/Alamy RF Long cylinder (2-D) Bullet (axisymmetric) Car (3-D) (a) (b) (c) cen96537_ch11_611-666.indd 614 14/01/17 3:15 pm 615 CHAPTER 11 and vibration. But in some cases drag produces a beneficial effect and we try to maximize it. Friction, for example, is a “life saver” in the brakes of automobiles. Likewise, it is the drag that makes it possible for people to parachute, for pollens to fly to distant locations, and for us all to enjoy the waves of the oceans and the relaxing movements of the leaves of trees. A stationary fluid exerts only normal pressure forces on the surface of a body immersed in it. A moving fluid, however, also exerts tangential shear forces on the surface because of the no-slip condition caused by viscous effects. Both of these forces, in general, have components in the direction of flow, and thus the drag force is due to the combined effects of pressure and wall shear forces in the flow direction. The components of the pressure and wall shear forces in the direction normal to the flow tend to move the body in that direction, and their sum is called lift. For two-dimensional flows, the resultant of the pressure and shear forces can be split into two components: one in the direction of flow, which is the drag force, and another in the direction normal to flow, which is the lift, as shown in Fig. 11–5. For three-dimensional flows, there is also a side force component in the direction normal to the page that tends to move the body in that direction. The fluid forces may also generate moments and cause the body to rotate. The moment about the flow direction is called the rolling moment, the moment about the lift direction is called the yawing moment, and the moment about the side force direction is called the pitching moment. For bodies that possess symmetry about the lift–drag plane such as cars, airplanes, and ships, the time-averaged side force, yawing moment, and rolling moment are zero when the wind and wave forces are aligned with the body. What remain for such bodies are the drag and lift forces and the pitching moment. For axisym metric bodies aligned with the flow, such as a bullet, the only time-averaged force exerted by the fluid on the body is the drag force. The pressure and shear forces acting on a differential area dA on the sur face are PdA and 𝜏w dA, respectively. The differential drag force and the lift force acting on dA in two-dimensional flow are (Fig. 11–5) dFD = −P dA cos 𝜃 + 𝜏w dA sin 𝜃 (11–1) and dFL = −P dA sin 𝜃 −𝜏w dA cos 𝜃 (11–2) where 𝜃 is the angle that the outer normal of dA makes with the positive flow direction. The total drag and lift forces acting on the body are deter mined by integrating Eqs. 11–1 and 11–2 over the entire surface of the body, Drag force: FD = ∫A dFD = ∫A (−P cos 𝜃 + 𝜏w sin 𝜃 ) dA (11–3) and Lift force: FL = ∫A dFL = −∫A (P sin 𝜃 + 𝜏w cos 𝜃 ) dA (11–4) These are the equations used to predict the net drag and lift forces on bodies when the flow is simulated on a computer (Chap. 15). However, when we perform experimental analyses, Eqs. 11–3 and 11–4 are not practical since the detailed distributions of pressure and shear forces are difficult to obtain FIGURE 11–4 High winds knock down trees, power lines, and even people as a result of the drag force.
70 hp 60 mi/h 50 hp 60 mi/h FIGURE 11–3 It is much easier to force a streamlined body than a blunt body through a fluid.
cen96537_ch11_611-666.indd 615 14/01/17 3:15 pm 616 EXTERNAL FLOW: DRAG AND LIFT by measurements. Fortunately, this information is often not needed. Usually all we need to know is the resultant drag force and lift acting on the entire body, which can be measured directly and easily in a wind tunnel. Equations 11–1 and 11–2 show that both the skin friction (wall shear) and pressure, in general, contribute to the drag and the lift. In the special case of a thin flat plate aligned parallel to the flow direction, the drag force depends on the wall shear only and is independent of pressure since 𝜃 = 90°. When the flat plate is placed normal to the flow direction, however, the drag force depends on the pressure only and is independent of wall shear since the shear stress in this case acts in the direction normal to flow and 𝜃 = 0° (Fig. 11–6). If the flat plate is tilted at an angle relative to the flow direc tion, then the drag force depends on both the pressure and the shear stress. The wings of airplanes are shaped and positioned specifically to generate lift with minimal drag. This is done by maintaining an angle of attack during cruising, as shown in Fig. 11–7. Both lift and drag are strong functions of the angle of attack, as we discuss later in this chapter. The pressure difference between the top and bottom surfaces of the wing generates an upward force that tends to lift the wing and thus the airplane to which it is connected. For slender bodies such as wings, the shear force acts nearly parallel to the flow direction, and thus its contribution to the lift is small. The drag force for such slender bodies is mostly due to shear forces (the skin friction). The drag and lift forces depend on the density 𝜌 of the fluid, the upstream velocity V, and the size, shape, and orientation of the body, among other things, and it is not practical to list these forces for a variety of situations. Instead, it is more convenient to work with appropriate dimensionless numbers that represent the drag and lift characteristics of the body. These numbers are the drag coefficient CD, and the lift coefficient CL, and they are defined as Drag coefficient: CD = FD 1 2 ρV 2A (11–5) Lift coefficient: CL = FL 1 2 ρV 2A (11–6) where A is ordinarily the frontal area (the area projected on a plane normal to the direction of flow) of the body. In other words, A is the area seen by a person looking at the body from the direction of the approaching fluid. The frontal area of a cylinder of diameter D and length L, for example, is A = LD.
In lift and drag calculations of some thin bodies, such as airfoils, A is taken to be the planform area, which is the area seen by a person looking at the body from above in a direction normal to the body. The drag and lift coef ficients are primarily functions of the shape of the body, but in some cases they also depend on the Reynolds number and the surface roughness. The term 1 2𝜌V 2 in Eqs. 11– 5 and 11–6 is the dynamic pressure. The local drag and lift coefficients vary along the surface as a result of the changes in the velocity boundary layer in the flow direction. We are usually interested in the drag and lift forces for the entire surface, which can be deter mined using the average drag and lift coefficients. Therefore, we present correlations for both local (identified with the subscript x) and average drag and lift coefficients. When relations for local drag and lift coefficients for a Boundary layer (a) (b) x u y τ τ τ w w w High pressure Low pressure Wall shear + + + + + + + + – – – – – – – – FIGURE 11–6 (a) Drag force acting on a flat plate parallel to the flow depends on wall shear only. (b) Drag force acting on a flat plate normal to the flow depends on the pressure only and is independent of the wall shear, which acts normal to the free-stream flow.
FL FD = FR cos FL = FR sin ϕ ϕ ϕ FR FD θ w A P (absolute) PdA w dA Outer normal n τ τ FIGURE 11–5 The pressure and viscous forces acting on a two-dimensional body and the resultant lift and drag forces.
cen96537_ch11_611-666.indd 616 14/01/17 3:15 pm 617 CHAPTER 11 surface of length L are available, the average drag and lift coefficients for the entire surface are determined by integration from CD = 1 L ∫ L 0 CD, x dx (11–7) and CL = 1 L ∫ L 0 CL, x dx (11–8) The forces acting on a falling body are usually the drag force, the buoyant force, and the weight of the body. When a body is dropped into the atmosphere or a lake, it first accelerates under the influence of its weight. The motion of the body is resisted by the drag force, which acts in the direction opposite to motion. As the velocity of the body increases, so does the drag force. This continues until all the forces balance each other and the net force acting on the body (and thus its acceleration) is zero. Then the velocity of the body remains constant during the rest of its fall if the properties of the fluid in the path of the body remain essentially constant. This is the maximum velocity a falling body can attain and is called the terminal velocity (Fig. 11–8). W Vterminal FD FD = W – FB (No acceleration) FB FIGURE 11–8 During a free fall, a body reaches its terminal velocity when the drag force equals the weight of the body minus the buoyant force.
Wind tunnel 60 mi/h FD FIGURE 11–9 Schematic for Example 11–1.
Lift Drag V FIGURE 11–7 Airplane wings are shaped and positioned to generate sufficient lift during flight while keeping drag at a minimum. Pressures above and below atmospheric pressure are indicated by plus and minus signs, respectively.
EXAMPLE 11–1 Measuring the Drag Coefficient of a Car The drag coefficient of a car at the design conditions of 1 atm, 70°F, and 60 mi/h is to be determined experimentally in a large wind tunnel in a full-scale test (Fig. 11–9). The frontal area of the car is 22.26 ft2. If the force acting on the car in the flow direction is measured to be 68 lbf, determine the drag coefficient of this car.
SOLUTION The drag force acting on a car is measured in a wind tunnel. The drag coefficient of the car at test conditions is to be determined.
Assumptions 1 The flow of air is steady and incompressible. 2 The cross section of the tunnel is large enough to simulate free flow over the car. 3 The bottom of the tunnel is also moving at the speed of air to approximate actual driving conditions or this effect is negligible.
Properties The density of air at 1 atm and 70°F is 𝜌 = 0.07489 lbm/ft3.
Analysis The drag force acting on a body and the drag coefficient are given by FD = CD A ρV 2 2 and CD = 2FD ρAV 2 where A is the frontal area. Substituting and noting that 1 mi/h = 1.467 ft/s, the drag coefficient of the car is determined to be CD = 2 × (68 lbf) (0.07489 lbm/ft3)(22.26 ft2)(60 × 1.467 ft/s)2 ( 32.2 lbm·ft/s2 1 lbf ) = 0.34 Discussion Note that the drag coefficient depends on the design conditions, and its value may be different at different conditions such as the Reynolds number. Therefore, the published drag coefficients of different vehicles can be compared meaningfully only if they are determined under dynamically similar conditions or if Reynolds number independence is demonstrated (Chap. 7). This shows the importance of developing standard testing procedures.
cen96537_ch11_611-666.indd 617 14/01/17 3:16 pm 618 EXTERNAL FLOW: DRAG AND LIFT 11–3 ■ FRICTION AND PRESSURE DRAG As mentioned in Section 11–2, the drag force is the net force exerted by a fluid on a body in the direction of flow due to the combined effects of wall shear and pressure forces. It is often instructive to separate the two effects, and study them separately. The part of drag that is due directly to wall shear stress 𝜏w is called the skin friction drag (or just friction drag FD, friction) since it is caused by fric tional effects, and the part that is due directly to pressure P is called the pressure drag (also called the form drag because of its strong dependence on the form or shape of the body). The friction and pressure drag coeffi cients are defined as CD, friction = FD, friction 1 2ρV 2A and CD, pressure = FD, pressure 1 2ρV 2A (11–9) When the friction and pressure drag coefficients (based on the same area A) or forces are available, the total drag coefficient or drag force is determined by simply adding them, CD = CD, friction + CD, pressure and FD = FD, friction + FD, pressure (11–10) The friction drag is the component of the wall shear force in the direction of flow, and thus it depends on the orientation of the body as well as the magnitude of the wall shear stress 𝜏w. The friction drag is zero for a flat surface normal to the flow, and maximum for a flat surface parallel to the flow since the friction drag in this case equals the total shear force on the surface. Therefore, for parallel flow over a flat surface, the drag coefficient is equal to the friction drag coefficient, or simply the friction coefficient. Friction drag is a strong function of viscosity, and increases with increas ing viscosity. The Reynolds number is inversely proportional to the viscosity of the fluid. Therefore, the contribution of friction drag to total drag for blunt bodies is less at higher Reynolds numbers and may be negligible at very high Reynolds numbers. The drag in such cases is mostly due to pressure drag. At low Reynolds numbers, most drag is due to friction drag. This is especially the case for highly streamlined bodies such as airfoils. The fric tion drag is also proportional to the surface area. Therefore, bodies with a larger surface area experience a larger friction drag. Large commercial airplanes, for example, reduce their total surface area and thus their drag by retracting their wing extensions when they reach cruising altitudes to save fuel. The friction drag coefficient is independent of surface roughness in laminar flow, but is a strong function of surface roughness in turbulent flow due to surface roughness elements protruding further into the bound ary layer. The friction drag coefficient is analogous to the friction factor in pipe flow discussed in Chap. 8, and its value depends on the flow regime. The pressure drag is proportional to the frontal area and to the difference between the pressures acting on the front and back of the immersed body. Therefore, the pressure drag is usually dominant for blunt bodies, small for streamlined bodies such as airfoils, and zero for thin flat plates paral lel to the flow (Fig. 11–10). The pressure drag becomes most significant FIGURE 11–10 Drag is due entirely to friction drag for a flat plate parallel to the flow; it is due entirely to pressure drag for a flat plate normal to the flow; and it is due to both (but mostly pressure drag) for a cylinder normal to the flow. The total drag coefficient CD is lowest for a parallel flat plate, highest for a vertical flat plate, and in between (but close to that of a vertical flat plate) for a cylinder. From G. M. Homsy et al., “Multi-Media Fluid Mechanics,” Cambridge Univ. Press (2001). Image © Stanford University (2000). Reprinted by permission.
cen96537_ch11_611-666.indd 618 14/01/17 3:16 pm 619 CHAPTER 11 when the velocity of the fluid is too high for the fluid to be able to follow the curvature of the body, and thus the fluid separates from the body at some point and creates a very low pressure region in the back. The pres sure drag in this case is due to the large pressure difference between the front and back sides of the body.
Reducing Drag by Streamlining The first thought that comes to mind to reduce drag is to streamline a body in order to reduce flow separation and thus to reduce pressure drag. Even car salespeople are quick to point out the low drag coefficients of their cars, owing to streamlining. But streamlining has opposite effects on pressure and friction drag forces. It decreases pressure drag by delay ing boundary layer separation and thus reducing the pressure difference between the front and back of the body and increases the friction drag by increasing the surface area. The end result depends on which effect domi nates. Therefore, any optimization study to reduce the drag of a body must consider both effects and must attempt to minimize the sum of the two, as shown in Fig. 11–11. The minimum total drag occurs at D/L = 0.25 for the case shown in Fig. 11–11. For the case of a circular cylinder with the same thickness as the streamlined shape of Fig. 11–11, the drag coef ficient would be about five times as much. Therefore, it is possible to reduce the drag of a cylindrical component to nearly one-fifth by the use of proper fairings. The effect of streamlining on the drag coefficient is described best by considering long elliptical cylinders with different aspect (or length-to- thickness) ratios L/D, where L is the length in the flow direction and D is the thickness, as shown in Fig. 11–12. Note that the drag coefficient decreases drastically as the ellipse becomes slimmer. For the special case of L/D = 1 (a circular cylinder), the drag coefficient is CD ≅ 1 at this Reynolds number. As the aspect ratio is decreased and the cylinder resembles a flat plate, the drag coefficient increases to 1.9, the value for a flat plate normal to flow. Note that the curve becomes nearly flat for aspect ratios greater than about 4. Therefore, for a given diameter D, elliptical shapes with an aspect ratio of about L/D ≅ 4 usually offer a good compromise between the total drag coefficient and length L. The reduction in the drag coefficient at high aspect ratios is primarily due to the boundary layer staying attached to the surface longer and the result ing pressure recovery. The pressure drag on an elliptical cylinder with an aspect ratio of 4 or greater is negligible (less than 2 percent of total drag at this Reynolds number). As the aspect ratio of an elliptical cylinder is increased by flattening it (i.e., decreasing D while holding L constant), the drag coefficient starts increasing and tends to infinity as L/D → ∞ (i.e., as the ellipse resem bles a flat plate parallel to flow). This is due to the frontal area, which appears in the denominator in the definition of CD, approaching zero. It does not mean that the drag force increases drastically (actually, the drag force decreases) as the body becomes flat. This shows that the frontal area is inappropriate for use in the drag force relations for slim bodies such as thin airfoils and flat plates. In such cases, the drag coefficient is defined 0.12 0.10 0.08 0.06 0.04 0.02 0 0 0.2 D/L 0.4 0.1 0.3 Total drag Pressure drag Friction drag L D V ρV2bD FD 1 2 CD = FIGURE 11–11 The variation of friction, pressure, and total drag coefficients of a two-dimensional streamlined strut with thickness-to-chord length ratio for Re = 4 × 104. Note that CD for airfoils and other thin bodies is based on planform area bL rather than frontal area bD, where b is the width of the 2-D body into the page.
Data from Abbott and von Doenhoff (1959).
Circular cylinder D L V Flat plate normal to flow FD 2bD ρV 1 2 CD = VD v Re = = 105 0 2.5 2.0 1.5 1.0 0.5 0 1 2 3 4 6 5 CD L/D FIGURE 11–12 The variation of the drag coefficient of a long elliptical cylinder with aspect ratio. Here CD is based on the frontal area bD where b is the width of the 2-D body into the page.
Data from Blevins (1984).
cen96537_ch11_611-666.indd 619 14/01/17 3:16 pm 620 EXTERNAL FLOW: DRAG AND LIFT on the basis of the planform area, which is simply the surface area of one side (top or bottom) of a flat plate parallel to the flow. This is quite appro priate since for slim bodies the drag is almost entirely due to friction drag, which is proportional to the surface area. Streamlining has the added benefit of reducing vibration and noise. Streamlining should be considered only for bluff bodies that are subjected to high-velocity fluid flow (and thus high Reynolds numbers) for which flow separation is a real possibility. It is not necessary for bodies that typi cally involve low Reynolds number flows (e.g., creeping flows in which Re < 1) as discussed in Chap. 10, since the drag in those cases is almost entirely due to friction drag, and streamlining would only increase the sur face area and thus the total drag. Therefore, careless streamlining may actu ally increase drag instead of decreasing it.
Flow Separation When driving on country roads, it is a common safety measure to slow down at sharp turns in order to avoid being thrown off the road. Many driv ers have learned the hard way that a car refuses to comply when forced to turn curves at excessive speeds. We can view this phenomenon as “the separation of cars” from roads. This phenomenon is also observed when fast vehicles jump off hills. At low velocities, the wheels of the vehicle always remain in contact with the road surface. But at high velocities, the vehicle is too fast to follow the curvature of the road and takes off at the hill, losing contact with the road. A fluid acts much the same way when forced to flow over a curved surface at high velocities. A fluid follows the front portion of the curved surface with no problem, but it has difficulty remaining attached to the surface on the back side. At sufficiently high velocities, the fluid stream detaches itself from the surface of the body. This is called flow separation (Fig. 11–13). Flow can separate from a surface even if it is fully submerged in a liquid or immersed in a gas (Fig. 11–14). The location of the separation point depends on several factors such as the Reynolds number, the surface roughness, and the level of fluctuations in the free stream, and it is usually difficult to predict exactly where separation will occur unless there are sharp corners or abrupt changes in the shape of the solid surface. When a fluid separates from a body, it forms a separated region between the body and the fluid stream. This low-pressure region behind the body where recirculating and backflows occur is called the separated region. The larger the separated region, the larger the pressure drag. The effects of flow separation are felt far downstream in the form of reduced velocity (relative to the upstream velocity). The region of flow trailing the body where the effects of the body on velocity are felt is called the wake (Fig. 11–15). The separated region comes to an end when the two flow streams reattach. Therefore, the separated region is an enclosed volume, whereas the wake keeps growing behind the body until the fluid in the wake region regains its velocity and the velocity profile becomes nearly flat again. Viscous and rotational effects are the most significant in the boundary layer, the sepa rated region, and the wake.
Wake region FIGURE 11–15 Flow separation and the wake region for flow over a tennis ball.
NASA and the Cislunar Aerospace, Inc.
Separation point FIGURE 11–13 Flow separation in a waterfall.
Separated flow region Separation point Reattachment point FIGURE 11–14 Flow separation over a backward-facing step along a wall.
cen96537_ch11_611-666.indd 620 14/01/17 3:16 pm 621 CHAPTER 11 The occurrence of separation is not limited to bluff bodies. Complete separation over the entire back surface may also occur on a streamlined body such as an airplane wing at a sufficiently large angle of attack (larger than about 15° for most airfoils), which is the angle the incoming fluid stream makes with the chord (the line that connects the nose and the trailing edge) of the wing. Flow separation on the top surface of a wing reduces lift drastically and may cause the airplane to stall. Stalling has been blamed for many airplane accidents and loss of efficiencies in tur bomachinery (Fig. 11–16). Note that drag and lift are strongly dependent on the shape of the body, and any effect that causes the shape to change has a profound effect on the drag and lift. For example, snow accumulation and ice formation on airplane wings may change the shape of the wings sufficiently to cause significant loss of lift. This phenomenon has caused many airplanes to lose altitude and crash and many others to abort takeoff. Therefore, it has become a routine safety measure to check for ice or snow buildup on critical components of airplanes before takeoff in bad weather. This is especially important for air planes that have waited a long time on the runway before takeoff because of heavy traffic. An important consequence of flow separation is the formation and shed ding of circulating fluid structures, called vortices, in the wake region. The periodic generation of these vortices downstream is referred to as vortex shedding. This phenomenon usually occurs during normal flow over long cylinders or spheres for Re ≳ 90. The vibrations generated by vortices near the body may cause the body to resonate to dangerous levels if the frequency of the vortices is close to the natural frequency of the body—a situation that must be avoided in the design of equipment that is subjected to high-velocity fluid flow such as the wings of airplanes and suspended bridges subjected to steady high winds.
11–4 ■ DRAG COEFFICIENTS OF COMMON GEOMETRIES The concept of drag has important consequences in daily life, and the drag behavior of various natural and human-made bodies is characterized by their drag coefficients measured under typical operating conditions. Although drag is caused by two different effects (friction and pressure), it is usu ally difficult to determine them separately. Besides, in most cases, we are interested in the total drag rather than the individual drag components, and thus usually the total drag coefficient is reported. The determination of drag coefficients has been the topic of numerous studies (mostly experimental), and there is a huge amount of drag coefficient data in the literature for just about any geometry of practical interest. The drag coefficient, in general, depends on the Reynolds number, espe cially for Reynolds numbers below about 104. At higher Reynolds num bers, the drag coefficients for most geometries remain essentially constant (Fig. 11–17). This is due to the flow at high Reynolds numbers becoming fully turbulent. However, this is not the case for rounded bodies such as (a) 5° (b) 15° (c) 30° FL FL FL FD FD FD FIGURE 11–16 At large angles of attack (usually larger than 15°), flow may separate completely from the top surface of an airfoil, reducing lift drastically and causing the airfoil to stall.
From G. M. Homsy et al., “Multi-Media Fluid Mechanics,” Cambridge Univ. Press (2001). Image © Stanford University (2000). Reprinted by permission.
cen96537_ch11_611-666.indd 621 14/01/17 3:16 pm 622 EXTERNAL FLOW: DRAG AND LIFT circular cylinders and spheres, as we discuss later in this section. The reported drag coefficients are usually applicable only to flows at high Reynolds numbers. The drag coefficient exhibits different behavior in the low (creeping), moderate (laminar), and high (turbulent) regions of the Reynolds number. The inertia effects are negligible in low Reynolds number flows (Re ≲ 1), called creeping flows (Chap. 10), and the fluid wraps around the body smoothly. The drag coefficient in this case is inversely proportional to the Reynolds number, and for a sphere it is determined to be Sphere: CD = 24 Re (Re ≲ 1) (11–11) Then the drag force acting on a spherical object at low Reynolds numbers becomes FD = CD A ρV 2 2 = 24 Re A ρV 2 2 = 24 ρVD/𝜇 𝜋D2 4 ρV 2 2 = 3𝜋𝜇VD (11–12) which is known as Stokes law, after British mathematician and physicist G. G. Stokes (1819–1903). This relation shows that at very low Reynolds numbers, the drag force acting on spherical objects is proportional to the diameter, the velocity, and the viscosity of the fluid. This relation is often applicable to dust particles in the air and suspended solid particles in water. The drag coefficients for low Reynolds number flows past some other geometries are given in Fig. 11–18. Note that at low Reynolds num bers, the shape of the body does not have a major influence on the drag coefficient. The drag coefficients for various two- and three-dimensional bodies are given in Tables 11–1 and 11–2 for large Reynolds numbers. We make sev eral observations from these tables about the drag coefficient at high Reyn olds numbers. First of all, the orientation of the body relative to the direc tion of flow has a major influence on the drag coefficient. For example, the drag coefficient for flow over a hemisphere is 0.4 when the spherical side faces the flow, but it increases threefold to 1.2 when the flat side faces the flow (Fig. 11–19). For blunt bodies with sharp corners, such as flow over a rectangular block or a flat plate normal to the flow, separation occurs at the edges of the front and back surfaces, with no significant change in the character of flow. Therefore, the drag coefficient of such bodies is nearly indepen dent of the Reynolds number. Note that the drag coefficient of a long rectangular rod is reduced almost by half from 2.2 to 1.2 by rounding the corners.
Biological Systems and Drag The concept of drag also has important consequences for biological sys tems. For example, the bodies of fish, especially the ones that swim fast for long distances (such as dolphins), are highly streamlined to minimize CD = 24/Re CD = 20.4/Re CD = 13.6/Re CD = 22.2/Re D Sphere Hemisphere Circular disk (normal to flow) Circular disk (parallel to flow) D D D V V V V FIGURE 11–18 Drag coefficients CD at low Reynolds numbers (Re ≲ 1 where Re = VD/𝜈 and A = 𝜋D2/4).
101 102 103 104 105 106 2.0 1.5 1.0 0.5 0 CD Re Disk V FIGURE 11–17 The drag coefficients for most geometries (but not all) remain essentially constant at Reynolds numbers above about 104.
cen96537_ch11_611-666.indd 622 14/01/17 3:16 pm 623 CHAPTER 11 drag (the drag coefficient of dolphins based on the wetted skin area is about 0.0035, comparable to the value for a flat plate in turbulent flow). So it is no surprise that we build submarines that mimic large fish. Tropi cal fish with fascinating beauty and elegance, on the other hand, swim short distances only. Obviously grace, not high speed and drag, was the primary consideration in their design. Birds teach us a lesson on drag reduction by extending their beak forward and folding their feet backward TABLE 11–1 Drag coefficients CD of various two-dimensional bodies for Re > 104 based on the frontal area A = bD, where b is the length in direction normal to the page (for use in the drag force relation FD = CD A𝜌V 2/2 where V is the upstream velocity) Square rod Rectangular rod L/D CD 0.0 1.9 0.1 1.9 0.5 2.5 1.0 2.2 2.0 1.7 3.0 1.3 Corresponds to thin plate L/D CD 0.5 1.2 1.0 0.9 2.0 0.7 4.0 0.7 Circular rod (cylinder) Elliptical rod CD L/D Laminar Turbulent 2 0.60 0.20 4 0.35 0.15 8 0.25 0.10 Equilateral triangular rod Semicircular shell Semicircular rod V V CD = 1.2 D CD = 1.7 D D Sharp corners: CD = 2.2 D V V Round corners (r/D = 0.2): CD = 1.2 r D D L L V V Sharp corners: Round front edge: D V Laminar: CD = 1.2 Turbulent: CD = 0.3 D L V D V V CD = 1.5 CD = 2.0 D CD = 2.3 D CD = 1.2 D V V cen96537_ch11_611-666.indd 623 14/01/17 3:16 pm 624 EXTERNAL FLOW: DRAG AND LIFT (continues) TABLE 11–2 Representative drag coefficients CD for various three-dimensional bodies based on the frontal area for Re > 104 unless stated otherwise (for use in the drag force relation FD = CD A𝜌V 2/2 where V is the upstream velocity) Cube, A = D 2 Thin circular disk, A = 𝜋D 2/4 Cone (for 𝜃 = 30°), A = 𝜋D 2/4 Sphere, A = 𝜋D 2/4 Ellipsoid, A = 𝜋D 2/4 CD L/D Laminar Turbulent Re ≲ 2 × 105 Re ≳ 2 × 106 0.75 0.5 0.2 1 0.5 0.2 2 0.3 0.1 4 0.3 0.1 8 0.2 0.1 Hemisphere, A = 𝜋D 2/4 Finite cylinder, vertical, A = LD Finite cylinder, horizontal, A = 𝜋D2/4 L/D CD L/D CD 1 0.6 0.5 1.1 2 0.7 1 0.9 5 0.8 2 0.9 10 0.9 4 0.9 40 1.0 8 1.0 ∞ 1.2 Values are for laminar flow (Re ≲ 2 × 105) Streamlined body, A = 𝜋D 2/4 Parachute, A = 𝜋D 2/4 Tree, A = frontal area V, m/s CD 10 0.4–1.2 20 0.3–1.0 30 0.2–0.7 CD = 1.3 D V A = frontal area D V CD = 1.05 D V CD = 1.1 D V CD = 0.5 D L V CD = 0.4 D CD = 1.2 D V V L D V D L V Laminar: Re 2 105 CD = 0.5 Turbulent: Re 2 106 CD = 0.2 D V See Fig. 11–36 for CD vs. Re for smooth and rough spheres.
D V CD = 0.04 V L D Rectangular plate, A = LD CD = 1.10 + 0.02 (L/D + D/L) for 1/30 < (L/D) < 30 cen96537_ch11_611-666.indd 624 14/01/17 3:16 pm 625 CHAPTER 11 during flight (Fig. 11–20). Airplanes, which look somewhat like large birds, retract their wheels after takeoff in order to reduce drag and thus fuel consumption. The flexible structure of plants enables them to reduce drag at high winds by changing their shapes. Large flat leaves, for example, curl into a low-drag conical shape at high wind speeds, while tree branches cluster to reduce drag. Flexible trunks bend under the influence of the wind to reduce drag, and the bending moment is lowered by reducing frontal area. If you watch the Olympic games, you have probably observed many instances of conscious effort by the competitors to reduce drag. Some exam ples: During 100-m running, the runners hold their fingers together and straight and move their hands parallel to the direction of motion to reduce the drag on their hands. Swimmers with long hair cover their head with a tight and smooth cover to reduce head drag. They also wear well-fitting one-piece swimming suits. Horse and bicycle riders lean forward as much as they can to reduce drag (by reducing both the drag coefficient and frontal area). Speed skiers do the same thing.
Drag Coefficients of Vehicles The term drag coefficient is commonly used in various areas of daily life. Car manufacturers try to attract consumers by pointing out the low drag coefficients of their cars (Fig. 11–21). The drag coefficients of vehicles range from about 1.0 for large semitrailers to 0.4 for minivans and 0.3 for TABLE 11–2 (Continued) Person (average) Bikes Semitrailer, A = frontal area Automotive, A = frontal area High-rise buildings, A = frontal area V A = 5.5 ft2 = 0.51 m2 CD = 1.1 Racing: A = 3.9 ft2 = 0.36 m2 CD = 0.9 Drafting: A = 3.9 ft2 = 0.36 m2 CD = 0.50 With fairing: A = 5.0 ft2 = 0.46 m2 CD = 0.12 Upright: CD = 0.9 CD = 0.5 Standing: CD A = 9 ft2 = 0.84 m2 Sitting: CD A = 6 ft2 = 0.56 m2 CEN_4 Without fairing: CD = 0.96 With fairing: CD = 0.76 Minivan: CD = 0.4 Passenger car or sports car: CD = 0.3 CD ≈ 1.0 to 1.4 V CD = 0.4 A hemisphere at two different orientations for Re > 104 CD = 1.2 V V FIGURE 11–19 The drag coefficient of a body may change drastically by changing the body’s orientation (and thus shape) relative to the direction of flow.
cen96537_ch11_611-666.indd 625 14/01/17 3:16 pm 626 EXTERNAL FLOW: DRAG AND LIFT passenger cars. In general, the more blunt the vehicle, the higher the drag coefficient. Installing a fairing reduces the drag coefficient of tractor-trailer rigs by about 20 percent by making the frontal surface more streamlined. As a rule of thumb, the percentage of fuel savings due to reduced drag is about half the percentage of drag reduction at highway speeds. When the effect of the road on air motion is disregarded, the ideal shape of a vehicle is the basic teardrop, with a drag coefficient of about 0.1 for the turbulent flow case. But this shape needs to be modified to accommo date several necessary external components such as wheels, mirrors, axles, and door handles. Also, the vehicle must be high enough for comfort and there must be a minimum clearance from the road. Further, a vehicle cannot be too long to fit in garages and parking spaces. Controlling the mate rial and manufacturing costs requires minimizing or eliminating any “dead” volume that cannot be utilized. The result is a shape that resembles more a box than a teardrop, and this was the shape of early cars with a drag coef ficient of about 0.8 in the 1920s. This wasn’t a problem in those days since the velocities were low, fuel was cheap, and drag was not a major design consideration. The average drag coefficients of cars dropped to about 0.70 in the 1940s, to 0.55 in the 1970s, to 0.45 in the 1980s, and to 0.30 in the 1990s as a result of improved manufacturing techniques for metal form ing and paying more attention to the shape of the car and streamlining (Fig. 11–22). The drag coefficient for well-built racing cars is about 0.2, but this is achieved after making the comfort of drivers a secondary consideration. Noting that the theoretical lower limit of CD is about 0.1 and the value for racing cars is 0.2, it appears that there is only a little room for further improvement in the drag coefficient of passenger cars from the current value of about 0.3. The drag coefficient of a Mazda 3, for example, is 0.29. For trucks and buses, the drag coefficient can be reduced further by optimizing the front and rear contours (by rounding, for example) to the extent it is practical while keeping the overall length of the vehicle the same. When traveling as a group, a sneaky way of reducing drag is drafting, a phenomenon well known by bicycle riders and car racers. It involves ap proach ing a moving body from behind and being drafted into the low-pressure re gion in the rear of the body. The drag coefficient of a racing bicyclist, for ex ample, is reduced from 0.9 to 0.5 (Table 11–2) by drafting, as also shown in Fig. 11–23. We also can help reduce the overall drag of a vehicle and thus fuel con sumption by being more conscientious drivers. For example, drag force is proportional to the square of velocity. Therefore, driving over the speed limit on the highways not only increases the chances of getting speed ing tickets or getting into an accident, but it also increases the amount of fuel consumption per mile. Therefore, driving at moderate speeds is safe and economical. Also, anything that extends from the car, even an arm, increases the drag coefficient. Driving with the windows rolled down also increases the drag and fuel consumption. At highway speeds, a driver can often save fuel in hot weather by running the air conditioner instead of driving with the windows rolled down. For many low-drag automobiles, the turbulence and additional drag generated by open windows consume FIGURE 11–21 This sleek-looking Toyota Prius has a drag coefficient of 0.26—one of the lowest for a passenger car.
© Hannu Liivaar/Alamy Stock Photo FIGURE 11–20 Birds teach us a lesson on drag reduction by extending their beak forward and folding their feet backward during flight.
© Photodisc/Getty Images RF cen96537_ch11_611-666.indd 626 14/01/17 3:16 pm 627 CHAPTER 11 more fuel than does the air conditioner, but this is not the case for high-drag vehicles.
Superposition The shapes of many bodies encountered in practice are not simple. But such bodies can be treated conveniently in drag force calculations by considering them to be composed of two or more simple bodies. A satellite dish mounted on a roof with a cylindrical bar, for example, can be considered to be a combination of a hemispherical body and a cylinder. Then the drag coefficient of the body can be determined approximately by using superposition. Such a simplistic approach does not account for the effects of components on each other, and thus the results obtained should be interpreted accordingly.
FIGURE 11–22 Streamlines around an aerodynamically designed modern car closely resemble the streamlines around the car in the ideal potential flow (assumes negligible friction), except near the rear end, resulting in a low drag coefficient.
From G. M. Homsy et al., “Multi-Media Fluid Mechanics,” Cambridge Univ. Press (2001). Image © Stanford Univ. (2000) and Sigurdur D. Thoroddsen. Reprinted by permission.
FIGURE 11–23 The drag coefficients of bodies following other moving bodies closely is reduced considerably due to drafting (i.e., entering into the low pressure region created by the body in front).
© Getty Images RF FIGURE 11–24 Schematic for Example 11–2.
EXAMPLE 11–2 Effect of Frontal Area on Fuel Efficiency of a Car Two common methods of improving fuel efficiency of a vehicle are to reduce the drag coefficient and the frontal area of the vehicle. Consider a car (Fig. 11–24) whose width (W ) and height (H) are 1.85 m and 1.70 m, respec tively, with a drag coefficient of 0.30. Determine the amount of fuel and money saved per year as a result of reducing the car height to 1.55 m while keeping its width the same. Assume the car is driven 18,000 km a year at an average speed of 95 km/h. Take the density and price of gasoline to be 0.74 kg/L and $0.95/L, respectively. Also take the density of air to be 1.20 kg/m3, the heating value of gasoline to be 44,000 kJ/kg, and the overall efficiency of the car’s drive train to be 30 percent.
SOLUTION The frontal area of a car is reduced by redesigning it. The resulting fuel and money savings per year are to be determined.
Assumptions 1 The car is driven 18,000 km a year at an average speed of 95 km/h. 2 The effect of reduction of the frontal area on the drag coefficient is negligible.
Properties The densities of air and gasoline are given to be 1.20 kg/m3 and 0.74 kg/L, respectively. The heating value of gasoline is given to be 44,000 kJ/kg. Analysis The drag force acting on a body is FD = CD A ρV 2 2 where A is the frontal area of the body. The drag force acting on the car before redesigning is FD = 0.3(1.85 × 1.70 m2) (1.20 kg/m3)(95 km/h)2 2 ( 1 m/s 3.6 km/h) 2 ( 1 N 1 kg·m/s2) = 394 N cen96537_ch11_611-666.indd 627 14/01/17 3:16 pm 628 EXTERNAL FLOW: DRAG AND LIFT Example 11–2 is indicative of the tremendous amount of effort put into redesigning various parts of cars such as the window moldings, the door handles, the windshield, and the front and rear ends in order to reduce aero dynamic drag. For a car moving on a level road at constant speed, the power developed by the engine is used to overcome rolling resistance, friction between moving components, aerodynamic drag, and driving the auxiliary equipment. The aerodynamic drag is negligible at low speeds, but becomes significant at speeds above about 30 mi/h. Reduction of the frontal area of Noting that work is force times distance, the amount of work done to overcome this drag force and the required energy input for a distance of 18,000 km are Wdrag = FD L = (394 N)(18,000 km/year)( 1000 m 1 km ) ( 1 kJ 1000 N·m) = 7.092 × 106 kJ/year Ein = Wdrag 𝜂car = 7.092 × 106 kJ/year 0.30 = 2.364 × 107 kJ/year The amount and the cost of the fuel that supplies this much energy are Amount of fuel = mfuel ρfuel = Ein/HV ρfuel = (2.364 × 107 kJ/year)/(44,000 kJ/kg) 0.74 kg/L = 726 L/year Cost = (Amount of fuel)(Unit cost) = (726 L/year)($0.95/L) = $690/year That is, the car uses about 730 liters of gasoline at a total cost of about $690 per year to overcome the drag. The drag force and the work done to overcome it are directly proportional to the frontal area. Then the percent reduction in the fuel consumption due to reducing the frontal area is equal to the percent reduction in the frontal area: Reduction ratio = A −Anew A = H −Hnew H = 1.70−1.55 1.70 = 0.0882 Amount reduction = (Reduction ratio)(Amount) Fuel reduction = 0.0882(726 L/year) = 64 L/year Cost reduction = (Reduction ratio)(Cost) = 0.0882($690/year) = $61/year Therefore, reducing the car’s height reduces the fuel consumption due to drag by nearly 9 percent.
Discussion Answers are given to 2 significant digits. This example dem onstrates that significant reductions in drag and fuel consumption can be achieved by reducing the frontal area of a vehicle as well as its drag coefficient.
cen96537_ch11_611-666.indd 628 14/01/17 3:16 pm 629 CHAPTER 11 the cars (to the dislike of tall drivers) has also contributed greatly to the reduction of drag and fuel consumption.
11–5 ■ PARALLEL FLOW OVER FLAT PLATES Consider the flow of a fluid over a flat plate, as shown in Fig. 11–25. Surfaces that are slightly contoured (such as turbine blades) also can be approximated as flat plates with reasonable accuracy. The x-coordinate is measured along the plate surface from the leading edge of the plate in the direction of the flow, and y is measured from the surface in the normal direction. The fluid approaches the plate in the x-direction with a uniform velocity V, which is equivalent to the velocity over the plate away from the surface. For the sake of discussion, we consider the fluid to consist of adjacent layers piled on top of each other. The velocity of the particles in the first fluid layer adjacent to the plate is zero because of the no-slip condition. This motionless layer slows down the particles of the neighboring fluid layer as a result of friction between the particles of these two adjoining fluid layers at different velocities. This fluid layer then slows down the molecules of the next layer, and so on. Thus, the presence of the plate is felt up to some nor mal distance 𝛿 from the plate beyond which the free-stream velocity remains virtually unchanged. As a result, the x-component of the fluid velocity, u, varies from 0 at y = 0 to nearly V (typically 0.99V) at y = 𝛿 (Fig. 11–26). The region of the flow above the plate bounded by 𝛿 in which the effects of the viscous shearing forces caused by fluid viscosity are felt is called the velocity boundary layer. The boundary layer thickness 𝛿 is typically defined as the distance y from the surface at which u = 0.99V. The hypothetical line of u = 0.99V divides the flow over a plate into two regions: the boundary layer region, in which the viscous effects and the velocity changes are significant, and the irrotational flow region, in which the frictional effects are negligible and the velocity remains essentially constant.
Laminar boundary layer Transition region Turbulent boundary layer y x 0 xcr Turbulent layer Overlap layer Buffer layer Viscous sublayer V V V Boundary layer thickness, δ FIGURE 11–25 The development of the boundary layer for flow over a flat plate, and the different flow regimes.
Not to scale.
cen96537_ch11_611-666.indd 629 14/01/17 3:16 pm 630 EXTERNAL FLOW: DRAG AND LIFT For parallel flow over a flat plate, the pressure drag is zero, and thus the drag coefficient is equal to the friction drag coefficient, or simply the friction coefficient (Fig. 11–27). That is, Flat plate: CD = CD, friction = Cf (11–13) Once the average friction coefficient Cf is available, the drag (or friction) force over the surface is determined from Friction force on a flat plate: FD = Ff = 1 2 Cf AρV 2 (11–14) where A is the surface area of the plate exposed to fluid flow. When both sides of a thin plate are subjected to flow, A becomes the total area of the top and bottom surfaces. Note that both the average friction coefficient Cf and the local friction coefficient Cf, x, in general, vary with location along the surface. Typical average velocity profiles in laminar and turbulent flow are sketched in Fig. 11–25. Note that the velocity profile in turbulent flow is much fuller than that in laminar flow, with a sharp drop near the surface. The turbulent boundary layer can be considered to consist of four regions, characterized by the distance from the wall. The very thin layer next to the wall where viscous effects are dominant is the viscous sublayer. The velocity profile in this layer is very nearly linear, and the flow is nearly parallel. Next to the viscous sublayer is the buffer layer, in which turbulent effects are becoming significant, but the flow is still dominated by viscous effects. Above the buf fer layer is the overlap layer, in which the turbulent effects are much more significant, but still not dominant. Above that is the turbulent (or outer) layer in which turbulent effects dominate over viscous effects. Note that the turbulent boundary layer profile on a flat plate closely resembles the bound ary layer profile in fully developed turbulent pipe flow (Chap. 8). The transition from laminar to turbulent flow depends on the surface geometry, surface roughness, upstream velocity, surface temperature, and the type of fluid, among other things, and is best characterized by the Reynolds number. The Reynolds number at a distance x from the leading edge of a flat plate is expressed as Rex = ρVx 𝜇 = Vx 𝜈 (11–15) where V is the upstream velocity and x is the characteristic length of the geometry, which, for a flat plate, is the length of the plate in the flow direc tion. Note that unlike pipe flow, the Reynolds number varies for a flat plate along the flow, reaching ReL = VL /𝜈 at the end of the plate. For any point on a flat plate, the characteristic length is the distance x of the point from the leading edge in the flow direction. For flow over a smooth flat plate, transition from laminar to turbulent begins at about Re ≅ 1 × 105, but does not become fully turbulent before the Reynolds number reaches much higher values, typically around 3 × 106 (Chap. 10). In engineering analysis, a generally accepted value for the criti cal Reynolds number is Rex, cr = ρVxcr 𝜇 = 5 × 105 The actual value of the engineering critical Reynolds number for a flat plate may vary somewhat from about 105 to 3 × 106 depending on the surface Relative velocities of fluid layers 0.99V Zero velocity at the surface V δ V FIGURE 11–26 The development of a boundary layer on a surface is due to the no-slip condition and friction.
Flow over a flat plate ρV 2 2 FD, pressure = 0 FD = FD, friction = Ff = Cf A CD, pressure = 0 CD = CD, friction = Cf FIGURE 11–27 For parallel flow over a flat plate, the pressure drag is zero, and thus the drag coefficient is equal to the friction coefficient and the drag force is equal to the friction force.
cen96537_ch11_611-666.indd 630 14/01/17 3:16 pm 631 CHAPTER 11 roughness, the turbulence level, and the variation of pressure along the sur face, as discussed in more detail in Chap. 10.
Friction Coefficient The friction coefficient for laminar flow over a flat plate can be determined theoretically by solving the conservation of mass and linear momentum equations numerically (Chap. 10). For turbulent flow, however, it must be determined experimentally and expressed by empirical correlations. The local friction coefficient varies along the surface of the flat plate as a result of the changes in the velocity boundary layer in the flow direction. We are usually interested in the drag force on the entire surface, which can be determined using the average friction coefficient. But sometimes we are also interested in the drag force at a certain location, and in such cases, we need to know the local value of the friction coefficient. With this in mind, we present correlations for both local (identified with the subscript x) and average fric tion coefficients over a flat plate for laminar, turbulent, and combined laminar and turbulent flow conditions. Once the local values are available, the average friction coefficient for the entire plate is determined by integration as Cf = 1 L ∫ L 0 Cf, x dx (11–16) Based on analysis, the boundary layer thickness and the local friction coef ficient at location x for laminar flow over a flat plate were determined in Chap. 10 to be Laminar: 𝛿 = 4.91x Re 1/2 x and Cf, x = 0.664 Re 1/2 x Rex ≲ 5 × 105 (11–17) The corresponding relations for turbulent flow are Turbulent: 𝛿 = 0.38x Re 1/5 x and Cf, x = 0.059 Re 1/5 x 5 × 105 ≲ Rex ≲ 107 (11–18) where x is the distance from the leading edge of the plate and Rex = Vx/𝜈 is the Reynolds number at location x. Note that Cf, x is proportional to 1/Rex 1/2 and thus to x−1/2 for laminar flow and it is proportional to x−1/5 for turbulent flow. In either case, Cf, x is infinite at the leading edge (x = 0), and therefore Eqs. 11–17 and 11–18 are not valid close to the leading edge. The variation of the boundary layer thickness 𝛿 and the friction coefficient Cf, x along a flat plate is sketched in Fig. 11–28. The local friction coefficients are higher in turbulent flow than they are in laminar flow because of the intense mix ing that occurs in the turbulent boundary layer. Note that Cf, x reaches its highest values when the flow becomes fully turbulent, and then decreases by a factor of x−1/5 in the flow direction, as shown in the figure. The average friction coefficient over the entire plate is determined by substituting Eqs. 11–17 and 11–18 into Eq. 11–16 and performing the inte grations (Fig. 11–29). We get Laminar: Cf = 1.33 Re 1/2 L ReL ≲ 5 × 105 (11–19) Turbulent: Cf = 0.074 Re 1/5 L 5 × 105 ≲ ReL ≲ 107 (11–20) Laminar Transition Turbulent Cf, x Cf,x V δ x FIGURE 11–28 The variation of the local friction coefficient for flow over a flat plate. Note that the vertical scale of the boundary layer is greatly exaggerated in this sketch.
1 –– L 0 L Cf V = Cf,x dx 1 –– L = dx 0.664 ––––– Rex 1/2 1/2 = 1.328 –––––– ReL = dx 0.664 ––––– L –1/2 = 2 × 0.664 –––––––– L –1/2 0 L = 0.664 ––––– L –1/2 x1/2 1 –– 2 0 L 0 L x – ––– v vL – ––– v ∫ ∫ ∫ V V ( ) ( ) ( ) FIGURE 11–29 The average friction coefficient over a surface is determined by integrating the local friction coefficient over the entire surface. The values shown here are for a laminar flat plate boundary layer.
cen96537_ch11_611-666.indd 631 14/01/17 3:16 pm 632 EXTERNAL FLOW: DRAG AND LIFT The first of these relations gives the average friction coefficient for the entire plate when the flow is laminar over the entire plate. The second rela tion gives the average friction coefficient for the entire plate only when the flow is turbulent over the entire plate, or when the laminar flow region of the plate is negligibly small relative to the turbulent flow region (that is, xcr ≪ L where the length of the plate xcr over which the flow is laminar is determined from Recr = 5 × 105 = Vxcr/𝜈). In some cases, a flat plate is sufficiently long for the flow to become turbulent, but not long enough to disregard the laminar flow region. In such cases, the average friction coefficient over the entire plate is determined by performing the integration in Eq. 11–16 over two parts: the laminar region 0 ≤ x ≤ xcr and the turbulent region xcr < x ≤ L as Cf = 1 L(∫ xcr 0 Cf, x, laminar dx + ∫ L xcr Cf, x, turbulent dx) (11–21) Note that we included the transition region with the turbulent region. Again taking the critical Reynolds number to be Recr = 5 × 105 and performing these integrations after substituting the indicated expressions, the average friction coefficient over the entire plate is determined to be Cf = 0.074 Re 1/5 L −1742 ReL 5 × 105 ≲ ReL ≲ 107 (11–22) The constants in this relation would be different for different critical Reynolds numbers. Also, the surfaces are assumed to be smooth, and the free stream to be of very low turbulence intensity. For laminar flow, the friction coefficient depends on only the Reynolds number, and the surface roughness has no effect. For turbulent flow, however, surface roughness causes the friction coefficient to increase severalfold, to the point that in the fully rough turbulent regime the friction coefficient is a function of surface roughness alone and is independent of the Reynolds number (Fig. 11–30). This is analogous to flow in pipes. A curve fit of experimental data for the average friction coefficient in this regime is given by Schlichting (1979) as Fully rough turbulent regime: Cf = (1.89 −1.62 log 𝜀 L) −2.5 (11–23) where 𝜀 is the surface roughness and L is the length of the plate in the flow direction. In the absence of a better one, this relation can be used for turbu lent flow on rough surfaces for Re > 106, especially when 𝜀/L > 10−4. Friction coefficients Cf for parallel flow over smooth and rough flat plates are plotted in Fig. 11–31 for both laminar and turbulent flows. Note that Cf increases severalfold with roughness in turbulent flow. Also note that Cf is independent of the Reynolds number in the fully rough region. This chart is the flat-plate analog of the Moody chart for pipe flows.
EXAMPLE 11–3 Drag Force on a Train Surface The top surface of the passenger car of a train moving at a velocity of 95 km/h is 2.1 m wide and 8 m long (Fig. 11–32). If the outdoor air is at 1 atm and 25°C, determine the drag force acting on the top surface of the car. Ignore any upstream boundary layer from the car(s) in front of this one; in other words, let the boundary layer start at the front edge of the top of the car. Relative Friction Roughness, Coefficient, 𝜀/L Cf 0.0 0.0029 1 × 10−5 0.0032 1 × 10−4 0.0049 1 × 10−3 0.0084 Smooth surface for Re = 107. Others calculated from Eq. 11–23 for fully rough flow.
FIGURE 11–30 For turbulent flow, surface roughness may cause the friction coefficient to increase severalfold.
0.014 0.012 0.010 0.008 0.006 0.004 0 105 108 ReL Laminar Turbulent smooth Transition 200 500 1000 2000 104 2 × 104 2 × 105 0.002 106 107 109 Cf 5 × 104 L= 300 106 Fully rough 5000 ε FIGURE 11–31 Friction coefficient for parallel flow over smooth and rough flat plates.
Data from White (2010).
cen96537_ch11_611-666.indd 632 14/01/17 3:16 pm 633 CHAPTER 11 11–6 ■ FLOW OVER CYLINDERS AND SPHERES Flow over cylinders and spheres is frequently encountered in practice. For example, the tubes in a shell-and-tube heat exchanger involve both internal flow through the tubes and external flow over the tubes, and both flows must be considered in the analysis of the heat exchanger. Also, many sports such as soccer, tennis, and golf involve flow over spherical balls. The characteristic length for a circular cylinder or sphere is taken to be the external diameter D. Thus, the Reynolds number is defined as Re = VD/𝜈 where V is the uniform velocity of the fluid as it approaches the cylinder or sphere. The critical Reynolds number for flow across a circular cylinder or sphere is about Recr ≅ 2 × 105. That is, the boundary layer remains laminar for about Re ≲ 2 × 105, is transitional for 2 × 105 ≲ Re ≲ 2 × 106, and becomes fully turbulent for Re ≳ 2 × 106. Cross-flow over a cylinder exhibits complex flow patterns, as shown in Fig. 11–33. The fluid approaching the cylinder branches out and encircles the cylinder, forming a boundary layer that wraps around the cylinder. The fluid particles on the midplane strike the cylinder at the stagnation point, bringing the fluid to a complete stop and thus raising the pressure at that point. The pressure decreases in the flow direction while the fluid velocity increases.
FIGURE 11–32 Schematic for Example 11–3.
95 km/h Air 25°C SOLUTION A train is cruising at a specified velocity. The drag force acting on the top surface of a passenger car of the train is to be determined.
Assumptions 1 The air flow is steady and incompressible. 2 The critical Reynolds number is Recr = 5 × 105. 3 Air is an ideal gas. 4 The top surface of the train is smooth (in reality it can be rough). 5 The air is calm (no significant winds).
Properties The density and kinematic viscosity of air at 1 atm and 25°C are 𝜌 = 1.184 kg/m3 and 𝜈 = 1.562 × 10−5 m2/s.
Analysis The Reynolds number is ReL = VL 𝜈= (95/3.6) m/s 1.562 × 10−5 m2/s = 1.352 × 107 which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow, and the friction coefficient is determined to be Cf = 0.074 Re 1/5 L −1742 ReL = 0.074 (1.352 × 107)1/5 − 1742 1.352 × 107 = 0.002645 Noting that the pressure drag is zero and thus CD = Cf for a flat plate, the drag force acting on the surface becomes FD = Cf AρV 2 2 = (0.002645)[(8 × 2.1) m2] (1.184 kg/m3)[(95/3.6) m/s]2 2 ( 1 N 1 kg·m/s2) = 18.3 N Discussion Note that we can solve this problem using the turbulent flow relation (instead of the combined laminar-turbulent flow relation) without much loss in accuracy since the Reynolds number is much greater than the critical value. Also, the actual drag force will probably be greater because of surface roughness effects. In a real train, however, the upstream boundary layers would significantly affect the drag on the car in question, perhaps even decreasing the drag on this car. cen96537_ch11_611-666.indd 633 14/01/17 3:16 pm 634 EXTERNAL FLOW: DRAG AND LIFT At very low upstream velocities (Re ≲ 1), the fluid completely wraps around the cylinder and the two arms of the fluid meet on the rear side of the cylinder in an orderly manner. Thus, the fluid follows the curvature of the cylinder. At higher velocities, the fluid still hugs the cylinder on the frontal side, but it is too fast to remain attached to the surface as it approaches the top (or bottom) of the cylinder. As a result, the boundary layer detaches from the surface, forming a separation region behind the cylinder. Flow in the wake region is characterized by periodic vortex formation and pressures much lower than the stagnation point pressure. The nature of the flow across a cylinder or sphere strongly affects the total drag coefficient CD. Both the friction drag and the pressure drag can be sig nificant. The high pressure in the vicinity of the stagnation point and the low pressure on the opposite side in the wake produce a net force on the body in the direction of flow. The drag force is primarily due to friction drag at low Reynolds numbers (Re ≲ 10) and to pressure drag at high Reynolds numbers (Re ≳ 5000). Both effects are significant at intermediate Reynolds numbers. The average drag coefficients CD for cross-flow over a smooth single circular cylinder and a sphere are given in Fig. 11–34. The curves exhibit different behaviors in different ranges of Reynolds numbers: • For Re ≲ 1, we have creeping flow (Chap. 10), and the drag coefficient decreases with increasing Reynolds number. For a sphere, it is CD = 24/Re. There is no flow separation in this regime.
• At about Re ≅ 10, separation starts occurring on the rear of the body with vortex shedding starting at about Re ≅ 90. The region of separation increases with increasing Reynolds number up to about Re ≅ 103. At this point, the drag is mostly (about 95 percent) due to pressure drag. The drag coefficient continues to decrease with increasing Reynolds number in this range of 10 ≲ Re ≲ 103. (A decrease in the drag coefficient does not necessarily indicate a decrease in drag. The drag force is proportional to the square of the velocity, and the increase in velocity at higher Reynolds numbers usually more than offsets the decrease in the drag coefficient.) Some useful empirical relations for drag coefficient over a sphere in this range of Reynolds numbers are: CD = 24 Re (1 + 0.0916 Re) for 0.1 < Re < 5 CD = 24 Re (1 + 0.158 Re2/3) for 5 < Re < 1000 FIGURE 11–33 Laminar boundary layer separation with a turbulent wake; flow over a circular cylinder at Re = 2000.
Courtesy of ONERA. Photo by Werlé.
cen96537_ch11_611-666.indd 634 14/01/17 3:16 pm 635 CHAPTER 11 • In the moderate range of 103 ≲ Re ≲ 105, the drag coefficient remains relatively constant. This behavior is characteristic of bluff bodies. The flow in the boundary layer is laminar in this range, but the flow in the separated region past the cylinder or sphere is highly turbulent with a wide turbulent wake.
• There is a sudden drop in the drag coefficient somewhere in the range of 105 ≲ Re ≲ 106 (usually, at about 2 × 105). This large reduction in CD is due to the flow in the boundary layer becoming turbulent, which moves the separation point further on the rear of the body, reducing the size of the wake and thus the magnitude of the pressure drag. This is in contrast to streamlined bodies, which experience an increase in the drag coef ficient (mostly due to friction drag) when the boundary layer becomes turbulent.
• There is a “transitional” regime for 2 × 105 ≲ Re ≲ 2 × 106, in which CD dips to a minimum value and then slowly rises to its final turbulent value. Flow separation occurs at about 𝜃 ≅ 80° (measured from the front stag nation point of a cylinder) when the boundary layer is laminar and at about 𝜃 ≅ 140° when it is turbulent (Fig. 11–35). The delay of separation in turbulent flow is caused by the rapid fluctuations of the fluid in the transverse direction, which enables the turbulent boundary layer to travel farther along the surface before separation occurs, resulting in a narrower wake and a smaller pressure drag. Keep in mind that turbulent flow has a fuller velocity profile as compared to the laminar case, and thus it requires a stronger adverse pressure gradient to overcome the additional momen tum close to the wall. In the range of Reynolds numbers where the flow changes from laminar to turbulent, even the drag force FD decreases as the velocity (and thus the Reynolds number) increases. This results in a sud den decrease in drag of a flying body (sometimes called the drag crisis) and instabilities in flight.
400 200 100 60 40 20 10 6 4 2 1 0.6 0.4 0.2 0.1 0.06 CD 10–1 100 101 102 103 104 Re 105 106 Smooth sphere Smooth cylinder FIGURE 11–34 Average drag coefficient for cross-flow over a smooth circular cylinder and a smooth sphere.
Data from H. Schlichting.
FIGURE 11–35 Flow visualization of flow over (a) a smooth sphere at Re = 15,000, and (b) a sphere at Re = 30,000 with a trip wire. The delay of boundary layer separation is clearly seen by comparing the two photographs.
Courtesy of ONERA. Photo by Werlé.
(a) (b) cen96537_ch11_611-666.indd 635 14/01/17 3:16 pm 636 EXTERNAL FLOW: DRAG AND LIFT Effect of Surface Roughness We mentioned earlier that surface roughness, in general, increases the drag coefficient in turbulent flow. This is especially the case for stream lined bodies. For blunt bodies such as a circular cylinder or sphere, how ever, an increase in the surface roughness may actually decrease the drag coefficient, as shown in Fig. 11–36 for a sphere. This is done by tripping the boundary layer into turbulence at a lower Reynolds number, and thus delaying flow separation, causing the fluid to close in behind the body, narrowing the wake, and reducing pressure drag considerably. This results in a much smaller drag coefficient and thus drag force for a rough-surfaced cylinder or sphere in a certain range of Reynolds number compared to a smooth one of identical size at the same velocity. At Re = 2 × 105, for example, CD ≅ 0.1 for a rough sphere with 𝜀/D = 0.0015, whereas CD ≅ 0.5 for a smooth one. Therefore, the drag coefficient in this case is reduced by a factor of 5 by simply roughening the surface. Note, however, that at Re = 106, CD ≅ 0.4 for a very rough sphere while CD ≅ 0.1 for the smooth one. Obviously, roughening the sphere in this case increases the drag by a factor of 4 (Fig. 11–37). The preceding discussion shows that roughening the surface can be used to great advantage in reducing drag, but it can also backfire on us if we are not careful—specifically, if we do not operate in the right range of the Reynolds number. With this consideration, golf balls are intentionally roughened to induce turbulence at a lower Reynolds number to take advantage of the sharp drop in the drag coefficient at the onset of turbulence in the boundary layer (the typical velocity range of golf balls is 15 to 150 m/s, and the Reynolds number is less than 4 × 105). The critical Reynolds number of dimpled golf balls is about 4 × 104. The occurrence of turbulent flow at this Reynolds number reduces the drag coefficient of a golf ball by about half, as shown in Fig. 11–36. For a given hit, this means a longer distance for the ball. Experi enced golfers also give the ball a spin during the hit, which helps the rough 0.6 0.5 0.4 0.3 0.2 0.1 0 VD v Re = 4 × 104 4 × 105 4 × 106 105 2 × 105 106 ε D ε Golf ball FD 2 1 2 D2 4 π CD = ρV = 1.25 × 10–2 D ε = 5 × 10–3 D ε = 1.5 × 10–3 D ε = 0 (smooth) = relative roughness D FIGURE 11–36 The effect of surface roughness on the drag coefficient of a sphere.
Data from Blevins (1984).
cen96537_ch11_611-666.indd 636 14/01/17 3:16 pm 637 CHAPTER 11 ball develop a lift and thus travel higher and farther. A similar argument can be given for a tennis ball. For a table tennis ball, however, the speeds are slower and the ball is smaller—it never reaches the turbulent range. There fore, the surfaces of table tennis balls are smooth. Once the drag coefficient is available, the drag force acting on a body in cross-flow is determined from Eq. 11–5 where A is the frontal area (A = LD for a cylinder of length L and A = 𝜋D2/4 for a sphere). It should be kept in mind that free-stream turbulence and disturbances by other bodies in the flow (such as flow over tube bundles) may affect the drag coeffi cient significantly.
CD Smooth Rough Surface, Re Surface 𝜀/D = 0.0015 2 × 105 0.5 0.1 106 0.1 0.4 FIGURE 11–37 Surface roughness may increase or decrease the drag coefficient of a spherical object, depending on the value of the Reynolds number.
River 30 m Pipe FIGURE 11–38 Schematic for Example 11–4.
EXAMPLE 11–4 Drag Force Acting on a Pipe in a River A 2.2-cm-outer-diameter pipe is to span across a river at a 30-m-wide section while being completely immersed in water (Fig. 11–38). The average flow velocity of water is 4 m/s and the water temperature is 15°C. Determine the drag force exerted on the pipe by the river.
SOLUTION A pipe is submerged in a river. The drag force that acts on the pipe is to be determined.
Assumptions 1 The outer surface of the pipe is smooth so that Fig. 11–34 can be used to determine the drag coefficient. 2 Water flow in the river is steady. 3 The direction of water flow is normal to the pipe. 4 Turbulence in river flow is not considered.
Properties The density and dynamic viscosity of water at 15°C are 𝜌 = 999.1 kg/m3 and 𝜇 = 1.138 × 10−3 kg/m·s.
Analysis Noting that D = 0.022 m, the Reynolds number is Re = VD 𝜈 = ρVD 𝜇 = (999.1 kg/m3)(4 m/s)(0.022 m) 1.138 × 10−3 kg/m·s = 7.73 × 104 The drag coefficient corresponding to this value is, from Fig. 11–34, CD = 1.0.
Also, the frontal area for flow past a cylinder is A = LD. Then the drag force acting on the pipe becomes FD = CD A ρV 2 2 = 1.0(30 × 0.022 m2) (999.1 kg/m3)(4 m/s)2 2 ( 1 N 1 kg·m/s2) = 5275 N ≅5300 N Discussion Note that this force is equivalent to the weight of a mass over 500 kg. Therefore, the drag force the river exerts on the pipe is equivalent to hang ing a total of over 500 kg in mass on the pipe supported at its ends 30 m apart. The necessary precautions should be taken if the pipe cannot support this force. If the river were to flow at a faster speed or if turbulent fluctuations in the river were more significant, the drag force would be even larger. Unsteady forces on the pipe might then be significant.
cen96537_ch11_611-666.indd 637 14/01/17 3:16 pm 638 EXTERNAL FLOW: DRAG AND LIFT 11–7 ■ LIFT Lift was defined earlier as the component of the net force (due to viscous and pressure forces) that is perpendicular to the flow direction, and the lift coefficient was expressed in Eq. 11 –6 as CL = FL 1 2 ρV 2A where A in this case is normally the planform area, which is the area that would be seen by a person looking at the body from above in a direction normal to the body, and V is the upstream velocity of the fluid (or, equiva lently, the velocity of a flying body in a quiescent fluid). For an airfoil of width (or span) b and chord length c (the length between the leading and trailing edges), the planform area is A = bc. The distance between the two ends of a wing or airfoil is called the wingspan or just the span. For an aircraft, the wingspan is taken to be the total distance between the tips of the two wings, which includes the width of the fuselage between the wings (Fig. 11–39). The average lift per unit planform area FL/A is called the wing loading, which is simply the ratio of the weight of the aircraft to the plan form area of the wings (since lift equals weight when flying at constant altitude). Airplane flight is based on lift, and thus developing a better understand ing of lift as well as improving the lift characteristics of bodies have been the focus of numerous studies. Our emphasis in this section is on devices such as airfoils that are specifically designed to generate lift while keeping the drag at a minimum. But it should be kept in mind that some devices such as spoilers and inverted airfoils on racing cars are designed for the opposite purpose of avoiding lift or even generating negative lift to improve traction and control (some early race cars actually “took off” at high speeds as a result of the lift produced, which alerted the engineers to come up with ways to reduce lift in their design). For devices that are intended to generate lift such as airfoils, the contri bution of viscous effects to lift is usually negligible since the bodies are streamlined, and wall shear is parallel to the surfaces of such devices and thus nearly normal to the direction of lift (Fig. 11–40). Therefore, lift in practice can be approximated as due entirely to the pressure distribution on the surfaces of the body, and thus the shape of the body has the primary influence on lift. Then the primary consideration in the design of airfoils is minimizing the average pressure at the upper surface while maximizing it at the lower surface. The Bernoulli equation can be used as a guide in identify ing the high- and low-pressure regions: Pressure is low at locations where the flow velocity is high, and pressure is high at locations where the flow velocity is low. Also, at moderate angles of attack, lift is practically inde pendent of the surface roughness since roughness affects the wall shear, not the pressure. The contribution of shear to lift is significant only for very small (lightweight) bodies that fly at low velocities (and thus low Reynolds numbers). Noting that the contribution of viscous effects to lift is negligible, we should be able to determine the lift acting on an airfoil by simply integrating the pressure distribution around the airfoil. The pressure changes in the flow Planform area, bc Angle of attack Chord, c Span, b FL FD α FIGURE 11–39 Definition of various terms associated with an airfoil.
FL V α FD Direction of wall shear Direction of lift FIGURE 11–40 For airfoils, the contribution of viscous effects to lift is usually negligible since wall shear is parallel to the surfaces and thus nearly normal to the direction of lift.
cen96537_ch11_611-666.indd 638 14/01/17 3:16 pm 639 CHAPTER 11 direction along the surface, but it remains essentially constant through the boundary layer in a direction normal to the surface (Chap. 10). Therefore, it seems reasonable to ignore the very thin boundary layer on the airfoil and calculate the pressure distribution around the airfoil from the relatively simple potential flow theory (zero vorticity, irrotational flow) for which net viscous forces are zero for flow past an airfoil. The flow fields obtained from such calculations are sketched in Fig. 11–41 for both symmetrical and nonsymmetrical airfoils by ignoring the thin boundary layer. At zero angle of attack, the lift produced by the symmet rical airfoil is zero, as expected because of symmetry, and the stagnation points are at the leading and trailing edges. For the nonsymmetrical airfoil, which is at a small angle of attack, the front stagnation point has moved down below the leading edge, and the rear stagnation point has moved up to the upper surface close to the trailing edge. To our surprise, the lift pro duced is calculated again to be zero—a clear contradiction of experimental observations and measurements. Obviously, the theory needs to be modified to bring it in line with the observed phenomenon. The source of inconsistency is the rear stagnation point being at the upper surface instead of the trailing edge. This requires the lower side fluid to make a nearly U-turn and flow around the sharp trailing edge toward the stagnation point while remaining attached to the surface, which is a physi cal impossibility since the observed phenomenon is the separation of flow at sharp turns (imagine a car attempting to make this turn at high speed). Therefore, the lower side fluid separates smoothly off the trailing edge, and the upper side fluid responds by pushing the rear stagnation point down stream. In fact, the stagnation point at the upper surface moves all the way to the trailing edge. This way the two flow streams from the top and the bottom sides of the airfoil meet at the trailing edge, yielding a smooth flow downstream parallel to the sharp trailing edge. Lift is generated because the flow velocity at the top surface is higher, and thus the pressure on that sur face is lower due to the Bernoulli effect. The potential flow theory and the observed phenomenon can be reconciled as follows: Flow starts out as predicted by theory, with no lift, but the lower fluid stream separates at the trailing edge when the velocity reaches a certain value. This forces the separated upper fluid stream to close in at the trailing edge, initiating clockwise circulation around the airfoil. This clockwise circu lation increases the velocity of the upper stream while decreasing that of the lower stream, causing lift. A starting vortex of opposite sign (counterclock wise circulation) is then shed downstream (Fig. 11–42), and smooth stream lined flow is established over the airfoil. When the potential flow theory is modified by the addition of an appropriate amount of circulation to move the stagnation point down to the trailing edge, excellent agreement is obtained between theory and experiment for both the flow field and the lift. It is desirable for airfoils to generate the most lift while producing the least drag. Therefore, a measure of performance for airfoils is the lift-to-drag ratio, which is equivalent to the ratio of the lift-to-drag coefficients CL/CD. This information is provided either by plotting CL versus CD for dif ferent values of the angle of attack (a lift–drag polar) or by plotting the ratio CL/CD versus the angle of attack. The latter is done for a particular airfoil design in Fig. 11–43. Note that the CL/CD ratio increases with the angle of (a) Irrotational flow past a symmetrical airfoil (zero lift) Stagnation points (b) Irrotational flow past a nonsymmetrical airfoil (zero lift) Stagnation points (c) Actual flow past a nonsymmetrical airfoil (positive lift) Stagnation points FIGURE 11–41 Irrotational and actual flow past symmetrical and nonsymmetrical two-dimensional airfoils.
Clockwise circulation Counterclockwise circulation Starting vortex FIGURE 11–42 Shortly after a sudden increase in angle of attack, a counterclockwise starting vortex is shed from the airfoil, while clockwise circulation appears around the airfoil, causing lift to be generated.
cen96537_ch11_611-666.indd 639 14/01/17 3:16 pm 640 EXTERNAL FLOW: DRAG AND LIFT attack until the airfoil stalls, and the value of the lift-to-drag ratio can be of the order of 100 for a two-dimensional airfoil. One obvious way to change the lift and drag characteristics of an airfoil is to change the angle of attack. On an airplane, for example, the entire plane is pitched up to increase lift, since the wings are fixed relative to the fuse lage. Another approach is to change the shape of the airfoil by the use of movable leading edge and trailing edge flaps, as is commonly done in mod ern large aircraft (Fig. 11–44). The flaps are used to alter the shape of the wings during takeoff and landing to maximize lift and to enable the aircraft to land or take off at low speeds. The increase in drag during this takeoff and landing is not much of a concern because of the relatively short time periods involved. Once at cruising altitude, the flaps are retracted, and the wing is returned to its “normal” shape with minimal drag coefficient and adequate lift coefficient to minimize fuel consumption while cruising at a constant altitude. Note that even a small lift coefficient can generate a large lift force during normal operation because of the large cruising velocities of aircraft and the proportionality of lift to the square of flow velocity. The effects of flaps on the lift and drag coefficients are shown in Fig. 11–45 for an airfoil. Note that the maximum lift coefficient increases from about 1.5 for the airfoil with no flaps to 3.5 for the double-slotted flap case. But also note that the maximum drag coefficient increases from about 0.06 for the air foil with no flaps to about 0.3 for the double-slotted flap case. This is a five fold increase in the drag coefficient, and the engines must work much harder to provide the necessary thrust to overcome this drag. The angle of attack of the flaps can be increased to maximize the lift coefficient. Also, the flaps extend the chord length, and thus enlarge the wing area A. The Boeing 727 uses a triple-slotted flap at the trailing edge and a slot at the leading edge. The minimum flight velocity is determined from the requirement that the total weight W of the aircraft be equal to lift and CL = CL, max. That is, W = FL = 1 2CL, max ρV 2 min A → Vmin = √ 2W ρCL, max A (11–24) For a given weight, the landing or takeoff speed can be minimized by maxi mizing the product of the lift coefficient and the wing area, CL, maxA. One way of doing that is to use flaps, as already discussed. Another way is to control the boundary layer, which can be accomplished simply by leaving flow sections (slots) between the flaps, as shown in Fig. 11–46. Slots are used to prevent the separation of the boundary layer from the upper surface of the wings and the flaps. This is done by allowing air to move from the high-pressure region under the wing into the low-pressure region at the top surface. Note that the lift 120 100 80 60 40 20 0 –20 –40 –4 –8 0 α, degrees 4 8 Stall ––– CL CD NACA 64(1) – 412 airfoil Re = 7 × 105 FIGURE 11–43 The variation of the lift-to-drag ratio with angle of attack for a two-dimensional airfoil.
Data from Abbott, von Doenhoff, and Stivers (1945).
FIGURE 11–44 The lift and drag characteristics of an airfoil during takeoff and landing are changed by changing the shape of the airfoil by the use of movable flaps.
Photos by Yunus Çengel.
(a) Flaps extended (landing) (b) Flaps retracted (cruising) cen96537_ch11_611-666.indd 640 14/01/17 3:16 pm 641 CHAPTER 11 coefficient reaches its maximum value CL = CL, max, and thus the flight velocity reaches its minimum, at stall conditions, which is a region of unstable opera tion and must be avoided. The Federal Aviation Administration (FAA) does not allow operation below 1.2 times the stall speed for safety. Another thing we notice from this equation is that the minimum velocity for takeoff or landing is inversely proportional to the square root of den sity. Noting that air density decreases with altitude (by about 15 percent at 1500 m), longer runways are required at airports at higher altitudes such as Denver to accommodate higher minimum takeoff and landing velocities. The situation becomes even more critical on hot summer days since the den sity of air is inversely proportional to temperature. The development of efficient (low-drag) airfoils was the subject of intense experimental investigations in the 1930s. These airfoils were standardized by the National Advisory Committee for Aeronautics (NACA, which is now NASA), and extensive lists of data on lift coefficients were reported. The variation of the lift coefficient CL with angle of attack for two 2-D (infinite span) airfoils (NACA 0012 and NACA 2412) is given in Fig. 11–47. We make the following observations from this figure: • The lift coefficient increases almost linearly with angle of attack 𝛼, reaches a maximum at about 𝛼 = 16°, and then starts to decrease sharply. This decrease of lift with further increase in the angle of attack is called stall, and it is caused by flow separation and the formation of a wide wake region over the top surface of the airfoil. Stall is highly undesirable since it also increases drag.
• At zero angle of attack (𝛼 = 0°), the lift coefficient is zero for symmetrical airfoils but nonzero for nonsymmetrical ones with greater curvature at the top surface. Therefore, planes with symmetrical wing sections must fly with their wings at higher angles of attack in order to produce the same lift.
• The lift coefficient is increased by severalfold by adjusting the angle of attack (from 0.25 at 𝛼 = 0° for the nonsymmetrical airfoil to 1.25 at 𝛼 = 10°).
• The drag coefficient also increases with angle of attack, often exponentially (Fig. 11–48). Therefore, large angles of attack should be used sparingly for short periods of time for fuel efficiency.
0 5 –5 20 10 3.48 CLmax CL CD Angle of attack, α (degrees) 15 0 0.15 0.05 0.30 0.25 0.10 0.20 3.5 3.0 2.5 2.0 1.5 1.0 0.5 CL 3.5 3.0 2.5 2.0 1.5 1.0 0.5 Slotted flap Clean (no flap) Double-slotted flap Double-slotted flap 2.67 Slotted flap 1.52 Clean (no flap) FIGURE 11–45 Effect of flaps on the lift and drag coefficients of an airfoil.
Data from Abbott and von Doenhoff, for NACA 23012 (1959).
Wing Slot Flap FIGURE 11–46 A flapped airfoil with a slot to prevent the separation of the boundary layer from the upper surface and to increase the lift coefficient.
cen96537_ch11_611-666.indd 641 14/01/17 3:16 pm 642 EXTERNAL FLOW: DRAG AND LIFT Finite-Span Wings and Induced Drag For airplane wings and other airfoils of finite span, the end effects at the tips become important because of the fluid leakage between the lower and upper surfaces. The pressure difference between the lower surface (high-pressure region) and the upper surface (low-pressure region) drives the fluid at the tips upward while the fluid is swept toward the back because of the relative motion between the fluid and the wing. This results in a swirling motion that spirals along the flow, called the tip vortex, at the tips of both wings. Vortices are also formed along the airfoil between the tips of the wings. These distributed vortices collect toward the edges after being shed from the trailing edges of the wings and combine with the tip vorti ces to form two streaks of powerful trailing vortices along the tips of the wings (Fig. 11–49). Trailing vortices generated by large aircraft persist for a long time for long distances (over 10 km) before they gradually disappear due to viscous dissipation. Such vortices and the accompanying downdraft are strong enough to cause a small aircraft to lose control and flip over if it flies through the wake of a larger aircraft. Therefore, following a large aircraft closely (within 10 km) poses a real danger for smaller aircraft. This issue is the controlling factor that governs the spacing of aircraft at take off, which limits the flight capacity at airports. In nature, this effect is used to advantage by birds that migrate in V-formation by utilizing the updraft generated by the bird in front. It has been determined that the birds in a typical flock can fly to their destination in V-formation with one-third less energy. Military jets also occasionally fly in V-formation for the same reason (Fig. 11–50). Tip vortices that interact with the free stream impose forces on the wing tips in all directions, including the flow direction. The component of the force in the flow direction adds to drag and is called induced drag. The total drag of a wing is then the sum of the induced drag (3-D effects) and the drag of the airfoil section (2-D effects).
–5 0 5 Angle of attack, α (degrees) 10 15 20 2.00 1.50 1.00 0.50 0 –0.50 CL V α V α NACA 0012 section NACA 2412 section FIGURE 11–47 The variation of the lift coefficient with angle of attack for a symmetrical and a nonsymmetrical airfoil.
Data from Abbott (1945, 1959).
0.020 0.016 0.012 0.008 0.004 0 CD 0 4 α 8 12 Angle of attack, α (degrees) 16 20 NACA 23015 section V FIGURE 11–48 The variation of the drag coefficient of an airfoil with angle of attack.
Data from Abbott and von Doenhoff (1959).
cen96537_ch11_611-666.indd 642 14/01/17 3:16 pm 643 CHAPTER 11 The ratio of the square of the average span of an airfoil to the planform area is called the aspect ratio. For an airfoil with a rectangular planform of chord c and span b, it is expressed as AR = b2 A = b2 bc = b c (11–25) Therefore, the aspect ratio is a measure of how (relatively) narrow an airfoil is in the flow direction. The lift coefficient of wings, in general, increases while the drag coefficient decreases with increasing aspect ratio. This is because a long narrow wing (large aspect ratio) has a shorter tip length and thus smaller tip losses and smaller induced drag than a short and wide wing of the same planform area. Therefore, bodies with large aspect ratios fly more efficiently, but they are less maneuverable because of their larger moment of inertia (owing to the greater distance from the center). Bodies with smaller aspect ratios maneuver better since the wings are closer to the central part. So it is no surprise that fighter planes (and fighter birds like falcons) have short and wide wings while large commercial planes (and soaring birds like albatrosses) have long and narrow wings. The end effects can be minimized by attaching endplates or winglets at the tips of the wings perpendicular to the top surface. The endplates func tion by blocking some of the leakage around the wing tips, which results in a considerable reduction in the strength of the tip vortices and the induced drag. Wing tip feathers on birds fan out for the same purpose (Fig. 11–51).
Lift Generated by Spinning You have probably experienced giving a spin to a tennis ball or making a drop shot on a tennis or ping-pong ball by giving a fore spin in order to alter the lift characteristics and cause the ball to produce a more desirable trajec tory and bounce of the shot. Golf, soccer, and baseball players also utilize spin in their games. The phenomenon of producing lift by the rotation of a solid body is called the Magnus effect after the German scientist Heinrich Magnus (1802–1870), who was the first to study the lift of rotating bod ies, which is illustrated in Fig. 11–52 for the simplified case of irrotational (potential) flow. When the ball is not spinning, the lift is zero because of top–bottom symmetry. But when the cylinder is rotated about its axis, the cylinder drags some fluid around because of the no-slip condition and the flow field reflects the superposition of the spinning and nonspinning flows. The stagnation points shift down, and the flow is no longer symmetric about the horizontal plane that passes through the center of the cylinder. The aver age pressure on the upper half is less than the average pressure on the lower half because of the Bernoulli effect, and thus there is a net upward force (lift) acting on the cylinder. A similar argument can be given for the lift generated on a spinning ball. The effect of the rate of rotation on the lift and drag coefficients of a smooth sphere is shown in Fig. 11–53. Note that the lift coefficient strongly depends on the rate of rotation, especially at low angular velocities. The effect of the rate of rotation on the drag coefficient is small. Roughness also affects the drag and lift coefficients. In a certain range of Reynolds number, roughness produces the desirable effect of increasing the lift coefficient while decreasing FIGURE 11–49 Trailing vortices visualized in various ways: (a) Smoke streaklines in a wind tunnel show vortex cores leaving the trailing edge of a rectangular wing; (b) Four contrails initially formed by condensation of water vapor in the low pressure region behind the jet engines eventually merge into the two counter-rotating trailing vortices that persist very far downstream; (c) A crop duster flies through smoky air which swirls around in one of the tip vortices from the aircraft’s wing.
(a) (b) (c) (a) From Head, Malcolm R. 1982 in Flow Visualization II, W. Merzkirch. Ed., 399–403, Washington: Hemisphere; (b) © Geostock/Getty Images RF;(c) NASA Langley Research Center cen96537_ch11_611-666.indd 643 14/01/17 3:16 pm 644 EXTERNAL FLOW: DRAG AND LIFT the drag coefficient. Therefore, golf balls with the right amount of roughness travel higher and farther than smooth balls for the same hit.
FIGURE 11–50 (a) Geese flying in their characteristic V-formation to save energy. (b) Military jets imitating nature.
(a) © Corbis RF (b) © Charles Smith/Corbis RF FIGURE 11–51 Induced drag is reduced by (a) wing tip feathers on bird wings and (b) endplates or other disruptions on airplane wings.
(a) © Ken Canning/Getty Images RF; (b) Courtesy of Schempp-Hirth Flugzeugbau GmbH (b) Winglets are used on this sailplane to reduce induced drag.
(a) A bald eagle with its wing feathers fanned out during flight.
EXAMPLE 11–5 Lift and Drag of a Commercial Airplane A commercial airplane has a total mass of 70,000 kg and a wing planform area of 150 m2 (Fig. 11–54). The plane has a cruising speed of 558 km/h and a cruis ing altitude of 12,000 m, where the air density is 0.312 kg/m3. The plane has dou ble-slotted flaps for use during takeoff and landing, but it cruises with all flaps retracted. Assuming the lift and the drag characteristics of the wings can be approx imated by NACA 23012 (Fig. 11–45), determine (a) the minimum safe speed for takeoff and landing with and without extending the flaps, (b) the angle of attack to cruise steadily at the cruising altitude, and (c) the power that needs to be supplied to provide enough thrust to overcome wing drag.
SOLUTION The cruising conditions of a passenger plane and its wing characteristics are given. The minimum safe landing and takeoff speeds, the angle of attack during cruising, and the power required are to be determined.
Assumptions 1 The drag and lift produced by parts of the plane other than the wings, such as the fuselage are not considered. 2 The wings are assumed to be two-dimensional airfoil sections, and the tip effects of the wings are not considered. 3 The lift and the drag characteristics of the wings are approximated by NACA 23012 so that Fig. 11–45 is applicable. 4 The average density of air on the ground is 1.20 kg/m3.
Properties The density of air is 1.20 kg/m3 on the ground and 0.312 kg/m3 at cruising altitude. The maximum lift coefficient CL, max of the wing is 3.48 and 1.52 with and without flaps, respectively (Fig. 11–45).
Analysis (a) The weight and cruising speed of the airplane are W = mg = (70,000 kg)(9.81 m/s2)( 1 N 1 kg·m/s2) = 686,700 N V = (558 km/h)( 1 m/s 3.6 km/h) = 155 m/s The minimum velocities corresponding to the stall conditions without and with flaps, respectively, are obtained from Eq. 11–24, Vmin 1 = √ 2W ρCL, max 1 A = √ 2(686,700 N) (1.2 kg/m3)(1.52)(150 m2) ( 1 kg·m/s2 1 N ) = 70.9 m/s Vmin 2 = √ 2W ρCL, max 2 A = √ 2(686,700 N) (1.2 kg/m3)(3.48)(150 m2) ( 1 kg·m/s2 1 N ) = 46.8 m/s Then the “safe” minimum velocities to avoid the stall region are obtained by multiplying the values above by 1.2: Without flaps: Vmin 1, safe = 1.2Vmin 1 = 1.2(70.9 m/s) = 85.1 m/s = 306 km/h With flaps: Vmin 2, safe = 1.2Vmin 2 = 1.2(46.8 m/s) = 56.2 m/s = 202 km/h since 1 m/s = 3.6 km/h. Note that the use of flaps allows the plane to take off and land at considerably lower velocities, and thus on a shorter runway.
cen96537_ch11_611-666.indd 644 14/01/17 3:16 pm 645 CHAPTER 11 Stagnation points Stagnation points High velocity, low pressure Lift Low velocity, high pressure (b) Potential flow over a rotating cylinder (a) Potential flow over a stationary cylinder FIGURE 11–52 Generation of lift on a rotating circular cylinder for the case of “idealized” potential flow (the actual flow involves flow separation in the wake region).
(b) When an aircraft is cruising steadily at a constant altitude, the lift must be equal to the weight of the aircraft, FL = W. Then the lift coefficient is CL = FL 1 2 ρ V 2A = 686,700 N 1 2 (0.312 kg/m3)(155 m/s)2(150 m2) ( 1 kg·m/s2 1 N ) = 1.22 For the case with no flaps, the angle of attack corresponding to this value of CL is determined from Fig. 11–45 to be 𝛼 ≅ 10°.
(c) When the aircraft is cruising steadily at a constant altitude, the net force acting on the aircraft is zero, and thus thrust provided by the engines must be equal to the drag force. The drag coefficient corresponding to the cruising lift coefficient of 1.22 is determined from Fig. 11–45 to be CD ≅ 0.03 for the case with no flaps. Then the drag force acting on the wings becomes FD = CD A ρV2 2 = (0.03)(150 m2) (0.312 kg/m3)(155 m/s)2 2 ( 1 kN 1000 kg·m/s2) = 16.9 kN Noting that power is force times velocity (distance per unit time), the power required to overcome this drag is equal to the thrust times the cruising velocity: Power = Thrust × Velocity = FDV = (16.9 kN)(155 m/s)( 1 kW 1 kN·m/s) = 2620 kW Therefore, the engines must supply 2620 kW of power to overcome the drag on the wings during cruising. For a propulsion efficiency of 30 percent (i.e., 30 percent of the energy of the fuel is utilized to propel the aircraft), the plane requires energy input at a rate of 8730 kJ/s.
Discussion The power determined is the power to overcome the drag that acts on the wings only and does not include the drag that acts on the remaining parts of the aircraft (the fuselage, the tail, etc.). Therefore, the total power required during cruising will be much greater. Also, it does not consider induced drag, which can be dominant during takeoff when the angle of attack is high (Fig. 11–45 is for a 2-D airfoil, and does not include 3-D effects).
cen96537_ch11_611-666.indd 645 14/01/17 3:16 pm 646 EXTERNAL FLOW: DRAG AND LIFT 0.8 0.6 0.4 0.2 00 1 2 3 4 5 D Smooth sphere ω VD Re = = 6 × 104 CD, CL V FD D2 CD = 1 2 π 4 ρV2 FL D2 CL = 1 2 π 4 ρV2 V 1 2 ωD/ v FIGURE 11–53 The variation of lift and drag coefficients of a smooth sphere with the nondimensional rate of rotation for Re = VD/𝜈 = 6 × 104.
Data from Goldstein (1938).
558 km/h 70,000 kg 12,000 m 150 m2, double-flapped FIGURE 11–54 Schematic for Example 11–5.
EXAMPLE 11–6 Effect of Spin on a Tennis Ball A tennis ball with a mass of 0.125 lbm and a diameter of 2.52 in is hit at 45 mi/h with a backspin of 4800 rpm (Fig. 11–55). Determine if the ball will fall or rise under the combined effect of gravity and lift due to spinning shortly after being hit in air at 1 atm and 80°F.
SOLUTION A tennis ball is hit with a backspin. It is to be determined whether the ball will fall or rise after being hit.
Assumptions 1 The surface of the ball is smooth enough for Fig. 11–53 to be applicable (this is a stretch for a tennis ball). 2 The ball is hit horizontally so that it starts its motion horizontally.
Properties The density and kinematic viscosity of air at 1 atm and 80°F are 𝜌 = 0.07350 lbm/ft3 and 𝜈 = 1.697 × 10−4 ft2/s.
Analysis The ball is hit horizontally, and thus it would normally fall under the effect of gravity without the spin. The backspin generates a lift, and the ball will rise if the lift is greater than the weight of the ball. The lift is determined from FL = CL A ρV 2 2 where A is the frontal area of the ball, which is A = 𝜋D 2/4. The translational and angular velocities of the ball are V = (45 mi/h)( 5280 ft 1 mi ) ( 1 h 3600 s) = 66 ft/s 𝜔= (4800 rev/min)( 2𝜋 rad 1 rev ) ( 1 min 60 s ) = 502 rad/s Then, the nondimensional rate of rotation is 𝜔D 2V = (502 rad/s)(2.52/12 ft) 2(66 ft/s) = 0.80 rad From Fig. 11–53, the lift coefficient corresponding to this value is CL = 0.21. Then the lift force acting on the ball is FL = (0.21) 𝜋(2.52/12 ft)2 4 (0.0735 lbm/ft3)(66 ft/s)2 2 ( 1 lbf 32.2 lbm·ft/s2) = 0.036 lbf The weight of the ball is W = mg = (0.125 lbm)(32.2 ft/s2)( 1 lbf 32.2 lbm·ft/s2) = 0.125 lbf which is more than the lift. Therefore, the ball will drop under the combined effect of gravity and lift due to spinning with a net force of 0.125 − 0.036 = 0.089 lbf.
Discussion This example shows that the ball can be hit much farther by giv ing it a backspin. Note that a topspin has the opposite effect (negative lift) and speeds up the drop of the ball to the ground. Also, the Reynolds number for this problem is 8 × 104, which is sufficiently close to the 6 × 104 for which Fig. 11–53 is prepared.
cen96537_ch11_611-666.indd 646 14/01/17 3:17 pm 647 CHAPTER 11 No discussion on lift and drag would be complete without mention ing the contributions of Wilbur (1867–1912) and Orville (1871–1948) Wright. The Wright Brothers are truly the most impressive engineering team of all time. Self-taught, they were well informed of the contem porary theory and practice in aeronautics. They both corresponded with other leaders in the field and published in technical journals. While they cannot be credited with developing the concepts of lift and drag, they used them to achieve the first powered, manned, heavier-than-air, con trolled flight (Fig. 11–56). They succeeded, while so many before them failed, because they evaluated and designed parts separately. Before the Wrights, experimenters were building and testing whole airplanes. While intuitively appealing, the approach did not allow the determina tion of how to make the craft better. When a flight lasts only a moment, you can only guess at the weakness in the design. Thus, a new craft did not necessarily perform any better than its predecessor. Testing was sim ply one belly flop followed by another. The Wrights changed all that. They studied each part using scale and full-size models in wind tun nels and in the field. Well before the first powered flyer was assem bled, they knew the area required for their best wing shape to support a plane carrying a man and the engine horsepower required to provide adequate thrust with their improved impeller. The Wright Brothers not only showed the world how to fly, they showed engineers how to use the equations presented here to design even better aircraft.
Flying in Nature! Birds, insects, and other flying animals use oscillatory and undulatory motions to generate lift and thrust in a manner that produces high agility and also efficiency. Although Leonardo da Vinci studied the flight of birds and designed a flapping-wing device a long time ago, this topic is still fresh and still complicated. The classic aerodynamic theory is appropriate only for the fixed-wing aircraft or gliding animals, where lift is obtained primarily by attached flow over a nominally rigid wing and fails to predict the generated lift on insect wings. The paradox that bumblebees can fly, despite that aerodynamic theory suggests that they cannot generate enough lift, has been largely resolved through the discovery of unsteady mecha nisms of lift generation and the dominant leading-edge vortex formed on their flapping wings. Ellington et al.8 were the first researchers to 45 mi/h 4800 rpm Ball m = 0.125 lbm FIGURE 11–55 Schematic for Example 11–6.
FIGURE 11–56 The Wright Brothers take flight at Kitty Hawk.
Library of Congress Prints & Photographs Division [LC-DIG-ppprs-00626]. Also keep in mind that although some spin may increase the distance traveled by a ball, there is an optimal spin that is a function of launch angle, as most golfers are now more aware. Too much spin decreases distance by introducing more induced drag.
This section was contributed by Azar Eslam Panah, Penn State University at Berks cen96537_ch11_611-666.indd 647 14/01/17 3:17 pm 648 EXTERNAL FLOW: DRAG AND LIFT FIGURE 11–57 Production of a vortex loop with cross sections of the loop core.
© Adrian Thomas, University of Oxford Free stream U Leading edge vortex Deflection of free stream Tip vortex core Direction of vortex convection ϕ ≈ 6 × 4 mm ϕ ≈ 2 mm ϕ ≈ 8 mm Clockwise vorticity Roll up of counterclockwise cross-stream vortex filaments Counterclockwise vorticity FIGURE 11–58 German President Christian Wulff flies the smart bird “Silvermoewe” from the company Fetsco in Esslingen, Germany.
© EPA European Pressphoto Agency b.v./Alamy FIGURE 11–59 Gliding Juvenile red-tailed hawk.
© Dr. Susa H. Stonedahl, St. Ambrose University.
visualize a tornado-like vortex that ran outward along the leading edge of each wing, by releasing smoke around the wings of a hawkmoth. In a similar study by Bomphrey et al.,6 they showed the complete vortex loop induced by the moth highlighting the main wake elements includ ing a leading edge vortex (LEV) (Fig. 11.57). These LEV structures on top of the wing generate lift beyond the values predicted by traditional aerodynamics. For many years after Ellington, researchers have studied the aerodynamics of birds/insects to quantify the extra lift and build robots that can fly like a real animal. Recently, Festo Company has designed and built an ultralight but powerful flight model which can start, fly, and land autonomously— with no additional drive mechanism as presented in Fig. 11.58. Why do we build robots like that or smaller? Well, as a practical application, you can send them inside collapsed buildings as first responders to locate injured people or maybe look for chemical leaks. These flying robots can also advance our biological understanding of flight. For example, they can be used to test specific hypotheses that concern stability and control in gusts and other ambient disturbances.15 Although we have made progress build ing these intriguing flapping robots, they cannot fly robustly in turbulent environments whereas animals can. As stated by Shyy et al.,18 “[robots] fly commercially or recreationally, but animals fly professionally.” Combining flapping patterns, body center of gravity, and tail adjustment, natural flyers can maneuver instantaneously on windy days and track targets precisely. A brief summary of different flapping modes found in nature is described below.19 Gliders When the birds stop flapping and keep their wings stretched out as shown in Fig. 11.59, they can generate only lift. This regime is known as “gliding.” A glider animal tilts its direction of motion slightly downward relative to the air; thus this gliding angle directly controls the lift-to-drag ratio. With the exception of hummingbirds, all flying birds can glide to some extent. As a rule, the smaller the bird, the shorter the distance it can glide and the faster it sinks. For example, a pigeon with 0.9 kg weight descends about 10 m during a glide of approximately 90 m; while a golden eagle with 4.3 kg weight can glide 170 m with the same loss in height. Eagles are capable of nonstop flapping flight but they usually spend less than 2 minutes per hour doing it to save energy.
Large Birds To generate thrust and lift for forward flight, large birds like pelicans and geese make use of a vertical (plunging) motion in combination with rota tional motion of the wing during a downstroke (downward wing motion). In contrast, the upstroke (upward wing motion) is relatively passive and the wing is rotated to a near zero effective angle of attack (Fig. 11.60). The wing size and flight velocity are large enough for flight in this regime. Flapping frequencies are typically quite low, the amplitude tends to be high, and the flow field is effectively attached throughout the flapping motion. cen96537_ch11_611-666.indd 648 14/01/17 3:17 pm 649 CHAPTER 11 FIGURE 11–60 American White Pelican in forward flight during (a) the onset of down stroke and (b) onset of upstroke motions.
© Dr. Forrest Stonedahl, Augustana College (b) (a) FIGURE 11–61 (a) Hovering hummingbird and (b) wing orientation in the various stages of flapping.
(a) © Susa H. Stonedahl, St. Ambrose University. Reprinted by permission.
(a) (b) Medium Birds Medium-sized birds such as pigeons use a combination of plunging, pitch ing, and rotating motion to generate lift and thrust in forward and hovering flight. Here, the stroke plane is rotated to produce a net force acting in the vertical direction. During the upstroke the wings are folded for minimum effective area; thus it is effectively passive. In this regime, the combina tion of wing size and speed of motion is sufficient to produce the neces sary forces in the traditional manner (attached flow) during the downstroke. However, extreme maneuvers are likely to require genuine unsteady and separated flow aerodynamics.
Small Birds and Insects When the wing size decreases to such small scales as in insects or hum mingbirds, the flight speed also decreases, and to some extent this leads to an increase in flapping frequency. In this regime, the stroke plane is almost horizontal and the wing is pitched at either end of the stroke (Fig. 11.61). This gives an effective positive angle of attack during both the upstroke and downstroke as the downstroke alone is insufficient to generate the necessary lift. The wing now effectively works in a propeller-type motion similar to a helicopter. Based on the different species and their size, hummingbirds can flap their wings at a frequency of 12–90 Hz during hovering. Such small flyers cannot glide as a result of the poor lift coefficients available at low Reynolds numbers. Even at high flapping frequencies, the effective Reynolds number is too low to ensure good lift coefficients; however, in this regime the LEV plays a major role in lift generation. Micro Air Vehicles (MAVs) and small birds share a direct connection in size, speed, and flight regime. Aerospace engineers are eagerly interested in the relationship of MAVs’ modern applications and substantive fundamental problems. How Do Birds/Insects Use Flapping Wing Motion to Fly?
Let’s invoke Newton’s second and third laws here to explain this: the wing pushes back on the air; the air accelerates backward due to the applied force; an equal and opposite reaction on the wing results in an applied forward force to the bird to overcome its drag. During each flapping cycle, the wing produces vortex structures which are responsible for the aerodynamic forces on the body. How do these structures form? Vorticity (local rotation of the fluid) is created on the surface of the wing. It separates at high angles of attack and rolls itself up into discrete vortices. Many birds and insects create large trailing edge vortices with their wings when they move through air. They also create the leading edge vortex (LEV) which is the most impor tant feature of the flow created by flapping wings because it significantly enhances lift forces. This also bears some resemblance to the phenomenon known as “dynamic stall,” which is observed on conventional engineered systems such as helicopter rotors and wind turbine blades. How does this flow separation increase lift? As explained by Sane,16 the fluid stream going over the wing separates while increasing its angle of attack and the LEV occupies the separation zone above the wing. Because the flow reattaches, the fluid continues to flow smoothly from the trail ing edge maintaining the Kutta condition. Subsequently, a suction force cen96537_ch11_611-666.indd 649 14/01/17 3:17 pm 650 EXTERNAL FLOW: DRAG AND LIFT FIGURE 11–62 (a) Wake topology of a pitching panel in absence of the LEV and (b) vorticity shed from a plunging panel in presence of the LEV using dye flow visualization technique.
© Dr. Azar Eslam Panah, Penn State University (b) (a) develops normal to the wing in the separation zone, resulting in substantial enhancement of lift. Other unsteady lift enhancement mechanisms are “clap and fling” motion (originally described by Weis-Fogh), Wagner effect, rota tional lift, and wake capture.16,18 As soon as the vortices separate from the body, they are left behind in the wake like a footprint of the bird or insect. The vorticity shed from oscillat ing bodies can significantly alter the aerodynamic forces on the body. Of equal importance, the shed vorticity field can impact aerodynamic loads on other bodies present in the wake, and thus understanding the evolution of the shed vorticity can provide insight into flow interactions within bird flocks and fish schools, as well as the control of bio-robotic vehicles. While much is known about the vortex topology of these 2-D and 3-D flows, there is still not a framework for a robust model by which the complete vortex structure in the wake may be predicted nor quantitative models for predict ing the strengths of the various constituents of the wakes. Some interest ing patterns arise in the wake of a pitching panel when the LEV is absent as well as a plunging panel when the LEV exists, which are displayed in Fig. 11.62 for limited parameter domains. As a wing oscillates, it sheds a highly complex, three-dimensional, and unsteady system of interconnected vortices. The aerodynamic performance of hovering insects and birds is largely explained by the presence of a stably attached LEV on top of their wings. Studying these vortex structures can help to better develop aerody namic models for flapping wings over a large range in size.
Can We Expect to See Large Airplanes with Flapping Wings in Near Future? Flapping flight is fascinating but very complicated and its complexity needs to be translated to a robust aerodynamic model in order to improve vehicle design. We already have a basic understanding of how birds and insects fly but we know little about flapping flight during aggressive maneuvers. Therefore, we need passionate researchers to investigate the flapping wing aerodynamics as rigorous and enthusiastic as aeronautical engineers who studied the fixed wing aircraft in the 1920s and 1930s.19 Next time you look at birds flying in the sky, try to ponder how to create a large-scale flapping-wing vehicle that can fly well despite complicated air flow. Maybe you can fulfill Leonardo da Vinci’s imagination!
SUMMARY In this chapter, we study flow of fluids over immersed bodies with emphasis on the resulting lift and drag forces. A fluid may exert forces and moments on a body in and about vari ous directions. The force a flowing fluid exerts on a body in the flow direction is called drag while that in the direction normal to the flow is called lift. The part of drag that is due directly to wall shear stress 𝜏w is called the skin friction drag since it is caused by frictional effects, and the part that is due directly to pressure P is called the pressure drag or form drag because of its strong dependence on the form or shape of the body. The drag coefficient CD and the lift coefficient CL are dimensionless numbers that represent the drag and the lift characteristics of a body and are defined as CD = FD 1 2ρV 2A and CL = FL 1 2ρV 2A cen96537_ch11_611-666.indd 650 14/01/17 3:17 pm 651 CHAPTER 11 where A is usually the frontal area (the area projected on a plane normal to the direction of flow) of the body. For plates and airfoils, A is taken to be the planform area, which is the area that would be seen by a person looking at the body from directly above. The drag coefficient, in general, depends on the Reynolds number, especially for Reynolds numbers below 104. At higher Reynolds numbers, the drag coefficients for many geometries remain essentially constant. A body is said to be streamlined if a conscious effort is made to align its shape with the anticipated streamlines in the flow in order to reduce drag. Otherwise, a body (such as a building) tends to block the flow and is said to be bluff. At sufficiently high velocities, the fluid stream detaches itself from the surface of the body. This is called flow separation. When a fluid stream separates from the body, it forms a separated region between the body and the fluid stream. Separation may also occur on a streamlined body such as an airplane wing at a sufficiently large angle of attack, which is the angle the incoming fluid stream makes with the chord (the line that connects the nose and the end) of the body. Flow separation on the top surface of a wing reduces lift drastically and may cause the airplane to stall. The region of flow above a surface in which the effects of the viscous shearing forces caused by fluid viscosity are felt is called the velocity boundary layer or just the boundary layer. The thickness of the boundary layer, 𝛿, is defined as the distance from the surface at which the velocity is 0.99V. The hypothetical line of velocity 0.99V divides the flow over a plate into two regions: the boundary layer region, in which the viscous effects and the velocity changes are significant, and the irrotational outer flow region, in which the frictional effects are negligible and the velocity remains essentially constant. For external flow, the Reynolds number is expressed as ReL = ρVL 𝜇 = VL 𝜈 where V is the upstream velocity and L is the characteristic length of the geometry, which is the length of the plate in the flow direction for a flat plate and the diameter D for a cylinder or sphere. The average friction coefficients over an entire flat plate are Laminar flow: Cf = 1.33 Re1/2 L ReL ≲ 5 × 105 Turbulent flow: Cf = 0.074 Re1/5 L 5 × 105 ≲ ReL ≲ 107 If the flow is approximated as laminar up to the engineer ing critical number of Recr = 5 × 105, and then turbulent beyond, the average friction coefficient over the entire flat plate becomes Cf = 0.074 Re1/5 L −1742 ReL 5 × 105 ≲ ReL ≲ 107 A curve fit of experimental data for the average friction coefficient in the fully rough turbulent regime is Rough surface: Cf = (1.89 −1.62 log 𝜀 L) −2.5 where 𝜀 is the surface roughness and L is the length of the plate in the flow direction. In the absence of a better one, this relation can be used for turbulent flow on rough surfaces for Re > 106, especially when 𝜀/L > 10−4. Surface roughness, in general, increases the drag coefficient in turbulent flow. For bluff bodies such as a circular cylin der or sphere, however, an increase in the surface roughness may decrease the drag coefficient. This is done by tripping the flow into turbulence at a lower Reynolds number, and thus causing the fluid to close in behind the body, narrowing the wake and reducing pressure drag considerably. It is desirable for airfoils to generate the most lift while producing the least drag. Therefore, a measure of perfor mance for airfoils is the lift-to-drag ratio, CL/CD. The minimum safe flight velocity of an aircraft is deter mined from Vmin = √ 2W ρCL, max A For a given weight, the landing or takeoff speed can be min imized by maximizing the product of the lift coefficient and the wing area, CL, maxA. For airplane wings and other airfoils of finite span, the pressure difference between the lower and the upper sur faces drives the fluid at the tips upward. This results in swirling eddies, called tip vortices. Tip vortices that inter act with the free stream impose forces on the wing tips in all directions, including the flow direction. The component of the force in the flow direction adds to drag and is called induced drag. The total drag of a wing is then the sum of the induced drag (3-D effects) and the drag of the airfoil section (2-D effects). It is observed that lift develops when a cylinder or sphere in flow is rotated at a sufficiently high rate. The phenome non of producing lift by the rotation of a solid body is called the Magnus effect. Some external flows, complete with flow details including plots of velocity fields, are solved using computational fluid dynamics, and presented in Chap. 15.
cen96537_ch11_611-666.indd 651 14/01/17 3:17 pm 652 EXTERNAL FLOW: DRAG AND LIFT 325 μm FIGURE 11–64 Microelectrokinetic actuator array (MEKA-5) with 25,600 individual actuators at 325-μm spacing for full-scale hydronautical drag reduction. Close-up of a single unit cell (top) and partial view of the full array (bottom).
Title 250 × 250 actuators Basic Unit Cell 6 × 6 actuators w/DSP Sensor/Actuator Element 1 sensor + 1 actuator FIGURE 11–63 Drag-reducing microactuator arrays on the hull of a submarine. Shown is the system architecture with tiles composed of unit cells containing sensors and actuators.
Guest Author: Werner J. A. Dahm, The University of Michigan A reduction of just a few percent in the drag that acts on an air vehicle, a naval surface vehicle, or an undersea vehicle can translate into large reduc tions in fuel weight and operating costs, or increases in vehicle range and payload. One approach to achieve such drag reduction is to actively control naturally occurring streamwise vortices in the viscous sublayer of the tur bulent boundary layer at the vehicle surface. The thin viscous sublayer at the base of any turbulent boundary layer is a powerful nonlinear system, capable of amplifying small microactuator-induced perturbations into large reductions in the vehicle drag. Numerous experimental, computational, and theoretical studies have shown that reductions of 15 to 25 percent in the wall shear stress are possible by properly controlling these sublayer structures. The challenge has been to develop large, dense arrays of microactuators that can manipulate these structures to achieve drag reduction on practical aero nautical and hydronautical vehicles (Fig. 11–63). The sublayer structures are typically a few hundred microns, and thus well matched to the scale of microelectromechanical systems (MEMS). Figure 11–64 shows an example of one type of such microscale actua tor array based on the electrokinetic principle that is potentially suitable for active sublayer control on real vehicles. Electrokinetic flow provides a way to move small amounts of fluid on very fast time scales in very small devices. The actuators impulsively displace a fixed volume of fluid between the wall and the viscous sublayer in a manner that counteracts the effect of the sublayer vortices. A system architecture based on independent unit cells, appropriate for large arrays of such microactuators, provides greatly reduced control processing requirements within individual unit cells, which consist of a relatively small number of individual sensors and actuators. Fundamental consideration of the scaling principles governing electroki netic flow, as well as the sublayer structure and dynamics and microfab rication technologies, have been used to develop and produce full-scale electrokinetic microactuator arrays that can meet many of the requirements for active sublayer control of turbulent boundary layers under real-vehicle conditions. Such microelectrokinetic actuator (MEKA) arrays, when fabricated with wall shear stress sensors also based on microelectromechanical systems fabrication, may in the future allow engineers to achieve dramatic reductions in the drag acting on practical aeronautical and hydronautical vehicles.
References Diez-Garias, F. J., Dahm, W. J. A., and Paul, P. H., “Microactuator Arrays for Sublayer Control in Turbulent Boundary Layers Using the Electrokinetic Principle,” AIAA Paper No. 2000-0548, AIAA, Washington, DC, 2000.
Diez, F. J., and Dahm, W. J. A., “Electrokinetic Microactuator Arrays and System Architecture for Active Sublayer Control of Turbulent Boundary Layers,” AIAA Journal, Vol. 41, pp. 1906–1915, 2003.
APPLICATION SPOTLIGHT ■ Drag Reduction cen96537_ch11_611-666.indd 652 14/01/17 3:17 pm 653 CHAPTER 11 REFERENCES AND SUGGESTED READING 1. I. H. Abbott. “The Drag of Two Streamline Bodies as Affected by Protuberances and Appendages,” NACA Report 451, 1932.
2. I. H. Abbott and A. E. von Doenhoff. Theory of Wing Sections, Including a Summary of Airfoil Data. New York: Dover, 1959.
3. I. H. Abbott, A. E. von Doenhoff, and L. S. Stivers. “Summary of Airfoil Data,” NACA Report 824, Langley Field, VA, 1945.
4. J. D. Anderson. Fundamentals of Aerodynamics, 5th ed. New York: McGraw-Hill, 2010.
5. R. D. Blevins. Applied Fluid Dynamics Handbook. New York: Van Nostrand Reinhold, 1984.
6. R. J. Bomphrey, Lawson, N. J., Taylor, G. K., Thomas, A. L. R., Application of digital particle image velocim etry to insect aerodynamics: measurement of the leading-edge vortex and near wake of a Hawkmoth, Experiments in Fluids, 40, 546–554, 2006.
7. S. W. Churchill and M. Bernstein. “A Correlating Equa tion for Forced Convection from Gases and Liquids to a Circular Cylinder in Cross Flow,” Journal of Heat Transfer 99, pp. 300–306, 1977.
8. C. P. Ellington, van den Berg, C., Willmott, A. P., and Thomas, A. L. R., Leading-edge vortices in insect flight, Nature 384, 626–630, 1996.
9. S. Goldstein. Modern Developments in Fluid Dynamics. London: Oxford Press, 1938.
10. J. Happel and H. Brenner. Low Reynolds Number Hydrodynamics with Special Applications to Particulate Media. Norwell, MA: Kluwer Academic Publishers, 2003.
11. S. F. Hoerner. Fluid-Dynamic Drag. [Published by the author.] Library of Congress No. 64, 1966.
12. J. D. Holmes. Wind Loading of Structures 2nd ed. London: Spon Press (Taylor and Francis), 2007.
13. G. M. Homsy, H. Aref, K. S. Breuer, S. Hochgreb, J. R. Koseff, B. R. Munson, K. G. Powell, C. R. Roberston, and S. T. Thoroddsen. Multi-Media Fluid Mechanics (CD) 2nd ed. Cambridge University Press, 2004.
14. W. H. Hucho. Aerodynamics of Road Vehicles 4th ed. London: Butterworth-Heinemann, 1998.
15. D. Lentink, Flying like a fly, Nature 498, 306–307, 2013.
16. S. P. Sane, The aerodynamics of insect flight, The Journal of Experimental Biology 206, 4191–4208 4191, 2003.
17. H. Schlichting. Boundary Layer Theory, 7th ed. New York: McGraw-Hill, 1979.
18. W. Shyy, Lian, Y., Tang, T., Viieru, D., and Liu, H., Aerodynamics of Low Reynolds Number Flyers, Cam bridge University Press, 2008.
19. Unsteady Aerodynamics for Micro Air Vehicles, RTO-TR-AVT-149, 2010.
20. M. Van Dyke. An Album of Fluid Motion. Stanford, CA: The Parabolic Press, 1982.
21. J. Vogel. Life in Moving Fluids, 2nd ed. Boston: Willard Grand Press, 1994.
22. F. M. White. Fluid Mechanics, 7th ed. New York: McGraw-Hill, 2010.
PROBLEMS Drag, Lift, and Drag Coefficients 11–1C What is drag? What causes it? Why do we usually try to minimize it?
11–2C What is lift? What causes it? Does wall shear con tribute to the lift?
11–3C Which bicyclist is more likely to go faster: one who keeps his head and his body in the most upright position or one who leans down and brings his body closer to his knees? Why?
11–4C Which car is more likely to be more fuel-efficient: one with sharp corners or one that is contoured to resemble an ellipse? Why?
11–5C Define the frontal area of a body subjected to exter nal flow. When is it appropriate to use the frontal area in drag and lift calculations?
11–6C Define the planform area of a body subjected to external flow. When is it appropriate to use the planform area in drag and lift calculations?
Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and SI users can ignore them. Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software.
cen96537_ch11_611-666.indd 653 14/01/17 3:17 pm 654 EXTERNAL FLOW: DRAG AND LIFT 11–7C What is the difference between the upstream veloc ity and the free-stream velocity? For what types of flow are these two velocities equal to each other?
11–8C What is the difference between streamlined and bluff bodies? Is a tennis ball a streamlined or bluff body?
11–9C Name some applications in which a large drag is desired.
11–10C During flow over a given body, the drag force, the upstream velocity, and the fluid density are measured. Explain how you would determine the drag coefficient. What area would you use in the calculations?
11–11C During flow over a given slender body such as a wing, the lift force, the upstream velocity, and the fluid density are measured. Explain how you would deter mine the lift coefficient. What area would you use in the calculations?
11–12C What is terminal velocity? How is it determined?
11–13C What is the difference between skin friction drag and pressure drag? Which is usually more significant for slender bodies such as airfoils?
11–14C What is the effect of surface roughness on the friction drag coefficient in laminar and turbulent flows?
11–15C What is the effect of streamlining on (a) friction drag and (b) pressure drag? Does the total drag acting on a body necessarily decrease as a result of streamlining? Explain.
11–16C What is flow separation? What causes it? What is the effect of flow separation on the drag coefficient?
11–17C What is drafting? How does it affect the drag coef ficient of the drafted body?
11–18C Consider laminar flow over a flat plate. How does the local friction coefficient change with position?
11–19C In general, how does the drag coefficient vary with the Reynolds number at (a) low and moderate Reynolds num bers and (b) at high Reynolds numbers (Re > 104)?
11–20C Fairings are attached to the front and back of a cylindrical body to make it look more streamlined. What is the effect of this modification on the (a) friction drag, (b) pres sure drag, and (c) total drag? Assume the Reynolds number is high enough so that the flow is turbulent for both cases.
FIGURE P11–20C Fairings Cylinder V 11–21 A car is moving at a constant velocity of 110 km/h. Determine the upstream velocity to be used in fluid flow analysis if (a) the air is calm, (b) wind is blowing against the direction of motion of the car at 30 km/h, and (c) wind is blowing in the same direction of motion of the car at 30 km/h.
11–22 The resultant of the pressure and wall shear forces acting on a body is measured to be 430 N, making 30° with the direction of flow. Determine the drag and the lift forces acting on the body.
30° FR = 430 N V FIGURE P11–22 11–23 During a high Reynolds number experiment, the total drag force acting on a spherical body of diameter D = 12 cm subjected to airflow at 1 atm and 5°C is measured to be 5.2 N. The pressure drag acting on the body is calculated by integrat ing the pressure distribution (measured by the use of pressure sensors throughout the surface) to be 4.9 N. Determine the friction drag coefficient of the sphere. Answer: 0.0115 11–24 The drag coefficient of a car at the design conditions of 1 atm, 25°C, and 90 km/h is to be determined experimen tally in a large wind tunnel in a full-scale test. The height and width of the car are 1.25 m and 1.65 m, respectively. If the horizontal force acting on the car is measured to be 220 N, determine the total drag coefficient of this car. Answer: 0.29 11–25E To reduce the drag coefficient and thus to improve the fuel efficiency, the frontal area of a car is to be reduced. Determine the amount of fuel and money saved per year as a result of reducing the frontal area from 20 to 13 ft2. Assume the car is driven 12,000 mi a year at an average speed of 55 mi/h. Take the density and price of gasoline to be 50 lbm/ft3 and $3.10/gal, respectively; the density of air to be 0.075 lbm/ft3, the heating value of gasoline to be 20,000 Btu/lbm; and the overall efficiency of the drive train to be 30 percent.
11–26E Reconsider Prob. 11–25E. Using appropriate software, investigate the effect of frontal area on the annual fuel consumption of the car. Let the frontal area vary from 10 to 30 ft2 in increments of 2 ft2. Tabulate and plot the results.
11–27 A circular sign has a diameter of 50 cm and is sub jected to normal winds up to 150 km/h at 10°C and 100 kPa. Determine the drag force acting on the sign. Also determine the bending moment at the bottom of its pole whose height from the ground to the bottom of the sign is 1.5 m. Disregard the drag on the pole.
cen96537_ch11_611-666.indd 654 14/01/17 3:17 pm 655 CHAPTER 11 1.5 m 150 km/h SIGN FIGURE P11–27 11–28 Suzy likes to drive with a silly sun ball on her car antenna. The frontal area of the ball is A = 2.08 × 10−3 m2.
As gas prices rise, her husband is concerned that she is wasting fuel because of the additional drag on the ball. He runs a quick test in the wind tunnel at his university and measures the drag coefficient to be CD = 0.87 at nearly all air speeds. Estimate how many liters of fuel she wastes per year by having this ball on her antenna. Use the following additional information: She drives about 15,000 km per year at an average speed of 20.8 m/s. The overall car effi ciency is 0.312, 𝜌fuel = 0.802 kg/L, and the heating value of the fuel is 44,020 kJ/kg. Use standard air properties. Is the amount of wasted fuel significant?
Photo by Suzanne Cimbala.
FIGURE P11–28 11–29 Advertisement signs are commonly carried by taxi cabs for additional income, but they also increase the fuel cost. Consider a sign that consists of a 0.30-m-high, 0.9-m-wide, and 0.9-m-long rectangular block mounted on top of a taxi cab such that the sign has a frontal area of 0.3 m by 0.9 m from all four sides. Determine the increase in the annual fuel cost of this taxicab due to this sign. Assume the taxicab is driven 60,000 km a year at an average speed of 50 km/h and the overall efficiency of the drive train is 28 percent. Take the density, unit price, and heating value of gasoline to be 0.72 kg/L, $1.10/L, and 42,000 kJ/kg, respectively, and the density of air to be 1.25 kg/m3.
TAXI Pa’s Pizza FIGURE P11–29 11–30E At highway speeds, about half of the power generated by the car’s engine is used to overcome aerodynamic drag, and thus the fuel consumption is nearly proportional to the drag force on a level road. Determine the percentage increase in fuel consumption of a car per unit time when a person who normally drives at 55 mi/h now starts driving at 70 mi/h.
11–31 A submarine can be treated as an ellipsoid with a diameter of 5 m and a length of 25 m. Determine the power required for this submarine to cruise horizontally and steadily at 40 km/h in seawater whose density is 1025 kg/m3. Also determine the power required to tow this submarine in air whose density is 1.30 kg/m3. Assume the flow is turbulent in both cases.
40 km/h Submarine FIGURE P11–31 11–32 A 70-kg bicyclist is riding her 15-kg bicycle down hill on a road with a slope of 8° without pedaling or braking. The bicyclist has a frontal area of 0.45 m2 and a drag coef ficient of 1.1 in the upright position, and a frontal area of 0.4 m2 and a drag coefficient of 0.9 in the racing position. Disregarding the rolling resistance and friction at the bear ings, determine the terminal velocity of the bicyclist for both positions. Take the air density to be 1.25 kg/m3. Answers: 70 km/h, 82 km/h 11–33 A wind turbine with two or four hollow hemispheri cal cups connected to a pivot is commonly used to measure wind speed. Consider a wind turbine with four 8-cm-diameter cups with a center-to-center distance of 40 cm, as shown in Fig. P11–33. The pivot is stuck as a result of some malfunc tion, and the cups stop rotating. For a wind speed of 15 m/s and air density of 1.25 kg/m3, determine the maximum torque this turbine applies on the pivot.
cen96537_ch11_611-666.indd 655 14/01/17 3:17 pm 656 EXTERNAL FLOW: DRAG AND LIFT 40 cm FIGURE P11–33 11–34 Reconsider Prob. 11–33. Using appropriate software, investigate the effect of wind speed on the torque applied on the pivot. Let the wind speed vary from 0 to 50 m/s in increments of 5 m/s. Tabulate and plot the results.
11–35 During steady motion of a vehicle on a level road, the power delivered to the wheels is used to overcome aero dynamic drag and rolling resistance (the product of the rolling resistance coefficient and the weight of the vehicle), assum ing the friction at the bearings of the wheels is negligible. Consider a car that has a total mass of 950 kg, a drag coef ficient of 0.32, a frontal area of 1.8 m2, and a rolling resis tance coefficient of 0.04. The maximum power the engine can deliver to the wheels is 80 kW. Determine (a) the speed at which the rolling resistance is equal to the aerodynamic drag force and (b) the maximum speed of this car. Take the air density to be 1.20 kg/m3.
11–36 Reconsider Prob. 11–35. Using appropriate software, investigate the effect of car speed on the required power to overcome (a) rolling resistance, (b) the aerodynamic drag, and (c) their combined effect. Let the car speed vary from 0 to 150 km/h in increments of 15 km/h. Tabulate and plot the results.
11–37E Bill gets a job delivering pizzas. The pizza com pany makes him mount a sign on the roof of his car. The frontal area of the sign is A = 0.612 ft2, and he estimates the drag coefficient to be CD = 0.94 at nearly all air speeds. Estimate how much additional money it costs Bill per year in fuel to drive with the sign on his roof compared to without the sign. Use the following additional information: He drives about 10,000 miles per year at an average speed of 45 mph. The overall car efficiency is 0.332, 𝜌fuel = 50.2 lbm/ft3, and the heating value of the fuel is 1.53 × 107 ft . lbf/lbm. The fuel costs $3.50 per gallon. Use standard air properties. Be careful with unit conversions.
11–38 A 0.80-m-diameter, 1.2-m-high garbage can is found in the morning tipped over due to high winds during the night. Assuming the average density of the garbage inside to be 150 kg/m3 and taking the air density to be 1.25 kg/m3, estimate the wind velocity during the night when the can was tipped over. Take the drag coefficient of the can to be 0.7. Answer: 135 km/h 11–39 An 8-mm-diameter plastic sphere whose density is 1150 kg/m3 is dropped into water at 20°C. Determine the ter minal velocity of the sphere in water.
11–40 A 7-m-diameter hot air balloon that has a total mass of 350 kg is standing still in air on a windless day. The balloon is suddenly subjected to 40 km/h winds. Deter mine the initial acceleration of the balloon in the horizontal direction.
11–41E The drag coefficient of a vehicle increases when its windows are rolled down or its sunroof is opened. A sports car has a frontal area of 18 ft2 and a drag coefficient of 0.32 when the windows and sunroof are closed. The drag coeffi cient increases to 0.41 when the sunroof is open. Determine the additional power consumption of the car when the sunroof is opened at (a) 35 mi/h and (b) 70 mi/h. Take the density of air to be 0.075 lbm/ft3.
Sunroof closed Sunroof open CD = 0.32 CD = 0.41 FIGURE P11–41E 11–42 To reduce the drag coefficient and thus to improve the fuel efficiency of cars, the design of side rearview mirrors has changed drastically in recent decades from a simple circular plate to a streamlined shape. Determine the amount of fuel and money saved per year as a result of replacing a 13-cm-diameter flat mirror by one with a hemi spherical back, as shown in the figure. Assume the car is cen96537_ch11_611-666.indd 656 14/01/17 3:17 pm 657 CHAPTER 11 driven 21,000 km a year at an average speed of 80 km/h. Take the density and price of gasoline to be 0.75 kg/L and $0.90/L, respectively; the heating value of gasoline to be 44,000 kJ/kg; and the overall efficiency of the drive train to be 30 percent.
D = 13 cm D = 13 cm Flat mirror 80 km/h Rounded mirror 80 km/h FIGURE P11–42 11–43 During major windstorms, high vehicles such as RVs and semis may be thrown off the road and boxcars off their tracks, especially when they are empty and in open areas. Consider a 6000-kg semi that is 10 m long, 2.5 m high, and 2 m wide. The distance between the bottom of the truck and the road is 0.8 m. Now the truck is exposed to winds from its side surface. Determine the wind velocity that will tip the truck over to its side. Take the air density to be 1.1 kg/m3 and assume the weight to be uniformly distributed.
10 m 0.8 m 2 m 2.5 m FIGURE P11–43 Flow over Flat Plates 11–44C What does the friction coefficient represent in flow over a flat plate? How is it related to the drag force acting on the plate?
11–45C What fluid property is responsible for the develop ment of the velocity boundary layer? What is the effect of the velocity on the thickness of the boundary layer?
11–46C How is the average friction coefficient determined in flow over a flat plate?
11–47E Light oil at 75°F flows over a 17-ft-long flat plate with a free-stream velocity of 5 ft/s. Determine the total drag force per unit width of the plate.
11–48 The local atmospheric pressure in Denver, Colorado (elevation 1610 m) is 83.4 kPa. Air at this pressure and at 25°C flows with a velocity of 9 m/s over a 2.5-m × 5-m flat plate. Determine the drag force acting on the top surface of the plate if the air flows parallel to the (a) 5-m-long side and (b) the 2.5-m-long side.
11–49E Air at 70°F flows over a 10-ft-long flat plate at 25 ft/s. Determine the local friction coeffi cient at intervals of 1 ft and plot the results against the dis tance from the leading edge.
11–50 Consider laminar flow of a fluid over a flat plate. Now the free-stream velocity of the fluid is tripled. Deter mine the change in the drag force on the plate. Assume the flow to remain laminar. Answer: A 5.20-fold increase 11–51E Consider a refrigeration truck traveling at 70 mi/h at a location where the air is at 1 atm and 80°F. The refrig erated compartment of the truck can be considered to be a 9-ft-wide, 8-ft-high, and 20-ft-long rectangular box. Assum ing the airflow over the entire outer surface to be turbulent and attached (no flow separation), determine the drag force acting on the top and side surfaces and the power required to overcome this drag.
20 ft Refrigeration truck 8 ft Air, 80°F V = 70 mi/h FIGURE P11–51E 11–52E Reconsider Prob. 11–51E. Using appropriate software, investigate the effect of truck speed on the total drag force acting on the top and side surfaces, and the power required to overcome it. Let the truck speed vary from 0 to 100 mi/h in increments of 10 mi/h. Tabulate and plot the results. 11–53 Air at 25°C and 1 atm is flowing over a long flat plate with a velocity of 8 m/s. Determine the distance from the leading edge of the plate where the flow becomes turbulent, and the thickness of the boundary layer at that location.
11–54 Repeat Prob. 11–53 for water.
cen96537_ch11_611-666.indd 657 14/01/17 3:17 pm 658 EXTERNAL FLOW: DRAG AND LIFT 11–55 During a winter day, wind at 70 km/h, 5°C, and 1 atm is blowing parallel to a 4-m-high and 15-m-long wall of a house. Approximating the wall surfaces as smooth, deter mine the friction drag acting on the wall. What would your answer be if the wind velocity has doubled? How realistic is it to treat the flow over side wall surfaces as flow over a flat plate? Answers: 35 N, 125 N Air 5°C 70 km/h 15 m 4 m FIGURE P11–55 11–56 The weight of a thin flat plate 50 cm × 50 cm in size is balanced by a counterweight that has a mass of 2 kg, as shown in Fig. P11–56. Now a fan is turned on, and air at 1 atm and 25°C flows downward over both surfaces of the plate (front and back in the sketch) with a free-stream velocity of 10 m/s. Determine the mass of the counter weight that needs to be added in order to balance the plate in this case.
50 cm 50 cm Plate Air 25°C, 10 m/s FIGURE P11–56 11–57 The forming section of a plastics plant puts out a continuous sheet of plastic that is 1.2 m wide and 2 mm thick at a rate of 18 m/min. The sheet is subjected to airflow at a velocity of 4 m/s on both top and bottom surfaces normal to the direction of motion of the sheet. The width of the air cooling section is such that a fixed point on the plastic sheet passes through that section in 2 s. Using properties of air at 1 atm and 60°C, determine the drag force the air exerts on the plastic sheet in the direction of airflow.
Air 4 m/s Plastic sheet 18 m/min FIGURE P11–57 Flow across Cylinders and Spheres 11–58C In flow over bluff bodies such as a cylinder, how does the pressure drag differ from the friction drag?
11–59C Why is flow separation in flow over cylinders delayed when the boundary layer is turbulent?
11–60C In flow over cylinders, why does the drag coef ficient suddenly drop when the boundary layer becomes turbulent? Isn’t turbulence supposed to increase the drag coefficient instead of decreasing it?
11–61 A 5-mm-diameter electrical transmission line is exposed to windy air. Determine the drag force exerted on a 160-m-long section of the wire during a windy day when the air is at 1 atm and 15°C and the wind is blowing across the transmission line at 50 km/h.
11–62 A long 5-cm-diameter steam pipe passes through some area open to the wind. Determine the drag force acting on the pipe per unit of its length when the air is at 1 atm and 10°C and the wind is blowing across the pipe at a speed of 50 km/h.
11–63 Consider 0.8-cm-diameter hail that is falling freely in atmospheric air at 1 atm and 5°C. Determine the terminal velocity of the hail. Take the density of hail to be 910 kg/m3.
11–64E A 1.2-in-outer-diameter pipe is to span across a river at a 115-ft-wide section while being completely immersed in water. The average flow velocity of the water is 8 ft/s, and its temperature is 70°F. Determine the drag force exerted on the pipe by the river. Answer: 748 lbf 11–65 A 2-m-long, 0.2-m-diameter cylindrical pine log (den sity = 513 kg/m3) is suspended by a crane in the horizontal position. The log is subjected to normal winds of 40 km/h at 5°C and 88 kPa. Disregarding the weight of the cable and its cen96537_ch11_611-666.indd 658 14/01/17 3:17 pm 659 CHAPTER 11 drag, determine the angle 𝜃 the cable will make with the hori zontal and the tension on the cable.
40 km/h 2 m 0.2 m θ FIGURE P11–65 11–66 A 0.12-mm-diameter dust particle whose density is 2.1 g/cm3 is observed to be suspended in the air at 1 atm and 20°C at a fixed point. Estimate the updraft velocity of air motion at that location. Assume Stokes law to be applicable. Is this a valid assumption? Answer: 0.90 m/s 11–67E A person extends his uncovered arms into the windy air outside at 1 atm and 60°F and 25 mi/h in order to feel nature closely. Treating the arm as a 2-ft-long and 4-in-diameter cylinder, determine the combined drag force on both arms. Answer: 2.12 lbf Air 60°F, 25 mi/h FIGURE P11–67E 11–68 One of the popular demonstrations in science museums involves the suspension of a ping-pong ball by an upward air jet. Children are amused by the ball always com ing back to the center when it is pushed by a finger to the side of the jet. Explain this phenomenon using the Bernoulli equation. Also determine the velocity of air if the ball has a mass of 3.1 g and a diameter of 4.2 cm. Assume the air is at 1 atm and 25°C.
Ball Air jet FIGURE P11–68 11–69 Dust particles of diameter 0.06 mm and density 1.6 g/cm3 are unsettled during high winds and rise to a height of 200 m by the time things calm down. Estimate how long it takes for the dust particles to fall back to the ground in still air at 1 atm and 30°C, and their velocity. Disregard the initial transient period during which the dust particles accel erate to their terminal velocity, and assume Stokes law to be applicable.
Lift 11–70C What is stall? What causes an airfoil to stall? Why are commercial aircraft not allowed to fly at conditions near stall?
11–71C Air is flowing past a nonsymmetrical airfoil at zero angle of attack. Is the (a) lift and (b) drag acting on the airfoil zero or nonzero?
11–72C Air is flowing past a symmetrical airfoil at zero angle of attack. Is the (a) lift and (b) drag acting on the air foil zero or nonzero?
11–73C Both the lift and the drag of an airfoil increase with an increase in the angle of attack. In general, which increases at a higher rate, the lift or the drag?
11–74C Why are flaps used at the leading and trailing edges of the wings of large aircraft during takeoff and land ing? Can an aircraft take off or land without them?
11–75C Why is the contribution of viscous effects to lift usually negligible for airfoils?
11–76C Air is flowing past a symmetrical airfoil at an angle of attack of 5°. Is the (a) lift and (b) drag acting on the airfoil zero or nonzero?
cen96537_ch11_611-666.indd 659 14/01/17 3:17 pm 660 EXTERNAL FLOW: DRAG AND LIFT 11–77C Air is flowing past a spherical ball. Is the lift exerted on the ball zero or nonzero? Answer the same ques tion if the ball is spinning.
11–78C What is the effect of wing tip vortices (the air cir culation from the lower part of the wings to the upper part) on the drag and the lift?
11–79C What is induced drag on wings? Can induced drag be minimized by using long and narrow wings or short and wide wings?
11–80C Explain why endplates or winglets are added to some airplane wings.
11–81C How do flaps affect the lift and the drag of wings?
11–82E A 2.4-in-diameter smooth ball rotating at 500 rpm is dropped in a water stream at 60°F flowing at 4 ft/s. Deter mine the lift and the drag force acting on the ball when it is first dropped in the water.
11–83 Consider an aircraft that takes off at 260 km/h when it is fully loaded. If the weight of the aircraft is increased by 10 percent as a result of overloading, determine the speed at which the overloaded aircraft will take off. Answer: 273 km/h 11–84 Consider an airplane whose takeoff speed is 220 km/h and that takes 15 s to take off at sea level. For an airport at an elevation of 1600 m (such as Denver), determine (a) the takeoff speed, (b) the takeoff time, and (c) the additional run way length required for this airplane. Assume constant accel eration for both cases.
220 km/h FIGURE P11–84 11–85 A jumbo jet airplane has a mass of about 400,000 kg when fully loaded with over 400 passengers and takes off at a speed of 250 km/h. Determine the takeoff speed when the airplane has 150 empty seats. Assume each passenger with luggage is 140 kg and the wing and flap settings are main tained the same. Answer: 243 km/h 11–86 Reconsider Prob. 11–85. Using appropriate software, investigate the effect of passenger count on the takeoff speed of the aircraft. Let the number of passengers vary from 0 to 500 in increments of 50. Tabulate and plot the results. 11–87 A tennis ball with a mass of 57 g and a diameter of 6.4 cm is hit with an initial velocity of 105 km/h and a backspin of 4200 rpm. Determine if the ball falls or rises under the combined effect of gravity and lift due to spinning shortly after hitting. Assume air is at 1 atm and 25°C.
4200 rpm 105 km/h FIGURE P11–87 11–88 A small aircraft has a wing area of 40 m2, a lift coef ficient of 0.45 at takeoff settings, and a total mass of 4000 kg. Determine (a) the takeoff speed of this aircraft at sea level at standard atmospheric conditions, (b) the wing loading, and (c) the required power to maintain a constant cruising speed of 360 km/h for a cruising drag coefficient of 0.035.
11–89 The NACA 64(l)–412 airfoil has a lift-to-drag ratio of 50 at 0° angle of attack, as shown in Fig. 11–43. At what angle of attack does this ratio increase to 80?
11–90 Consider a light plane that has a total weight of 11,000 N and a wing area of 39 m2 and whose wings resem ble the NACA 23012 airfoil with no flaps. Using data from Fig. 11–45, determine the takeoff speed at an angle of attack of 5° at sea level. Also determine the stall speed. Answers: 99.7 km/h, 62.7 km/h 11–91 A small airplane has a total mass of 1800 kg and a wing area of 42 m2. Determine the lift and drag coeffi cients of this airplane while cruising at an altitude of 4000 m at a constant speed of 280 km/h and generating 190 kW of power.
11–92 An airplane has a mass of 48,000 kg, a wing area of 300 m2, a maximum lift coefficient of 3.2, and a cruising drag coefficient of 0.03 at an altitude of 12,000 m. Determine cen96537_ch11_611-666.indd 660 14/01/17 3:17 pm 661 CHAPTER 11 (a) the takeoff speed at sea level, assuming it is 20 percent over the stall speed, and (b) the thrust that the engines must deliver for a cruising speed of 900 km/h.
11–93E An airplane is consuming fuel at a rate of 7 gal/ min when cruising at a constant altitude of 10,000 ft at con stant speed. Assuming the drag coefficient and the engine efficiency to remain the same, determine the rate of fuel con sumption at an altitude of 30,000 ft at the same speed.
Review Problems 11–94 A 1.2-m-external-diameter spherical tank is located outdoors at 1 atm and 25°C and is subjected to winds at 48 km/h. Determine the drag force exerted on it by the wind. Answer: 16.7 N 11–95E A 7-ft-diameter spherical tank completely sub merged in freshwater is being towed by a ship at 8 ft/s. Assuming turbulent flow, determine the required towing power.
11–96 A 2-m-high, 4-m-wide rectangular advertisement panel is attached to a 4-m-wide, 0.15-m-high rectangular con crete block (density = 2300 kg/m3) by two 5-cm-diameter, 4-m-high (exposed part) poles, as shown in Fig. P11–96. If the sign is to withstand 150 km/h winds from any direction, determine (a) the maximum drag force on the panel, (b) the drag force acting on the poles, and (c) the minimum length L of the concrete block for the panel to resist the winds. Take the density of air to be 1.30 kg/m3.
0.15 m Concrete 4 m 4 m 4 m 2 m L FIGURE P11–96 11–97 A plastic boat whose bottom surface can be approxi mated as a 1.5-m-wide, 2-m-long flat surface is to move through water at 15°C at speeds up to 45 km/h. Determine the friction drag exerted on the boat by the water and the power needed to overcome it.
45 km/h FIGURE P11–97 11–98 Reconsider Prob. 11–97. Using appropriate software, investigate the effect of boat speed on the drag force acting on the bottom surface of the boat, and the power needed to overcome it. Let the boat speed vary from 0 to 100 km/h in increments of 10 km/h. Tabulate and plot the results.
11–99 Stokes law can be used to determine the viscosity of a fluid by dropping a spherical object in it and measuring the terminal velocity of the object in that fluid. This can be done by plotting the distance traveled against time and observing when the curve becomes linear. During such an experiment a 3-mm-diameter glass ball (𝜌 = 2500 kg/m3) is dropped into a fluid whose density is 875 kg/m3, and the terminal velocity is measured to be 0.15 m/s. Disregarding the wall effects, determine the viscosity of the fluid.
0.15 m/s Glass ball FIGURE P11–99 11–100E The passenger compartment of a minivan travel ing at 50 mi/h in ambient air at 1 atm and 80°F is modeled as a 4.5-ft-high, 6-ft-wide, and 11-ft-long rectangular box. The airflow over the exterior surfaces is assumed to be turbulent because of the intense vibrations involved. Determine the drag force acting on the top and the two side surfaces of the van and the power required to overcome it.
cen96537_ch11_611-666.indd 661 14/01/17 3:17 pm 662 EXTERNAL FLOW: DRAG AND LIFT Air 50 mi/h 80°F FIGURE P11–100E 11–101E A commercial airplane has a total mass of 150,000 lbm and a wing planform area of 1700 ft2. The plane has a cruising speed of 625 mi/h and a cruising alti tude of 38,000 ft where the air density is 0.0208 lbm/ft3. The plane has double-slotted flaps for use during takeoff and landing, but it cruises with all flaps retracted. Assum ing the lift and drag characteristics of the wings can be approximated by NACA 23012, determine (a) the minimum safe speed for takeoff and landing with and without extend ing the flaps, (b) the angle of attack to cruise steadily at the cruising altitude, and (c) the power that needs to be sup plied to provide enough thrust to overcome drag. Take the air density on the ground to be 0.075 lbm/ft3.
11–102 An automotive engine can be approximated as a 0.4-m-high, 0.60-m-wide, and 0.7-m-long rectangular block. The ambient air is at 1 atm and 15°C. Determine the drag force acting on the bottom surface of the engine block as the car travels at a velocity of 120 km/h. Assume the flow to be turbulent over the entire surface because of the constant agitation of the engine block. Answer: 1.22 N Air 120 km/h 15°C Engine block FIGURE P11–102 11–103 A paratrooper and his 7-m-diameter parachute weigh 1200 N. Taking the average air density to be 1.2 kg/m3, deter mine the terminal velocity of the paratrooper. Answer: 6.3 m/s 7 m 1200 N FIGURE P11–103 11–104 It is proposed to meet the water needs of a recrea tional vehicle (RV) by installing a 3-m-long, 0.5-m-diameter cylindrical tank on top of the vehicle. Determine the addi tional power requirement of the RV at a speed of 80 km/h when the tank is installed such that its circular surfaces face (a) the front and back (as sketched) and (b) the sides of the RV. Assume atmospheric conditions are 87 kPa and 20°C. Answers: (a) 1.05 kW, (b) 6.77 kW 3 m 0.5 m FIGURE P11–104 11–105 A 9-cm-diameter smooth sports ball has a veloc ity of 36 km/h during a typical hit. Determine the percent increase in the drag coefficient if the ball is given a spin of 3500 rpm in air at 1 atm and 25°C.
11–106 Calculate the thickness of the boundary layer during flow over a 2.5-m-long flat plate at intervals of 25 cm and plot the boundary layer over the plate for the flow of (a) air, (b) water, and (c) engine oil at 1 atm and 20°C at an upstream velocity of 3 m/s.
cen96537_ch11_611-666.indd 662 14/01/17 3:17 pm 663 CHAPTER 11 11–107 Consider a blimp that can be approximated as a 3-m diameter, 8-m long ellipsoid and is connected to the ground. On a windless day, the rope tension due to the net buoyancy effect is measured to be 120 N. Determine the rope tension when there are 50 km/h winds blowing along the blimp (parallel to the blimp axis).
FIGURE P11–107 11–108 A 17,000-kg tractor-trailer rig has a frontal area of 9.2 m2, a drag coefficient of 0.96, a rolling resistance coef ficient of 0.05 (multiplying the weight of a vehicle by the rolling resistance coefficient gives the rolling resistance), a bearing friction resistance of 350 N, and a maximum speed of 110 km/h on a level road during steady cruising in calm weather with an air density of 1.25 kg/m3. Now a fairing is installed to the front of the rig to suppress separation and to streamline the flow to the top surface, and the drag coeffi cient is reduced to 0.76. Determine the maximum speed of the rig with the fairing. Answer: 133 km/h 11–109 During an experiment, three aluminum balls (𝜌s = 2600 kg/m3) having diameters 2, 4, and 10 mm, respectively, are dropped into a tank filled with glycerin at 22°C (𝜌f = 1274 kg/m3 and 𝜇 = 1 kg/m·s). The termi nal settling velocities of the balls are measured to be 3.2, 12.8, and 60.4 mm/s, respectively. Compare these values with the velocities predicted by Stokes law for drag force FD = 3𝜋𝜇DV, which is valid for very low Reynolds numbers (Re ≪ 1). Determine the error involved for each case and assess the accuracy of Stokes law. 11–110 Repeat Prob. 11–109 by considering the more general form of Stokes law expressed as FD = 3𝜋𝜇DV + (9𝜋/16)𝜌V 2D2 where 𝜌 is the fluid density.
11–111 A small aluminum ball with D = 2 mm and 𝜌s = 2700 kg/m3 is dropped into a large container filled with oil at 40°C (𝜌f = 876 kg/m3 and 𝜇 = 0.2177 kg/m·s). The Reynolds number is expected to be low and thus Stokes law for drag force FD = 3𝜋𝜇DV to be applicable. Show that the variation of velocity with time can be expressed as V = (a/b) (1 − e−bt ) where a = g(1 − 𝜌f /𝜌s) and b = 18𝜇/(𝜌sD2). Plot the variation of velocity with time, and calculate the time it takes for the ball to reach 99 percent of its terminal velocity.
11–112 Engine oil at 40°C is flowing over a long flat plate with a velocity of 6 m/s. Determine the distance xcr from the leading edge of the plate where the flow becomes turbulent, and calculate and plot the thickness of the boundary layer over a length of 2xcr.
11–113 The cylindrical chimney of a factory has an external diameter of 1.1 m and is 20 m high. Determine the bending moment at the base of the chimney when winds at 110 km/h are blowing across it. Take the atmospheric conditions to be 20°C and 1 atm.
11–114 There are several equations and approximations for calculating the terminal settling velocity Vt of small spheri cal particles of diameter Dp in quiescent air. The simplest case is to use Stokes approximation, namely CD = 24/Re. For very small particles (typically less than a micron), how ever, rarified gas effects come into play, and these effects are accounted for by using the Cunningham correction factor, C = 1 + Kn[2.514 + 0.80 exp(−0.55 Kn )]where Kn = 𝜆/Dp is the Knudsen number and 𝜆 is the mean free path of the air molecules. The drag coefficient is modified (corrected) by dividing by C, i.e., replacing CD by CD/C. At Reynolds num bers greater than about 0.1, Stokes approximation for drag coefficient must also be corrected, and some useful piecewise expressions for drag coefficient are • CD = 24 Re for Re < 0.1 • CD = 24 Re (1 + 0.0916 Re) for 0.1 < Re < 5 • CD = 24 Re (1 + 0.158 Re2/3) for 5 < Re < 1000 Consider spherical particles of density 1000 kg/m3 settling in quiescent air at 25oC and atmospheric pressure, at which 𝜆 = 0.06704 microns and 𝜌air = 1.184 kg/m3. On the same plot, draw three curves of Vt vs. Dp: • Best estimate: Use appropriate equation for CD, depend ing on Reynolds number; apply Cunningham correction factor, and iterate as necessary to obtain the best possible estimate of drag coefficient and terminal settling velocity.
• Stokes approximation with Cunningham correction: Use CD = 24/Re regardless of the value of Re; apply Cunningham correction factor and iterate as necessary to obtain the Stokes flow estimate of terminal settling velocity.
• Best estimate of CD, but no Cunningham correction: Use appropriate equation for CD, depending on Reynolds cen96537_ch11_611-666.indd 663 14/01/17 3:17 pm 664 EXTERNAL FLOW: DRAG AND LIFT number; ignore the Cunningham correction factor (set C = 1), and iterate as necessary to obtain the best estimate of terminal settling velocity when C is ignored.
For consistency, make your plot a log-log plot, with Dp rang ing from 10–3 to 103 microns on the horizontal axis, and Vt ranging from 10–8 to 101 m/s on the vertical axis. Print out your plot, and make sure the axes and data sets are clearly labeled with a legend or other label. Finally, summarize and comment: Specifically, for what range of particle diameter is it okay to ignore the Cunningham correction factor? For what range of particle diameter is it okay to use the Stokes approximation?
Fundamentals of Engineering (FE) Exam Problems 11–115 Which quantities are physical phenomena associ ated with fluid flow over bodies?
I. Drag force acting on automobiles II. The lift developed by airplane wings III. Upward draft of rain and snow IV. Power generated by wind turbines (a) I and II (b) I and III (c) II and III (d ) I, II, and III (e) I, II, III, and IV 11–116 The sum of the components of the pressure and wall shear forces in the direction normal to the flow is called (a) Drag (b) Friction (c) Lift (d ) Bluff (e) Blunt 11–117 The manufacturer of a car reduces the drag coef ficient of the car from 0.38 to 0.33 as a result of some modifications in its shape and design. If, on average, the aerodynamic drag accounts for 20 percent of the fuel consumption, the percent reduction in the fuel consumption of the car due to reducing the drag coefficient is (a) 15% (b) 13% (c) 6.6% (d ) 2.6% (e) 1.3% 11–118 A person is driving his motorcycle at a speed of 90 km/h in air at 20°C. The frontal area of the motorcycle and driver is 0.75 m2. If the drag coefficient under these condi tions is estimated to be 0.90, the drag force acting on the car in the flow direction is (a) 379 N (b) 204 N (c) 254 N (d ) 328 N (e) 420 N 11–119 A car is moving at a speed of 70 km/h in air at 20°C. The frontal area of the car is 2.4 m2. If the drag force acting on the car in the flow direction is 205 N, the drag coefficient of the car is (a) 0.312 (b) 0.337 (c) 0.354 (d ) 0.375 (e) 0.391 11–120 The region of flow trailing the body where the effects of the body are felt is called (a) Wake (b) Separated region (c) Stall (d ) Vortice (e) Irrotational 11–121 The turbulent boundary layer can be considered to consist of four regions. Which choice is not one of them?
(a) Buffer layer (b) Overlap layer (c) Transition layer (d ) Viscous layer (e) Turbulent layer 11–122 Air at 30°C flows over a 3.0-cm-outer-diameter, 45-m-long pipe with a velocity of 6 m/s. Calculate the drag force exerted on the pipe by the air. (Air properties at 30°C are: 𝜌 = 1.164 kg/m3, 𝜈 = 1.608 × 10−5 m2/s.) (a) 19.3 N (b) 36.8 N (c) 49.3 N (d ) 53.9 N (e) 60.1 N 11–123 Water at 10°C flows over a 4.8-m-long flat plate with a velocity of 1.15 m/s. If the width of the plate is 6.5 m, calculate the average friction coefficient over the entire plate. (Water properties at 10°C are: 𝜌 = 999.7 kg/m3, 𝜇 = 1.307 × 10−3 kg/m·s.) (a) 0.00288 (b) 0.00295 (c) 0.00309 (d ) 0.00302 (e) 0.00315 11–124 Water at 10°C flows over a 1.1-m-long flat plate with a velocity of 0.55 m/s. If the width of the plate is 2.5 m, calculate the drag force acting on the top side of the plate. (Water properties at 10°C are: 𝜌 = 999.7 kg/m3, 𝜇 = 1.307 × 10−3 kg/m·s.) (a) 0.46 N (b) 0.81 N (c) 2.75 N (d ) 4.16 N (e) 6.32 N 11–125 A 0.8-m-outer-diameter spherical tank is com pletely submerged in a flowing water stream at a velocity of 2.5 m/s. Calculate the drag force acting on the tank. (Water properties are: 𝜌 = 998.0 kg/m3, 𝜇 = 1.002 × 10−3 kg/m·s.) (a) 878 N (b) 627 N (c) 545 N (d ) 356 N (e) 220 N 11–126 An airplane is cruising at a velocity of 950 km/h in air whose density is 0.526 kg/m3. The airplane has a wing planform area of 90 m2. The lift and drag coefficients on cruising conditions are estimated to be 2.0 and 0.06, respec tively. The power that needs to be supplied to provide enough trust to overcome wing drag is (a) 21,500 kW (b) 19,300 kW (c) 23,600 kW (d ) 25,200 kW (e) 26,100 kW 11–127 An airplane has a total mass of 35,000 kg and a wing planform area of 65 m2. The airplane is cruising at 10,000 m altitude with a velocity of 1100 km/h. The density of air on cruising altitude is 0.414 kg/m3. The lift coefficient of this airplane at the cruising altitude is (a) 0.273 (b) 0.290 (c) 0.456 (d ) 0.874 (e) 1.22 11–128 An airplane has a total mass of 18,000 kg and a wing planform area of 40 m2. The density of air at the ground is 1.2 kg/m3. The maximum lift coefficient is 3.48. The minimum safe speed for takeoff and landing while extending the flaps is (a) 199 km/h (b) 211 km/h (c) 225 km/h (d ) 240 km/h (e) 257 km/h Design and Essay Problems 11–129 Write a report on the history of the reduction of the drag coefficients of cars and obtain the drag coefficient data for some recent car models from the catalogs of car manufacturers or from the Internet. cen96537_ch11_611-666.indd 664 14/01/17 3:17 pm 665 CHAPTER 11 11–130 Write a report on the flaps used at the leading and trailing edges of the wings of large commercial aircraft. Discuss how the flaps affect the drag and lift coefficients during takeoff and landing.
11–131 Discuss how to calculate drag force in unsteady flow.
11–132 Large commercial airplanes cruise at high altitudes (up to about 40,000 ft) to save fuel. Discuss how flying at high altitudes reduces drag and saves fuel. Also discuss why small planes fly at relatively low altitudes. 11–133 Many drivers turn off their air conditioners and roll down the car windows in hopes of saving fuel. But it is claimed that this apparent “free cooling” actually increases the fuel consumption of some cars. Investigate this matter and write a report on which practice saves gasoline under what conditions.
11–134 Consider a car driving along at absolute speed V against a head wind (wind blowing opposite to the direction of the moving car) at absolute speed Uwind, noting that Uwind < V. Everyone should agree that the aerodynamic drag force on the car must be calculated using the relative speed, V – (–Uwind) = V + Uwind since that is the speed of the air actually “felt” by the car. A question comes up, however, when calculating the engine power required to overcome this aerodynamic drag, where power is equal to force times veloc ity. There are two proposed methods: Method A: Multiply aerodynamic drag force by the car’s speed relative to the wind.
Method B: Multiply aerodynamic drag force by the car’s actual (absolute) speed.
Make arguments for both methods and then choose which one is correct. Discuss.
Uwind FD V FIGURE P11–134 11–135 Consider the boundary layer growing on a flat plate aligned with the freestream flow. Plot local skin friction coefficient Cf ,x as a function of Reynolds number Rex for a laminar flat plate boundary layer for Reynolds number in the range 102 < Rex < 105. For consistency, use a log scale for the horizontal (Rex) axis, and use a linear scale for the verti cal (Cf ,x) axis. Then consider the two flat plates shown here, in which one dimension is twice as long as the other. Which orientation, (a) or (b), produces the greater drag? Explain.
(a) (b) x x V V FIGURE P11–135 cen96537_ch11_611-666.indd 665 14/01/17 3:17 pm This page intentionally left blank 12 CHAPTER 667 OBJECTIVES When you finish reading this chapter, you should be able to ■ ■ Appreciate the consequences of compressibility in gas flow ■ ■ Understand why a nozzle must have a diverging section to accelerate a gas to supersonic speeds ■ ■ Predict the occurrence of shocks and calculate property changes across a shock wave ■ ■ Understand the effects of friction and heat transfer on compressible flows High-speed color schlieren image of the bursting of a toy balloon overfilled with compressed air. This 1-microsecond exposure captures the shattered bal loon skin and reveals the bubble of compressed air inside beginning to expand. The balloon burst also drives a weak spherical shock wave, visible here as a circle surrounding the balloon. The silhouette of the photographer’s hand on the air valve can be seen at center right. © G.S. Settles, Gas Dynamics Lab, Penn State University. Used with permission.
C O M P R E S S I BLE FLOW F or the most part, we have limited our consideration so far to flows for which density variations and thus compressibility effects are neg ligible. In this chapter, we lift this limitation and consider flows that involve significant changes in density. Such flows are called compressible flows, and they are frequently encountered in devices that involve the flow of gases at very high speeds. Compressible flow combines fluid dynam ics and thermodynamics in that both are necessary to the development of the required theoretical background. In this chapter, we develop the general relations associated with compressible flows for an ideal gas with constant specific heats.
We start this chapter by reviewing the concepts of stagnation state, speed of sound, and Mach number for compressible flows. The relationships between the static and stagnation fluid properties are developed for isentro pic flows of ideal gases, and they are expressed as functions of specific heat ratios and the Mach number. The effects of area changes for one-dimen sional isentropic subsonic and supersonic flows are discussed. These effects are illustrated by considering the isentropic flow through converging and converging–diverging nozzles. The concept of shock waves and the variation of flow properties across normal and oblique shock waves are discussed. Finally, we consider the effects of friction and heat transfer on compressible flows and develop relations for property changes.
cen96537_ch12_667-732.indd 667 29/12/16 6:21 pm 668 COMPRESSIBLE FLOW 12–1 ■ STAGNATION PROPERTIES When analyzing control volumes, we find it very convenient to combine the internal energy and the flow energy of a fluid into a single term, enthalpy, defined per unit mass as h = u + P/𝜌. Whenever the kinetic and poten tial energies of the fluid are negligible, as is often the case, the enthalpy represents the total energy of a fluid. For high-speed flows, such as those encountered in jet engines (Fig. 12–1), the potential energy of the fluid is still negligible, but the kinetic energy is not. In such cases, it is convenient to combine the enthalpy and the kinetic energy of the fluid into a single term called stagnation (or total) enthalpy h0, defined per unit mass as h0 = h + V 2 2 (kJ/kg) (12–1) When the potential energy of the fluid is negligible, the stagnation enthalpy represents the total energy of a flowing fluid stream per unit mass. Thus it simplifies the thermodynamic analysis of high-speed flows.
Throughout this chapter the ordinary enthalpy h is referred to as the static enthalpy, whenever necessary, to distinguish it from the stagnation enthalpy. Notice that the stagnation enthalpy is a combination property of a fluid, just like the static enthalpy, and these two enthalpies are identical when the kinetic energy of the fluid is negligible.
Consider the steady flow of a fluid through a duct such as a nozzle, dif fuser, or some other flow passage where the flow takes place adiabatically and with no shaft or electrical work, as shown in Fig. 12–2. Assuming the fluid experiences little or no change in its elevation and its potential energy, the energy balance relation (E .
in = E .
out) for this single-stream steady-flow device reduces to h1 + V 2 1 2 = h2 + V 2 2 2 (12–2) or h01 = h02 (12–3) That is, in the absence of any heat and work interactions and any changes in potential energy, the stagnation enthalpy of a fluid remains constant during a steady-flow process. Flows through nozzles and diffusers usually satisfy these conditions, and any increase in fluid velocity in these devices creates an equivalent decrease in the static enthalpy of the fluid.
If the fluid were brought to a complete stop, then the velocity at state 2 would be zero and Eq. 12–2 would become h1 + V 2 1 2 = h2 = h02 Thus the stagnation enthalpy represents the enthalpy of a fluid when it is brought to rest adiabatically.
During a stagnation process, the kinetic energy of a fluid is converted to enthalpy (internal energy + flow energy), which results in an increase in the fluid temperature and pressure. The properties of a fluid at the stagnation state are called stagnation properties (stagnation temperature, stagnation (a) FIGURE 12–2 Steady flow of a fluid through an adiabatic duct.
h02 = h1 V1 h01 h01 h2 V2 Control volume Turbine Combustion chamber Exhaust nozzle Fan Compressors FIGURE 12–1 Aircraft and jet engines involve high speeds, and thus the kinetic energy term should always be considered when analyzing them.
(a) © Corbis RF; (b) Photo courtesy of United Technologies Corporation/Pratt & Whitney. Used by permission. All rights reserved.
(b) cen96537_ch12_667-732.indd 668 29/12/16 6:21 pm 669 CHAPTER 12 pressure, stagnation density, etc.). The stagnation state and the stagnation properties are indicated by the subscript 0.
The stagnation state is called the isentropic stagnation state when the stagnation process is reversible as well as adiabatic (i.e., isentropic). The entropy of a fluid remains constant during an isentropic stagnation process. The actual (irreversible) and isentropic stagnation processes are shown on an h-s diagram in Fig. 12–3. Notice that the stagnation enthalpy of the fluid (and the stagnation temperature if the fluid is an ideal gas) is the same for both cases. However, the actual stagnation pressure is lower than the isentropic stagnation pressure since entropy increases during the actual stagnation process as a result of fluid friction. Many stagnation processes are approximated to be isentropic, and isentropic stagnation prop erties are simply referred to as stagnation properties.
When the fluid is approximated as an ideal gas with constant specific heats, its enthalpy can be replaced by cpT and Eq. 12–1 is expressed as cpT0 = cpT + V 2 2 or T0 = T + V 2 2cp (12–4) Here, T0 is called the stagnation (or total) temperature, and it represents the temperature an ideal gas attains when it is brought to rest adiabatically. The term V 2/2cp corresponds to the temperature rise during such a process and is called the dynamic temperature. For example, the dynamic tem perature of air flowing at 100 m/s is (100 m/s)2/(2 × 1.005 kJ/kg·K) = 5.0 K. Therefore, when air at 300 K and 100 m/s is brought to rest adiabatically (at the tip of a temperature probe, for example), its temperature rises to the stagnation value of 305 K (Fig. 12–4). Note that for low-speed flows, the stagnation and static (or ordinary) temperatures are practically the same. But for high-speed flows, the temperature measured by a stationary probe placed in the fluid (the stagnation temperature) may be significantly higher than the static temperature of the fluid.
The pressure a fluid attains when brought to rest isentropically is called the stagnation pressure P0. For ideal gases with constant specific heats, P0 is related to the static pressure of the fluid by P0 P = ( T0 T ) k/(k−1) (12–5) By noting that 𝜌 = 1/v and using the isentropic relation Pv k = P0v0 k, the ratio of the stagnation density to static density is expressed as ρ0 ρ = ( T0 T ) 1/(k−1) (12–6) When stagnation enthalpies are used, there is no need to refer explicitly to kinetic energy. Then the energy balance E .
in = E .
out for a single-stream, steady-flow device can be expressed as qin + win + (h01 + gz1) = qout + wout + (h02 + gz2) (12–7) FIGURE 12–3 The actual state, actual stagnation state, and isentropic stagnation state of a fluid on an h-s diagram.
s Actual state h Isentropic stagnation state P0 P0, act Actual stagnation state h V 0 h P 2 2 FIGURE 12–4 The temperature of an ideal gas flowing at a velocity V rises by V 2/2cp when it is brought to a complete stop.
Temperature rise during stagnation Air 100 m/s 305 K 300 K cen96537_ch12_667-732.indd 669 29/12/16 6:21 pm 670 COMPRESSIBLE FLOW where h01 and h02 are the stagnation enthalpies at states 1 and 2, respectively. When the fluid is an ideal gas with constant specific heats, Eq. 12–7 becomes (qin −qout) + (win −wout) = cp(T02 −T01) + g(z2 −z1) (12–8) where T01 and T02 are the stagnation temperatures.
Notice that kinetic energy terms do not explicitly appear in Eqs. 12–7 and 12–8, but the stagnation enthalpy terms account for their contribution.
EXAMPLE 12–1 Compression of High-Speed Air in an Aircraft An aircraft is flying at a cruising speed of 250 m/s at an altitude of 5000 m where the atmospheric pressure is 54.05 kPa and the ambient air temperature is 255.7 K. The ambient air is first decelerated in a diffuser before it enters the com pressor (Fig. 12–5). Approximating both the diffuser and the compressor to be isentropic, determine (a) the stagnation pressure at the compressor inlet and (b) the required compressor work per unit mass if the stagnation pressure ratio of the com pressor is 8.
SOLUTION High-speed air enters the diffuser and the compressor of an aircraft. The stagnation pressure of the air and the compressor work input are to be deter mined.
Assumptions 1 Both the diffuser and the compressor are isentropic. 2 Air is an ideal gas with constant specific heats at room temperature.
Properties The constant-pressure specific heat cp and the specific heat ratio k of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 Analysis (a) Under isentropic conditions, the stagnation pressure at the com pressor inlet (diffuser exit) can be determined from Eq. 12–5. However, first we need to find the stagnation temperature T01 at the compressor inlet. Under the stated assumptions, T01 is determined from Eq. 12–4 to be T01 = T1 + V 2 1 2cp = 255.7 K + (250 m/s)2 (2)(1.005 kJ/kg·K) ( 1 kJ/kg 1000 m2/s2) = 286.8 K Then from Eq. 12–5, P01 = P1( T01 T1 ) k/(k−1) = (54.05 kPa)( 286.8 K 255.7 K) 1.4/(1.4 −1) = 80.77 kPa That is, the temperature of air would increase by 31.1°C and the pressure by 26.72 kPa as air is decelerated from 250 m/s to zero velocity. These increases in the tem perature and pressure of air are due to the conversion of the kinetic energy into enthalpy.
(b) To determine the compressor work, we need to know the stagnation tem perature of air at the compressor exit T02. The stagnation pressure ratio across the compressor P02/P01 is specified to be 8. Since the compression process is approximated as isentropic, T02 can be determined from the ideal-gas isentropic relation (Eq. 12–5): T02 = T01( P02 P01) (k−1)/k = (286.8 K)(8)(1.4 −1)/1.4 = 519.5 K Compressor T1 P01 T01 P02 T02 = 255.7 K V1 = 250 m/s P1 = 54.05 kPa Diffuser Aircraft engine FIGURE 12–5 Schematic for Example 12–1.
cen96537_ch12_667-732.indd 670 29/12/16 6:21 pm 671 CHAPTER 12 12–2 ■ ONE-DIMENSIONAL ISENTROPIC FLOW An important parameter in the study of compressible flow is the speed of sound c, which was shown in Chap. 2 to be related to other fluid properties as c = √(∂P/∂ρ)s (12–9) or c = √k(∂P/∂ρ)T (12–10) For an ideal gas it simplifies to c = √kRT (12–11) where k is the specific heat ratio of the gas and R is the specific gas constant. The ratio of the speed of the flow to the speed of sound is the dimensionless Mach number Ma, Ma = V c (12–12) During fluid flow through many devices such as nozzles, diffusers, and turbine blade passages, flow quantities vary primarily in the flow direction only, and the flow can be approximated as one-dimensional isentropic flow with good accuracy. Therefore, it merits special consideration. Before pre senting a formal discussion of one-dimensional isentropic flow, we illustrate some important aspects of it with an example.
FIGURE 12–6 Schematic for Example 12–2.
1400 Stagnation region: 1400 kPa 200°C CO2 1000 767 200 P, kPa m = 3.00 kg/s ⋅ Disregarding potential energy changes and heat transfer, the compressor work per unit mass of air is determined from Eq. 12–8: win = cp(T02 −T01) = (1.005 kJ/kg·K)(519.5 K −286.8 K) = 233.9 kJ/kg Thus the work supplied to the compressor is 233.9 kJ/kg.
Discussion Notice that using stagnation properties automatically accounts for any changes in the kinetic energy of a fluid stream.
EXAMPLE 12–2 Gas Flow through a Converging–Diverging Duct Carbon dioxide flows steadily through a varying cross-sectional area duct such as a nozzle shown in Fig. 12–6 at a mass flow rate of 3.00 kg/s. The carbon dioxide enters the duct at a pressure of 1400 kPa and 200°C with a low velocity, and it expands in the nozzle to an exit pressure of 200 kPa. The duct is designed so that the flow can be approximated as isentropic. Determine the density, velocity, flow area, and Mach number at each location along the duct that corresponds to an overall pressure drop of 200 kPa.
SOLUTION Carbon dioxide enters a varying cross-sectional area duct at specified conditions. The flow properties are to be determined along the duct.
cen96537_ch12_667-732.indd 671 29/12/16 6:21 pm 672 COMPRESSIBLE FLOW Assumptions 1 Carbon dioxide is an ideal gas with constant specific heats at room temperature. 2 Flow through the duct is steady, one-dimensional, and isentro pic.
Properties For simplicity we use cp = 0.846 kJ/kg·K and k = 1.289 throughout the calculations, which are the constant-pressure specific heat and specific heat ratio values of carbon dioxide at room temperature. The gas constant of carbon dioxide is R = 0.1889 kJ/kg·K.
Analysis We note that the inlet temperature is nearly equal to the stagnation temperature since the inlet velocity is small. The flow is isentropic, and thus the stagnation temperature and pressure throughout the duct remain constant. Therefore, T0 ≅T1 = 200°C = 473 K and P0 ≅P1 = 1400 kPa To illustrate the solution procedure, we calculate the desired properties at the location where the pressure is 1200 kPa, the first location that corresponds to a pressure drop of 200 kPa. From Eq. 12–5, T = T0( P P0) (k−1)/k = (473 K)( 1200 kPa 1400 kPa) (1.289−1)/1.289 = 457 K From Eq. 12–4, V = √2cp(T0 −T) = √2(0.846 kJ/kg·K)(473 K −457 K)( 1000 m2/s3 1 kJ/kg ) = 164.5 m/s ≅164 m/s From the ideal-gas relation, ρ = P RT = 1200 kPa (0.1889 kPa·m3/kg·K)(457 K) = 13.9 kg/m3 From the mass flow rate relation, A = m · ρV = 3.00 kg/s (13.9 kg/m3)(164.5 m/s) = 13.1 × 10−4 m2 = 13.1 cm2 From Eqs. 12–11 and 12–12, c = √kRT = √(1.289)(0.1889 kJ/kg·K)(457 K)( 1000 m2/s2 1 kJ/kg ) = 333.6 m/s Ma = V c = 164.5 m/s 333.6 m/s = 0.493 The results for the other pressure steps are summarized in Table 12–1 and are plotted in Fig. 12–7.
Discussion Note that as the pressure decreases, the temperature and speed of sound decrease while the fluid velocity and Mach number increase in the flow direction. The density decreases slowly at first and rapidly later as the fluid velocity increases.
FIGURE 12–7 Variation of normalized fluid properties and cross-sectional area along a duct as the pressure drops from 1400 to 200 kPa.
200 400 600 800 1000 Ma ρ A, Ma, ρ, T, V 1200 1400 P, kPa T A V Flow direction cen96537_ch12_667-732.indd 672 29/12/16 6:21 pm 673 CHAPTER 12 We note from Example 12–2 that the flow area decreases with decreasing pressure down to a critical-pressure value where the Mach number is unity, and then it begins to increase with further reductions in pressure. The Mach number is unity at the location of smallest flow area, called the throat (Fig. 12–8). Note that the velocity of the fluid keeps increasing after pass ing the throat although the flow area increases rapidly in that region. This increase in velocity past the throat is due to the rapid decrease in the fluid density. The flow area of the duct considered in this example first decreases and then increases. Such ducts are called converging–diverging nozzles. These nozzles are used to accelerate gases to supersonic speeds and should not be confused with Venturi nozzles, which are used strictly for incom pressible flow. The first use of such a nozzle occurred in 1893 in a steam turbine designed by a Swedish engineer, Carl G. B. de Laval (1845–1913), and therefore converging–diverging nozzles are often called Laval nozzles.
Variation of Fluid Velocity with Flow Area It is clear from Example 12–2 that the couplings among the velocity, den sity, and flow areas for isentropic duct flow are rather complex. In the remainder of this section we investigate these couplings more thoroughly, and we develop relations for the variation of static-to-stagnation property ratios with the Mach number for pressure, temperature, and density. We begin our investigation by seeking relationships among the pressure, tem perature, density, velocity, flow area, and Mach number for one-dimensional isentropic flow. Consider the mass balance for a steady-flow process: m · = ρAV = constant Differentiating and dividing the resultant equation by the mass flow rate, we obtain dρ ρ + dA A + dV V = 0 (12–13) Neglecting the potential energy, the energy balance for an isentropic flow with no work interactions is expressed in differential form as (Fig. 12–9) dP ρ + V dV = 0 (12–14) TABLE 12–1 Variation of fluid properties in flow direction in the duct described in Example 12–2 for m ⋅ = 3 kg/s = constant P, kPa T, K V, m/s 𝜌, kg/m3 c, m/s A, cm2 Ma 1400 473 0 15.7 339.4 ∞ 0 1200 457 164.5 13.9 333.6 13.1 0.493 1000 439 240.7 12.1 326.9 10.3 0.736 800 417 306.6 10.1 318.8 9.64 0.962 767 413 317.2 9.82 317.2 9.63 1.000 600 391 371.4 8.12 308.7 10.0 1.203 400 357 441.9 5.93 295.0 11.5 1.498 200 306 530.9 3.46 272.9 16.3 1.946 767 kPa is the critical pressure where the local Mach number is unity.
FIGURE 12–8 The cross section of a nozzle at the smallest flow area is called the throat.
Fluid Converging–diverging nozzle Throat Converging nozzle Throat Fluid FIGURE 12–9 Derivation of the differential form of the energy equation for steady isentropic flow.
0 (isentropic) dP CONSERVATION OF ENERGY (steady flow, w = 0, q = 0, Δpe = 0) h1 + V 2 2 1 V 2 2 = h2 + 2 or h + V 2 2 = constant Differentiate, dh + V dV = 0 Also, = dh – v dP dh = v dP ρ ρ = 1 Substitute, dP + V dV = 0 T ds cen96537_ch12_667-732.indd 673 29/12/16 6:21 pm 674 COMPRESSIBLE FLOW This relation is also the differential form of Bernoulli’s equation when changes in potential energy are negligible, which is a form of Newton’s sec ond law of motion for steady-flow control volumes. Combining Eqs. 12–13 and 12–14 gives dA A = dP ρ ( 1 V 2 −dρ dP) (12–15) Rearranging Eq. 12–9 as (∂𝜌/∂P)s = 1/c2 and substituting into Eq. 12–15 yield dA A = dP ρV 2 (1 −Ma2) (12–16) This is an important relation for isentropic flow in ducts since it describes the variation of pressure with flow area. We note that A, 𝜌, and V are posi tive quantities. For subsonic flow (Ma < 1), the term 1 − Ma2 is positive; and thus dA and dP must have the same sign. That is, the pressure of the fluid must increase as the flow area of the duct increases and must decrease as the flow area of the duct decreases. Thus, at subsonic velocities, the pressure decreases in converging ducts (subsonic nozzles) and increases in diverging ducts (subsonic diffusers).
In supersonic flow (Ma > 1), the term 1 − Ma2 is negative, and thus dA and dP must have opposite signs. That is, the pressure of the fluid must increase as the flow area of the duct decreases and must decrease as the flow area of the duct increases. Thus, at supersonic velocities, the pressure decreases in diverging ducts (supersonic nozzles) and increases in converg ing ducts (supersonic diffusers).
Another important relation for the isentropic flow of a fluid is obtained by substituting 𝜌V = −dP/dV from Eq. 12–14 into Eq. 12–16: dA A = −dV V (1 −Ma2) (12–17) This equation governs the shape of a nozzle or a diffuser in subsonic or supersonic isentropic flow. Noting that A and V are positive quantities, we conclude the following: For subsonic flow (Ma < 1), dA dV < 0 For supersonic flow (Ma > 1), dA dV > 0 For sonic flow (Ma = 1), dA dV = 0 Thus the proper shape of a nozzle depends on the highest velocity desired relative to the sonic velocity. To accelerate a fluid, we must use a converging nozzle at subsonic velocities and a diverging nozzle at supersonic velocities. The velocities encountered in most familiar applications are well below the sonic velocity, and thus it is natural that we visualize a nozzle as a con verging duct. However, the highest velocity we can achieve by a converging nozzle is the sonic velocity, which occurs at the exit of the nozzle. If we extend the converging nozzle by further decreasing the flow area, in hopes of accelerating the fluid to supersonic velocities, as shown in Fig. 12–10, Ma Ma Ma B = 1 (sonic) Attachment A < 1 B A P0, T0 P0, T0 A = 1 (sonic) A Converging nozzle Converging nozzle FIGURE 12–10 We cannot attain supersonic velocities by extending the converging section of a converging nozzle. Doing so will only move the sonic cross section farther downstream and decrease the mass flow rate.
cen96537_ch12_667-732.indd 674 29/12/16 6:21 pm 675 CHAPTER 12 we are up for disappointment. Now the sonic velocity will occur at the exit of the converging extension, instead of the exit of the original nozzle, and the mass flow rate through the nozzle will decrease because of the reduced exit area.
Based on Eq. 12–16, which is an expression of the conservation of mass and energy principles, we must add a diverging section to a converging nozzle to accelerate a fluid to supersonic velocities. The result is a converging– diverging nozzle. The fluid first passes through a subsonic (converging) section, where the Mach number increases as the flow area of the nozzle decreases, and then reaches the value of unity at the nozzle throat. The fluid continues to accelerate as it passes through a supersonic (diverging) section. Noting that m · = ρAV for steady flow, we see that the large decrease in density makes acceleration in the diverging section possible. An example of this type of flow is the flow of hot combustion gases through a nozzle in a gas turbine.
The opposite process occurs in the engine inlet of a supersonic aircraft. The fluid is decelerated by passing it first through a supersonic diffuser, which has a flow area that decreases in the flow direction. Ideally, the flow reaches a Mach number of unity at the diffuser throat. The fluid is further decelerated in a subsonic diffuser, which has a flow area that increases in the flow direction, as shown in Fig. 12–11.
Property Relations for Isentropic Flow of Ideal Gases Next we develop relations between the static properties and stagnation properties of an ideal gas in terms of the specific heat ratio k and the Mach number Ma. We assume the flow is isentropic and the gas has constant specific heats.
FIGURE 12–11 Variation of flow properties in subsonic and supersonic nozzles and diffusers.
Subsonic nozzle (a) Subsonic flow Ma < 1 Ma < 1 Supersonic diffuser Supersonic nozzle Ma > 1 Ma > 1 Subsonic diffuser (b) Supersonic flow P decreases V increases Ma increases T decreases ρ decreases P decreases V increases Ma increases T decreases ρ decreases P increases V decreases Ma decreases T increases ρ increases P increases V decreases Ma decreases T increases ρ increases cen96537_ch12_667-732.indd 675 29/12/16 6:21 pm 676 COMPRESSIBLE FLOW The temperature T of an ideal gas anywhere in the flow is related to the stagnation temperature T0 through Eq. 12–4: T0 = T + V 2 2cp or T0 T = 1 + V 2 2cpT Noting that cp = kR/(k − 1), c2 = kRT, and Ma = V/c, we see that V 2 2cpT = V 2 2[kR/(k −1)]T = ( k −1 2 ) V 2 c 2 = ( k −1 2 )Ma2 Substitution yields T0 T = 1 + ( k −1 2 )Ma2 (12–18) which is the desired relation between T0 and T.
The ratio of the stagnation to static pressure is obtained by substituting Eq. 12–18 into Eq. 12–5: P0 P = [1 + ( k −1 2 )Ma2 ] k/(k−1) (12–19) The ratio of the stagnation to static density is obtained by substituting Eq. 12–18 into Eq. 12–6: ρ0 ρ = [1 + ( k −1 2 )Ma2 ] 1/(k−1) (12–20) Numerical values of T/T0, P/P0, and 𝜌/𝜌0 are listed versus the Mach number in Table A–13 for k = 1.4, which are very useful for practical compressible flow calculations involving air.
The properties of a fluid at a location where the Mach number is unity (the throat) are called critical properties, and the ratios in Eqs. (12–18) through (12–20) are called critical ratios when Ma = 1 (Fig. 12–12). It is standard practice in the analysis of compressible flow to let the superscript asterisk () represent the critical values. Setting Ma = 1 in Eqs. 12–18 through 12–20 yields T T = 2 k + 1 (12–21) P P0 = ( 2 k + 1) k/(k−1) (12–22) ρ ρ0 = ( 2 k + 1) 1/(k−1) (12–23) These ratios are evaluated for various values of k and are listed in Table 12–2. The critical properties of compressible flow should not be con fused with the thermodynamic properties of substances at the critical point (such as the critical temperature Tcr and critical pressure Pcr).
FIGURE 12–12 When Mat = 1, the properties at the nozzle throat are the critical properties.
Subsonic nozzle Supersonic nozzle T0 P0 ρ0 ρ T P (if Mat = 1) ρ T P (Mat = 1) Throat Throat T0 P0 ρ0 cen96537_ch12_667-732.indd 676 29/12/16 6:21 pm 677 CHAPTER 12 TABLE 12–2 The critical-pressure, critical-temperature, and critical-density ratios for isentropic flow of some ideal gases Superheated steam, k = 1.3 Hot products of combustion, k = 1.33 Air, k = 1.4 Monatomic gases, k = 1.667 P P0 0.5457 0.5404 0.5283 0.4871 T T0 0.8696 0.8584 0.8333 0.7499 ρ ρ0 0.6276 0.6295 0.6340 0.6495 EXAMPLE 12–3 Critical Temperature and Pressure in Gas Flow Calculate the critical pressure and temperature of carbon dioxide for the flow condi tions described in Example 12–2 (Fig. 12–13).
SOLUTION For the flow discussed in Example 12–2, the critical pressure and temperature are to be calculated.
Assumptions 1 The flow is steady, adiabatic, and one-dimensional. 2 Carbon diox ide is an ideal gas with constant specific heats.
Properties The specific heat ratio of carbon dioxide at room temperature is k = 1.289.
Analysis The ratios of critical to stagnation temperature and pressure are deter mined to be T T0 = 2 k + 1 = 2 1.289 + 1 = 0.8737 P P0 = ( 2 k + 1) k/(k−1) = ( 2 1.289 + 1) 1.289/(1.289 −1) = 0.5477 Noting that the stagnation temperature and pressure are, from Example 12–2, T0 = 473 K and P0 = 1400 kPa, we see that the critical temperature and pressure in this case are T = 0.8737T0 = (0.8737)(473 K) = 413 K P = 0.5477P0 = (0.5477)(1400 kPa) = 767 kPa Discussion Note that these values agree with those listed in the 5th row of Table 12–1, as expected. Also, property values other than these at the throat would indicate that the flow is not critical, and the Mach number is not unity.
12–3 ■ ISENTROPIC FLOW THROUGH NOZZLES Converging or converging–diverging nozzles are found in many engineering applications including steam and gas turbines, aircraft and spacecraft propul sion systems, and even industrial blasting nozzles and torch nozzles. In this section we consider the effects of back pressure (i.e., the pressure applied T P CO2 T0 = 473 K P0 = 1.4 MPa FIGURE 12–13 Schematic for Example 12–3.
cen96537_ch12_667-732.indd 677 29/12/16 6:21 pm 678 COMPRESSIBLE FLOW at the nozzle discharge region) on the exit velocity, the mass flow rate, and the pressure distribution along the nozzle.
Converging Nozzles Consider the subsonic flow through a converging nozzle as shown in Fig. 12–14. The nozzle inlet is attached to a reservoir at pressure Pr and temperature Tr. The reservoir is sufficiently large so that the nozzle inlet velocity is negligi ble. Since the fluid velocity in the reservoir is zero and the flow through the nozzle is approximated as isentropic, the stagnation pressure and stagnation temperature of the fluid at any cross section through the nozzle are equal to the reservoir pressure and temperature, respectively.
Now we begin to reduce the back pressure and observe the resulting effects on the pressure distribution along the length of the nozzle, as shown in Fig. 12–14. If the back pressure Pb is equal to P1, which is equal to Pr, there is no flow and the pressure distribution is uniform along the nozzle. When the back pressure is reduced to P2, the exit plane pressure Pe also drops to P2. This causes the pressure along the nozzle to decrease in the flow direction.
When the back pressure is reduced to P3 (= P, which is the pressure required to increase the fluid velocity to the speed of sound at the exit plane or throat), the mass flow reaches a maximum value and the flow is said to be choked. Further reduction of the back pressure to level P4 or below does not result in additional changes in the pressure distribution, or anything else along the nozzle length.
Under steady-flow conditions, the mass flow rate through the nozzle is constant and is expressed as m · = ρAV = ( P RT)A(Ma√kRT) = PAMa√ k RT Solving for T from Eq. 12–18 and for P from Eq. 12–19 and substituting, m · = AMaP0√k/(RT0) 1 + (k −1)Ma2/2/[2(k−1)] (12–24) Thus the mass flow rate of a particular fluid through a nozzle is a function of the stagnation properties of the fluid, the flow area, and the Mach number. Equation 12–24 is valid at any cross section, and thus m . can be evaluated at any location along the length of the nozzle.
For a specified flow area A and stagnation properties T0 and P0, the maxi mum mass flow rate can be determined by differentiating Eq. 12–24 with respect to Ma and setting the result equal to zero. It yields Ma = 1. Since the only location in a nozzle where the Mach number can be unity is the location of minimum flow area (the throat), the mass flow rate through a nozzle is a maximum when Ma = 1 at the throat. Denoting this area by A, we obtain an expression for the maximum mass flow rate by substituting Ma = 1 in Eq. 12–24: m · max = AP0√ k RT0 ( 2 k + 1) (k+1)/[2(k−1)] (12–25) FIGURE 12–14 The effect of back pressure on the pressure distribution along a converging nozzle.
x Lowest exit pressure P/P0 Reservoir Pe x P Vr = 0 Pr = P0 Tr = T0 0 1 5 4 3 2 1 Pb (Back pressure) P0 Pb = P0 Pb = P Pb < P Pb > P Pb = 0 cen96537_ch12_667-732.indd 678 29/12/16 6:21 pm 679 CHAPTER 12 Thus, for a particular ideal gas, the maximum mass flow rate through a nozzle with a given throat area is fixed by the stagnation pressure and temperature of the inlet flow. The flow rate can be controlled by changing the stagna tion pressure or temperature, and thus a converging nozzle can be used as a flowmeter. The flow rate can also be controlled, of course, by varying the throat area. This principle is very important for chemical processes, medical devices, flowmeters, and anywhere the mass flux of a gas must be known and controlled.
A plot of m . versus Pb/P0 for a converging nozzle is shown in Fig. 12–15. Notice that the mass flow rate increases with decreasing Pb/P0, reaches a maximum at Pb = P, and remains constant for Pb/P0 values less than this critical ratio. Also illustrated on this figure is the effect of back pressure on the nozzle exit pressure Pe. We observe that Pe = { Pb for Pb ≥P P for Pb < P To summarize, for all back pressures lower than the critical pressure P, the pressure at the exit plane of the converging nozzle Pe is equal to P, the Mach number at the exit plane is unity, and the mass flow rate is the maximum (or choked) flow rate. Because the velocity of the flow is sonic at the throat for the maximum flow rate, a back pressure lower than the critical pressure can not be sensed in the nozzle upstream flow and does not affect the flow rate.
The effects of the stagnation temperature T0 and stagnation pressure P0 on the mass flow rate through a converging nozzle are illustrated in Fig. 12–16 where the mass flow rate is plotted against the static-to-stagnation pressure ratio at the throat Pt/P0. An increase in P0 (or a decrease of T0) will increase the mass flow rate through the converging nozzle; a decrease in P0 (or an increase in T0) will decrease it. We could also conclude this by carefully observing Eqs. 12–24 and 12–25.
A relation for the variation of flow area A through the nozzle relative to throat area A can be obtained by combining Eqs. 12–24 and 12–25 for the same mass flow rate and stagnation properties of a particular fluid. This yields A A = 1 Ma[( 2 k + 1) (1 + k −1 2 Ma2)] (k+1)/[2(k−1)] (12–26) Table A–13 gives values of A/A as a function of the Mach number for air (k = 1.4). There is one value of A/A for each value of the Mach number, but there are two possible values of the Mach number for each value of A/A—one for subsonic flow and another for supersonic flow. Another parameter sometimes used in the analysis of one-dimensional isentropic flow of ideal gases is Ma, which is the ratio of the local velocity to the speed of sound at the throat: Ma = V c (12–27) Equation 12–27 can also be expressed as Ma = V c c c = Ma c c = Ma√kRT √kRT = Ma√ T T FIGURE 12–15 The effect of back pressure Pb on the mass flow rate m . and the exit pressure Pe of a converging nozzle.
P 0 1.0 5 4 3 2 1 1.0 P P0 P0 P0 max 5 4 3 2 1 1.0 m .
m .
Pb Pe / P0 P0 P0 P Pb FIGURE 12–16 The variation of the mass flow rate through a nozzle with inlet stagnation properties.
0 Pt Decrease in P0, 1.0 m P0 P0 P0, T0 increase in T0, or both Increase in P0, decrease in T0, or both Mat = 1 Mat < 1 ⋅ P cen96537_ch12_667-732.indd 679 29/12/16 6:21 pm 680 COMPRESSIBLE FLOW where Ma is the local Mach number, T is the local temperature, and T is the critical temperature. Solving for T from Eq. 12–18 and for T from Eq. 12–21 and substituting, we get Ma = Ma√ k + 1 2 + (k −1)Ma2 (12–28) Values of Ma are also listed in Table A–13 versus the Mach number for k = 1.4 (Fig. 12–17). Note that the parameter Ma differs from the Mach number Ma in that Ma is the local velocity nondimensionalized with respect to the sonic velocity at the throat, whereas Ma is the local velocity nondimensionalized with respect to the local sonic velocity. (Recall that the sonic velocity in a nozzle varies with temperature and thus with location.) EXAMPLE 12–4 Isentropic Flow of Air in a Nozzle Air enters a nozzle at 30 psia, 630 R, and a velocity of 450 ft/s (Fig. 12–18). Approximating the flow as isentropic, determine the pressure and temperature of air at a location where the air velocity equals the speed of sound. What is the ratio of the area at this location to the entrance area?
SOLUTION Air enters a nozzle at specified temperature, pressure, and velocity. The exit pressure, exit temperature, and exit-to-inlet area ratio are to be determined for a Mach number of Ma = 1 at the exit.
Assumptions 1 Air is an ideal gas with constant specific heats at room tempera ture. 2 Flow through the nozzle is approximated as steady, one-dimensional, and isentropic.
Properties The properties of air are k = 1.4 and cp = 0.240 Btu/lbm·R (Table A–1E).
Analysis The properties of the fluid at the location where Ma = 1 are the critical properties, denoted by superscript . We first determine the stagnation temperature and pressure, which remain constant throughout the nozzle since the flow is isentropic. T0 = T + V 2 i 2cp = 630 R + (450 ft/s)2 2(0.240 Btu/lbm·R)( 1 Btu/lbm 25,037 ft2/s2) = 646.9 R P0 = Pi( T0 Ti ) k/(k−1) = (30 psia)( 646.9 K 630 K ) 1.4/(1.4−1) = 32.9 psia From Table A–13 (or from Eqs. 12–18 and 12–19) at Ma = 1, we read T/T0 = 0.8333 P/P0 = 0.5283 Thus, T = 0.8333T0 = 0.8333(646.9 R) = 539 R P = 0.5283P0 = 0.5283(32.9 psia) = 17.4 psia Also, ci = √kRTi = √(1.4)(0.06855 Btu/lbm·R)(630 R) ( 25,037 ft2/s2 1 Btu/lbm ) = 1230 ft/s Mai = Vi ci = 450 ft/s 1230 ft/s = 0.3657 FIGURE 12–17 Various property ratios for isentropic flow through nozzles and diffusers are listed in Table A–13 for k = 1.4 (air) for convenience.
Ma T0 0.90 A T P0 ρ0 P ρ A Ma .
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1.00 1.10 0.9146 1.0000 1.0812 .
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30 psia 630 R 450 ft/s Ma = 1 Air nozzle FIGURE 12–18 Schematic for Example 12–4.
cen96537_ch12_667-732.indd 680 29/12/16 6:22 pm 681 CHAPTER 12 From Table A–13 at this Mach number we read Ai /A = 1.7426. Thus the ratio of the throat area to the nozzle inlet area is A Ai = 1 1.7426 = 0.574 Discussion If we solve this problem using the relations for compressible isentropic flow, the results would be identical to at least three significant digits.
EXAMPLE 12–5 Air Loss from a Flat Tire Air in an automobile tire is maintained at a pressure of 220 kPa (gage) in an environment where the atmospheric pressure is 94 kPa. The air in the tire is at the ambient temperature of 25°C. A 4-mm-diameter leak develops in the tire as a result of an accident (Fig. 12–19). Approximating the flow as isentropic determine the initial mass flow rate of air through the leak.
SOLUTION A leak develops in an automobile tire as a result of an accident. The initial mass flow rate of air through the leak is to be determined. Assumptions 1 Air is an ideal gas with constant specific heats. 2 Flow of air through the hole is isentropic.
Properties The specific gas constant of air is R = 0.287 kPa⋅m3/kg⋅K. The specific heat ratio of air at room temperature is k = 1.4.
Analysis The absolute pressure in the tire is P = Pgage + Patm = 220 + 94 = 314 kPa The critical pressure is (from Table 12–2) P = 0.5283Po = (0.5283)(314 kPa) = 166 kPa > 94 kPa Therefore, the flow is choked, and the velocity at the exit of the hole is the sonic speed. Then the flow properties at the exit become ρ0 = P0 RT0 = 314 kPa (0.287 kPa·m3/kg·K)(298 K) = 3.671 kg/m3 ρ = ρ( 2 k + 1) 1/(k−1) = (3.671 kg/m3)( 2 1.4 + 1) 1/(1.4 −1) = 2.327 kg/m3 T = 2 k + 1 T0 = 2 1.4 + 1 (298 K) = 248.3 K V = c = √kRT = √(1.4)(0.287 kJ/kg·K)( 1000 m2/s2 1 kJ/kg )(248.3 K) = 315.9 m/s Then the initial mass flow rate through the hole is m · = ρAV = (2.327 kg/m3)𝜋(0.004 m)2/4 = 0.00924 kg/s = 0.554 kg/min Discussion The mass flow rate decreases with time as the pressure inside the tire drops.
FIGURE 12–19 Schematic for Example 12–5.
Air T = 25°C Pg = 220 kPa cen96537_ch12_667-732.indd 681 29/12/16 6:22 pm 682 COMPRESSIBLE FLOW Converging–Diverging Nozzles When we think of nozzles, we ordinarily think of flow passages whose cross-sectional area decreases in the flow direction. However, the highest velocity to which a fluid can be accelerated in a converging nozzle is lim ited to the sonic velocity (Ma = 1), which occurs at the exit plane (throat) of the nozzle. Accelerating a fluid to supersonic velocities (Ma > 1) can be accomplished only by attaching a diverging flow section to the subsonic nozzle at the throat. The resulting combined flow section is a converging– diverging nozzle, which is standard equipment in supersonic aircraft and rocket propulsion (Fig. 12–20). Forcing a fluid through a converging–diverging nozzle is no guarantee that the fluid will be accelerated to a supersonic velocity. In fact, the fluid may find itself decelerating in the diverging section instead of accelerating if the back pressure is not in the right range. The state of the nozzle flow is determined by the overall pressure ratio Pb /P0. Therefore, for given inlet conditions, the flow through a converging–diverging nozzle is governed by the back pressure Pb, as will be explained. Consider the converging–diverging nozzle shown in Fig. 12–21. A fluid enters the nozzle with a low velocity at stagnation pressure P0. When Pb = P0 (case A), there is no flow through the nozzle. This is expected since the flow in a nozzle is driven by the pressure difference between the nozzle inlet and the exit. Now let us examine what happens as the back pressure is lowered.
1. When P0 > Pb > PC, the flow remains subsonic throughout the nozzle, and the mass flow is less than that for choked flow. The fluid velocity increases in the first (converging) section and reaches a maximum at the throat (but Ma < 1). However, most of the gain in velocity is lost in the second (diverging) section of the nozzle, which acts as a diffuser. The pressure Nozzle Fuel Oxidizer Combustion chamber FIGURE 12–20 Converging–diverging nozzles are commonly used in rocket engines to provide high thrust.
(Right) Courtesy NASA cen96537_ch12_667-732.indd 682 29/12/16 6:22 pm 683 CHAPTER 12 decreases in the converging section, reaches a minimum at the throat, and increases at the expense of velocity in the diverging section.
2. When Pb = PC, the throat pressure becomes P and the fluid achieves sonic velocity at the throat. But the diverging section of the nozzle still acts as a diffuser, slowing the fluid to subsonic velocities. The mass flow rate that was increasing with decreasing Pb also reaches its maximum value. Recall that P is the lowest pressure that can be obtained at the throat, and the sonic velocity is the highest velocity that can be achieved with a converging nozzle. Thus, lowering Pb further has no influence on the fluid flow in the converging part of the nozzle or the mass flow rate through the nozzle. However, it does influence the character of the flow in the diverging section.
3. When PC > Pb > PE, the fluid that achieved a sonic velocity at the throat continues accelerating to supersonic velocities in the diverging section as the pressure decreases. This acceleration comes to a sudden stop, however, as a normal shock develops at a section between the throat and the exit FIGURE 12–21 The effects of back pressure on the flow through a converging–diverging nozzle.
0 x Subsonic flow at nozzle exit (shock in nozzle) Exit Supersonic flow at nozzle exit (no shock in nozzle) PA A Ma Throat Inlet B C D } } } Subsonic flow at nozzle exit (no shock) x Vi ≅ 0 P0 Pe P0 E, F, G Shock in nozzle Sonic flow at throat Throat Exit Inlet 0 1 Sonic flow at throat Shock in nozzle E, F, G x Subsonic flow at nozzle exit (shock in nozzle) Supersonic flow at nozzle exit (no shock in nozzle) A B C D } } Subsonic flow at nozzle exit (no shock) P Throat } PB PC PD PE PG PF P Pb Pb cen96537_ch12_667-732.indd 683 29/12/16 6:22 pm 684 COMPRESSIBLE FLOW plane, which causes a sudden drop in velocity to subsonic levels and a sudden increase in pressure. The fluid then continues to decelerate further in the remaining part of the converging–diverging nozzle. Flow through the shock is highly irreversible, and thus it cannot be approximated as isentropic. The normal shock moves downstream away from the throat as Pb is decreased, and it approaches the nozzle exit plane as Pb approaches PE.
When Pb = PE, the normal shock forms at the exit plane of the nozzle. The flow is supersonic through the entire diverging section in this case, and it can be approximated as isentropic. However, the fluid velocity drops to subsonic levels just before leaving the nozzle as it crosses the normal shock. Normal shock waves are discussed in Section 12–4.
4. When PE > Pb > 0, the flow in the diverging section is supersonic, and the fluid expands to PF at the nozzle exit with no normal shock forming within the nozzle. Thus, the flow through the nozzle can be approximated as isentropic. When Pb = PF, no shocks occur within or outside the nozzle. When Pb < PF, irreversible mixing and expansion waves occur downstream of the exit plane of the nozzle. When Pb > PF, however, the pressure of the fluid increases from PF to Pb irreversibly in the wake of the nozzle exit, creating what are called oblique shocks.
FIGURE 12–22 Schematic for Example 12–6.
At = 20 cm2 T0 = 800 K Mae = 2 P0 = 1.0 MPa Vi ≅ 0 EXAMPLE 12–6 Airflow through a Converging–Diverging Nozzle Air enters a converging–diverging nozzle, shown in Fig. 12–22, at 1.0 MPa and 800 K with negligible velocity. The flow is steady, one-dimensional, and isentropic with k = 1.4. For an exit Mach number of Ma = 2 and a throat area of 20 cm2, determine (a) the throat conditions, (b) the exit plane conditions, includ ing the exit area, and (c) the mass flow rate through the nozzle.
SOLUTION Air flows through a converging–diverging nozzle. The throat and the exit conditions and the mass flow rate are to be determined.
Assumptions 1 Air is an ideal gas with constant specific heats at room tempera ture. 2 Flow through the nozzle is steady, one-dimensional, and isentropic.
Properties The specific heat ratio of air is given to be k = 1.4. The gas constant of air is 0.287 kJ/kg⋅K.
Analysis The exit Mach number is given to be 2. Therefore, the flow must be sonic at the throat and supersonic in the diverging section of the nozzle. Since the inlet velocity is negligible, the stagnation pressure and stagnation temperature are the same as the inlet temperature and pressure, P0 = 1.0 MPa and T0 = 800 K. Assum ing ideal-gas behavior, the stagnation density is ρ0 = P0 RT0 = 1000 kPa (0.287 kPa·m3/kg·K)(800 K) = 4.355 kg/m3 (a) At the throat of the nozzle Ma = 1, and from Table A–13 we read P P0 = 0.5283 T T0 = 0.8333 ρ ρ0 = 0.6339 Thus, P = 0.5283P0 = (0.5283)(1.0 MPa) = 0.5283 MPa cen96537_ch12_667-732.indd 684 29/12/16 6:22 pm 685 CHAPTER 12 T = 0.8333T0 = (0.8333)(800 K) = 666.6 K ρ = 0.6339ρ0 = (0.6339)(4.355 kg/m3) = 2.761 kg/m3 Also, V = c = √kRT = √(1.4)(0.287 kJ/kg·K)(666.6 K) ( 1000 m2/s2 1 kJ/kg ) = 517.5 m/s (b) Since the flow is isentropic, the properties at the exit plane can also be calcu lated by using data from Table A–13. For Ma = 2 we read Pe P0 = 0.1278 Te T0 = 0.5556 ρe ρ0 = 0.2300 Mae = 1.6330 Ae A = 1.6875 Thus, Pe = 0.1278P0 = (0.1278)(1.0 MPa) = 0.1278 MPa Te = 0.5556T0 = (0.5556)(800 K) = 444.5 K ρe = 0.2300ρ0 = (0.2300)(4.355 kg/m3) = 1.002 kg/m3 Ae = 1.6875A = (1.6875)(20 cm2) = 33.75 cm2 and Ve = Maec = (1.6330)(517.5 m/s) = 845.1 m/s The nozzle exit velocity could also be determined from Ve = Maece, where ce is the speed of sound at the exit conditions: Ve = Maece = Mae√kRTe = 2√(1.4)(0.287 kJ/kg·K)(444.5 K) ( 1000 m2/s2 1 kJ/kg ) = 845.2 m/s (c) Since the flow is steady, the mass flow rate of the fluid is the same at all sec tions of the nozzle. Thus it may be calculated by using properties at any cross sec tion of the nozzle. Using the properties at the throat, we find that the mass flow rate is m · = ρAV = (2.761 kg/m3)(20 × 10−4 m2)(517.5 m/s) = 2.86 kg/s Discussion Note that this is the highest possible mass flow rate that can flow through this nozzle for the specified inlet conditions.
12–4 ■ SHOCK WAVES AND EXPANSION WAVES We discussed in Chap. 2 that sound waves are caused by infinitesimally small pressure disturbances, and they travel through a medium at the speed of sound. We have also seen in the present chapter that for some back pres sure values, abrupt changes in fluid properties occur in a very thin sec tion of a converging–diverging nozzle under supersonic flow conditions, cen96537_ch12_667-732.indd 685 29/12/16 6:22 pm 686 COMPRESSIBLE FLOW creating a shock wave. It is of interest to study the conditions under which shock waves develop and how they affect the flow.
Normal Shocks First we consider shock waves that occur in a plane normal to the direction of flow, called normal shock waves. The flow process through the shock wave is highly irreversible and cannot be approximated as being isentropic. Next we follow the footsteps of Pierre Laplace (1749–1827), G. F. Bernhard Riemann (1826–1866), William Rankine (1820–1872), Pierre Henry Hugo niot (1851–1887), Lord Rayleigh (1842–1919), and G. I. Taylor (1886–1975) and develop relationships for the flow properties before and after the shock. We do this by applying the conservation of mass, momentum, and energy rela tions as well as some property relations to a stationary control volume that contains the shock, as shown in Fig. 12–23. The normal shock waves are extremely thin, so the entrance and exit flow areas for the control volume are approximately equal (Fig. 12–24). We assume steady flow with no heat and work interactions and no poten tial energy changes. Denoting the properties upstream of the shock by the subscript 1 and those downstream of the shock by 2, we have the following: Conservation of mass: ρ1AV1 = ρ2AV2 (12–29) or ρ1V1 = ρ2V2 Conservation of energy: h1 + V 2 1 2 = h2 + V 2 2 2 (12–30) or h01 = h02 (12–31) Linear momentum equation: Rearranging Eq. 12–14 and integrating yield A(P1 −P2) = m · (V2 −V1) (12–32) Increase of entropy: s2 −s1 ≥0 (12–33) We can combine the conservation of mass and energy relations into a single equation and plot it on an h-s diagram, using property relations. The resul tant curve is called the Fanno line, and it is the locus of states that have the same value of stagnation enthalpy and mass flux (mass flow per unit flow area). Likewise, combining the conservation of mass and momentum equa tions into a single equation and plotting it on the h-s diagram yield a curve called the Rayleigh line. Both these lines are shown on the h-s diagram in Fig. 12–25. As proved later in Example 12–7, the points of maximum entropy on these lines (points a and b) correspond to Ma = 1. The state on the upper part of each curve is subsonic and on the lower part supersonic.
The Fanno and Rayleigh lines intersect at two points (points 1 and 2), which represent the two states at which all three conservation equations are satisfied. One of these (state 1) corresponds to the state before the shock, FIGURE 12–23 Control volume for flow across a normal shock wave.
Control volume Flow Ma1 > 1 P1 V1 s1 Shock wave P2 V2 h2 h1 s2 ρ1 ρ2 Ma2 < 1 FIGURE 12–24 Schlieren image of a normal shock in a Laval nozzle. The Mach number in the nozzle just upstream (to the left) of the shock wave is about 1.3. Boundary layers distort the shape of the normal shock near the walls and lead to flow separation beneath the shock.
© G.S. Settles, Gas Dynamics Lab, Penn State University. Used with permission.
cen96537_ch12_667-732.indd 686 29/12/16 6:22 pm 687 CHAPTER 12 and the other (state 2) corresponds to the state after the shock. Note that the flow is supersonic before the shock and subsonic afterward. Therefore the flow must change from supersonic to subsonic if a shock is to occur. The larger the Mach number before the shock, the stronger the shock will be. In the limiting case of Ma = 1, the shock wave simply becomes a sound wave. Notice from Fig. 12–25 that entropy increases, s2 > s1. This is expected since the flow through the shock is adiabatic but irreversible.
The conservation of energy principle (Eq. 12–31) requires that the stag nation enthalpy remain constant across the shock; h01 = h02. For ideal gases h = h(T), and thus T01 = T02 (12–34) That is, the stagnation temperature of an ideal gas also remains constant across the shock. Note, however, that the stagnation pressure decreases across the shock because of the irreversibilities, while the ordinary (static) temperature rises drastically because of the conversion of kinetic energy into enthalpy due to a large drop in fluid velocity (see Fig. 12–26).
We now develop relations between various properties before and after the shock for an ideal gas with constant specific heats. A relation for the ratio of the static temperatures T2/T1 is obtained by applying Eq. 12–18 twice: T01 T1 = 1 + ( k −1 2 )Ma 2 1 and T02 T2 = 1 + ( k −1 2 )Ma2 2 Dividing the first equation by the second one and noting that T01 = T02, we have T2 T1 = 1 + Ma 2 1 (k −1)/2 1 + Ma 2 2 (k −1)/2 (12–35) From the ideal-gas equation of state, ρ1 = P1 RT1 and ρ2 = P2 RT2 Substituting these into the conservation of mass relation 𝜌1V1 = 𝜌2V2 and noting that Ma = V/c and c = √kRT, we have T2 T1 = P2V2 P1V1 = P2Ma2c2 P1Ma1c1 = P2Ma2√T2 P1Ma1√T1 = ( P2 P1) 2 ( Ma2 Ma1) 2 (12–36) Combining Eqs. 12–35 and 12–36 gives the pressure ratio across the shock: Fanno line: P2 P1 = Ma1√1 + Ma 2 1 (k −1)/2 Ma2√1 + Ma 2 2 (k −1)/2 (12–37) Equation 12–37 is a combination of the conservation of mass and energy equations; thus, it is also the equation of the Fanno line for an ideal gas with constant specific heats. A similar relation for the Rayleigh line is obtained by combining the conservation of mass and momentum equations. From Eq. 12–32, FIGURE 12–25 The h-s diagram for flow across a normal shock.
Ma = 1 0 s SHOCK WAVE Subsonic flow h a h01 = h02 P02 P01 Ma = 1 b (Ma < 1) 2 Supersonic flow (Ma > 1) 1 s1 s2 2 2 Fanno line Rayleigh line h01 h2 h1 h02 V 2 1 V 2 2 FIGURE 12–26 Variation of flow properties across a normal shock in an ideal gas.
Normal shock P P0 V Ma T T0 ρ s increases decreases decreases decreases increases remains constant increases increases cen96537_ch12_667-732.indd 687 29/12/16 6:22 pm 688 COMPRESSIBLE FLOW P1 −P2 = m · A (V2 −V1) = ρ2V 2 2 −ρ1V 2 1 However, ρV 2 = ( P RT)(Ma c)2 = ( P RT) (Ma√kRT)2 = Pk Ma2 Thus, P1(1 + kMa 2 1 ) = P2(1 + kMa 2 2 ) or Rayleigh line: P2 P1 = 1 + kMa 2 1 1 + kMa 2 2 (12–38) Combining Eqs. 12–37 and 12–38 yields Ma 2 2 = Ma 2 1 + 2/(k −1) 2Ma 2 1 k/(k −1) −1 (12–39) This represents the intersections of the Fanno and Rayleigh lines and relates the Mach number upstream of the shock to that downstream of the shock.
The occurrence of shock waves is not limited to supersonic nozzles only. This phenomenon is also observed at the engine inlet of supersonic aircraft, where the air passes through a shock and decelerates to subsonic veloci ties before entering the diffuser of the engine (Fig. 12–27). Explosions also produce powerful expanding spherical normal shocks, which can be very destructive (Fig. 12–28).
FIGURE 12–27 The air inlet of a supersonic fighter jet is designed such that a shock wave at the inlet decelerates the air to subsonic velocities, increasing the pressure and temperature of the air before it enters the engine.
© StockTrek/Getty Images RF FIGURE 12–28 Schlieren image of the blast wave (expanding spherical normal shock) produced by the explosion of a firecracker. The shock expanded radially outward in all directions at a supersonic speed that decreased with radius from the center of the explosion. A microphone sensed the sudden change in pressure of the passing shock wave and triggered the microsecond flashlamp that exposed the photograph.
© G.S. Settles, Gas Dynamics Lab, Penn State University. Used with permission.
cen96537_ch12_667-732.indd 688 29/12/16 6:22 pm 689 CHAPTER 12 Various flow property ratios across the shock are listed in Table A–14 for an ideal gas with k = 1.4. Inspection of this table reveals that Ma2 (the Mach number after the shock) is always less than 1 and that the larger the super sonic Mach number before the shock, the smaller the subsonic Mach number after the shock. Also, we see that the static pressure, temperature, and den sity all increase after the shock while the stagnation pressure decreases.
The entropy change across the shock is obtained by applying the entropy-change equation for an ideal gas across the shock: s2 −s1 = cP ln T2 T1 −R ln P2 P1 (12–40) which can be expressed in terms of k, R, and Ma1 by using the relations developed earlier in this section. A plot of nondimensional entropy change across the normal shock (s2 − s1)/R versus Ma1 is shown in Fig. 12–29. Since the flow across the shock is adiabatic and irreversible, the second law of thermodynamics requires that the entropy increase across the shock wave. Thus, a shock wave cannot exist for values of Ma1 less than unity where the entropy change would be negative. For adiabatic flows, shock waves can exist only for supersonic flows, Ma1 > 1.
FIGURE 12–29 Entropy change across a normal shock.
0 IMPOSSIBLE Subsonic flow before shock Ma1 = 1 Ma1 Supersonic flow before shock (s2 – s1)/R s2 – s1 > 0 s2 – s1 < 0 EXAMPLE 12–7 The Point of Maximum Entropy on the Fanno Line Show that the point of maximum entropy on the Fanno line (point a of Fig. 12–25) for the adiabatic steady flow of a fluid in a duct corresponds to the sonic velocity, Ma = 1.
SOLUTION It is to be shown that the point of maximum entropy on the Fanno line for steady adiabatic flow corresponds to sonic velocity.
Assumption The flow is steady, adiabatic, and one-dimensional.
Analysis In the absence of any heat and work interactions and potential energy changes, the steady-flow energy equation reduces to h + V 2 2 = constant Differentiating yields dh + V dV = 0 For a very thin shock with negligible change of duct area across the shock, the steady-flow continuity (conservation of mass) equation is expressed as ρV = constant Differentiating, we have ρ dV + V dρ = 0 Solving for dV gives dV = −V dρ ρ cen96537_ch12_667-732.indd 689 29/12/16 6:22 pm 690 COMPRESSIBLE FLOW Combining this with the energy equation, we have dh −V 2 dρ ρ = 0 which is the equation for the Fanno line in differential form. At point a (the point of maximum entropy) ds = 0. Then from the second T ds relation (T ds = dh − v dP ) we have dh = v dP = dP/𝜌. Substituting yields dP ρ −V 2 dρ ρ = 0 at s = constant Solving for V, we have V = ( ∂P ∂ρ) s 1/2 which is the relation for the speed of sound, Eq. 12–9. Thus V = c and the proof is complete.
T1 = 444.5 K Ma1 = 2 P01 = 1.0 MPa P1 = 0.1278 MPa ρ1 = 1.002 kg/m3 Shock wave 1 2 m = 2.86 kg/s · FIGURE 12–30 Schematic for Example 12–8.
EXAMPLE 12–8 Shock Wave in a Converging–Diverging Nozzle If the air flowing through the converging–diverging nozzle of Example 12–6 expe riences a normal shock wave at the nozzle exit plane (Fig. 12–30), determine the following after the shock: (a) the stagnation pressure, static pressure, static tem perature, and static density; (b) the entropy change across the shock; (c) the exit velocity; and (d ) the mass flow rate through the nozzle. Approximate the flow as steady, one-dimensional, and isentropic with k = 1.4 from the nozzle inlet to the shock location.
SOLUTION Air flowing through a converging–diverging nozzle experiences a normal shock at the exit. The effect of the shock wave on various properties is to be determined.
Assumptions 1 Air is an ideal gas with constant specific heats at room tempera ture. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. 3 The shock wave occurs at the exit plane.
Properties The constant-pressure specific heat and the specific heat ratio of air are cp = 1.005 kJ/kg·K and k = 1.4. The gas constant of air is 0.287 kJ/kg⋅K.
Analysis (a) The fluid properties at the exit of the nozzle just before the shock (denoted by subscript 1) are those evaluated in Example 12–6 at the nozzle exit to be P01 = 1.0 MPa P1 = 0.1278 MPa T1 = 444.5 K ρ1 = 1.002 kg/m3 The fluid properties after the shock (denoted by subscript 2) are related to those before the shock through the functions listed in Table A–14. For Ma1 = 2.0, we read Ma2 = 0.5774 P02 P01 = 0.7209 P2 P1 = 4.5000 T2 T1 = 1.6875 ρ2 ρ1 = 2.6667 Then the stagnation pressure P02, static pressure P2, static temperature T2, and static density 𝜌2 after the shock are cen96537_ch12_667-732.indd 690 29/12/16 6:22 pm 691 CHAPTER 12 P02 = 0.7209P01 = (0.7209)(1.0 MPa) = 0.721 MPa P2 = 4.5000P1 = (4.5000)(0.1278 MPa) = 0.575 MPa T2 = 1.6875T1 = (1.6875)(444.5 K) = 750 K ρ2 = 2.6667ρ1 = (2.6667)(1.002 kg/m3) = 2.67 kg/m3 (b) The entropy change across the shock is s2 −s1 = cρ ln T2 T1 −R ln P2 P1 = (1.005 kJ/kg·K) ln (1.6875) − (0.287 kJ/kg·K) ln (4.5000) = 0.0942 kJ/kg·K Thus, the entropy of the air increases as it passes through a normal shock, which is highly irreversible.
(c) The air velocity after the shock is determined from V2 = Ma2c2, where c2 is the speed of sound at the exit conditions after the shock: V2 = Ma2c2 = Ma2√kRT2 = (0.5774)√(1.4)(0.287 kJ/kg·K)(750.1 K)( 1000 m2/s2 1 kJ/kg ) = 317 m/s (d ) The mass flow rate through a converging–diverging nozzle with sonic conditions at the throat is not affected by the presence of shock waves in the nozzle. Therefore, the mass flow rate in this case is the same as that determined in Example 12–6: m · = 2.86 kg/s Discussion This result can easily be verified by using property values at the nozzle exit after the shock at all Mach numbers significantly greater than unity.
Example 12–8 illustrates that the stagnation pressure and velocity decrease while the static pressure, temperature, density, and entropy increase across the shock (Fig. 12–31). The rise in the temperature of the fluid downstream of a shock wave is of major concern to the aerospace engineer because it creates heat transfer problems on the leading edges of wings and nose cones of space reentry vehicles and the recently proposed hypersonic space planes. Overheating, in fact, led to the tragic loss of the space shuttle Columbia in February of 2003 as it was reentering earth’s atmosphere.
Oblique Shocks Not all shock waves are normal shocks (perpendicular to the flow direction). For example, when the space shuttle travels at supersonic speeds through the atmosphere, it produces a complicated shock pattern consisting of inclined FIGURE 12–31 When a lion tamer cracks his whip, a weak spherical shock wave forms near the tip and spreads out radially; the pressure inside the expanding shock wave is higher than ambient air pressure, and this is what causes the crack when the shock wave reaches the lion’s ear.
© Joshua Ets-Hokin/Getty Images RF cen96537_ch12_667-732.indd 691 29/12/16 6:22 pm 692 COMPRESSIBLE FLOW shock waves called oblique shocks (Fig. 12–32). As you can see, some por tions of an oblique shock are curved, while other portions are straight.
First, we consider straight oblique shocks, like that produced when a uni form supersonic flow (Ma1 > 1) impinges on a slender, two-dimensional wedge of half-angle 𝛿 (Fig. 12–33). Since information about the wedge can not travel upstream in a supersonic flow, the fluid “knows” nothing about the wedge until it hits the nose. At that point, since the fluid cannot flow through the wedge, it turns suddenly through an angle called the turning angle or deflection angle 𝜃. The result is a straight oblique shock wave, aligned at shock angle or wave angle 𝛽, measured relative to the oncoming flow (Fig. 12–34). To conserve mass, 𝛽 must obviously be greater than 𝛿. Since the Reynolds number of supersonic flows is typically large, the boundary layer growing along the wedge is very thin, and we ignore its effects. The flow therefore turns by the same angle as the wedge; namely, deflection angle 𝜃 is equal to wedge half-angle 𝛿. If we take into account the displacement thickness effect of the boundary layer (Chap. 10), the deflection angle 𝜃 of the oblique shock turns out to be slightly greater than wedge half-angle 𝛿.
Like normal shocks, the Mach number decreases across an oblique shock, and oblique shocks are possible only if the upstream flow is supersonic. However, unlike normal shocks, in which the downstream Mach number is always subsonic, Ma2 downstream of an oblique shock can be subsonic, sonic, or supersonic, depending on the upstream Mach number Ma1 and the turning angle.
We analyze a straight oblique shock in Fig. 12–34 by decomposing the velocity vectors upstream and downstream of the shock into normal and tangential components, and considering a small control volume around the shock. Upstream of the shock, all fluid properties (velocity, density, pres sure, etc.) along the lower left face of the control volume are identical to those along the upper right face. The same is true downstream of the shock. Therefore, the mass flow rates entering and leaving those two faces cancel each other out, and conservation of mass reduces to ρ1V1, n A = ρ2V2, n A → ρ1V1, n = ρ2V2, n (12–41) FIGURE 12–32 Schlieren image of a small model of the space shuttle orbiter being tested at Mach 3 in the supersonic wind tunnel of the Penn State Gas Dynamics Lab. Several oblique shocks are seen in the air surrounding the spacecraft.
© G.S. Settles, Gas Dynamics Lab, Penn State University. Used with permission.
FIGURE 12–33 An oblique shock of shock angle 𝛽 formed by a slender, two-dimensional wedge of half-angle 𝛿. The flow is turned by deflection angle 𝜃 downstream of the shock, and the Mach number decreases.
δ β θ Ma1 Ma1 Ma2 Oblique shock cen96537_ch12_667-732.indd 692 29/12/16 6:22 pm 693 CHAPTER 12 where A is the area of the control surface that is parallel to the shock. Since A is the same on either side of the shock, it has dropped out of Eq. 12–41.
As you might expect, the tangential component of velocity (parallel to the oblique shock) does not change across the shock, i.e., V1, t = V2, t. This is easily proven by applying the tangential momentum equation to the control volume.
When we apply conservation of momentum in the direction normal to the oblique shock, the only forces are pressure forces, and we get P1A −P2A = ρV2, n AV2, n −ρV1, n AV1, n → P1 −P2 = ρ2V 2 2, n −ρ1V 2 1, n (12–42) Finally, since there is no work done by the control volume and no heat trans fer into or out of the control volume, stagnation enthalpy does not change across an oblique shock, and conservation of energy yields h01 = h02 = h0 → h1 + 1 2 V 2 1, n + 1 2 V 2 1, t = h2 + 1 2 V 2 2, n + 1 2 V 2 2, t But since V1, t = V2, t, this equation reduces to h1 + 1 2 V 2 1, n = h2 + 1 2 V 2 2, n (12–43) Careful comparison reveals that the equations for conservation of mass, momentum, and energy (Eqs. 12–41 through 12–43) across an oblique shock are identical to those across a normal shock, except that they are written in terms of the normal velocity component only. Therefore, the normal shock relations derived previously apply to oblique shocks as well, but must be writ ten in terms of Mach numbers Ma1, n and Ma2, n normal to the oblique shock. This is most easily visualized by rotating the velocity vectors in Fig. 12–34 by angle 𝜋/2 − 𝛽, so that the oblique shock appears to be vertical (Fig. 12–35). Trigonometry yields Ma1, n = Ma1 sin 𝛽 and Ma2, n = Ma2 sin(𝛽−𝜃) (12–44) where Ma1, n = V1, n /c1 and Ma2, n = V2, n /c2. From the point of view shown in Fig. 12–35, we see what looks like a normal shock, but with some super posed tangential flow “coming along for the ride.” Thus, All the equations, shock tables, etc., for normal shocks apply to oblique shocks as well, provided that we use only the normal components of the Mach number.
In fact, you may think of normal shocks as special oblique shocks in which shock angle 𝛽 = 𝜋/2, or 90°. We recognize immediately that an oblique shock can exist only if Ma1, n > 1 and Ma2, n < 1. The normal shock equations appropriate for oblique shocks in an ideal gas are summa rized in Fig. 12–36 in terms of Ma1, n.
For known shock angle 𝛽 and known upstream Mach number Ma1, we use the first part of Eq. 12–44 to calculate Ma1, n, and then use the normal shock tables (or their corresponding equations) to obtain Ma2, n. If we also knew the deflection angle 𝜃, we could calculate Ma2 from the second part of Eq. 12–44. But, in a typical application, we know either 𝛽 or 𝜃, but not both. Fortunately, a bit more algebra provides us with a relationship between 𝜃, FIGURE 12–34 Velocity vectors through an oblique shock of shock angle 𝛽 and deflection angle 𝜃.
V2, n Oblique shock Control volume V1, n V1 V2 V1, t P1 P2 V2, t θ β FIGURE 12–35 The same velocity vectors of Fig. 12–34, but rotated by angle 𝜋/2 − 𝛽, so that the oblique shock is vertical. Normal Mach numbers Ma1, n and Ma2, n are also defined.
V1, n P1 P2 V1 V1, t Ma1, n > 1 Ma2, n < 1 Oblique shock V2, n V2, t θ β β – θ V2 cen96537_ch12_667-732.indd 693 29/12/16 6:22 pm 694 COMPRESSIBLE FLOW 𝛽, and Ma1. We begin by noting that tan 𝛽 = V1, n /V1, t and tan(𝛽 − 𝜃) = V2, n /V2, t (Fig. 12–35). But since V1, t = V2, t, we combine these two expres sions to yield V2, n V1, n = tan(𝛽−𝜃) tan 𝛽 = 2 + (k −1)Ma 2 1, n (k + 1)Ma 2 1, n = 2 + (k −1)Ma 2 1 sin2 𝛽 (k + 1)Ma 2 1 sin2 𝛽 (12–45) where we have also used Eq. 12–44 and the fourth equation of Fig. 12–36. We apply trigonometric identities for cos 2𝛽 and tan(𝛽 − 𝜃), namely, cos 2𝛽= cos2 𝛽−sin2 𝛽 and tan(𝛽−𝜃) = tan 𝛽−tan 𝜃 1 + tan 𝛽 tan 𝜃 After some algebra, Eq. 12–45 reduces to The 𝜃-𝛽-Ma relationship: tan 𝜃= 2 cot 𝛽(Ma 2 1 sin2 𝛽−1) Ma 2 1 (k + cos 2𝛽) + 2 (12–46) Equation 12–46 provides deflection angle 𝜃 as a unique function of shock angle 𝛽, specific heat ratio k, and upstream Mach number Ma1. For air (k = 1.4), we plot 𝜃 versus 𝛽 for several values of Ma1 in Fig. 12–37. We note that this plot is often presented with the axes reversed (𝛽 versus 𝜃) in compressible flow textbooks, since, physically, shock angle 𝛽 is determined by deflection angle 𝜃.
Much can be learned by studying Fig. 12–37, and we list some observa tions here: • Figure 12–37 displays the full range of possible shock waves at a given free-stream Mach number, from the weakest to the strongest. For any value of Mach number Ma1 greater than 1, the possible values of 𝜃 range from 𝜃 = 0° at some value of 𝛽 between 0 and 90°, to a maximum value 𝜃 = 𝜃max at an intermediate value of 𝛽, and then back to 𝜃 = 0° at 𝛽 = 90°. Straight oblique shocks for 𝜃 or 𝛽 outside of this range cannot and do not exist. At Ma1 = 1.5, for example, straight oblique shocks cannot exist in FIGURE 12–36 Relationships across an oblique shock for an ideal gas in terms of the normal component of upstream Mach number Ma1, n.
P02 P01 = (k + 1)Ma1, n 2 2 + (k – 1)Ma1, n 2 ] ] k/(k – 1) (k + 1) 2k Ma 2 1, n – k + 1 1/(k – 1) T2 T1 = [2 + (k – 1)Ma1, n 2 ] 2k Ma 1, n 2 – k + 1 (k + 1)2Ma1, n 2 ρ2 ρ1 = V 1, n V 2, n = (k + 1)Ma1, n 2 2 + (k – 1)Ma1, n 2 P2 P1 = 2k Ma1 2 n – k + 1 k + 1 Ma 2, n = (k – 1)Ma 2 n + 2 2k Ma1, n 2 – k + 1 h01 = h02 T01 = T02 √ [ [ , , 1 FIGURE 12–37 The dependence of straight oblique shock deflection angle 𝜃 on shock angle 𝛽 for several values of upstream Mach number Ma1. Calculations are for an ideal gas with k = 1.4. The dashed red line connects points of maximum deflection angle (𝜃 = 𝜃max). Weak oblique shocks are to the left of this line, while strong oblique shocks are to the right of this line. The dashed green line connects points where the downstream Mach number is sonic (Ma2 = 1). Supersonic downstream flow (Ma2 > 1) is to the left of this line, while subsonic downstream flow (Ma2 < 1) is to the right of this line.
0 10 20 30 40 β, degrees θ, degrees Ma2 > 1 Ma2 = 1 Ma1 ∞ θ = θmax Weak 50 1.2 1.5 2 3 10 5 60 70 80 90 0 10 20 30 40 50 Strong Ma2 < 1 cen96537_ch12_667-732.indd 694 29/12/16 6:22 pm 695 CHAPTER 12 air with shock angle 𝛽 less than about 42°, nor with deflection angle 𝜃 greater than about 12°. If the wedge half-angle is greater than 𝜃max, the shock becomes curved and detaches from the nose of the wedge, forming what is called a detached oblique shock or a bow wave (Fig. 12–38). The shock angle 𝛽 of the detached shock is 90° at the nose, but 𝛽 decreases as the shock curves downstream. Detached shocks are much more com plicated than simple straight oblique shocks to analyze. In fact, no simple solutions exist, and prediction of detached shocks requires computational methods (Chap. 15).
• Similar oblique shock behavior is observed in axisymmetric flow over cones, as in Fig. 12–39, although the 𝜃-𝛽-Ma relationship for axisymmetric flows differs from that of Eq. 12–46.
• When supersonic flow impinges on a blunt (or bluff) body—a body without a sharply pointed nose, the wedge half-angle 𝛿 at the nose is 90°, and an attached oblique shock cannot exist, regardless of Mach number. In fact, a detached oblique shock occurs in front of all such blunt-nosed bodies, whether two-dimensional, axisymmetric, or fully three-dimensional. For example, a detached oblique shock is seen in front of the space shuttle model in Fig. 12–32 and in front of a sphere in Fig. 12–40.
• While 𝜃 is a unique function of Ma1 and 𝛽 for a given value of k, there are two possible values of 𝛽 for 𝜃 < 𝜃max. The dashed red line in Fig. 12–37 passes through the locus of 𝜃max values, dividing the shocks into weak oblique shocks (the smaller value of 𝛽) and strong oblique shocks (the larger value of 𝛽). At a given value of 𝜃, the weak shock is more common and is “preferred” by the flow unless the downstream pressure conditions are high enough for the formation of a strong shock.
• For a given upstream Mach number Ma1, there is a unique value of 𝜃 for which the downstream Mach number Ma2 is exactly 1. The dashed green line in Fig. 12–37 passes through the locus of values where Ma2 = 1. To the left of this line, Ma2 > 1, and to the right of this line, Ma2 < 1. Down stream sonic conditions occur on the weak shock side of the plot, with 𝜃 very close to 𝜃max. Thus, the flow downstream of a strong oblique shock is always subsonic (Ma2 < 1). The flow downstream of a weak oblique shock remains supersonic, except for a narrow range of 𝜃 just below 𝜃max, where it is subsonic, although it is still called a weak oblique shock.
FIGURE 12–38 A detached oblique shock occurs upstream of a two-dimensional wedge of half-angle 𝛿 when 𝛿 is greater than the maximum possible deflection angle 𝜃. A A shock of this kind is called a bow wave because of its resemblance to the water wave that forms at the bow of a ship.
Ma1 Detached oblique shock δ > θmax FIGURE 12–39 Still frames from schlieren video graphy illustrating the detachment of an oblique shock from a cone with increasing cone half-angle 𝛿 in air at Mach 3. At (a) 𝛿 = 20° and (b) 𝛿 = 40°, the oblique shock remains attached, but by (c) 𝛿 = 60°, the oblique shock has detached, forming a bow wave.
© G.S. Settles, Gas Dynamics Lab, Penn State University. Used with permission. (a) (b) (c) Ma1 δ δ = 20° δ = 40° δ = 60° cen96537_ch12_667-732.indd 695 29/12/16 6:22 pm 696 COMPRESSIBLE FLOW • As the upstream Mach number approaches infinity, straight oblique shocks become possible for any 𝛽 between 0 and 90°, but the maximum possible turning angle for k = 1.4 (air) is 𝜃max ≅ 45.6°, which occurs at 𝛽 = 67.8°. Straight oblique shocks with turning angles above this value of 𝜃max are not possible, regardless of the Mach number.
• For a given value of upstream Mach number, there are two shock angles where there is no turning of the flow (𝜃 = 0°): the strong case, 𝛽 = 90°, corresponds to a normal shock, and the weak case, 𝛽 = 𝛽min, represents the weakest possible oblique shock at that Mach number, which is called a Mach wave. Mach waves are caused, for example, by very small non-uniformities on the walls of a supersonic wind tunnel (several can be seen in Figs. 12–32 and 12–39). Mach waves have no effect on the flow, since the shock is vanishingly weak. In fact, in the limit, Mach waves are isentropic. The shock angle for Mach waves is a unique function of the Mach number and is given the symbol 𝜇, not to be confused with the coefficient of viscosity. Angle 𝜇 is called the Mach angle and is found by setting 𝜃 equal to zero in Eq. 12–46, solving for 𝛽 = 𝜇, and taking the smaller root. We get Mach angle: 𝜇= sin−1(1/Ma1) (12–47) Since the specific heat ratio appears only in the denominator of Eq. 12–46, 𝜇 is independent of k. Thus, we can estimate the Mach number of any supersonic flow simply by measuring the Mach angle and applying Eq. 12–47.
Prandtl–Meyer Expansion Waves We now address situations where supersonic flow is turned in the opposite direction, such as in the upper portion of a two-dimensional wedge at an angle of attack greater than its half-angle 𝛿 (Fig. 12–41). We refer to this type of flow as an expanding flow, whereas a flow that produces an oblique shock may be called a compressing flow. As previously, the flow changes direc tion to conserve mass. However, unlike a compressing flow, an expanding flow does not result in a shock wave. Rather, a continuous expanding region called an expansion fan appears, composed of an infinite number of Mach waves called Prandtl–Meyer expansion waves. In other words, the flow does not turn suddenly, as through a shock, but gradually—each successive Mach wave turns the flow by an infinitesimal amount. Since each individual expan sion wave is nearly isentropic, the flow across the entire expansion fan is also nearly isentropic. The Mach number downstream of the expansion increases (Ma2 > Ma1), while pressure, density, and temperature decrease, just as they do in the supersonic (expanding) portion of a converging–diverging nozzle.
Prandtl–Meyer expansion waves are inclined at the local Mach angle 𝜇, as sketched in Fig. 12–41. The Mach angle of the first expansion wave is easily determined as 𝜇1 = sin−1(1/Ma1). Similarly, 𝜇2 = sin−1(1/Ma2), where we must be careful to measure the angle relative to the new direction of flow downstream of the expansion, namely, parallel to the upper wall of the wedge in Fig. 12–41 if we neglect the influence of the boundary layer along the wall. But how do we determine Ma2? It turns out that the turning angle 𝜃 across the FIGURE 12–40 Color schlieren image of Mach 3.0 flow from left to right over a sphere. A curved shock wave called a bow shock forms in front of the sphere and curves downstream.
© G.S. Settles, Gas Dynamics Lab, Penn State University. Used with permission.
FIGURE 12–41 An expansion fan in the upper portion of the flow formed by a two-dimensional wedge at an angle of attack in a supersonic flow. The flow is turned by angle 𝜃, and the Mach number increases across the expansion fan. Mach angles upstream and downstream of the expansion fan are indicated. Only three expansion waves are shown for simplicity, but in fact, there are an infinite number of them. (An oblique shock is also present in the bottom portion of this flow.) δ θ Ma1 > 1 μ1 μ2 Ma2 Expansion waves Oblique shock cen96537_ch12_667-732.indd 696 29/12/16 6:22 pm 697 CHAPTER 12 expansion fan can be calculated by integration, making use of the isentropic flow relationships. For an ideal gas, the result is (Anderson, 2003), Turning angle across an expansion fan: 𝜃= 𝜈(Ma2) −𝜈(Ma1) (12–48) where 𝜈(Ma) is an angle called the Prandtl–Meyer function (not to be con fused with the kinematic viscosity), 𝜈(Ma) = √ k + 1 k −1 tan−1(√ k −1 k + 1 (Ma2 −1)) −tan−1(√Ma2 −1) (12–49) Note that 𝜈(Ma) is an angle, and can be calculated in either degrees or radians. Physically, 𝜈(Ma) is the angle through which the flow must expand, starting with 𝜈 = 0 at Ma = 1, in order to reach a supersonic Mach number, Ma > 1.
To find Ma2 for known values of Ma1, k, and 𝜃, we calculate 𝜈(Ma1) from Eq. 12–49, 𝜈(Ma2) from Eq. 12–48, and then Ma2 from Eq. 12–49, noting that the last step involves solving an implicit equation for Ma2. Since there is no heat transfer or work, and the flow can be approximated as isentropic through the expansion, T0 and P0 remain constant, and we use the isentropic flow relations derived previously to calculate other flow properties down stream of the expansion, such as T2, 𝜌2, and P2.
Prandtl–Meyer expansion fans also occur in axisymmetric supersonic flows, as in the corners and trailing edges of a cone-cylinder (Fig. 12–42). Some very complex and, to some of us, beautiful interactions involving both shock waves and expansion waves occur in the supersonic jet produced by an “overexpanded” nozzle, as in Fig. 12–43. When such patterns are visible in the exhaust of a jet engine, pilots refer to it as a “tiger tail.” Analysis of such flows is beyond the scope of the present text; interested readers are referred to compressible flow textbooks such as Thompson (1972), Leipmann and Roshko (2001), and Anderson (2003).
FIGURE 12–42 (a) Mach 3 flow over an axisymmetric cone of 10-degree half-angle. The boundary layer becomes turbulent shortly downstream of the nose, generating Mach waves that are visible in the color schlieren image. (b) A similar pattern is seen in this color schlieren image for Mach 3 flow over an 11-degree 2-D wedge. Expansion waves are seen at the corners where the wedge flattens out.
© G.S. Settles, Gas Dynamics Lab, Penn State University. Used with permission.
(a) (b) cen96537_ch12_667-732.indd 697 29/12/16 6:22 pm 698 COMPRESSIBLE FLOW FIGURE 12–43 The complex interactions between shock waves and expansion waves in an “overexpanded” supersonic jet. (a) The flow is visualized by a schlieren-like differential interferogram. (b) Color shlieren image. (c) Tiger tail shock pattern.
(a) Reproduced courtesy of the French-German Research Institute of Saint Louis, ISL. (b) © G.S. Settles, Gas Dynamics Lab, Penn State University. Used with permission. (c) Photo Courtesy Joint Strike Fighter Program, Department of Defense and Pratt & Whitney.
(a) (b) (c) EXAMPLE 12–9 Estimation of the Mach Number from Mach Lines Estimate the Mach number of the free-stream flow upstream of the space shuttle in Fig. 12–32 from the figure alone. Compare with the known value of Mach number provided in the figure caption.
cen96537_ch12_667-732.indd 698 29/12/16 6:22 pm 699 CHAPTER 12 FIGURE 12–44 Two possible oblique shock angles, (a) 𝛽weak and (b) 𝛽strong, formed by a two-dimensional wedge of half-angle 𝛿 = 10°.
Ma1 Strong shock δ = 10° βstrong (a) (b) Ma1 Weak shock δ = 10° βweak EXAMPLE 12–10 Oblique Shock Calculations Supersonic air at Ma1 = 2.0 and 75.0 kPa impinges on a two-dimensional wedge of half-angle 𝛿 = 10° (Fig. 12–44). Calculate the two possible oblique shock angles, 𝛽weak and 𝛽strong, that could be formed by this wedge. For each case, calculate the pressure and Mach number downstream of the oblique shock, compare, and discuss.
SOLUTION We are to calculate the shock angle, Mach number, and pressure downstream of the weak and strong oblique shock formed by a two-dimensional wedge.
Assumptions 1 The flow is steady. 2 The boundary layer on the wedge is very thin.
Properties The fluid is air with k = 1.4.
Analysis Because of assumption 2, we approximate the oblique shock deflection angle as equal to the wedge half-angle, i.e., 𝜃 ≅ 𝛿 = 10°. With Ma1 = 2.0 and 𝜃 = 10°, we solve Eq. 12–46 for the two possible values of oblique shock angle 𝛽: 𝛽weak = 39.3° and 𝛽strong = 83.7°. From these values, we use the first part of Eq. 12–44 to calculate upstream normal Mach number Ma1, n, Weak shock: Ma1, n = Ma1 sin 𝛽 → Ma1, n = 2.0 sin 39.3° = 1.267 and Strong shock: Ma1, n = Ma1 sin 𝛽 → Ma1, n = 2.0 sin 83.7° = 1.988 We substitute these values of Ma1, n into the second equation of Fig. 12–36 to cal culate the downstream normal Mach number Ma2, n. For the weak shock, Ma2, n = 0.8032, and for the strong shock, Ma2, n = 0.5794. We also calculate the downstream pressure for each case, using the third equation of Fig. 12–36, which gives Weak shock: P2 P1 = 2k Ma 2 1, n −k + 1 k + 1 → P2 = (75.0 kPa) 2(1.4)(1.267)2 −1.4 + 1 1.4 + 1 = 128 kPa and Strong shock: P2 P1 = 2k Ma 2 1, n −k + 1 k + 1 → P2 = (75.0 kPa) 2(1.4)(1.988)2 −1.4 + 1 1.4 + 1 = 333 kPa SOLUTION We are to estimate the Mach number from a figure and compare it to the known value.
Analysis Using a protractor, we measure the angle of the Mach lines in the free-stream flow: 𝜇 ≅ 19°. The Mach number is obtained from Eq. 12–47, 𝜇= sin−1( 1 Ma1) → Ma1 = 1 sin 19° → Ma1 = 3.07 Our estimated Mach number agrees with the experimental value of 3.0 ± 0.1.
Discussion The result is independent of the fluid properties.
cen96537_ch12_667-732.indd 699 29/12/16 6:22 pm 700 COMPRESSIBLE FLOW Finally, we use the second part of Eq. 12–44 to calculate the downstream Mach number, Weak shock: Ma2 = Ma2, n sin(𝛽−𝜃) = 0.8032 sin(39.3° −10°) = 1.64 and Strong shock: Ma2 = Ma2, n sin(𝛽−𝜃) = 0.5794 sin(83.7° −10°) = 0.604 The changes in Mach number and pressure across the strong shock are much greater than the changes across the weak shock, as expected.
Discussion Since Eq. 12–46 is implicit in 𝛽, we solve it by an iterative approach or with an equation solver. For both the weak and strong oblique shock cases, Ma1, n is supersonic and Ma2, n is subsonic. However, Ma2 is supersonic across the weak oblique shock, but subsonic across the strong oblique shock. We could also use the normal shock tables in place of the equations, but with loss of precision.
FIGURE 12–45 An expansion fan caused by the sudden expansion of a wall with 𝛿 = 10°.
Ma1 = 2.0 Ma2 δ = 10° θ EXAMPLE 12–11 Prandtl–Meyer Expansion Wave Calculations Supersonic air at Ma1 = 2.0 and 230 kPa flows parallel to a flat wall that suddenly expands by 𝛿 = 10° (Fig. 12–45). Ignoring any effects caused by the boundary layer along the wall, calculate downstream Mach number Ma2 and pressure P2.
SOLUTION We are to calculate the Mach number and pressure downstream of a sudden expansion along a wall.
Assumptions 1 The flow is steady. 2 The boundary layer on the wall is very thin.
Properties The fluid is air with k = 1.4.
Analysis Because of assumption 2, we approximate the total deflection angle as equal to the wall expansion angle, i.e., 𝜃 ≅ 𝛿 = 10°. With Ma1 = 2.0, we solve Eq. 12–49 for the upstream Prandtl–Meyer function, 𝜈(Ma) = √ k + 1 k −1 tan−1 (√ k −1 k + 1 (Ma2 −1)) −tan−1(√Ma2 −1) = √ 1.4 + 1 1.4 −1 tan−1(√ 1.4 −1 1.4 + 1 (2.02 −1)) −tan−1(√2.02 −1) = 26.38° Next, we use Eq. 12–48 to calculate the downstream Prandtl–Meyer function, 𝜃= 𝜈(Ma2) −𝜈(Ma1) → 𝜈(Ma2) = 𝜃+ 𝜈(Ma1) = 10° + 26.38° = 36.38° Ma2 is found by solving Eq. 12–49, which is implicit—an equation solver is helpful. We get Ma2 = 2.38. There are also compressible flow calcula tors on the Internet that solve these implicit equations, along with both normal and oblique shock equations; e.g., see www.aoe.vt.edu/~devenpor/ aoe3114/calc.html.
cen96537_ch12_667-732.indd 700 29/12/16 6:22 pm 701 CHAPTER 12 We use the isentropic relations to calculate the downstream pressure, P2 = P2/P0 P1/P0 P1 = [1 + ( k −1 2 )Ma 2 2 ] −k/(k−1) [1 + ( k −1 2 )Ma 2 1 ] −k/(k−1) (230 kPa) = 126 kPa Since this is an expansion, Mach number increases and pressure decreases, as expected.
Discussion We could also solve for downstream temperature, density, etc., using the appropriate isentropic relations.
FIGURE 12–46 Many practical compressible flow problems involve combustion, which may be modeled as heat gain through the duct wall.
Fuel nozzles or spray bars Flame holders Air inlet 12–5 ■ DUCT FLOW WITH HEAT TRANSFER AND NEGLIGIBLE FRICTION (RAYLEIGH FLOW) So far we have limited our consideration mostly to isentropic flow, also called reversible adiabatic flow since it involves no heat transfer and no irreversibilities such as friction. Many compressible flow problems encoun tered in practice involve chemical reactions such as combustion, nuclear reactions, evaporation, and condensation as well as heat gain or heat loss through the duct wall. Such problems are difficult to analyze exactly since they may involve significant changes in chemical composition during flow, and the conversion of latent, chemical, and nuclear energies to thermal energy (Fig. 12–46).
The essential features of such complex flows can still be captured by a simple analysis by modeling the generation or absorption of thermal energy as heat transfer through the duct wall at the same rate and disregarding any changes in chemical composition. This simplified problem is still too com plicated for an elementary treatment of the topic since the flow may involve friction, variations in duct area, and multidimensional effects. In this sec tion, we limit our consideration to one-dimensional flow in a duct of con stant cross-sectional area with negligible frictional effects.
Consider steady one-dimensional flow of an ideal gas with constant spe cific heats through a constant-area duct with heat transfer, but with neg ligible friction. Such flows are referred to as Rayleigh flows after Lord Rayleigh (1842–1919). The conservation of mass, momentum, and energy equations for the control volume shown in Fig. 12–47 are written as follows: Continuity equation Noting that the duct cross-sectional area A is constant, the relation m .
1 = m .
2 or 𝜌1A1V1 = 𝜌2 A2V2 reduces to ρ1V1 = ρ2V2 (12–50) x-Momentum equation Noting that the frictional effects are negligible and thus there are no shear forces, and assuming there are no external and body forces, the momentum equation ∑F → = ∑ out 𝛽m · V → −∑ in 𝛽m · V → FIGURE 12–47 Control volume for flow in a constant-area duct with heat transfer and negligible friction.
P1, T1, ρ1 P2, T2, ρ2 V1 Control volume Q .
V2 cen96537_ch12_667-732.indd 701 29/12/16 6:22 pm 702 COMPRESSIBLE FLOW in the flow (or x-) direction becomes a balance between static pressure forces and momentum transfer. Noting that the flows are high speed and turbulent and we are ignoring friction, the momentum flux correc tion factor is approximately 1 (𝛽 ≅ 1) and thus can be neglected. Then, P1A1 −P2A2 = m · V2 −m · V1 → P1 −P2 = (ρ2V2)V2 −(ρ1V1)V1 or P1 + ρ1V 2 1 = P2 + ρ2V 2 2 (12–51) Energy equation The control volume involves no shear, shaft, or other forms of work, and the potential energy change is negligible. If the rate of heat transfer is Q . and the heat transfer per unit mass of fluid is q = Q . /m ., the steady-flow energy balance E .
in = E .
out becomes Q . + m ·(h1 + V 2 1 2 ) = m ·(h2 + V 2 2 2 ) → q + h1 + V 2 1 2 = h2 + V 2 2 2 (12–52) For an ideal gas with constant specific heats, ∆h = cp ∆T, and thus q = cp(T2 −T1) + V 2 2 −V 2 1 2 (12–53) or q = h02 −h01 = cp(T02 −T01) (12–54) Therefore, the stagnation enthalpy h0 and stagnation temperature T0 change during Rayleigh flow (both increase when heat is transferred to the fluid and thus q is positive, and both decrease when heat is trans ferred from the fluid and thus q is negative).
Entropy change In the absence of any irreversibilities such as friction, the entropy of a system changes by heat transfer only: it increases with heat gain, and decreases with heat loss. Entropy is a property and thus a state function, and the entropy change of an ideal gas with constant specific heats during a change of state from 1 to 2 is given by s2 −s1 = cp ln T2 T1 −R ln P2 P1 (12–55) The entropy of a fluid may increase or decrease during Rayleigh flow, depending on the direction of heat transfer.
Equation of state Noting that P = 𝜌RT, the properties P, 𝜌, and T of an ideal gas at states 1 and 2 are related to each other by P1 ρ1T1 = P2 ρ2T2 (12–56) Consider a gas with known properties R, k, and cp. For a specified inlet state 1, the inlet properties P1, T1, 𝜌1, V1, and s1 are known. The five exit cen96537_ch12_667-732.indd 702 29/12/16 6:22 pm 703 CHAPTER 12 properties P2, T2, 𝜌2, V2, and s2 can be determined from Eqs. 12–50, 12–51, 12–53, 12–55, and 12–56 for any specified value of heat transfer q. When the velocity and temperature are known, the Mach number can be deter mined from Ma = V/c = V/√kRT.
Obviously there is an infinite number of possible downstream states 2 corresponding to a given upstream state 1. A practical way of determining these downstream states is to assume various values of T2, and calculate all other properties as well as the heat transfer q for each assumed T2 from Eqs. 12–50 through 12–56. Plotting the results on a T-s diagram gives a curve passing through the specified inlet state, as shown in Fig. 12–48. The plot of Rayleigh flow on a T-s diagram is called the Rayleigh line, and several important observations can be made from this plot and the results of the calculations: 1. All the states that satisfy the conservation of mass, momentum, and energy equations as well as the property relations are on the Rayleigh line. Therefore, for a given initial state, the fluid cannot exist at any downstream state outside the Rayleigh line on a T-s diagram. In fact, the Rayleigh line is the locus of all physically attainable downstream states corresponding to an initial state.
2. Entropy increases with heat gain, and thus we proceed to the right on the Rayleigh line as heat is transferred to the fluid. The Mach number is Ma = 1 at point a, which is the point of maximum entropy (see Example 12–12 for proof ). The states on the upper arm of the Rayleigh line above point a are subsonic, and the states on the lower arm below point a are supersonic. Therefore, a process proceeds to the right on the Rayleigh line with heat addition and to the left with heat rejection regardless of the initial value of the Mach number.
3. Heating increases the Mach number for subsonic flow, but decreases it for supersonic flow. The flow Mach number approaches Ma = 1 in both cases (from 0 in subsonic flow and from ∞ in supersonic flow) during heating.
4. It is clear from the energy balance q = cp(T02 − T01) that heating increases the stagnation temperature T0 for both subsonic and supersonic flows, and cooling decreases it. (The maximum value of T0 occurs at Ma = 1.) This is also the case for the static temperature T ex cept for the narrow Mach number range of 1/√k < Ma < 1 in subsonic flow (see Example 12–12). Both temperature and the Mach number increase with heating in subsonic flow, but T reaches a maximum Tmax at Ma = 1/√k (which is 0.845 for air), and then decreases. It may seem peculiar that the temperature of a fluid drops as heat is transferred to it. But this is no more peculiar than the fluid velocity increasing in the diverging section of a converging–diverging nozzle. The cooling effect in this region is due to the large increase in the fluid velocity and the accompanying drop in temperature in accordance with the relation T0 = T + V2/2cp. Note also that heat rejection in the region 1/√k < Ma < 1 causes the fluid temperature to increase (Fig. 12–49).
5. The momentum equation P + KV = constant, where K = 𝜌V = constant (from the continuity equation), reveals that velocity and static pressure FIGURE 12–48 T-s diagram for flow in a constant-area duct with heat transfer and negligible friction (Rayleigh flow).
Mab = 1/ k Maa = 1 Ma < 1 Ma > 1 Cooling (Ma 0) Cooling (Ma ∞) Heating (Ma 1) Heating (Ma 1) Tmax smax s T a b √ FIGURE 12–49 During heating, fluid temperature always increases if the Rayleigh flow is supersonic, but the temperature may actually drop if the flow is subsonic.
T01 Supersonic flow Heating T02 > T01 T1 T2 > T1 T01 Subsonic flow Heating T02 > T01 T1 T2 > T1 or T2 < T1 cen96537_ch12_667-732.indd 703 29/12/16 6:22 pm 704 COMPRESSIBLE FLOW have opposite trends. Therefore, static pressure decreases with heat gain in subsonic flow (since velocity and the Mach number increase), but increases with heat gain in supersonic flow (since velocity and the Mach number decrease).
6. The continuity equation 𝜌V = constant indicates that density and velocity are inversely proportional. Therefore, density decreases with heat transfer to the fluid in subsonic flow (since velocity and the Mach number increase), but increases with heat gain in supersonic flow (since velocity and the Mach number decrease).
7. On the left half of Fig. 12–48, the lower arm of the Rayleigh line is steeper than the upper arm (in terms of s as a function of T), which indicates that the entropy change corresponding to a specified tem perature change (and thus a given amount of heat transfer) is larger in supersonic flow. The effects of heating and cooling on the properties of Rayleigh flow are listed in Table 12–3. Note that heating or cooling has opposite effects on most properties. Also, the stagnation pressure decreases during heating and increases during cooling regardless of whether the flow is subsonic or supersonic.
TABLE 12–3 The effects of heating and cooling on the properties of Rayleigh flow Heating Cooling Property Subsonic Supersonic Subsonic Supersonic Velocity, V Increase Decrease Decrease Increase Mach number, Ma Increase Decrease Decrease Increase Stagnation temperature, T0 Increase Increase Decrease Decrease Temperature, T Increase for Ma < 1/k1/2 Increase Decrease for Ma < 1/k1/2 Decrease Decrease for Ma > 1/k1/2 Increase for Ma > 1/k1/2 Density, 𝜌 Decrease Increase Increase Decrease Stagnation pressure, P0 Decrease Decrease Increase Increase Pressure, P Decrease Increase Increase Decrease Entropy, s Increase Increase Decrease Decrease FIGURE 12–50 The T-s diagram of Rayleigh flow considered in Example 12–12.
smax Tmax Ma < 1 = 0 Ma > 1 ( ) T a a b b ds dT s = 0 ( ) dT ds EXAMPLE 12–12 Extrema of Rayleigh Line Consider the T-s diagram of Rayleigh flow, as shown in Fig. 12–50. Using the differential forms of the conservation equations and property relations, show that the Mach number is Maa = 1 at the point of maximum entropy (point a), and Mab = 1√k at the point of maximum temperature (point b).
SOLUTION It is to be shown that Maa = 1 at the point of maximum entropy and Mab = 1√k at the point of maximum temperature on the Rayleigh line.
cen96537_ch12_667-732.indd 704 29/12/16 6:22 pm 705 CHAPTER 12 Assumptions The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid.
Analysis The differential forms of the continuity (𝜌V = constant), momentum [rearranged as P + (𝜌V )V = constant], ideal gas (P = 𝜌RT ), and enthalpy change (∆h = cp ∆T ) equations are expressed as ρV = constant → ρ dV + V dρ = 0 → dρ ρ = −dV V (1) P + (ρV)V = constant → dP + (ρV) dV = 0 → dP dV = −ρV (2) P = ρRT → dP = ρR dT + RT dρ → dP P = dT T + dρ ρ (3) The differential form of the entropy change relation (Eq. 12–40) of an ideal gas with constant specific heats is ds = cp dT T − R dP P (4) Substituting Eq. 3 into Eq. 4 gives ds = cp dT T −R( dT T + dρ ρ ) = (cp −R) dT T −R dρ ρ = R k −1 dT T −R dρ ρ (5) since cp − R = cv → kcv − R = cv → cv = R /(k − 1) Dividing both sides of Eq. 5 by dT and combining with Eq. 1, ds dT = R T(k −1) + R V dV dT (6) Dividing Eq. 3 by d V and combining it with Eqs. 1 and 2 give, after rearranging, dT dV = T V −V R (7) Substituting Eq. 7 into Eq. 6 and rearranging, ds dT = R T(k −1) + R T −V 2/R = R(kRT −V 2) T(k −1)(RT −V 2) (8) Setting ds/dT = 0 and solving the resulting equation R(kRT − V2) = 0 for V give the velocity at point a to be Va = √kRTa and Maa = Va ca = √kRTa √kRTa = 1 (9) Therefore, sonic conditions exist at point a, and thus the Mach number is 1.
Setting dT /ds = (ds /dT )−1 = 0 and solving the resulting equation T (k − 1) × (RT − V 2) = 0 for velocity at point b give Vb = √RTb and Mab = Vb cb = √RTb √kRTb = 1 √k (10) Therefore, the Mach number at point b is Mab = 1√k. For air, k = 1.4 and thus Mab = 0.845.
cen96537_ch12_667-732.indd 705 29/12/16 6:22 pm 706 COMPRESSIBLE FLOW Discussion Note that in Rayleigh flow, sonic conditions are reached as the entropy reaches its maximum value, and maximum temperature occurs during sub sonic flow.
EXAMPLE 12–13 Effect of Heat Transfer on Flow Velocity Starting with the differential form of the energy equation, show that the flow veloc ity increases with heat addition in subsonic Rayleigh flow, but decreases in super sonic Rayleigh flow.
SOLUTION It is to be shown that flow velocity increases with heat addition in subsonic Rayleigh flow and that the opposite occurs in supersonic flow.
Assumptions 1 The assumptions associated with Rayleigh flow are valid. 2 There are no work interactions and potential energy changes are negligible.
Analysis Consider heat transfer to the fluid in the differential amount of 𝛿q. The differential forms of the energy equations are expressed as 𝛿 q = dh0 = d(h + V 2 2 ) = cp dT + V dV (1) Dividing by cpT and factoring out d V/V give 𝛿 q cpT = dT T + V dV cpT = dV V ( V dV dT T + (k −1)V 2 kRT ) (2) where we also used cp = kR/ (k − 1). Noting that Ma2 = V2/c2 = V2/kRT and using Eq. 7 for dT/dV from Example 12–12 give 𝛿 q cpT = dV V ( V T ( T V −V R) + (k −1)Ma2) = dV V (1−V 2 TR + k Ma2 −Ma2) (3) Canceling the two middle terms in Eq. 3 since V2/TR = k Ma2 and rearranging give the desired relation, dV V = 𝛿 q cpT 1 (1 −Ma2) (4) In subsonic flow, 1 − Ma2 > 0 and thus heat transfer and velocity change have the same sign. As a result, heating the fluid (𝛿q > 0) increases the flow velocity while cooling decreases it. In supersonic flow, however, 1 − Ma2 < 0 and heat transfer and velocity change have opposite signs. As a result, heating the fluid (𝜹q > 0) decreases the flow velocity while cooling increases it (Fig. 12–51).
Discussion Note that heating the fluid has the opposite effect on flow velocity in subsonic and supersonic Rayleigh flows.
FIGURE 12–51 Heating increases the flow velocity in subsonic flow, but decreases it in supersonic flow.
Supersonic flow V1 V2 < V1 Subsonic flow δq δq V1 V2 > V1 Property Relations for Rayleigh Flow It is often desirable to express the variations in properties in terms of the Mach number Ma. Noting that Ma = V/c = V/√kRT and thus V = Ma√kRT, ρV2 = ρkRTMa2 = kPMa2 (12–57) cen96537_ch12_667-732.indd 706 29/12/16 6:22 pm 707 CHAPTER 12 since P = 𝜌RT. Substituting into the momentum equation (Eq. 12–51) gives P1 + kP1Ma1 2 = P2 + kP2Ma2 2, which can be rearranged as P2 P1 = 1 + kMa 2 1 1 + kMa 2 2 (12–58) Again utilizing V = Ma√kRT, the continuity equation 𝜌1V1 = 𝜌2V2 is expressed as ρ1 ρ2 = V2 V1 = Ma2√kRT2 Ma1√kRT1 = Ma2√T2 Ma1√T1 (12–59) Then the ideal-gas relation (Eq. 12–56) becomes T2 T1 = P2 P1 ρ1 ρ2 = ( 1 + kMa 2 1 1 + kMa 2 2 ) ( Ma2√T2 Ma1√T1 ) (12–60) Solving Eq. 12–60 for the temperature ratio T2/T1 gives T2 T1 = ( Ma2(1 + kMa 2 1 ) Ma1(1 + kMa 2 2 )) 2 (12–61) Substituting this relation into Eq. 12–59 gives the density or velocity ratio as ρ2 ρ1 = V1 V2 = Ma 2 1 (1 + kMa 2 2 ) Ma 2 2 (1 + kMa 2 1 ) (12–62) Flow properties at sonic conditions are usually easy to determine, and thus the critical state corresponding to Ma = 1 serves as a convenient reference point in compressible flow. Taking state 2 to be the sonic state (Ma2 = 1, and superscript is used) and state 1 to be any state (no subscript), the property relations in Eqs. 12–58, 12–61, and 12–62 reduce to (Fig. 12–52) P P = 1 + k 1 + kMa2 T T = ( Ma(1 + k) 1 + kMa2 ) 2 and V V = ρ ρ = (1 + k)Ma2 1 + kMa2 (12–63) Similar relations can be obtained for dimensionless stagnation tempera ture and stagnation pressure as follows: T0 T 0 = T0 T T T T T 0 = (1 + k −1 2 Ma2) ( Ma(1 + k) 1 + kMa2 ) 2 (1 + k −1 2 ) −1 (12–64) which simplifies to T0 T0 = (k + 1)Ma2[2 + (k −1)Ma2] (1 + kMa2)2 (12–65) Also, P0 P 0 = P0 P P P P P 0 = (1 + k −1 2 Ma2) k/(k−1) ( 1 + k 1 + kMa2) (1 + k −1 2 ) −k/(k−1) (12–66) FIGURE 12–52 Summary of relations for Rayleigh flow.
V V ρ ρ = = (1 + k)Ma2 1 + kMa2 P P = 1 + k 1 + kMa2 T T =( Ma(1 + k) 1 + kMa2) 2 P0 P0 = k + 1 1 + kMa2( 2 + (k – 1)Ma2 k + 1 ) k/(k – 1) T 0 T0= (k + 1)Ma2[2 + (k – 1)Ma2] (1 + kMa2)2 cen96537_ch12_667-732.indd 707 29/12/16 6:22 pm 708 COMPRESSIBLE FLOW which simplifies to P 0 P 0 = k + 1 1 + kMa2 ( 2 + (k −1)Ma2 k + 1 ) k/(k−1) (12–67) The five relations in Eqs. 12–63, 12–65, and 12–67 enable us to calculate the dimensionless pressure, temperature, density, velocity, stagnation tem perature, and stagnation pressure for Rayleigh flow of an ideal gas with a specified k for any given Mach number. Representative results are given in tabular and graphical form in Table A–15 for k = 1.4.
Choked Rayleigh Flow It is clear from the earlier discussions that subsonic Rayleigh flow in a duct may accelerate to sonic velocity (Ma = 1) with heating. What hap pens if we continue to heat the fluid? Does the fluid continue to acceler ate to supersonic velocities? An examination of the Rayleigh line indi cates that the fluid at the critical state of Ma = 1 cannot be accelerated to supersonic velocities by heating. Therefore, the flow is choked. This is analogous to not being able to accelerate a fluid to supersonic velocities in a converging nozzle by simply extending the converging flow section. If we keep heating the fluid, we will simply move the critical state further downstream and reduce the flow rate since fluid density at the critical state will now be lower. Therefore, for a given inlet state, the correspond ing critical state fixes the maximum possible heat transfer for steady flow (Fig. 12–53). That is, qmax = h 0 −h01 = cp(T 0 −T01) (12–68) Further heat transfer causes choking and thus the inlet state to change (e.g., inlet velocity will decrease), and the flow no longer follows the same Rayleigh line. Cooling the subsonic Rayleigh flow reduces the velocity, and the Mach number approaches zero as the temperature approaches absolute zero. Note that the stagnation temperature T0 is maximum at the critical state of Ma = 1.
In supersonic Rayleigh flow, heating decreases the flow velocity. Further heating simply increases the temperature and moves the critical state farther downstream, resulting in a reduction in the mass flow rate of the fluid. It may seem like supersonic Rayleigh flow can be cooled indefinitely, but it turns out that there is a limit. Taking the limit of Eq. 12–65 as the Mach number approaches infinity gives limMa→∞ T0 T 0 = 1 −1 k2 (12–69) which yields T0/T0 = 0.49 for k = 1.4. Therefore, if the critical stagnation temperature is 1000 K, air cannot be cooled below 490 K in Rayleigh flow. Physically this means that the flow velocity reaches infinity by the time the temperature reaches 490 K—a physical impossibility. When supersonic flow cannot be sustained, the flow undergoes a normal shock wave and becomes subsonic.
FIGURE 12–53 For a given inlet state, the maximum possible heat transfer occurs when sonic conditions are reached at the exit state.
T01 Rayleigh flow Choked flow qmax T02 = T01 T1 T2 = T cen96537_ch12_667-732.indd 708 29/12/16 6:22 pm 709 CHAPTER 12 EXAMPLE 12–14 Rayleigh Flow in a Tubular Combustor A combustion chamber consists of tubular combustors of 15-cm diameter. Compressed air enters the tubes at 550 K, 480 kPa, and 80 m/s (Fig. 12–54). Fuel with a heating value of 42,000 kJ/kg is injected into the air and is burned with an air–fuel mass ratio of 40. Approximating combustion as a heat transfer process to air, determine the temperature, pressure, velocity, and Mach number at the exit of the combustion chamber.
SOLUTION Fuel is burned in a tubular combustion chamber with compressed air. The exit temperature, pressure, velocity, and Mach number are to be determined.
Assumptions 1 The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid. 2 Combustion is complete, and it is treated as a heat addition process, with no change in the chemi cal composition of the flow. 3 The increase in mass flow rate due to fuel injection is disregarded.
Properties We take the properties of air to be k = 1.4, cp = 1.005 kJ/kg·K, and R = 0.287 kJ/kg·K.
Analysis The inlet density and mass flow rate of air are ρ1 = P1 RT1 = 480 kPa (0.287 kJ/kg·K)(550 K) = 3.041 kg/m3 m · air = ρ1A1V1 = (3.041 kg/m3) 𝜋(0.15 m)2/4 = 4.299 kg/s The mass flow rate of fuel and the rate of heat transfer are m · fuel = m · air AF = 4.299 kg/s 40 = 0.1075 kg/s Q . = m · fuel HV = (0.1075 kg/s)(42,000 kJ/kg) = 4514 kW q = Q .
m · air = 4514 kJ/s 4.299 kg/s = 1050 kJ/kg The stagnation temperature and Mach number at the inlet are T01 = T1 + V 2 1 2cp = 550 K + (80 m/s)2 2(1.005 kJ/kg·K) ( 1 kJ/kg 1000 m2/s2) = 553.2 K c1 = √kRT1 = √(1.4)(0.287 kJ/kg·K)(550 K)( 1000 m2/s2 1 kJ/kg ) = 470.1 m/s Ma1 = V1 c1 = 80 m/s 470.1 m/s = 0.1702 The exit stagnation temperature is, from the energy equation q = cp (T02 − T01), T02 = T01 + q cp = 553.2 K + 1050 kJ/kg 1.005 kJ/kg·K = 1598 K FIGURE 12–54 Schematic of the combustor tube analyzed in Example 12–14.
Combustor tube P1 = 480 kPa P2, T2, V2 T1 = 550 K V1 = 80 m/s Q .
cen96537_ch12_667-732.indd 709 29/12/16 6:22 pm 710 COMPRESSIBLE FLOW The maximum value of stagnation temperature T0 occurs at Ma = 1, and its value can be determined from Table A–15 or from Eq. 12–65. At Ma1 = 0.1702 we read T0 /T0 = 0.1291. Therefore, T 0 = T01 0.1291 = 553.2 K 0.1291 = 4284 K The stagnation temperature ratio at the exit state and the Mach number corresponding to it are, from Table A–15, T02 T 0 = 1598 K 4284 K = 0.3730 → Ma2 = 0.3142 ≅0.314 The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are (Table A–15): Ma1 = 0.1702: T1 T = 0.1541 P1 P = 2.3065 V1 V = 0.0668 Ma2 = 0.3142: T2 T = 0.4389 P2 P = 2.1086 V2 V = 0.2082 Then the exit temperature, pressure, and velocity are determined to be T2 T1 = T2/T T1/T = 0.4389 0.1541 = 2.848 → T2 = 2.848T1 = 2.848(550 K) = 1570 K P2 P1 = P2/P P1/P = 2.1086 2.3065 = 0.9142 →P2 = 0.9142P1 = 0.9142(480 kPa) = 439 kPa V2 V1 = V2/V V1/V = 0.2082 0.0668 = 3.117 → V2 = 3.117V1 = 3.117(80 m/s) = 249 m/s Discussion Note that the temperature and velocity increase and pressure decreases during this subsonic Rayleigh flow with heating, as expected. This prob lem can also be solved using appropriate relations instead of tabulated values, which can likewise be coded for convenient computer solutions.
12–6 ■ ADIABATIC DUCT FLOW WITH FRICTION (FANNO FLOW) Wall friction associated with high-speed flow through short devices with large cross-sectional areas such as large nozzles is often negligible, and flow through such devices can be approximated as being frictionless. But wall friction is significant and should be considered when studying flows through long flow sections, such as long ducts, especially when the cross-sectional area is small. In this section we consider compressible flow with significant wall friction but negligible heat transfer in ducts of constant cross-sectional area.
Consider steady, one-dimensional, adiabatic flow of an ideal gas with con stant specific heats through a constant-area duct with significant frictional effects. Such flows are referred to as Fanno flows. The conservation of cen96537_ch12_667-732.indd 710 29/12/16 6:22 pm 711 CHAPTER 12 mass, momentum, and energy equations for the control volume shown in Fig. 12–55 are written as follows: Continuity equation Noting that the duct cross-sectional area A is constant (and thus A1 = A2 = Ac), the relation m .
1 = m .
2 or 𝜌1A1V1 = 𝜌2A2V2 reduces to ρ1V1 = ρ2V2 → ρV = constant (12–70) x-Momentum equation Denoting the friction force exerted on the fluid by the inner surface of the duct by Ffriction and assuming there are no other external and body forces, the momentum equation ∑F → = ∑ out 𝛽m · V → −∑ in 𝛽m · V → in the flow direction can be expressed as P1A −P2 A −Ffriction = m · V2 −m · V1 → P1 −P2 −Ffriction A = (ρ2V2)V2 −(ρ1V1)V1 where even though there is friction at the walls, and the velocity profiles are not uniform, we approximate the momentum flux correction factor 𝛽 as 1 for simplicity since the flow is usually fully developed and turbulent. The equation is rewritten as P1 + ρ1V 2 1 = P2 + ρ2V 2 2 + Ffriction A (12–71) Energy equation The control volume involves no heat or work interactions and the potential energy change is negligible. Then the steady-flow energy balance E .
in = E .
out becomes h1 + V 2 1 2 = h2 + V 2 2 2 → h01 = h02 → h0 = h + V2 2 = constant (12–72) For an ideal gas with constant specific heats, ∆h = cp ∆T and thus T1 + V 2 1 2cp = T2 + V 2 2 2cp → T01 = T02 → T0 = T + V 2 2cp = constant (12–73) Therefore, the stagnation enthalpy h0 and stagnation temperature T0 remain constant during Fanno flow.
Entropy change In the absence of any heat transfer, the entropy of a system can be changed only by irreversibilities such as friction, whose effect is always to increase entropy. Therefore, the entropy of the fluid must increase during Fanno flow. The entropy change in this case is equivalent to entropy increase or entropy generation, and for an ideal gas with constant specific heats it is expressed as s2 −s1 = cp ln T2 T1 −R ln P2 P1 > 0 (12–74) FIGURE 12–55 Control volume for adiabatic flow in a constant-area duct with friction.
V1 V2 Control volume Ffriction x P2, T2, ρ2 P1, T1, ρ1 A1 = A2 = A cen96537_ch12_667-732.indd 711 29/12/16 6:22 pm 712 COMPRESSIBLE FLOW Equation of state Noting that P = 𝜌RT, the properties P, 𝜌, and T of an ideal gas at states 1 and 2 are related to each other by P1 ρ1T1 = P2 ρ2T2 (12–75) Consider a gas with known properties R, k, and cp flowing in a duct of constant cross-sectional area A. For a specified inlet state 1, the inlet prop erties P1, T1, 𝜌1, V1, and s1 are known. The five exit properties P2, T2, 𝜌2, V2, and s2 can be determined from Eqs. 12–70 through 12–75 for any specified value of the friction force Ffriction. Knowing the velocity and temperature, we can also determine the Mach number at the inlet and the exit from the rela tion Ma = V/c = V√kRT.
Obviously there is an infinite number of possible downstream states 2 corresponding to a given upstream state 1. A practical way of determining these downstream states is to assume various values of T2, and calculate all other properties as well as the friction force for each assumed T2 from Eqs. 12–70 through 12–75. Plotting the results on a T-s diagram gives a curve passing through the specified inlet state, as shown in Fig. 12–56. The plot of Fanno flow on a T-s diagram is called the Fanno line, and several important observations can be made from this plot and the results of calculations: 1. All states that satisfy the conservation of mass, momentum, and energy equations as well as the property relations are on the Fanno line. Therefore, for a given inlet state, the fluid cannot exist at any downstream state outside the Fanno line on a T-s diagram. In fact, the Fanno line is the locus of all possible downstream states corresponding to an initial state. Note that if there were no friction, the flow properties would have remained constant along the duct during Fanno flow.
2. Friction causes entropy to increase, and thus a process always proceeds to the right along the Fanno line. At the point of maximum entropy, the Mach number is Ma = 1. All states on the upper part of the Fanno line are subsonic, and all states on the lower part are supersonic.
3. Friction increases the Mach number for subsonic Fanno flow, but decreases it for supersonic Fanno flow. The Mach number approaches unity (Ma = 1) in both cases.
4. The energy balance requires that stagnation temperature T0 = T + V 2/2cp remain constant during Fanno flow. But the actual temperature may change. Velocity increases and thus temperature decreases during subsonic flow, but the opposite occurs during supersonic flow (Fig. 12–57).
5. The continuity equation 𝜌V = constant indicates that density and velocity are inversely proportional. Therefore, the effect of friction is to decrease density in subsonic flow (since velocity and Mach number increase), but to increase it in supersonic flow (since velocity and Mach number decrease).
The effects of friction on the properties of Fanno flow are listed in Table 12–4. Note that frictional effects on most properties in subsonic flow are opposite to those in supersonic flow. However, the effect of friction is to always decrease stagnation pressure, regardless of whether the flow is FIGURE 12–56 T-s diagram for adiabatic frictional flow in a constant-area duct (Fanno flow). Numerical values are for air with k = 1.4 and inlet conditions of T1 = 500 K, P1 = 600 kPa, V1 = 80 m/s, and an assigned value of s1 = 0.
0 0.1 Ma = 1 0.2 0.3 200 300 400 500 1 T, K s, kJ/kg . K Ma < 1 and s = smax Ma > 1 FIGURE 12–57 Friction causes the Mach number to increase and the temperature to decrease in subsonic Fanno flow, but it does the opposite in supersonic Fanno flow.
Subsonic flow Ffriction T2 < T1 T2 > T1 T1 Ma2 > Ma1 Ma1 Supersonic flow Ffriction T1 Ma2 < Ma1 Ma1 cen96537_ch12_667-732.indd 712 29/12/16 6:22 pm 713 CHAPTER 12 subsonic or supersonic. But friction has no effect on stagnation temperature since friction simply causes the mechanical energy to be converted to an equivalent amount of thermal energy.
Property Relations for Fanno Flow In compressible flow, it is convenient to express the variation of properties in terms of Mach number, and Fanno flow is no exception. However, Fanno flow involves the friction force, which is proportional to the square of the velocity even when the friction factor is constant. But in compressible flow, velocity varies significantly along the flow, and thus it is necessary to per form a differential analysis to account for the variation of the friction force properly. We begin by obtaining the differential forms of the conservation equations and property relations.
Continuity equation The differential form of the continuity equation is obtained by differentiating the continuity relation 𝜌V = constant and rearranging, ρ dV + V dρ = 0 → dρ ρ = −dV V (12–76) x-Momentum equation Noting that m .
1 = m .
2 = m . = 𝜌AV and A1 = A2 = A, applying the momentum equation ∑F → = ∑ out 𝛽m · V → −∑ in 𝛽m · V → to the differential control volume in Fig. 12–58 gives PAc −(P + dP)A −𝛿 Ffriction = m · (V + dV) −m · V where we have again approximated the momentum flux correction factor 𝛽 as 1. This equation simplifies to −dPA −𝛿 Ffriction = ρAV dV or dP + 𝛿 Ffriction A + ρV dV = 0 (12–77) FIGURE 12–58 Differential control volume for adiabatic flow in a constant-area duct with friction.
V V + dV ρ ρ + dρ dx Differential control volume δFfriction P T P + dP T + dT A1 = A2 = A x TABLE 12–4 The effects of friction on the properties of Fanno flow Property Subsonic Supersonic Velocity, V Increase Decrease Mach number, Ma Increase Decrease Stagnation temperature, T0 Constant Constant Temperature, T Decrease Increase Density, 𝜌 Decrease Increase Stagnation pressure, P0 Decrease Decrease Pressure, P Decrease Increase Entropy, s Increase Increase cen96537_ch12_667-732.indd 713 29/12/16 6:22 pm 714 COMPRESSIBLE FLOW The friction force is related to the wall shear stress 𝜏w and the local friction factor fx by 𝛿 Ffriction = 𝜏w dAs = 𝜏w p dx = ( fx 8 ρV 2) 4A Dh dx = fx 2 A dx Dh ρV 2 (12–78) where dx is the length of the flow section, p is the perimeter, and Dh = 4A/p is the hydraulic diameter of the duct (note that Dh reduces to ordinary diameter D for a duct of circular cross section). Substituting, dP + ρV 2 fx 2Dh dx + ρV dV = 0 (12–79) Noting that V = Ma√kRT and P = 𝜌RT, we have 𝜌V 2 = 𝜌kRTMa2 = kPMa2 and 𝜌V = kPMa2/V. Substituting into Eq. 12–79, 1 kMa2 dP P + fx 2Dh dx + dV V = 0 (12–80) Energy equation Noting that cp = kR/(k − 1) and V 2 = Ma2kRT, the energy equation T0 = constant or T + V 2/2cp = constant is expressed as T0 = T (1 + k −1 2 Ma2) = constant (12–81) Differentiating and rearranging give dT T = − 2(k −1)Ma2 2 + (k −1)Ma2 dMa Ma (12–82) which is an expression for the differential change in temperature in terms of a differential change in Mach number.
Mach number The Mach number relation for ideal gases can be expressed as V 2 = Ma2kRT. Differentiating and rearranging give 2V dV = 2MakRT dMa + kRMa2 dT → (12–83) 2V dV = 2 V2 Ma dMa + V2 T dT Dividing each term by 2V 2 and rearranging, dV V = dMa Ma + 1 2 dT T (12–84) Combining Eq. 12–84 with Eq. 12–82 gives the velocity change in terms of the Mach number as dV V = dMa Ma − (k −1)Ma2 2 + (k −1)Ma2 dMa Ma or dV V = 2 2 + (k −1)Ma2 dMa Ma (12–85) Ideal gas The differential form of the ideal-gas equation is obtained by differentiating the equation P = 𝜌RT, dP = ρR dT + RT dρ → dP P = dT T + dρ ρ (12–86) cen96537_ch12_667-732.indd 714 29/12/16 6:22 pm 715 CHAPTER 12 Combining with the continuity equation (Eq. 12–76) gives dP P = dT T −dV V (12–87) Now combining with Eqs. 12–82 and 12–84 gives dP P = −2 + 2(k −1)Ma2 2 + (k −1)Ma2 dMa Ma (12–88) which is an expression for differential changes in P with Ma.
Substituting Eqs. 12–85 and 12–88 into 12–80 and simplifying give the differential equation for the variation of the Mach number with x as fx Dh dx = 4(1 −Ma2) kMa3 [2 + (k −1)Ma2] dMa (12–89) Considering that all Fanno flows tend to Ma = 1, it is again convenient to use the critical point (i.e., the sonic state) as the reference point and to express flow properties relative to the critical point properties, even if the actual flow never reaches the critical point. Integrating Eq. 12–89 from any state (Ma = Ma and x = x) to the critical state (Ma = 1 and x = xcr) gives fL Dh = 1 −Ma2 kMa2 + k + 1 2k ln (k + 1)Ma2 2 + (k −1)Ma2 (12–90) where f is the average friction factor between x and xcr, which is assumed to be constant, and L = xcr − x is the channel length required for the Mach number to reach unity under the influence of wall friction. Therefore, L represents the distance between a given section where the Mach number is Ma and a section (an imaginary section if the duct is not long enough to reach Ma = 1) where sonic conditions occur (Fig. 12–59).
Note that the value of fL/Dh is fixed for a given Mach number, and thus values of fL/Dh can be tabulated versus Ma for a specified k. Also, the value of duct length L needed to reach sonic conditions (or the “sonic length”) is inversely proportional to the friction factor. Therefore, for a given Mach number, L is large for ducts with smooth surfaces and small for ducts with rough surfaces.
The actual duct length L between two sections where the Mach numbers are Ma1 and Ma2 can be determined from fL Dh = ( fL Dh ) 1 −( fL Dh ) 2 (12–91) The average friction factor f, in general, is different in different parts of the duct. If f is approximated as constant for the entire duct (including the hypo thetical extension part to the sonic state), then Eq. 12–91 simplifies to L = L 1 −L 2 ( f = constant) (12–92) Therefore, Eq. 12–90 can be used for short ducts that never reach Ma = 1 as well as long ones with Ma = 1 at the exit. The friction factor depends on the Reynolds number Re = 𝜌VDh/𝜇, which varies along the duct, and the roughness ratio 𝜀/Dh of the surface. The variation of Re is mild, however, since 𝜌V = constant (from continuity), and any change in Re is due to the variation of viscosity with temperature. FIGURE 12–59 The length L represents the distance between a given section where the Mach number is Ma and a real or imaginary section where Ma = 1.
Hypothetical duct extension to sonic state Sonic state as reference point Ma Ma = 1 P T T P V V x L L cen96537_ch12_667-732.indd 715 29/12/16 6:22 pm 716 COMPRESSIBLE FLOW Therefore, it is a reasonable approximation to evaluate f from the Moody chart or Colebrook equation discussed in Chap. 8 at the average Reynolds number and to treat it as a constant. This is the case for subsonic flow since the temperature changes involved are relatively small. The treatment of the friction factor for supersonic flow is beyond the scope of this text. The Colebrook equation is implicit in f, and thus it is more convenient to use the explicit Haaland relation expressed as 1 √f ≅−1.8 log[ 6.9 Re + ( 𝜀/D 3.7 ) 1.11 ] (12–93) The Reynolds numbers encountered in compressible flow are typically high, and at very high Reynolds numbers (fully rough turbulent flow) the friction factor is independent of the Reynolds number. For Re → ∞, the Colebrook equation reduces to 1√f = −2.0 log [(𝜀/Dh)/3.7].
Relations for other flow properties can be determined similarly by inte grating the dP/P, dT/T, and dV/V relations from Eqs. 12–79, 12–82, and 12–85, respectively, from any state (no subscript and Mach number Ma) to the sonic state (with a superscript asterisk and Ma = 1) with the following results (Fig. 12–60): P P = 1 Ma ( k + 1 2 + (k −1)Ma2) 1/2 (12–94) T T = k + 1 2 + (k −1)Ma2 (12–95) V V = ρ ρ = Ma( k + 1 2 + (k −1)Ma2) 1/2 (12–96) A similar relation can be obtained for the dimensionless stagnation pres sure as follows: P0 P 0 = P 0 P P P P P 0 =(1 + k −1 2 Ma2) k/(k−1) 1 Ma( k + 1 2 +(k −1)Ma2) 1/2 ( 1 + k −1 2 ) −k/(k−1) which simplifies to P0 P 0 = ρ0 ρ 0 = 1 Ma( 2 + (k −1)Ma2 k + 1 ) (k+1)/[2(k−1)] (12–97) Note that the stagnation temperature T0 is constant for Fanno flow, and thus T0 /T0 = 1 everywhere along the duct.
Equations 12–90 through 12–97 enable us to calculate the dimensionless pres sure, temperature, density, velocity, stagnation pressure, and fL/Dh for Fanno flow of an ideal gas with a specified k for any given Mach number. Representa tive results are given in tabular and graphical form in Table A–16 for k = 1.4.
Choked Fanno Flow It is clear from the previous discussions that friction causes subsonic Fanno flow in a constant-area duct to accelerate toward sonic velocity, and the Mach number becomes exactly unity at the exit for a certain duct length. This duct length is referred to as the maximum length, the sonic length, or the critical length, and is denoted by L. You may be curious to know what happens if we extend the duct length beyond L. In particular, does the flow accelerate to FIGURE 12–60 Summary of relations for Fanno flow.
fL Dh = 1 – Ma2 kMa2 + k + 1 2k ln (k + 1)Ma2 2 + (k – 1)Ma2 V V = ρ ρ = Ma k + 1 2 + (k – 1)Ma2 1/2 P P = 1 Ma k + 1 2 + (k – 1)Ma2 1/2 T T = k + 1 2 + (k – 1)Ma2 P0 P0 = ρ0 ρ0 = 1 Ma 2 + (k – 1)Ma2 k + 1 (k + 1)/[2(k – 1)] ( ) ( ) ( ) cen96537_ch12_667-732.indd 716 29/12/16 6:22 pm 717 CHAPTER 12 supersonic velocities? The answer to this question is a definite no since at Ma = 1 the flow is at the point of maximum entropy, and proceeding along the Fanno line to the supersonic region would require the entropy of the fluid to decrease—a violation of the second law of thermodynamics. (Note that the exit state must remain on the Fanno line to satisfy all conservation require ments.) Therefore, the flow is choked. This again is analogous to not being able to accelerate a gas to supersonic velocities in a converging nozzle by simply extending the converging flow section. If we extend the duct length beyond L anyway, we simply move the critical state further downstream and reduce the flow rate. This causes the inlet state to change (e.g., inlet velocity decreases), and the flow shifts to a different Fanno line. Further increase in duct length further decreases the inlet velocity and thus the mass flow rate.
Friction causes supersonic Fanno flow in a constant-area duct to decel erate and the Mach number to decrease toward unity. Therefore, the exit Mach number again becomes Ma = 1 if the duct length is L, as in subsonic flow. But unlike subsonic flow, increasing the duct length beyond L cannot choke the flow since it is already choked. Instead, it causes a normal shock to occur at such a location that the continuing subsonic flow becomes sonic again exactly at the duct exit (Fig. 12–61). As the duct length increases, the location of the normal shock moves further upstream. Eventually, the shock occurs at the duct inlet. Further increase in duct length moves the shock to the diverging section of the converging–diverging nozzle that originally gen erates the supersonic flow, but the mass flow rate still remains unaffected since the mass flow rate is fixed by the sonic conditions at the throat of the nozzle, and it does not change unless the conditions at the throat change.
Ma > 1 Converging– diverging nozzle Normal shock Duct inlet Duct exit Ma < 1 Ma = 1 FIGURE 12–61 If duct length L is greater than L, supersonic Fanno flow is always sonic at the duct exit. Extending the duct will only move the location of the normal shock further upstream.
P1 = 150 kPa T1 = 300 K Ma1 = 0.4 Ma2 = 1 T P V D = 3 cm L1 FIGURE 12–62 Schematic for Example 12–15.
EXAMPLE 12–15 Choked Fanno Flow in a Duct Air enters a 3-cm-diameter smooth adiabatic duct at Ma1 = 0.4, T1 = 300 K, and P1 = 150 kPa (Fig. 12–62). If the Mach number at the duct exit is 1, determine the duct length and temperature, pressure, and velocity at the duct exit. Also determine the percentage of stagnation pressure lost in the duct.
SOLUTION Air enters a constant-area adiabatic duct at a specified state and leaves at the sonic state. The duct length, exit temperature, pressure, velocity, and the percentage of stagnation pressure lost in the duct are to be determined.
Assumptions 1 The assumptions associated with Fanno flow (i.e., steady, fric tional flow of an ideal gas with constant properties through a constant cross-sec tional area adiabatic duct) are valid. 2 The friction factor is constant along the duct.
Properties We take the properties of air to be k = 1.4, cp = 1.005 kJ/kg·K, R = 0.287 kJ/kg·K, and 𝜈 = 1.58 × 10−5 m2/s.
Analysis We first determine the inlet velocity and the inlet Reynolds number, c1 = √kRT1 = √(1.4)(0.287 kJ/kg·K)(300 K)( 1000 m2/s2 1 kJ/kg ) = 347 m/s V1 = Ma1c1 = 0.4(347 m/s) = 139 m/s Re1 = V1D 𝜈 = (139 m/s)(0.03 m) 1.58 × 10−5 m2/s = 2.637 × 105 cen96537_ch12_667-732.indd 717 29/12/16 6:22 pm 718 COMPRESSIBLE FLOW The friction factor is determined from the Colebrook equation, 1 √f = −2.0 log( 𝜀/D 3.7 + 2.51 Re√f ) → 1 √f = −2.0 log( 0 3.7 + 2.51 2.637 × 105√f ) Its solution is f = 0.0148 The Fanno flow functions corresponding to the inlet Mach number of 0.4 are (Table A–16): P01 P 0 = 1.5901 T1 T = 1.1628 P1 P = 2.6958 V1 V = 0.4313 fL 1 D = 2.3085 Noting that denotes sonic conditions, which exist at the exit state, the duct length and the exit temperature, pressure, and velocity are determined to be L 1 = 2.3085D f = 2.3085(0.03 m) 0.0148 = 4.68 m T = T1 1.1628 = 300 K 1.1628 = 258 K P = P1 2.6958 = 150 kPa 2.6958 = 55.6 kPa V = V1 0.4313 = 139 m/s 0.4313 = 322 m/s Thus, for the given friction factor, the duct length must be 4.68 m for the Mach number to reach Ma = 1 at the duct exit. The fraction of inlet stagnation pressure P01 lost in the duct due to friction is P01 −P 0 P01 = 1 −P 0 P01 = 1 − 1 1.5901 = 0.371 or 37.1% Discussion This problem can also be solved using appropriate relations instead of tabulated values for the Fanno functions. Also, we determined the friction fac tor at the inlet conditions and assumed it to remain constant along the duct. To check the validity of this assumption, we calculate the friction factor at the outlet conditions. It can be shown that the friction factor at the duct outlet is 0.0121—a drop of 18 percent, which is large. Therefore, we should repeat the calcula tions using the average value of the friction factor (0.0148 + 0.0121)/2 = 0.0135. This would give the duct length to be L1 = 2.3085(0.03m)/0.0135 = 5.13 m, and we take this to be the required duct length.
EXAMPLE 12–16 Exit Conditions of Fanno Flow in a Duct Air enters a 27-m-long 5-cm-diameter adiabatic duct at V1 = 85 m/s, T1 = 450 K, and P1 = 220 kPa (Fig. 12–63). The average friction factor for the duct is estimated to be 0.023. Determine the Mach number at the duct exit and the mass flow rate of air.
SOLUTION Air enters a constant-area adiabatic duct of given length at a specified state. The exit Mach number and the mass flow rate are to be determined.
Hypothetical duct extension to sonic state Exit Ma2 Ma = 1 T T1 = 450 K P1 = 220 kPa P V1 = 85 m/s V x L = 27 m L1 L2 FIGURE 12–63 Schematic for Example 12–16.
cen96537_ch12_667-732.indd 718 29/12/16 6:22 pm 719 CHAPTER 12 Assumptions 1 The assumptions associated with Fanno flow (i.e., steady, frictional flow of an ideal gas with constant properties through a constant cross-sectional area adiabatic duct) are valid. 2 The friction factor is constant along the duct.
Properties We take the properties of air to be k = 1.4, cp = 1.005 kJ/kg·K, and R = 0.287 kJ/kg·K.
Analysis The first thing we need to know is whether the flow is choked at the exit or not. Therefore, we first determine the inlet Mach number and the corre sponding value of the function fL/Dh, c1 = √kRT1 = √(1.4)(0.287 kJ/kg·K)(450 K)( 1000 m2/s2 1 kJ/kg ) = 425 m/s Ma1 = V1 c1 = 85 m/s 425 m/s = 0.200 Corresponding to this Mach number we read, from Table A–16, (fL/Dh)1 = 14.5333. Also, using the actual duct length L, we have fL Dh = (0.023)(27 m) 0.05 m = 12.42 < 14.5333 Therefore, flow is not choked and the exit Mach number is less than 1. The function fL/Dh at the exit state is calculated from Eq. 12–91, ( fL Dh ) 2 = ( fL Dh ) 1 −fL Dh = 14.5333 −12.42 = 2.1133 The Mach number corresponding to this value of fL/D is 0.42, obtained from Table A–16. Therefore, the Mach number at the duct exit is Ma2 = 0.420 The mass flow rate of air is determined from the inlet conditions to be ρ1 = P1 RT1 = 220 kPa (0.287 kJ/kg·K)(450 K) ( 1 kJ 1 kPa·m3) = 1.703 kg/m3 m · air = ρ1A1V1 = (1.703 kg/m3) [𝜋(0.05 m)2/4] (85 m/s) = 0.284 kg/s Discussion Note that it takes a duct length of 27 m for the Mach number to increase from 0.20 to 0.42, but only 4.6 m to increase from 0.42 to 1. Therefore, the Mach number rises at a much higher rate as sonic conditions are approached. To gain some insight, let’s determine the lengths corresponding to fL/Dh values at the inlet and the exit states. Noting that f is assumed to be constant for the entire duct, the maximum (or sonic) duct lengths at the inlet and exit states are Lmax, 1 = L 1 = 14.5333 Dh f = 14.5333 0.05 m 0.023 = 31.6 m Lmax, 2 = L 2 = 2.1133 Dh f = 2.1133 0.05 m 0.023 = 4.59 m (or, Lmax, 2 = Lmax, 1 − L = 31.6 − 27 = 4.6 m). Therefore, the flow would reach sonic conditions if a 4.6-m-long section were added to the existing duct.
cen96537_ch12_667-732.indd 719 29/12/16 6:22 pm 720 COMPRESSIBLE FLOW FIGURE 12–65 Shadowgram of the swept interaction generated by a fin mounted on a flat plate at Mach 3.5. The oblique shock wave generated by the fin (at top of image) bifurcates into a “𝜆-foot” beneath which the boundary layer separates and rolls up. The airflow through the 𝜆-foot above the separation zone forms a supersonic “jet” that curves downward and impinges upon the wall. This three-dimensional interaction required a special optical technique known as conical shadowgraphy to visualize the flow.
© Photo by F.S. Alvi and Gary S. Settles. Used with permission.
FIGURE 12–64 Normal shock wave above the wing of an L-1011 commercial jet aircraft in transonic flight, made visible by background distortion of low clouds over the Pacific Ocean.
NASA Dryden Research Center. Photo by Carla Thomas.
Guest Author: Gary S. Settles, The Pennsylvania State University Shock waves and boundary layers are among nature’s most incompatible phenomena. Boundary layers, as described in Chap. 10, are susceptible to separation from aerodynamic surfaces wherever strong adverse pressure gradients occur. Shock waves, on the other hand, produce very strong adverse pressure gradients, since a finite rise in static pressure occurs across a shock wave over a negligibly short streamwise distance. Thus, when a boundary layer encounters a shock wave, a complicated flow pattern develops and the boundary layer often separates from the surface to which it was attached.
There are important cases in high-speed flight and wind tunnel testing where such a clash is unavoidable. For example, commercial jet transport aircraft cruise in the bottom edge of the transonic flow regime, where the airflow over their wings actually goes supersonic and then returns to subsonic flow through a normal shock wave (Fig. 12–64). If such an aircraft flies significantly faster than its design cruise Mach number, serious aerodynamic disturbances arise due to shock-wave/boundary-layer interactions causing flow separation on the wings. This phenomenon thus limits the speed of passenger aircraft around the world. Some military aircraft are designed to avoid this limit and fly super sonically, but shock-wave/boundary-layer interactions are still limiting factors in their engine air inlets.
The interaction of a shock wave and a boundary layer is a type of viscous–inviscid interaction in which the viscous flow in the boundary layer encounters the essentially inviscid shock wave generated in the free stream. The boundary layer is slowed and thickened by the shock and may sepa rate. The shock, on the other hand, bifurcates when flow separation occurs (Fig. 12–65). Mutual changes in both the shock and the boundary layer con tinue until an equilibrium condition is reached. Depending upon boundary conditions, the interaction can vary in either two or three dimensions and may be steady or unsteady.
Such a strongly interacting flow is difficult to analyze, and no simple solu tions exist. Moreover, in most of the problems of practical interest, the bound ary layer in question is turbulent. Modern computational methods are able to predict many features of these flows by supercomputer solutions of the Reynolds-averaged Navier–Stokes equations. Wind tunnel experiments play a key role in guiding and validating such computations. Overall, the shock-wave/ boundary-layer interaction has become one of the pacing problems of modern fluid dynamics research.
References Knight, D. D., et al., “Advances in CFD Prediction of Shock Wave Turbulent Boundary Layer Interactions,” Progress in Aerospace Sciences 39(2-3), pp. 121–184, 2003.
Alvi, F. S., and Settles, G. S., “Physical Model of the Swept Shock Wave/Boundary-Layer Interaction Flowfield,” AIAA Journal 30, pp. 2252–2258, Sept. 1992.
APPLICATION SPOTLIGHT ■ Shock-Wave/Boundary-Layer Interactions cen96537_ch12_667-732.indd 720 29/12/16 6:22 pm 721 CHAPTER 12 SUMMARY In this chapter the effects of compressibility on gas flow are examined. When dealing with compressible flow, it is convenient to combine the enthalpy and the kinetic energy of the fluid into a single term called stagnation (or total) enthalpy h0, defined as h0 = h + V 2 2 The properties of a fluid at the stagnation state are called stagnation properties and are indicated by the subscript zero. The stagnation temperature of an ideal gas with con stant specific heats is T0 = T + V 2 2cp which represents the temperature an ideal gas would attain if it is brought to rest adiabatically. The stagnation proper ties of an ideal gas are related to the static properties of the fluid by P0 P = ( T0 T ) k/(k−1) and ρ0 ρ = ( T0 T ) 1/(k−1) The velocity at which an infinitesimally small pressure wave travels through a medium is the speed of sound. For an ideal gas it is expressed as c = √( ∂P ∂ρ) s = √kRT The Mach number is the ratio of the actual velocity of the fluid to the speed of sound at the same state: Ma = V c The flow is called sonic when Ma = 1, subsonic when Ma < 1, supersonic when Ma > 1, hypersonic when Ma > > 1, and transonic when Ma ≅ 1.
Nozzles whose flow area decreases in the flow direction are called converging nozzles. Nozzles whose flow area first decreases and then increases are called converging–diverging nozzles. The location of the smallest flow area of a nozzle is called the throat. The highest velocity to which a fluid can be accelerated in a converging nozzle is the sonic velocity. Accelerating a fluid to supersonic velocities is possible only in converging–diverging nozzles. In all supersonic converging– diverging nozzles, the flow velocity at the throat is the velocity of sound.
The ratios of the stagnation to static properties for ideal gases with constant specific heats can be expressed in terms of the Mach number as T0 T = 1 + ( k −1 2 )Ma2 P0 P = [1 + ( k −1 2 )Ma2 ] k/(k−1) and ρ0 ρ = [1 + ( k −1 2 )Ma2 ] 1/(k−1) When Ma = 1, the resulting static-to-stagnation property ratios for the temperature, pressure, and density are called critical ratios and are denoted by the superscript asterisk: T T0 = 2 k + 1 P P0 = ( 2 k + 1) k/(k−1) and ρ ρ0 = ( 2 k + 1) 1/(k−1) The pressure outside the exit plane of a nozzle is called the back pressure. For all back pressures lower than P, the pressure at the exit plane of the converging nozzle is equal to P, the Mach number at the exit plane is unity, and the mass flow rate is the maximum (or choked) flow rate.
In some range of back pressure, the fluid that achieved a sonic velocity at the throat of a converging–diverging nozzle and is accelerating to supersonic velocities in the diverging section experiences a normal shock, which causes a sud den rise in pressure and temperature and a sudden drop in velocity to subsonic levels. Flow through the shock is highly irreversible, and thus it cannot be approximated as isentro pic. The properties of an ideal gas with constant specific heats before (subscript 1) and after (subscript 2) a shock are related by T01 = T02 Ma2 = √ (k −1)Ma 2 1 + 2 2kMa 2 1 −k + 1 T2 T1 = 2 + Ma 2 1 (k −1) 2 + Ma 2 2 (k −1) and P2 P1 = 1 + kMa 2 1 1 + kMa 2 2 = 2kMa 2 1 −k + 1 k + 1 These equations also hold across an oblique shock, provided that the component of the Mach number normal to the oblique shock is used in place of the Mach number.
Steady one-dimensional flow of an ideal gas with constant specific heats through a constant-area duct with heat transfer and negligible friction is referred to as Rayleigh flow. The property relations and curves for Rayleigh flow are given in Table A–15. Heat transfer during Rayleigh flow are deter mined from q = cp(T02 −T01) = cp(T2 −T1) + V 2 2 −V 2 1 2 cen96537_ch12_667-732.indd 721 29/12/16 6:22 pm 722 COMPRESSIBLE FLOW Steady, frictional, and adiabatic flow of an ideal gas with constant specific heats through a constant-area duct is referred to as Fanno flow. The channel length required for the Mach number to reach unity under the influence of wall friction is denoted by L and is expressed as fL Dh = 1 −Ma2 kMa2 + k + 1 2k ln (k + 1)Ma2 2 + (k −1)Ma2 where f is the average friction factor. The duct length between two sections where the Mach numbers are Ma1 and Ma2 is determined from fL Dh = ( fL Dh ) 1 −( fL Dh ) 2 During Fanno flow, the stagnation temperature T0 remains constant. Other property relations and curves for Fanno flow are given in Table A–16.
This chapter provides an overview of compressible flow and is intended to motivate the interested student to under take a more in-depth study of this exciting subject. Some compressible flows are analyzed in Chap. 15 using computa tional fluid dynamics.
REFERENCES AND SUGGESTED READING 1. J. D. Anderson. Modern Compressible Flow with Histori cal Perspective, 3rd ed. New York: McGraw-Hill, 2003.
2. Y. A. Çengel and M. A. Boles. Thermodynamics: An Engineering Approach, 8th ed. New York: McGraw-Hill Education, 2015.
3. H. Cohen, G. F. C. Rogers, and H. I. H. Saravanamuttoo. Gas Turbine Theory, 3rd ed. New York: Wiley, 1987.
4. W. J. Devenport. Compressible Aerodynamic Calculator,
5. R. W. Fox and A. T. McDonald. Introduction to Fluid Mechanics, 8th ed. New York: Wiley, 2011.
6. H. Liepmann and A. Roshko. Elements of Gas Dynamics, Dover Publications, Mineola, NY, 2001.
7. C. E. Mackey, responsible NACA officer and curator. Equations, Tables, and Charts for Compressible Flow. NACA Report 1135.
8. A. H. Shapiro. The Dynamics and Thermodynamics of Compressible Fluid Flow, vol. 1. New York: Ronald Press Company, 1953.
9. P. A. Thompson. Compressible-Fluid Dynamics, New York: McGraw-Hill, 1972.
10. United Technologies Corporation. The Aircraft Gas Turbine and Its Operation, 1982.
11. M. Van Dyke, An Album of Fluid Motion. Stanford, CA: The Parabolic Press, 1982.
12. F. M. White. Fluid Mechanics, 7th ed. New York: McGraw-Hill, 2010.
PROBLEMS Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and SI users can ignore them. Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software.
Stagnation Properties 12–1C What is dynamic temperature?
12–2C A high-speed aircraft is cruising in still air. How does the temperature of air at the nose of the aircraft differ from the temperature of air at some distance from the aircraft?
12–3C In air-conditioning applications, the temperature of air is measured by inserting a probe into the flow stream. Thus, the probe actually measures the stagnation temperature. Does this cause any significant error?
12–4C How and why is the stagnation enthalpy h0 defined? How does it differ from ordinary (static) enthalpy?
12–5 Calculate the stagnation temperature and pressure for the following substances flowing through a duct: (a) helium at 0.25 MPa, 50°C, and 290 m/s; (b) nitrogen at 0.15 MPa, 50°C, and 300 m/s; and (c) steam at 0.1 MPa, 350°C, and 340 m/s.
12–6 Determine the stagnation temperature and stagnation pressure of air that is flowing at 36 kPa, 238 K, and 325 m/s. Answers: 291 K, 72.4 kPa cen96537_ch12_667-732.indd 722 29/12/16 6:22 pm 723 CHAPTER 12 12–7 Air flows through a device such that the stagnation pressure is 0.4 MPa, the stagnation temperature is 400°C, and the velocity is 520 m/s. Determine the static pressure and temperature of the air at this state. Answers: 545 K, 0.184 MPa 12–8E Steam flows through a device with a stagnation pressure of 120 psia, a stagnation temperature of 700°F, and a velocity of 900 ft/s. Assuming ideal-gas behavior, determine the static pressure and temperature of the steam at this state.
12–9 Air enters a compressor with a stagnation pressure of 100 kPa and a stagnation temperature of 35°C, and it is com pressed to a stagnation pressure of 900 kPa. Assuming the compression process to be isentropic, determine the power input to the compressor for a mass flow rate of 0.04 kg/s. Answer: 10.8 kW 12–10 Products of combustion enter a gas turbine with a stagnation pressure of 0.90 MPa and a stagnation temperature of 840°C, and they expand to a stagnation pressure of 100 kPa. Taking k = 1.33 and R = 0.287 kJ/kg·K for the products of combustion, and assuming the expansion process to be isen tropic, determine the power output of the turbine per unit mass flow.
One-Dimensional Isentropic Flow 12–11C Is it possible to accelerate a gas to a supersonic velocity in a converging nozzle? Explain.
12–12C A gas at a specified stagnation temperature and pressure is accelerated to Ma = 2 in a converging–diverging nozzle and to Ma = 3 in another nozzle. What can you say about the pressures at the throats of these two nozzles?
12–13C A gas initially at a subsonic velocity enters an adi abatic diverging duct. Discuss how this affects (a) the veloc ity, (b) the temperature, (c) the pressure, and (d ) the density of the fluid.
12–14C A gas initially at a supersonic velocity enters an adiabatic converging duct. Discuss how this affects (a) the velocity, (b) the temperature, (c) the pressure, and (d ) the density of the fluid.
12–15C A gas initially at a subsonic velocity enters an adiabatic converging duct. Discuss how this affects (a) the velocity, (b) the temperature, (c) the pressure, and (d ) the density of the fluid.
12–16C A gas initially at a supersonic velocity enters an adiabatic diverging duct. Discuss how this affects (a) the velocity, (b) the temperature, (c) the pressure, and (d ) the density of the fluid.
12–17C Consider a converging nozzle with sonic speed at the exit plane. Now the nozzle exit area is reduced while the nozzle inlet conditions are maintained constant. What will happen to (a) the exit velocity and (b) the mass flow rate through the nozzle?
12–18 Consider a large commercial airplane cruising at a speed of 1050 km/h in air at an altitude of 10 km where the standard air temperature is −50°C. Determine if the speed of this airplane is subsonic or supersonic.
12–19 Calculate the critical temperature, pressure, and den sity of (a) air at 200 kPa, 100°C, and 325 m/s, and (b) helium at 200 kPa, 60°C, and 300 m/s.
12–20E Air at 25 psia, 320°F, and Mach number Ma = 0.7 flows through a duct. Calculate the velocity and the stag nation pressure, temperature, and density of air. Answers: 958 ft/s, 856 R, 34.7 psia, 0.109 lbm/ft3 12–21 Air enters a converging–diverging nozzle at a pressure of 1200 kPa with negligible velocity. What is the lowest pressure that can be obtained at the throat of the nozzle? Answer: 634 kPa 12–22 Helium enters a converging–diverging nozzle at 0.7 MPa, 800 K, and 100 m/s. What are the lowest temperature and pressure that can be obtained at the throat of the nozzle?
12–23 In March 2004, NASA successfully launched an experimental supersonic-combustion ramjet engine (called a scramjet) that reached a record-setting Mach number of 7. Taking the air temperature to be −50°C, determine the speed of this engine. Answer: 7550 km/h 12–24E Reconsider the scram jet engine discussed in Prob. 12–23. Determine the speed of this engine in miles per hour corresponding to a Mach number of 7 in air at a temperature of −45°F.
12–25 Air at 200 kPa, 100°C, and Mach number Ma = 0.8 flows through a duct. Calculate the velocity and the stagna tion pressure, temperature, and density of the air.
12–26 Reconsider Prob. 12–25. Using appropriate software, study the effect of Mach numbers in the range 0.1 to 2 on the velocity, stagnation pressure, tem perature, and density of air. Plot each parameter as a function of the Mach number.
12–27 An aircraft is designed to cruise at Mach number Ma = 1.1 at 12,000 m where the atmospheric temperature is 236.15 K. Determine the stagnation temperature on the leading edge of the wing.
Isentropic Flow through Nozzles 12–28C How does the parameter Ma differ from the Mach number Ma?
12–29C What would happen if we tried to further accelerate a supersonic fluid with a diverging diffuser?
cen96537_ch12_667-732.indd 723 29/12/16 6:22 pm 724 COMPRESSIBLE FLOW 12–30C Is it possible to accelerate a fluid to supersonic velocities with a velocity other than the sonic velocity at the throat? Explain 12–31C Consider a converging nozzle and a converging– diverging nozzle having the same throat areas. For the same inlet conditions, how would you compare the mass flow rates through these two nozzles?
12–32C Consider gas flow through a converging nozzle with specified inlet conditions. We know that the highest velocity the fluid can have at the nozzle exit is the sonic velocity, at which point the mass flow rate through the nozzle is a maximum. If it were possible to achieve hypersonic velocities at the nozzle exit, how would it affect the mass flow rate through the nozzle?
12–33C Consider subsonic flow in a converging nozzle with fixed inlet conditions. What is the effect of dropping the back pressure to the critical pressure on (a) the exit velocity, (b) the exit pressure, and (c) the mass flow rate through the nozzle?
12–34C Consider the isentropic flow of a fluid through a converging–diverging nozzle with a subsonic velocity at the throat. How does the diverging section affect (a) the velocity, (b) the pressure, and (c) the mass flow rate of the fluid?
12–35C Consider subsonic flow in a converging nozzle with specified conditions at the nozzle inlet and critical pres sure at the nozzle exit. What is the effect of dropping the back pressure well below the critical pressure on (a) the exit velocity, (b) the exit pressure, and (c) the mass flow rate through the nozzle?
12–36C What would happen if we attempted to decelerate a supersonic fluid with a diverging diffuser?
12–37 Nitrogen enters a converging–diverging nozzle at 700 kPa and 400 K with a negligible velocity. Determine the critical velocity, pressure, temperature, and density in the nozzle.
12–38 Air enters a converging–diverging nozzle at 1.2 MPa with a negligible velocity. Approximating the flow as isentro pic, determine the back pressure that would result in an exit Mach number of 1.8. Answer: 209 kPa 12–39 An ideal gas flows through a passage that first con verges and then diverges during an adiabatic, reversible, steady-flow process. For subsonic flow at the inlet, sketch the variation of pressure, velocity, and Mach number along the length of the nozzle when the Mach number at the minimum flow area is equal to unity.
12–40 Repeat Prob. 12–39 for supersonic flow at the inlet.
12–41 Explain why the maximum flow rate per unit area for a given ideal gas depends only on P0 /√T0. For an ideal gas with k = 1.4 and R = 0.287 kJ/kg·K, find the constant a such that m · /A = aP0 /√T 0.
12–42 For an ideal gas obtain an expression for the ratio of the speed of sound where Ma = 1 to the speed of sound based on the stagnation temperature, c/c0.
12–43 An ideal gas with k = 1.4 is flowing through a noz zle such that the Mach number is 1.6 where the flow area is 45 cm2. Approximating the flow as isentropic, determine the flow area at the location where the Mach number is 0.8.
12–44 Repeat Prob. 12–43 for an ideal gas with k = 1.33.
12–45 Air enters a nozzle at 0.5 MPa, 420 K, and a veloc ity of 110 m/s. Approximating the flow as isentropic, deter mine the pressure and temperature of air at a location where the air velocity equals the speed of sound. What is the ratio of the area at this location to the entrance area? Answers: 355 K, 278 kPa, 0.428 12–46 Repeat Prob. 12–45 assuming the entrance velocity is negligible.
12–47 Air at 900 kPa and 400 K enters a converging nozzle with a negligible velocity. The throat area of the nozzle is 10 cm2. Approximating the flow as isentropic, calculate and plot the exit pressure, the exit velocity, and the mass flow rate versus the back pressure Pb for 0.9 ≥ Pb ≥ 0.1 MPa.
12–48 Reconsider Prob. 12–47. Using appropriate software, solve the problem for the inlet condi tions of 0.8 MPa and 1200 K.
12–49E Air enters a converging–diverging nozzle of a supersonic wind tunnel at 150 psia and 100°F with a low velocity. The flow area of the test section is equal to the exit area of the nozzle, which is 5 ft2. Calculate the pressure, tem perature, velocity, and mass flow rate in the test section for a Mach number Ma = 2. Explain why the air must be very dry for this application. Answers: 19.1 psia, 311 R, 1729 ft/s, 1435 lbm/s Shock Waves and Expansion Waves 12–50C Can the Mach number of a fluid be greater than 1 after a normal shock wave? Explain.
12–51C What do the states on the Fanno line and the Rayleigh line represent? What do the intersection points of these two curves represent?
12–52C It is claimed that an oblique shock can be analyzed like a normal shock provided that the normal component of velocity (normal to the shock surface) is used in the analysis. Do you agree with this claim?
12–53C How does the normal shock affect (a) the fluid velocity, (b) the static temperature, (c) the stagnation temper ature, (d ) the static pressure, and (e) the stagnation pressure?
12–54C How do oblique shocks occur? How do oblique shocks differ from normal shocks?
cen96537_ch12_667-732.indd 724 29/12/16 6:22 pm 725 CHAPTER 12 12–55C For an oblique shock to occur, does the upstream flow have to be supersonic? Does the flow downstream of an oblique shock have to be subsonic?
12–56C Are the isentropic relations of ideal gases appli cable for flows across (a) normal shock waves, (b) oblique shock waves, and (c) Prandtl–Meyer expansion waves?
12–57C Consider supersonic airflow approaching the nose of a two-dimensional wedge and experiencing an oblique shock. Under what conditions does an oblique shock detach from the nose of the wedge and form a bow wave? What is the numerical value of the shock angle of the detached shock at the nose?
12–58C Consider supersonic flow impinging on the rounded nose of an aircraft. Is the oblique shock that forms in front of the nose an attached or a detached shock? Explain.
12–59C Can a shock wave develop in the converging section of a converging–diverging nozzle? Explain.
12–60 Air enters a normal shock at 26 kPa, 230 K, and 815 m/s. Calculate the stagnation pressure and Mach number upstream of the shock, as well as pressure, temperature, velocity, Mach number, and stagnation pressure downstream of the shock.
12–61 Calculate the entropy change of air across the nor mal shock wave in Prob. 12–60. Answer: 0.242 kJ/kg·K 12–62 Air enters a converging–diverging nozzle with low velocity at 2.4 MPa and 120°C. If the exit area of the nozzle is 3.5 times the throat area, what must the back pressure be to produce a normal shock at the exit plane of the nozzle? Answer: 0.793 MPa 12–63 What must the back pressure be in Prob. 12–62 for a normal shock to occur at a location where the cross-sectional area is twice the throat area?
12–64E Air flowing steadily in a nozzle experiences a nor mal shock at a Mach number of Ma = 2.5. If the pressure and temperature of air are 10.0 psia and 440.5 R, respectively, upstream of the shock, calculate the pressure, temperature, velocity, Mach number, and stagnation pressure downstream of the shock. Compare these results to those for helium undergoing a normal shock under the same conditions.
12–65E Reconsider Prob. 12–64E. Using appropriate software, study the effects of both air and helium flowing steadily in a nozzle when there is a normal shock at a Mach number in the range 2 < Ma1 < 3.5. In addi tion to the required information, calculate the entropy change of the air and helium across the normal shock. Tabulate the results in a parametric table.
12–66 Air enters a converging–diverging nozzle of a super sonic wind tunnel at 1 MPa and 300 K with a low velocity. If a normal shock wave occurs at the exit plane of the noz zle at Ma = 2.4, determine the pressure, temperature, Mach number, velocity, and stagnation pressure after the shock wave. Answers: 448 kPa, 284 K, 0.523, 177 m/s, 540 kPa 12–67 Using appropriate software, calculate and plot the entropy change of air across the normal shock for upstream Mach numbers between 0.5 and 1.5 in increments of 0.1. Explain why normal shock waves can occur only for upstream Mach numbers greater than Ma = 1.
12–68 Consider supersonic airflow approaching the nose of a two-dimensional wedge at a Mach number of 3. Using Fig. 12–37, determine the minimum shock angle and the maximum deflection angle a straight oblique shock can have.
12–69 Air flowing at 32 kPa, 240 K, and Ma1 = 3.6 is forced to undergo an expansion turn of 15°. Determine the Mach number, pressure, and temperature of air after the expansion. Answers: 4.81, 6.65 kPa, 153 K 12–70 Consider the supersonic flow of air at upstream conditions of 70 kPa and 260 K and a Mach number of 2.4 over a two-dimensional wedge of half-angle 10°. If the axis of the wedge is tilted 25° with respect to the upstream air-flow, determine the downstream Mach number, pressure, and temperature above the wedge. Answers: 3.105, 23.8 kPa, 191 K Ma1 = 2.4 Ma2 25° 10° FIGURE P12–70 12–71 Reconsider Prob. 12–70. Determine the downstream Mach number, pressure, and temperature below the wedge for a strong oblique shock for an upstream Mach number of 5.
12–72E Air at 14 psia, 40°F, and a Mach number of 2.0 is forced to turn upward by a ramp that makes an 8° angle off the flow direction. As a result, a weak oblique shock forms. Determine the wave angle, Mach number, pressure, and tem perature after the shock.
12–73 Air flowing at 40 kPa, 210 K, and a Mach number of 3.4 impinges on a two-dimensional wedge of half-angle 8°. Determine the two possible oblique shock angles, 𝛽weak and 𝛽strong, that could be formed by this wedge. For each case, calculate the pressure and Mach number downstream of the oblique shock.
cen96537_ch12_667-732.indd 725 29/12/16 6:22 pm 726 COMPRESSIBLE FLOW 12–74 Air flowing steadily in a nozzle experiences a normal shock at a Mach number of Ma = 2.6. If the pressure and tem perature of air are 58 kPa and 270 K, respectively, upstream of the shock, calculate the pressure, temperature, velocity, Mach number, and stagnation pressure downstream of the shock. Compare these results to those for helium undergoing a normal shock under the same conditions.
12–75 Calculate the entropy changes of air and helium across the normal shock wave in Prob. 12–74.
12–76 For an ideal gas flowing through a normal shock, develop a relation for V2/V1 in terms of k, Ma1, and Ma2.
Duct Flow with Heat Transfer and Negligible Friction (Rayleigh Flow) 12–77C What is the characteristic aspect of Rayleigh flow? What are the main assumptions associated with Ray leigh flow?
12–78C On a T-s diagram of Rayleigh flow, what do the points on the Rayleigh line represent?
12–79C What is the effect of heat gain and heat loss on the entropy of the fluid during Rayleigh flow?
12–80C Consider subsonic Rayleigh flow of air with a Mach number of 0.92. Heat is now transferred to the fluid and the Mach number increases to 0.95. Does the tempera ture T of the fluid increase, decrease, or remain constant dur ing this process? How about the stagnation temperature T0?
12–81C What is the effect of heating the fluid on the flow velocity in subsonic Rayleigh flow? Answer the same ques tions for supersonic Rayleigh flow.
12–82C Consider subsonic Rayleigh flow that is acceler ated to sonic velocity (Ma = 1) at the duct exit by heating. If the fluid continues to be heated, will the flow at duct exit be supersonic, subsonic, or remain sonic?
12–83 Argon gas enters a constant cross-sectional area duct at Ma1 = 0.2, P1 = 320 kPa, and T1 = 400 K at a rate of 0.85 kg/s. Disregarding frictional losses, determine the highest rate of heat transfer to the argon without reducing the mass flow rate.
12–84E Air is heated as it flows through a 6 in × 6 in square duct with negligible friction. At the inlet, air is at T1 = 700 R, P1 = 80 psia, and V1 = 260 ft/s. Determine the rate at which heat must be transferred to the air to choke the flow at the duct exit, and the entropy change of air during this process.
12–85 Compressed air from the compressor of a gas turbine enters the combustion chamber at T1 = 700 K, P1 = 600 kPa, and Ma1 = 0.2 at a rate of 0.3 kg/s. Via combustion, heat is transferred to the air at a rate of 150 kJ/s as it flows through the duct with negligible friction. Determine the Mach number at the duct exit, and the drop in stagnation pressure P01 − P02 during this process. Answers: 0.271, 12.7 kPa 12–86 Repeat Prob. 12–85 for a heat transfer rate of 300 kJ/s.
12–87E Air flows with negligible friction through a 6-in-diameter duct at a rate of 9 lbm/s. The temperature and pressure at the inlet are T1 = 800 R and P1 = 30 psia, and the Mach number at the exit is Ma2 = 1. Determine the rate of heat transfer and the pressure drop for this section of the duct.
12–88 Air is heated as it flows subsonically through a duct. When the amount of heat transfer reaches 67 kJ/kg, the flow is observed to be choked, and the velocity and the static pres sure are measured to be 680 m/s and 270 kPa. Disregarding frictional losses, determine the velocity, static temperature, and static pressure at the duct inlet.
12–89 Air enters a rectangular duct at T1 = 285 K, P1 = 390 kPa, and Ma1 = 2. Heat is transferred to the air in the amount of 55 kJ/kg as it flows through the duct. Disregarding frictional losses, determine the temperature and Mach number at the duct exit. Answers: 372 K, 1.63 Air 55 kJ/kg P1 = 390 kPa T1 = 285 K Ma1 = 2 FIGURE P12–89 12–90 Repeat Prob. 12–89 assuming air is cooled in the amount of 55 kJ/kg.
12–91 Consider supersonic flow of air through a 7-cm-diameter duct with negligible friction. Air enters the duct at Ma1 = 1.8, P01 = 140 kPa, and T01 = 600 K, and it is decelerated by heat ing. Determine the highest temperature that air can be heated by heat addition while the mass flow rate remains constant.
12–92 Air enters an approximately frictionless duct with V1 = 70 m/s, T1 = 600 K, and P1 = 350 kPa. Letting the exit temperature T2 vary from 600 to 5000 K, evaluate the entropy change at intervals of 200 K, and plot the Rayleigh line on a T-s diagram.
Adiabatic Duct Flow with Friction (Fanno Flow) 12–93C On a T-s diagram of Fanno flow, what do the points on the Fanno line represent?
12–94C What is the characteristic aspect of Fanno flow? What are the main approximations associated with Fanno flow?
12–95C What is the effect of friction on the entropy of the fluid during Fanno flow?
12–96C What is the effect of friction on flow velocity in subsonic Fanno flow? Answer the same question for super sonic Fanno flow.
cen96537_ch12_667-732.indd 726 29/12/16 6:22 pm 727 CHAPTER 12 12–97C Consider supersonic Fanno flow that is deceler ated to sonic velocity (Ma = 1) at the duct exit as a result of frictional effects. If the duct length is increased further, will the flow at the duct exit be supersonic, subsonic, or remain sonic? Will the mass flow rate of the fluid increase, decrease, or remain constant as a result of increasing the duct length?
12–98C Consider supersonic Fanno flow of air with an inlet Mach number of 1.8. If the Mach number decreases to 1.2 at the duct exit as a result of friction, does the (a) stagnation tem perature T0, (b) stagnation pressure P0, and (c) entropy s of the fluid increase, decrease, or remain constant during this process?
12–99C Consider subsonic Fanno flow accelerated to sonic velocity (Ma = 1) at the duct exit as a result of frictional effects. If the duct length is increased further, will the flow at the duct exit be supersonic, subsonic, or remain sonic? Will the mass flow rate of the fluid increase, decrease, or remain constant as a result of increasing the duct length?
12–100C Consider subsonic Fanno flow of air with an inlet Mach number of 0.70. If the Mach number increases to 0.90 at the duct exit as a result of friction, will the (a) stagnation temperature T0, (b) stagnation pressure P0, and (c) entropy s of the fluid increase, decrease, or remain constant during this process?
12–101 Air enters a 12-m-long, 5-cm-diameter adiabatic duct at V1 = 70 m/s, T1 = 500 K, and P1 = 300 kPa. The average friction factor for the duct is estimated to be 0.023. Determine the Mach number at the duct exit, the exit velocity, and the mass flow rate of air.
12–102 Air enters a 5-cm-diameter, 4-m-long adiabatic duct with inlet conditions of Ma1 = 2.8, T1 = 380 K, and P1 = 80 kPa. It is observed that a normal shock occurs at a location 3 m from the inlet. Taking the average friction factor to be 0.007, determine the velocity, temperature, and pressure at the duct exit. Answers: 572 m/s, 813 K, 328 kPa FIGURE P12–102 L1 = 3 m P1 = 80 kPa T1 = 380 K Ma1 = 2.8 Normal shock 12–103E Helium gas with k = 1.667 enters a 6-in-diameter duct at Ma1 = 0.2, P1 = 60 psia, and T1 = 600 R. For an average friction factor of 0.025, determine the maximum duct length that will not cause the mass flow rate of helium to be reduced. Answer: 291 ft 12–104 Air enters a 12-cm-diameter adiabatic duct at Ma1 = 0.4, T1 = 550 K, and P1 = 200 kPa. The average friction factor for the duct is estimated to be 0.021. If the Mach number at the duct exit is 0.8, determine the duct length, temperature, pressure, and velocity at the duct exit.
L P1 = 200 kPa T1 = 550 K Ma1 = 0.4 Ma2 = 0.8 FIGURE P12–104 12–105 Air enters a 12-cm-diameter adiabatic duct with inlet conditions of V1 = 150 m/s, T1 = 500 K, and P1 = 200 kPa. For an average friction factor of 0.014, determine the duct length from the inlet where the inlet velocity doubles. Also determine the pressure drop along that section of the duct.
12–106E Air flows through a 6-in-diameter, 50-ft-long adi abatic duct with inlet conditions of V1 = 500 ft/s, T01 = 650 R, and P1 = 50 psia. For an average friction factor of 0.02, determine the velocity, temperature, and pressure at the exit of the duct.
12–107 Consider subsonic airflow through a 20-cm-diameter adiabatic duct with inlet conditions of T1 = 330 K, P1 = 180 kPa, and Ma1 = 0.1. Taking the average friction factor to be 0.02, determine the duct length required to accelerate the flow to a Mach number of unity. Also, calculate the duct length at Mach number intervals of 0.1, and plot the duct length against the Mach number for 0.1 ≤ Ma ≤ 1. Discuss the results.
12–108 Repeat Prob. 12–107 for helium gas.
12–109 Air in a room at T0 = 300 K and P0 = 100 kPa is drawn steadily by a vacuum pump through a 1.4-cm-diameter, 35-cm-long adiabatic tube equipped with a con verging nozzle at the inlet. The flow in the nozzle section can be approximated as isentropic, and the average friction factor for the duct can be taken to be 0.018. Determine the maximum mass flow rate of air that can be sucked through this tube and the Mach number at the tube inlet. Answers: 0.0305 kg/s, 0.611 L = 35 cm P0 = 100 kPa T0 = 300 K Vacuum pump D = 1.4 cm FIGURE P12–109 cen96537_ch12_667-732.indd 727 29/12/16 6:22 pm 728 COMPRESSIBLE FLOW 12–110 Repeat Prob. 12–109 for a friction factor of 0.025 and a tube length of 1 m.
12–111 Argon gas with k = 1.667, cp = 0.5203 kJ/kg·K, and R = 0.2081 kJ/kg·K enters an 8-cm-diameter adiabatic duct with V1 = 70 m/s, T1 = 520 K, and P1 = 350 kPa. Taking the average friction factor to be 0.005 and letting the exit temperature T2 vary from 540 K to 400 K, evaluate the entropy change at intervals of 10 K, and plot the Fanno line on a T-s diagram.
Review Problems 12–112 Quiescent carbon dioxide at 900 kPa and 500 K is accelerated isentropically to a Mach number of 0.6. Deter mine the temperature and pressure of the carbon dioxide after acceleration. Answers: 475 K, 728 kPa 12–113 The thrust developed by the engine of a Boeing 777 is about 380 kN. Assuming choked flow in the nozzles, determine the mass flow rate of air through the nozzle. Take the ambient conditions to be 215 K and 35 kPa.
12–114 A stationary temperature probe inserted into a duct where air is flowing at 190 m/s reads 85°C. What is the actual temperature of the air? Answer: 67.0°C 12–115 Nitrogen enters a steady-flow heat exchanger at 150 kPa, 10°C, and 100 m/s, and it receives heat in the amount of 150 kJ/kg as it flows through it. The nitrogen leaves the heat exchanger at 100 kPa with a velocity of 200 m/s. Determine the stagnation pressure and temperature of the nitrogen at the inlet and exit states.
12–116E Air flowing at 8 psia, 480 R, and Ma1 = 2.0 is forced to undergo a compression turn of 15°. Determine the Mach number, pressure, and temperature of the air after the compression.
12–117 A subsonic airplane is flying at a 5000-m altitude where the atmospheric conditions are 54 kPa and 256 K. A Pitot static probe measures the difference between the static and stagnation pressures to be 16 kPa. Calculate the speed of the airplane and the flight Mach number. Answers: 199 m/s, 0.620 12–118 Air enters a 5.5-cm-diameter adiabatic duct with inlet conditions of Ma1 = 2.2, T1 = 250 K, and P1 = 60 kPa, and exits at a Mach number of Ma2 = 1.8. Taking the average friction factor to be 0.03, determine the velocity, temperature, and pressure at the exit.
12–119 Helium enters a nozzle at 0.5 MPa, 600 K, and a velocity of 120 m/s. Assuming isentropic flow, determine the pressure and temperature of helium at a location where the velocity equals the speed of sound. What is the ratio of the area at this location to the entrance area?
12–120 Repeat Problem 12–119 assuming the entrance velocity is negligible.
12–121 Nitrogen enters a duct with varying flow area at 400 K, 100 kPa, and a Mach number of 0.3. Assuming a steady, isentropic flow, determine the temperature, pressure, and Mach number at a location where the flow area has been reduced by 20 percent.
12–122 Repeat Prob. 12–121 for an inlet Mach number of 0.5.
12–123 Nitrogen enters a converging–diverging nozzle at 620 kPa and 310 K with a negligible velocity, and it experi ences a normal shock at a location where the Mach number is Ma = 3.0. Calculate the pressure, temperature, velocity, Mach number, and stagnation pressure downstream of the shock. Compare these results to those of air undergoing a normal shock at the same conditions.
12–124 An aircraft flies with a Mach number Ma1 = 0.9 at an altitude of 7000 m where the pressure is 41.1 kPa and the temperature is 242.7 K. The diffuser at the engine inlet has an exit Mach number of Ma2 = 0.3. For a mass flow rate of 50 kg/s, determine the static pressure rise across the diffuser and the exit area.
12–125 Consider an equimolar mixture of oxygen and nitrogen. Determine the critical temperature, pressure, and density for stagnation temperature and pressure of 550 K and 350 kPa.
12–126E Helium expands in a nozzle from 220 psia, 740 R, and negligible velocity to 15 psia. Calculate the throat and exit areas for a mass flow rate of 0.2 lbm/s, assuming the nozzle is isentropic. Why must this nozzle be converging–diverging?
12–127 Helium expands in a nozzle from 0.8 MPa, 500 K, and negligible velocity to 0.1 MPa. Calculate the throat and exit areas for a mass flow rate of 0.34 kg/s, assuming the nozzle is isentropic. Why must this nozzle be converging– diverging? Answers: 5.96 cm2, 8.97 cm2 12–128 In compressible flow, velocity measurements with a Pitot probe can be grossly in error if relations developed for incompressible flow are used. Therefore, it is essential that compressible flow relations be used when evaluating flow velocity from Pitot probe measurements. Consider super sonic flow of air through a channel. A probe inserted into the flow causes a shock wave to occur upstream of the probe, and it measures the stagnation pressure and temperature to be 620 kPa and 340 K, respectively. If the static pressure upstream is 110 kPa, determine the flow velocity.
Shock wave P1 = 110 kPa P02 = 620 kPa T02 = 340 K FIGURE P12–128 cen96537_ch12_667-732.indd 728 29/12/16 6:22 pm 729 CHAPTER 12 12–129 Using appropriate software and the relations given in Table A–14, generate the one-dimensional normal shock functions by varying the upstream Mach number from 1 to 10 in increments of 0.5 for air with k = 1.4.
12–130 Repeat Prob. 12–129 for methane with k = 1.3.
12–131 Air in a room at T0 = 290 K and P0 = 90 kPa is to be drawn by a vacuum pump through a 3-cm-diameter, 2-m-long adiabatic tube equipped with a converging nozzle at the inlet. The flow in the nozzle section can be approximated as isentropic. The static pressure is measured to be 87 kPa at the tube inlet and 55 kPa at the tube exit. Determine the mass flow rate of air through the duct, the air velocity at the duct exit, and the average friction factor for the duct.
12–132 Derive an expression for the speed of sound based on van der Waals’ equation of state P = RT(v − b) − a/v2. Using this relation, determine the speed of sound in carbon dioxide at 80°C and 320 kPa, and compare your result to that obtained by assuming ideal-gas behavior. The van der Waals constants for carbon dioxide are a = 364.3 kPa·m6/kmol2 and b = 0.0427 m3/kmol.
12–133 Consider supersonic airflow through a 12-cm-diameter adiabatic duct with inlet conditions of T1 = 500 K, P1 = 80 kPa, and Ma1 = 3. Taking the aver age friction factor to be 0.03, determine the duct length required to decelerate the flow to a Mach number of unity. Also, calculate the duct length at Mach number intervals of 0.25, and plot the duct length against the Mach number for 1 ≤ Ma ≤ 3. Discuss the results.
12–134 Air is heated as it flows subsonically through a 10 cm × 10 cm square duct. The properties of air at the inlet are maintained at Ma1 = 0.6, P1 = 350 kPa, and T1 = 420 K at all times. Disregarding frictional losses, determine the highest rate of heat transfer to the air in the duct without affecting the inlet conditions. Answer: 716 kW P1 = 350 kPa T1 = 420 K Ma1 = 0.6 Qmax FIGURE P12–134 12–135 Repeat Prob. 12–134 for helium.
12–136 Air is accelerated as it is heated in a duct with neg ligible friction. Air enters at V1 = 125 m/s, T1 = 400 K, and P1 = 35 kPa and the exits at a Mach number of Ma2 = 0.8. Determine the heat transfer to the air, in kJ/kg. Also deter mine the maximum amount of heat transfer without reducing the mass flow rate of air.
12–137 Air at sonic conditions and at static temperature and pressure of 340 K and 250 kPa, respectively, is to be accelerated to a Mach number of 1.6 by cooling it as it flows through a channel with constant cross-sectional area. Disre garding frictional effects, determine the required heat transfer from the air, in kJ/kg. Answer: 47.5 kJ/kg 12–138 Combustion gases with an average specific heat ratio of k = 1.33 and a gas constant of R = 0.280 kJ/kg⋅K enter a 10-cm-diameter adiabatic duct with inlet conditions of Ma1 = 2, T1 = 510 K, and P1 = 180 kPa. If a normal shock occurs at a location 2 m from the inlet, determine the velocity, temperature, and pressure at the duct exit. Take the average friction factor of the duct to be 0.010.
12–139 Air is cooled as it flows through a 30-cm-diameter duct. The inlet conditions are Ma1 = 1.2, T01 = 350 K, and P01 = 240 kPa and the exit Mach number is Ma2 = 2.0. Disre garding frictional effects, determine the rate of cooling of air.
12–140 Using appropriate software and the relations in Table A–13, calculate the one-dimensional compressible flow functions for an ideal gas with k = 1.667, and present your results by duplicating Table A–13.
12–141 Using appropriate software and the relations in Table A–14, calculate the one-dimensional normal shock functions for an ideal gas with k = 1.667, and present your results by duplicating Table A–14.
12–142 Air is flowing through a 6-cm-diameter adiabatic duct with inlet conditions of V1 = 120 m/s, T1 = 400 K, and P1 = 100 kPa and an exit Mach num ber of Ma2 = 1. To study the effect of duct length on the mass flow rate and the inlet velocity, the duct is now extended until its length is doubled while P1 and T1 are held constant. Taking the average friction factor to be 0.02, calculate the mass flow rate, and the inlet velocity, for various extension lengths, and plot them against the extension length. Discuss the results.
12–143 Using appropriate software, determine the shape of a converging–diverging nozzle for air for a mass flow rate of 3 kg/s and inlet stagnation condi tions of 1400 kPa and 200°C. Approximate the flow as isentropic. Repeat the calculations for 50-kPa increments of pressure drop to an exit pressure of 100 kPa. Plot the nozzle to scale. Also, calculate and plot the Mach number along the nozzle.
12–144 Steam at 6.0 MPa and 700 K enters a converg ing nozzle with a negligible velocity. The noz zle throat area is 8 cm2. Approximating the flow as isentro pic, plot the exit pressure, the exit velocity, and the mass flow rate through the nozzle versus the back pressure Pb for 6.0 ≥ Pb ≥ 3.0 MPa. Treat the steam as an ideal gas with k = 1.3, cp = 1.872 kJ/kg·K, and R = 0.462 kJ/kg·K.
cen96537_ch12_667-732.indd 729 29/12/16 6:22 pm 730 COMPRESSIBLE FLOW 12–145 Find the expression for the ratio of the stagnation pressure after a shock wave to the static pressure before the shock wave as a function of k and the Mach number upstream of the shock wave Ma1.
12–146 Using appropriate software and the relations given in Table A–13, calculate the one-dimensional isentropic compressible-flow functions by varying the upstream Mach number from 1 to 10 in increments of 0.5 for air with k = 1.4.
12–147 Repeat Prob. 12–146 for methane with k = 1.3.
12–148 Plot the mass flow parameter m · √RT0 /(AP0) versus the Mach number for k = 1.2, 1.4, and 1.6 in the range of 0 ≤ Ma ≤ 1.
12–149 Obtain Eq. 12–10 by starting with Eq. 12–9 and using the cyclic rule and the thermodynamic property relations cp T = ( ∂s ∂T) P and cv T = ( ∂s ∂T) v .
12–150 For ideal gases undergoing isentropic flows, obtain expressions for P/P, T/T, and 𝜌/𝜌 as functions of k and Ma.
12–151 Using Eqs. 12–4, 12–13, and 12–14, verify that for the steady flow of ideal gases dT0/T = dA/A + (1 − Ma2) dV/V. Explain the effect of heating and area changes on the velocity of an ideal gas in steady flow for (a) subsonic flow and (b) supersonic flow.
12–152 Air flows through a converging–diverging nozzle in which the exit area is 2.896 times the throat area. Upstream of the nozzle entrance, the velocity is negligibly small and the pressure and temperature are 2.0 MPa and 150°C, respectively. Calculate the back pressure (just out side the nozzle) such that a normal shock sits right at the nozzle exit plane.
Fundamentals of Engineering (FE) Exam Problems 12–153 An aircraft is cruising in still air at 5°C at a veloc ity of 400 m/s. The air temperature at the nose of the aircraft where stagnation occurs is (a) 5°C (b) 25°C (c) 55°C (d ) 80°C (e) 85°C 12–154 Air is flowing in a wind tunnel at 25°C, 95 kPa, and 250 m/s. The stagnation pressure at the location of a probe inserted into the flow section is (a) 184 kPa (b) 98 kPa (c) 161 kPa (d ) 122 kPa (e) 135 kPa 12–155 Air is flowing in a wind tunnel at 12°C and 66 kPa at a velocity of 190 m/s. The Mach number of the flow is (a) 0.56 m/s (b) 0.65 m/s (c) 0.73 m/s (d ) 0.87 m/s (e) 1.7 m/s 12–156 An aircraft is reported to be cruising in still air at −20°C and 40 kPa at a Mach number of 0.86. The velocity of the aircraft is (a) 91 m/s (b) 220 m/s (c) 186 m/s (d ) 274 m/s (e) 378 m/s 12–157 Consider a converging nozzle with a low velocity at the inlet and sonic velocity at the exit plane. Now the nozzle exit diameter is reduced by half while the nozzle inlet tem perature and pressure are maintained the same. The nozzle exit velocity will (a) remain the same (b) double (c) quadruple (d ) go down by half (e) go down by one-fourth 12–158 Carbon dioxide enters a converging–diverging noz zle at 60 m/s, 310°C, and 300 kPa, and it leaves the nozzle at a supersonic velocity. The velocity of carbon dioxide at the throat of the nozzle is (a) 125 m/s (b) 225 m/s (c) 312 m/s (d ) 353 m/s (e) 377 m/s 12–159 Argon gas is approaching a converging–diverging nozzle with a low velocity at 20°C and 150 kPa, and it leaves the nozzle at a supersonic velocity. If the cross-sectional area of the throat is 0.015 m2, the mass flow rate of argon through the nozzle is (a) 0.47 kg/s (b) 1.7 kg/s (c) 2.6 kg/s (d ) 6.6 kg/s (e) 10.2 kg/s 12–160 Air is approaching a converging–diverging nozzle with a low velocity at 12°C and 200 kPa, and it leaves the nozzle at a supersonic velocity. The velocity of air at the throat of the nozzle is (a) 338 m/s (b) 309 m/s (c) 280 m/s (d ) 256 m/s (e) 95 m/s 12–161 Consider gas flow through a converging–diverging nozzle. Of the five following statements, select the one that is incorrect: (a) The fluid velocity at the throat can never exceed the speed of sound.
(b) If the fluid velocity at the throat is below the speed of sound, the diversion section will act like a diffuser.
(c) If the fluid enters the diverging section with a Mach number greater than one, the flow at the nozzle exit will be supersonic.
(d ) There will be no flow through the nozzle if the back pres sure equals the stagnation pressure.
(e) The fluid velocity decreases, the entropy increases, and stagnation enthalpy remains constant during flow through a normal shock.
12–162 Combustion gases with k = 1.33 enter a converging nozzle at stagnation temperature and pressure of 350°C and 400 kPa, and are discharged into the atmospheric air at 20°C and 100 kPa. The lowest pressure that will occur within the nozzle is cen96537_ch12_667-732.indd 730 29/12/16 6:22 pm 731 CHAPTER 12 (a) 13 kPa (b) 100 kPa (c) 216 kPa (d ) 290 kPa (e) 315 kPa Design and Essay Problems 12–163 Find out if there is a supersonic wind tunnel on your campus. If there is, obtain the dimensions of the wind tunnel and the temperatures and pressures as well as the Mach number at several locations during operation. For what typical experiments is the wind tunnel used?
12–164 Assuming you have a thermometer and a device to measure the speed of sound in a gas, explain how you can determine the mole fraction of helium in a mixture of helium gas and air.
12–165 Design a 1-m-long cylindrical wind tunnel whose diameter is 25 cm operating at a Mach number of 1.8. Atmo spheric air enters the wind tunnel through a converging– diverging nozzle where it is accelerated to supersonic velocities. Air leaves the tunnel through a converging– diverging diffuser where it is decelerated to a very low velocity before entering the fan section. Disregard any irre versibilities. Specify the temperatures and pressures at several locations as well as the mass flow rate of air at steady-flow conditions. Why is it often necessary to dehumidify the air before it enters the wind tunnel?
P0 Ma = 1.8 D = 25 cm T0 FIGURE P12–165 12–166 In your own words, write a summary of the dif ferences between incompressible flow, subsonic flow, and supersonic flow.
cen96537_ch12_667-732.indd 731 29/12/16 6:22 pm cen96537_ch12_667-732.indd 732 29/12/16 6:22 pm This page intentionally left blank 13 CHAPTER 733 OBJECTIVES When you finish reading this chapter, you should be able to ■ ■ Understand how flow in open channels differs from pressur ized flow in pipes ■ ■ Learn the different flow regimes in open channels and their characteristics ■ ■ Predict if hydraulic jumps are to occur during flow, and calculate the fraction of energy dissipated during hydraulic jumps ■ ■ Understand how flow rates in open channels are measured using sluice gates and weirs O P E N -C H A N NE L FLOW O pen-channel flow implies flow in a channel open to the atmo sphere, but flow in a conduit is also open-channel flow if the liq uid does not fill the conduit completely, and thus there is a free surface. An open-channel flow involves liquids only (typically water or wastewater) exposed to a gas (usually air, which is at atmospheric pressure). Flow in pipes is driven by gravity and/or a pressure difference, whereas flow in a channel is driven naturally by gravity. Water flow in a river, for example, is driven by the upstream and downstream elevation difference. The flow rate in an open channel is established by the dynamic balance between gravity and friction. Inertia of the flowing liquid also becomes important in unsteady flow. The free surface coincides with the hydraulic grade line (HGL) and the pressure is constant along the free surface. But the height of the free surface from the channel bottom and thus all dimensions of the flow cross section along the channel are not known a priori—they change along with average flow velocity. In this chapter we present the basic principles of open-channel flows and the associated correlations for steady one-dimensional flow in channels of com mon cross sections. Detailed information can be obtained from several books written on the topic, some of which are listed in the references.
In this photograph, a branch of the Okavango River meanders through the Okavango Delta, Botswana.
© Getty Images/Flicker RF cen96537_ch13_733-792.indd 733 14/01/17 3:20 pm 734 open-channel flow 13–1 ■ CLASSIFICATION OF OPEN-CHANNEL FLOWS Open-channel flow refers to the flow of liquids in channels open to the atmosphere or in partially filled conduits and is characterized by the pres ence of a liquid–gas interface called the free surface (Fig. 13–1). Most natu ral flows encountered in practice, such as the flow of water in creeks, rivers, and floods, as well as the draining of rainwater off highways, park ing lots, and roofs are open-channel flows. Human-made open-channel flow systems include irrigation systems, sewer lines, drainage ditches, and gutters, and the design of such systems is an important application area of engineering. In an open channel, the flow velocity is zero at the side and bottom surfaces because of the no-slip condition, and maximum at the midplane for sym metric geometries, typically somewhat below the free surface, as shown in Fig. 13–2. (Because of secondary flows that occur even in straight channels when they are narrow, the maximum axial velocity occurs below the free sur face, typically within the top 25 percent of depth.) Furthermore, flow velocity also varies in the flow direction in most cases. Therefore, the velocity dis tribution (and thus flow) in open channels is, in general, three-dimensional. In engineering practice, however, the equations are written in terms of the average velocity at a cross section of the channel. Since the average velocity varies only with streamwise distance x, V is a one-dimensional variable. The one-dimensionality makes it possible to solve significant real-world problems in a simple manner by hand calculations, and we restrict our consideration in this chapter to flows with one-dimensional average velocity. Despite its sim plicity, the one-dimensional equations provide remarkably accurate results and are commonly used in practice. The no-slip condition on the channel walls gives rise to velocity gradients, and wall shear stress 𝜏w develops along the wetted surfaces. The wall shear stress varies along the wetted perimeter at a given cross section and offers resistance to flow. The magnitude of this resistance depends on the viscos ity of the fluid as well as the velocity gradients at the wall surface, which in turn depend on wall roughness. Open-channel flows are also classified as being steady or unsteady. A flow is said to be steady if there is no change with time at a given location. The representative quantity in open-channel flows is the flow depth (or alter nately, the average velocity), which may vary along the channel. The flow is said to be steady if the flow depth does not vary with time at any given location along the channel (although it may vary from one location to another). Otherwise, the flow is unsteady. In this chapter we deal with steady flow only.
Uniform and Varied Flows Flow in open channels is also classified as being uniform or nonuniform (also called varied), depending on how the flow depth y (the distance of the free surface from the bottom of the channel measured in the vertical direc tion) varies along the channel. The flow in a channel is said to be uniform if the flow depth (and thus the average velocity) remains constant. Other wise, the flow is said to be nonuniform or varied, indicating that the flow (a) (b) FIGURE 13–1 Natural and human-made open-channel flows are characterized by a free surface open to the atmosphere.
(a) © Doug Sherman/Geofile RF; (b) © Corbis RF 2.0 1.5 1.0 0.5 FIGURE 13–2 Typical constant axial velocity contours in an open channel of trapezoidal cross section; values are relative to the average velocity.
cen96537_ch13_733-792.indd 734 14/01/17 3:20 pm 735 CHAPTER 13 depth varies with distance in the flow direction. Uniform flow conditions are commonly encountered in practice in long straight sections of channels with constant slope, constant roughness, and constant cross section. In open channels of constant slope and constant cross section, the liquid accelerates until the head loss due to frictional effects equals the elevation drop. The liquid at this point reaches its terminal velocity, and uniform flow is established. The flow remains uniform as long as the slope, cross section, and surface roughness of the channel remain unchanged. The flow depth in uniform flow is called the normal depth yn, which is an important charac teristic parameter for open-channel flows (Fig. 13–3). The presence of an obstruction in the channel, such as a gate or a change in slope or cross section, causes the flow depth to vary, and thus the flow to become varied or nonuniform. Such varied flows are common in both natural and human-made open channels such as rivers, irrigation systems, and sewer lines. The varied flow is called rapidly varied flow (RVF) if the flow depth changes markedly over a relatively short distance in the flow direction (such as the flow of water past a partially open gate or over a falls), and gradually varied flow (GVF) if the flow depth changes gradu ally over a long distance along the channel. A gradually varied flow region typically occurs between rapidly varied and uniform flow regions, as shown in Fig. 13–4. In gradually varied flows, we can work with the one-dimensional average velocity just as we can with uniform flows. However, average velocity is not always the most useful or most appropriate parameter for rapidly varying flows. Therefore, the analysis of rapidly varied flows is rather complicated, especially when the flow is unsteady (such as the breaking of waves on the shore). For a known discharge rate, the flow height in a gradually varied flow region (i.e., the profile of the free surface) in a specified open chan nel can be determined in a step-by-step manner by starting the analysis at a cross section where the flow conditions are known, and evaluating head loss, elevation drop, and then the average velocity for each step.
Laminar and Turbulent Flows in Channels Like pipe flow, open-channel flow can be laminar, transitional, or turbulent, depending on the value of the Reynolds number expressed as Re = ρVRh 𝜇 = VRh 𝜈 (13–1) V = constant Slope: S0 = constant y = yn = constant Uniform flow FIGURE 13–3 For uniform flow in an open channel, the flow depth y and the average flow velocity V remain constant.
GVF UF RVF GVF UF FIGURE 13–4 Uniform flow (UF), gradually varied flow (GVF), and rapidly varied flow (RVF) in an open channel.
cen96537_ch13_733-792.indd 735 14/01/17 3:20 pm 736 open-channel flow Here V is the average liquid velocity, 𝜈 is the kinematic viscosity, and Rh is the hydraulic radius defined as the ratio of the cross-sectional flow area Ac and the wetted perimeter p, Hydraulic radius: Rh = Ac p (m) (13–2) Considering that open channels come with rather irregular cross sections, the hydraulic radius serves as the characteristic dimension and brings uni formity to the treatment of open channels. Also, the Reynolds number is constant for the entire uniform flow section of an open channel. You might expect that the hydraulic radius would be defined as half the hydraulic diameter, but this is unfortunately not the case. Recall that the hydraulic diameter Dh for pipe flow is defined as Dh = 4Ac/p so that the hydraulic diameter reduces to the pipe diameter for circular pipes. The relation between hydraulic radius and hydraulic diameter is Hydraulic diameter: Dh = 4Ac p = 4Rh (13–3) So, we see that the hydraulic radius is in fact one-fourth, rather than one-half, of the hydraulic diameter (Fig. 13–5). Therefore, a Reynolds number based on the hydraulic radius is one-fourth of the Reynolds number based on hydraulic diameter as the characteris tic dimension. So it will come as no surprise that the flow is laminar for Re ≲ 2000 in pipe flow, but for Re ≲ 500 in open-channel flow. Also, open-channel flow is usually turbulent for Re ≳ 2500 and transitional for 500 ≲ Re ≲ 2500. Laminar flow is encountered when a thin layer of water (such as the rainwater draining off a road or parking lot) flows at a low velocity. The kinematic viscosity of water at 20°C is 1.00 × 10−6 m2/s, and the average flow velocity in open channels is usually above 0.5 m/s. Also, the hydraulic radius is usually greater than 0.1 m. Therefore, the Reynolds num ber associated with water flow in open channels is typically above 50,000, and thus the flow is almost always turbulent. Note that the wetted perimeter includes the sides and the bottom of the channel in contact with the liquid—it does not include the free surface and the parts of the sides exposed to air. For example, the wetted perimeter and the cross-sectional flow area for a rectangular channel of height h and width b containing water of depth y are p = b + 2y and Ac = yb, respectively. Then, Rectangular channel: Rh = Ac p = yb b + 2y = y 1 + 2y/b (13–4) As another example, the hydraulic radius for the drainage of water of depth y off a parking lot of width b is (Fig. 13–6) Liquid layer of thickness y: Rh = Ac p = yb b + 2y ≅yb b ≅y (13–5) since b ≫ y. Therefore, the hydraulic radius for the flow of a liquid film over a large surface is simply the thickness of the liquid layer.
I’ve known since grade school that radius is half of diameter. Now they tell me that hydraulic radius is one-fourth of hydraulic diameter!
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FIGURE 13–5 The relationship between the hydraulic radius and hydraulic diameter is not what you might expect.
cen96537_ch13_733-792.indd 736 14/01/17 3:20 pm 737 CHAPTER 13 13–2 ■ FROUDE NUMBER AND WAVE SPEED Open-channel flow is also classified as subcritical, critical, or supercritical, depending on the value of the dimensionless Froude number mentioned in Chap. 7 and defined as Froude number: Fr = V √gLc (13–6) where g is the gravitational acceleration, V is the average liquid velocity at a cross section, and Lc is the characteristic length. Lc is taken to be the flow depth y for wide rectangular channels, and Fr = V/√gy. The Froude number is an important parameter that governs the character of flow in open channels. The flow is classified as Fr < 1 Subcritical or tranquil flow Fr = 1 Critical flow (13–7) Fr > 1 Supercritical or rapid flow y R R θ Ac = R2(θ – sin θ cos θ) θ – sin θ cos θ 2θ θ Ac p p = 2Rθ Rh = R = y(b + y/tan θ) b + 2y/sin θ Ac p Rh = = yb b + 2y Ac p y 1 + 2y/b Rh = = = y b yb b + 2y y ≪ b (b) Trapezoidal channel (d) Liquid film of thickness y (a) Circular channel (θ in rad) (c) Rectangular channel Ac p yb b Rh = y = ≅ ≅ b y b FIGURE 13–6 Hydraulic radius relations for various open-channel geometries.
cen96537_ch13_733-792.indd 737 14/01/17 3:20 pm 738 open-channel flow This resembles the classification of compressible flow with respect to the Mach number: subsonic for Ma < 1, sonic for Ma = 1, and supersonic for Ma > 1 (Fig. 13–7). Indeed, the denominator of the Froude number has the dimensions of velocity, and it represents the speed c0 at which a small dis turbance travels in still liquid, as shown later in this section. Therefore, in analogy to the Mach number, the Froude number is expressed as the ratio of the flow speed to the wave speed, Fr = V/c0, just as the Mach number is expressed as the ratio of the flow speed to the sound speed, Ma = V/c. The Froude number can also be thought of as the square root of the ratio of inertia (or dynamic) force to gravity force (or weight). This is demon strated by multiplying both the numerator and the denominator of the square of the Froude number V 2/gLc by 𝜌A, where 𝜌 is density and A is a represen tative area, which gives Fr2 = V 2 gLc ρA ρA = 2(1 2ρV 2A) mg ∝Inertia force Gravity force (13–8) Here LcA represents volume, 𝜌LcA is the mass of this fluid volume, and mg is the weight. The numerator is twice the inertial force 1 2𝜌V 2A, which can be thought of as the dynamic pressure 1 2𝜌V 2 times the cross-sectional area, A. Therefore, the flow in an open channel is dominated by inertial forces when the Froude number is large and by gravity forces when the Froude number is small. It follows that at low flow velocities (Fr < 1), a small disturbance trav els upstream (with a velocity c0 − V relative to a stationary observer) and affects the upstream conditions. This is called subcritical or tranquil flow. But at high flow velocities (Fr > 1), a small disturbance cannot travel upstream (in fact, the wave is washed downstream at a velocity of V − c0 relative to a stationary observer) and thus the upstream conditions cannot be influenced by the downstream conditions. This is called supercritical or rapid flow, and the flow in this case is controlled by the upstream con ditions. Therefore, a surface wave travels upstream when Fr < 1, is swept downstream when Fr > 1, and appears frozen on the surface when Fr = 1. Also, when the water is shallow compared to the wavelength of the disturbance, the surface wave speed increases with flow depth y, and thus a surface distur bance propagates much faster in deep channels than it does in shallow ones. Consider the flow of a liquid in an open rectangular channel of cross-sectional area Ac with volume flow rate V ·. When the flow is critical, Fr = 1 and the average flow velocity is V = √gyc, where yc is the critical depth. Noting that V · = AcV = Ac√gyc, the critical depth is expressed as Critical depth (general): yc = V · 2 gA 2 c (13–9) For a rectangular channel of width b we have Ac = byc, and the critical depth relation reduces to Critical depth (rectangular): yc = ( V · 2 gb2) 1/3 (13–10) The liquid depth is y > yc for subcritical flow and y < yc for supercritical flow (Fig. 13–8).
Compressible Flow Open-Channel Flow Ma = V/c Fr = V/c0 Ma < 1 Subsonic Fr < 1 Subcritical Ma = 1 Sonic Fr = 1 Critical Ma > 1 Supersonic Fr > 1 Supercritical V = speed of flow c = √kRT = speed of sound (ideal gas) c0 = √gy = speed of wave (liquid) FIGURE 13–7 Analogy between the Mach number for compressible flow and the Froude number for open-channel flow.
Subcritical flow: y > yc Supercritical flow: y < yc yc y yc y FIGURE 13–8 Definitions of subcritical flow and supercritical flow in terms of critical depth.
cen96537_ch13_733-792.indd 738 14/01/17 3:20 pm 739 CHAPTER 13 As in compressible flow, a liquid can accelerate from subcritical to supercritical flow. Of course, it can also decelerate from supercritical to subcritical flow, and it can do so by undergoing a shock. The shock in this case is called a hydraulic jump, which corresponds to a normal shock in compressible flow. Therefore, the analogy between open-channel flow and compressible flow is remarkable.
Speed of Surface Waves We are all familiar with the waves forming on the free surfaces of oceans, lakes, rivers, and even swimming pools. The surface waves can be very high, like the ones we see on the oceans, or barely noticeable. Some are smooth; some break on the surface. A basic understanding of wave motion is necessary for the study of certain aspects of open-channel flow, and here we present a brief description. A detailed treatment of wave motion can be found in numerous books written on the subject. An important parameter in the study of open-channel flow is the wave speed c0, which is the speed at which a surface disturbance travels through a liquid. Consider a long, wide channel that initially contains a still liquid of height y. One end of the channel is moved with speed 𝛿V, generating a surface wave of height 𝛿y propagating at a speed of c0 into the still liquid, as shown in Fig. 13–9a. Now consider a control volume that encloses the wave front and moves with it, as shown in Fig. 13–9b. To an observer traveling with the wave front, the liquid to the right appears to be moving toward the wave front with speed c0 and the liquid to the left appears to be moving away from the wave front with speed c0 − 𝛿V. Of course the observer would think the con trol volume that encloses the wave front (and herself or himself) is station ary, and he or she would be witnessing a steady-flow process. The steady-flow mass balance m . 1 = m . 2 (or the continuity relation) for this control volume of width b is expressed as ρc0 yb = ρ(c0 −𝛿V)(y + 𝛿y)b → 𝛿V = c0 𝛿y y + 𝛿y (13–11) We make the following approximations: (1) the velocity is nearly constant across the channel and thus the momentum flux correction factors (𝛽1 and 𝛽2) are one, (2) the distance across the wave is short and thus friction at the bottom surface and air drag at the top are negligible, (3) the dynamic effects are negligible and thus the pressure in the liquid varies hydrostatically; in terms of gage pressure, P1, avg = 𝜌gh1, avg = 𝜌g(y/2) and P2, avg = 𝜌gh2, avg = 𝜌g( y + 𝛿y)/2, (4) the mass flow rate is constant with m .
1 = m .
2 = 𝜌c0yb, and (5) there are no external forces or body forces and thus the only forces acting on the control volume in the horizontal x-direction are the pressure forces. Then, the momentum equation ∑F ›= ∑ out 𝛽m · V ›−∑ in 𝛽m · V › in the x-direction becomes a balance between hydrostatic pressure forces and momentum transfer, P2, avg A2 −P1, avg A1 = m · (−V2) −m · (−V1) (13–12) Moving plate Moving wavefront (a) Generation and propagation of a wave Still liquid y c0 δy δV y Control volume (b) Control volume relative to an observer traveling with the wave, with gage pressure distributions shown c0 c0 – δV δy ρgy ρg(y + δy) (1) (2) FIGURE 13–9 The generation and analysis of a wave in an open channel.
cen96537_ch13_733-792.indd 739 14/01/17 3:21 pm 740 open-channel flow Note that both the inlet and the outlet average velocities are negative since they are in the negative x-direction. Substituting, ρg(y + 𝛿y)2b 2 −ρgy2b 2 = ρc0yb(−c0 + 𝛿V) −ρc0yb(−c0) (13–13) or, g(1 + 𝛿y 2y) 𝛿y = c0 𝛿V (13–14) Combining the momentum and continuity relations and rearranging give c 2 0 = gy(1 + 𝛿y y ) (1 + 𝛿y 2y) (13–15) Therefore, the wave speed c0 is proportional to the wave height 𝛿y. For infinitesimal surface waves, 𝛿y ≪ y and thus Infinitesimal surface waves: c0 = √gy (13–16) Therefore, the speed of infinitesimal surface waves is proportional to the square root of liquid depth. Again note that this analysis is valid only for shallow liquid bodies, such as those encountered in open channels. Other wise, the wave speed is independent of liquid depth for deep bodies of liquid, such as the oceans. The wave speed can also be determined by using the energy balance relation instead of the momentum equation together with the continuity relation. Note that the waves eventually die out because of the viscous effects that are neglected in the analysis. Also, for flow in channels of nonrectangular cross section, the hydraulic depth defined as yh = Ac/Lt where Lt is the top width of the flow section should be used in the calcula tion of Froude number in place of the flow depth y. For a half-full circular channel, for example, the hydraulic depth is yh = (𝜋R2/2)/2R = 𝜋R/4. We know from experience that when a rock is thrown into a lake, the con centric waves that form propagate evenly in all directions and vanish after some distance. But when the rock is thrown into a river, the upstream side of the wave moves upstream if the flow is tranquil or subcritical (V < c0), moves downstream if the flow is rapid or supercritical (V > c0), and remains stationary at the location where it is formed if the flow is critical (V = c0). You may be wondering why we pay so much attention to flow being subcritical or supercritical. The reason is that the character of the flow is strongly influenced by this phenomenon. For example, a rock at the riverbed may cause the water level at that location to rise or to drop, depending on whether the flow is subcritical or supercritical. Also, the liquid level drops gradually in the flow direction in subcritical flow, but a sudden rise in liq uid level, called a hydraulic jump, may occur in supercritical flow (Fr > 1) as the flow decelerates to subcritical (Fr < 1) velocities. This phenomenon can occur downstream of a sluice gate as shown in Fig. 13–10. The liquid approaches the gate with a subcritical velocity, but the upstream liquid level is sufficiently high to accelerate the liquid to a supercritical level as it passes through the gate (just like a gas flowing in a converging–diverging nozzle). But if the downstream section of the channel is not sufficiently sloped down, it cannot maintain this supercritical veloc ity, and the liquid jumps up to a higher level with a larger cross-sectional area, and thus to a lower subcritical velocity. Finally, the flow in rivers, canals, and Subcritical flow Sluice gate Hydraulic jump Supercritical flow Subcritical flow (a) FIGURE 13–10 (a) Flow through a sluice gate. (b) A hydraulic jump observed in a water channel.
(b) © Girish Kumar Rajan (b) cen96537_ch13_733-792.indd 740 14/01/17 3:21 pm 741 CHAPTER 13 irrigation systems is typically subcritical. But the flow past sluice gates and spillways is typically supercritical. You can create a beautiful hydraulic jump the next time you wash dishes (Fig. 13–11). Let the water from the faucet hit the middle of a dinner plate. As the water spreads out radially, its depth decreases and the flow is super critical. Eventually, a hydraulic jump occurs, which you can see as a sudden increase in water depth. Try it!
13–3 ■ SPECIFIC ENERGY Consider the flow of a liquid in a channel at a cross section where the flow depth is y, the average flow velocity is V, and the elevation of the bottom of the channel at that location relative to some reference datum is z. For simplicity, we ignore the variation of liquid speed over the cross section and assume the speed to be V everywhere. The total mechanical energy of this liquid in the channel in terms of heads is expressed as (Fig. 13–12) H = z + P ρg + V 2 2g = z + y + V 2 2g (13–17) where z is the elevation head, P/𝜌g = y is the gage pressure head, and V2/2g is the velocity or dynamic head. The total energy as expressed in Eq. 13–17 is not a realistic representation of the true energy of a flowing fluid since the choice of the reference datum and thus the value of the elevation head z is rather arbitrary. The intrinsic energy of a fluid at a cross section is repre sented more realistically if the reference datum is taken to be the bottom of the channel so that z = 0 there. Then the total mechanical energy of a fluid in terms of heads becomes the sum of the pressure and dynamic heads. The sum of the pressure and dynamic heads of a liquid in an open channel is called the specific energy Es and is expressed as (Bakhmeteff, 1932) Es = y + V 2 2g (13–18) as shown in Fig. 13–12. Consider flow in an open channel of rectangular cross section and of constant width b. Noting that the volume flow rate is V · = AcV = ybV, the average flow velocity is V = V .
yb (13–19) Substituting into Eq. 13–18, the specific energy becomes Es = y + V . 2 2gb2y2 (13–20) This equation is very instructive as it shows the variation of the specific energy with flow depth. During steady flow in an open channel the flow rate is constant, and a plot of Es versus y for constant V · and b is given in Fig. 13–13. We observe the following from this figure: • The distance from a point on the vertical y-axis to the curve represents the specific energy at that y-value. The part between the Es = y line and the curve corresponds to dynamic head (or kinetic energy head) of the liquid, and the remaining part to pressure head (or potential energy head).
FIGURE 13–11 A hydraulic jump can be observed on a dinner plate when (a) it is right-side-up, but not when (b) it is upside down.
Photo by Po-Ya Abel Chuang. Used by permission.
(a) (b) cen96537_ch13_733-792.indd 741 14/01/17 3:21 pm 742 open-channel flow • The specific energy tends to infinity as y → 0 (due to the velocity approaching infinity), and it becomes equal to flow depth y for large values of y (due to the velocity and thus the kinetic energy becoming very small). The specific energy reaches a minimum value Es, min at some intermediate point, called the critical point, characterized by the critical depth yc and critical velocity Vc. The minimum specific energy is also called the critical energy.
• There is a minimum specific energy Es, min required to support the specified flow rate V ·. Therefore, Es cannot be below Es, min for a given V ·.
• A horizontal line intersects the specific energy curve at one point only, and thus a fixed value of flow depth corresponds to a fixed value of specific energy. This is expected since the velocity has a fixed value when V ·, b, and y are specified. However, for Es > Es, min, a vertical line intersects the curve at two points, indicating that a flow can have two different depths (and thus two different velocities) corresponding to a fixed value of specific energy. These two depths are called alternate depths. For flow through a sluice gate with negligible frictional losses (and thus Es = constant), the upper depth corresponds to the upstream flow, and the lower depth to the downstream flow (Fig. 13–14).
• A small change in specific energy near the critical point causes a large difference between alternate depths and may cause violent fluctuations in flow level. Therefore, operation near the critical point should be avoided in the design of open channels. The value of the minimum specific energy and the critical depth at which it occurs is determined by differentiating Es from Eq. 13–20 with respect to y for constant b and V ·, and setting the derivative equal to zero: dEs dy = d dy (y + V .
2 2gb2y2) = 1 − V .
2 gb2y3 = 0 (13–21) Solving for y, which is the critical flow depth yc, gives yc = ( V . 2 gb 2) 1/3 (13–22) The flow rate at the critical point can be expressed as V · = ycbVc. Substitut ing, the critical velocity is determined to be Vc = √gyc (13–23) which is the wave speed. The Froude number at this point is Fr = V √gy = Vc √gyc = 1 (13–24) indicating that the point of minimum specific energy is indeed the critical point, and the flow becomes critical when the specific energy reaches its minimum value. It follows that the flow is subcritical at lower flow velocities and thus higher flow depths (the upper arm of the curve in Fig. 13–13), supercritical at higher velo cities and thus lower flow depths (the lower arm of the curve), and critical at the critical point (the point of minimum specific energy).
z y Es V2 2g Energy line Reference datum FIGURE 13–12 The specific energy Es of a liquid in an open channel is the total mechanical energy (expressed as a head) relative to the bottom of the channel.
y Es Es, min Es = y Subcritical flow, Fr < 1 Fr = 1 Critical depth Supercritical flow, Fr > 1 yc y V2 2g .
V = constant FIGURE 13–13 Variation of specific energy Es with depth y for a specified flow rate.
y1 V1 Sluice gate V2 y2 FIGURE 13–14 A sluice gate illustrates alternate depths—the deep liquid upstream of the sluice gate and the shallow liquid downstream of the sluice gate.
cen96537_ch13_733-792.indd 742 14/01/17 3:21 pm 743 CHAPTER 13 Noting that Vc = √gyc, the minimum (or critical) specific energy can be expressed in terms of the critical depth alone as Es, min = yc + V 2 c 2g = yc + gyc 2g = 3 2 yc (13–25) In uniform flow, the flow depth and the flow velocity, and thus the spe cific energy, remain constant since Es = y + V 2/2g. The head loss is made up by the decline in elevation (the channel is sloped downward in the flow direction). In nonuniform flow, however, the specific energy may increase or decrease, depending on the slope of the channel and the frictional losses. If the decline in elevation across a flow section is more than the head loss in that section, for example, the specific energy increases by an amount equal to the difference between elevation drop and head loss. The specific energy concept is a particularly useful tool when studying varied flows.
EXAMPLE 13–1 Character of Flow and Alternate Depth Water is flowing steadily in a 0.4-m-wide rectangular open channel at a rate of 0.2 m3/s (Fig. 13–15). If the flow depth is 0.15 m, determine the flow velocity and if the flow is subcritical or supercritical. Also determine the alternate flow depth if the character of flow were to change.
SOLUTION Water flow in a rectangular open channel is considered. The character of flow, the flow velocity, and the alternate depth are to be determined. Assumptions The specific energy is constant.
Analysis The average flow velocity is determined from V = V .
Ac = V .
yb = 0.2 m3/s (0.15 m)(0.4 m) = 3.33 m/s The critical depth for this flow is yc = ( V . 2 gb2) 1/3 = ( (0.2 m3/s)2 (9.81 m/s2)(0.4 m)2) 1/3 = 0.294 m Therefore, the flow is supercritical since the actual flow depth is y = 0.15 m, and y < yc. Another way to determine the character of flow is to calculate the Froude number, Fr = V √gy = 3.33 m/s √(9.81 m/s2)(0.15 m) = 2.75 Again the flow is supercritical since Fr > 1. The specific energy for the given conditions is Es1 = y1 + V .
2 2gb2y 2 1 = (0.15 m) + (0.2 m3/s)2 2(9.81 m/s2)(0.4 m)2(0.15 m)2 = 0.7163 m Then the alternate depth is determined from Es1 = Es2 to be Es2 = y2 + V .
2 2gb2y 2 2 → 0.7163 m = y2 + (0.2 m3/s)2 2(9.81 m/s2)(0.4 m)2y 2 2 Solving for y2 gives the alternate depth to be y2 = 0.69 m. Therefore, if the character of flow were to change from supercritical to subcritical while holding the specific energy constant, the flow depth would rise from 0.15 to 0.69 m.
0.2 m3/s 0.15 m 0.4 m FIGURE 13–15 Schematic for Example 13–1.
cen96537_ch13_733-792.indd 743 14/01/17 3:21 pm 744 open-channel flow 13–4 ■ CONSERVATION OF MASS AND ENERGY EQUATIONS Open-channel flows involve liquids whose densities are nearly constant, and thus the one-dimensional steady-flow conservation of mass equation is ex pressed as V .
= AcV = constant (13–26) That is, the product of the flow cross section and the average flow velocity remains constant throughout the channel. Equation 13–26 between two sec tions along the channel is expressed as Continuity equation: Ac1V1 = Ac2V2 (13–27) which is identical to the steady-flow conservation of mass equation for liquid flow in a pipe. Note that both the flow cross section and the average flow velocity may vary during flow, but, as stated, their product remains constant. To determine the total energy of a liquid flowing in an open channel rela tive to a reference datum, as shown in Fig. 13–16, consider a point A in the liquid at a distance a from the free surface (and thus a distance y − a from the channel bottom). Noting that the elevation, pressure (hydrostatic pressure relative to the free surface), and velocity at point A are zA = z + (y − a), PA = 𝜌ga, and VA = V, respectively, the total energy of the liquid in terms of heads is HA = zA + PA ρg + V 2 A 2g = z + (y −a) + ρga ρg + V 2 2g = z + y + V 2 2g (13–28) which is independent of the location of the point A at a cross section. There fore, the total mechanical energy of a liquid at any cross section of an open channel can be expressed in terms of heads as H = z + y + V 2 2g (13–29) where y is the flow depth, z is the elevation of the channel bottom, and V is the average flow velocity. Then the one-dimensional energy equation for open-channel flow between an upstream section 1 and a downstream section 2 is written as Energy equation: z1 + y1 + V 2 1 2g = z2 + y2 + V 2 2 2g + hL (13–30) The head loss hL due to frictional effects is expressed as in pipe flow as hL = f L Dh V 2 2g = f L Rh V 2 8g (13–31) where f is the average friction factor and L is the length of channel between sections 1 and 2. The relation Dh = 4Rh should be observed when using the hydraulic radius instead of the hydraulic diameter.
Discussion Note that if the water underwent a hydraulic jump at constant spe cific energy (the frictional losses being equal to the drop in elevation), the flow depth would rise to 0.69 m, assuming of course that the side walls of the channel are high enough.
z A y y – a a V2 V 2g Energy line H = z + y + Reference datum V2 2g FIGURE 13–16 The total energy of a liquid flowing in an open channel.
cen96537_ch13_733-792.indd 744 14/01/17 3:21 pm 745 CHAPTER 13 Flow in open channels is gravity driven, and thus a typical channel is slightly sloped down. The slope of the bottom of the channel is expressed as S0 = tan 𝛼 = z1 −z2 x2 −x1 ≅ z1 −z2 L (13–32) where 𝛼 is the angle the channel bottom makes with the horizontal. In general, the bottom slope S0 is very small, and thus the channel bottom is nearly horizontal. Therefore, L ≅ x2 − x1, where x is the distance in the horizontal direction. Also, the flow depth y, which is measured in the verti cal direction, can be taken to be the depth normal to the channel bottom with negligible error. If the channel bottom is straight so that the bottom slope is constant, the vertical drop between sections 1 and 2 can be expressed as z1 −z2 = S0L. Then the energy equation (Eq. 13–30) becomes Energy equation: y1 + V 2 1 2g + S0L = y2 + V 2 2 2g + hL (13–33) This equation has the advantage that it is independent of a reference datum for elevation. In the design of open-channel systems, the bottom slope is selected such that it provides adequate elevation drop to overcome the frictional head loss and thus to maintain flow at the desired rate. Therefore, there is a close con nection between the head loss and the bottom slope, and it makes sense to express the head loss as a slope (or the tangent of an angle). This is done by defining a friction slope as Friction slope: Sf = hL L (13–34) Then the energy equation is written as Energy equation: y1 + V 2 1 2g = y2 + V 2 2 2g + (Sf −S0)L (13–35) Note that the friction slope is equal to the bottom slope when the head loss is equal to the elevation drop. That is, Sf = S0 when hL = z1 − z2. Figure 13–17 also shows the energy line, which is a distance z + y + V 2/2g (total mechanical energy of the liquid expressed as a head) above the horizontal reference datum. The energy line is typically sloped down like the channel itself as a result of frictional losses, the vertical drop being equal to the head loss hL and thus the slope being the same as the friction slope. Note that if there were no head loss, the energy line would be hori zontal even when the channel is not. The elevation and velocity heads (z + y and V 2/2g) would then be able to convert to each other during flow in this case, but their sum would remain constant.
13–5 ■ UNIFORM FLOW IN CHANNELS We mentioned in Section 13–1 that flow in a channel is called uniform flow if the flow depth (and thus the average flow velocity since V · = AcV = constant in steady flow) remains constant. Uniform flow conditions are commonly encountered in practice in long straight runs of channels with constant slope, α z V 2 V2 V1 L 2g x1 x2 y1 y2 z1 z2 Energy line Horizontal reference datum (1) (2) Slope: S0 = constant x hL 1 2g V 2 2 FIGURE 13–17 The total energy of a liquid at two sections of an open channel.
cen96537_ch13_733-792.indd 745 14/01/17 3:21 pm 746 open-channel flow constant cross section, and constant surface lining. In the design of open channels, it is very desirable to have uniform flow in the majority of the system since this means having a channel of constant wall height, which is easier to design and build. The flow depth in uniform flow is called the normal depth yn, and the average flow velocity is called the uniform-flow velocity V0. The flow remains uniform as long as the slope, cross section, and surface roughness of the channel remain unchanged (Fig. 13–18). When the bottom slope is increased, the flow velocity increases and the flow depth decreases. There fore, a new uniform flow is established with a new (lower) flow depth. The opposite occurs if the bottom slope is decreased. During flow in open channels of constant slope S0, constant cross section Ac, and constant surface friction factor f, the terminal velocity is reached and thus uniform flow is established when the head loss equals the elevation drop. Therefore, hL = f L Dh V 2 2g or S0L = f L Rh V 2 0 8g (13–36) since hL = S0L in uniform flow and Dh = 4Rh. Solving the second relation for V0, the uniform-flow velocity and the flow rate are determined to be V0 = C√S0Rh and V .
= CAc√S0Rh (13–37) where C = √8g/f (13–38) is called the Chezy coefficient. Equations 13–37 and the coefficient C are named in honor of the French engineer Antoine Chezy (1718–1798), who first proposed a similar relationship in about 1769. The Chezy coefficient is a dimensional quantity, and its value ranges from about 30 m1/2/s for small channels with rough surfaces to 90 m1/2/s for large channels with smooth surfaces (or, 60 ft1/2/s to 160 ft1/2/s in English units). The Chezy coefficient can be determined in a straightforward manner from Eq. 13–38 by first determining the friction factor f as done for pipe flow in Chap. 8 from the Moody chart or the Colebrook equation for the fully rough turbulent limit (Re → ∞), f = [2.0 log(14.8Rh /𝜀 )]−2 (13–39) Here, 𝜀 is the mean surface roughness. Note that open-channel flow is typi cally turbulent, and the flow is fully developed by the time uniform flow is established. Therefore, it is reasonable to use the friction factor relation for fully developed turbulent flow. Also, at large Reynolds numbers, the friction factor curves corresponding to specified relative roughness are nearly hori zontal, and thus the friction factor is independent of the Reynolds number. The flow in that region is called fully rough turbulent flow (Chap. 8). Since the introduction of the Chezy equations, considerable effort has been devoted by numerous investigators to the development of simpler empirical relations for the average velocity and flow rate. The most widely used equation was developed independently by the Frenchman Philippe-Gaspard Gauckler (1826–1905) in 1868 and the Irishman Robert Manning (1816–1897) in 1889.
α z x1 x2 y1 y2 z1 z2 (1) (2) Slope: S0 = tan α = constant Head loss = elevation loss hL = z1 – z2 = S0L V1 = V2 = V0 y1 = y2 = yn x x2 – x1 = L cosα ≅ L FIGURE 13–18 In uniform flow, the flow depth y, the average flow velocity V, and the bottom slope S0 remain constant, and the head loss equals the elevation loss, hL = z1 − z2 = Sf L = S0 L.
cen96537_ch13_733-792.indd 746 14/01/17 3:21 pm 747 CHAPTER 13 Both Gauckler and Manning made recommendations that the constant in the Chezy equation be expressed as C = a n R 1/6 h (13–40) where n is called the Manning coefficient, whose value depends on the roughness of the channel surfaces. Substituting into Eqs. 13–37 gives the fol lowing empirical relations known as the Manning equations (also referred to as Gauckler–Manning equations since they were first proposed by Gauckler) for the uniform-flow velocity and the flow rate, Uniform flow: V0 = a n R 2/3 h S 1/2 0 and V .
= a n AcR 2/3 h S 1/2 0 (13–41) The factor a is a dimensional constant whose value in SI units is a = 1 m1/3/s. Noting that 1 m = 3.2808 ft, its value in English units is a = 1 m1/3/s = (3.2808 ft)1/3/s = 1.486 ft1/3/s (13–42) Note that the bottom slope S0 and the Manning coefficient n are dimension less quantities, and Eqs. 13–41 give the velocity in m/s and the flow rate in m3/s in SI units when Rh is expressed in m. (The corresponding units in English units are ft/s and ft3/s when Rh is expressed in ft.) Experimentally determined values of n are given in Table 13–1 for numer ous natural and artificial channels. More extensive tables are available in the literature. Note that the value of n varies from 0.010 for a glass channel to 0.150 for a floodplain laden with trees (15 times that of a glass channel). There is considerable uncertainty in the value of n, especially in natural channels, as you would expect, since no two channels are exactly alike. The scatter can be 20 percent or more. Nevertheless, coefficient n is approxi mated as being independent of the size and shape of the channel—it varies only with the surface roughness.
Critical Uniform Flow Flow through an open channel becomes critical flow when the Froude number Fr = 1 and thus the flow speed equals the wave speed Vc = √gyc, where yc is the critical flow depth, defined previously (Eq. 13–9). When the volume flow rate V · , the channel slope S0, and the Manning coefficient n are known, the normal flow depth yn can be determined from the Man ning equation (Eq. 13–41). However, since Ac and Rh are both functions of yn, the equation often ends up being implicit in yn and requires a numerical (or trial and error) approach to solve. If yn = yc, the flow is uniform critical flow, and bottom slope S0 equals the critical slope Sc in this case. When flow depth yn is known instead of the flow rate V · , the flow rate can be determined from the Manning equation and the critical flow depth from Eq. 13–9. Again the flow is critical only if yn = yc. During uniform critical flow, S0 = Sc and yn = yc. Replacing V · and S0 in the Manning equation by V .
= Ac√gyc and Sc, respectively, and solving for Sc gives the following general relation for the critical slope, Critical slope (general): Sc = gn2yc a2R 4/3 h (13–43) TABLE 13–1 Mean values of the Manning coefficient n for water flow in open channels From Chow (1959).
Wall Material n A. Artificially lined channels Glass 0.010 Brass 0.011 Steel, smooth 0.012 Steel, painted 0.014 Steel, riveted 0.015 Cast iron 0.013 Concrete, finished 0.012 Concrete, unfinished 0.014 Wood, planed 0.012 Wood, unplaned 0.013 Clay tile 0.014 Brickwork 0.015 Asphalt 0.016 Corrugated metal 0.022 Rubble masonry 0.025 B. Excavated earth channels Clean 0.022 Gravelly 0.025 Weedy 0.030 Stony, cobbles 0.035 C. Natural channels Clean and straight 0.030 Sluggish with deep pools 0.040 Major rivers 0.035 Mountain streams 0.050 D. Floodplains Pasture, farmland 0.035 Light brush 0.050 Heavy brush 0.075 Trees 0.150 The uncertainty in n can be ±20 percent or more.
cen96537_ch13_733-792.indd 747 14/01/17 3:21 pm 748 open-channel flow For film flow or flow in a wide rectangular channel with b ≫ yc, Eq. 13–43 simplifies to Critical slope (b ≫ yc): Sc = gn2 a2y 1/3 c (13–44) This equation gives the slope necessary to maintain a critical flow of depth yc in a wide rectangular channel having a Manning coefficient of n.
Superposition Method for Nonuniform Perimeters The surface roughness and thus the Manning coefficient for most natural and some human-made channels vary along the wetted perimeter and even along the channel. A river, for example, may have a stony bottom for its reg ular bed but a surface covered with bushes for its extended floodplain. There are several methods for solving such problems, either by finding an effective Manning coefficient n for the entire channel cross section, or by consider ing the channel in subsections and applying the superposition principle. For example, a channel cross section can be divided into N subsections, each with its own uniform Manning coefficient and flow rate. When determining the perimeter of a section, only the wetted portion of the boundary for that section is considered, and the imaginary boundaries are ignored. The flow rate through the channel is the sum of the flow rates through all the sections, as illustrated in Example 13–4.
EXAMPLE 13–2 Flow Rate in Open Channel in Uniform Flow Water flows uniformly half-full in a 2-m-diameter circular channel that is laid on a grade of 1.5 m/km (Fig. 13–19). If the channel is constructed of finished concrete, determine the flow rate of the water.
SOLUTION Water flows uniformly half-full in a circular finished-concrete channel. For a given bottom slope, the flow rate is to be determined.
Assumptions 1 The flow is steady and uniform. 2 The bottom slope is constant. 3 The roughness coefficient is constant along the channel.
Properties The Manning coefficient for an open channel of finished concrete is n = 0.012 (Table 13–1).
Analysis The flow area, wetted perimeter, and hydraulic radius of the channel are Ac = 𝜋R2 2 = 𝜋(1 m)2 2 = 1.571 m2 p = 2𝜋R 2 = 2𝜋(1 m) 2 = 3.142 m Rh = Ac P = 𝜋R2/2 𝜋R = R 2 = 1 m 2 = 0.50 m R = 1 m θ = π/2 FIGURE 13–19 Schematic for Example 13–2 cen96537_ch13_733-792.indd 748 14/01/17 3:21 pm 749 CHAPTER 13 y b = 4 ft V = 51 ft3/s .
FIGURE 13–20 Schematic for Example 13–3.
Then the flow rate can be determined from Manning’s equation to be V .
= a n AcR 2/3 h S 1/2 0 = 1 m1/3/s 0.012 (0.571 m2)(0.50 m)2/3(1.5/1000)1/2 = 3.19 m3/s Discussion Note that the flow rate in a given channel is a strong function of the bottom slope. If the slope were doubled to 3.0 m/km, all else being equal, the flow rate would increase to 4.52 m3/s, a factor of √2.
EXAMPLE 13–3 The Height of a Rectangular Channel Water is to be transported in an unfinished-concrete rectangular channel with a bot tom width of 4 ft at a rate of 51 ft3/s. The terrain is such that the channel bottom drops 2 ft per 1000 ft length. Determine the minimum height of the channel under uniform-flow conditions (Fig. 13–20). What would your answer be if the bottom drop is just 1 ft per 1000 ft length?
SOLUTION Water is flowing in an unfinished-concrete rectangular channel with a specified bottom width. The minimum channel height corresponding to a speci fied flow rate is to be determined.
Assumptions 1 The flow is steady and uniform. 2 The bottom slope is constant. 3 The roughness of the wetted surface of the channel and thus the friction coeffi cient are constant.
Properties The Manning coefficient for an open channel with unfinished-con crete surfaces is n = 0.014.
Analysis The cross-sectional area, perimeter, and hydraulic radius of the channel are Ac = by = (4 ft)y p = b + 2y = (4 ft) + 2y Rh = Ac p = 4y 4 + 2y The bottom slope of the channel is S0 = 2/1000 = 0.002. Using the Manning equation, the flow rate through the channel is expressed as V .
= a n AcR 2/3 h S 1/2 0 51 ft3/s = 1.486 ft1/3/s 0.014 (4y ft2)( 4y 4 + 2y ft) 2/3 (0.002)1/2 which is a nonlinear equation in y. Using an equation solver or an iterative approach, the flow depth is determined to be y = 2.5 ft If the bottom drop were just 1 ft per 1000 ft length, the bottom slope would be S0 = 0.001, and the flow depth would be y = 3.3 ft.
Discussion Note that y is the flow depth, and thus this is the minimum value for the channel height. Also, there is considerable uncertainty in the value of the Man ning coefficient n, and this should be considered when deciding the height of the channel to be built.
cen96537_ch13_733-792.indd 749 14/01/17 3:21 pm 750 open-channel flow 6 m 3 m 2 m 8 m Light brush n2 = 0.050 Clean natural channel n1 = 0.030 s 1 2 FIGURE 13–21 Schematic for Example 13–4.
EXAMPLE 13–4 Channels with Nonuniform Roughness Water flows in a channel whose bottom slope is 0.003 and whose cross section is shown in Fig. 13–21. The dimensions and the Manning coefficients for the surfaces of different subsections are also given on the figure. Determine the flow rate through the channel and the effective Manning coefficient for the channel.
SOLUTION Water is flowing through a channel with nonuniform sur face properties. The flow rate and the effective Manning coefficient are to be determined.
Assumptions 1 The flow is steady and uniform. 2 The bottom slope is constant. 3 The Manning coefficients do not vary along the channel.
Analysis The channel involves two parts with different roughnesses, and thus it is appropriate to divide the channel into two subsections as indi cated in Fig. 13–21. The flow rate for each subsection is determined from the Manning equation, and the total flow rate is determined by adding them up. The side length of the triangular channel is s = √32 + 32 = 4.243 m. Then the flow area, perimeter, and hydraulic radius for each subsection and the entire channel become Subsection 1: Ac1 = 21 m2 p1 = 10.486 m Rh1 = Ac1 p1 = 21 m2 10.486 m = 2.00 m Subsection 2: Ac2 = 16 m2 p2 = 10 m Rh2 = Ac2 p2 = 16 m2 10 m = 1.60 m Entire channel: Ac = 37 m2 p = 20.486 m Rh = Ac p = 37 m2 20.486 m = 1.806 m Using the Manning equation for each subsection, the total flow rate through the channel is determined to be V .
= V .
1 + V .
2 = a n1 Ac1R 2/3 h1 S 1/2 0 + a n2 Ac2R 2/3 h2 S 1/2 0 = (1 m1/3/s) (21 m2)(2 m)2/3 0.030 + (16 m2)(1.60 m)2/3 0.050 1/2 = 84.8 m3/s ≅85 m3/s cen96537_ch13_733-792.indd 750 14/01/17 3:21 pm 751 CHAPTER 13 13–6 ■ BEST HYDRAULIC CROSS SECTIONS Open-channel systems are usually designed to transport a liquid to a loca tion at a lower elevation at a specified rate under the influence of gravity at the lowest possible cost. Noting that no energy input is required, the cost of an open-channel system consists primarily of the initial construction cost, which is proportional to the physical size of the system. Therefore, for a given channel length, the perimeter of the channel is representative of the system cost, and it should be kept to a minimum in order to minimize the size and thus the cost of the system. From another perspective, resistance to flow is due to wall shear stress 𝜏w and the wall area, which is equivalent to the wetted perimeter per unit chan nel length. Therefore, for a given flow cross-sectional area Ac, the smaller the wetted perimeter p, the smaller the resistance force, and thus the larger the average velocity and the flow rate. From yet another perspective, for a specified channel geometry with a specified bottom slope S0 and surface lining (and thus the roughness coefficient n), the flow velocity is given by the Manning formula as V = aR 2/3 h S 1/2 0 /n.
Therefore, the flow velocity increases with the hydraulic radius, and the hydraulic radius must be maximized (and thus the perimeter must be mini mized since Rh = Ac/p) in order to maximize the average flow velocity or the flow rate per unit cross-sectional area. Thus we conclude the following: The best hydraulic cross section for an open channel is the one with the maximum hydraulic radius or, equivalently, the one with the minimum wetted perimeter for a specified cross-sectional area. The shape with the minimal perimeter per unit area is a circle. Therefore, on the basis of minimum flow resistance, the best cross section for an open channel is a semicircular one (Fig. 13–22). However, it is usually cheaper to construct an open channel with straight sides (such as channels with trap ezoidal or rectangular cross sections) instead of semicircular ones, and the general shape of the channel may be specified a priori. Thus it makes sense to analyze each geometric shape separately for the best cross section. As a motivational example, consider a rectangular channel of finished concrete (n = 0.012) of width b and flow depth y with a bottom slope of 1° (Fig. 13–23). To determine the effects of the aspect ratio y/b on the hydraulic radius Rh and the flow rate V · for a cross-sectional area of 1 m2, Rh and V · are R y FIGURE 13–22 The best hydraulic cross section for an open channel is a semicircular one since it has the minimum wetted perimeter for a specified cross-sectional area, and thus the minimum flow resistance.
y b FIGURE 13–23 A rectangular open channel of width b and flow depth y. For a given cross-sectional area, the highest flow rate occurs when y = b/2.
Knowing the total flow rate, the effective Manning coefficient for the entire channel is determined from the Manning equation, neff = aAcR 2/3 h S 1/2 0 V .
= (1 m1/3/s)(37 m2)(1.806 m)2/3(0.003)1/2 84.8 m3/s = 0.035 Discussion The effective Manning coefficient neff of the channel turns out to lie between the two n values, as expected. The weighted average of the Manning coef ficient of the channel is navg = (n1p1 + n2p2)/p = 0.040, which is quite different than neff. Therefore, using a weighted average Manning coefficient for the entire channel may be tempting, but it would not be so accurate.
cen96537_ch13_733-792.indd 751 14/01/17 3:21 pm 752 open-channel flow evaluated from the Manning formula. The results are tabulated in Table 13–2 and plotted in Fig. 13–24 for aspect ratios from 0.1 to 5. We observe from this table and the plot that the flow rate V · increases as the flow aspect ratio y/b is increased, reaches a maximum at y/b = 0.5, and then starts to decrease (the numerical values for V · can also be interpreted as the flow velocities in m/s since Ac = 1 m2). We see the same trend for the hydraulic radius, but the opposite trend for the wetted perimeter p. These results confirm that the best cross section for a given shape is the one with the maximum hydraulic radius, or equivalently, the one with the minimum perimeter.
TABLE 13–2 Variation of the hydraulic radius Rh and the flow rate V · with aspect ratio y /b for a rectangular channel with Ac = 1 m2, S0 = tan 1°, and n = 0.012 Aspect Channel Flow Hydraulic Flow Rate Ratio Width Depth Perimeter Radius V ·, y/b b, m y, m p, m Rh, m m3/s 0.1 3.162 0.316 3.795 0.264 4.53 0.2 2.236 0.447 3.130 0.319 5.14 0.3 1.826 0.548 2.921 0.342 5.39 0.4 1.581 0.632 2.846 0.351 5.48 0.5 1.414 0.707 2.828 0.354 5.50 0.6 1.291 0.775 2.840 0.352 5.49 0.7 1.195 0.837 2.869 0.349 5.45 0.8 1.118 0.894 2.907 0.344 5.41 0.9 1.054 0.949 2.951 0.339 5.35 1.0 1.000 1.000 3.000 0.333 5.29 1.5 0.816 1.225 3.266 0.306 5.00 2.0 0.707 1.414 3.536 0.283 4.74 3.0 0.577 1.732 4.041 0.247 4.34 4.0 0.500 2.000 4.500 0.222 4.04 5.0 0.447 2.236 4.919 0.203 3.81 0 3.75 4.15 4.55 4.95 5.35 5.75 1 2 3 Aspect ratio r = y/b Flow rate V, m3/s .
4 5 FIGURE 13–24 Variation of the flow rate in a rectangular channel with aspect ratio r = y/b for Ac = 1 m2 and S0 = tan 1°.
cen96537_ch13_733-792.indd 752 14/01/17 3:21 pm 753 CHAPTER 13 Rectangular Channels Consider liquid flow in an open channel of rectangular cross section of width b and flow depth y. The cross-sectional area and the wetted perimeter at a flow section are Ac = yb and p = b + 2y (13–45) Solving the first relation of Eq. 13–45 for b and substituting it into the second relation give p = Ac y + 2y (13–46) Now we apply the criterion that the best hydraulic cross section for an open channel is the one with the minimum wetted perimeter for a given cross-sectional area. Taking the derivative of p with respect to y while holding Ac constant gives dp dy = −Ac y2 + 2 = −by y2 + 2 = −b y + 2 (13–47) Setting dp/dy = 0 and solving for y, the criterion for the best hydraulic cross section is determined to be Best hydraulic cross section (rectangular channel): y = b 2 (13–48) Therefore, a rectangular open channel should be designed such that the liq uid height is half the channel width to minimize flow resistance or to maxi mize the flow rate for a given cross-sectional area. This also minimizes the perimeter and thus the construction costs. This result confirms the finding from Table 13–2 that y = b/2 gives the best cross section.
Trapezoidal Channels In practice, many man-made water channels are not rectangular or round, but trapezoidal (Fig. 13–25a). Consider liquid flow in an open channel of trapezoidal cross section of bottom width b, flow depth y, and trapezoid angle 𝜃 measured from the horizontal, as shown in Fig. 13–25b. The cross-sectional area and the wetted perimeter at a flow section are Ac = (b + y tan 𝜃 )y and p = b + 2y sin 𝜃 (13–49) Solving the first relation of Eq. 13–49 for b and substituting it into the second relation give p = Ac y − y tan 𝜃 + 2y sin 𝜃 (13–50) Taking the derivative of p with respect to y while holding Ac and 𝜃 constant gives dp dy = −Ac y2 − 1 tan 𝜃 + 2 sin 𝜃 = −b + y/tan 𝜃 y − 1 tan 𝜃 + 2 sin 𝜃 (13–51) Setting dp/dy = 0 and solving for y, the criterion for the best hydraulic cross section for any specified trapezoid angle 𝜃 is determined to be Best hydraulic cross section (trapezoidal channel): y = b sin 𝜃 2(1 −cos 𝜃 ) (13–52) FIGURE 13–25 (a) Partially filled trapezoidal channel. (b) Parameters for a trapezoidal channel.
(a) Photo by John M. Cimbala.
y θ b Rh = = Ac p y(b + y/tan θ) b + 2y/sin θ s (b) (b) (a) cen96537_ch13_733-792.indd 753 14/01/17 3:21 pm 754 open-channel flow For the special case of 𝜃 = 90° (a rectangular channel), this relation reduces to y = b/2, as expected. The hydraulic radius Rh for a trapezoidal channel can be expressed as Rh = Ac p = y(b + y/tan 𝜃 ) b + 2y/sin 𝜃 = y(b sin 𝜃 + y cos 𝜃 ) b sin 𝜃 + 2y (13–53) Rearranging Eq. 13–52 as bsin 𝜃 = 2y(1 − cos 𝜃), substituting into Eq. 13–53 and simplifying, the hydraulic radius for a trapezoidal channel with the best cross section becomes Hydraulic radius for the best cross section: Rh = y 2 (13–54) Therefore, the hydraulic radius is half the flow depth for trapezoidal chan nels with the best cross section regardless of the trapezoid angle 𝜃. Similarly, the trapezoid angle for the best hydraulic cross section is deter mined by taking the derivative of p (Eq. 13–50) with respect to 𝜃 while holding Ac and y constant, setting dp/d𝜃 = 0, and solving the resulting equa tion for 𝜃. This gives Best trapezoid angle: 𝜃 = 60° (13–55) Substituting the best trapezoid angle 𝜃 = 60° into the best hydraulic cross-section relation y = b sin 𝜃/(2 − 2 cos 𝜃) gives Best flow depth for 𝜃 = 60°: y = √3 2 b (13–56) Then the length of the side edge of the flow section and the flow area become s = y sin 60° = b√3/2 √3/2 = b (13–57) p = 3b (13–58) Ac = (b + y tan 𝜃 )y = (b + b√3/2 tan 60°)(b√3/2) = 3√3 4 b2 (13–59) since tan 60° = √3. Therefore, the best cross section for trapezoidal chan nels is half of a hexagon (Fig. 13–26). This is not surprising since a hexagon closely approximates a circle, and a half-hexagon has the least perimeter per unit cross-sectional area of all trapezoidal channels. Best hydraulic cross sections for other channel shapes can be determined in a similar manner. For example, the best hydraulic cross section for a circular channel of diameter D can be shown to be y = D/2.
EXAMPLE 13–5 Best Cross Section of an Open Channel Water is to be transported at a rate of 2 m3/s in uniform flow in an open channel whose surfaces are asphalt lined. The bottom slope is 0.001. Determine the dimensions of the best cross section if the shape of the channel is (a) rectangular and (b) trapezoidal (Fig. 13–27).
y b b Rh = b = y 2 3 4 b = 3 2 Ac = b2 3 4 3 60° √ √ √ FIGURE 13–26 The best cross section for trapezoidal channels is half of a hexagon.
cen96537_ch13_733-792.indd 754 14/01/17 3:21 pm 755 CHAPTER 13 13–7 ■ GRADUALLY VARIED FLOW To this point we considered uniform flow during which the flow depth y and the flow velocity V remain constant. In this section we consider gradually varied flow (GVF), which is a form of steady nonuniform flow characterized by gradual variations in flow depth and velocity (small slopes and no abrupt changes) and a free surface that always remains smooth (no discontinuities or zigzags). Flows that involve rapid changes in flow depth and velocity, called rapidly varied flows (RVF), are considered in Section 13–8. A change y b b b = 3 2 5 b 2 y = b b 2 60° FIGURE 13–27 Schematic for Example 13–5.
FIGURE 13–28 Many man-made water channels are trapezoidal in shape because of low construction cost and good performance. © Pixtal/AGE Fotostock RF SOLUTION Water is to be transported in an open channel at a specified rate. The best channel dimensions are to be determined for rectangular and trapezoidal shapes.
Assumptions 1 The flow is steady and uniform. 2 The bottom slope is constant. 3 The roughness of the wetted surface of the channel and thus the friction coefficient are constant.
Properties The Manning coefficient for an open channel with asphalt lining is n = 0.016.
Analysis (a) The best cross section for a rectangular channel occurs when the flow height is half the channel width, y = b/2. Then the cross-sectional area, perim eter, and hydraulic radius of the channel are Ac = by = b2 2 p = b + 2y = 2b Rh = Ac p = b 4 Substituting into the Manning equation, V .
= a n AcR 2/3 h S 1/2 0 → b = ( 2nV .
42/3 a√S0 ) 3/8 = ( 2(0.016)(2 m3/s)42/3 (1 m1/3/s)√0.001 ) 3/8 which gives b = 1.84 m. Therefore, Ac = 1.70 m2, p = 3.68 m, and the dimensions of the best rectangular channel are b = 1.84 m and y = 0.92 m (b) The best cross section for a trapezoidal channel occurs when the trapezoid angle is 60° and flow height is y = b√3/2. Then, Ac = y(b + b cos 𝜃 ) = 0.5√3b2(1 + cos 60°) = 0.75√3b2 p = 3b Rh = y 2 = √3 4 b Substituting into the Manning equation, V .
= a n AcR 2/3 h S 1/2 0 → b = ( (0.016)(2 m3/s) 0.75√3(√3/4)2/3(1 m1/3/s)√0.001) 3/8 which yields b = 1.12 m. Therefore, Ac = 1.64 m2, p = 3.37 m, and the dimensions of the best trapezoidal channel are b = 1.12 m y = 0.973 m and 𝜃 = 60° Discussion Note that the trapezoidal cross section is better since it has a smaller perimeter (3.37 m versus 3.68 m) and thus lower construction cost. This is why many man-made waterways are trapezoidal in shape (Fig. 13–28). However, the average velocity through the trapezoidal channel is larger since Ac is smaller.
cen96537_ch13_733-792.indd 755 14/01/17 3:21 pm 756 open-channel flow in the bottom slope or cross section of a channel or an obstruction in the path of flow may cause the uniform flow in a channel to become gradually or rapidly varied flow. Rapidly varied flows occur over a short section of the channel with rel atively small surface area, and thus frictional losses associated with wall shear are negligible. Head losses in RVF are highly localized and are due to intense agitation and turbulence. Losses in GVF, on the other hand, are primarily due to frictional effects along the channel and can be determined from the Manning formula. In gradually varied flow, the flow depth and velocity vary slowly, and the free surface is stable. This makes it possible to formulate the variation of flow depth along the channel on the basis of the conservation of mass and energy principles and to obtain relations for the profile of the free surface. In uniform flow, the slope of the energy line is equal to the slope of the bot tom surface. Therefore, the friction slope equals the bottom slope, Sf = S0. In gradually varied flow, however, these slopes are different (Fig. 13–29). Consider steady flow in a rectangular open channel of width b, and assume any variation in the bottom slope and water depth to be rather gradual. We again write the equations in terms of average velocity V and approximate the pressure distribution as hydrostatic. From Eq. 13–17, the total head of the liquid at any cross section is H = zb + y + V 2/2g, where zb is the vertical distance of the bottom surface from the reference datum. Differentiating H with respect to x gives dH dx = d dx (zb + y + V 2 2g) = dzb dx + dy dx + V g dV dx (13–60) But H is the total energy of the liquid and thus dH/dx is the slope of the energy line (a negative quantity), which is equal to the negative of the friction slope, as shown in Fig. 13–29. Also, dzb/dx is the negative of the bottom slope. Therefore, dH dx = −dhL dx = −Sf and dzb dx = −S0 (13–61) Substituting Eqs. 13–61 into Eq. 13–60 gives S0 −Sf = dy dx + V g dV dx (13–62) The conservation of mass equation for steady flow in a rectangular channel is V · = ybV = constant. Differentiating with respect to x gives 0 = bV dy dx + yb dV dx → dV dx = −V y dy dx (13–63) Substituting Eq. 13–63 into Eq. 13–62 and noting that V/√gy is the Froude number, S0 −Sf = dy dx −V 2 gy dy dx = dy dx −Fr2 dy dx (13–64) Solving for dy/dx gives the desired relation for the rate of change of flow depth (or the surface profile) in gradually varied flow in an open channel, The GVF equation: dy dx = S0 −Sf 1 −Fr2 (13–65) z, H V2 V + dV y + dy V 2g x x + dx y zb dx zb + dzb Energy line, H Friction slope Sf Horizontal Horizontal reference datum Bottom slope S0 x dhL (V + dV)2 2g FIGURE 13–29 Variation of properties over a differential flow section in an open channel under conditions of gradually varied flow (GVF).
cen96537_ch13_733-792.indd 756 14/01/17 3:21 pm 757 CHAPTER 13 which is analogous to the variation of flow area as a function of the Mach number in compressible flow. This relation is derived for a rectangular chan nel, but it is also valid for channels of other constant cross sections provided that the Froude number is expressed accordingly. An analytical or numerical solution of this differential equation gives the flow depth y as a function of x for a given set of parameters, and the function y(x) is the surface profile. The general trend of flow depth—whether it increases, decreases, or remains constant along the channel—depends on the sign of dy/dx, which depends on the signs of the numerator and the denominator of Eq. 13–65. The Froude number is always positive and so is the friction slope Sf (except for the idealized case of flow with negligible frictional effects for which both hL and Sf are zero). The bottom slope S0 is positive for down ward-sloping sections (typically the case), zero for horizontal sections, and negative for upward-sloping sections of a channel (adverse flow). The flow depth increases when dy/dx > 0, decreases when dy/dx < 0, and remains constant (and thus the free surface is parallel to the channel bottom, as in uniform flow) when dy/dx = 0 and thus S0 = Sf (Fig. 13–30). For specified values of S0 and Sf, the term dy/dx may be positive or negative, depending on whether the Froude number is less than or greater than 1. Therefore, the flow behavior is opposite in subcritical and supercritical flows. For S0 − Sf > 0, for example, the flow depth increases in the flow direction in subcritical flow, but it decreases in supercritical flow. The determination of the sign of the denominator 1 − Fr2 is easy: it is positive for subcritical flow (Fr < 1), and negative for supercritical flow (Fr > 1). But the sign of the numerator depends on the relative magnitudes of S0 and Sf. Note that the friction slope Sf is always positive, and its value is equal to the channel slope S0 in uniform flow, y = yn. The friction slope is a quantity that varies with streamwise distance, and is calculated from the Manning equation, based upon the depth at each streamwise location, as demonstrated in Example 13–6. Noting that head loss increases with increasing velocity, and that the velocity is inversely proportional to flow depth for a given flow rate, Sf > S0 and thus S0 − Sf < 0 when y < yn, and Sf < S0 and thus S0 − Sf > 0 when y > yn. The numerator S0 − Sf is always negative for horizontal (S0 = 0) and upward-sloping (S0 < 0) channels, and thus the flow depth decreases in the flow direction during subcritical flows in such channels.
Liquid Surface Profiles in Open Channels, y(x) Open-channel systems are designed and built on the basis of the projected flow depths along the channel. Therefore, it is important to be able to pre dict the flow depth for a specified flow rate and specified channel geometry. A plot of flow depth versus downstream distance is the surface profile y(x) of the flow. The general characteristics of surface profiles for gradually var ied flow depend on the bottom slope and flow depth relative to the critical and normal depths. A typical open channel involves various sections of different bottom slopes S0 and different flow regimes, and thus various sections of different surface profiles. For example, the general shape of the surface profile in a downward-sloping section of a channel is different than that in an upward-sloping section. Likewise, the profile in subcritical flow is different than the FIGURE 13–30 A slow-moving river of approximately constant depth and cross section, such as the Chicago River shown here, is an example of uniform flow with S0 ≈ Sf and dy/dx ≈ 0.
© Hisham F. Ibrahim/Getty Images RF cen96537_ch13_733-792.indd 757 14/01/17 3:21 pm 758 open-channel flow profile in supercritical flow. Unlike uniform flow that does not involve iner tial forces, gradually varied flow involves acceleration and deceleration of liquid, and the surface profile reflects the dynamic balance between liquid weight, shear force, and inertial effects. Each surface profile is identified by a letter that indicates the slope of the channel and by a number that indicates flow depth relative to the critical depth yc and normal depth yn. The slope of the channel can be steep (S), critical (C), mild (M), horizontal (H), or adverse (A) (Fig. 13–31). The chan nel slope is said to be mild if yn > yc, steep if yn < yc, critical if yn = yc, horizontal if S0 = 0 (zero bottom slope), and adverse if S0 < 0 (negative slope). Note that a liquid flows uphill in an open channel that has an adverse slope. The classification of a channel section depends on the flow rate and the channel cross section as well as the slope of the channel bottom. A channel section that is classified to have a mild slope for one flow may have a steep slope for another flow, and even a critical slope for a third flow. Therefore, we need to calculate the critical depth yc and the normal depth yn before we can assess the slope. The number designation indicates the initial position of the liquid surface for a given channel slope relative to the surface levels in critical and uniform flows, as shown in Fig. 13–32. A surface profile is designated by 1 if the flow depth is above both critical and normal depths (y > yc and y > yn), by 2 if the flow depth is between the two (yn > y > yc or yn < y < yc), and by 3 if the flow depth is below both the critical and normal depths (y < yc and y < yn). There fore, three different profiles are possible for a specified type of channel slope. But for channels with zero or adverse slopes, type 1 flow cannot exist since the flow can never be uniform in horizontal and upward channels, and thus normal depth is not defined. Also, type 2 flow does not exist for channels with critical slope since normal and critical depths are identical in this case. The five classes of slopes and the three types of initial positions discussed give a total of 12 distinct configurations for surface profiles in GVF, all tabu lated and sketched in Table 13–3. The Froude number is also given for each case, with Fr > 1 for y < yc, as well as the sign of the slope dy/dx of the sur face profile determined from Eq. 13–65, dy/dx = (S0 − Sf)/(1 − Fr2). Note that dy/dx > 0, and thus the flow depth increases in the flow direction when both S0 − Sf and 1 − Fr2 are positive or negative. Otherwise dy/dx < 0 and the flow depth decreases. In type 1 flows, the flow depth increases in the flow direc tion and the surface profile approaches the horizontal plane asymptotically. In type 2 flows, the flow depth decreases and the surface profile approaches the lower of yc or yn. In type 3 flows, the flow depth increases and the surface profile approaches the lower of yc or yn. These trends in surface profiles con tinue as long as there is no change in bottom slope or roughness. Consider the case in Table 13–3 designated M1 (mild channel slope and y > yn > yc). The flow is subcritical since y > yc and thus Fr < 1 and 1 − Fr2 > 0. Also, Sf < S0 and thus S0 − Sf > 0 since y > yn, and thus the flow velocity is less than the velocity in normal flow. Therefore, the slope of the surface profile dy/dx = (S0 − Sf)/(1 − Fr2) > 0, and the flow depth y increases in the flow direction. But as y increases, the flow velocity decreases, and thus Sf and Fr approach zero. Consequently, dy/dx approaches S0 and the rate of increase in flow depth becomes equal to the channel slope. This requires the surface profile to become horizontal at Horizontal A S H M C Mild Steep Critical Adverse FIGURE 13–31 Designation of the letters S, C, M, H, and A for liquid surface profiles for different types of slopes.
3 Channel bottom Free surface in critical flow Free surface in uniform flow yn yc y 2 1 FIGURE 13–32 Designation of the numbers 1, 2, and 3 for liquid surface profiles based on the value of the flow depth relative to the normal and critical depths.
cen96537_ch13_733-792.indd 758 14/01/17 3:21 pm 759 CHAPTER 13 TABLE 13–3 Classification of surface profiles in gradually varied flow. The vertical scale is greatly exaggerated.
Channel Profile Froude Profile Surface Slope Notation Flow Depth Number Slope Profile Steep (S) yc > yn S1 y > yc Fr < 1 dy dx > 0 S0 < Sc S2 yn < y < yc Fr > 1 dy dx < 0 S3 y < yn Fr > 1 dy dx > 0 Critical (C) yc = yn C1 y > yc Fr < 1 dy dx > 0 S0 < Sc C3 y < yc Fr > 1 dy dx > 0 Mild (M) yc < yn M1 y > yn Fr < 1 dy dx > 0 S0 < Sc M2 yc < y < yn Fr < 1 dy dx < 0 M3 y < yc Fr > 1 dy dx > 0 Horizontal (H) yn → ∞ H2 y > yc Fr < 1 dy dx < 0 S0 = 0 H3 y < yc Fr > 1 dy dx > 0 Adverse (A) S0 < 0 A2 y > yc Fr < 1 dy dx < 0 yn: does A3 y < yc Fr > 1 dy dx > 0 not exist yn yc Channel bottom, S0 > Sc S1 S2 S3 Horizontal yc = yn Channel bottom, S0 = Sc C1 C3 Horizontal yc Channel bottom, S0 < 0 A2 A3 yc Channel bottom, S0 = 0 H2 H3 Normal depth Surface profile y(x) Starting point Horizontal yn M1 yc Critical depth Channel bottom, S0 < Sc M2 M3 M3 M2 M1 H2 A2 A3 A2 H2 H3 S2 S1 S1 S3 C1 C3 M3 M2 M1 H2 A2 A3 A2 H2 H3 S2 S1 S1 S3 C1 C3 M3 M2 M1 H2 A2 A3 A2 H2 H3 S2 S1 S1 S3 C1 C3 M3 M2 M1 H2 A2 A3 A2 H2 H3 S2 S1 S1 S3 C1 C3 M3 M2 M1 H2 A2 A3 A2 H2 H3 S2 S1 S1 S3 C1 C3 cen96537_ch13_733-792.indd 759 14/01/17 3:21 pm 760 open-channel flow large y. Then we conclude that the M1 surface profile first rises in the flow direction and then tends to a horizontal asymptote. As y → yc in subcritical flow (such as M2, H2, and A2), we have Fr → 1 and 1 − Fr2 → 0, and thus the slope dy/dx tends to negative infinity. But as y → yc in supercritical flow (such as M3, H3, and A3), we have Fr → 1 and 1 − Fr2 → 0, and thus the slope dy/dx, which is a positive quantity, tends to infinity. That is, the free surface rises almost vertically and the flow depth increases very rapidly. This cannot be sustained physically, and the free surface breaks down. The result is a hydraulic jump. The one-dimensional approximation is no longer applicable when this happens.
Some Representative Surface Profiles A typical open-channel system involves several sections of different slopes, with connections called transitions, and thus the overall surface profile of the flow is a continuous profile made up of the individual profiles described earlier. Some representative surface profiles commonly encountered in open channels, including some composite profiles, are given in Fig. 13–33. For each case, the change in surface profile is caused by a change in channel geometry such as an abrupt change in slope or an obstruction in the flow such as a sluice gate. More composite profiles can be found in specialized books listed in the references. A point on a surface profile represents the flow height at that point that satisfies the mass, momentum, and energy conservation relations. Note that dy/dx ≪ 1 and S0 ≪ 1 in gradually varied flow, and the slopes of both the channels and the surface profiles in these sketches are highly exaggerated for better visualization. Many channels and surface profiles would appear nearly horizontal if drawn to scale. Figure 13–33a shows the surface profile for gradually varied flow in a channel with mild slope and a sluice gate. The subcritical upstream flow (note that the flow is subcritical since the slope is mild) slows down as it approaches the gate (such as a river approaching a dam) and the liquid level rises. The flow past the gate is supercritical (since the height of the opening is less than the critical depth). Therefore, the surface profile is M1 before the gate and M3 after the gate prior to the hydraulic jump. A section of an open channel may have a negative slope and involve uphill flow, as shown in Fig. 13–33b. Flow with an adverse slope cannot be maintained unless the inertia forces overcome the gravity and viscous forces that oppose the fluid motion. Therefore, an uphill channel section must be followed by a downhill section or a free outfall. For subcritical flow with an adverse slope approaching a sluice gate, the flow depth decreases as the gate is approached, yielding an A2 profile. Flow past the gate is typically supercritical, yielding an A3 profile prior to the hydraulic jump. The open-channel section in Fig. 13–33c involves a slope change from steep to less steep. The flow velocity in the less steep part is lower (a smaller elevation drop to drive the flow), and thus the flow depth is higher when uniform flow is established again. Noting that uniform flow with steep slope must be supercritical ( y < yc), the flow depth increases from the initial to the new uniform level smoothly through an S3 profile. Figure 13–33d shows a composite surface profile for an open channel that involves various flow sections. Initially the slope is mild, and the flow is uniform and subcritical. Then the slope changes to steep, and the flow cen96537_ch13_733-792.indd 760 14/01/17 3:21 pm 761 CHAPTER 13 becomes supercritical when uniform flow is established. The critical depth occurs at the break in grade. The change of slope is accompanied by a smooth decrease in flow depth through an M2 profile at the end of the mild section, and through an S2 profile at the beginning of the steep section. In the horizontal section, the flow depth increases first smoothly through an H3 profile, and then rapidly during a hydraulic jump. The flow depth then decreases through an H2 profile as the liquid accelerates toward the end yn1 yc yc yn2 (a) Flow through a sluice gate in an open channel with mild slope (b) Flow through a sluice gate in an open channel with adverse slope and free outfall (c) Uniform supercritical flow changing from steep to less steep slope (d) Uniform subcritical flow changing from mild to steep to horizontal slope with free outfall Mild Adverse Uniform flow Uniform flow Less steep Horizontal Free outfall Hydraulic jump Uniform flow Uniform flow H3 H2 M2 S2 Steep Mild Steep Uniform flow Uniform flow Hydraulic jump Hydraulic jump M1 A2 A3 A2 M3 yn2 yn2 yn2 yc yc yn1 yc yn1 y < yn2 S3 FIGURE 13–33 Some common surface profiles encountered in open-channel flow. All flows are from left to right.
cen96537_ch13_733-792.indd 761 14/01/17 3:21 pm 762 open-channel flow of the channel to a free outfall. The flow becomes critical before reaching the end of the channel, and the outfall controls the upstream flow past the hydraulic jump. The outfalling flow stream is supercritical. Note that uniform flow cannot be established in a horizontal channel since the gravity force has no component in the flow direction, and the flow is inertia-driven.
Numerical Solution of Surface Profile The prediction of the surface profile y(x) is an important part of the design of open-channel systems. A good starting point for the determination of the surface profile is the identification of the points along the channel, called the control points, at which the flow depth can be calculated from a knowl edge of flow rate. For example, the flow depth at a section of a rectangular channel where critical flow occurs, called the critical point, is determined from yc = (V · 2/gb2)1/3. The normal depth yn, which is the flow depth reached when uniform flow is established, also serves as a control point. Once flow depths at control points are available, the surface profile upstream or down stream is determined usually by numerical integration of the nonlinear dif ferential equation (Eq. 13–65, repeated here) dy dx = S0 −Sf 1 −Fr2 (13–66) The friction slope Sf is determined from the uniform-flow conditions, and the Froude number from a relation appropriate for the channel cross section.
EXAMPLE 13–6 Gradually Varied Flow with M1 Surface Profile Gradually varied flow of water in a wide rectangular channel with a per-unit-width flow rate of 1 m3/s⋅m and a Manning coefficient of n = 0.02 is considered. (Fig. 13-34) The slope of the channel is 0.001, and at the location x = 0, the flow depth is measured to be 0.8 m. (a) Determine the normal and critical depths of the flow and classify the water surface profile, and (b) calculate the flow depth y at x = 1000 m by integrating the GVF equation numerically over the range 0 ≤ x ≤ 1000 m. Repeat part (b) to obtain the flow depths for different x values, and plot the surface profile.
SOLUTION Gradually varied flow of water in a wide rectangular channel is considered. The normal and critical flow depths, the flow type, and the flow depth at a specified location are to be determined, and the surface profile is to be plotted.
Assumptions 1 The channel is wide, and the flow is gradually varied. 2 The bottom slope is constant. 3 The roughness of the wetted surface of the channel and thus the friction coefficient are constant.
Properties The Manning coefficient of the channel is given to be n = 0.02.
Analysis (a) The channel is said to be wide, and thus the hydraulic radius is equal to the flow depth, Rh ≅ y. Knowing the flow rate per unit width (b = 1 m), the normal depth is determined from the Manning equation to be V · = a nAcR 2/3 h S 1/2 0 = a n (yb)y2/3S 1/2 0 = a n by5/3S 1/2 0 yn = ( (V ·/b)n aS 1/2 0 ) 3/5 = ( (1 m2/s)(0.02) (1 m1/3/s)(0.001)1/2) 3/5 = 0.76 m y y0 = 0.8 m 0 Bottom slope, S0 = 0.001 FIGURE 13–34 Schematic for Example 13–6.
cen96537_ch13_733-792.indd 762 14/01/17 3:21 pm 763 CHAPTER 13 Distance along the channel, m 0 100 200 300 400 500 600 700 800 900 1000 Water depth, m 0.80 0.82 0.86 0.90 0.96 1.03 1.10 1.18 1.26 1.35 1.44 y, m 1000 800 400 x, m 200 0 600 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.0 y yn yc FIGURE 13–35 Flow depth and surface profile for the GVF problem discussed in Example 13–6.
The critical depth for this flow is yc = V · 2 gA 2 c = V · 2 g(by)2 →yc = ( (V ·/b)2 g ) 1/3 = ( (1 m2/s)2 (9.81 m/s2)) 1/3 = 0.47 m Noting that yc < yn < y at x = 0, we see from Table 13–3 that the water surface profile during this GVF is classified as M1.
(b) Knowing the initial condition y (0) = 0.8 m, the flow depth y at any x location is determined by numerical integration of the GVF equation dy dx = S0 −Sf 1 −Fr2 where the Froude number for a wide rectangular channel is Fr = V √gy = V ·/by √gy = V ·/b √gy3 and the friction slop is determined from the uniform-flow equation by setting S0 = Sf, V · = a nby5/3S 1/2 f →Sf = ( (V ·/b)n ay5/3 ) 2 = (V ·/b)2n2 a2y10/3 Substituting, the GVF equation for a wide rectangular channel becomes dy dx = S0 −(V ·/b)2n2/(a2y10/3) 1 −(V ·/b)2/(gy3) which is highly nonlinear, and thus it is difficult (if not impossible) to integrate analytically. Fortunately, nowadays solving nonlinear differential equations by inte grating such nonlinear equations numerically using a program like EES or Matlab is easy. With this mind, the solution of the nonlinear first order differential equation subject to the initial condition y (x1) = y1 is ex pressed as y = y1 + ∫ x2 x1 f(x,y)dx where f(x,y) = S0 −(V ·/b)2n2/(a2y10/3) 1 −(V ·/b)2/(gy3) and where y = y(x) is the water depth at the specified location x. For given numeri cal values, this problem can be solved using EES as follows: Vol = 1 “m^3/s, volume flow rate per unit width, b = 1 m” b = 1 “m, width of channel” n = 0.02 “Manning coefficient” S_0 = 0.001 “slope of channel” g = 9.81 “gravitational acceleration, m/s^2” x1 = 0; y1=0.8 “m, initial condition” x2 = 1000 “m, length of channel” f_xy = (S_0-((Vol/b)^2n^2/y^(10/3)))/(1-(Vol/b)^2/(gy^3)) “the GVF equation to be integrated” y = y 1 + integral(f_xy, x, x1, x2) “integral equation with automatic step size.” Copying the mini program above into a blank EES screen and calculating gives the water depth at a location of 1000 m, y(x2) = y(1000 m) = 1.44 m cen96537_ch13_733-792.indd 763 14/01/17 3:21 pm 764 open-channel flow Note that the built-in function “integral” performs integrations numeri cally between specified limits using an automatically adjusted step size. Water depths at different locations along the channel are obtained by repeating the calculations at different x2 values. Plotting the results gives the surface pro file, as shown in Fig. 13–35. Using the curve-fit feature of EES, we can even curve-fit the flow depth data into the following second-order polynomial, yapprox(x) = 0.7930 + 0.0002789x + 3.7727 × 10−7x2 It can be shown that the flow depth results obtained from this curve-fit formula do not differ from tabulated data by more than 1 percent.
Discussion The graphical result confirms the quantitative prediction from Table 13–3 that an M1 profile should yield increasing water depth in the downstream direction. This problem can also be solved using other programs, like Matlab, using the code given in Fig. 13–36.
EXAMPLE 13–7 Classification of Channel Slope Water is flowing uniformly in a rectangular open channel with unfinished-concrete surfaces. The channel width is 6 m, the flow depth is 2 m, and the bottom slope is 0.004. Determine if the channel should be classified as mild, critical, or steep for this flow (Fig. 13–37).
SOLUTION Water is flowing uniformly in an open channel. It is to be deter mined whether the channel slope is mild, critical, or steep for this flow.
Assumptions 1 The flow is steady and uniform. 2 The bottom slope is constant. 3 The roughness of the wetted surface of the channel and thus the friction coeffi cient are constant.
Properties The Manning coefficient for an open channel with unfinished-con crete surfaces is n = 0.014.
Analysis The cross-sectional area, perimeter, and hydraulic radius are Ac = yb = (2 m)(6 m) = 12 m2 p = b + 2y = 6 m + 2(2 m) = 10 m Rh = Ac p = 12 m2 10 m = 1.2 m The flow rate is determined from the Manning equation to be V .
= a n AcR 2/3 h S 1/2 0 = 1 m1/3/s 0.014 (12 m2)(1.2 m)2/3(0.004)1/2 = 61.2 m3/s Noting that the flow is uniform, the specified flow rate is the normal depth and thus y = yn = 2 m. The critical depth for this flow is yc = V . 2 gA 2 c = (61.2 m3/s)2 (9.81 m/s2)(12 m2)2 = 2.65 m This channel at these flow conditions is classified as steep since yn < yc, and the flow is supercritical.
Discussion If the flow depth were greater than 2.65 m, the channel slope would be said to be mild. Therefore, the bottom slope alone is not sufficient to classify a downhill channel as being mild, critical, or steep.
clear all domain=[0 1000]; % limits on integral s0=.001; % channel slope n=.02; % Manning roughness q=1; % per-unit-width flowrate g=9.81; % gravity (SI) y0=.8; % initial condition on depth [X,Y]=ode45(‘simple_flow_derivative’, [domain(1) domain (end)],y0, [],s0,n,q,g,domain); plot (X, Y, ‘k’) axis([0 1000 0 max(Y)]) xlabel(‘x (m)’);ylabel(‘y (m)’); function yprime=simple_flow_ derivative(x,y,flag,s0, n,q,g, (domain) yprime=(s0-n.^2q.^2./y.^(10/3))./(1- q.^2/g./y.^3); FIGURE 13–36 A Matlab program for solving the GVF problem of Example 13–6.
5 b 2 b = 6 m S0 = 0.004 y = 2 m FIGURE 13–37 Schematic for Example 13–7.
cen96537_ch13_733-792.indd 764 14/01/17 3:21 pm 765 CHAPTER 13 13–8 ■ RAPIDLY VARIED FLOW AND THE HYDRAULIC JUMP Recall that flow in open channels is called rapidly varied flow (RVF) if the flow depth changes markedly over a relatively short distance in the flow direction (Fig. 13–38). Such flows occur in sluice gates, broad- or sharp-crested weirs, waterfalls, and the transition sections of channels for expan sion and contraction. A change in the cross section of the channel is one cause of rapidly varied flow. But some rapidly varied flows, such as flow through a sluice gate, occur even in regions where the channel cross section is constant. Rapidly varied flows are typically complicated by the fact that they may involve significant multidimensional and transient effects, backflows, and flow separation (Fig. 13–39). Therefore, rapidly varied flows are usually studied experimentally or numerically. But despite these complexities, it is still possible to analyze some rapidly varied flows using the one-dimensional flow approximation with reasonable accuracy. The flow in steep channels may be supercritical, and the flow must change to subcritical if the channel can no longer sustain supercritical flow due to a reduced slope of the channel or increased frictional effects. Any such change from supercritical to subcritical flow occurs through a hydraulic jump. A hydraulic jump involves considerable mixing and agitation, and thus a significant amount of mechanical energy dissipation. Consider steady flow through a control volume that encloses the hydrau lic jump, as shown in Fig. 13–39. To make a simple analysis possible, we make the following approximations: 1. The velocity is nearly constant across the channel at sections 1 and 2, and therefore the momentum-flux correction factors are 𝛽1 = 𝛽2 ≅ 1. 2. The pressure in the liquid varies hydrostatically, and we consider gage pressure only since atmospheric pressure acts on all surfaces and its effect cancels out. 3. The wall shear stress and its associated losses are negligible relative to the losses that occur during the hydraulic jump due to the intense agitation. 4. The channel is wide and horizontal. 5. There are no external or body forces other than gravity. For a channel of width b, the conservation of mass relation m .
2 = m .
1 is expressed as 𝜌y1bV1 = 𝜌y2bV2 or y1V1 = y2V2 (13–67) Noting that the only forces acting on the control volume in the horizontal x-direction are the pressure forces, the momentum equation ∑F → = ∑ out 𝛽m · V → − ∑ in 𝛽m · V → in the x-direction becomes a balance between hydrostatic pressure forces and momentum transfer, P1, avg A1 −P2, avg A2 = m · V2 −m · V 1 (13–68) FIGURE 13–38 Rapidly varied flow occurs when there is a sudden change in flow, such as an abrupt change in cross section.
FIGURE 13–39 When riding the rapids, a kayaker encounters several features of both gradually varied flow (GVF) and rapidly varied flow (RVF), with the latter being more exciting.
© Karl Weatherly/Getty Images RF cen96537_ch13_733-792.indd 765 14/01/17 3:21 pm 766 open-channel flow where P1, avg = 𝜌gy1/2 and P2, avg = 𝜌gy2/2. For a channel width of b, we have A1 = y1b, A2 = y2b, and m . = m .
2 = m .
1 = 𝜌A1V1 = 𝜌y1bV1. Substituting and simplifying, the momentum equation reduces to y 2 1 −y 2 2 = 2y1V1 g (V2 −V1) (13–69) Eliminating V2 by using V2 = ( y1/y2)V1 from Eq. 13–67 gives y 2 1 −y 2 2 = 2y1V1 2 gy2 (y1 −y2) (13–70) Canceling the common factor y1 − y2 from both sides and rearranging give ( y2 y1) 2 + y2 y1 −2Fr 2 1 = 0 (13–71) where Fr1 = V1/√gy1. This is a quadratic equation for y2/y1, and it has two roots—one negative and one positive. Noting that y2/y1 cannot be negative since both y2 and y1 are positive quantities, the depth ratio y2/y1 is deter mined to be Depth ratio: y2 y1 = 0.5(−1 + √1 + 8Fr 2 1 ) (13–72) The energy equation (Eq. 13–30) for this horizontal flow section is y1 + V 2 1 2g = y2 + V 2 2 2g + hL (13–73) Noting that V2 = (y1/y2)V1 and Fr1 = V1/√gy1, the head loss associated with a hydraulic jump is expressed as hL = y1 −y2 + V 2 1 −V 2 2 2g = y1 −y2 + y1Fr 2 1 2 (1 −y 2 1 y 2 2 ) (13–74) The energy line for a hydraulic jump is shown in Fig. 13–40. The drop in the energy line across the jump represents the head loss hL associated with the jump. For given values of Fr1 and y1, the downstream flow depth y2 and the head loss hL can be calculated from Eqs. 13–72 and 13–74, respectively. Plotting hL against Fr1 would reveal that hL becomes negative for Fr1 < 1, which is impossible (it would correspond to negative entropy generation, which would be a violation of the second law of thermodynamics). Thus we conclude that the upstream flow must be supercritical (Fr1 > 1) for a hydraulic jump to occur. In other words, it is impossible for subcritical flow to undergo a hydraulic jump. This is analogous to gas flow having to be supersonic (Mach number greater than 1) to undergo a shock wave. Head loss is a measure of the mechanical energy dissipated via inter nal fluid friction, and head loss is usually undesirable as it represents the mechanical energy wasted. But sometimes hydraulic jumps are designed in conjunction with stilling basins and spillways of dams, and it is desir able to waste as much of the mechanical energy as possible to minimize the mechanical energy of the water and thus its potential to cause damage. This is done by first producing supercritical flow by converting high pressure to high linear velocity, and then allowing the flow to agitate and dissipate part x Es Es1 Es2 = y2 + 2g y ρgy1 hL y1 y2 V2 V 1 Energy line Control volume (1) (2) 1 2 Subcritical Supercritical ρgy2 V 2 2 FIGURE 13–40 Schematic and flow depth-specific energy diagram for a hydraulic jump (specific energy decreases).
cen96537_ch13_733-792.indd 766 14/01/17 3:21 pm 767 CHAPTER 13 of its kinetic energy as it breaks down and decelerates to a subcritical veloc ity. Therefore, a measure of performance of a hydraulic jump is its fraction of energy dissipation. The specific energy of the liquid before the hydraulic jump is Es1 = y1 + V1 2 /2g. Then the energy dissipation ratio (Fig. 13–41) is defined as Energy dissipation ratio = hL Es1 = hL y1 + V 2 1 /2g = hL y1(1 + Fr 2 1 /2) (13–75) The fraction of energy dissipation ranges from just a few percent for weak hydraulic jumps (Fr1 < 2) to 85 percent for strong jumps (Fr1 > 9). Unlike a normal shock in gas flow, which occurs at a cross section and thus has negligible thickness, the hydraulic jump occurs over a considerable channel length. In the Froude number range of practical interest, the length of the hydraulic jump is observed to be 4 to 7 times the downstream flow depth y2. Experimental studies indicate that hydraulic jumps can be classified into five categories as shown in Table 13–4, depending primarily on the value of the upstream Froude number Fr1. For Fr1 somewhat higher than 1, the liquid rises slightly during the hydraulic jump, producing standing waves. At larger Fr1, highly damaging oscillating waves occur. The desirable range of Froude numbers is 4.5 < Fr1 < 9, which produces stable and well-balanced steady waves with high levels of energy dissipation within the jump. Hydraulic jumps with Fr1 > 9 produce very rough waves. The depth ratio y2/y1 ranges from slightly over 1 for undular jumps that are mild and involve small rises in surface level to over 12 for strong jumps that are rough and involve high rises in surface level. In this section we limit our consideration to wide horizontal rectangular channels so that edge and gravity effects are negligible. Hydraulic jumps in nonrectangular and sloped channels behave similarly, but the flow charac teristics and thus the relations for depth ratio, head loss, jump length, and dissipation ratio are different.
hL y1 y2 V2 V1 2g Energy line Dissipation ratio = = hL Es1 hL y1 + V 2/2g (1) (2) 2g 1 V 2 1 V 2 2 FIGURE 13–41 The energy dissipation ratio represents the fraction of mechanical energy dissipated during a hydraulic jump.
hL V1 = 7 m/s V2 y1 = 0.8 m y2 Energy line (1) (2) FIGURE 13–42 Schematic for Example 13–8.
EXAMPLE 13–8 Hydraulic Jump Water discharging into a 10-m-wide rectangular horizontal channel from a sluice gate is observed to have undergone a hydraulic jump. The flow depth and velocity before the jump are 0.8 m and 7 m/s, respectively. Determine (a) the flow depth and the Froude number after the jump, (b) the head loss and the energy dissi pation ratio, and (c) the wasted power production potential due to the hydraulic jump (Fig. 13–42).
SOLUTION Water at a specified depth and velocity undergoes a hydraulic jump in a horizontal channel. The depth and Froude number after the jump, the head loss and the dissipation ratio, and the wasted power potential are to be determined.
Assumptions 1 The flow is steady or quasi-steady. 2 The channel is sufficiently wide so that the end effects are negligible.
Properties The density of water is 1000 kg/m3.
cen96537_ch13_733-792.indd 767 14/01/17 3:21 pm 768 open-channel flow TABLE 13–4 Classification of hydraulic jumps Source: U.S. Bureau of Reclamation (1955). Depth Fraction of Upstream Ratio Energy Surface Fr1 y2/y1 Dissipation Description Profile <1 1 0 Impossible jump. Would violate the second law of thermodynamics. 1–1.7 1–2 <5% Undular jump (or standing wave). Small rise in surface level. Low energy dissipation. Surface rollers develop near Fr = 1.7. 1.7–2.5 2–3.1 5–15% Weak jump. Surface rising smoothly, with small rollers. Low energy dissipation. 2.5–4.5 3.1–5.9 15–45% Oscillating jump. Pulsations caused by jets entering at the bottom generate large waves that can travel for miles and damage earth banks. Should be avoided in the design of stilling basins. 4.5–9 5.9–12 45–70% Steady jump. Stable, well-balanced, and insensitive to downstream conditions. Intense eddy motion and high level of energy dissipation within the jump. Recommended range for design. >9 >12 70–85% Strong jump. Rough and intermittent. Very effective energy dissipation, but may be uneconomical compared to other designs because of the larger water heights involved.
Analysis (a) The Froude number before the hydraulic jump is Fr1 = V1 √gy1 = 7 m/s √(9.81 m/s2)(0.8 m) = 2.50 which is greater than 1. Therefore, the flow is indeed supercritical before the jump. The flow depth, velocity, and Froude number after the jump are y2 = 0.5y1(−1 + √1 + 8Fr 2 1 ) = 0.5(0.8 m)(−1 + √1 + 8 × 2.502) = 2.46 m V2 = y1 y2 V1 = 0.8 m 2.46 m (7 m/s) = 2.28 m/s Fr2 = V2 √gy2 = 2.28 m/s √(9.81 m/s2)(2.46 m) = 0.464 cen96537_ch13_733-792.indd 768 14/01/17 3:21 pm 769 CHAPTER 13 FIGURE 13–43 A weir is a flow control device in which the water flows over the obstruction.
(a) © Design Pics RF/The Irish Image Collection/Getty Images RF; (b) Photo courtesy of Bryan Lewis.
(a) (b) Note that the flow depth triples and the Froude number reduces to about one-fifth after the jump.
(b) The head loss is determined from the energy equation to be hL = y1 −y2 + V 2 1 −V 2 2 2g = (0.8 m) −(2.46 m) + (7 m/s)2 −(2.28 m/s)2 2(9.81 m/s2) = 0.572 m The specific energy of water before the jump and the dissipation ratio are Es1 = y1 + V 2 1 2g = (0.8 m) + (7 m/s)2 2(9.81 m/s2) = 3.30 m Dissipation ratio = hL Es1 = 0.572 m 3.30 m = 0.173 Therefore, 17.3 percent of the available head (or mechanical energy) of the liquid is wasted (converted to thermal energy) as a result of frictional effects during this hydraulic jump.
(c) The mass flow rate of water is m · = ρV .
= ρby1V1 = (1000 kg/m3)(0.8 m)(10 m)(7 m/s) = 56,000 kg/s Then the power dissipation corresponding to a head loss of 0.572 m becomes E .
dissipated = m · ghL = (56,000 kg/s)(9.81 m/s2)(0.572 m)( 1 N 1 kg·m/s2) = 314,000 N·m/s = 314 kW Discussion The results show that the hydraulic jump is a highly dissipative process, wasting 314 kW of power production potential in this case. That is, if the water were routed to a hydraulic turbine instead of being released from the sluice gate, up to 314 kW of power could be generated. But this potential is converted to useless thermal energy instead of useful power, causing a water temperature rise of ΔT = E .
dissipated m · cp = 314 kJ/s (56,000 kg/s)(4.18 kJ/kg·°C) = 0.0013°C Note that a 314-kW resistance heater would cause the same temperature rise for water flowing at a rate of 56,000 kg/s.
13–9 ■ FLOW CONTROL AND MEASUREMENT The flow rate in pipes and ducts is controlled by various kinds of valves. Liquid flow in open channels, however, is not confined, and thus the flow rate is controlled by partially blocking the channel. This is done by either allowing the liquid to flow over the obstruction or under it. An obstruction that allows the liquid to flow over it is called a weir (Fig. 13–43), and an obstruction with an adjustable opening at the bottom that allows the liquid to flow underneath it is called an underflow gate. Such devices can be used to control the flow rate through the channel as well as to measure it.
cen96537_ch13_733-792.indd 769 14/01/17 3:21 pm 770 open-channel flow Underflow Gates There are numerous types of underflow gates to control the flow rate, each with certain advantages and disadvantages. Underflow gates are located at the bottom of a wall, dam, or an open channel. Two common types of such gates, the sluice gate and the drum gate, are shown in Fig. 13–44. A sluice gate is typically vertical and has a plane surface, whereas a drum gate has a circular cross section with a streamlined surface. When the gate is partially opened, the upstream liquid accelerates as it approaches the gate, reaches critical speed at the gate, and accelerates further to supercritical speeds past the gate. Therefore, an underflow gate is analo gous to a converging–diverging nozzle in gas dynamics. The discharge from an underflow gate is called a free outflow if the liquid jet streaming out of the gate is open to the atmosphere (Fig. 13–44a), and it is called a drowned (or submerged) outflow if the discharged liquid flashes back and submerges the jet (Fig. 13–44b). In drowned flow, the liquid jet undergoes a hydraulic jump, and thus the downstream flow is subcritical. Also, drowned outflow in volves a high level of turbulence and backflow, and thus a large head loss hL. The flow depth-specific energy diagram for flow through underflow gates with free and drowned outflow is given in Fig. 13–45. Note that the specific energy remains constant for idealized gates with negligible frictional effects (from point 1 to point 2a), but decreases for actual gates. The downstream is supercriti cal for a gate with free outflow (point 2b), but subcritical for one with drowned outflow (point 2c) since a drowned outflow also involves a hydraulic jump to subcritical flow, which involves considerable mixing and energy dissipation. Approximating the frictional effects as negligible and the upstream (or reservoir) velocity to be low, it can be shown by using the Bernoulli equa tion that the discharge velocity of a free jet is (see Chap. 5 for details) V = √2gy1 (13–76) The frictional effects can be accounted for by modifying this relation with a discharge coefficient Cd. Then the discharge velocity at the gate and the flow rate become V = Cd√2gy1 and V .
= Cdba√2gy1 (13–77) where b and a are the width and the height of the gate opening, respectively.
y1 a V1 (b) Sluice gate with drowned outflow Sluice gate V2 y2 y1 a V1 (a) Sluice gate with free outflow Sluice gate Vena contracta V2 y2 y1 V1 (c) Drum gate V2 y2 Drum FIGURE 13–44 Common types of underflow gates to control flow rate.
Es Es1 = y1 + Es1 = Es2a 2g y 2a 2b 1 2c Subcritical flow Drowned outflow Frictionless gate Supercritical flow V 2 1 FIGURE 13–45 Schematic and flow depth-specific energy diagram for flow through underflow gates.
cen96537_ch13_733-792.indd 770 14/01/17 3:21 pm 771 CHAPTER 13 The discharge coefficient Cd = 1 for idealized flow, but Cd < 1 for actual flow through the gates. Experimentally determined values of Cd for under flow gates are plotted in Fig. 13–46 as functions of the contraction coeffi cient y2/a and the depth ratio y1/a. Note that most values of Cd for free out flow from a vertical sluice gate range between 0.5 and 0.6. The Cd values drop sharply for drowned outflow, as expected, and the flow rate decreases for the same upstream conditions. For a given value of y1/a, the value of Cd decreases with increasing y2/a.
0 0 0.1 0.2 0.3 0.4 0.5 0.6 2 4 6 8 y1/a Cd 10 12 14 16 Free outflow Drowned outflow y2/a = 2 3 4 5 6 7 8 FIGURE 13–46 Discharge coefficients for drowned and free discharge from underflow gates.
Data from Henderson, Open Channel Flow, 1st Edition, © 1966. Pearson Education, Inc., Upper Saddle River, NJ.
y1 = 3 m a = 0.25 m y2 = 1.5 m Sluice gate FIGURE 13–47 Schematic for Example 13–9.
EXAMPLE 13–9 Sluice Gate with Drowned Outflow Water is released from a 3-m-deep reservoir into a 6-m-wide open channel through a sluice gate with a 0.25-m-high opening at the channel bottom. The flow depth after all turbulence subsides is measured to be 1.5 m. Determine the rate of discharge (Fig. 13–47).
SOLUTION Water is released from a reservoir through a sluice gate into an open channel. For specified flow depths, the rate of discharge is to be determined.
Assumptions 1 The flow is steady in the mean. 2 The channel is sufficiently wide so that the end effects are negligible.
Analysis The depth ratio y1/a and the contraction coefficient y2/a are y1 a = 3 m 0.25 m = 12 and y2 a = 1.5 m 0.25 m = 6 cen96537_ch13_733-792.indd 771 14/01/17 3:21 pm 772 open-channel flow Overflow Gates Recall that the total mechanical energy of a liquid at any cross section of an open channel can be expressed in terms of heads as H = zb + y + V 2/2g, where y is the flow depth, zb is the elevation of the channel bottom, and V is the average flow velocity. During flow with negligible frictional effects (head loss hL = 0), the total mechanical energy remains constant, and the one-dimensional energy equation for open-channel flow between upstream section 1 and downstream section 2 is written as zb1 + y1 + V 2 1 2g = zb2 + y2 + V 2 2 2g or Es1 = Δzb + Es2 (13–78) where Es = y + V 2/2g is the specific energy and ∆zb = zb2 − zb1 is the elevation of the bottom point of flow at section 2 relative to that at section 1. Therefore, the specific energy of a liquid stream increases by |∆zb| during downhill flow (note that ∆zb is negative for channels inclined down), decreases by ∆zb during uphill flow, and remains constant during horizontal flow. (The specific energy also decreases by hL for all cases if the frictional effects are not negligible.) For a channel of constant width b, V · = AcV = byV = constant in steady flow and V = V ·/Ac. Then the specific energy becomes Es = y + V .
2 2gb2y2 (13–79) The variation of the specific energy Es with flow depth y for steady flow in a channel of constant width b is replotted in Fig. 13–48. This diagram is extremely valuable as it shows the allowable states during flow. Once the upstream conditions at a flow section 1 are specified, the state of the liquid at any section 2 on an Es–y diagram must fall on a point on the specific energy curve that passes through point 1.
Flow over a Bump with Negligible Friction Now consider steady flow with negligible friction over a bump of height ∆zb in a horizontal channel of constant width b, as shown in Fig. 13–47. The energy equation in this case is, from Eq. 13–78, Es2 = Es1 −Δzb (13–80) Therefore, the specific energy of the liquid decreases by ∆zb as it flows over the bump, and the state of the liquid on the Es–y diagram shifts to the left The corresponding discharge coefficient is determined from Fig. 13–46 to be Cd = 0.47. Then the discharge rate becomes V .
= Cdba√2gy1 = 0.47(6 m)(0.25 m)√2(9.81 m/s2)(3 m) = 5.41 m3/s Discussion In the case of free flow, the discharge coefficient would be Cd = 0.59, with a corresponding flow rate of 6.78 m3/s. Therefore, the flow rate decreases considerably when the outflow is drowned.
Es yc Emin Es = y V 2 2g y y Supercritical flow, Fr > 1 Fr = 1 Critical depth Subcritical flow, Fr < 1 V = constant .
FIGURE 13–48 Variation of specific energy Es with depth y for a specified flow rate in a channel of constant width.
cen96537_ch13_733-792.indd 772 14/01/17 3:21 pm 773 CHAPTER 13 by ∆zb, as shown in Fig. 13–49. The conservation of mass equation for a channel of large width is y2V2 = y1V1 and thus V2 = (y1/y2)V1. Then the spe cific energy of the liquid over the bump can be expressed as Es2 = y2 + V 2 2 2g → Es1 −Δzb = y2 + V 2 1 2g y 2 1 y 2 2 (13–81) Rearranging, y 3 2 −(Es1 −Δzb)y 2 2 + V 2 1 2g y 2 1 = 0 (13–82) which is a third-degree polynomial equation in y2 and thus has three solu tions. Disregarding the negative solution, it appears that the flow depth over the bump can have two values. Now the curious question is, does the liquid level rise or drop over the bump? Our intuition says the entire liquid body will follow the bump and thus the liquid surface will rise over the bump, but this is not necessarily so. Noting that specific energy is the sum of the flow depth and dynamic head, either scenario is possible, depending on how the velocity changes. The Es–y diagram in Fig. 13–49 gives us the definite answer: If the flow before the bump is subcritical (state 1a), the flow depth y2 decreases (state 2a). If the decrease in flow depth is greater than the bump height (i.e., y1 − y2 > ∆zb), the free surface is suppressed. But if the flow is supercritical as it approaches the bump (state 1b), the flow depth rises over the bump (state 2b), creating a bump along the free surface. The situation is reversed if the channel has a depression of depth ∆zb instead of a bump: The specific energy in this case increases (so that state 2 is to the right of state 1 on the Es–y diagram) since ∆zb is negative. There fore, the flow depth increases if the approach flow is subcritical and decreases if it is supercritical. Now let’s reconsider flow over a bump with negligible friction, as dis cussed earlier. As the height of the bump ∆zb is increased, point 2 (either 2a or 2b for sub- or supercritical flow) continues shifting to the left on the Es–y diagram, until finally reaching the critical point. That is, the flow over the bump is critical when the bump height is ∆zc = Es1 − Esc = Es1 − Emin, and the specific energy of the liquid reaches its minimum level. The question that comes to mind is, what happens if the bump height is increased further? Does the specific energy of the liquid continue decreas ing? The answer to this question is a resounding no since the liquid is already at its minimum energy level, and its energy cannot decrease any further. In other words, the liquid is already at the furthest left point on the Es–y diagram, and no point further left can satisfy conservation of mass and energy and the momentum equation. Therefore, the flow must remain critical. The flow at this state is said to be choked. In gas dynamics, this is analogous to the flow in a converging nozzle accelerating as the back pres sure is lowered, and reaching the speed of sound at the nozzle exit when the back pressure reaches the critical pressure. But the nozzle exit velocity remains at the sonic level no matter how much the back pressure is lowered. Here again, the flow is choked.
Es Emin = Ec y Δzb Supercritical flow Subcritical flow 2b 2a 1a 1b V2 V1 y1 y2 Δzb Subcritical upstream flow Supercritical upstream flow Bump FIGURE 13–49 Schematic and flow depth-specific energy diagram for flow over a bump for subcritical and supercritical upstream flows.
cen96537_ch13_733-792.indd 773 14/01/17 3:21 pm 774 open-channel flow Broad-Crested Weir The discussions on flow over a high bump can be summarized as follows: The flow over a sufficiently high obstruction in an open channel is always critical. Such obstructions placed intentionally in an open channel to mea sure the flow rate are called weirs. Therefore, the flow velocity over a suf ficiently broad weir is the critical velocity, which is expressed as V = √gyc, where yc is the critical depth. Then the flow rate over a weir of width b is ex pressed as V .
= AcV = ycb√gyc = bg1/2y 3/2 c (13–83) A broad-crested weir is a rectangular block of height Pw and length Lw that has a horizontal crest over which critical flow occurs (Fig. 13–50). The upstream head above the top surface of the weir is called the weir head and is denoted by H. To obtain a relation for the critical depth yc in terms of weir head H, we write the energy equation between a section upstream and a section over the weir for flow with negligible friction as H + Pw + V 2 1 2g = yc + Pw + V 2 c 2g (13–84) Cancelling Pw from both sides and substituting Vc = √gyc give yc = 2 3 (H + V 2 1 2g) (13–85) Substituting into Eq. 13–83, the flow rate for this idealized flow case with negligible friction is determined to be V .
ideal = b√g( 2 3) 3/2 (H + V1 2 2g ) 3/2 (13–86) This relation shows the functional dependence of the flow rate on the flow parameters, but it overpredicts the flow rate by several percent because it does not consider the frictional effects. These effects are typically accounted for by modifying the theoretical relation (Eq. 13–86) with an experimentally determined weir discharge coefficient Cwd as Broad-crested weir: V .
= Cwd, broadb√g( 2 3) 3/2 (H + V 2 1 2g ) 3/2 (13–87) where reasonably accurate values of discharge coefficients for broad-crested weirs can be obtained from (Chow, 1959) Cwd, broad = 0.65 √1 + H/Pw (13–88) More accurate but complicated relations for Cwd, broad are also available in the literature (e.g., Ackers, 1978). Also, the upstream velocity V1 is usu ally very low, and it can be disregarded. This is especially the case for high weirs. Then the flow rate is approximated as Broad-crested weir with low V1: V .
≅Cwd, broadb√g( 2 3) 3/2 H3/2 (13–89) It should always be kept in mind that the basic requirement for the use of Eqs. 13–87 to 13–89 is the establishment of critical flow above the weir, V1 Pw H Discharge Vc Lw yc Broad-crested weir FIGURE 13–50 Flow over a broad-crested weir.
cen96537_ch13_733-792.indd 774 14/01/17 3:21 pm 775 CHAPTER 13 and this puts some limitations on the weir length Lw. If the weir is too long (Lw > 12H), wall shear effects dominate and cause the flow over the weir to be subcritical. If the weir is too short (Lw < 2H), the liquid may not be able to accelerate to critical velocity. Based on observations, the proper length of the broad-crested weir is 2H < Lw < 12H. Note that a weir that is too long for one flow may be too short for another flow, depending on the value of the weir head H. Therefore, the range of flow rates should be known before a weir can be selected.
Sharp-Crested Weirs A sharp-crested weir is a vertical plate placed in a channel that forces the liquid to flow through an opening to measure the flow rate. The type of the weir is characterized by the shape of the opening. A vertical thin plate with a straight top edge is referred to as rectangular weir since the cross sec tion of the flow over it is rectangular; a weir with a triangular opening is referred to as a triangular weir; etc. Upstream flow is subcritical and becomes critical as it approaches the weir. The liquid continues to accelerate and discharges as a supercritical flow stream that resembles a free jet. The reason for acceleration is the steady decline in the elevation of the free surface, and the conversion of this ele vation head into velocity head. The flow-rate correlations given below are based on the free overfall of liquid discharge past the weir, called a nappe, being clear from the weir. It may be necessary to ventilate the space under the nappe to assure atmospheric pressure underneath. Empirical relations for drowned weirs are also available. Consider the flow of a liquid over a sharp-crested weir placed in a hori zontal channel, as shown in Fig. 13–51. For simplicity, the velocity upstream of the weir is approximated as being nearly constant through vertical cross section 1. The total energy of the upstream liquid expressed as a head rela tive to the channel bottom is the specific energy, which is the sum of the flow depth and the velocity head. That is, y1 + V1 2 /2g, where y1 = H + Pw. The flow over the weir is not one-dimensional since the liquid undergoes large changes in velocity and direction over the weir. But the pressure within the nappe is atmospheric. A simple relation for the variation of liquid velocity over the weir is ob tained by assuming negligible friction and writing the Bernoulli equation between a point in upstream flow (point 1) and a point over the weir at a distance h from the upstream liquid level as H + Pw + V 2 1 2g = (H + Pw −h) + u 2 2 2g (13–90) Cancelling the common terms and solving for u2, the idealized velocity dis tribution over the weir is determined to be u2 = √2gh + V 2 1 (13–91) In reality, the liquid surface level drops somewhat over the weir as the liquid starts its free overfall (the drawdown effect at the top) and the flow separation at the top edge of the weir further narrows the nappe (the contraction effect at the bottom). As a result, the flow height over the weir is considerably smaller than H. When the drawdown and contraction effects are disregarded y x V1 u2(h) h Nappe Weir (2) (1) H Pw 2 1 FIGURE 13–51 Flow over a sharp-crested weir.
cen96537_ch13_733-792.indd 775 14/01/17 3:21 pm 776 open-channel flow for simplicity, the flow rate is obtained by integrating the product of the flow velocity and the differential flow area over the entire flow area, V .
= ∫Ac u2 dAc2 = ∫ H h=0 √2gh + V 2 1 w dh (13–92) where w is the width of the flow area at distance h from the upstream free surface. In general, w is a function of h. But for a rectangular weir, w = b, which is constant. Then the integration can be performed easily, and the flow rate for a rectangular weir for idealized flow with negligible friction and negli gible drawdown and contraction effects is determined to be V .
ideal = 2 3 b√2g[(H + V 2 1 2g ) 3/2 −( V 2 1 2g ) 3/2 ] (13–93) When the weir height is large relative to the weir head (Pw ≫ H), the upstream velocity V1 is low and the upstream velocity head can be neglected. That is, V1 2/2g ≪ H. Then, V .
ideal, rec ≅2 3 b√2gH3/2 (13–94) Therefore, the flow rate can be determined from knowledge of two geometric quantities: the crest width b and the weir head H, which is the vertical dis tance between the weir crest and the upstream free surface. This simplified analysis gives the general form of the flow-rate relation, but it needs to be modified to account for the frictional and surface tension effects, which play a secondary role, as well as the drawdown and contrac tion effects. Again this is done by multiplying the ideal flow-rate relations by an experimentally determined weir discharge coefficient Cwd. Then the flow rate for a sharp-crested rectangular weir is expressed as Sharp-crested rectangular weir: V .
rec = Cwd, rec 2 3 b√2gH3/2 (13–95) where, from Ref. 1 (Ackers, 1978), Cwd, rec = 0.598 + 0.0897 H Pw for H Pw ≤2 (13–96) This formula is applicable over a wide range of upstream Reynolds number defined as Re = V1H/𝜈. More precise but also more complex correlations are also available in the literature. Note that Eq. 13–95 is valid for full-width rectangular weirs. If the width of the weir is less than the channel width so that the flow is forced to contract, an additional coefficient for contraction correction should be incorporated to properly account for this effect. Another type of sharp-crested weir commonly used for flow measurement is the triangular weir (also called the V-notch weir) shown in Fig. 13–52. The triangular weir has the advantage that it maintains a high weir head H even for small flow rates because of the decreasing flow area with decreasing H, and thus it can be used to measure a wide range of flow rates accurately. From geometric consideration, the notch width can be expressed as w = 2(H − h) tan(𝜃/2), where 𝜃 is the V-notch angle. Substituting into Eq. 13–92 and performing the integration give the ideal flow rate for a triangular weir to be V · ideal, tri = 8 15 tan( 𝜃 2)√2gH5/2 (13–97) Upstream free surface Weir plate θ H w h Pw FIGURE 13–52 A triangular (or V-notch) sharp-crested weir plate geometry. The view is from downstream looking upstream.
cen96537_ch13_733-792.indd 776 14/01/17 3:21 pm 777 CHAPTER 13 where we again neglected the upstream velocity head. The frictional and other dissipative effects are again accounted for conveniently by multiplying the ideal flow rate by a weir discharge coefficient. Then the flow rate for a sharp-crested triangular weir becomes Sharp-crested triangular weir: V .
= Cwd, tri 8 15 tan( 𝜃 2)√2gH5/2 (13–98) where the values of Cwd, tri typically range between 0.58 and 0.62. Therefore, the fluid friction, the constriction of flow area, and other dissipative effects cause the flow rate through the V-notch to decrease by about 40 percent compared to the ideal case. For most practical cases (H > 0.2 m and 45° < 𝜃 < 120°), the value of the weir discharge coefficient is about Cwd, tri = 0.58. More precise values are available in the literature.
Es Es1 Es2 y2 y1 y Δzb Subcritical flow 2 1 V1 = 1.2 m/s y1 = 0.80 m y2 Δzb = 0.15 m Depression over the bump Bump FIGURE 13–53 Schematic and flow depth-specific energy diagram for Example 13–10.
EXAMPLE 13–10 Subcritical Flow over a Bump Water flowing in a wide horizontal open channel encounters a 15-cm-high bump at the bottom of the channel. If the flow depth is 0.80 m and the velocity is 1.2 m/s before the bump, determine if the water surface is depressed over the bump (Fig. 13–53) and if so, by how much.
SOLUTION Water flowing in a horizontal open channel encounters a bump. It will be determined if the water surface is depressed over the bump.
Assumptions 1 The flow is steady. 2 Frictional effects are negligible so that there is no dissipation of mechanical energy. 3 The channel is sufficiently wide so that the end effects are negligible.
Analysis The upstream Froude number and the critical depth are Fr1 = V1 √gy1 = 1.2 m/s √(9.81 m2/s)(0.80 m) = 0.428 yc =( V .
2 gb2) 1/3 = ( (by1V1)2 gb2 ) 1/3 = ( y 2 1 V 2 1 g ) 1/3 = ( (0.8 m)2(1.2 m/s)2 9.81 m/s2 ) 1/3 = 0.455 m The flow is subcritical since Fr < 1 and therefore the flow depth decreases over the bump. The upstream specific energy is Es1 = y1 + V 2 1 2g = (0.80 m) + (1.2 m/s)2 2(9.81 m/s2) = 0.873 m The flow depth over the bump is determined from y 3 2 −(Es1 −Δzb)y 2 2 + V 2 1 2g y 2 1 = 0 Substituting, y 3 2 −(0.873 −0.15 m)y 2 2 + (1.2 m/s)2 2(9.81 m/s2) (0.80 m)2 = 0 or y 3 2 −0.723y 2 2 + 0.0470 = 0 cen96537_ch13_733-792.indd 777 14/01/17 3:21 pm 778 open-channel flow Using an equation solver, the three roots of this equation are determined to be 0.59 m, 0.36 m, and −0.22 m. We discard the negative solution as physi cally impossible. We also eliminate the solution 0.36 m since it is less than the critical depth, and it can occur only in supercritical flow. Thus the only meaningful solution for flow depth over the bump is y2 = 0.59 m. Then the distance of the water surface over the bump from the channel bottom is ∆zb + y2 = 0.15 + 0.59 = 0.74 m, which is less than y1 = 0.80 m. Therefore, the water surface is depressed over the bump in the amount of Depression = y1 −( y2 + Δzb) = 0.80 −(0.59 + 0.15) = 0.06 m Discussion Note that having y2 < y1 does not necessarily indicate that the water surface is depressed (it may still rise over the bump). The surface is depressed over the bump only when the difference y1 − y2 is larger than the bump height ∆zb. Also, the actual value of depression may be different than 0.06 m because of the frictional effects that are neglected in the analysis.
EXAMPLE 13–11 Measuring Flow Rate by a Weir The flow rate of water in a 5-m-wide horizontal open channel is being measured with a 0.60-m-high sharp-crested rectangular weir of equal width. If the water depth upstream is 1.5 m, determine the flow rate of water (Fig. 13–54).
SOLUTION The water depth upstream of a horizontal open channel equip ped with a sharp-crested rectangular weir is measured. The flow rate is to be deter mined.
Assumptions 1 The flow is steady. 2 The upstream velocity head is negligible. 3 The channel is sufficiently wide so that the end effects are negligible.
Analysis The weir head is H = y1 −Pw = 1.5 −0.60 = 0.90 m The discharge coefficient of the weir is Cwd, rec = 0.598 + 0.0897 H Pw = 0.598 + 0.0897 0.90 0.60 = 0.733 The condition H/Pw < 2 is satisfied since 0.9/0.6 = 1.5. Then the water flow rate through the channel becomes V .
rec = Cwd, rec 2 3 b√2gH3/2 = (0.733) 2 3 (5 m)√2(9.81 m/s2)(0.90 m)3/2 = 9.24 m3/s Discussion The upstream velocity and the upstream velocity head are V1 = V .
by1 = 9.24 m3/s (5 m)(1.5 m) = 1.23 m/s and V 2 1 2g = (1.23 m/s)2 2(9.81 m/s2) = 0.077 m This is 8.6 percent of the weir head, which is significant. When the upstream veloc ity head is considered, the flow rate becomes 10.2 m3/s, which is about 10 per cent higher than the value determined. Therefore, it is good practice to consider the upstream velocity head unless the weir height Pw is very large relative to the weir head H.
V1 Sharp-crested rectangular weir y1 = 1.5 m P w = 0.60 m b = 5 m FIGURE 13–54 Schematic for Example 13–11.
cen96537_ch13_733-792.indd 778 14/01/17 3:21 pm 779 CHAPTER 13 FIGURE 13–55 A scour hole developed around this bridge pier near San Diego during high flows in the river channel.
Photo by Peggy Johnson, Penn State, used by permission.
FIGURE 13–56 Scour that developed around the bridge foundation during a 50 year flood in 1996 caused this bridge to fail in central PA. A temporary metal bridge was placed across the opening while a new bridge was being designed.
Photo by Peggy Johnson, Penn State, used by permission.
Guest Author: Peggy A. Johnson, Penn State University Bridge scour is the most common cause of bridge failure in the United States (Wardhana and Hadipriono, 2003). Bridge scour is the erosion of a stream or river channel bed in the vicinity of a bridge, including erosion around the bridge piers and abutments as well as the erosion and lowering of the entire channel bed. Scour around bridge foundations has been a leading cause of bridge failure for the nearly 400,000 bridges over waterways in the United States. (Figs. 13-55 and 13-56) A few recent examples of the damage that can be caused by high flows in rivers at bridges illustrate the magnitude of the problem. During the 1993 flood in the upper Mississippi and lower Mis souri river basins, at least 22 of the 28 bridge failures were due to scour, at an estimated cost of more than $8 million (Kamojjala et al., 1994). During the “Super Flood” in Tennessee in 2010 in which more than 30 counties were declared major disaster areas, flooding in Tennessee’s rivers caused scour and embankment erosion at 587 bridges and resulted in the closure of more than 50 bridges. In the fall of 2011, Hurricane Irene and Tropical Storm Lee in the mid-Atlantic and northeast United States caused flooding in rivers that resulted in numerous bridge failures and damage to bridges due to scour. The mechanics of scour at bridge piers has been studied in laboratories and computer models. The primary mechanism is thought to be due to a “horse shoe” vortex that forms during floods as an adverse pressure gradient caused by the pier drives a portion of the approach flow downward just ahead of the pier (Arneson et al., 2012). The rate of erosion of the scour hole is directly associ ated with the magnitude of the downflow, which is directly related to the veloc ity of the approaching river flow. The strong vortex lifts the sediment out of the hole and deposits it downstream in the wake vortex. The result is a deep hole upstream of the bridge pier that can cause the bridge foundation to become unstable. Protecting bridge piers over rivers and streams against the damaging flood waters remains a major challenge for states across the country. Flood flows in channels have enormous capacity to move sediment and rock; thus, traditional protection, such as riprap, is often not sufficient. There has been considerable research on the use of vanes and similar structures in the river channel to help direct the flow around the bridge piers and abutments and provide a smoother transition of the flow through the bridge opening (Johnson et al., 2010).
References Arneson, L. A., Zevenbergen, L. W., Lagasse, P. F., and Clopper, P. E. (2012). Hydraulic Engineering Circular 18, Evaluating Scour at Bridges. Federal Highway Administration Report FHWA-HIF-12-003, HEC-18, Washington, D.C.
Johnson, P. A., Sheeder, S. A., and Newlin, J. T. (2010). Waterway transitions at US bridges. Water and Environment Journal, 24 (2010), 274–281.
Kamojjala, S., Gattu, N. P., Parola, A. C., and Hagerty, D. J. (1994). “Analysis of 1993 Upper Mississippi flood highway infrastructure damage,” in ASCE Proceedings of the First International Conference of Water Resources Engineering, San Antonio, TX, pp. 1061–1065.
Wardhana, K., and Hadipriono, F. C. (2003). ASCE Journal of Performance of Constructed Facilities, 17(3), 144–150.
APPLICATION SPOTLIGHT ■ Bridge Scour cen96537_ch13_733-792.indd 779 14/01/17 3:21 pm 780 open-channel flow SUMMARY Open-channel flow refers to the flow of liquids in channels open to the atmosphere or in partially filled conduits. The flow in a channel is said to be uniform if the flow depth (and thus the average velocity) remains constant. Otherwise, the flow is said to be nonuniform or varied. The hydraulic radius is defined as Rh = Ac/p. The dimensionless Froude number is defined as Fr = V √gLc = V √gy The flow is classified as subcritical for Fr < 1, critical for Fr = 1, and supercritical for Fr > 1. Flow depth in critical flow is called the critical depth and is expressed as yc = V .
2 gA 2 c or yc = ( V .
2 gb2) 1/3 where b is the channel width for wide channels. The speed at which a surface disturbance travels through a liquid of depth y is the wave speed c0, which is expressed as c0 = √gy. The total mechanical energy of a liquid in a channel is expressed in terms of heads as H = zb + y + V 2 2g where zb is the elevation head, P/𝜌g = y is the pressure head, and V 2/2g is the velocity head. The sum of the pressure and dynamic heads is called the specific energy Es, Es = y + V 2 2g The conservation of mass equation is Ac1V1 = Ac2V2. The energy equation is expressed as y1 + V 2 1 2g + S0L = y2 + V 2 2 2g + hL Here hL is the head loss and S0 = tan 𝜃 is the bottom slope of a channel. The friction slope is defined as Sf = hL/L. The flow depth in uniform flow is called the normal depth yn, and the average flow velocity is called the uniform-flow velocity V0. The velocity and flow rate in uniform flow are given by V0 = a nR 2/3 h S 1/2 0 and V .
= a nAcR 2/3 h S 1/2 0 where n is the Manning coefficient whose value depends on the roughness of the channel surfaces, and a = 1 m1/3/s = (3.2808 ft)1/3/s = 1.486 ft1/3/s. If yn = yc, the flow is uni form critical flow, and the bottom slope S0 equals the critical slope Sc expressed as Sc = gn2yc a2R 4/3 h which simplifies to Sc = gn2 a2y 1/3 c for film flow or flow in a wide rectangular channel with b ≫ yc. The best hydraulic cross section for an open channel is the one with the maximum hydraulic radius, or equivalently, the one with the minimum wetted perimeter for a specified cross-sectional area. The criteria for best hydraulic cross sec tion for a rectangular channel is y = b/2. The best cross sec tion for trapezoidal channels is half of a hexagon. In gradually varied flow (GVF), the flow depth changes gradually and smoothly with downstream distance. The surface profile y(x) is calculated by integrating the GVF equation, dy dx = S0 −Sf 1 −Fr2 In rapidly varied flow (RVF), the flow depth changes markedly over a relatively short distance in the flow direc tion. Any change from supercritical to subcritical flow occurs through a hydraulic jump, which is a highly dissipa tive process. The depth ratio y2/y1, head loss, and energy dis sipation ratio during hydraulic jump are expressed as y2 y1 = 0.5(−1 + √1 + 8Fr 2 1 ) hL = y1 −y2 + V 2 1 −V 2 2 2g = y1 −y2 + y1Fr 2 1 2 (1 −y 2 1 y 2 2 ) Dissipation ratio = hL Es1 = hL y1 + V 2 1 /2g = hL y1(1 + Fr 2 1 /2) An obstruction that allows the liquid to flow over it is called a weir, and an obstruction with an adjustable opening at the bottom that allows the liquid to flow underneath it is called an underflow gate. The flow rate through a sluice gate is given by V .
= Cdba√2gy1 where b and a are the width and the height of the gate open ing, respectively, and Cd is the discharge coefficient, which accounts for the frictional effects. A broad-crested weir is a rectangular block that has a hor izontal crest over which critical flow occurs. The upstream head above the top surface of the weir is called the weir head, H. The flow rate is expressed as V .
= Cwd, broadb√g( 2 3) 3/2 (H + V 2 1 2g) 3/2 cen96537_ch13_733-792.indd 780 14/01/17 3:21 pm 781 CHAPTER 13 where the discharge coefficient is Cwd, broad = 0.65 √1 + H/Pw The flow rate for a sharp-crested rectangular weir is expressed as V .
rec = Cwd, rec 2 3 b√2gH3/2 where Cwd, rec = 0.598 + 0.0897 H Pw for H Pw ≤2 For a sharp-crested triangular weir, the flow rate is given as V .
= Cwd, tri 8 15 tan( 𝜃 2)√2gH 5/2 where the values of Cwd, tri typically range between 0.58 and 0.62. Open-channel analysis is commonly used in the design of sewer systems, irrigation systems, floodways, and dams. Some open-channel flows are analyzed in Chap. 15 using computational fluid dynamics (CFD).
REFERENCES AND SUGGESTED READING 1. P. Ackers et al. Weirs and Flumes for Flow Measurement. New York: Wiley, 1978.
2. B. A. Bakhmeteff. Hydraulics of Open Channels. New York: McGraw-Hill, 1932.
3. M. H. Chaudhry. Open Channel Flow. Upper Saddle River, NJ: Prentice Hall, 1993.
4. V. T. Chow. Open Channel Hydraulics. New York: McGraw-Hill, 1959.
5. R. H. French. Open Channel Hydraulics. New York: McGraw-Hill, 1985.
6. F. M. Henderson. Open Channel Flow. New York: Macmillan, 1966.
7. C. C. Mei. The Applied Dynamics of Ocean Surface Waves. New York: Wiley, 1983.
8. U. S. Bureau of Reclamation. “Research Studies on Stilling Basins, Energy Dissipaters, and Associated Appurtenances,” Hydraulic Lab Report Hyd.-399, June 1, 1955.
PROBLEMS Classification, Froude Number, and Wave Speed 13–1C What is the driving force for flow in an open chan nel? How is the flow rate in an open channel established?
13–2C How does open-channel flow differ from internal flow?
13–3C How does the pressure change along the free surface in an open-channel flow?
13–4C What causes the flow in an open channel to be var ied (or nonuniform)? How does rapidly varied flow differ from gradually varied flow?
13–5C What is normal depth? Explain how it is established in open channels.
13–6C How does uniform flow differ from nonuniform flow in open channels? In what kind of channels is uniform flow observed?
13–7C Given the average flow velocity and the flow depth, explain how you would determine if the flow in open chan nels is tranquil, critical, or rapid.
13–8C The flow in an open channel is observed to have undergone a hydraulic jump. Is the flow upstream from the jump necessarily supercritical? Is the flow downstream from the jump necessarily subcritical?
13–9C What is critical depth in open-channel flow? For a given average flow velocity, how is it determined?
13–10C What is the Froude number? How is it defined? What is its physical significance?
13–11C In open channels, how is hydraulic radius defined? Knowing the hydraulic radius, how can the hydraulic diameter of the channel be determined?
13–12 Water at 20°C flows in a partially full 4-m-diameter circular channel at an average velocity of 2 m/s. If the maxi mum water depth is 1 m, determine the hydraulic radius, the Reynolds number, and the flow regime.
Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and SI users can ignore them. Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software.
cen96537_ch13_733-792.indd 781 14/01/17 3:21 pm 782 open-channel flow 1 m R = 2 m FIGURE P13–12 13–13 Consider the flow of water in a wide channel. Deter mine the speed of a small disturbance in the flow if the flow depth is (a) 50 cm and (b) 100 cm. What would your answer be if the fluid were oil?
13–14 Water at 15°C is flowing uniformly in a 2-m-wide rectangular channel at an average velocity of 1.5 m/s. If the water depth is 24 cm, determine whether the flow is subcriti cal or supercritical. Answer: subcritical 13–15 After heavy rain, water flows on a concrete surface at an average velocity of 1.3 m/s. If the water depth is 2 cm, determine whether the flow is subcritical or supercritical.
13–16 Water at 20°C is flowing uniformly in a wide rectan gular channel at an average velocity of 1.5 m/s. If the water depth is 0.16 m, determine (a) whether the flow is lami nar or turbulent and (b) whether the flow is subcritical or supercritical.
13–17 Water at 10°C flows in a 3-m-diameter circular chan nel half-full at an average velocity of 2.5 m/s. Determine the hydraulic radius, the Reynolds number, and the flow regime (laminar or turbulent).
13–18 Repeat Prob. 13–17 for a channel diameter of 2 m.
13–19 A single wave is initiated in a sea by a strong jolt during an earthquake. Taking the average water depth to be 2 km and the density of seawater to be 1.030 kg/m3, deter mine the speed of propagation of this wave.
Specific Energy and the Energy Equation 13–20C Consider steady flow of water through two identi cal open rectangular channels at identical flow rates. If the flow in one channel is subcritical and in the other supercriti cal, can the specific energies of the water in these two chan nels be identical? Explain.
13–21C Consider steady flow of a liquid through a wide rectangular channel. It is claimed that the energy line of flow is parallel to the channel bottom when the frictional losses are negligible. Do you agree?
13–22C Consider steady one-dimensional flow through a wide rectangular channel. Someone claims that the total mechanical energy of the fluid at the free surface of a cross section is equal to that of the fluid at the channel bottom of the same cross section. Do you agree? Explain.
13–23C How is the total mechanical energy of a fluid dur ing steady one-dimensional flow through a wide rectangular channel expressed in terms of heads? How is it related to the specific energy of the fluid?
13–24C How is the specific energy of a fluid flowing in an open channel defined in terms of heads?
13–25C For a given flow rate through an open channel, the variation of specific energy with flow depth is studied. One person claims that the specific energy of the fluid will be minimum when the flow is critical, but another person claims that the specific energy will be minimum when the flow is subcritical. What is your opinion?
13–26C Consider steady supercritical flow of water through an open rectangular channel at a constant flow rate. Someone claims that the larger is the flow depth, the larger the specific energy of water. Do you agree? Explain.
13–27C During steady and uniform flow through an open channel of rectangular cross section, a person claims that the specific energy of the fluid remains constant. A second per son claims that the specific energy decreases along the flow because of the frictional effects and thus head loss. With which person do you agree? Explain.
13–28C How is the friction slope defined? Under what conditions is it equal to the bottom slope of an open channel?
13–29 Water at 10°C flows in a 6-m-wide rectangular chan nel at a depth of 0.55 m and a flow rate of 12 m3/s. Determine (a) the critical depth, (b) whether the flow is subcritical or supercritical, and (c) the alternate depth. Answers: (a) 0.742 m, (b) supercritical, (c) 1.03 m 13–30E Water at 65°F flows at a depth of 1.4 ft with an average velocity of 20 ft/s in a wide rectangular channel. Determine (a) the Froude number, (b) the critical depth, and (c) whether the flow is subcritical or supercritical. What would your response be if the flow depth were 0.2 ft?
13–31E Repeat Prob. 13–30E for an average velocity of 10 ft/s.
13–32 Water flows through a 2-m-wide rectangular chan nel with an average velocity of 5 m/s. If the flow is critical, determine the flow rate of water. Answer: 25.5 m3/s 13–33 Water at 20°C flows at a depth of 0.4 m with an average velocity of 4 m/s in a rectangular channel. Determine the specific energy of the water and whether the flow is sub critical or supercritical.
13–34 Water flows half-full through a hexagonal channel of bottom width 2 m at a rate of 60 m3/s. Determine (a) the average velocity and (b) whether the flow is subcritical and supercritical.
13–35 Repeat Prob. 13–34 for a flow rate of 30 m3/s.
13–36 Water flows half-full through a 38-cm-diameter steel channel at an average velocity of 2.3 m/s. Determine the volume flow rate and whether the flow is subcritical or supercritical.
cen96537_ch13_733-792.indd 782 14/01/17 3:21 pm 783 CHAPTER 13 13–37 Water flows steadily in a 1.75-m-wide rectangular channel at a rate of 0.85 m3/s. If the flow depth is 0.40 m, determine the flow velocity and if the flow is subcritical or supercritical. Also determine the alternate flow depth if the character of flow were to change.
Uniform Flow and Best Hydraulic Cross Sections 13–38C Which is a better hydraulic cross section for an open channel: one with a small or a large hydraulic radius?
13–39C Which is the best hydraulic cross section for an open channel: (a) circular, (b) rectangular, (c) trapezoidal, or (d) triangular?
13–40C The best hydraulic cross section for a rectangular open channel is one whose fluid height is (a) half, (b) twice, (c) equal to, or (d) one-third the channel width.
13–41C The best hydraulic cross section for a trapezoidal channel of base width b is one for which the length of the side edge of the flow section is (a) b, (b) b/2, (c) 2b, or (d) √3b.
13–42C When is the flow in an open channel said to be uniform? Under what conditions will the flow in an open channel remain uniform?
13–43C During uniform flow in an open channel, someone claims that the head loss can be determined by simply multi plying the bottom slope by the channel length. Can it be this simple? Explain.
13–44C Consider uniform flow through a wide rectangular channel. If the bottom slope is increased, the flow depth will (a) increase, (b) decrease, or (c) remain constant.
13–45 Water is flowing uniformly in a finished-concrete channel of trapezoidal cross section with a bottom width of 0.8 m, trapezoid angle of 50°, and a bottom angle of 0.4°. If the flow depth is measured to be 0.52 m, determine the flow rate of water through the channel.
y = 0.52 m θ = 50° b = 0.8 m FIGURE P13–45 13–46E A 3-ft-diameter semicircular channel made of unfinished concrete is to transport water to a distance of 1 mi uniformly. If the flow rate is to reach 90 ft3/s when the channel is full, determine the minimum elevation difference across the channel.
13–47 A trapezoidal channel with a bottom width of 6 m, free surface width of 12 m, and flow depth of 1.6 m discharges water at a rate of 80 m3/s. If the surfaces of the channel are lined with asphalt (n = 0.016), determine the elevation drop of the channel per kilometer. Answer: 6.75 m 1.6 m 12 m 6 m FIGURE P13–47 13–48 Reconsider Prob. 13–47. If the maximum flow height the channel can accommodate is 3.2 m, determine the maximum flow rate through the channel.
13–49 Consider water flow through two identical channels with square flow sections of 4 m × 4 m. Now the two channels are combined, forming a 8-m-wide channel. The flow rate is adjusted so that the flow depth remains constant at 4 m. Determine the percent increase in flow rate as a result of combining the channels.
4 m 4 m 4 m 4 m FIGURE P13–49 13–50 Water is to be transported at a rate of 10 m3/s in uniform flow in an open channel whose surfaces are asphalt lined. The bottom slope is 0.0015. Determine the dimensions of the best cross section if the shape of the channel is (a) circular of diameter D, (b) rectangular of bottom width b, and (c) trap e zoidal of bottom width b.
13–51E Water is to be transported in a cast iron rectangular channel with a bottom width of 6 ft at a rate of 70 ft3/s. The terrain is such that the channel bottom drops 2.1 ft per 1000 ft length. Determine the minimum height of the channel under uniform-flow conditions.
cen96537_ch13_733-792.indd 783 14/01/17 3:21 pm 784 open-channel flow y b = 6 ft V = 70 ft3/s .
FIGURE P13–51E 13–52 A clean-earth trapezoidal channel with a bottom width of 2.5 m and a side surface slope of 1:1 is to drain water uniformly at a rate of 14 m3/s to a distance of 1 km. If the flow depth is not to exceed 1.2 m, determine the required elevation drop. Answer: 7.02 m 13–53 A water draining system with a constant slope of 0.0025 is to be built of three circular channels made of fin ished concrete. Two of the channels have a diameter of 1.8 m and drain into the third channel. If all channels are to run half-full and the losses at the junction are negligible, deter mine the diameter of the third channel. Answer: 2.33 m 13–54 Water flows in a channel whose bottom slope is 0.002 and whose cross section is as shown in Fig. P13–54. The dimensions and the Manning coefficients for the sur faces of different subsections are also given on the figure. Determine the flow rate through the channel and the effective Manning coefficient for the channel.
6 m 1.5 m 2 m 2 m 10 m Light brush n2 = 0.050 Concrete channel n1 = 0.014 1 2 FIGURE P13–54 13–55 A 2-m-internal-diameter circular steel storm drain (n = 0.012) is to discharge water uniformly at a rate of 12 m3/s to a distance of 1 km. If the maximum depth is to be 1.5 m, determine the required elevation drop.
1.5 m R = 1 m FIGURE P13–55 13–56 A cast iron V-shaped water channel shown in Fig. P13–56 has a bottom slope of 0.5°. For a flow depth of 0.75 m at the center, determine the discharge rate in uniform flow. Answer: 1.03 m3/s 20° 20° 0.75 m FIGURE P13–56 13–57 Consider uniform flow in an asphalt-lined rect angular channel with a flow area of 2 m2 and a bottom slope of 0.0003. By varying the depth-to-width ratio y/b from 0.1 to 2.0, calculate and plot the flow rate, and con firm that the best flow cross section occurs when the flow depth-to-width ratio is 0.5.
13–58E A rectangular channel with a bottom slope of 0.0004 is to be built to transport water at a rate of 1500 ft3/s. Determine the best dimensions of the channel if it is to be made of (a) unfinished concrete and (b) finished concrete. Answer: (a) 21.5 ft × 10.7 ft, (b) 20.3 ft × 10.1 ft 13–59E Repeat Prob. 13–58E for a flow rate of 750 ft3/s.
13–60 A trapezoidal channel made of unfinished concrete has a bottom slope of 1°, base width of 5 m, and a side sur face slope of 1:1, as shown in Fig. P13–49. For a flow rate of 25 m3/s, determine the normal depth h.
45° 45° 5 m h FIGURE P13–60 13–61 Repeat Prob. 13–60 for a weedy excavated earth channel with n = 0.030.
13–62 Consider uniform flow through an open channel lined with bricks with a Manning coefficient of n = 0.015. If the Manning coefficient doubles (n = 0.030) as a result of some algae growth on surfaces while the flow cross section remains constant, the flow rate will (a) double, (b) decrease by a factor of √2, (c) remain unchanged, (d) decrease by half, or (e) decrease by a factor of 21/3.
13–63 During uniform flow in open channels, the flow velocity and the flow rate can be determined from the Manning equations expressed as V0 = (a/n)Rh 2/3S0 1/2 and V · = (a/n)AcRh 2/3S0 1/2. What is the value and dimension of the con stant a in these equations in SI units? Also, explain how the cen96537_ch13_733-792.indd 784 14/01/17 3:21 pm 785 CHAPTER 13 Manning coefficient n can be determined when the friction factor f is known.
13–64 Show that for uniform critical flow, the general critical slope relation Sc = gn2yc a2R 4/3 h reduces to Sc = gn2 a2y 1/3 c for film flow with b ≫ yc.
Gradually and Rapidly Varied Flows and Hydraulic Jump 13–65C Is it possible for subcritical flow to undergo a hydraulic jump? Explain.
13–66C How does nonuniform or varied flow differ from uniform flow?
13–67C Someone claims that frictional losses associated with wall shear on surfaces can be neglected in the analysis of rapidly varied flow, but should be considered in the analy sis of gradually varied flow. Do you agree with this claim? Justify your answer.
13–68C Consider steady flow of water in an upward-sloped channel of rectangular cross section. If the flow is supercriti cal, the flow depth will (a) increase, (b) remain constant, or (c) decrease in the flow direction.
13–69C How does gradually varied flow (GVF) differ from rapidly varied flow (RVF)?
13–70C Why is the hydraulic jump sometimes used to dis sipate mechanical energy? How is the energy dissipation ratio for a hydraulic jump defined?
13–71C Consider steady flow of water in a horizontal channel of rectangular cross section. If the flow is subcriti cal, the flow depth will (a) increase, (b) remain constant, or (c) decrease in the flow direction.
13–72C Consider steady flow of water in a downward-sloped channel of rectangular cross section. If the flow is subcritical and the flow depth is greater than the normal depth ( y > yn), the flow depth will (a) increase, (b) remain constant, or (c) decrease in the flow direction.
13–73C Consider steady flow of water in a horizontal channel of rectangular cross section. If the flow is supercriti cal, the flow depth will (a) increase, (b) remain constant, or (c) decrease in the flow direction.
13–74C Consider steady flow of water in a downward-sloped channel of rectangular cross section. If the flow is subcritical and the flow depth is less than the normal depth ( y < yn), the flow depth will (a) increase, (b) remain con stant, or (c) decrease in the flow direction.
13–75 Water is flowing in a 90° V-shaped cast iron channel with a bottom slope of 0.0018 at a rate of 3 m3/s. Determine if the slope of this channel should be classified as mild, criti cal, or steep for this flow. Answer: mild 13–76 Consider uniform water flow in a wide brick channel of slope 0.4°. Determine the range of flow depth for which the channel is classified as being steep.
13–77E Consider the flow of water through a 12-ft-wide unfinished-concrete rectangular channel with a bottom slope of 0.5°. If the flow rate is 300 ft3/s, determine if the slope of this channel is mild, critical, or steep. Also, for a flow depth of 3 ft, classify the surface profile while the flow develops.
13–78 Water flows uniformly in a rectangular channel with finished-concrete surfaces. The channel width is 3 m, the flow depth is 1.2 m, and the bottom slope is 0.002. Determine if the channel should be classified as mild, critical, or steep for this flow.
y = 1.2 m b = 3 m FIGURE P13–78 13–79 Consider the flow of water in a 10-m-wide channel at a rate of 70 m3/s and a flow depth of 0.80 m. The water now undergoes a hydraulic jump, and the flow depth after the jump is measured to be 2.4 m. Determine the mechanical power wasted during this jump. Answer: 1.28 MW 13–80 The flow depth and velocity of water after undergo ing a hydraulic jump are measured to be 1.1 m and 1.75 m/s, respectively. Determine the flow depth and velocity before the jump, and the fraction of mechanical energy dissipated.
13–81E Water flowing in a wide channel at a depth of 2 ft and a velocity of 40 ft/s undergoes a hydraulic jump. Deter mine the flow depth, velocity, and Froude number after the jump, and the head loss associated with the jump.
13–82 Consider uniform flow of water in a wide rectangular channel with a per-unit-width flow rate of 1.5 m3/s⋅m and a Manning coefficient of 0.03. The slope of the channel is 0.0005. (a) Calculate the normal and critical depths of the flow and determine if the uniform flow is subcritical or supercritical. (b) Next, a dam is installed (at x = 0) in order to impound a reservoir of water upstream. This raises the water surface profile upstream, creating a “backwater” curve (Fig. P13–82). The new water depth just upstream of the dam is 2.5 m. Determine how far upstream of the dam the “reservoir” extends. You may con sider the reservoir boundary to be the point at which the water depth is within 5% of the original uniform water depth. Answer: (b) 3500 m cen96537_ch13_733-792.indd 785 14/01/17 3:21 pm 786 open-channel flow yn Before yn After x = 0 FIGURE P13–82 13–83 Water flowing in a wide horizontal channel at a flow depth of 56 cm and an average velocity of 9 m/s undergoes a hydraulic jump. Determine the head loss associated with the hydraulic jump.
13–84 Water discharging into a 9-m-wide rectangular horizontal channel from a sluice gate is observed to have undergone a hydraulic jump. The flow depth and velocity before the jump are 1.2 m and 11 m/s, respectively. Deter mine (a) the flow depth and the Froude number after the jump, (b) the head loss and the dissipation ratio, and (c) the mechanical energy dissipated by the hydraulic jump.
V1 = 11 m/s V2 y1 = 1.2 m y2 (1) (2) FIGURE P13–84 13–85 During a hydraulic jump in a wide channel, the flow depth increases from 1.1 to 3.3 m. Determine the veloci ties and Froude numbers before and after the jump, and the energy dissipation ratio.
13–86 Consider gradually varied flow over a bump in a wide channel, as shown in Fig. P13–86. The initial flow velocity is 0.75 m/s, the initial flow depth is 1m, the Manning parameter is 0.02, and the elevation of the chan nel bottom is prescribed to be zb = ∆zb exp[−0.001(x − 100)2] where the maximum bump height ∆zb is equal to 0.15m and the crest of the bump is located at x = 100m. (a) Calculate and plot the critical depth of the flow and (where it exists) the normal depth of the flow. (b) Integrate the GVF equation over the range 0 ≤ x ≤ 200m, and comment on the observed behavior of the free surface in light of the classification scheme presented in Table 13–3.
y1 = 1 m 100 m x = 0 FIGURE P13–86 13–87 Consider a wide rectangular water channel with a per-unit-width flow rate of 5 m3/s⋅m and a Manning coefficient of n = 0.02. The channel is com prised of a 100 m length having a slope of S01 = 0.01 fol lowed by a 100 m length having a slope of S02 = 0.02. (a) Calculate the normal and critical depths for the two channel segments. (b) Given an initial water depth of 1.25 m, calculate and graph the water surface profile over the full 200 m extent of the channel. Also classify the two channel segments (M1, A2, etc.).
y0 x = 0 FIGURE P13–87 13–88 Repeat Problem 13–87 for the case of an initial water depth of 0.75m instead of 1.25m.
13–89 While the GVF equation cannot be used to pre dict a hydraulic jump directly, it can be coupled with the ideal hydraulic jump depth ratio equation in order to help locate the position at which a jump will occur in a chan nel. Consider a jump created in a wide (Rh ≈ y) horizontal (S0 = 0) laboratory flume having a length of 3m and a cen96537_ch13_733-792.indd 786 14/01/17 3:21 pm 787 CHAPTER 13 Manning coefficient of 0.009. The supercritical flow under the head gate has an initial depth of 0.01 m at x = 0. The tail gate results in an overflow depth of 0.08 m at x = 3m. The per-unit-width flow rate is 0.025 m3/s⋅m. (a) Calculate the critical depth of the flow and verify that the initial and final flows are supercritical and subcritical, respectively. (b) Deter mine the location of the hydraulic jump. Hint: Integrate the GVF equation from x = 0 to a “guessed” location of the jump, apply the jump depth-ratio equation, and integrate the GVF equation using this new initial condition from the jump loca tion to x = 3 m. If you do not obtain the desired overflow depth, try a new jump location. Answer: (b) 1.80m yf = 0.08 m y0 = 0.01 m x = 0 x = 3 Jump FIGURE P13–89 13–90E Consider gradually varied flow of water in a 20-ft wide rectangular channel with a flow rate of 300 ft3/s and a Manning coefficient of 0.008. The slope of the channel is 0.01, and at the location x = 0, the mean flow speed is measured to be 5.2 ft/s. Determine the classification of the water surface profile, and, by integrating the GVF equation numerically, calculate the flow depth y at (a) x = 500 ft, (b) 1000 ft, and (c) 2000 ft.
V0 = 5.2 ft/s y0 y x 0 S0 = 0.01 FIGURE P13–90E 13–91 Consider gradually varied flow of water in a wide rectangular irrigation channel with a per-unit-width flow rate of 5 m3/s⋅m, a slope of 0.01, and a Man ning coefficient of 0.02. The flow is initially at uniform depth. At a given location, x = 0, the flow enters a 200 m length of channel where lack of maintenance has re sulted in a channel roughness of 0.03. Following this stretch of channel, the roughness returns to the initial (maintained) value. (a) Calculate the normal and critical depths of the flow for the two distinct segments. (b) Numerically solve the gradu ally varied flow equation over the range 0 ≤ x ≤ 400 m. Plot your solution (i.e., y vs. x) and comment about the behavior of the water surface.
FIGURE P13–91 yn1 Rough surface Smoother surface 200 m 0 x 200 m 13–92 Consider the gradually varied flow equation, dy dx = S0 −Sf 1 −Fr2 For the case of a wide rectangular channel, show that this can be reduced to the following form, which explicitly shows the importance of the relationship between y, yn, and yc: dy dx = S0[1 −( yn/y)10/3] 1 −( yc/y)3 Flow Control and Measurement in Channels 13–93C What is the basic principle of operation of a broad-crested weir used to measure flow rate through an open channel?
13–94C What is a sharp-crested weir? On what basis are the sharp-crested weirs classified?
13–95C For sluice gates, how is the discharge coefficient Cd defined? What are typical values of Cd for sluice gates with free outflow? What is the value of Cd for the idealized frictionless flow through the gate?
13–96C Consider steady frictionless flow over a bump of height ∆z in a horizontal channel of constant width b. Will the flow depth y increase, decrease, or remain constant as the fluid flows over the bump? Assume the flow to be subcritical.
13–97C Consider the flow of a liquid over a bump during subcritical flow in an open channel. The specific energy and the flow depth decrease over the bump as the bump height is increased. What will the character of flow be when the spe cific energy reaches its minimum value? Will the flow become supercritical if the bump height is increased even further?
13–98C Draw a flow depth-specific energy diagram for flow through underwater gates, and indicate the flow through the gate for cases of (a) frictionless gate, (b) sluice gate with free outflow, and (c) sluice gate with drowned outflow (including the hydraulic jump back to subcritical flow).
13–99 Consider uniform water flow in a wide rectangular channel with a depth of 2 m made of unfinished concrete laid on a slope of 0.0022. Determine the flow rate of water per meter cen96537_ch13_733-792.indd 787 14/01/17 3:21 pm 788 open-channel flow width of channel. Now water flows over a 15-cm-high bump. If the water surface over the bump remains flat (no rise or drop), determine the change in discharge rate of water per meter width of the channel. (Hint: Investigate if a flat surface over the bump is physically possible.) 13–100 Consider the uniform flow of water in a wide chan nel with a velocity of 10 m/s and flow depth of 0.65 m. Now water flows over a 30-cm-high bump. Determine the change (increase or decrease) in the water surface level over the bump. Also determine if the flow over the bump is sub- or supercritical.
13–101 Water is released from a 12-m-deep reservoir into a 6-m-wide open channel through a sluice gate with a 1-m-high opening at the channel bottom. If the flow depth downstream from the gate is measured to be 3 m, determine the rate of discharge through the gate.
a = 1 m Sluice gate y2 = 3 m y1 = 12 m FIGURE P13–101 13–102E A full-width sharp-crested weir is to be used to measure the flow rate of water in a 7-ft-wide rectangular channel. The maximum flow rate through the channel is 180 ft3/s, and the flow depth upstream from the weir is not to exceed 3 ft. Determine the appropriate height of the weir.
13–103 The flow rate of water in a 10-m-wide horizontal channel is being measured using a 1.3-m-high sharp-crested rectangular weir that spans across the channel. If the water depth upstream is 3.4 m, determine the flow rate of water. Answer: 66.8 m3/s V1 Sharp-crested rectangular weir y1 = 3.4 m Pw = 1.3 m FIGURE P13–103 13–104 Repeat Prob. 13–103 for the case of a weir height of 1.6 m.
13–105 Water flows over a 1.5-m-high sharp-crested rect angular weir. The flow depth upstream of the weir is 2.5 m, and water is discharged from the weir into an unfinished-concrete channel of equal width where uniform-flow con ditions are established. If no hydraulic jump is to occur in the downstream flow, determine the maximum slope of the downstream channel.
13–106E Water flows through a sluice gate with a 1.1-ft-high opening and is discharged with free outflow. If the upstream flow depth is 5 ft, determine the flow rate per unit width and the Froude number downstream the gate.
13–107E Repeat Prob. 13–106E for the case of a drowned gate with a downstream flow depth of 4 ft.
13–108 Water is to be discharged from an 8-m-deep lake into a channel through a sluice gate with a 5-m wide and 0.6-m-high opening at the bottom. If the flow depth down stream from the gate is measured to be 4 m, determine the rate of discharge through the gate.
13–109 Water flowing in a wide channel encounters a 22-cm-high bump at the bottom of the channel. If the flow depth is 1.2 m and the velocity is 2.5 m/s before the bump, determine if the flow is choked over the bump, and discuss.
V1 = 2.5 m/s y1 = 1.2 m y2 Δzb = 0.22 m Depression over the bump Bump FIGURE P13–109 13–110E Consider water flow through a wide channel at a flow depth of 8 ft. Now water flows through a sluice gate with a 1-ft-high opening, and the freely discharged outflow subsequently undergoes a hydraulic jump. Disregarding any losses associated with the sluice gate itself, determine the flow depth and velocities before and after the jump, and the fraction of mechanical energy dissipated during the jump.
13–111 The flow rate of water flowing in a 5-m-wide chan nel is to be measured with a sharp-crested triangular weir 0.5 m above the channel bottom with a notch angle of 80°. If the flow depth upstream from the weir is 1.5 m, determine the flow rate of water through the channel. Take the weir dis charge coefficient to be 0.60. Answer: 1.19 m3/s cen96537_ch13_733-792.indd 788 14/01/17 3:22 pm 789 CHAPTER 13 0.5 m 5 m Weir plate 1 m 80° Upstream free surface FIGURE P13–111 13–112 Repeat Prob. 13–111 for an upstream flow depth of 0.90 m.
13–113 A sharp-crested triangular weir with a notch angle of 80° is used to measure the discharge rate of water from a large lake into a spillway. If a weir with half the notch angle (𝜃 = 40°) is used instead, determine the percent reduction in the flow rate. Assume the water depth in the lake and the weir discharge coefficient remain unchanged.
13–114 A 0.80-m-high broad-crested weir is used to mea sure the flow rate of water in a 5-m-wide rectangular chan nel. The flow depth well upstream from the weir is 1.8 m. Determine the flow rate through the channel and the mini mum flow depth above the weir.
Discharge 1.8 m 0.80 m Broad-crested weir FIGURE P13–114 13–115 Repeat Prob. 13–114 for an upstream flow depth of 1.4 m.
13–116 Consider uniform water flow in a wide channel made of unfinished concrete laid on a slope of 0.0019. Now water flows over a 20-cm-high bump. If the flow over the bump is exactly critical (Fr = 1), determine the flow rate and the flow depth over the bump per meter width. Answers: 29.2 m3/s, 4.39 m y1 y2 Δzb = 20 cm Bump Slope = 0.0019 FIGURE P13–116 13–117 Consider water flow over a 0.80-m-high suffi ciently long broad-crested weir. If the minimum flow depth above the weir is measured to be 0.50 m, determine the flow rate per meter width of channel and the flow depth upstream of the weir.
Review Problems 13–118 A rectangular channel is 4.0 m wide and 7.0 m deep, but it is filled with water up to only 6.0 m from the bot tom. The channel walls are made of corrugated metal. Calcu late the hydraulic diameter.
13–119 Consider the same rectangular channel as in the previous problem (4.0 m wide, filled with water to 6.0 m, and walls of corrugated metal). If the slope is 0.65°, estimate the volume flow rate of the water.
13–120 Water flows in a canal at an average velocity of 6 m/s. Determine if the flow is subcritical or supercritical for flow depths of (a) 0.2 m, (b) 2 m, and (c) 1.63 m.
13–121 Water at 15°C flows at a depth of 0.25 m with an average velocity of 7 m/s in a rectangular channel. Determine (a) the critical depth, (b) the alternate depth, and (c) the min imum specific energy.
13–122 A trapezoidal channel with a bottom width of 4 m and a side slope of 45° discharges water at a rate of 18 m3/s. If the flow depth is 0.6 m, determine if the flow is subcritical or supercritical.
13–123 A trapezoidal channel with brick lining has a bottom slope of 0.001 and a base width of 4 m, and the side surfaces are angled 25° from the horizontal, as shown in Fig. P13–123. If the normal depth is measured to be 1.5 m, estimate the flow rate of water through the channel. Answer: 22.5 m3/s 25˚ 25˚ 4 m 1.5 m FIGURE P13–123 13–124 Water flows through a 1.8-m-wide rectangular chan nel with a Manning coefficient of n = 0.012. If the water is 0.7 m deep and the bottom slope of the channel is 0.6°, determine the rate of discharge of the channel in uniform flow.
13–125 A rectangular channel with a bottom width of 7 m discharges water at a rate of 45 m3/s. Determine the flow depth below which the flow is supercritical. Answer: 1.62 m 13–126 Consider a 1-m-internal-diameter water channel made of finished concrete (n = 0.012). The channel slope is 0.002. For a flow depth of 0.36 m at the center, determine the flow rate of water through the channel. Answer: 0.322 m3/s cen96537_ch13_733-792.indd 789 14/01/17 3:22 pm 790 open-channel flow 0.36 m R = 0.5 m FIGURE P13–126 13–127 Reconsider Prob. 13–126. By varying the flow depth-to-radius ratio y/R from 0.1 to 1.9 while holding the flow area constant and evaluating the flow rate, show that the best cross section for flow through a cir cular channel occurs when the channel is half-full. Tabulate and plot your results.
13–128 Water is discharged from a dam into a wide spill way to avoid overflow and to reduce the risk of flooding. A large fraction of the destructive power of the water is dissi pated by a hydraulic jump during which the water depth rises from 0.70 to 5.0 m. Determine the velocities of water before and after the jump, and the mechanical power dissipated per meter width of the spillway.
13–129 Water flows in a channel whose bottom slope is 0.5° and whose cross section is as shown in Fig. P13–129. The dimensions and the Manning coefficients for the sur faces of different subsections are also given on the figure. Determine the flow rate through the channel and the effective Manning coefficient for the channel.
6 m 1 m 1 m 10 m Heavy brush n2 = 0.075 Clean earth channel n1 = 0.022 FIGURE P13–129 13–130 Consider two identical channels, one rectangular of bottom width b and one circular of diameter D, with identi cal flow rates, bottom slopes, and surface linings. If the flow height in the rectangular channel is also b and the circular channel is flowing half-full, determine the relation between b and D.
13–131 The flow rate of water in a 6-m-wide rectangular channel is to be measured using a 1.3-m-high sharp-crested rectangular weir that spans across the channel. If the head above the weir crest is 0.70 m upstream from the weir, deter mine the flow rate of water.
13–132E A rectangular channel with unfinished concrete surfaces is to be built to discharge water uniformly at a rate of 200 ft3/s. For the case of best cross section, determine the bottom width of the channel if the available vertical drop is (a) 5 and (b) 10 ft per mile. Answers: (a) 8.58 ft, (b) 7.54 ft 13–133E Repeat Prob. 13–132E for the case of a trapezoi dal channel of best cross section.
13–134E Consider two identical 15-ft-wide rectangular channels each equipped with a 3-ft-high full-width weir, except that the weir is sharp-crested in one channel and broad-crested in the other. For a flow depth of 5 ft in both channels, determine the flow rate through each channel. Answers: 149 ft3/s, 66.0 ft3/s 13–135 A 5-m-wide rectangular channel lined with finished concrete is to be designed to trans port water to a distance of 1 km at a rate of 12 m3/s. Using appropriate software, investigate the effect of bottom slope on flow depth (and thus on the required channel height). Let the bottom angle vary from 0.5 to 10° in increments of 0.5°. Tab ulate and plot the flow depth against the bottom angle, and discuss the results.
13–136 Repeat Prob. 13–135 for a trapezoidal channel that has a base width of 5 m and a side surface angle of 45°.
13–137 In practice, the V-notch is commonly used to measure flow rate in open channels. Using the idealized Torricelli’s equation V = √2g(H −y) for velocity, develop a relation for the flow rate through the V-notch in terms of the angle 𝜃. Also, show the variation of the flow rate with 𝜃 by evaluating the flow rate for 𝜃 = 25, 40, 60, and 75°, and plotting the results.
θ H = 25 cm y FIGURE P13–137 13–138 Water flows uniformly half-full in a 3.2-m-diameter circular channel laid with a slope of 0.004. If the flow rate of water is measured to be 4.5 m3/s, determine the Manning coefficient of the channel and the Froude number. Answers: 0.0487, 0.319 13–139 Consider water flow through a wide rectangular channel undergoing a hydraulic jump. Show that the ratio of the Froude numbers before and after the jump can be expressed in terms of flow depths y1 and y2 before and after the jump, respectively, as Fr1/Fr2 = √(y2/y1)3 13–140 A sluice gate with free outflow is used to control the discharge rate of water through a channel. Determine the flow rate per unit width when the gate is raised to yield a gap of cen96537_ch13_733-792.indd 790 14/01/17 3:22 pm 791 CHAPTER 13 50 cm and the upstream flow depth is measured to be 2.8 m. Also determine the flow depth and the velocity downstream.
13–141 Water flowing in a wide channel at a flow depth of 45 cm and an average velocity of 6.5 m/s undergoes a hydrau lic jump. Determine the fraction of the mechanical energy of the fluid dissipated during this jump. Answer: 27.1 percent 13–142 Water flowing through a sluice gate undergoes a hydraulic jump, as shown in Fig. P13–142. The velocity of the water is 1.25 m/s before reaching the gate and 4 m/s after the jump. Determine the flow rate of water through the gate per meter of width, the flow depths y1 and y2, and the energy dissipation ratio of the jump.
y3 = 3 m y1 y2 V1 = 1.25 m/s V3 = 4 m/s Sluice gate FIGURE P13–142 13–143 Repeat Prob. 13–142 for a velocity of 3.2 m/s after the hydraulic jump.
13–144 Water is discharged from a 5-m-deep lake into a fin ished concrete channel with a bottom slope of 0.004 through a sluice gate with a 0.7-m-high opening at the bottom. Shortly after supercritical uniform-flow conditions are established, the water undergoes a hydraulic jump. Determine the flow depth, velocity, and Froude number after the jump. Disregard the bottom slope when analyzing the hydraulic jump.
13–145 Consider the flow of water through a parabolic notch shown in Fig. P13–145. Develop a relation for the flow rate, and calculate its numerical value for the ideal case in which the flow velocity is given by Torricelli’s equation V = √2g(H −y). Answer: 0.123 m3/s y x H = 0.5 m b = 0.4 m y = cx2 FIGURE P13–145 13–146 Water flowing in a wide horizontal channel approaches a 20-cm-high bump with a velocity of 1.25 m/s and a flow depth of 1.8 m. Determine the velocity, flow depth, and Froude number over the bump.
V2 y2 V1 = 1.25 m/s 20 cm y1 = 1.8 m FIGURE P13–146 13–147 Reconsider Prob. 13–146. Determine the bump height for which the flow over the bump is critical (Fr = 1).
13–148 Consider water flow through a V-shaped channel. Determine the angle 𝜃 the channel makes from the horizontal for which the flow is most efficient.
θ θ y FIGURE P13–148 Fundamentals of Engineering (FE) Exam Problems 13–149 If the flow depth remains constant in an open-channel flow, the flow is called (a) Uniform flow (b) Steady flow (c) Varied flow (d ) Unsteady flow (e) Laminar flow 13–150 Which choices are examples of open-channel flow?
I. Flow of water in rivers II. Draining of rainwater off highways III. Upward draft of rain and snow IV. Sewer lines (a) I and II (b) I and III (c) II and III (d ) I, II, and IV (e) I, II, III, and IV 13–151 Consider water flow in a rectangular open channel of height 2 m and width 5 m containing water of depth 1 m. The hydraulic radius for this flow is (a) 0.71 m (b) 0.82 m (c) 0.94 m (d ) 1.1 m (e) 1.3 m 13–152 Water flows in a rectangular open channel of width 5 m at a rate of 7.5 m3/s. The critical depth for this flow is (a) 5 m (b) 2.5 m (c) 1.5 m (d ) 0.96 m (e) 0.61 m 13–153 Water flows in a rectangular open channel of width 0.6 m at a rate of 0.25 m3/s. If the flow depth is 0.2 m, what is the alternate flow depth if the character of flow were to change?
(a) 0.2 m (b) 0.26 m (c) 0.35 m (d ) 0.6 m (e) 0.8 m 13–154 Water flows in a 9-m-wide rectangular open chan nel at a rate of 55 m3/s. If the flow depth is 2.4 m, the Froude number is (a) 0.682 (b) 0.787 (c) 0.525 (d ) 1.00 (e) 2.65 cen96537_ch13_733-792.indd 791 14/01/17 3:22 pm 792 open-channel flow 13–155 Water flows in a clean and straight natural chan nel of rectangular cross section with a bottom width of 0.75 m and a bottom slope angle of 0.6°. If the flow depth is 0.15 m, the flow rate of water through the channel is (a) 0.0317 m3/s (b) 0.05 m3/s (c) 0.0674 m3/s (d ) 0.0866 m3/s (e) 1.14 m3/s 13–156 Water is to be transported in a finished-concrete rectangular channel with a bottom width of 1.2 m at a rate of 5 m3/s. The channel bottom drops 1 m per 500 m length. The minimum height of the channel under uniform-flow conditions is (a) 1.9 m (b) 1.5 m (c) 1.2 m (d ) 0.92 m (e) 0.60 m 13–157 Water is to be transported in a 4-m-wide rectan gular open channel. The flow depth to maximize the flow rate is (a) 1 m (b) 2 m (c) 4 m (d ) 6 m (e) 8 m 13–158 Water is to be transported in a clay tile lined rectangular channel at a rate of 0.8 m3/s. The channel bot tom slope is 0.0015. The width of the channel for the best cross section is (a) 0.68 m (b) 1.33 m (c) 1.63 m (d ) 0.98 m (e) 1.15 m 13–159 Water is to be transported in a clay tile lined trapezoidal channel at a rate of 0.6 m3/s. The channel bot tom slope is 0.0015. The width of the channel for the best cross section is (a) 0.48 m (b) 0.63 m (c) 0.70 m (d ) 0.82 m (e) 0.97 m 13–160 Water flows uniformly in a finished-concrete rectangular channel with a bottom width of 0.85 m. The flow depth is 0.4 m and the bottom slope is 0.003. The channel should be classified as (a) Steep (b) Critical (c) Mild (d ) Horizontal (e) Adverse 13–161 Water discharges into a rectangular horizontal channel from a sluice gate and undergoes a hydraulic jump. The channel is 25-m-wide and the flow depth and velocity before the jump are 2 m and 9 m/s, respectively. The flow depth after the jump is (a) 1.26 m (b) 2 m (c) 3.61 m (d ) 4.83 m (e) 6.55 m 13–162 Water discharges into a rectangular horizontal channel from a sluice gate and undergoes a hydraulic jump. The flow depth and velocity before the jump are 1.25 m and 8.5 m/s, respectively. The percentage available head loss due to the hydraulic jump is (a) 4.7% (b) 7.2% (c) 8.8% (d ) 13.5% (e) 16.3% 13–163 Water discharges into a 7-m-wide rectangular hori zontal channel from a sluice gate and undergoes a hydraulic jump. The flow depth and velocity before the jump are 0.65 m and 5 m/s, respectively. The wasted power potential due to the hydraulic jump is (a) 158 kW (b) 112 kW (c) 67.3 kW (d ) 50.4 kW (e) 37.6 kW 13–164 The flow rate of water in a 3-m-wide horizontal open channel is being measured with a 0.25-m-high sharp-crested rectangular weir of equal width. If the water depth upstream is 0.9 m, the flow rate of water is (a) 1.75 m3/s (b) 2.22 m3/s (c) 2.84 m3/s (d ) 3.86 m3/s (e) 5.02 m3/s 13–165 Water is released from a 0.8-m-deep reservoir into a 4-m-wide open channel through a sluice gate with a 0.1-m-high opening at the channel bottom. The flow depth after all turbulence subsides is 0.5 m. The rate of discharge is (a) 0.92 m3/s (b) 0.79 m3/s (c) 0.66 m3/s (d ) 0.47 m3/s (e) 0.34 m3/s Design and Essay Problems 13–166 Using catalogs or websites, obtain information from three different weir manufacturers. Compare the differ ent weir designs, and discuss the advantages and disadvan tages of each design. Indicate the applications for which each design is best suited.
13–167 Consider water flow in the range of 10 to 15 m3/s through a horizontal section of a 5-m-wide rectangular channel. A rectangular or triangular thin-plate weir is to be installed to measure the flow rate. If the water depth is to remain under 2 m at all times, specify the type and dimen sions of an appropriate weir. What would your response be if the flow range were 0 to 15 m3/s?
cen96537_ch13_733-792.indd 792 14/01/17 3:22 pm 14 CHAPTER 793 T U R B O M AC H I N E RY I n this chapter we discuss the basic principles of a common and impor tant application of fluid mechanics, turbomachinery. First we classify turbomachines into two broad categories, pumps and turbines. Then we discuss both of these turbomachines in more detail, mostly qualitatively, explaining the basic principles of their operation. We emphasize prelimi nary design and overall performance of turbomachines rather than detailed design. In addition, we discuss how to properly match the requirements of a fluid flow system to the performance characteristics of a turbomachine. A significant portion of this chapter is devoted to turbomachinery scaling laws—a practical application of dimensional analysis. We show how the scaling laws are used in the design of new turbomachines that are geometri cally similar to existing ones.
OBJECTIVES When you finish reading this chapter, you should be able to ■ ■ Identify various types of pumps and turbines, and understand how they work ■ ■ Apply dimensional analysis to design new pumps or turbines that are geometrically similar to existing pumps or turbines ■ ■ Perform basic vector analysis of the flow into and out of pumps and turbines ■ ■ Use specific speed for preliminary design and selection of pumps and turbines The jet engines on modern commercial airplanes are highly complex turbomachines that include both pump (compressor) and turbine sections.
© Stockbyte/Punchstock RF cen96537_ch14_793-884.indd 793 29/12/16 4:37 pm 794 TURBOMACHINERY 14–1 ■ CLASSIFICATIONS AND TERMINOLOGY There are two broad categories of turbomachinery, pumps and turbines. The word pump is a general term for any fluid machine that adds energy to a fluid. Some authors call pumps energy absorbing devices since energy is supplied to them, and they transfer most of that energy to the fluid, usually via a rotating shaft (Fig. 14–1a). The increase in fluid energy is usually felt as an increase in the pressure of the fluid. Turbines, on the other hand, are energy producing devices—they extract energy from the fluid and transfer most of that energy to some form of mechanical energy output, typically in the form of a rotating shaft (Fig. 14–1b). The fluid at the outlet of a turbine suffers an energy loss, typically in the form of a loss of pressure. An ordinary person may think that the energy supplied to a pump increases the speed of fluid passing through the pump and that a turbine extracts energy from the fluid by slowing it down. This is not necessar ily the case. Consider a control volume surrounding a pump (Fig. 14–2). We assume steady conditions. By this we mean that neither the mass flow rate nor the rotational speed of the rotating blades changes with time. (The detailed flow field near the rotating blades inside the pump is not steady of course, but control volume analysis is not concerned with details inside the control volume.) By conservation of mass, we know that the mass flow rate into the pump must equal the mass flow rate out of the pump. If the flow is incompressible, the volume flow rates at the inlet and outlet must be equal as well. Furthermore, if the diameter of the outlet is the same as that of the inlet, conservation of mass requires that the average speed across the outlet must be identical to the average speed across the inlet. In other words, the pump does not necessarily increase the speed of the fluid passing through it; rather, it increases the pressure of the fluid. Of course, if the pump were turned off, there might be no flow at all. So, the pump does increase fluid speed compared to the case of no pump in the system. However, in terms of changes from the inlet to the outlet across the pump, fluid speed is not necessarily increased. (The output speed may even be lower than the input speed if the outlet diameter is larger than that of the inlet.) The purpose of a pump is to add energy to a fluid, resulting in an increase in fluid pressure, not necessarily an increase of fluid speed across the pump.
An analogous statement is made about the purpose of a turbine: The purpose of a turbine is to extract energy from a fluid, resulting in a decrease of fluid pressure, not necessarily a decrease of fluid speed across the turbine. Fluid machines that move liquids are called pumps, but there are several other names for machines that move gases (Fig. 14–3). A fan is a gas pump with relatively low pressure rise and high flow rate. Examples include ceil ing fans, house fans, and propellers. A blower is a gas pump with relatively moderate to high pressure rise and moderate to high flow rate. Examples include centrifugal blowers and squirrel cage blowers in automobile ven tilation systems, furnaces, and leaf blowers. A compressor is a gas pump designed to deliver a very high pressure rise, typically at low to moderate flow rates. Examples include air compressors that run pneumatic tools and Pump Flow out Energy supplied, Eout > Ein (a) Flow in Ein Eout Turbine Flow out Energy extracted, Eout < Ein (b) Flow in Ein Eout ω ω FIGURE 14–1 (a) A pump supplies energy to a fluid, while (b) a turbine extracts energy from a fluid.
Pump P out V out Pin Vin Din Control volume Dout ω FIGURE 14–2 For the case of steady flow, conservation of mass requires that the mass flow rate out of a pump must equal the mass flow rate into the pump; for incompressible flow with equal inlet and outlet cross-sectional areas (Dout = Din), we conclude that Vout = Vin, but Pout > Pin.
P Low Medium High High Medium Low Fan Blower Compressor Δ V FIGURE 14–3 When used with gases, pumps are called fans, blowers, or compressors, depending on the relative values of pressure rise and volume flow rate.
cen96537_ch14_793-884.indd 794 29/12/16 4:37 pm 795 CHAPTER 14 inflate tires at automobile service stations, and refrigerant compressors used in heat pumps, refrigerators, and air conditioners. Pumps and turbines in which energy is supplied or extracted by a rotat ing shaft are properly called turbomachines, since the Latin prefix turbo means “spin.” Not all pumps or turbines utilize a rotating shaft, however. The hand-operated air pump you use to inflate the tires of your bicycle is a prime example (Fig. 14–4a). The up and down reciprocating motion of a plunger or piston replaces the rotating shaft in this type of pump, and it is more proper to call it simply a fluid machine instead of a turboma chine. An old-fashioned well pump operates in a similar manner to pump water instead of air (Fig. 14–4b). Nevertheless, the words turbomachine and turbomachinery are often used in the literature to refer to all types of pumps and turbines regardless of whether they utilize a rotating shaft or not. Fluid machines may also be broadly classified as either positive-displace ment machines or dynamic machines, based on the manner in which energy transfer occurs. In positive-displacement machines, fluid is directed into a closed volume. Energy transfer to the fluid is accomplished by movement of the boundary of the closed volume, causing the volume to expand or contract, thereby sucking fluid in or squeezing fluid out, respectively. Your heart is a good example of a positive-displacement pump (Fig. 14–5a). It is designed with one-way valves that open to let blood in as heart cham bers expand, and other one-way valves that open as blood is pushed out of those chambers when they contract. An example of a positive-displacement turbine is the common water meter in your house (Fig. 14–5b), in which water forces itself into a closed chamber of expanding volume connected to an output shaft that turns as water enters the chamber. The boundary of the volume then collapses, turning the output shaft some more, and letting the water continue on its way to your sink, shower, etc. The water meter records each 360° rotation of the output shaft, and the meter is precisely calibrated to the known volume of fluid in the chamber.
FIGURE 14–5 (a) The human heart is an example of a positive-displacement pump; blood is pumped by expansion and contraction of heart chambers called ventricles. (b) The common water meter in your house is an example of a positive-displacement turbine; water fills and exits a chamber of known volume for each revolution of the output shaft.
(b) Courtesy of Badger Meter, Inc.
Used by permission.
Pulmonary valve Inferior vena cava Tricuspid valve Superior vena cava Pulmonary artery Right atrium Left atrium Left ventricle Right ventricle Aorta Pulmonary vein Mitral valve Aortic valve (a) (b) FIGURE 14–4 Not all pumps have a rotating shaft; (a) energy is supplied to this manual tire pump by the up and down motion of a person’s arm to pump air; (b) a similar mechanism is used to pump water with an old-fashioned well pump.
(a) Photo by Andrew Cimbala.
(b) © Bear Dancer Studios/Mark Dierker.
(a) (b) cen96537_ch14_793-884.indd 795 29/12/16 4:37 pm 796 TURBOMACHINERY In dynamic machines, there is no closed volume; instead, rotating blades supply or extract energy to or from the fluid. For pumps, these rotating blades are called impeller blades, while for turbines, the rotating blades are called runner blades or buckets. Examples of dynamic pumps include enclosed pumps and ducted pumps (those with casings around the blades such as the water pump in your car’s engine), and open pumps (those with out casings such as the ceiling fan in your house, the propeller on an air plane, or the rotor on a helicopter). Examples of dynamic turbines include enclosed turbines, such as the hydroturbine that extracts energy from water in a hydroelectric dam, and open turbines such as the wind turbine that extracts energy from the wind (Fig. 14–6).
14–2 ■ PUMPS Some fundamental parameters are used to analyze the performance of a pump. The mass flow rate m . of fluid through the pump is an obvious primary pump performance parameter. For incompressible flow, it is more common to use volume flow rate rather than mass flow rate. In the tur bomachinery industry, volume flow rate is called capacity and is simply mass flow rate divided by fluid density, Volume flow rate (capacity): V .
= m · ρ (14–1) The performance of a pump is characterized additionally by its net head H, defined as the change in Bernoulli head between the inlet and outlet of the pump, Net head: H = ( P ρg + V 2 2g + z) out −( P ρg + V 2 2g + z) in (14–2) The dimension of net head is length, and it is often listed as an equivalent column height of water, even for a pump that is not pumping water. For the case in which a liquid is being pumped, the Bernoulli head at the inlet is equivalent to the energy grade line at the inlet, EGLin, obtained by aligning a Pitot probe in the center of the flow as illustrated in Fig. 14–7. The energy grade line at the outlet EGLout is obtained in the same manner, as also illustrated in the figure. In the general case, the outlet of the pump may be at a different elevation than the inlet, and its diameter and average speed may not be the same as those at the inlet. Regardless of these differ ences, net head H is equal to the difference between EGLout and EGLin, Net head for a liquid pump: H = EGLout −EGLin Consider the special case of incompressible flow through a pump in which the inlet and outlet diameters are identical, and there is no change in elevation. Equation 14–2 reduces to Special case with Dout = Din and zout = zin: H = Pout −Pin ρg For this simplified case, net head is simply the pressure rise across the pump expressed as a head (column height of the fluid).
FIGURE 14–6 A wind turbine is a good example of a dynamic machine of the open type; air turns the blades, and the output shaft drives an electric generator.
The Wind Turbine Company. Used by permission.
cen96537_ch14_793-884.indd 796 29/12/16 4:37 pm 797 CHAPTER 14 Net head is proportional to the useful power actually delivered to the fluid. It is traditional to call this power the water horsepower, even if the fluid being pumped is not water, and even if the power is not measured in units of horsepower. By dimensional reasoning, we must multiply the net head of Eq. 14–2 by mass flow rate and gravitational acceleration to obtain dimensions of power. Thus, Water horsepower: W .
water horsepower = m · gH = ρgV .
H (14–3) All pumps suffer from irreversible losses due to friction, internal leakage, flow separation on blade surfaces, turbulent dissipation, etc. Therefore, the mechanical energy supplied to the pump must be larger than W .
water horsepower. In pump terminology, the external power supplied to the pump is called the brake horsepower, which we abbreviate as bhp. For the typical case of a rotating shaft supplying the brake horsepower, Brake horsepower: bhp = W .
shaft = 𝜔 Tshaft (14–4) where 𝜔 is the rotational speed of the shaft (rad/s) and Tshaft is the torque supplied to the shaft. We define pump efficiency 𝜂pump as the ratio of useful power to supplied power, Pump efficiency: 𝜂pump = W .
water horsepower W .
shaft = W .
water horsepower bhp = ρgV .
H 𝜔 Tshaft (14–5) Pump Performance Curves and Matching a Pump to a Piping System The maximum volume flow rate through a pump occurs when its net head is zero, H = 0; this flow rate is called the pump’s free delivery. The free delivery condition is achieved when there is no flow restriction at the pump inlet or outlet—in other words when there is no load on the pump. At this operating point, V . is large, but H is zero; the pump’s efficiency is zero because the pump is doing no useful work, as is clear from Eq. 14–5. At the other extreme, the shutoff head is the net head that occurs when the volume flow rate is zero, V . = 0, and is achieved when the outlet port of the pump is blocked off. Under these conditions, H is large but V . is zero; the pump’s efficiency (Eq. 14–5) is again zero, because the pump is doing no useful work. Between these two extremes, from shutoff to free delivery, the pump’s net head may increase from its shutoff value somewhat as the flow rate increases, but H must eventually decrease to zero as the volume flow rate increases to its free delivery value. The pump’s efficiency reaches its maximum value somewhere between the shutoff condition and the free delivery condition; this operating point of maximum efficiency is appropri ately called the best efficiency point (BEP), and is notated by an asterisk (H, V .
, bhp). Curves of H, 𝜂pump, and bhp as functions of V . are called pump performance curves (or characteristic curves, Chap. 8); typical curves at one rotational speed are plotted in Fig. 14–8. The pump perfor mance curves change with rotational speed. It is important to realize that for steady conditions, a pump can operate only along its performance curve. Thus, the operating point of a piping ω H Datum plane (z = 0) Pump EGLout EGLin bhp P out V out Pin Vin zin zout Din Dout FIGURE 14–7 The net head of a pump, H, is defined as the change in Bernoulli head from inlet to outlet; for a liquid, this is equivalent to the change in the energy grade line, H = EGLout − EGLin, rela tive to some arbitrary datum plane; bhp is the brake horsepower, the external power supplied to the pump.
cen96537_ch14_793-884.indd 797 29/12/16 4:37 pm 798 TURBOMACHINERY system is determined by matching system requirements (required net head) to pump performance (available net head). In a typical application, Hrequired and Havailable match at one unique value of flow rate—this is the operating point or duty point of the system.
The steady operating point of a piping system is established at the volume flow rate where Hrequired = Havailable.
For a given piping system with its major and minor losses, elevation changes, etc., the required net head increases with volume flow rate. On the other hand, the available net head of most pumps decreases with flow rate, as in Fig. 14–8, at least over the majority of its recommended operating range. Hence, the system curve and the pump performance curve intersect as sketched in Fig. 14–9, and this establishes the operating point. If we are lucky, the operating point is at or near the best efficiency point of the pump. In most cases, however, as illustrated in Fig. 14–9, the pump does not run at its optimum efficiency. If efficiency is of major concern, the pump should be carefully selected (or a new pump should be designed) such that the operating point is as close to the best efficiency point as possible. In some cases it may be possible to change the shaft rotation speed so that an exist ing pump can operate much closer to its design point (best efficiency point). There are unfortunate situations where the system curve and the pump performance curve intersect at more than one operating point. This can occur when a pump that has a dip in its net head performance curve is mated to a system that has a fairly flat system curve, as illustrated in Fig. 14–10. Although rare, such situations are possible and should be avoided, because the system may “hunt” for an operating point, leading to an unsteady-flow situation. It is fairly straightforward to match a piping system to a pump, once we realize that the term for useful pump head (hpump, u) that we used in the head form of the energy equation (Chap. 5) is the same as the net head (H) used in the present chapter. Consider, for example, a general piping system with elevation change, major and minor losses, and fluid acceleration (Fig. 14–11). We begin by solving the energy equation for the required net head Hrequired, Hrequired = hpump, u = P2 −P1 ρg + 𝛼2V2 2 −𝛼1V1 2 2g + (z2 −z1) + hL, total (14–6) where we assume that there is no turbine in the system, although that term can be added back in, if necessary. We have also included the kinetic energy correction factors in Eq. 14–6 for greater accuracy, even though it is com mon practice in the turbomachinery industry to ignore them (𝛼1 and 𝛼2 are often assumed to be unity since the flow is turbulent). Equation 14–6 is evaluated from the inlet of the piping system (point 1, upstream of the pump) to the outlet of the piping system (point 2, down stream of the pump). Equation 14–6 agrees with our intuition, because it tells us that the useful pump head delivered to the fluid does four things: • It increases the static pressure of the fluid from point 1 to point 2 (first term on the right).
• It increases the dynamic pressure (kinetic energy) of the fluid from point 1 to point 2 (second term on the right).
0 0 Shutoff head Free delivery BEP bhp bhp H H ηpump H, ηpump, or bhp V V ∙ ∙ FIGURE 14–8 Typical pump performance curves for a centrifugal pump with backward-inclined blades; the curve shapes for other types of pumps may differ, and the curves change as shaft rotation speed is changed.
0 0 System curve Pump performance curve Operating point BEP Havailable H Hrequired V ∙ FIGURE 14–9 The operating point of a piping system is established as the volume flow rate where the system curve and the pump performance curve intersect.
cen96537_ch14_793-884.indd 798 29/12/16 4:37 pm 799 CHAPTER 14 • It raises the elevation (potential energy) of the fluid from point 1 to point 2 (third term on the right).
• It overcomes irreversible head losses in the piping system (last term on the right).
In a general system, the change in static pressure, dynamic pressure, and elevation may be either positive or negative, while irreversible head losses are always positive. In many mechanical and civil engineering problems in which the fluid is a liquid, the elevation term is important, but when the fluid is a gas, such as in ventilation and air pollution control problems, the elevation term is almost always negligible. To match a pump to a system, and to determine the operating point, we equate Hrequired of Eq. 14–6 to Havailable, which is the (typically known) net head of the pump as a function of volume flow rate.
Operating point: Hrequired = Havailable (14–7) The most common situation is that an engineer selects a pump that is some what heftier than actually required. The volume flow rate through the piping system is then a bit larger than needed, and a valve or damper is installed in the line so that the flow rate can be decreased as necessary.
H Possible operating points Havailable Hrequired 0 0 V ∙ FIGURE 14–10 Situations in which there can be more than one unique operating point should be avoided. In such cases a different pump should be used.
z2 – z1 z1 V2 V1 ≅ 0 Pump Valve Valve z2 1 2 Reservoir FIGURE 14–11 Equation 14–6 emphasizes the role of a pump in a piping system; namely, it increases (or decreases) the static pressure, dynamic pressure, and elevation of the fluid, and it overcomes irreversible losses.
EXAMPLE 14–1 Operating Point of a Fan in a Ventilation System A local ventilation system (hood and exhaust duct) is used to remove air and contaminants produced by a dry-cleaning operation (Fig. 14–12). The duct is round and is constructed of galvanized steel with longitudinal seams and with joints every 30 in (0.76 m). The inner diameter (ID) of the duct is D = 9.06 in (0.230 m), and its total length is L = 44.0 ft (13.4 m). There are five CD3-9 elbows along the duct. The equivalent roughness height of this duct is 0.15 mm, and each elbow has a minor (local) loss coefficient of KL = C0 = 0.21. Note the notation C0 for minor loss coefficient, commonly used in the ventilation industry (ASHRAE, 2001). To ensure adequate ventilation, the minimum required volume flow rate through the duct is V . = 600 cfm (cubic feet per minute), or 0.283 m3/s at 25°C. Literature from the hood manufacturer lists the hood entry loss coeffi cient as 1.3 based on duct velocity. When the damper is fully open, its loss coef ficient is 1.8. A centrifugal fan with 9.0-in inlet and outlet diameters is available. Its performance data are shown in Table 14–1, as listed by the manufacturer. Predict the operating point of this local ventilation system, and draw a plot of required and available fan pressure rise as functions of volume flow rate. Is the chosen fan adequate?
SOLUTION We are to estimate the operating point for a given fan and duct system and to plot required and available fan pressure rise as functions of volume flow rate. We are then to determine if the selected fan is adequate.
Assumptions 1 The flow is steady. 2 The concentration of contaminants in the air is low; the fluid properties are those of air alone. 3 The flow at the outlet is fully developed turbulent pipe flow with 𝛼 = 1.05.
Properties For air at 25°C, 𝜈 = 1.562 × 10−5 m2/s and 𝜌 = 1.184 kg/m3. Standard atmospheric pressure is Patm = 101.3 kPa.
cen96537_ch14_793-884.indd 799 29/12/16 4:37 pm 800 TURBOMACHINERY Analysis We apply the steady energy equation in head form (Eq. 14–6) from point 1 in the stagnant air region in the room to point 2 at the duct outlet, Hrequired = P2 −P1 ρg + 𝛼2V2 2 −𝛼1V1 2 2g + (z2 −z1) + hL, total (1) In Eq. 1 we may ignore the air speed at point 1 since it was chosen (wisely) far enough away from the hood inlet so that the air is nearly stagnant. At point 1, we let P1 = Patm. At point 2, P2 is then equal to Patm − 𝜌g (z2 − z1) since the jet discharges into stagnant outside air at higher elevation z2 on the roof of the building. Thus, the pressure terms cancel with the elevation terms, and Eq. 1 reduces to Required net head: Hrequired = 𝛼2V2 2 2g + hL, total (2) The total head loss in Eq. 2 is a combination of major and minor losses and depends on volume flow rate. Since the duct diameter is constant, Total irreversible head loss: hL, total = (f L D + ∑KL) V 2 2g (3) The dimensionless roughness factor is 𝜀/D = (0.15 mm)/(230 mm) = 6.52 × 10−4. The Reynolds number of the air flowing through the duct is Reynolds number: Re = DV 𝜈 = D 𝜈 4V .
𝜋 D2 = 4V .
𝜈𝜋 D (4) The Reynolds number varies with volume flow rate. At the minimum required flow rate, the air speed through the duct is V = V2 = 6.81 m/s, and the Reynolds number is Re = 4(0.283 m3/s) (1.562 × 10−5 m2/s)𝜋 (0.230 m) = 1.00 × 105 From the Moody chart (or the Colebrook equation) at this Reynolds number and roughness factor, the friction factor is f = 0.0209. The sum of all the minor loss coefficients is Minor losses: ∑KL = 1.3 + 5(0.21) + 1.8 = 4.15 (5) Substituting these values at the minimum required flow rate into Eq. 2, the required net head of the fan at the minimum flow rate is Hrequired = (𝛼2 + f L D + ∑KL) V 2 2g = (1.05 + 0.0209 13.4 m 0.230 m + 4.15) (6.81 m/s)2 2(9.81 m/s2) = 15.2 m of air (6) Note that the head is expressed naturally in units of equivalent column height of the pumped fluid, which is air in this case. We convert to an equivalent column height of water by multiplying by the ratio of air density to water density, Damper Fan Hood z2 z1 2 1 V ∙ FIGURE 14–12 The local ventilation system for Example 14–1, showing the fan and all minor losses.
Note that the head data are listed as inches of water, even though air is the fluid. This is common practice in the ventilation industry.
TABLE 14–1 Manufacturer’s performance data for the fan of Example 14–1 V .
, cfm Havailable, inches H2O 0 250 500 750 1000 1200 0.90 0.95 0.90 0.75 0.40 0.0 cen96537_ch14_793-884.indd 800 29/12/16 4:37 pm 801 CHAPTER 14 It is common practice in the pump industry to offer several choices of impeller diameter for a single pump casing. There are several reasons for this: (1) to save manufacturing costs, (2) to enable capacity increase by simple impeller replacement, (3) to standardize installation mountings, and (4) to enable reuse of equipment for a different application. When plotting the performance of such a “family” of pumps, pump manufacturers do not plot separate curves of H, 𝜂pump, and bhp for each impeller diameter in the form sketched in Fig. 14–8. Instead, they prefer to combine the performance curves of an entire family of pumps of different impeller diameters onto a single plot (Fig. 14–14). Specifically, they plot a curve of H as a function of V . for each impeller diameter in the same way as in Fig. 14–8, but cre ate contour lines of constant efficiency, by drawing smooth curves through points that have the same value of 𝜂pump for the various choices of impel ler diameter. Contour lines of constant bhp are often drawn on the same plot in similar fashion. An example is provided in Fig. 14–15 for a family of centrifugal pumps manufactured by Taco, Inc. In this case, five impeller diameters are available, but the identical pump casing is used for all five options. As seen in Fig. 14–15, pump manufacturers do not always plot their pumps’ performance curves all the way to free delivery. This is because the pumps are usually not operated there due to the low values of net head and efficiency. If higher values of flow rate and/or net head are required, the customer should step up to the next larger casing size, or consider using additional pumps in series or parallel. It is clear from the performance plot of Fig. 14–15 that for a given pump casing, the larger the impeller, the higher the maximum achievable effi ciency. Why then would anyone buy the smaller impeller pump? To answer this question, we must recognize that the customer’s application requires a certain combination of flow rate and net head. If the requirements match a particular impeller diameter, it may be more cost effective to sacrifice pump efficiency in order to satisfy these requirements. Hrequired, inches of water = Hrequired, air ρair ρwater = (15.2 m) 1.184 kg/m3 998.0 kg/m3 ( 1 in 0.0254 m) = 0.709 inches of water (7) We repeat the calculations at several values of volume flow rate, and compare to the available net head of the fan in Fig. 14–13. The operating point is at a volume flow rate of about 650 cfm, at which both the required and available net head equal about 0.83 inches of water. We conclude that the chosen fan is more than adequate for the job.
Discussion The purchased fan is somewhat more powerful than required, yielding a higher flow rate than necessary. The difference is small and is acceptable; the butterfly damper valve could be partially closed to cut back the flow rate to 600 cfm if necessary. For safety reasons, it is clearly better to oversize than undersize a fan when used with an air pollution control system.
1 0.5 0.3 0.1 0 0 200 400 800 1000 0.6 0.9 0.8 0.7 0.4 0.2 600 H, inches H2O Operating point Havailable Hrequired ∙ V, cfm FIGURE 14–13 Net head as a function of volume flow rate for the ventilation system of Example 14–1. The point where the available and required values of H intersect is the operating point.
0 0 BEP, ηpump = 85% 50% 60% 70% 75% 80% 80% 70% D4 D3 D2 D1 H ∙ V FIGURE 14–14 Typical pump performance curves for a family of centrifugal pumps of the same casing diameter but different impeller diameters.
cen96537_ch14_793-884.indd 801 29/12/16 4:37 pm 802 TURBOMACHINERY EXAMPLE 14–2 Selection of Pump Impeller Size A washing operation at a power plant requires 370 gallons per minute (gpm) of water. The required net head is about 24 ft at this flow rate. A newly hired engineer looks through some catalogs and decides to purchase the 8.25-in impeller option of the Taco Model 4013 FI Series centrifugal pump of Fig. 14–15. If the pump operates at 1160 rpm, as specified in the performance plot, she reasons, its performance curve intersects 370 gpm at H = 24 ft. The chief engineer, who is very concerned about efficiency, glances at the performance curves and notes that the efficiency of this pump at this operating point is only 70 percent. He sees that the 12.75-in impeller option achieves a higher efficiency (about 76.5 percent) at the same flow rate. He notes that a throttle valve can be installed downstream of the pump to increase the required net head so that the pump operates at this higher efficiency. He asks the junior engineer to justify her choice of impeller diameter. Namely, he asks her to calculate which impeller option (8.25-in or 12.75-in) would need the least amount of electricity to operate (Fig. 14–16). Perform the comparison and discuss.
120 60 0 0 Flow in gallons per minute 1000 80 100 40 20 100 200 300 400 500 600 700 800 900 Head in feet 5 10 15 20 25 30 35 40 45 50 55 60 25 20 15 10 5 0 L/s 0 2 4 6 8 10 250 200 150 100 50 0 Head in meters Head in kilopascals 30 25 20 15 10 5 0 Feet NPSH Curve no. 2313 Min. Imp. Bio. 6.75" Size 5 × 4 × 12.75 kPa Model 4013 Fl & Cl Series 1160 RPM 12.75" (241.3mm) REQUIRED NPSH 11.25" (229mm) 9.75" (216mm) 8.25" (203mm) 6.75" (111mm) Curves based on clear water with specific gravity of 1.0 2HP(1.5kW) 3HP(2.2kW) 5HP(3.7kW) 7.5HP(6.6kW) 10HP(7.5kW) 15HP(11.2kW) 50% 55% 60% 65% 70% 72% 74% 76% 78% 80% 78% 76% 74% 72% 70% 65% 60% 55% 50% FIGURE 14–15 Example of a manufacturer’s performance plot for centrifugal pumps. Each pump has the same casing, but a different impeller diameter.
Courtesy of Taco, Inc., Cranston, RI. Used by permission. Taco® is a registered trademark of Taco, Inc.
cen96537_ch14_793-884.indd 802 29/12/16 4:37 pm 803 CHAPTER 14 Pump Cavitation and Net Positive Suction Head When pumping liquids, it is possible for the local pressure inside the pump to fall below the vapor pressure of the liquid, Pv. (Pv is also called the saturation pressure Psat and is listed in thermodynamics tables as a func tion of saturation temperature.) When P < Pv, vapor-filled bubbles called cavitation bubbles appear. In other words, the liquid boils locally, typically on the suction side of the rotating impeller blades where the pressure is lowest (Fig. 14–17). After the cavitation bubbles are formed, they are transported through the pump to regions where the pressure is higher, causing rapid col lapse of the bubbles. It is this collapse of the bubbles that is undesirable, since it causes noise, vibration, reduced efficiency, and most importantly, damage to the impeller blades. Repeated bubble collapse near a blade surface leads to pitting or erosion of the blade and eventually catastrophic blade failure. To avoid cavitation, we must ensure that the local pressure everywhere inside the pump stays above the vapor pressure. Since pressure is most easily SOLUTION For a given flow rate and net head, we are to calculate which impeller size uses the least amount of power, and we are to discuss our results.
Assumptions 1 The water is at 70°F. 2 The flow requirements (volume flow rate and head) are constant.
Properties For water at 70°F, 𝜌 = 62.30 lbm/ft3.
Analysis From the contours of brake horsepower that are shown on the perfor mance plot of Fig. 14–15, the junior engineer estimates that the pump with the smaller impeller requires about 3.2 hp from the motor. She verifies this estimate by using Eq. 14–5, Required bhp for the 8.25-in impeller option: bhp = ρgV .
H 𝜂pump = (62.30 lbm/ft3)(32.2 ft/s2)(370 gal/min)(24 ft) 0.70 × ( 0.1337 ft3 gal ) ( lbf 32.2 lbm·ft/s2) ( 1 min 60 s ) ( hp·s 550 ft·lbf) = 3.20 hp Similarly, the larger-diameter impeller option requires Required bhp for the 12.75-in impeller option: bhp = 8.78 hp using the operating point of that pump, namely, V . = 370 gpm, H = 72.0 ft, and 𝜂pump = 76.5 percent (Fig. 14–15). Clearly, the smaller-diameter impeller option is the better choice in spite of its lower efficiency, because it uses less than half the power.
Discussion Although the larger impeller pump would operate at a somewhat higher value of efficiency, it would deliver about 72 ft of net head at the required flow rate. This is overkill, and the throttle valve would be required to make up the difference between this net head and the required flow head of 24 ft of water. A throttle valve does nothing more than waste mechanical energy, however; so the gain in efficiency of the pump is more than offset by losses through the throttle valve. If the flow head or capacity requirements increase at some time in the future, a larger impeller can be purchased for the same casing.
Is she trying to tell me that the less efficient pump can actually save on energy costs?
FIGURE 14–16 In some applications, a less efficient pump from the same family of pumps may require less energy to operate. An even better choice, however, would be a pump whose best efficiency point occurs at the required operating point of the pump, but such a pump is not always commercially available.
cen96537_ch14_793-884.indd 803 29/12/16 4:37 pm 804 TURBOMACHINERY measured (or estimated) at the inlet of the pump, cavitation criteria are typi cally specified at the pump inlet. It is useful to employ a flow parameter called net positive suction head (NPSH), defined as the difference between the pump’s inlet stagnation pressure head and the vapor pressure head, Net positive suction head: NPSH = ( P ρg + V 2 2g) pump inlet − Pv ρg (14–8) Pump manufacturers test their pumps for cavitation in a pump test facility by varying the volume flow rate and inlet pressure in a controlled manner. Specifically, at a given flow rate and liquid temperature, the pressure at the pump inlet is slowly lowered until cavitation occurs somewhere inside the pump. The value of NPSH is calculated using Eq. 14–8 and is recorded at this operating condition. The process is repeated at several other flow rates, and the pump manufacturer then publishes a performance parameter called the required net positive suction head (NPSHrequired), defined as the mini mum NPSH necessary to avoid cavitation in the pump. The measured value of NPSHrequired varies with volume flow rate, and therefore NPSHrequired is often plotted on the same pump performance curve as net head (Fig. 14–18). When expressed properly in units of head of the liquid being pumped, NPSHrequired is independent of the type of liquid. However, if the required net positive suction head is expressed for a particular liquid in pressure units such as pascals or psi, the engineer must be careful to convert this pressure to the equivalent column height of the actual liquid being pumped. Note that since NPSHrequired is usually much smaller than H over the majority of the performance curve, it is often plotted on a separate expanded vertical axis for clarity (see Fig. 14–15) or as contour lines when being shown for a family of pumps. NPSHrequired typically increases with volume flow rate, although for some pumps it decreases with V . at low flow rates where the pump is not operating very efficiently, as sketched in Fig. 14–18. In order to ensure that a pump does not cavitate, the actual or available NPSH must be greater than NPSHrequired. It is important to note that the value of NPSH varies not only with flow rate, but also with liquid temperature, since Pv is a function of temperature. NPSH also depends on the type of liq uid being pumped, since there is a unique Pv versus T curve for each liquid. Since irreversible head losses through the piping system upstream of the inlet increase with flow rate, the pump inlet stagnation pressure head decreases with flow rate. Therefore, the value of NPSH decreases with V .
, as sketched in Fig. 14–19. By identifying the volume flow rate at which the curves of actual NPSH and NPSHrequired intersect, we estimate the maximum volume flow rate that can be delivered by the pump without cavitation (Fig. 14–19).
EXAMPLE 14–3 Maximum Flow Rate to Avoid Pump Cavitation The 11.25-in impeller option of the Taco Model 4013 FI Series centrifugal pump of Fig. 14–15 is used to pump water at 25°C from a reservoir whose surface is 4.0 ft above the centerline of the pump inlet (Fig. 14–20). The piping system from the reservoir to the pump consists of 10.5 ft of cast iron pipe with an ID of 4.0 in and an average inner roughness height of 0.02 in. There are several minor losses: a sharp-edged inlet (KL = 0.5), three flanged smooth 90° regular elbows Pressure side Suction side Impeller blade Cavitation bubbles form Cavitation bubbles collapse ω FIGURE 14–17 Cavitation bubbles forming and collapsing on the suction side of an impeller blade.
NPSHrequired Head H 0 0 ∙ V FIGURE 14–18 Typical pump performance curve in which net head and required net positive suction head are plotted versus volume flow rate.
NPSH H No cavitation Head NPSH required Cavitation 0 0 Vmax ∙ ∙ V FIGURE 14–19 The volume flow rate at which the actual NPSH and the required NPSH intersect represents the maximum flow rate that can be delivered by the pump without the occurrence of cavitation.
cen96537_ch14_793-884.indd 804 29/12/16 4:37 pm 805 CHAPTER 14 (KL = 0.3 each), and a fully open flanged globe valve (KL = 6.0). Estimate the maximum volume flow rate (in units of gpm) that can be pumped without cav itation. If the water were warmer, would this maximum flow rate increase or decrease? Why? Discuss how you might increase the maximum flow rate while still avoiding cavitation.
SOLUTION For a given pump and piping system we are to estimate the maxi mum volume flow rate that can be pumped without cavitation. We are also to discuss the effect of water temperature and how we might increase the maximum flow rate. Assumptions 1 The flow is steady. 2 The liquid is incompressible. 3 The flow at the pump inlet is turbulent and fully developed, with 𝛼 = 1.05.
Properties For water at T = 25°C, 𝜌 = 997.0 kg/m3, 𝜇 = 8.91 × 10−4 kg/m · s, and Pv = 3.169 kPa. Standard atmospheric pressure is Patm = 101.3 kPa.
Analysis We apply the steady energy equation in head form along a streamline from point 1 at the reservoir surface to point 2 at the pump inlet, P1 ρg + 𝛼1V1 2 2g + z1 + hpump, u = P2 ρg + 𝛼2V2 2 2g + z2 + hturbine, e + hL, total (1) In Eq. 1 we have ignored the water speed at the reservoir surface (V1 ≅ 0). There is no turbine in the piping system. Also, although there is a pump in the system, there is no pump between points 1 and 2; hence the pump head term also drops out. We solve Eq. 1 for P2/𝜌g, which is the pump inlet pressure expressed as a head, Pump inlet pressure head: P2 ρg = Patm ρg + (z1 −z2) −𝛼2V2 2 2g −hL, total (2) Note that in Eq. 2, we have recognized that P1 = Patm since the reservoir surface is exposed to atmospheric pressure. The available net positive suction head at the pump inlet is obtained from Eq. 14–8. After substitution of Eq. 2, we get Available NPSH: NPSH = Patm −Pv ρg + (z1 −z2) −hL, total −(𝛼2 −1)V2 2 2g (3) Since we know Patm, Pv, and the elevation difference, all that remains is to estimate the total irreversible head loss through the piping system, which depends on volume flow rate. Since the pipe diameter is constant, Irreversible head loss: hL, total = (f L D + ∑KL) V 2 2g (4) The rest of the problem is most easily solved on a computer. For a given volume flow rate, we calculate speed V and Reynolds number Re. From Re and the known pipe roughness, we use the Moody chart (or the Colebrook equation) to obtain friction factor f. The sum of all the minor loss coefficients is Minor losses: ∑KL = 0.5 + 3 × 0.3 + 6.0 = 7.4 (5) We make one calculation by hand for illustrative purposes. At V . = 400 gpm (0.02523 m3/s), the average speed of water through the pipe is V = V .
A = 4V .
𝜋 D2 = 4(0.02523 m3/s) 𝜋 (4.0 in)2 ( 1 in 0.0254 m) 2 = 3.112 m/s (6) z2 Pump Inlet piping system Valve 2 z1 1 Reservoir FIGURE 14–20 Inlet piping system from the reservoir (1) to the pump inlet (2) for Example 14–3.
cen96537_ch14_793-884.indd 805 29/12/16 4:37 pm 806 TURBOMACHINERY Pumps in Series and Parallel When faced with the need to increase volume flow rate or pressure rise by a small amount, you might consider adding an additional smaller pump in series or in parallel with the original pump. While series or parallel arrange ment is acceptable for some applications, arranging dissimilar pumps in series or in parallel may lead to problems, especially if one pump is much larger than the other (Fig. 14–22). A better course of action is to increase the original pump’s speed and/or input power (larger electric motor), replace the impeller with a larger one, or replace the entire pump with a larger one. The logic for this decision can be seen from the pump performance curves, realizing that pressure rise and volume flow rate are related. Arranging which produces a Reynolds number of Re = 𝜌VD/𝜇 = 3.538 × 105. At this Reynolds number, and with roughness factor 𝜀/D = 0.005, the Colebrook equa tion yields f = 0.0306. Substituting the given properties, along with f, D, L, and Eqs. 4, 5, and 6, into Eq. 3, we calculate the available net positive suction head at this flow rate, NPSH = (101,300 −3169) N/m2 (997.0 kg/m3)(9.81 m/s2)( kg·m/s2 N ) + 1.219 m −(0.0306 10.5 ft 0.3333 ft + 7.4 −(1.05 −1)) (3.112 m/s)2 2(9.81 m/s2) = 7.148 m = 23.5 ft (7) The required net positive suction head is obtained from Fig. 14–15. At our example flow rate of 400 gpm, NPSHrequired is just above 4.0 ft. Since the actual NPSH is much higher than this, we need not worry about cavitation at this flow rate. We use EES (or a spreadsheet) to calculate NPSH as a function of volume flow rate, and the results are plotted in Fig. 14–21. It is clear from this plot that at 25°C, cavitation occurs at flow rates above approximately 600 gpm—close to the free delivery. If the water were warmer than 25°C, the vapor pressure would increase, the vis cosity would decrease, and the density would decrease slightly. The calculations are repeated at T = 60°C, at which 𝜌 = 983.3 kg/m3, 𝜇 = 4.67 × 10−4 kg/m · s, and Pv = 19.94 kPa. The results are also plotted in Fig. 14–21, where we see that the maxi mum volume flow rate without cavitation decreases with temperature (to about 555 gpm at 60°C). This decrease agrees with our intuition, since warmer water is already closer to its boiling point from the start. Finally, how can we increase the maximum flow rate? Any modification that increases the available NPSH helps. We can raise the height of the reservoir surface (to increase the hydrostatic head). We can reroute the piping so that only one elbow is necessary and replace the globe valve with a ball valve (to decrease the minor losses). We can increase the diameter of the pipe and decrease the surface roughness (to decrease the major losses). In this particular problem, the minor losses have the greatest influence, but in many problems, the major losses are more significant, and increasing the pipe diameter is most effective. That is one reason why many centrifugal pumps have a larger inlet diameter than outlet diameter.
Discussion Note that NPSHrequired does not depend on water temperature, but the actual or available NPSH decreases with temperature (Fig. 14–21).
30 5 0 300 400 500 700 10 25 20 15 600 NPSH, ft ∙ V, gpm Available NPSH, 60°C Available NPSH, 25°C No cavitation, T = 25°C Required NPSH No cavitation, T = 60°C FIGURE 14–21 Net positive suction head as a function of volume flow rate for the pump of Example 14–3 at two temperatures. Cavitation is predicted to occur at flow rates greater than the point where the available and required values of NPSH intersect.
(a) (b) FIGURE 14–22 Arranging two very dissimilar pumps in (a) series or (b) parallel can some times lead to problems.
cen96537_ch14_793-884.indd 806 29/12/16 4:37 pm 807 CHAPTER 14 dissimilar pumps in series may create problems because the volume flow rate through each pump must be the same, but the overall pressure rise is equal to the pressure rise of one pump plus that of the other. If the pumps have widely different performance curves, the smaller pump may be forced to operate beyond its free delivery flow rate, whereupon it acts like a head loss, reducing the total volume flow rate. Arranging dissimilar pumps in parallel may create problems because the overall pressure rise must be the same, but the net volume flow rate is the sum of that through each branch. If the pumps are not sized properly, the smaller pump may not be able to handle the large head imposed on it, and the flow in its branch could actu ally be reversed; this would inadvertently reduce the overall pressure rise. In either case, the power supplied to the smaller pump would be wasted. Keeping these cautions in mind, there are many applications where two or more similar (usually identical) pumps are operated in series or in parallel. When operated in series, the combined net head is simply the sum of the net heads of each pump (at a given volume flow rate), Combined net head for n pumps in series: Hcombined = ∑ n i=1 Hi (14–9) Equation 14–9 is illustrated in Fig. 14–23 for three pumps in series. In this example, pump 3 is the strongest and pump 1 is the weakest. The shutoff head of the three pumps combined in series is equal to the sum of the shut off head of each individual pump. For low values of volume flow rate, the net head of the three pumps in series is equal to H1 + H2 + H3. Beyond the free delivery of pump 1 (to the right of the first vertical dashed red line in Fig. 14–23), pump 1 should be shut off and bypassed. Otherwise it would be running beyond its maximum designed operating point, and the pump or its motor could be damaged. Furthermore, the net head across this pump would be negative as previously discussed, contributing to a net loss in the system. With pump 1 bypassed, the combined net head becomes H2 + H3. Similarly, beyond the free delivery of pump 2, that pump should also be shut off and bypassed, and the combined net head is then equal to H3 alone, as indicated to the right of the second vertical dashed gray line in Fig. 14–23. 0 0 H Combined net head Pump 1 should be shut off and bypassed Pump 1 Pump 2 Pump 3 Pump 2 should be shut off and bypassed H1 + H2 + H3 H2 + H3 H3 only Shutoff head of combined pumps Free delivery of combined pumps ∙ V FIGURE 14–23 Pump performance curve (dark blue) for three dissimilar pumps in series. At low values of volume flow rate, the combined net head is equal to the sum of the net head of each pump by itself. However, to avoid pump damage and loss of combined net head, any individual pump should be shut off and bypassed at flow rates larger than that pump’s free delivery, as indicated by the vertical dashed red lines. If the three pumps were identical, it would not be necessary to turn off any of the pumps, since the free delivery of each pump would occur at the same volume flow rate.
cen96537_ch14_793-884.indd 807 29/12/16 4:37 pm 808 TURBOMACHINERY In this case, the combined free delivery is the same as that of pump 3 alone, assuming that the other two pumps are bypassed. When two or more identical (or similar) pumps are operated in parallel, their individual volume flow rates (rather than net heads) are summed, Combined capacity for n pumps in parallel: V .
combined = ∑ n i=1 V .
i (14–10) As an example, consider the same three pumps, but arranged in parallel rather than in series. The combined pump performance curve is shown in Fig. 14–24. The free delivery of the three combined pumps is equal to the sum of the free delivery of each individual pump. For low values of net head, the capacity of the three pumps in parallel is equal to V .
1 + V .
2 + V .
3. Above the shutoff head of pump 1 (above the first horizontal dashed red line in Fig. 14–24), pump 1 should be shut off and its branch should be blocked (with a valve). Otherwise it would be running beyond its maximum designed operating point, and the pump or its motor could be damaged. Fur thermore, the volume flow rate through this pump would be negative as pre viously discussed, contributing to a net loss in the system. With pump 1 shut off and blocked, the combined capacity becomes V .
2 + V .
3. Similarly, above the shutoff head of pump 2, that pump should also be shut off and blocked. The combined capacity is then equal to V .
3 alone, as indicated above, the second horizontal dashed gray line in Fig. 14–24. In this case, the combined shutoff head is the same as that of pump 3 alone, assuming that the other two pumps are shut off and their branches are blocked. In practice, several pumps may be combined in parallel to deliver a large volume flow rate (Fig. 14–25). Examples include banks of pumps used to circulate water in cooling towers and chilled water loops (Wright, 1999). Ideally all the pumps should be identical so that we don’t need to worry about shutting any of them off (Fig. 14–24). Also, it is wise to install check valves in each branch so that when a pump needs to be shut down Pump 3 0 0 H Shutoff head of combined pumps Free delivery of combined pumps V V3 only V2 + V3 Pump 1 should be shut off V1 + V2 + V3 Pump 1 Pump 2 Combined capacity Pump 2 should be shut off ∙ ∙ ∙ ∙ ∙ ∙ ∙ FIGURE 14–24 Pump performance curve (dark blue) for three pumps in parallel. At a low value of net head, the combined capacity is equal to the sum of the capacity of each pump by itself. However, to avoid pump damage and loss of combined capacity, any individual pump should be shut off at net heads larger than that pump’s shutoff head, as indicated by the horizontal dashed gray lines. That pump’s branch should also be blocked with a valve to avoid reverse flow. If the three pumps were identical, it would not be necessary to turn off any of the pumps, since the shutoff head of each pump would occur at the same net head.
cen96537_ch14_793-884.indd 808 29/12/16 4:37 pm 809 CHAPTER 14 (for maintenance or when the required flow rate is low), backflow through the pump is avoided. Note that the extra valves and piping required for a parallel pump network add additional head losses to the system; thus the overall performance of the combined pumps suffers somewhat.
Positive-Displacement Pumps People have designed numerous positive-displacement pumps throughout the centuries. In each design, fluid is sucked into an expanding volume and then pushed along as that volume contracts, but the mechanism that causes this change in volume differs greatly among the various designs. Some designs are very simple, like the flexible-tube peristaltic pump (Fig. 14–26a) that compresses a tube by small wheels, pushing the fluid along. (This mecha nism is somewhat similar to peristalsis in your esophagus or intestines, where muscles rather than wheels compress the tube.) Others are more com plex, using rotating cams with synchronized lobes (Fig. 14–26b), interlock ing gears (Fig. 14–26c), or screws (Fig. 14–26d). Positive-displacement pumps are ideal for high-pressure applications like pumping viscous liquids or thick slurries, and for applications where precise amounts of liquid are to be dispensed or metered, as in medical applications.
FIGURE 14–25 Several identical pumps are often run in a parallel configuration so that a large volume flow rate can be achieved when necessary. Three parallel pumps are shown.
Courtesy of Goulds Pumps, ITT Corporation.
cen96537_ch14_793-884.indd 809 29/12/16 4:37 pm 810 TURBOMACHINERY To illustrate the operation of a positive-displacement pump, we sketch four phases of half of a cycle of a simple rotary pump with two lobes on each rotor (Fig. 14–27). The two rotors are synchronized by an external gear box so as to rotate at the same angular speed, but in opposite directions. In the dia gram, the top rotor turns clockwise and the bottom rotor turns counterclock wise, sucking in fluid from the left and discharging it to the right. A white dot is drawn on one lobe of each rotor to help you visualize the rotation.
(a) (b) (c) (d) FIGURE 14–26 Examples of positive-displacement pumps: (a) flexible-tube peristaltic pump, (b) three-lobe rotary pump, (c) gear pump, and (d) double screw pump.
Adapted from F. M. White, Fluid Mechanics 4/e. Copyright © 1999. The McGraw-Hill Companies, Inc. In 45° 90° 135° Out 180° FIGURE 14–27 Four phases (one-eighth of a turn apart) in the operation of a two-lobe rotary pump, a type of positive-displacement pump. The blue region represents a chunk of fluid pushed through the top rotor, while the red region represents a chunk of fluid pushed through the bottom rotor, which rotates in the opposite direction. Flow is from left to right.
cen96537_ch14_793-884.indd 810 29/12/16 4:37 pm 811 CHAPTER 14 Gaps exist between the rotors and the housing and between the lobes of the rotors themselves, as illustrated (and exaggerated) in Fig. 14–27. Fluid can leak through these gaps, reducing the pump’s efficiency. High-viscosity fluids cannot penetrate the gaps as easily; hence the net head (and efficiency) of a rotary pump generally increases with fluid viscosity, as shown in Fig. 14–28. This is one reason why rotary pumps (and other types of positive-displacement pumps) are a good choice for pumping highly viscous fluids and slurries. They are used, for example, as automobile engine oil pumps and in the foods industry to pump heavy liquids like syrup, tomato paste, and chocolate, and slurries like soups. The pump performance curve (net head versus capacity) of a rotary pump is nearly vertical throughout its recommended operating range, since the capacity is fairly constant regardless of load at a given rotational speed (Fig. 14–28). However, as indicated by the dashed blue curve in Fig. 14–28, at very high values of net head, corresponding to very high pump outlet pres sure, leaks become more severe, even for high-viscosity fluids. In addition, the motor driving the pump cannot overcome the large torque caused by this high outlet pressure, and the motor begins to suffer stall or overload, which may burn out the motor. Therefore, rotary pump manufacturers do not rec ommend operation of the pump above a certain maximum net head, which is typically well below the shutoff head. The pump performance curves sup plied by the manufacturer often do not even show the pump’s performance outside of its recommended operating range. Positive-displacement pumps have many advantages over dynamic pumps. For example, a positive-displacement pump is better able to handle shear sensitive liquids since the induced shear is much less than that of a dynamic pump operating at similar pressure and flow rate. Blood is a shear sensitive liquid, and this is one reason why positive-displacement pumps are used for artificial hearts. A well-sealed positive-displacement pump can create a significant vacuum pressure at its inlet, even when dry, and is thus able to lift a liquid from several meters below the pump. We refer to this kind of pump as a self-priming pump (Fig. 14–29). Finally, the rotor(s) of a positive- displacement pump run at lower speeds than the rotor (impeller) of a dynamic pump at similar loads, extending the useful lifetime of seals, etc. There are some disadvantages of positive-displacement pumps as well. Their volume flow rate cannot be changed unless the rotation rate is changed. (This is not as simple as it sounds, since most AC electric motors are designed to operate at one or more fixed rotational speeds.) They cre ate very high pressure at the outlet side, and if the outlet becomes blocked, ruptures may occur or electric motors may overheat, as previously discussed. Overpressure protection (e.g., a pressure-relief valve) is often required for this reason. Because of their design, positive-displacement pumps sometimes deliver a pulsating flow, which may be unacceptable for some applications. Analysis of positive-displacement pumps is fairly straightforward. From the geometry of the pump, we calculate the closed volume (Vclosed) that is filled (and expelled) for every n rotations of the shaft. Volume flow rate is then equal to rotation rate n . times Vclosed divided by n, Volume flow rate, positive-displacement pump: V .
= n · Vclosed n (14–11) 0 0 Free delivery Maximum recommended net head Recommended operating range Increasing viscosity H Shutoff head ∙ V FIGURE 14–28 Comparison of the pump performance curves of a rotary pump operating at the same speed, but with fluids of various viscosities. To avoid motor overload the pump should not be oper ated in the shaded region.
Out Self-priming pump Hose In FIGURE 14–29 A pump that can lift a liquid even when the pump itself is “empty” is called a self-priming pump.
cen96537_ch14_793-884.indd 811 29/12/16 4:37 pm 812 TURBOMACHINERY Dynamic Pumps There are three main types of dynamic pumps that involve rotating blades called impeller blades or rotor blades, which impart momentum to the fluid. For this reason they are sometimes called rotodynamic pumps or simply rotary pumps (not to be confused with rotary positive-displacement pumps, which use the same name). There are also some nonrotary dynamic pumps, such as jet pumps and electromagnetic pumps; these are not dis cussed in this text. Rotary pumps are classified by the manner in which flow exits the pump: centrifugal flow, axial flow, and mixed flow (Fig. 14–31). In a centrifugal-flow pump, fluid enters axially (in the same direction as the axis of the rotating shaft) in the center of the pump, but is discharged radi ally (or tangentially) along the outer radius of the pump casing. For this rea son centrifugal pumps are also called radial-flow pumps. In an axial-flow pump, fluid enters and leaves axially, typically along the outer portion of the pump because of blockage by the shaft, motor, hub, etc. A mixed-flow pump is intermediate between centrifugal and axial, with the flow enter ing axially, not necessarily in the center, but leaving at some angle between radially and axially.
Centrifugal Pumps Centrifugal pumps and blowers can be easily identified by their snail-shaped casing, called the scroll (also called the volute) (Fig. 14–32). They are found all around your home—in dishwashers, hot tubs, clothes washers and dryers, EXAMPLE 14–4 Volume Flow Rate through a Positive-Displacement Pump A two-lobe rotary positive-displacement pump, similar to that of Fig. 14–27, moves 0.45 cm3 of SAE 30 motor oil in each lobe volume Vlobe, as sketched in Fig. 14–30. Calculate the volume flow rate of oil for the case where n . = 900 rpm.
SOLUTION We are to calculate the volume flow rate of oil through a positive-displacement pump for given values of lobe volume and rotation rate.
Assumptions 1 The flow is steady in the mean. 2 There are no leaks in the gaps between lobes or between lobes and the casing. 3 The oil is incompressible.
Analysis By studying Fig. 14–27, we see that for half of a rotation (180° for n = 0.5 rotations) of the two counter-rotating shafts, the total volume of oil pumped is Vclosed = 2Vlobe. The volume flow rate is then calculated from Eq. 14–11, V .
= n · Vclosed n = (900 rot/min) 2(0.45 cm3) 0.5 rot = 1620 cm3/min Discussion If there were leaks in the pump, the volume flow rate would be lower. The oil’s density is not needed for calculation of the volume flow rate. However, the higher the fluid density, the higher the required shaft torque and brake horsepower.
In Out ∙ V V ∙ V FIGURE 14–30 The two-lobe rotary pump of Example 14–4. Flow is from left to right.
(a) (b) (c) Impeller shroud Blade Flow out Flow in ω ω Blade Impeller shroud Impeller hub Flow out Flow out Flow in ω Blade Flow in FIGURE 14–31 The impeller (rotating portion) of the three main categories of dynamic pumps: (a) centrifugal flow, (b) mixed flow, and (c) axial flow.
cen96537_ch14_793-884.indd 812 29/12/16 4:37 pm 813 CHAPTER 14 hairdryers, vacuum cleaners, kitchen exhaust hoods, bathroom exhaust fans, leaf blowers, furnaces, etc. They are used in cars—the water pump in the engine, the air blower in the heater/air conditioner unit, etc. Centrifugal pumps are ubiquitous in industry as well; they are used in building ventilation sys tems, washing operations, cooling ponds and cooling towers, and in numerous other industrial operations in which fluids are pumped. A schematic diagram of a centrifugal pump is shown in Fig. 14–33. Note that a shroud often surrounds the impeller blades to increase blade stiff ness. In pump terminology, the rotating assembly that consists of the shaft, the hub, the impeller blades, and the impeller shroud is called the impeller or rotor. Fluid enters axially through the hollow middle portion of the pump (the eye), after which it encounters the rotating blades. It acquires tangen tial and radial velocity by momentum transfer with the impeller blades, and acquires additional radial velocity by so-called centrifugal forces, which are actually a lack of sufficient centripetal forces to sustain circular motion. The flow leaves the impeller after gaining both speed and pressure as it is flung radially outward into the scroll. As sketched in Fig. 14–33, the scroll is a snail-shaped diffuser whose purpose is to decelerate the fast-moving fluid leaving the trailing edges of the impeller blades, thereby further increasing the fluid’s pressure, and to combine and direct the flow from all the blade passages toward a common outlet. As mentioned previously, if the flow is steady in the mean, if the fluid is incompressible, and if the inlet and outlet diameters are the same, the average flow speed at the outlet is identical to that at the inlet. Thus, it is not necessarily the speed, but the pressure that increases from inlet to outlet through a centrifugal pump. There are three types of centrifugal pump that warrant discussion, based on impeller blade geometry, as sketched in Fig. 14–34: backward-inclined blades, radial blades, and forward-inclined blades. Centrifugal pumps with backward-inclined blades (Fig. 14–34a) are the most common. These yield the highest efficiency of the three because fluid flows into and out of the blade passages with the least amount of turning. Sometimes the blades are airfoil shaped, yielding similar performance but even higher efficiency. The pressure rise is intermediate between the other two types of centrifugal Impeller shroud Impeller Eye Vout, P out Pin Vin b1 r1 ω b2 Scroll Side view Frontal view Casing Shaft In Out r2 ω In Impeller blade FIGURE 14–33 Side view and frontal view of a typical centrifugal pump. Fluid enters axially in the middle of the pump (the eye), is flung around to the outside by the rotating blade assembly (impeller), is diffused in the expanding diffuser (scroll ), and is discharged out the side of the pump. We define r1 and r2 as the radial locations of the impeller blade inlet and outlet, respectively; b1 and b2 are the axial blade widths at the impeller blade inlet and outlet, respectively.
FIGURE 14–32 A typical centrifugal blower with its characteristic snail-shaped scroll or volute. Courtesy of the New York Blower Company, Willowbrook, IL. Used by permission.
cen96537_ch14_793-884.indd 813 29/12/16 4:37 pm 814 TURBOMACHINERY pumps. Centrifugal pumps with radial blades (also called straight blades, Fig. 14–34b) have the simplest geometry and produce the largest pressure rise of the three for a wide range of volume flow rates, but the pressure rise decreases rapidly after the point of maximum efficiency. Centrifugal pumps with forward-inclined blades (Fig. 14–34c) produce a pressure rise that is nearly constant, albeit lower than that of radial or backward-inclined blades, over a wide range of volume flow rates. Forward-inclined centrifugal pumps generally have more blades, but the blades are smaller, as sketched in Fig. 14–34c. Centrifugal pumps with forward-inclined blades generally have a lower maximum efficiency than do straight-bladed pumps. Radial and backward-inclined centrifugal pumps are preferred for applications where one needs to provide volume flow rate and pressure rise within a narrow range of values. If a wider range of volume flow rates and/or pressure rises are desired, the performance of radial pumps and backward-inclined pumps may not be able to satisfy the new requirements; these types of pumps are less forgiving (less robust). The performance of forward-inclined pumps is more forgiving and accommodates a wider variation, at the cost of lower efficiency and less pressure rise per unit of input power. If a pump is needed to produce large pressure rise over a wide range of volume flow rates, the forward-inclined centrifugal pump is attractive. Net head and brake horsepower performance curves for these three types of centrifugal pump are compared in Fig. 14–34d. The curves have been adjusted such that each pump achieves the same free delivery (maximum volume flow rate at zero net head). Note that these are qualitative sketches for comparison purposes only—actual measured performance curves may differ significantly in shape, depending on details of the pump design. For any inclination of the impeller blades (backward, radial, or forward), we can analyze the velocity vectors through the blades. The actual flow field is unsteady, fully three-dimensional, and perhaps compressible. For simplic ity in our analysis we consider steady flow in both the absolute reference frame and in the relative frame of reference rotating with the impeller. We consider only incompressible flow, and we consider only the radial or nor mal velocity component (subscript n) and the circumferential or tangential velocity component (subscript t) from blade inlet to blade outlet. We do not consider the axial velocity component (to the right in Fig. 14–35 and into the page in the frontal view of Fig. 14–33). In other words, although there is a nonzero axial component of velocity through the impeller, it does not enter our analysis. A close-up side view of a simplified centrifugal pump is sketched in Fig. 14–35, where we define V1, n and V2, n as the average normal components of velocity at radii r1 and r2, respectively. Although a gap is shown between the blade and the casing, we assume in our simplified analy sis that no leakage occurs in these gaps. The volume flow rate V . entering the eye of the pump passes through the circumferential cross-sectional area defined by width b1 at radius r1. Conser vation of mass requires that this same volume flow rate must pass through the circumferential cross-sectional area defined by width b2 at radius r2. Using the average normal velocity components V1, n and V2, n defined in Fig. 14–35, we write Volume flow rate: V .
= 2𝜋 r1b1V1, n = 2𝜋 r2b2V2, n (14–12) 0 0 bhp Radial Backward Forward H H or bhp (d) (c) (b) (a) ω ω ω ∙ V FIGURE 14–34 The three main types of centrifugal pumps are those with (a) backward-inclined blades, (b) radial blades, and (c) forward-inclined blades; (d) comparison of net head and brake horsepower performance curves for the three types of centrifugal pumps.
cen96537_ch14_793-884.indd 814 29/12/16 4:37 pm 815 CHAPTER 14 from which we obtain V2, n = V1, n r1b1 r2b2 (14–13) It is clear from Eq. 14–13 that V2, n may be less than, equal to, or greater than V1, n, depending on the values of b and r at the two radii. We sketch a close-up frontal view of one impeller blade in Fig. 14–36, where we show both radial and tangential velocity components. We have drawn a backward-inclined blade, but the same analysis holds for blades of any inclination. The inlet of the blade (at radius r1) moves at tangential veloc ity 𝜔r1. Likewise, the outlet of the blade moves at tangential velocity 𝜔r2. It is clear from Fig. 14–36 that these two tangential velocities differ not only in magnitude, but also in direction, because of the inclination of the blade. We define leading edge angle 𝛽1 as the blade angle relative to the reverse tangen tial direction at radius r1. In like manner we define trailing edge angle 𝛽2 as the blade angle relative to the reverse tangential direction at radius r2. We now make a significant simplifying approximation. We assume that the flow impinges on the blade parallel to the blade’s leading edge and exits the blade parallel to the blade’s trailing edge. In other words, We assume that the flow is everywhere tangent to the blade surface when viewed from a reference frame rotating with the blade.
At the inlet, this approximation is sometimes called the shockless entry condition, not to be confused with shock waves (Chap. 12). Rather, the terminology implies smooth flow into the impeller blade without a sudden turning “shock.” Inherent in this approximation is the assumption that there is no flow separation anywhere along the blade surface. If the centrifugal pump operates at or near its design conditions, this assumption is valid. However, when the pump operates far off design conditions, the flow may separate off the blade surface (typically on the suction side where there are adverse pressure gradients), and our simplified analysis breaks down. Velocity vectors V › 1, relative and V › 2, relative are drawn in Fig. 14–36 parallel to the blade surface, in accordance with our simplifying assumption. These are the velocity vectors seen from the relative reference frame of an observer moving with the rotating blade. When we vectorially add tangential veloc ity 𝜔r1 (the velocity of the blade at radius r1) to V › 1, relative by completing the parallelogram as sketched in Fig. 14–36, the resultant vector is the absolute fluid velocity V › 1 at the blade inlet. In exactly similar fashion, we obtain V › 2, the absolute fluid velocity at the blade outlet (also sketched in Fig. 14–36). For completeness, normal velocity components V1, n and V2, n are also shown in Fig. 14–36. Notice that these normal velocity components are indepen dent of which frame of reference we use, absolute or relative. To evaluate the torque on the rotating shaft, we apply the angular momen tum relation for a control volume, as discussed in Chap. 6. We choose a control volume surrounding the impeller blades, from radius r1 to radius r2, as sketched in Fig. 14–37. We also introduce in Fig. 14–37 angles 𝛼1 and 𝛼2, defined as the angle of departure of the absolute velocity vector from the normal direction at radii r1 and r2, respectively. In keeping with the con cept of treating a control volume like a “black box,” we ignore details of individual impeller blades. Instead we make the approximation that flow Impeller blade In Shaft V1, n V2, n b2 b1 Vin, Pin, r1 r2 Casing Scroll Out ω ∙ V FIGURE 14–35 Close-up side view of the simpli fied centrifugal flow pump used for elementary analysis of the velocity vectors; V1, n and V2, n are defined as the average normal (radial) compo nents of velocity at radii r1 and r2, respectively.
V2, n V2, t V1, n V2, relative V2 ωr1 ωr2 ω r2 β2 β2 β2 β1 β1 r1 V1 V1, relative V2, relative FIGURE 14–36 Close-up frontal view of the simpli fied centrifugal flow pump used for elementary analysis of the velocity vectors. Absolute velocity vectors of the fluid are shown as bold arrows. It is assumed that the flow is everywhere tangent to the blade surface when viewed from a reference frame rotating with the blade, as indicated by the relative velocity vectors.
cen96537_ch14_793-884.indd 815 29/12/16 4:37 pm 816 TURBOMACHINERY enters the control volume with uniform absolute velocity V › 1 around the entire circumference at radius r1 and exits with uniform absolute velocity V › 2 around the entire circumference at radius r2. Since moment of momentum is defined as the cross product r → × V ›, only the tangential components of V › 1 and V › 2 are relevant to the shaft torque. These are shown as V1, t and V2, t in Fig. 14–37. It turns out that shaft torque is equal to the change in moment of momentum from inlet to outlet, as given by the Euler turbomachine equation (also called Euler’s turbine formula), derived in Chap. 6, Euler turbomachine equation: Tshaft = ρV .
(r2V2, t −r1V1, t) (14–14) Or, in terms of angles 𝛼1 and 𝛼2 and the magnitudes of the absolute velocity vectors, Alternative form, Euler turbomachine equation: Tshaft = ρV .
(r2V2 sin 𝛼2 −r1V1 sin 𝛼1) (14–15) In our simplified analysis there are no irreversible losses. Hence, pump efficiency 𝜂pump = 1, implying that water horsepower W .
water horsepower and brake horsepower bhp are the same. Using Eqs. 14–3 and 14–4, bhp = 𝜔 Tshaft = ρ𝜔 V .
(r2V2, t −r1V1, t) = W .
water horsepower = ρgV .
H (14–16) which is solved for net head H, Net head: H = 1 g (𝜔 r2V2, t −𝜔 r1V1, t) (14–17) EXAMPLE 14–5 Idealized Blower Performance A centrifugal blower rotates at n . = 1750 rpm (183.3 rad/s). Air enters the impeller normal to the blades (𝛼1 = 0°) and exits at an angle of 40° from radial (𝛼2 = 40°) as sketched in Fig. 14–38. The inlet radius is r1 = 4.0 cm, and the inlet blade width b1 = 5.2 cm. The outlet radius is r2 = 8.0 cm, and the outlet blade width b2 = 2.3 cm. The volume flow rate is 0.13 m3/s. For the idealized case, i.e., 100 percent efficiency, calculate the net head produced by this blower in equivalent millimeters of water column height. Also calculate the required brake horsepower in watts.
SOLUTION We are to calculate the brake horsepower and net head of an ideal ized blower at a given volume flow rate and rotation rate.
Assumptions 1 The flow is steady in the mean. 2 There are no leaks in the gaps between rotor blades and blower casing. 3 The air flow is incompressible. 4 The efficiency of the blower is 100 percent (no irreversible losses).
Properties We take the density of air to be 𝜌air = 1.20 kg/m3.
Analysis Since the volume flow rate (capacity) is given, we calculate the normal velocity components at the inlet using Eq. 14–12, V1, n = V .
2𝜋 r1b1 = 0.13 m3/s 2𝜋 (0.040 m)(0.052 m) = 9.947 m/s (1) V1 = V1, n, and V1, t = 0, since 𝛼1 = 0°. Similarly, V2, n = 11.24 m/s, and V2, t = V2, n tan 𝛼2 = (11.24 m/s) tan(40°) = 9.435 m/s (2) V1, n V2, n V1, t r2 r1 α2 α1 V2, t V2 V1 Control volume Shaft Tshaft = torque supplied to shaft ω FIGURE 14–37 Control volume (shaded) used for angular momentum analysis of a centrifugal pump; absolute tangential velocity components V1, t and V2, t are labeled.
V2, n r2 r1 α2 V2, t V2 Control volume ω V1 FIGURE 14–38 Control volume and absolute velocity vectors for the centrifugal blower of Example 14–5. The view is along the blower axis.
cen96537_ch14_793-884.indd 816 29/12/16 4:37 pm 817 CHAPTER 14 In order to design the shape of the impeller blades, we must use trigonom etry to obtain expressions for V1, t and V2, t in terms of blade angles 𝛽1 and 𝛽2. Applying the law of cosines (Fig. 14–39) to the triangle in Fig. 14–36 formed by absolute velocity vector V › 2, relative velocity vector V › 2, relative, and the tangential velocity of the blade at radius r2 (of magnitude 𝜔r2) we get V2 2 = V 2 2, relative + 𝜔 2r2 2 −2𝜔 r2V2, relative cos 𝛽2 (14–18) But we also see from Fig. 14–36 that V2, relative cos 𝛽2 = 𝜔 r2 −V2, t Substitution of this equation into Eq. 14–18 yields 𝜔 r2V2, t = 1 2 (V2 2 −V 2 2, relative + 𝜔 2r2 2) (14–19) A similar equation results for the blade inlet (change all subscripts 2 in Eq. 14–19 to subscript 1). Substitution of these into Eq. 14–17 yields Net head: H = 1 2g [(V2 2 −V1 2) + (𝜔 2r2 2 −𝜔 2r1 2) −(V 2 2, relative −V 2 1, relative)] (14–20) In words, Eq. 14–20 states that in the ideal case (no irreversible losses), the net head is proportional to the change in absolute kinetic energy plus the rotor-tip kinetic energy change minus the change in relative kinetic energy from inlet to outlet of the impeller. Finally, equating Eq. 14–20 and Eq. 14–2, Now we use Eq. 14–17 to predict the net head, H = 𝜔 g (r2V2, t −r1V1, t ) = 183.3 rad/s 9.81 m/s2 (0.080 m)(9.435 m/s) = 14.1 m (3) Note that the net head of Eq. 3 is in meters of air, the pumped fluid. To convert to pressure in units of equivalent millimeters of water column, we multiply by the ratio of air density to water density, Hwater column = H ρair ρwater = (14.1 m) 1.20 kg/m3 998 kg/m3 ( 1000 mm 1 m ) = 17.0 mm of water (4) Finally, we use Eq. 14–16 to predict the required brake horsepower, bhp = ρgV .
H = (1.20 kg/m3)(9.81 m/s2)(0.13 m3/s)(14.1 m)( W·s kg·m2/s2) = 21.6 W (5) Discussion Note the unit conversion in Eq. 5 from kilograms, meters, and seconds to watts; this conversion turns out to be useful in many turbomachinery calculations. The actual net head delivered to the air will be lower than that predicted by Eq. 3 due to inefficiencies. Similarly, actual brake horsepower will be higher than that predicted by Eq. 5 due to inefficiencies in the blower, friction on the shaft, etc.
0 ⏟ Law of Cosines c2 = a2 + b2 – 2ab cos C A B C a b c FIGURE 14–39 The law of cosines is utilized in the analysis of a centrifugal pump.
cen96537_ch14_793-884.indd 817 29/12/16 4:37 pm 818 TURBOMACHINERY where we set subscript 2 as the outflow and subscript 1 as the inflow, we see that ( P ρg + V 2 relative 2g −𝜔 2r2 2g + z) out = ( P ρg + V 2 relative 2g −𝜔 2r2 2g + z) in (14–21) Note that we are not limited to analysis of only the inlet and outlet. In fact, we may apply Eq. 14–21 to any two radii along the impeller. In general then, we write an equation that is commonly called the Bernoulli equation in a rotating reference frame: P ρg + V 2 relative 2g −𝜔 2r2 2g + z = constant (14–22) We see that Eq. 14–22 is the same as the usual Bernoulli equation, except that since the speed used is the relative speed (in the rotating reference frame), an “extra” term (the third term on the left of Eq. 14–22) appears in the equation to account for rotational effects (Fig. 14–40). We emphasize that Eq. 14–22 is an approximation, valid only for the ideal case in which there are no irrevers ible losses through the impeller. Nevertheless, it is valuable as a first-order approximation for flow through the impeller of a centrifugal pump. We now examine Eq. 14–17, the equation for net head, more closely. Since the term containing V1, t carries a negative sign, we obtain the maxi mum H by setting V1, t to zero. (We are assuming that there is no mechanism in the eye of the pump that can generate a negative value of V1, t.) Thus, a first-order approximation for the design condition of the pump is to set V1, t = 0. In other words, we select the blade inlet angle 𝛽1 such that the flow into the impeller blade is purely radial from an absolute reference frame, and V1, n = V1. The velocity vectors at r = r1 in Fig. 14–36 are magnified and redrawn in Fig. 14–41. Using some trigonometry we see that V1, t = 𝜔 r1 − V1, n tan 𝛽1 (14–23) A similar expression is obtained for V2, t (replace subscript 1 by 2), or in fact for any radius between r1 and r2. When V1, t = 0 and V1, n = V1, 𝜔 r1 = V1, n tan 𝛽1 (14–24) Finally, combining Eq. 14–24 with Eq. 14–12, we have an expression for volume flow rate as a function of inlet blade angle 𝛽1 and rotational speed, V .
= 2𝜋 b1𝜔 r1 2 tan 𝛽1 (14–25) Equation 14–25 can be used for preliminary design of the impeller blade shape as illustrated by Example 14–6.
Vrelative V Absolute Rotating ω r r ω FIGURE 14–40 For the approximation of flow through an impeller with no irreversible losses, it is often more convenient to work with a relative frame of reference rotating with the impeller; in that case, the Bernoulli equation gets an addi tional term, as indicated in Eq. 14–22.
V1, t V1, n V1, relative V1 ωr1 ω β1 β1 r1 FIGURE 14–41 Close-up frontal view of the veloc ity vectors at the impeller blade inlet. The absolute velocity vector is shown as a bold arrow.
EXAMPLE 14–6 Preliminary Design of a Centrifugal Pump A centrifugal pump is being designed to pump liquid refrigerant R-134a at room temperature and atmospheric pressure. The impeller inlet and outlet radii are r1 = 100 and r2 = 180 mm, respectively (Fig. 14–42). The impeller inlet and outlet cen96537_ch14_793-884.indd 818 29/12/16 4:37 pm 819 CHAPTER 14 widths are b1 = 50 and b2 = 30 mm (into the page of Fig. 14–42). The pump is to deliver 0.25 m3/s of the liquid at a net head of 14.5 m when the impeller rotates at 1720 rpm. Design the blade shape for the case in which these operating conditions are the design conditions of the pump (V1, t = 0, as sketched in the figure); specifi cally, calculate angles 𝛽1 and 𝛽2, and discuss the shape of the blade. Also predict the horsepower required by the pump.
SOLUTION For a given flow rate, net head, and dimensions of a centrifugal pump, we are to design the blade shape (leading and trailing edge angles). We are also to estimate the horsepower required by the pump.
Assumptions 1 The flow is steady. 2 The liquid is incompressible. 3 There are no irreversible losses through the impeller. 4 This is only a preliminary design.
Properties For refrigerant R-134a at T = 20°C, vf = 0.0008157 m3/kg. Thus 𝜌 = 1/vf = 1226 kg/m3.
Analysis We calculate the required water horsepower from Eq. 14–3, W .
water horsepower = ρgV .
H = (1226 kg/m3)(9.81 m/s2)(0.25 m3/s)(14.5 m)( W·s kg·m2/s2) = 43,600 W The required brake horsepower will be greater than this in a real pump. However, in keeping with the approximations for this preliminary design, we assume 100 percent efficiency such that bhp is approximately equal to W .
water horsepower, bhp ≅W .
water horsepower = 43,600 W( hp 745.7 W) = 58.5 hp We report the final result to two significant digits in keeping with the precision of the given quantities; thus, bhp ≈ 59 horsepower. In all calculations with rotation, we need to convert the rotational speed from n . (rpm) to 𝜔 (rad/s), as illustrated in Fig. 14–43, 𝜔 = 1720 rot min ( 2𝜋 rad rot ) ( 1 min 60 s ) = 180.1 rad/s (1) We calculate the blade inlet angle using Eq. 14–25, 𝛽1 = arctan ( V .
2𝜋 b1𝜔 r1 2)= arctan ( 0.25 m3/s 2𝜋 (0.050 m)(180.1 rad/s)(0.10 m)2)= 23.8° We find 𝛽2 by utilizing the equations derived earlier for our elementary analysis. First, for the design condition in which V1, t = 0, Eq. 14–17 reduces to Net head: H = 1 g (𝜔 r2V2, t −𝜔 r1V1, t ) = 𝜔 r2V2, t g 0 from which we calculate the tangential velocity component, V2, t = gH 𝜔 r2 (2) ⏟ V2, n V2, relative V1 ωr1 ωr2 ω r2 β2 β2 β1 β1 r1 V1, relative V2 FIGURE 14–42 Relative and absolute velocity vectors and geometry for the centrifugal pump impeller design of Example 14–6.
CAUTION Always convert rotation rate from rpm to radians per second.
FIGURE 14–43 Proper unit conversion requires the units of rotation rate to be rad/s.
cen96537_ch14_793-884.indd 819 29/12/16 4:38 pm 820 TURBOMACHINERY Using Eq. 14–12, we calculate the normal velocity component, V2, n = V .
2𝜋 r2b2 (3) Next, we perform the same trigonometry used to derive Eq. 14–23, but on the trail ing edge of the blade rather than the leading edge. The result is V2, t = 𝜔 r2 − V2, n tan 𝛽2 from which we finally solve for 𝛽2, 𝛽2 = arctan ( V2, n 𝜔 r2 −V2, t) (4) After substitution of Eqs. 2 and 3 into Eq. 4, and insertion of the numerical values, we obtain 𝛽2 = 14.7° We report the final results to only two significant digits. Thus our preliminary design requires backward-inclined impeller blades with 𝛽1 ≅ 24° and 𝛽2 ≅ 15°. Once we know the leading and trailing edge blade angles, we design the detailed shape of the impeller blade by smoothly varying blade angle 𝛽 from 𝛽1 to 𝛽2 as radius increases from r1 to r2. As sketched in Fig. 14–44, the blade can be of various shapes while still keeping 𝛽1 ≅ 24° and 𝛽2 ≅ 15°, depending on how we vary 𝛽 with the radius. In the figure, all three blades begin at the same location (zero absolute angle) at radius r1; the leading edge angle for all three blades is 𝛽1 = 24°. The medium length blade (the brown one in Fig. 14–44) is constructed by varying 𝛽 linearly with r. Its trailing edge intercepts radius r2 at an absolute angle of approximately 93°. The longer blade (the black one in the figure) is con structed by varying 𝛽 more rapidly near r1 than near r2. In other words, the blade curvature is more pronounced near its leading edge than near its trailing edge. It intercepts the outer radius at an absolute angle of about 114°. Finally, the shortest blade (the blue blade in Fig. 14–44) has less blade curvature near its leading edge, but more pronounced curvature near its trailing edge. It intercepts r2 at an abso lute angle of approximately 77°. It is not immediately obvious which blade shape is best.
Discussion Keep in mind that this is a preliminary design in which irrevers ible losses are ignored. A real pump would have losses, and the required brake horsepower would be higher (perhaps 20 to 30 percent higher) than the value estimated here. In a real pump with losses, a shorter blade has less skin fric tion drag, but the normal stresses on the blade are larger because the flow is turned more sharply near the trailing edge where the velocities are largest; this may lead to structural problems if the blades are not very thick, especially when pumping dense liquids. A longer blade has higher skin friction drag, but lower normal stresses. In addition, you can see from a simple blade volume estimate in Fig. 14–44 that for the same number of blades, the longer the blades, the more flow blockage, since the blades are of finite thickness. In addition, the dis placement thickness effect of boundary layers growing along the blade surfaces (Chap. 10) leads to even more pronounced blockage for the long blades. Obvi ously some engineering optimization is required to determine the exact shape of the blade.
r1 r2 β2 β1 ω β2 β2 FIGURE 14–44 Three possible blade shapes for the centrifugal pump impeller design of Example 14–6. All three blades have leading edge angle 𝛽1 = 24° and trailing edge angle 𝛽2 = 15°, but differ in how 𝛽 is varied with the radius. The drawing is to scale.
cen96537_ch14_793-884.indd 820 29/12/16 4:38 pm 821 CHAPTER 14 How many blades should we use in an impeller? If we use too few blades, circulatory flow loss will be high. Circulatory flow loss occurs because there is a finite number of blades. Recall that in our preliminary analysis, we assume a uniform tangential velocity V2, t around the entire circumfer ence of the outlet of the control volume (Fig. 14–37). This is strictly cor rect only if we have an infinite number of infinitesimally thin blades. In a real pump, of course, the number of blades is finite, and the blades are not infinitesimally thin. As a result, the tangential component of the absolute velocity vector is not uniform, but drops off in the spaces between blades as illustrated in Fig. 14–45a. The net result is an effectively smaller value of V2, t, which in turn decreases the actual net head. This loss of net head (and pump efficiency) is called circulatory flow loss. On the other hand, if we have too many blades (as in Fig. 14–45b) there will be excessive flow blockage losses and losses due to the growing boundary layers, again lead ing to nonuniform flow speeds at the outer radius of the pump and lower net head and efficiency. These losses are called passage losses. The bottom line is that some engineering optimization is necessary in order to choose both the blade shape and number of blades. Such analysis is beyond the scope of the present text. A quick perusal through the turbomachinery literature shows that 11, 14, and 16 are common numbers of rotor blades for medium-sized centrifugal pumps. Once we have designed the pump for specified net head and flow rate (design conditions), we can estimate its net head at conditions away from design conditions. In other words, keeping b1, b2, r1, r2, 𝛽1, 𝛽2, and 𝜔 fixed, we vary the volume flow rate above and below the design flow rate. We have all the equations: Eq. 14–17 for net head H in terms of absolute tangential velocity components V1, t and V2, t, Eq. 14–23 for V1, t and V2, t as functions of absolute normal velocity components V1, n and V2, n, and Eq. 14–12 for V1, n and V2, n as functions of volume flow rate V .
. In Fig. 14–46 we combine these equations to generate a plot of H versus V . for the pump designed in Example 14–6. The solid blue line is the predicted performance, based on our preliminary analysis. The predicted performance curve is nearly linear with V . both above and below design conditions since the term 𝜔r1V1, t in Eq. 14–17 is small compared to the term 𝜔r2V2, t. Recall that at the pre dicted design conditions, we had set V1, t = 0. For volume flow rates higher than this, V1, t is predicted by Eq. 14–23 to be negative. In keeping with our previous assumptions, however, it is not possible to have negative values of V1, t. Thus, the slope of the predicted performance curve changes suddenly beyond the design conditions. Also sketched in Fig. 14–46 is the actual performance of this centrifugal pump. While the predicted performance is close to the actual performance at design conditions, the two curves deviate substantially away from design conditions. At all volume flow rates, the actual net head is lower than the predicted net head. This is due to irreversible effects such as friction along blade surfaces, leakage of fluid between the blades and the casing, prerota tion (swirl) of fluid in the region of the eye, flow separation on the leading edges of the blades (shock losses) or in the expanding portions of the flow passages, circulatory flow loss, passage loss, and irreversible dissipation of swirling eddies in the volute, among other things.
(b) ω (a) V2, t ω FIGURE 14–45 (a) A centrifugal pump impeller with too few blades leads to excessive circulatory flow loss—the tangential velocity at outer radius r2 is smaller in the gaps between blades than at the trailing edges of the blades (absolute tangential velocity vectors are shown). (b) On the other hand, since real im peller blades have finite thickness, an impeller with too many blades leads to passage losses due to excessive flow blockage and large skin friction drag (velocity vectors in a frame of reference rotating with the impeller are shown exiting one blade row). The bottom line is that pump engineers must optimize both blade shape and number of blades.
cen96537_ch14_793-884.indd 821 29/12/16 4:38 pm 822 TURBOMACHINERY Axial Pumps Axial pumps do not utilize so-called centrifugal forces. Instead, the impeller blades behave more like the wing of an airplane (Fig. 14–47), producing lift by changing the momentum of the fluid as they rotate. The rotor of a heli copter, for example, is a type of axial-flow pump (Fig. 14–48). The lift force on the blade is caused by pressure differences between the top and bottom surfaces of the blade, and the change in flow direction leads to downwash (a column of descending air) through the rotor plane. From a time-averaged perspective, there is a pressure jump across the rotor plane that induces a downward airflow (Fig. 14 –48). Imagine turning the rotor plane vertically; we now have a propeller (Fig. 14–49a). Both the helicopter rotor and the airplane propeller are exam ples of open axial-flow fans, since there is no duct or casing around the tips of the blades. The common window fan you install in your bedroom window in the summer operates under the same principles, but the goal is to blow air rather than to provide a force. Be assured, however, that there is a net force acting on the fan housing. If air is blown from left to right, the force on the fan acts to the left, and the fan is held down by the window sash. The casing around the house fan also acts as a short duct, which helps to direct the flow and eliminate some losses at the blade tips. The small cooling fan inside your computer is typically an axial-flow fan; it looks like a miniature window fan (Fig. 14–49b) and is an example of a ducted axial-flow fan. If you look closely at the airplane propeller blade in Fig. 14–49a, the rotor blade of a helicopter, the propeller blade of a radio-controlled model airplane, or even the blade of a well-designed window fan, you will notice some twist in the blade. Specifically, the airfoil at a cross section near the hub or root of the blade is at a higher pitch angle (𝜃) than the airfoil at a cross section near the tip, 𝜃root > 𝜃tip (Fig. 14–50). This is because the tangential speed of the blade increases linearly with radius, u𝜃= 𝜔 r (14–26) At a given radius then, the velocity V › relative of the air relative to the blade is estimated to first order as the vector sum of inlet velocity V › in and the nega tive of blade velocity V › blade, V › relative ≅ V › in − V › blade (14–27) where the magnitude of V › blade is equal to the tangential blade speed u𝜃, as given by Eq. 14–26. The direction of V › blade is tangential to the rotational path of the blade. At the blade position sketched in Fig. 14–50, V › blade is to the left. In Fig. 14–51 we compute V › relative graphically using Eq. 14–27 at two radii—the root radius and the tip radius of the rotor blade sketched in Fig. 14–50. As you can see, the relative angle of attack 𝛼 is the same in either case. In fact, the amount of twist is determined by setting pitch angle 𝜃 such that 𝛼 is the same at any radius. Note also that the magnitude of the relative velocity V › relative increases from the root to the tip. It follows that the dynamic pressure encountered by 30 25 20 15 H, m 10 5 0 0.2 0.25 , m3/s 0.3 Actual performance Predicted performance Irreversible losses Design conditions V ∙ FIGURE 14–46 Net head as a function of volume flow rate for the pump of Example 14-6. The difference between predicted and actual performance is due to unaccounted irreversibilities in the prediction.
FL Streamlines FIGURE 14–47 The blades of an axial-flow pump be have like the wing of an airplane. The air is turned downward by the wing as it generates lift force FL.
Lift Downwash Low P High P FIGURE 14–48 Downwash and pressure rise across the rotor plane of a helicopter, which is a type of axial-flow pump.
cen96537_ch14_793-884.indd 822 29/12/16 4:38 pm 823 CHAPTER 14 FIGURE 14–49 Axial-flow fans may be open or ducted: (a) a propeller is an open fan, and (b) a computer cooling fan is a ducted fan.
Photos by John M. Cimbala.
(a) (a) (b) (b) Hub Inlet direction ω Vin Root uθ = ωr θtip θroot Tip FIGURE 14–50 A well-designed rotor blade or propeller blade has twist, as shown by the blue cross-sectional slices through one of the three blades; blade pitch angle 𝜃 is higher at the root than at the tip because the tangential speed of the blade increases with radius.
FIGURE 14–51 Graphical computation of vector V › relative at two radii: (a) root, and (b) tip of the rotor blade sketched in Fig. 14–50.
Vrelative Vblade Vrelative Vin θroot θtip Vblade α α (a) (b) Vin cross sections of the blade increases with radius, and the lift force per unit width into the page in Fig. 14–51 also increases with radius. Propellers tend to be narrower at the root and wider toward the tip in order to take advan tage of the larger lift contribution available toward the tip. At the very tip, however, the blade is usually rounded off to avoid excessive induced drag (Chap. 11) that would exist if the blade were simply chopped off abruptly as in Fig. 14–50. Equation 14–27 is not exact for several reasons. First, the rotating motion of the rotor introduces some swirl to the airflow (Fig. 14–52). This reduces the effective tangential speed of the blade relative to the incoming air. Sec ond, since the hub of the rotor is of finite size, the air accelerates around it, causing the air speed to increase locally at cross sections of the blade close to the root. Third, the axis of the rotor or propeller may not be aligned exactly parallel to the incoming air. Finally, the air speed itself is not easily determined because it turns out that the air accelerates as it approaches the cen96537_ch14_793-884.indd 823 29/12/16 4:38 pm 824 TURBOMACHINERY whirling rotor. There are methods available to approximate these and other secondary effects, but they are beyond the scope of the present text. The first-order approximation given by Eq. 14–27 is adequate for preliminary rotor and propeller design, as illustrated in Example 14–7.
EXAMPLE 14–7 Calculation of Twist in an Airplane Propeller Suppose you are designing the propeller of a radio-controlled model airplane. The overall diameter of the propeller is 34.0 cm, and the hub assembly diameter is 5.5 cm (Fig. 14–53). The propeller rotates at 1700 rpm, and the airfoil chosen for the propeller cross section achieves its maximum efficiency at an angle of attack of 14°. When the airplane flies at 30 mi/h (13.4 m/s), calculate the blade pitch angle from the root to the tip of the blade such that 𝛼 = 14° everywhere along the propeller blade.
SOLUTION We are to calculate blade pitch angle 𝜃 from the root to the tip of the propeller such that the angle of attack is 𝛼 = 14° at every radius along the propeller blade.
Assumptions 1 The air at these low speeds is incompressible. 2 We neglect the secondary effects of swirl and acceleration of the air as it approaches the propeller; i.e., the magnitude of V → in is approximated to be equal to the speed of the aircraft. 3 The airplane flies level, such that the propeller axis is parallel to the incoming air velocity.
Analysis The velocity of the air relative to the blade is approximated to first order at any radius by using Eq. 14–27. A sketch of the velocity vectors at some arbitrary radius r is shown in Fig. 14–54. From the geometry we see that Pitch angle at arbitrary radius r: 𝜃= 𝛼+ 𝜙 (1) and 𝜙= arctan ∣V › in∣ ∣V › blade∣ = arctan ∣V › in∣ 𝜔 r (2) where we have also used Eq. 14–26 for the blade speed at radius r. At the root (r = Dhub/2 = 2.75 cm), Eq. 2 becomes 𝜃= 𝛼+ 𝜙= 14° + arctan [ 13.4 m/s (1700 rot/min)(0.0275 m)( 1 rot 2𝜋 rad) ( 60 s min)]= 83.9° Similarly, the pitch angle at the tip (r = Dpropeller/2 = 17.0 cm) is 𝜃= 𝛼+ 𝜙= 14° + arctan [ 13.4 m/s (1700 rot/min)(0.17 m)( 1 rot 2𝜋 rad) ( 60 s min)] = 37.9° At radii between the root and the tip, Eqs. 1 and 2 are used to calculate 𝜃 as a function of r. Results are plotted in Fig. 14–55.
Discussion The pitch angle is not linear because of the arctangent function in Eq. 2.
ω FIGURE 14–52 The rotating blades of a rotor or propeller induce swirl in the surrounding fluid.
ω Vblade Dhub Dpropeller Airplane nose Vin FIGURE 14–53 Setup for the design of the model airplane propeller of Example 14–7, not to scale.
α θ ϕ Vrelative Vin –Vblade FIGURE 14–54 Velocity vectors at some arbitrary radius r of the propeller of Example 14–7.
cen96537_ch14_793-884.indd 824 29/12/16 4:38 pm 825 CHAPTER 14 Airplane propellers have variable pitch, meaning that the pitch of the entire blade can be adjusted by rotating the blades through mechanical linkages in the hub. For example, when a propeller-driven airplane is sit ting at the airport, warming up its engines at high rpm, why does it not start moving? Well, for one thing, the brakes are being applied. But more importantly, propeller pitch is adjusted so that the average angle of attack of the airfoil cross sections is nearly zero—little or no net thrust is pro vided. While the airplane taxies to the runway, the pitch is adjusted so as to produce a small amount of thrust. As the plane takes off, the engine rpm is high, and the blade pitch is adjusted such that the propeller delivers maximum thrust. In many cases the pitch can even be adjusted “backward” (negative angle of attack) to provide reverse thrust to slow down the airplane after landing. We plot qualitative performance curves for a typical propeller fan in Fig. 14–56. Unlike centrifugal fans, brake horsepower tends to decrease with flow rate. In addition, the efficiency curve leans more to the right com pared to that of centrifugal fans (see Fig. 14–8). The result is that efficiency drops off rapidly for volume flow rates higher than that at the best effi ciency point. The net head curve also decreases continuously with flow rate (although there are some wiggles), and its shape is much different than that of a centrifugal flow fan. If the head requirements are not severe, propeller fans can be operated beyond the point of maximum efficiency to achieve higher volume flow rates. Since bhp decreases at high values of V .
, there is not a power penalty when the fan is run at high flow rates. For this reason it is tempting to install a slightly undersized fan and push it beyond its best efficiency point. At the other extreme, if operated below its maximum effi ciency point, the flow may be noisy and unstable, which indicates that the fan may be oversized (larger than necessary). For these reasons, it is usually best to run a propeller fan at, or slightly above, its maximum efficiency point. When used to move flow in a duct, a single-impeller axial-flow fan is called a tube-axial fan (Fig. 14–57a). In many practical engineering appli cations of axial-flow fans, such as exhaust fans in kitchens, building venti lation duct fans, fume hood fans, and automotive radiator cooling fans, the swirling flow produced by the rotating blades (Fig. 14–57a) is of no con cern. But the swirling motion and increased turbulence intensity can con tinue for quite some distance downstream, and there are applications where swirl (or its affiliated noise and turbulence) is highly undesirable. Examples include wind tunnel fans, torpedo fans, and some specialized mine shaft ventilation fans. There are two basic designs that largely eliminate swirl: A second rotor that rotates in the opposite direction can be added in series with the existing rotor to form a pair of counter-rotating rotor blades; such a fan is called a counter-rotating axial-flow fan (Fig. 14–57b). The swirl caused by the upstream rotor is cancelled by an opposite swirl caused by the downstream rotor. Alternatively, a set of stator blades can be added either upstream or downstream of the rotating impeller. As implied by their name, stator blades are stationary (nonrotating) guide vanes that simply redirect the fluid. An axial-flow fan with a set of rotor blades (the impeller or the rotor) and a set of stator blades called vanes (the stator) is called a vane-axial fan (Fig. 14–57c). The stator blade design of the vane-axial fan is 90 60 50 40 30 0 5 10 15 r, cm 20 70 80 θ, degrees Hub Tip FIGURE 14–55 Blade pitch angle as a function of radius for the propeller of Example 14–7.
0 0 ηfan H, ηfan, or bhp bhp H V ∙ FIGURE 14–56 Typical fan performance curves for a propeller (axial-flow) fan.
cen96537_ch14_793-884.indd 825 29/12/16 4:38 pm 826 TURBOMACHINERY much simpler and less expensive to implement than is the counter-rotating axial-flow fan design. The swirling fluid downstream of a tube-axial fan wastes kinetic energy and has a high level of turbulence; the vane-axial fan partially recovers this wasted kinetic energy and reduces the level of turbulence. Vane-axial fans are thus both quieter and more energy efficient than tube-axial fans. A prop erly designed counter-rotating axial-flow fan may be even quieter and more energy efficient. Furthermore, since there are two sets of rotating blades, a higher pressure rise can be obtained with the counter-rotating design. The construction of a counter-rotating axial-flow fan is more complex, of course, requiring either two synchronized motors or a gear box. Axial-flow fans can be either belt driven or direct drive. The motor of a direct-drive vane-axial fan is mounted in the middle of the duct. It is com mon practice (and good design) to use the stator blades to provide physi cal support for the motor. Photographs of a belt-driven tube-axial fan and a direct-drive vane-axial fan are provided in Fig. 14–58. The stator blades of the vane-axial fan can be seen behind (downstream of) the rotor blades in Fig. 14–58b. An alternative design is to place the stator blades upstream of the impeller, imparting preswirl to the fluid. The swirl caused by the rotat ing impeller blades then removes this preswirl. It is fairly straightforward to design the shape of the blades in all these axial-flow fan designs, at least to first order. For simplicity, we assume thin blades (e.g., blades made out of sheet metal) rather than airfoil-shaped blades. Consider, for example, a vane-axial flow fan with rotor blades upstream of stator blades (Fig. 14–59). The distance between the rotor and stator has been exaggerated in this figure to enable velocity vectors to be drawn between the blades. The hub radius of the stator is assumed to be the same as the hub radius of the rotor so that the cross-sectional area of flow remains constant. As we did previously with the propeller, we consider the cross section of one impeller blade as it passes vertically in front of us. Since there are multiple blades, the next blade passes by shortly thereafter. At a chosen radius r, we make the two-dimensional approximation that the Impeller Hub (a) (b) Motor ω Impeller 1 Impeller 2 Gear box (c) Hub Motor ω ω Impeller Stator Hub Motor ω v FIGURE 14–57 A tube-axial fan (a) imparts swirl to the exiting fluid, while (b) a counter-rotating axial-flow fan and (c) a vane-axial fan are designed to remove the swirl.
(a) FIGURE 14–58 Axial-flow fans: (a) a belt-driven tube-axial fan without stator blades, and (b) a direct-drive vane-axial fan with stator blades to reduce swirl and improve efficiency.
(a) © Penn Barry 2012. Used by permission. (b) © Howden Group Limited 2016.
Used with Permission.
(b) cen96537_ch14_793-884.indd 826 29/12/16 4:38 pm 827 CHAPTER 14 blades pass by as an infinite series of two-dimensional blades called a blade row or cascade. A similar assumption is made for the stator blades, even though they are stationary. Both blade rows are sketched in Fig. 14–59. In Fig. 14–59b, the velocity vectors are seen from an absolute reference frame, i.e., that of a fixed observer looking horizontally at the vane-axial flow fan. Flow enters from the left at speed Vin in the horizontal (axial) direction. The rotor blade row moves at constant speed 𝜔r vertically upward in this reference frame, as indicated. Flow is turned by these mov ing blades and leaves the trailing edge upward and to the right as indicated in Fig. 14–59b as vector V › rt. (The subscript notation indicates rotor trailing edge.) To find the magnitude and direction of V › rt, we redraw the blade rows and vectors in a relative reference frame (the frame of reference of the rotat ing rotor blade) in Fig. 14–59c. This reference frame is obtained by sub tracting the rotor blade velocity (adding a vector of magnitude 𝜔r pointing vertically downward) from all velocity vectors. As shown in Fig. 14–59c, the velocity vector relative to the leading edge of the rotor blade is V › in, relative, calculated as the vector sum of V › in and the downward vector of magnitude 𝜔r. We adjust the pitch of the rotor blade such that V › in, relative is parallel (tan gential) to the leading edge of the rotor blade at this cross section. Flow is turned by the rotor blade. We assume that the flow leaving the rotor blade is parallel to the blade’s trailing edge (from the relative refer ence frame), as sketched in Fig. 14–59c as vector V › rt, relative. We also know that the horizontal (axial) component of V › rt, relative must equal V › in in order to conserve mass. Note that we are assuming incompressible flow and constant flow area normal to the page in Fig. 14–59. Thus, the axial component of velocity must be everywhere equal to Vin. This piece of information estab lishes the magnitude of vector V › rt, relative, which is not the same as the magni tude of V › in, relative. Returning to the absolute reference frame of Fig. 14–59b, absolute velocity V › rt is calculated as the vector sum of V › rt, relative and the vertically upward vector of magnitude 𝜔r. Finally, the stator blade is designed such that V › rt is parallel to the leading edge of the stator blade. The flow is once again turned, this time by the sta tor blade. Its trailing edge is horizontal so that the flow leaves axially (with out any swirl). The final outflow velocity must be identical to the inflow velocity by conservation of mass if we assume incompressible flow and constant flow area normal to the page. In other words, V › out = V › in. For com pleteness, the outflow velocity in the relative reference frame is sketched in Fig. 14–59c. We also see that V › out, relative = V › in, relative. Now imagine repeating this analysis for all radii from the hub to the tip. As with the propeller, we would design our blades with some twist since the value of 𝜔r increases with radius. A modest improvement in efficiency can be gained at design conditions by using airfoils instead of sheet metal blades; the improvement is more significant at off-design conditions. If there are, say, seven rotor blades in a vane-axial fan, how many sta tor blades should there be? You might at first say seven so that the stator matches the rotor—but this would be a very poor design! Why? Because at the instant in time when one blade of the rotor passes directly in front of a stator blade, all six of its brothers would do the same. Each stator blade would simultaneously encounter the disturbed flow in the wake of a rotor blade. The resulting flow would be both pulsating and noisy, and the entire Rotor blade row (a) (c) (b) ω Stator blade row Vin r ωr ωr ωr ωr ωr Vin Vin Vin Vrt Vrt, relative Vrt, relative Vin, relative Vout, relative Vout Vout Vout FIGURE 14–59 Analysis of a vane-axial flow fan at ra dius r using the two-dimensional blade row approximation; (a) overall view, (b) absolute reference frame, and (c) reference frame relative to the rotating rotor blades (impeller).
cen96537_ch14_793-884.indd 827 29/12/16 4:38 pm 828 TURBOMACHINERY unit would vibrate severely. Instead, it is good design practice to choose the number of stator blades such that it has no common denominator with the number of rotor blades. Combinations like seven and eight, seven and nine, six and seven, or nine and eleven are good choices. Combinations like eight and ten (common denominator of two) or nine and twelve (common denominator of three) are not good choices. We plot the performance curves of a typical vane-axial flow fan in Fig. 14–60. The general shapes are very similar to those of a propeller fan (Fig. 14–56), and you are referred to the discussion there. After all, a vane-axial flow fan is really the same as a propeller fan or tube-axial flow fan except for the additional stator blades that straighten the flow and tend to smooth out the performance curves. As discussed previously, an axial-flow fan delivers high volume flow rate, but fairly low pressure rise. Some applications require both high flow rate and high pressure rise. In such cases, several stator–rotor pairs can be combined in series, typically with a common shaft and common hub (Fig. 14–61). When two or more rotor–stator pairs are combined like this we call it a multistage axial-flow pump. A blade row analysis similar to the one of Fig. 14–59 is applied to each successive stage. The details of the analysis can get complicated, however, because of compressibility effects and because the flow area from the hub to the tip may not remain constant. In a mul tistage axial-flow compressor, for example, the flow area decreases down stream. The blades of each successive stage get smaller as the air gets fur ther compressed. In a multistage axial-flow turbine, the flow area typically grows downstream as pressure is lost in each successive stage of the turbine. One well-known example of a turbomachine that utilizes both multistage axial-flow compressors and multistage axial-flow turbines is the turbofan engine used to power modern commercial airplanes. A cutaway schematic diagram of a turbofan engine is shown in Fig. 14–62. Some of the air passes Rotor 1 Rotor 2 Stator 1 ω Stator 2 Rotating hub Shaft FIGURE 14–61 A multistage axial-flow pump consists of two or more rotor–stator pairs.
Fan Bypass air Combustion chamber Low pressure turbine Exhaust High pressure turbine High pressure compressor Low pressure compressor FIGURE 14–62 Pratt & Whitney PW4000 turbofan engine; an example of a multistage axial-flow turbomachine.
Photo courtesy of United Technologies Corporation/Pratt & Whitney. Used by permission. All rights reserved.
0 0 bhp H ηfan H, ηfan, or bhp V ∙ FIGURE 14–60 Typical fan performance curves for a vane-axial flow fan.
cen96537_ch14_793-884.indd 828 29/12/16 4:38 pm 829 CHAPTER 14 through the fan, which delivers thrust much like a propeller. The rest of the air passes through a low-pressure compressor, a high-pressure compressor, a combustion chamber, a high-pressure turbine, and then finally a low-pressure turbine. The air and products of combustion are then exhausted at high speed to provide even more thrust. Computational fluid dynamics (CFD) codes are obviously quite useful in the design of such complex turboma chines (Chap. 15).
EXAMPLE 14–8 Design of a Vane-Axial Flow Fan for a Wind Tunnel A vane-axial flow fan is being designed to power a wind tunnel. There must not be any swirl in the flow downstream of the fan. It is decided that the stator blades should be upstream of the rotor blades (Fig. 14–63) to protect the impeller blades from damage by objects that might accidentally get blown into the fan. To reduce expenses, both the stator and rotor blades are to be constructed of sheet metal. The leading edge of each stator blade is aligned axially (𝛽sl = 0.0°) and its trailing edge is at angle 𝛽st = 60.0° from the axis as shown in the sketch. (The subscript notation “sl” indicates stator leading edge and “st” indicates stator trailing edge.) There are 16 stator blades. At design conditions, the axial-flow speed through the blades is 47.1 m/s, and the impeller rotates at 1750 rpm. At radius r = 0.40 m, calculate the leading and trailing edge angles of the rotor blade, and sketch the shape of the blade. How many rotor blades should there be?
SOLUTION For given flow conditions and stator blade shape at a given radius, we are to design the rotor blade. Specifically, we are to calculate the leading and trailing edge angles of the rotor blade and sketch its shape. We are also to decide how many rotor blades to construct.
Assumptions 1 The air is nearly incompressible. 2 The flow area between the hub and tip is constant. 3 Two-dimensional blade row analysis is appropriate.
Analysis First we analyze flow through the stator from an absolute reference frame, using the two-dimensional approximation of a cascade (blade row) of stator blades (Fig. 14–64). Flow enters axially (horizontally) and is turned 60.0° downward. Since the axial component of velocity must remain constant to conserve mass, the magnitude of the velocity leaving the trailing edge of the stator, V → st, is calculated as Vst = Vin cos 𝛽st = 47.1 m/s cos (60.0°) = 94.2 m/s (1) The direction of V → st is assumed to be that of the stator trailing edge. In other words, we assume that the flow turns nicely through the blade row and exits parallel to the trailing edge of the blade, as shown in Fig. 14–64. We convert V → st to the relative reference frame moving with the rotor blades. At a radius of 0.40 m, the tangential velocity of the rotor blades is u𝜃= 𝜔 r = (1750 rot/min)( 2𝜋 rad rot ) ( 1 min 60 s )(0.40 m) = 73.30 m/s (2) Rotor ω Stator Vin βst Hub and motor r ωr ? ?
Vout FIGURE 14–63 Schematic diagram of the vane-axial flow fan of Example 14–8. The stator precedes the rotor, and the shape of the rotor blade is unknown—it is to be designed.
Vin Stator blade row βst βst Vin Vst FIGURE 14–64 Velocity vector analysis of the stator blade row of the vane-axial flow fan of Example 14–8; absolute reference frame.
cen96537_ch14_793-884.indd 829 29/12/16 4:38 pm 830 TURBOMACHINERY 14–3 ■ PUMP SCALING LAWS Dimensional Analysis Turbomachinery provides a very practical example of the power and useful ness of dimensional analysis (Chap. 7). We apply the method of repeating variables to the relationship between gravity times net head (gH) and pump properties such as volume flow rate (V .
); some characteristic length, typi cally the diameter of the impeller blades (D); blade surface roughness height (𝜀); and impeller rotational speed (𝜔), along with fluid properties density (𝜌) and viscosity (𝜇). Note that we treat the group gH as one variable. Since the rotor blade row moves upward in Fig. 14–63, we add a downward veloc ity with magnitude given by Eq. 2 to translate V → st into the rotating reference frame sketched in Fig. 14–65. The angle of the leading edge of the rotor, 𝛽rl, is calculated by using trigonometry, 𝛽rl = arctan 𝜔 r + Vin tan 𝛽st Vin = arctan (73.30 m/s) + (47.1 m/s) tan (60.0°) 47.1 m/s = 73.09° (3) The air must now be turned by the rotor blade row in such a way that it leaves the trailing edge of the rotor blade at a zero angle (axially, no swirl) from an absolute reference frame. This determines the rotor’s trailing edge angle, 𝛽rt. Specifically, when we add an upward velocity of magnitude 𝜔r (Eq. 2) to the rel ative velocity exiting the trailing edge of the rotor, V → strt, relative, we convert back to the absolute reference frame, and obtain V → rt, the velocity leaving the rotor trailing edge. It is this velocity, V → rt, that must be axial (horizontal). Furthermore, to conserve mass, V → rt must equal V → in since we are assuming incompressible flow. Working backwards, we construct V → rt, relative in Fig. 14–66. Trigonometry reveals that 𝛽rt = arctan 𝜔 r Vin = arctan 73.30 m/s 47.1 m/s = 57.28° (4) We conclude that the rotor blade at this radius has a leading edge angle of about 73.1° (Eq. 3) and a trailing edge angle of about 57.3° (Eq. 4). A sketch of the rotor blade at this radius is provided in Fig. 14–65; the total curvature is small, being less than 16° from leading to trailing edge. Finally, to avoid interaction of the stator blade wakes with the rotor blade leading edges, we choose the number of rotor blades such that it has no common denominator with the number of stator blades. Since there are 16 stator blades, we pick a number like 13, 15, or 17 rotor blades. Choosing 14 would not be appropriate since it shares a common denominator of 2 with the number 16. Choosing 12 would be worse since it shares both 2 and 4 as common denominators.
Discussion We can repeat the calculation for all radii from hub to tip, completing the design of the entire rotor. There would be twist, as discussed previously.
ωr βrt βrl Vst Rotor blade row Vst, relative Vrt, relative FIGURE 14–65 Analysis of the stator trailing edge velocity of Example 14–8 as it impinges on the rotor leading edge; relative reference frame.
Vrt = Vin βrt ωr Vrt, relative FIGURE 14–66 Analysis of the rotor trailing edge velocity of Example 14–8; absolute reference frame.
cen96537_ch14_793-884.indd 830 29/12/16 4:38 pm 831 CHAPTER 14 The dimensionless Pi groups are shown in Fig. 14–67; the result is the fol lowing relationship involving dimensionless parameters: gH 𝜔 2D2 = function of ( V .
𝜔 D3, ρ𝜔 D2 𝜇 , 𝜀 D) (14–28) A similar analysis with input brake horsepower as a function of the same variables results in bhp ρ𝜔 3D5 = function of ( V .
𝜔 D3, ρ𝜔 D2 𝜇 , 𝜀 D) (14–29) The second dimensionless parameter (or П group) on the right side of both Eqs. 14–28 and 14–29 is obviously a Reynolds number since 𝜔D is a characteristic velocity, Re = ρ𝜔 D2 𝜇 The third П on the right is the nondimensional roughness parameter. The three new dimensionless groups in these two equations are given symbols and named as follows: Dimensionless pump parameters: CH = Head coefficient = gH 𝜔 2D2 CQ = Capacity coefficient = V .
𝜔 D3 (14–30) CP = Power coefficient = bhp ρ𝜔 3D5 Note the subscript Q in the symbol for capacity coefficient. This comes from the nomenclature found in many fluid mechanics and turbomachinery textbooks that Q rather than V . is the volume flow rate through the pump. We use the notation CQ for consistency with turbomachinery convention, even though we use V . for volume flow rate to avoid confusion with heat transfer. When pumping liquids, cavitation may be of concern, and we need another dimensionless parameter related to the required net positive suction head. Fortunately, we can simply substitute NPSHrequired in place of H in the dimensional analysis, since they have identical dimensions (length). The result is CNPSH = Suction head coefficient = gNPSHrequired 𝜔 2D2 (14–31) Other variables, such as gap thickness between blade tips and pump housing and blade thickness, can be added to the dimensional analysis if necessary. Fortunately, these variables typically are of only minor impor tance and are not considered here. In fact, you may argue that two pumps are not even strictly geometrically similar unless gap thickness, blade thick ness, and surface roughness scale geometrically.
FIGURE 14–67 Dimensional analysis of a pump.
ω D bhp gH = ƒ( , D, ε, ω, ρ, μ) k = n – j = 7 – 3 = 4 Π’s expected.
ε Π1 = gH ω2D2 V Π2 = ωD3 Π3 = μ ρωD2 Π4 = D ε V ∙ ∙ V ∙ V, ρ, μ ∙ cen96537_ch14_793-884.indd 831 29/12/16 4:38 pm 832 TURBOMACHINERY Relationships derived by dimensional analysis, such as Eqs. 14–28 and 14–29, are interpreted as follows: If two pumps, A and B, are geometri cally similar (pump A is geometrically proportional to pump B, although they may be of different sizes), and if the independent П’s are equal to each other (in this case if CQ, A = CQ, B, ReA = ReB, and 𝜀A/DA = 𝜀B/DB), then the dependent П’s are guaranteed to also be equal to each other as well. In particular, CH, A = CH, B from Eq. 14–28 and CP, A = CP, B from Eq. 14–29. If such conditions are established, the two pumps are said to be dynamically similar (Fig. 14–68). When dynamic similarity is achieved, the operating point on the pump performance curve of pump A and the corresponding operating point on the pump performance curve of pump B are said to be homologous. The requirement of equality of all three of the independent dimension less parameters can be relaxed somewhat. If the Reynolds numbers of both pump A and pump B exceed several thousand, turbulent flow conditions exist inside the pump. It turns out that for turbulent flow, if the values of ReA and ReB are not equal, but not too far apart, dynamic similarity between the two pumps is still a reasonable approximation. This fortunate condi tion is due to Reynolds number independence (Chap. 7). (Note that if the pumps operate in the laminar regime, or at low Re, the Reynolds number must usually remain as a scaling parameter.) In most cases of practical tur bomachinery engineering analysis, the effect of differences in the roughness parameter is also small, unless the roughness differences are large, as when one is scaling from a very small pump to a very large pump (or vice versa). Thus, for many practical problems, we may neglect the effect of both Re and 𝜀/D. Equations 14–28 and 14–29 then reduce to CH ≅function of CQ CP ≅function of CQ (14–32) As always, dimensional analysis cannot predict the shape of the functional relationships of Eq. 14–32, but once these relationships are obtained for a particular pump, they can be generalized for geometrically similar pumps that are of different diameters, operate at different rotational speeds and flow rates, and operate even with fluids of different density and viscosity. We transform Eq. 14–5 for pump efficiency into a function of the dimen sionless parameters of Eq. 14–30, 𝜂pump = ρ(V .
)(gH) bhp = ρ(𝜔 D3CQ)(𝜔 2D2CH) ρ𝜔 3D5CP = CQCH CP ≅function of CQ (14–33) Since 𝜂pump is already dimensionless, it is another dimensionless pump parameter all by itself. Note that since Eq. 14–33 reveals that 𝜂pump can be formed by the combination of three other П’s, 𝜂pump is not necessary for pump scaling. It is, however, certainly a useful parameter. Since CH, CP, and 𝜂pump are approximated as functions only of CQ, we often plot these three parameters as functions of CQ on the same plot, generating a set of nondimensional pump performance curves. An example is provided in Fig. 14–69 for the case of a typical centrifugal pump. The curve shapes for other types of pumps would, of course, be different. The simplified similarity laws of Eqs. 14–32 and 14–33 break down when the full-scale prototype is significantly larger than its model (Fig. 14–70); ωA DA HA bhpA Pump A VA, ρA, μA εA ωB DB HB bhpB Pump B εB ∙ VB, ρB, μB ∙ VA ∙ VB ∙ FIGURE 14–68 Dimensional analysis is useful for scaling two geometrically similar pumps. If all the dimensionless pump parameters of pump A are equivalent to those of pump B, the two pumps are dynamically similar.
CH CH CP CP ηpump BEP 0 0 CQ CQ FIGURE 14–69 When plotted in terms of dimensionless pump parameters, the performance curves of all pumps in a family of geometrically similar pumps collapse onto one set of nondimensional pump performance curves. Values at the best efficiency point are indicated by asterisks.
cen96537_ch14_793-884.indd 832 29/12/16 4:38 pm 833 CHAPTER 14 the prototype’s performance is generally better. There are several reasons for this: The prototype pump often operates at high Reynolds numbers that are not achievable in the laboratory. We know from the Moody chart that the friction factor decreases with Re, as does boundary layer thickness. Hence, the influence of viscous boundary layers is less significant as pump size increases, since the boundary layers occupy a less significant percentage of the flow path through the impeller. In addition, the relative roughness (𝜀/D) on the surfaces of the prototype impeller blades may be significantly smaller than that on the model pump blades unless the model surfaces are micropo lished. Finally, large full-scale pumps have smaller tip clearances relative to the blade diameter; therefore, tip losses and leakage are less significant. Some empirical equations have been developed to account for the increase in efficiency between a small model and a full-scale prototype. One such equation was suggested by Moody (1926) for turbines, but it can be used as a first-order correction for pumps as well, Moody efficiency correction equation for pumps: 𝜂pump, prototype ≅1 −(1 −𝜂pump, model)( Dmodel Dprototype) 1/5 (14–34) Pump Specific Speed Another useful dimensionless parameter called pump specific speed (NSp) is formed by a combination of parameters CQ and CH: Pump specific speed: NSp = CQ 1/2 CH 3/4 = (V .
/𝜔 D3)1/2 (gH/𝜔 2D2)3/4 = 𝜔 V .
1/2 (gH)3/4 (14–35) If all engineers watched their units carefully, NSp would always be listed as a dimensionless parameter. Unfortunately, practicing engineers have grown accustomed to using inconsistent units in Eq. 14–35, which renders the per fectly fine dimensionless parameter NSp into a cumbersome dimensional quantity (Fig. 14–71). Further confusion results because some engineers prefer units of rotations per minute (rpm) for rotational speed, while oth ers use rotations per second (Hz), the latter being more common in Europe. In addition, practicing engineers in the United States typically ignore the gravitational constant in the definition of NSp. In this book, we add sub scripts “Eur” or “US” to NSp in order to distinguish the dimensional forms of pump specific speed from the nondimensional form. In the United States, it is customary to write H in units of feet (net head expressed as an equiva lent column height of the fluid being pumped), V . in units of gallons per minute (gpm), and rotation rate in terms of n . (rpm) instead of 𝜔 (rad/s). Using Eq. 14–35 we define Pump specific speed, customary U.S. units: NSp, US = (n ·, rpm)(V .
, gpm)1/2 (H, ft)3/4 (14–36) In Europe it is customary to write H in units of meters (and to include g = 9.81 m/s2 in the equation), V . in units of m3/s, and rotation rate n . in Dprototype Prototype Dmodel Scale model ∙ Vmodel ∙ Vprototype FIGURE 14–70 When a small-scale model is tested to predict the performance of a full-scale prototype pump, the measured efficiency of the model is typically somewhat lower than that of the pro totype. Empirical correction equations such as Eq. 14–34 have been developed to account for the improvement of pump efficiency with pump size.
You did what? Why would you turn a dimensionless parameter into a dimensional quantity? That’s the exact opposite of what you should be doing!
FIGURE 14–71 Even though pump specific speed is a dimensionless parameter, it is common practice to write it as a dimensional quantity using an inconsistent set of units.
cen96537_ch14_793-884.indd 833 29/12/16 4:38 pm 834 TURBOMACHINERY units of rotations per second (Hz) instead of 𝜔 (rad/s) or n . (rpm). Using Eq. 14–35 we define Pump specific speed, customary European units: NSp, Eur = (n ·, Hz)(V .
, m3/s)1/2 (gH, m2/s2)3/4 (14–37) The conversions between these three forms of pump specific speed are pro vided as ratios for your convenience in Fig. 14–72. When you become a prac ticing engineer, you will need to be very careful that you know which form of pump specific speed is being used, although it may not always be obvious. Technically, pump specific speed could be applied at any operating condi tion and would just be another function of CQ. That is not how it is typically used, however. Instead, it is common to define pump specific speed at only one operating point, namely, the best efficiency point (BEP) of the pump. The result is a single number that characterizes the pump.
Pump specific speed is used to characterize the operation of a pump at its optimum conditions (best efficiency point) and is useful for preliminary pump selection and/or design.
As plotted in Fig. 14–73, centrifugal pumps perform optimally for NSp near 1, while mixed-flow and axial pumps perform best at NSp near 2 and 5, respec tively. It turns out that if NSp is less than about 1.5, a centrifugal pump is the best choice. If NSp is between about 1.5 and 3.5, a mixed-flow pump is a bet ter choice. When NSp is greater than about 3.5, an axial pump should be used. These ranges are indicated in Fig. 14–73 in terms of NSp, NSp, US, and NSp, Eur. Sketches of the blade types are also provided on the plot for reference.
Conversion ratios NSp NSp, US NSp, US = 2734 NSp NSp = 2π NSp, Eur NSp, Eur = NSp 1 2π NSp, Eur = 5.822 × 10–5 NSp, US NSp,US = 17,180 NSp, Eur = 3.658 × 10–4 FIGURE 14–72 Conversions between the dimension less, customary U.S., and customary European definitions of pump specific speed. Numerical values are given to four significant digits. The conversions for NSp, US assume standard earth gravity.
EXAMPLE 14–9 Using Pump Specific Speed for Preliminary Pump Design A pump is being designed to deliver 320 gpm of gasoline at room temperature. The required net head is 23.5 ft (of gasoline). It has already been determined that the pump shaft is to rotate at 1170 rpm. Calculate the pump specific speed in both nondimensional form and customary U.S. form. Based on your result, decide which kind of dynamic pump would be most suitable for this application.
SOLUTION We are to calculate pump specific speed and then determine whether a centrifugal, mixed-flow, or axial pump would be the best choice for this particular application.
Assumptions 1 The pump operates near its best efficiency point. 2 The maxi mum efficiency versus pump specific speed curve follows Fig. 14–73 reasonably well.
Analysis First, we calculate pump specific speed in customary U.S. units, NSp, US = (1170 rpm)(320 gpm)1/2 (23.5 ft)3/4 = 1960 (1) cen96537_ch14_793-884.indd 834 29/12/16 4:38 pm 835 CHAPTER 14 Affinity Laws We have developed dimensionless groups that are useful for relating any two pumps that are both geometrically similar and dynamically similar. It is convenient to summarize the similarity relationships as ratios. Some authors call these relationships similarity rules, while others call them affinity laws. For any two homologous states A and B, V .
B V .
A = 𝜔 B 𝜔 A( DB DA ) 3 (14–38a) Affinity laws: HB HA = ( 𝜔 B 𝜔 A) 2 ( DB DA) 2 (14–38b) bhpB bhpA = ρB ρA( 𝜔 B 𝜔 A) 3 ( DB DA ) 5 (14–38c) Equations 14–38 apply to both pumps and turbines. States A and B can be any two homologous states between any two geometrically similar turbomachines, or even between two homologous states of the same machine. Examples include changing rotational speed or pumping a different fluid with the same pump. For the simple case of a given pump in which 𝜔 is varied, but the same fluid is pumped, DA = DB, and 𝜌A = 𝜌B. In such a 1 0.7 0.5 0.1 0.2 1 NSp NSp, Eur 2 5 10 0.8 0.9 0.6 0.5 0.05 0.1 0.2 0.5 1 0.02 20,000 10,000 5000 2000 1000 500 ηmax Centrifugal Mixed Axial NSp, US FIGURE 14–73 Maximum efficiency as a function of pump specific speed for the three main types of dynamic pump. The hori zontal scales show nondimensional pump specific speed (NSp), pump specific speed in customary U.S. units (NSp, US), and pump specific speed in customary European units (NSp, Eur).
We convert to normalized pump specific speed using the conversion factor given in Fig. 14–72, NSp = NSp, US( NSp NSp, US) = 1960(3.658 × 10−4) = 0.717 (2) Using either Eq. 1 or 2, Fig. 14–73 shows that a centrifugal flow pump is the most suitable choice.
Discussion Notice that the properties of the fluid never entered our calculations. The fact that we are pumping gasoline rather than some other liquid like water is irrelevant. However, the brake horsepower required to run the pump does depend on the fluid density.
cen96537_ch14_793-884.indd 835 29/12/16 4:38 pm 836 TURBOMACHINERY case, Eqs. 14–38 reduce to the forms shown in Fig. 14–74. A mnemonic has been developed to help us remember the exponent on 𝜔, as indicated in the figure. Note also that anywhere there is a ratio of two rotational speeds (𝜔), we may substitute the appropriate values of rpm (n .) instead, since the con version is the same in both the numerator and the denominator. The pump affinity laws are quite useful as a design tool. In particular, suppose the performance curves of an existing pump are known, and the pump operates with reasonable efficiency and reliability. The pump manu facturer decides to design a new, larger pump for other applications, e.g., to pump a much heavier fluid or to deliver a substantially greater net head. Rather than starting from scratch, engineers often simply scale up an exist ing design. The pump affinity laws enable such scaling to be accomplished with a minimal amount of effort.
V: Volume flow rate H: Head VB ∙ VA ∙ ωB 1 ωA = nB ⋅ ⋅ nA 1 = HB HA ωB 2 ⋅ ωA = nB ⋅ nA 2 = P: Power bhpB bhpA ωB 3 ⋅ ωA = nB ⋅ nA 3 = ( ( ( ( ( ( ( ( ( ( ( ( FIGURE 14–74 When the affinity laws are applied to a single pump in which the only thing that is varied is shaft rotational speed 𝜔, or shaft rpm, n ., Eqs. 14–38 reduce to those shown above, for which a mnemonic can be used to help us remember the exponent on 𝜔 (or on n .): Very Hard Problems are as easy as 1, 2, 3.
14 4 0 0 0.5 1.5 ωB/ωA 2.5 6 12 10 8 2 1 2 bhpB bhpA HB HA FIGURE 14–75 When the speed of a pump is increased, net head increases rapidly; brake horsepower increases even more rapidly.
EXAMPLE 14–10 The Effects of Doubling Pump Speed Professor Seymour Fluids uses a small closed-loop water tunnel to perform flow visualization research. He would like to double the water speed in the test section of the tunnel and realizes that the least expensive way to do this is to double the rotational speed of the flow pump. What he doesn’t realize is how much more powerful the new electric motor will need to be! If Professor Fluids doubles the flow speed, by approximately what factor will the motor power need to be increased?
SOLUTION For a doubling of 𝜔, we are to calculate by what factor the power to the pump motor must increase.
Assumptions 1 The water remains at the same temperature. 2 After doubling pump speed, the pump runs at conditions homologous to the original conditions.
Analysis Since neither diameter nor density has changed, Eq. 14–38c reduces to Ratio of required shaft power: bhpB bhpA = ( 𝜔 B 𝜔 A ) 3 (1) Setting 𝜔B = 2𝜔A in Eq. 1 gives bhpB = 8bhpA. Thus the power to the pump motor must be increased by a factor of 8. A similar analysis using Eq. 14–38b shows that the pump’s net head increases by a factor of 4. As seen in Fig. 14–75, both net head and power increase rapidly as pump speed is increased.
Discussion The result is only approximate since we have not included any analy sis of the piping system. While doubling the flow speed through the pump increases available head by a factor of 4, doubling the flow speed through the water tunnel does not necessarily increase the required head of the system by the same factor of 4 (e.g., the friction factor decreases with the Reynolds number except at very high values of Re). In other words, our assumption 2 is not necessarily correct. The system will, of course, adjust to an operating point at which required and available heads match, but this point will not necessarily be homologous with the original operating point. Nevertheless, the approximation is useful as a first-order result. Professor Fluids may also need to be concerned with the possibility of cavitation at the higher speed.
cen96537_ch14_793-884.indd 836 29/12/16 4:38 pm 837 CHAPTER 14 200 140 20 0 10 9 5 4 3 2 1 0 0 200 400 600 800 40 160 180 60 100 120 80 H, cm (or η, %) bhp, W 6 7 8 V, cm3/s ⋅ bhp H ηpump FIGURE 14–76 Data points and smoothed dimensional pump performance curves for the water pump of Example 14–11.
EXAMPLE 14–11 Design of a New Geometrically Similar Pump After graduation, you work for a pump manufacturing company. One of your company’s best-selling products is a water pump, which we shall call pump A. Its impeller diameter is DA = 6.0 cm, and its performance data when operating at n .
A = 1725 rpm (𝜔A = 180.6 rad/s) are shown in Table 14–2. The marketing research department is recommending that the company design a new product, namely, a larger pump (which we shall call pump B) that will be used to pump liquid refrigerant R-134a at room temperature. The pump is to be designed such that its best efficiency point occurs as close as possible to a volume flow rate of V .
B = 2400 cm3/s and at a net head of HB = 450 cm (of R-134a). The chief engi neer (your boss) tells you to perform some preliminary analyses using pump scal ing laws to determine if a geometrically scaled-up pump could be designed and built to meet the given requirements. (a) Plot the performance curves of pump A in both dimensional and dimensionless form, and identify the best efficiency point. (b) Calculate the required pump diameter DB, rotational speed n .
B, and brake horsepower bhpB for the new product.
SOLUTION (a) For a given table of pump performance data for a water pump, we are to plot both dimensional and dimensionless performance curves and identify the BEP. (b) We are to design a new geometrically similar pump for refrigerant R-134a that operates at its BEP at given design conditions.
Assumptions 1 The new pump can be manufactured so as to be geometrically similar to the existing pump. 2 Both liquids (water and refrigerant R-134a) are incompressible. 3 Both pumps operate under steady conditions.
Properties At room temperature (20°C), the density of water is 𝜌water = 998.0 kg/m3 and that of refrigerant R-134a is 𝜌R-134a = 1226 kg/m3.
Analysis (a) First, we apply a second-order least-squares polynomial curve fit to the data of Table 14–2 to obtain smooth pump performance curves. These are plot ted in Fig. 14–76, along with a curve for brake horsepower, which is obtained from Eq. 14–5. A sample calculation, including unit conversions, is shown in Eq. 1 for the data at V .
A = 500 cm3/s, which is approximately the best efficiency point: bhpA = ρwatergV .
AHA 𝜂pump,A = (998.0 kg/m3)(9.81 m/s2)(500 cm3/s)(150 cm) 0.81 ( 1 m 100 cm) 4 ( W·s kg·m2/s2) = 9.07 W (1) Note that the actual value of bhpA plotted in Fig. 14–76 at V .
A = 500 cm3/s differs slightly from that of Eq. 1 due to the fact that the least-squares curve fit smoothes out scatter in the original tabulated data. Next we use Eqs. 14–30 to convert the dimensional data of Table 14–2 into non dimensional pump similarity parameters. Sample calculations are shown in Eqs. 2 through 4 at the same operating point as before (at the approximate location of the BEP). At V .
A = 500 cm3/s the capacity coefficient is approximately CQ = V .
𝜔 D3 = 500 cm3/s (180.6 rad/s)(6.0 cm)3 = 0.0128 (2) TABLE 14–2 Manufacturer’s performance data for a water pump operating at 1725 rpm and room temperature (Example 14–11) V .
, cm3/s H, cm 𝜂pump, % 100 180 32 200 185 54 300 175 70 400 170 79 500 150 81 600 95 66 700 54 38 Net head is in centimeters of water.
cen96537_ch14_793-884.indd 837 29/12/16 4:38 pm 838 TURBOMACHINERY The head coefficient at this flow rate is approximately CH = gH 𝜔 2D2 = (9.81 m/s2)(1.50 m) (180.6 rad/s)2(0.060 m)2 = 0.125 (3) Finally, the power coefficient at V .
A = 500 cm3/s is approximately CP = bhp ρ𝜔 3D5 = 9.07 W (998 kg/m3)(180.6 rad/s)3(0.060 m)5( kg·m2/s2 W·s ) = 0.00198 (4) These calculations are repeated (with the aid of a spreadsheet) at values of V .
A between 100 and 700 cm3/s. The curve-fitted data are used so that the normalized pump performance curves are smooth; they are plotted in Fig. 14–77. Note that 𝜂pump is plotted as a fraction rather than as a percentage. In addition, in order to fit all three curves on one plot with a single ordinate, and with the abscissa centered nearly around unity, we have multiplied CQ by 100, CH by 10, and CP by 100. You will find that these scaling factors work well for a wide range of pumps, from very small to very large. A vertical line at the BEP is also sketched in Fig. 14–77 from the smoothed data. The curve-fitted data yield the following nondimensional pump performance parameters at the BEP: CQ = 0.0112 CH = 0.133 CP = 0.00184 𝜂 pump = 0.812 (5) (b) We design the new pump such that its best efficiency point is homologous with the BEP of the original pump, but with a different fluid, a different pump diam eter, and a different rotational speed. Using the values identified in Eq. 5, we use Eqs. 14–30 to obtain the operating conditions of the new pump. Namely, since both V .
B and HB are known (design conditions), we solve simultaneously for DB and 𝜔B. After some algebra in which we eliminate 𝜔B, we calculate the design diameter for pump B, DB = ( V .
B 2CH (CQ)2gHB) 1/4 = ( (0.0024 m3/s)2(0.133) (0.0112)2(9.81 m/s2)(4.50 m)) 1/4 = 0.108 m (6) In other words, pump A needs to be scaled up by a factor of DB/DA = 10.8 cm/6.0 cm = 1.80. With the value of DB known, we return to Eqs. 14–30 to solve for 𝜔B, the design rotational speed for pump B, 𝜔 B = V .
B (CQ)DB 3 = 0.0024 m3/s (0.0112)(0.108 m)3 = 168 rad/s → n .
B = 1610 rpm (7) Finally, the required brake horsepower for pump B is calculated from Eqs. 14–30, bhpB = (CP)ρB𝜔 B 3DB 5 = (0.00184)(1226 kg/m3)(168 rad/s)3(0.108 m)5( W·s kg·m2/s2)= 160 W (8) An alternative approach is to use the affinity laws directly, eliminating some inter mediate steps. We solve Eqs. 14–38a and b for DB by eliminating the ratio 𝜔B/𝜔A. 1.6 0.6 0 0 0.5 1.5 CQ × 100 2 0.8 1.4 1.2 1 0.4 0.2 1 ηpump CH × 10 CP × 100 BEP FIGURE 14–77 Smoothed nondimensional pump performance curves for the pumps of Example 14–11; BEP is estimated as the operating point where 𝜂pump is a maximum.
cen96537_ch14_793-884.indd 838 29/12/16 4:38 pm 839 CHAPTER 14 14–4 ■ TURBINES Turbines have been used for centuries to convert freely available mechanical energy from rivers and wind into useful mechanical work, usually through a rotating shaft. Whereas the rotating part of a pump is called the impeller, the rotating part of a hydroturbine is called the runner. When the working fluid is water, the turbomachines are called hydraulic turbines or hydro turbines. When the working fluid is air, and energy is extracted from the wind, the machine is properly called a wind turbine. The word windmill should technically be applied only when the mechanical energy output is used to grind grain, as in ancient times (Fig. 14–79). However, most people use the word windmill to describe any wind turbine, whether used to grind grain, pump water, or generate electricity. In coal or nuclear power plants, the working fluid is usually steam; hence, the turbomachines that convert energy from the steam into mechanical energy of a rotating shaft are called steam turbines. A more generic name for turbines that employ a compress ible gas as the working fluid is gas turbine. (The turbine in a modern com mercial jet engine is a type of gas turbine.) In general, energy-producing turbines have somewhat higher overall effi ciencies than do energy-absorbing pumps. Large hydroturbines, for example, achieve overall efficiencies above 95 percent, while the best efficiency of large pumps is a little more than 90 percent. There are several reasons for this. First, pumps normally operate at higher rotational speeds than do turbines; therefore, shear stresses and frictional losses are higher. Second, conver sion of kinetic energy into flow energy (pumps) has inherently higher losses than does the reverse (turbines). You can think of it this way: Since pressure rises across a pump (adverse pressure gradient), but drops across a turbine (favorable pressure gradient), boundary layers are less likely to separate in a turbine than in a pump. Third, turbines (especially hydroturbines) are often much larger than pumps, and viscous losses become less important as size increases. Finally, while pumps often operate over a wide range of flow rates, most electricity-generating turbines run within a narrower operating range and at a controlled constant speed; they can therefore be designed to operate most We then plug in the known value of DA and the curve-fitted values of V .
A and HA at the BEP (Fig. 14–78). The result agrees with those calculated before. In a similar manner we can calculate 𝜔B and bhpB.
Discussion Although the desired value of 𝜔B has been calculated precisely, a practical issue is that it is difficult (if not impossible) to find an electric motor that rotates at exactly the desired rpm. Standard single-phase, 60-Hz, 120-V AC electric motors typically run at 1725 or 3450 rpm. Thus, we may not be able to meet the rpm requirement with a direct-drive pump. Of course, if the pump is belt-driven or if there is a gear box or a frequency controller, we can easily adjust the configuration to yield the desired rotation rate. Another option is that since 𝜔B is only slightly smaller than 𝜔A, we drive the new pump at standard motor speed (1725 rpm), providing a somewhat stronger pump than necessary. The disadvantage of this option is that the new pump would then operate at a point not exactly at the BEP.
From the affinity laws, DB = DA HA HB VB = (6.0 cm) = 10.8 cm 159.3 cm 450 cm 2400 cm3 s cm3 s 438 1/2 1/4 1/2 1/4 ∙ VA ∙ ( ( ( ( ( ( ( ( FIGURE 14–78 The affinity laws are manipulated to obtain an expression for the new pump diameter DB. 𝜔B and bhpB can be obtained in similar fashion (not shown).
FIGURE 14–79 An old windmill in Klostermolle, Vestervig, Denmark that was used in the 1800s to grind grain. (Note that the blades must be covered to function.) Modern “windmills” that generate electricity are more properly called wind turbines.
© OJPhotos/Alamy RF cen96537_ch14_793-884.indd 839 29/12/16 4:38 pm 840 TURBOMACHINERY efficiently at those conditions. In the United States, the standard AC electri cal supply is 60 Hz (3600 cycles per minute); thus most wind, water, and steam turbines operate at speeds that are natural fractions of this, namely, 7200 rpm divided by the number of poles on the generator, usually an even number. Large hydroturbines usually operate at low speeds like 7200/60 = 120 rpm or 7200/48 = 150 rpm. Gas turbines used for power generation run at much higher speeds, some up to 7200/2 = 3600 rpm! As with pumps, we classify turbines into two broad categories, positive displacement and dynamic. For the most part, positive-displacement turbines are small devices used for volume flow rate measurement, while dynamic turbines range from tiny to huge and are used for both flow measurement and power production. We provide details about both of these categories.
Positive-Displacement Turbines A positive-displacement turbine may be thought of as a positive-displacement pump running backward—as fluid pushes into a closed volume, it turns a shaft or displaces a reciprocating rod. The closed volume of fluid is then pushed out as more fluid enters the device. There is a net head loss through the positive-displacement turbine; in other words, energy is extracted from the flowing fluid and is turned into mechanical energy. However, positive-displacement turbines are generally not used for power production, but rather for flow rate or flow volume measurement. The most common example is the water meter in your house (Fig. 14–80). Many commercial water meters use a nutating disc that wobbles and spins as water flows through the meter. The disc has a sphere in its center with appro priate linkages that transfer the eccentric spinning motion of the nutating disc into rotation of a shaft. The volume of fluid that passes through the device per 360o rotation of the shaft is known precisely, and thus the total volume of water used is recorded by the device. When water is flowing at moderate speed from a spigot in your house, you can sometimes hear a bubbly sound coming from the water meter—this is the sound of the nutating disc wobbling inside the meter. There are, of course, other positive-displacement turbine designs, just as there are various designs of positive-displacement pumps.
Dynamic Turbines Dynamic turbines are used both as flow measuring devices and as power generators. For example, meteorologists use a three-cup anemometer to mea sure wind speed (Fig. 14–81a). Experimental fluid mechanics researchers use small turbines of various shapes (most of which look like small propel lers) to measure air speed or water speed (Chap. 8). In these applications, the shaft power output and the efficiency of the turbine are of little concern. Rather, these instruments are designed such that their rotational speed can be accurately calibrated to the speed of the fluid. Then, by electronically counting the number of blade rotations per second, the speed of the fluid is calculated and displayed by the device. A novel application of a dynamic turbine is shown in Fig. 14–81b. NASA researchers mounted turbines at the wing tips of a Piper PA28 research aircraft to extract energy from wing tip vortices (Chap. 11); the extracted energy was converted to electricity to be used for on-board power requirements.
FIGURE 14–80 The nutating disc fluid flowmeter is a type of positive-displacement turbine used to measure volume flow rate: (a) cutaway view and (b) diagram showing motion of the nutating disc. This type of flowmeter is commonly used as a water meter in homes. (a) Courtesy of Niagara Meters, Spartanburg, SC.
(a) (a) Shaft Linkage Flow out Flow in (b) ω Nutating disc cen96537_ch14_793-884.indd 840 29/12/16 4:38 pm 841 CHAPTER 14 In this chapter, we emphasize large dynamic turbines that are designed to produce electricity. Most of our discussion concerns hydroturbines that utilize the large elevation change across a dam to generate electricity, and wind tur bines that generate electricity from blades rotated by the wind. There are two basic types of dynamic turbine—impulse and reaction, each of which are dis cussed in some detail. Comparing the two power-producing dynamic turbines, impulse turbines require a higher head, but can operate with a smaller volume flow rate. Reaction turbines can operate with much less head, but require a higher volume flow rate.
Impulse Turbines In an impulse turbine, the fluid is sent through a nozzle so that most of its available mechanical energy is converted into kinetic energy. The high-speed jet then impinges on bucket-shaped vanes that transfer energy to the turbine shaft, as sketched in Fig. 14–82. The modern and most efficient type of impulse turbine was invented by Lester A. Pelton (1829–1908) in 1878, and the rotating wheel is now called a Pelton wheel in his honor. The buck ets of a Pelton wheel are designed so as to split the flow in half, and turn the flow nearly 180° around (with respect to a frame of reference moving with the bucket), as illustrated in Fig. 14–82b. According to legend, Pelton modeled the splitter ridge shape after the nostrils of a cow’s nose. A portion of the outermost part of each bucket is cut out so that the majority of the jet can pass through the bucket that is not aligned with the jet (bucket n + 1 in Fig. 14–82a) to reach the most aligned bucket (bucket n in Fig. 14–82a). In this way, the maximum amount of momentum from the jet is utilized. These details are seen in a photograph of a Pelton wheel (Fig. 14–83). Figure 14–84 shows a Pelton wheel in operation; the splitting and turning of the water jet is clearly seen. We analyze the power output of a Pelton wheel turbine by using the Euler turbomachine equation. The power output of the shaft is equal to 𝜔Tshaft, where Tshaft is given by Eq. 14–14, Euler turbomachine equation for a turbine: W .
shaft = 𝜔 Tshaft = ρ𝜔 V .
(r2V2, t −r1V1, t) (14–39) Shaft Splitter ridge rω β Vj Nozzle Bucket n + 1 (a) (b) Bucket n ω Vj –rω Vj –rω r FIGURE 14–82 Schematic diagram of a Pelton-type impulse turbine; the turbine shaft is turned when high-speed fluid from one or more jets impinges on buckets mounted to the turbine shaft. (a) Side view, absolute reference frame, and (b) bottom view of a cross section of bucket n, rotating reference frame.
FIGURE 14–81 Examples of dynamic turbines: (a) a typical three-cup anemometer used to measure wind speed, and (b) a Piper PA28 research airplane with turbines designed to extract energy from the wing tip vortices.
(a) © Matthias Engelien/Alamy RF. (b) NASA Langley Research Center.
(a) (a) (b) (b) cen96537_ch14_793-884.indd 841 29/12/16 4:38 pm 842 TURBOMACHINERY We must be careful of negative signs since this is an energy-producing rather than an energy-absorbing device. For turbines, it is conventional to define point 2 as the inlet and point 1 as the outlet. The center of the bucket moves at tangential velocity r𝜔, as illustrated in Fig. 14–82. We simplify the analysis by assuming that since there is an opening in the outermost part of each bucket, the entire jet strikes the bucket that happens to be at the direct bottom of the wheel at the instant of time under consideration (bucket n in Fig. 14–82a). Furthermore, since both the size of the bucket and the diameter of the water jet are small compared to the wheel radius, we approximate r1 FIGURE 14–83 A close-up view of a Pelton wheel showing the detailed design of the buckets; the electrical generator is on the right. This Pelton wheel is on display at the Waddamana Power Station Museum near Bothwell, Tasmania.
Courtesy of Hydro Tasmania, www.hydro.com.au. Used by permission.
FIGURE 14–84 A view from the bottom of an operating Pelton wheel illustrating the splitting and turning of the water jet in the bucket. The water jet enters from the left, and the Pelton wheel is turning to the right.
© ANDRITZ HYDRO GmbH cen96537_ch14_793-884.indd 842 29/12/16 4:38 pm 843 CHAPTER 14 and r2 as equal to r. Finally, we make the approximation that the water is turned through angle 𝛽 without losing any speed; in the relative frame of reference moving with the bucket, the relative exit speed is thus Vj − r𝜔 (the same as the relative inlet speed) as sketched in Fig. 14–82b. Return ing to the absolute reference frame, which is necessary for the application of Eq. 14–39, the tangential component of velocity at the inlet, V2, t, is simply the jet speed itself, Vj. We construct a velocity diagram in Fig. 14–85 as an aid in calculating the tangential component of absolute velocity at the out let, V1, t. After some trigonometry, which you can verify after noting that sin (𝛽 − 90°) = −cos 𝛽, V 1, t = r𝜔 + (V j −r𝜔 ) cos 𝛽 Upon substitution of this equation, Eq. 14–39 yields W .
shaft = ρr𝜔 V .
{ V j −[r𝜔 + (V j −r𝜔 )cos 𝛽]} which simplifies to Output shaft power: W .
shaft = ρr𝜔 V .
(V j −r𝜔 )(1 −cos 𝛽) (14–40) Obviously, the maximum power is achieved theoretically if 𝛽 = 180°. How ever, if that were the case, the water exiting one bucket would strike the back side of its neighbor coming along behind it, reducing the generated torque and power. It turns out that in practice, the maximum power is achieved by reducing 𝛽 to around 160° to 165°. The efficiency factor due to 𝛽 being less than 180° is Efficiency factor due to 𝛽: 𝜂𝛽= W .
shaft, actual W .
shaft, ideal = 1 −cos 𝛽 1 −cos (180°) (14–41) When 𝛽 = 160°, for example, 𝜂𝛽 = 0.97—a loss of only about 3 percent. Finally, we see from Eq. 14–40 that the shaft power output W .
shaft is zero if r𝜔 = 0 (wheel not turning at all). W .
shaft is also zero if r𝜔 = Vj (bucket moving at the jet speed). Somewhere in between these two extremes lies the optimum wheel speed. By setting the derivative of Eq. 14–40 with respect to r𝜔 to zero, we find that this occurs when r𝜔 = Vj / 2 (bucket moving at half the jet speed, as shown in Fig. 14–86). For an actual Pelton wheel turbine, there are other losses besides that reflected in Eq. 14–41: mechanical friction, aerodynamic drag on the buck ets, friction along the inside walls of the buckets, nonalignment of the jet and bucket as the bucket turns, backsplashing, and nozzle losses. Even so, the efficiency of a well-designed Pelton wheel turbine can approach 90 percent. In other words, up to 90 percent of the available mechanical energy of the water is converted to rotating shaft energy.
Reaction Turbines The other main type of energy-producing hydroturbine is the reaction turbine, which consists of fixed guide vanes called stay vanes, adjustable guide vanes called wicket gates, and rotating blades called runner blades (Fig. 14–87). Flow enters tangentially at high pressure, is turned toward Splitter ridge β rω Vj –rω Vj –rω V 1 V 1,t FIGURE 14–85 Velocity diagram of flow into and out of a Pelton wheel bucket. We translate outflow velocity from the moving reference frame to the absolute refer ence frame by adding the speed of the bucket (r𝜔) to the right.
Vj Vj 2 rω = —– Shaft Nozzle ω = —– r Vj 2r FIGURE 14–86 The theoretical maximum power achievable by a Pelton turbine occurs when the wheel rotates at 𝜔 = V j/(2r), i.e., when the bucket moves at half the speed of the water jet.
cen96537_ch14_793-884.indd 843 29/12/16 4:38 pm 844 TURBOMACHINERY the runner by the stay vanes as it moves along the spiral casing or volute, and then passes through the wicket gates with a large tangential velocity component. Momentum is exchanged between the fluid and the runner as the runner rotates, and there is a large pressure drop. Unlike the impulse turbine, the water completely fills the casing of a reaction turbine. For this reason, a reaction turbine generally produces more power than an impulse turbine of the same diameter, net head, and volume flow rate. The angle of the wicket gates is adjustable so as to control the volume flow rate through the runner. (In most designs the wicket gates can close on each other, cutting off the flow of water into the runner.) At design conditions the flow leaving the wicket gates impinges parallel to the runner blade leading edge (from a rotating frame of reference) to avoid shock losses. Note that in a good design, the number of wicket gates does not share a common denomi nator with the number of runner blades. Otherwise there would be severe vibration caused by simultaneous impingement of two or more wicket gate wakes onto the leading edges of the runner blades. For example, in Fig. 14–87 there are 17 runner blades and 20 wicket gates. These are typi cal numbers for many large reaction hydroturbines, as shown in the pho tographs in Figs. 14–89 and 14–90. The number of stay vanes and wicket gates is usually the same (there are 20 stay vanes in Fig. 14–87). This is not a problem since neither of them rotate, and unsteady wake interaction is not an issue. There are two main types of reaction turbine—Francis and Kaplan. The Francis turbine is somewhat similar in geometry to a centrifugal or mixed-flow pump, but with the flow in the opposite direction. Note, however, that a typical pump running backward would not be a very efficient turbine. The Francis turbine is named in honor of James B. Francis (1815–1892), who developed the design in the 1840s. In contrast, the Kaplan turbine is some what like an axial-flow fan running backward. If you have ever seen a win dow fan start spinning in the wrong direction when a gust of wind blows through the window, you can visualize the basic operating principle of a Kaplan turbine. The Kaplan turbine is named in honor of its inventor, Vik tor Kaplan (1876–1934). There are actually several subcategories of both Francis and Kaplan turbines, and the terminology used in the hydroturbine field is not always standard. Recall that we classify dynamic pumps according to the angle at which the flow exits the impeller blade—centrifugal (radial), mixed flow, or axial (see Fig. 14–31). In a similar but reversed manner, we classify reaction tur bines according to the angle that the flow enters the runner (Fig. 14–88). If the flow enters the runner radially as in Fig. 14–88a, the turbine is called a Francis radial-flow turbine (see also Fig. 14–87). If the flow enters the runner at some angle between radial and axial (Fig. 14–88b), the turbine is called a Francis mixed-flow turbine. The latter design is more common. Some hydroturbine engineers use the term “Francis turbine” only when there is a band on the runner as in Fig. 14–88b. Francis turbines are most suited for heads that lie between the high heads of Pelton wheel turbines and the low heads of Kaplan turbines. A typical large Francis turbine may have 16 or more runner blades and can achieve a turbine efficiency of 90 to 95 percent. If the runner has no band, and flow enters the runner partially turned, it is called a propeller mixed-flow turbine or simply a mixed-flow ρ • b1 r1 ω b2 Top view Side view Stay vanes Wicket gates Runner blades Band Shaft Draft tube Volute Out Vout, Pout Vin, Pin r2 In Out ω FIGURE 14–87 A reaction turbine differs significantly from an impulse turbine; instead of using water jets, a volute is filled with swirling water that drives the runner. For hydroturbine applications, the axis is typically vertical. Top and side views are shown, including the fixed stay vanes and adjustable wicket gates.
cen96537_ch14_793-884.indd 844 29/12/16 4:38 pm 845 CHAPTER 14 turbine (Fig. 14–88c). Finally, if the flow is turned completely axially before entering the runner (Fig. 14–88d ), the turbine is called an axial-flow turbine. The runners of an axial-flow turbine typically have only three to eight blades, a lot fewer than Francis turbines. Of these there are two types: Kaplan turbines and propeller turbines. Kaplan turbines are called double regulated because the flow rate is controlled in two ways—by turning the wicket gates and by adjusting the pitch on the runner blades. Propeller turbines are nearly identical to Kaplan turbines except that the blades are fixed (pitch is not adjustable), and the flow rate is regulated only by the wicket gates (single regulated). Compared to the Pelton and Francis tur bines, Kaplan turbines and propeller turbines are most suited for low head, high volume flow rate conditions. Their efficiencies rival those of Francis turbines and may be as high as 94 percent. Figure 14–89 is a photograph of the radial-flow runner of a Francis radial-flow turbine. The workers are shown to give you an idea of how large the runners are in a hydroelectric power plant. Figure 14–90 is a photograph of the mixed-flow runner of a Francis turbine, and Fig. 14–91 is a photograph of an axial-flow propeller turbine. The view is from the inlet (top). We sketch in Fig. 14–92 a typical hydroelectric dam that utilizes Francis reaction turbines to generate electricity. The overall or gross head Hgross is defined as the elevation difference between the reservoir surface upstream of the dam and the surface of the water exiting the dam, Hgross = zA − zE. If there were no irreversible losses anywhere in the system, the maximum amount of power that could be generated per turbine would be Ideal power production: W .
ideal = ρgV .
Hgross (14–42) Of course, there are irreversible losses throughout the system, so the power actually produced is lower than the ideal power given by Eq. 14–42.
ω ω ω ω (a) Hub Wicket gate (b) (c) (d) Crown Crown Hub Stay vane Band Band FIGURE 14–88 The distinguishing characteristics of the four subcategories of reaction turbines: (a) Francis radial flow, (b) Francis mixed flow, (c) propeller mixed flow, and (d ) propeller axial flow. The main difference between (b) and (c) is that Francis mixed-flow runners have a band that rotates with the runner, while propeller mixed-flow runners do not. There are two types of propeller mixed-flow turbines: Kaplan turbines have adjustable pitch blades, while propeller turbines do not. Note that the terminology used here is not universal among turbomachinery textbooks nor among hydroturbine manufacturers.
FIGURE 14–89 The runner of a Francis radial-flow turbine used at the Boundary hydroelectric power station on the Pend Oreille River north of Spokane, WA. There are 17 runner blades of outer diameter 18.5 ft (5.6 m). The turbine rotates at 128.57 rpm and produces 230 MW of power at a volume flow rate of 335 m3/s from a net head of 78 m.
© American Hydro Corporation. Used by permission.
cen96537_ch14_793-884.indd 845 29/12/16 4:38 pm 846 TURBOMACHINERY We follow the flow of water through the whole system of Fig. 14–92, defining terms and discussing losses along the way. We start at point A upstream of the dam where the water is still, at atmospheric pressure, and at its highest elevation, zA. Water flows at volume flow rate V . through a large tube through the dam called the penstock. Flow to the penstock can be cut off by closing a large gate valve called a head gate at the penstock inlet. If we were to insert a Pitot probe at point B at the end of the penstock just before the turbine, as illustrated in Fig. 14–92, the water in the tube would rise to a column height equal to the energy grade line EGLin at the inlet of the turbine. This column height is lower than the water level at point A, due to irreversible losses in the penstock and its inlet. The flow then passes through the turbine, which is connected by a shaft to the electric generator. Note that the electric generator itself has irreversible losses. From a fluid mechanics perspective, however, we are interested only in the losses through the turbine and downstream of the turbine. After passing through the turbine runner, the exiting fluid (point C) still has appreciable kinetic energy, and perhaps swirl. To recover some of this kinetic energy (which would otherwise be wasted), the flow enters an expanding area diffuser called a draft tube, which turns the flow horizon tally and slows down the flow speed, while increasing the pressure prior to discharge into the downstream water, called the tailrace. If we were to imagine another Pitot probe at point D (the exit of the draft tube), the water in the tube would rise to a column height equal to the energy grade line FIGURE 14–90 The runner of a Francis mixed-flow turbine used at the Smith Mountain hydroelectric power station in Roanoke, VA. There are 17 runner blades of outer diameter 20.3 ft (6.19 m). The turbine rotates at 100 rpm and produces 194 MW of power at a volume flow rate of 375 m3/s from a net head of 54.9 m.
© American Hydro Corporation. Used by permission.
FIGURE 14–91 The five-bladed propeller turbine used at the Warwick hydroelectric power station in Cordele, GA. There are five runner blades of outer diameter 12.7 ft (3.87 m). The turbine rotates at 100 rpm and produces 5.37 MW of power at a volume flow rate of 63.7 m3/s from a net head of 9.75 m.
© American Hydro Corporation. Used by permission.
cen96537_ch14_793-884.indd 846 29/12/16 4:38 pm 847 CHAPTER 14 labeled EGLout in Fig. 14–92. Since the draft tube is considered to be an integral part of the turbine assembly, the net head across the turbine is spec ified as the difference between EGLin and EGLout, Net head for a hydraulic turbine: H = EGLin −EGLout (14–43) In words, The net head of a turbine is defined as the difference between the energy grade line just upstream of the turbine and the energy grade line at the exit of the draft tube. At the draft tube exit (point D) the flow speed is significantly slower than that at point C upstream of the draft tube; however, it is finite. All the kinetic energy leaving the draft tube is dissipated in the tailrace. This represents an irreversible head loss and is the reason why EGLout is higher than the eleva tion of the tailrace surface, zE. Nevertheless, significant pressure recovery occurs in a well-designed draft tube. The draft tube causes the pressure at the outlet of the runner (point C) to decrease below atmospheric pressure, thereby enabling the turbine to utilize the available head most efficiently. In other words, the draft tube causes the pressure at the runner outlet to be lower than it would have been without the draft tube—increasing the change in pressure from the inlet to the outlet of the turbine. Designers must be careful, however, because subatmospheric pressures may lead to cavitation, which is undesirable for many reasons, as discussed previously. If we were interested in the net efficiency of the entire hydroelectric plant, we would define this efficiency as the ratio of actual electric power pro duced to ideal power (Eq. 14–42), based on gross head. Of more concern in this chapter is the efficiency of the turbine itself. By convention, turbine efficiency is based on net head H rather than gross head Hgross. Specifically, 𝜂turbine is defined as the ratio of brake horsepower output (actual turbine Generator Shaft Turbine Tailrace Draft tube Arbitrary datum plane (z = 0) B D E Power station Dam Head gate (open) Penstock A C Net head H Gross head Hgross EGLout EGLin zE zA V ∙ FIGURE 14–92 Typical setup and terminology for a hydroelectric plant that utilizes a Francis turbine to generate electricity; drawing not to scale. The Pitot probes are shown for illustrative purposes only.
cen96537_ch14_793-884.indd 847 29/12/16 4:38 pm 848 TURBOMACHINERY output shaft power) to water horsepower (power extracted from the water flowing through the turbine), Turbine efficiency: 𝜂turbine = W .
shaft W .
water horsepower = bhp ρgHV . (14–44) Note that turbine efficiency 𝜂turbine is the reciprocal of pump efficiency 𝜂pump, since bhp is the actual output instead of the required input (Fig. 14–93). Note also that we are considering only one turbine at a time in this discus sion. Most large hydroelectric power plants have several turbines arranged in parallel. This offers the power company the opportunity to turn off some of the turbines during times of low power demand and for maintenance. Hoover Dam in Boulder City, Nevada, for example, has 17 parallel turbines, 15 of which are identical large Francis turbines that can produce approxi mately 130 MW of electricity each (Fig. 14–94). The maximum gross head is 590 ft (180 m). The total peak power production of the power plant exceeds 2 GW (2000 MW). We perform preliminary design and analysis of turbines in the same way we did previously for pumps, using the Euler turbomachine equa tion and velocity diagrams. In fact, we keep the same notation, namely r1 for the inner radius and r2 for the outer radius of the rotating blades. For a turbine, however, the flow direction is opposite to that of a pump, so the inlet is at radius r2 and the outlet is at radius r1. For a first-order analysis we approximate the blades as being infinitesimally thin. We also assume that the blades are aligned such that the flow is always tangent to the blade surface, and we ignore viscous effects (boundary layers) at the surfaces. Higher-order corrections are best obtained with a computational fluid dynamics code. Consider for example the top view of the Francis turbine of Fig. 14–87. Velocity vectors are drawn in Fig. 14–95 for both the absolute reference η = efficiency = actual output required input ηpump = = Wwater horsepower ⋅ Wshaft ⋅ ρgHV ⋅ bhp ηturbine = = Wwater horsepower ⋅ Wshaft ⋅ bhp Thus, for a pump, and for a turbine, Efficiency is always defined as ρgHV ⋅ FIGURE 14–93 By definition, efficiency must always be less than unity. The efficiency of a turbine is the reciprocal of the efficiency of a pump.
FIGURE 14–94 (a) An aerial view of Hoover Dam and (b) the top (visible) portion of several of the parallel electric generators driven by hydraulic turbines at Hoover Dam.
(a) © Corbis RF (b) © Brand X Pictures RF (a) (a) (b) (b) cen96537_ch14_793-884.indd 848 29/12/16 4:38 pm 849 CHAPTER 14 frame and the relative reference frame rotating with the runner. Beginning with the stationary guide vane (thick black line in Fig. 14–95), the flow is turned so that it strikes the runner blade (thick brown line) at absolute velocity V › 2. But the runner blade is rotating counterclockwise, and at radius r2 it moves tangentially to the lower left at speed 𝜔r2. To translate into the rotating reference frame, we form the vector sum of V › 2 and the negative of 𝜔r2, as shown in the sketch. The resultant is vector V › 2, relative, which is parallel to the runner blade leading edge (angle 𝛽2 from the tangent line of circle r2). The tangential component V2, t of the absolute velocity vector V › 2 is required for the Euler turbomachine equation (Eq. 14–39). After some trigonometry, Runner leading edge: V2, t = 𝜔 r2 −V2, n tan 𝛽2 (14–45) Following the flow along the runner blade in the relative (rotating) ref erence frame, we see that the flow is turned such that it exits parallel to the trailing edge of the runner blade (angle 𝛽1 from the tangent line of circle r1). Finally, to translate back to the absolute reference frame we vec torially add V › 1, relative and blade speed 𝜔r1, which acts to the left as sketched in Fig. 14–96. The resultant is absolute vector V › 1. Since mass must be conserved, the normal components of the absolute velocity vectors V1, n and V2, n are related through Eq. 14–12, where axial blade widths b1 and b2 are defined in Fig. 14–87. After some trigonometry (which turns out to be identical to that at the leading edge), we generate an expression for the tangential component V1, t of absolute velocity vector V › 1 for use in the Euler turbomachine equation, Runner trailing edge: V1, t = 𝜔 r1 −V1, n tan 𝛽1 (14–46) Alert readers will notice that Eq. 14–46 for a turbine is identical to Eq. 14–23 for a pump. This is not just fortuitous, but results from the fact that the velocity vectors, angles, etc., are defined in the same way for a turbine as for a pump except that everything is flowing in the opposite direction. For some hydroturbine runner applications, high power/high flow opera tion can result in V1, t < 0. Here the runner blade turns the flow so much that the flow at the runner outlet rotates in the direction opposite to runner rotation, a situation called reverse swirl (Fig. 14–97). The Euler turboma chine equation predicts that maximum power is obtained when V1, t < 0, so we suspect that reverse swirl should be part of a good turbine design. In practice, however, it has been found that the best efficiency operation of most hydroturbines occurs when the runner imparts a small amount of with-rotation swirl to the flow exiting the runner (swirl in the same direc tion as runner rotation). This improves draft tube performance. A large amount of swirl (either reverse or with-rotation) is not desirable, because it leads to much higher losses in the draft tube. (High swirl velocities result in “wasted” kinetic energy.) Obviously, much fine tuning needs to be done in order to design the most efficient hydroturbine system (including the draft tube as an integral component) within imposed design constraints. Also keep in mind that the flow is three-dimensional; there is an axial component ωr2 ω V2, relative V2, t V2, n r1 r2 V2 β2 FIGURE 14–95 Relative and absolute velocity vectors and geometry for the outer radius of the runner of a Francis turbine. Absolute velocity vectors are bold.
ω V2, relative r1 ωr1 r2 V1 β1 V1, relative V2 FIGURE 14–96 Relative and absolute velocity vectors and geometry for the inner radius of the runner of a Francis turbine. Absolute velocity vectors are bold.
cen96537_ch14_793-884.indd 849 29/12/16 4:38 pm 850 TURBOMACHINERY of the velocity as the flow is turned into the draft tube, and there are differ ences in velocity in the circumferential direction as well. It doesn’t take long before you realize that computer simulation tools are enormously useful to turbine designers. In fact, with the help of modern CFD codes, the efficiency of hydroturbines has increased to the point where retrofits of old turbines in hydroelectric plants are economically wise and common. An example CFD output is shown in Fig. 14–98 for a Francis mixed-flow turbine.
ω Reverse swirl FIGURE 14–97 In some Francis mixed-flow turbines, high-power, high-volume flow rate conditions sometimes lead to reverse swirl, in which the flow exiting the runner swirls in the direction opposite to that of the runner itself, as sketched here. FIGURE 14–98 Visualization from CFD analysis of a hydroturbine including the wicket gates, runner, and draft tube. Five of the twenty wicket gates are hidden to better show the runner blades. The flow pathlines, which can be thought of as the paths that single molecules of water take as they flow through the turbine, are shown relative to the angular velocity of each stage and help visualize how the water enters and exits each component of the turbine. The surface pressure contours are colored by static pressure and show areas of high and low pressure on the wicket gates and runner blades. While numerical data of head, flow, and power are extracted from the CFD model during design iterations, this type of visualization is very useful to engineers as a qualitative analysis to identify areas that are susceptible to cavitation damage and unwanted vortices. © American Hydro Corporation. Used by permission.
EXAMPLE 14–12 Effect of Component Efficiencies on Plant Efficiency A hydroelectric power plant is being designed. The gross head from the reservoir to the tailrace is 1065 ft, and the volume flow rate of water through each turbine is 203,000 gpm at 70°F. There are 12 identical parallel turbines, each with an efficiency of 95.2 percent, and all other mechanical energy losses (through the penstock, etc.) are estimated to reduce the output by 3.5 percent. The generator itself has an efficiency of 94.5 percent. Estimate the electric power production from the plant in MW.
SOLUTION We are to estimate the power production from a hydroelectric plant.
Properties The density of water at T = 70°F is 62.30 lbm/ft3.
cen96537_ch14_793-884.indd 850 29/12/16 4:38 pm 851 CHAPTER 14 Analysis The ideal power produced by one hydroturbine is W .
ideal = ρgV .
Hgross = (62.30 lbm/ft3)(32.2 ft/s2)(203,000 gal/min)(1065 ft) × ( lbf·s2 32.2 lbm·ft) (0.1337 ft3 gal) ( 1.356 W ft·lbf/s ) ( 1 min 60 s ) ( 1 MW 106 W) = 40.70 MW But inefficiencies in the turbine, the generator, and the rest of the system reduce the actual electrical power output. For each turbine, W .
electrical = W .
ideal𝜂turbine𝜂generator𝜂other = (40.70 MW)(0.952)(0.945)(1 −0.035) = 35.3 MW Finally, since there are 12 turbines in parallel, the total power produced is W .
total electrical = 12 W .
electrical = 12(35.3 MW) = 424 MW Discussion A small improvement in any of the efficiencies ends up increasing the power output and it thus increases the power company’s profitability.
EXAMPLE 14–13 Hydroturbine Design A retrofit Francis radial-flow hydroturbine is being designed to replace an old turbine in a hydroelectric dam. The new turbine must meet the follow ing design restrictions in order to properly couple with the existing setup: The runner inlet radius is r2 = 8.20 ft (2.50 m) and its outlet radius is r1 = 5.80 ft (1.77 m). The runner blade widths are b2 = 3.00 ft (0.914 m) and b1 = 8.60 ft (2.62 m) at the inlet and outlet, respectively. The runner must rotate at n . = 120 rpm (𝜔 = 12.57 rad/s) to turn the 60-Hz electric generator. The wicket gates turn the flow by angle 𝛼2 = 33° from radial at the runner inlet, and the flow at the runner outlet is to have angle 𝛼1 between −10° and 10° from radial (Fig. 14–99) for proper flow through the draft tube. The volume flow rate at design conditions is 9.50 × 106 gpm (599 m3/s), and the gross head provided by the dam is Hgross = 303 ft (92.4 m). (a) Calculate the inlet and outlet runner blade angles 𝛽2 and 𝛽1, respectively, and predict the power output and required net head if irreversible losses are neglected for the case with 𝛼1 = 10° from radial (with-rotation swirl). (b) Repeat the calcula tions for the case with 𝛼1 = 0° from radial (no swirl). (c) Repeat the calculations for the case with 𝛼1 = −10° from radial (reverse swirl).
SOLUTION For a given set of hydroturbine design criteria we are to calculate runner blade angles, required net head, and power output for three cases—two with swirl and one without swirl at the runner outlet.
Assumptions 1 The flow is steady. 2 The fluid is water at 20°C. 3 The blades are infinitesimally thin. 4 The flow is everywhere tangent to the runner blades. 5 We neglect irreversible losses through the turbine.
Properties For water at 20°C, 𝜌 = 998.0 kg/m3.
α2 α1 r2 r1 Control volume ω V2 V1, n V1, t V2, t V2, n V1 FIGURE 14–99 Top view of the absolute velocities and flow angles associated with the runner of a Francis turbine being designed for a hydroelectric dam (Example 14–13). The control volume is from the inlet to the outlet of the runner.
cen96537_ch14_793-884.indd 851 29/12/16 4:38 pm 852 TURBOMACHINERY Analysis (a) We solve for the normal component of velocity at the inlet using Eq. 14–12, V2, n = V .
2𝜋 r2b2 = 599 m3/s 2𝜋 (2.50 m)(0.914 m) = 41.7 m/s (1) Using Fig. 14–99 as a guide, the tangential velocity component at the inlet is V2, t = V2, n tan 𝛼2 = (41.7 m/s) tan 33° = 27.1 m/s (2) We now solve Eq. 14–45 for the runner leading edge angle 𝛽2, 𝛽2 = arctan( V2, n 𝜔 r2 −V2, t) = arctan ( 41.7 m/s (12.57 rad/s)(2.50 m) −27.1 m/s)= 84.1° (3) Equations 1 through 3 are repeated for the runner outlet, with the following results: Runner outlet: V1, n = 20.6 m/s, V1, t = 3.63 m/s, 𝛽1 = 47.9° (4) The top view of this runner blade is sketched (to scale) in Fig. 14–100. Using Eqs. 2 and 4, the shaft output power is estimated from the Euler turbomachine equation, Eq. 14–39, W .
shaft = ρ𝜔 V · (r2V2, t −r1V1, t) = (998.0 kg/m3)(12.57 rads/s)(599 m3/s) × (2.50 m)(27.2 m/s) −(1.77 m)(3.63 m/s) = 461 MW = 6.18 × 105 hp (5) Finally, we calculate the required net head using Eq. 14–44, assuming that 𝜂turbine = 100 percent since we are ignoring irreversibilities, H = bhp ρgV . = 461 MW (998.0 kg/m3)(9.81 m/s2)(599 m3/s)( 106 kg·m2/s2 MW·s )= 78.6 m (6) (b) When we repeat the calculations with no swirl at the runner outlet (𝛼1 = 0°), the runner blade trailing edge angle reduces to 42.8°, and the output power increases to 509 MW (6.83 × 105 hp). The required net head increases to 86.8 m.
(c) When we repeat the calculations with reverse swirl at the runner outlet (𝛼1 = −10°), the runner blade trailing edge angle reduces to 38.5°, and the output power increases to 557 MW (7.47 × 105 hp). The required net head increases to 95.0 m. A plot of power and net head as a function of runner outlet flow angle 𝛼1 is shown in Fig. 14–101. You can see that both bhp and H increase with decreasing 𝛼1.
Discussion The theoretical output power increases by about 10 percent by elimi nating swirl from the runner outlet and by nearly another 10 percent when there is 10° of reverse swirl. However, the gross head available from the dam is only 92.4 m. Thus, the reverse swirl case of part (c) is clearly impossible, since the predicted net head is required to be greater than Hgross. Keep in mind that this is a preliminary design in which we are neglecting irreversibilities. The actual output power will be lower and the actual required net head will be higher than the values predicted here.
100 40 0 700 600 200 100 0 –20 –10 0 10 20 60 80 20 H, m bhp, MW 300 400 500 α1, degrees bhp H Hgross FIGURE 14–101 Ideal required net head and brake horsepower output as functions of runner outlet flow angle for the turbine of Example 14–13.
ω r1 α1 α2 β1 β2 r2 V2 V1 FIGURE 14–100 Sketch of the runner blade design of Example 14–13, top view. A guide vane and absolute velocity vectors are also shown.
cen96537_ch14_793-884.indd 852 29/12/16 4:39 pm 853 CHAPTER 14 Gas and Steam Turbines Most of our discussion so far has concerned hydroturbines. We now dis cuss turbines that are designed for use with gases, like combustion products or steam. In a coal or nuclear power plant, high-pressure steam is produced by a boiler and then sent to a steam turbine to produce electricity. Because of reheat, regeneration, and other efforts to increase overall efficiency, these steam turbines typically have two sections (high pressure and low pressure), each of which has multiple stages. A stage is defined as a row of stator blades and a row of rotor blades. Most power plant steam turbines are multistage axial-flow devices like that shown in Fig. 14–102. Not shown are the stator vanes (called nozzles) that direct the flow between each set of turbine blades (called buckets). Analysis of axial-flow turbines is very similar to that of axial-flow fans, as discussed in Section 14–2, and is not repeated here. Similar axial-flow turbines are used in jet aircraft engines (Fig. 14–62) and gas turbine generators (Fig. 14–103). A gas turbine generator is similar to a jet engine except that instead of providing thrust, the turbomachine is designed to transfer as much of the fuel’s energy as possible into the rotat ing shaft, which is connected to an electric generator. Gas turbines used for power generation are typically much larger than jet engines, of course, since they are ground-based. As with hydroturbines, a significant gain in effi ciency is realized as overall turbine size increases.
Wind Turbines As global demand for energy increases, the supply of fossil fuels dimin ishes and the price of energy continues to rise. To keep up with global energy demand, renewable sources of energy such as solar, wind, wave, tidal, hydroelectric, and geothermal must be tapped more extensively. In this section we concentrate on wind turbines used to generate electricity. We note the distinction between the terms windmill used for mechanical power generation (grinding grain, pumping water, etc.) and wind turbine used for electrical power generation, although technically both devices are turbines since they extract energy from the fluid. Although the wind is “free” and renewable, modern wind turbines are expensive and suffer from one obvi ous disadvantage compared to most other power generation devices—they produce power only when the wind is blowing, and the power output of a wind turbine is thus inherently unsteady. Furthermore and equally obvi ous is the fact that wind turbines need to be located where the wind blows, which is often far from traditional power grids, requiring construction of new high-voltage power lines. Nevertheless, wind turbines are expected to play an ever-increasing role in the global supply of energy for the foresee able future. Numerous innovative wind turbine designs have been proposed and tested over the centuries as sketched in Fig. 14–104. We generally categorize wind turbines by the orientation of their axis of rotation: horizontal axis wind turbines (HAWTs) and vertical axis wind turbines (VAWTs). An alterna tive way to categorize them is by the mechanism that provides torque to the rotating shaft: lift or drag. So far, none of the VAWT designs or drag-type FIGURE 14–103 The rotor assembly of the MS7001F gas turbine being lowered into the bottom half of the gas turbine casing. Flow is from right to left, with the upstream set of rotor blades (called blades) compris ing the multistage compressor and the downstream set of rotor blades (called buckets) comprising the multistage turbine. Compressor stator blades (called vanes) and turbine stator blades (called nozzles) can be seen in the bottom half of the gas turbine casing. This gas turbine spins at 3600 rpm and produces over 135 MW of power. Courtesy of GE Energy.
FIGURE 14–102 Turbine blades (buckets) in the rotating portion of the low pressure stage of a steam turbine used in power plants. High pressure superheated steam enters at the middle and is split evenly left and right to accommodate the large volume flow rate without requiring an enor mous casing. The curved shapes of the buckets are clearly seen in the stages on the left side. Pressure decreases across each stage, requiring progressively larger buckets in the direction of flow.
© Miss Kanithar Aiumla-OR/Shutterstock RF Much of the material for this section is condensed from Manwell et al. (2010), and the authors acknowledge Professors J. F. Manwell, J. G. McGowan, and A. L. Rogers for their help in reviewing this section.
cen96537_ch14_793-884.indd 853 29/12/16 4:39 pm 854 TURBOMACHINERY Horizontal axis turbines Single bladed Up-wind Multi-rotor Cross-wind Savonius Cross-wind paddles Diffuser Concentrator Unconfined vortex Counter-rotating blades Down-wind Enfield-Andreau Sail wing Three bladed U.S. farm windmill multi-bladed Bicycle multi-bladed Double bladed cen96537_ch14_793-884.indd 854 29/12/16 4:39 pm 855 CHAPTER 14 FIGURE 14–104 Various wind turbine designs and their categorization. Adapted from Manwell et al. (2010).
Primarily drag-type Savonius Plates Cupped Shield Multi-bladed Savonius Primarily lift-type Combinations Others - Darrieus Savonius / - Darrieus Split Savonius Deflector Sunlight Venturi Confined Vortex Magnus Airfoil ∆ - Darrieus Giromill Turbine ϕ ϕ Vertical axis turbines cen96537_ch14_793-884.indd 855 29/12/16 4:39 pm 856 TURBOMACHINERY designs has achieved the efficiency or success of the lift-type HAWT. This is why the vast majority of wind turbines being built around the world are of this type, often in clusters affectionately called wind farms (Fig. 14–105). For this reason, the lift-type HAWT is the only type of wind turbine dis cussed in any detail in this section. [See Manwell et al. (2010) for a detailed discussion as to why drag-type devices have inherently lower efficiency than lift-type devices.] Every wind turbine has a characteristic power performance curve; a typi cal one is sketched in Fig. 14–106, in which electrical power output is plot ted as a function of wind speed V at the height of the turbine’s axis. We identify three key locations on the wind-speed scale: • Cut-in speed is the minimum wind speed at which useful power can be generated.
• Rated speed is the wind speed that delivers the rated power, usually the maximum power.
• Cut-out speed is the maximum wind speed at which the wind turbine is designed to produce power. At wind speeds greater than the cut-out speed, the turbine blades are stopped by some type of braking mecha nism to avoid damage and for safety issues. The short section of dashed red line indicates the power that would be produced if cut-out were not implemented. The design of HAWT turbine blades includes tapering and twist to maxi mize performance and is similar to the design of axial flow fans (propel lers), as discussed in Section 14–2 and is not repeated here. The design of turbine blade twist, for example, is nearly identical to the design of propel ler blade twist, as in Example 14–7, and the blade pitch angle decreases from hub to tip in much the same manner as that of a propeller. While the fluid mechanics of wind turbine design is critical, the power performance curve also is influenced by the electrical generator, the gearbox, and struc tural issues. Inefficiencies appear in every component of course, as in all machines. We define the disk area A of a wind turbine as the area normal to the wind direction swept out by the turbine blades as they rotate (Fig. 14–107). The available wind power W .
available in the disk area is calculated as the rate of change of kinetic energy of the wind, W .
available = d(1 2mV 2) dt = 1 2 V 2 dm dt = 1 2 V 2 m · = 1 2 V 2ρVA = 1 2 ρV 3A (14–47) We notice immediately that the available wind power is proportional to the disk area—doubling the turbine blade diameter exposes the wind turbine to four times as much available wind power. For comparison of various wind turbines and locations, it is more useful to think in terms of the available wind power per unit area, which we call the wind power density, typically in units of W/m2, Wind power density: W .
available A = 1 2 ρV 3 (14–48) (a) (b) FIGURE 14–105 (a) Wind farms are popping up all over the world to help reduce the global demand for fossil fuels. (b) Some wind turbines are even being installed on buildings! (These three turbines are on a building at the Bahrain World Trade Center.) (a) © Digital Vision/Punchstock RF (b) © Adam Jan/AFP/Getty Images cen96537_ch14_793-884.indd 856 29/12/16 4:39 pm 857 CHAPTER 14 Thus, • The wind power density is directly proportional to air density—cold air has a larger wind power density than warm air blowing at the same speed, although this effect is not as significant as wind speed. • The wind power density is proportional to the cube of the wind speed— doubling the wind speed increases the wind power density by a factor of 8. It should be obvious then why wind farms are located where the wind speed is high!
Equation 14–48 is an instantaneous equation. As we all know, however, wind speed varies greatly throughout the day and throughout the year. For this reason, it is useful to define the average wind power density in terms of annual average wind speed V, based on hourly averages as Average wind power density: W .
available A = 1 2 ρavgV 3Ke (14–49) where Ke is a correction factor called the energy pattern factor. In prin ciple, it is analogous to the kinetic energy factor 𝛼 that we use in control volume analyses (Chap. 5). Ke is defined as Ke = 1 NV 3 ∑ N i=1 Vi 3 (14–50) where N = 8760, which is the number of hours in a year. As a general rule of thumb, a location is considered poor for construction of wind turbines if the average wind power density is less than about 100 W/m2, good if it is around 400 W/m2, and great if it is greater than about 700 W/m2. Other factors affect the choice of a wind turbine site, such as atmospheric tur bulence intensity, terrain, obstacles (buildings, trees, etc.), environmental impact, etc. See Manwell et al. (2010) for further details. For analysis purposes, we consider a given wind speed V and define the aerodynamic efficiency of a wind turbine as the fraction of available wind power that is extracted by the turbine blades. This efficiency is commonly called the power coefficient, CP, Power coefficient: Cp = W .
rotor shaft output W .
available = W .
rotor shaft output 1 2 ρV 3 A (14–51) It is fairly simple to calculate the maximum possible power coefficient for a wind turbine, and this was first done by Albert Betz (1885–1968) in the mid 1920s. We consider two control volumes surrounding the disk area—a large control volume and a small control volume—as sketched in Fig. 14–108, with upstream wind speed V taken as V1. The axisymmetric stream tube (enclosed by streamlines as drawn on the top and bottom of Fig. 14–108) can be thought of as forming an imaginary “duct” for the flow of air through the turbine. The control volume momen tum equation for the large control volume for steady flow is ∑F →= ∑ out 𝛽m · V →−∑ in 𝛽m · V → FIGURE 14–106 Typical qualitative wind-turbine power performance curve with definitions of cut-in, rated, and cut-out speeds.
Cut-in speed Rated speed Cut-out speed Wind speed, V • Welectrical FIGURE 14–107 The disk area of a wind turbine is defined as the swept area or frontal area of the turbine as “seen” by the oncoming wind, as sketched here in red. The disk area is (a) circular for a horizontal axis turbine and (b) diamond-shaped for a vertical axis turbine.
(a) © Construction Photography/Corbis RF (b) © Doug Sherman/GeoFile RF (a) (b) cen96537_ch14_793-884.indd 857 29/12/16 4:39 pm 858 TURBOMACHINERY and is analyzed in the streamwise (x) direction. Since locations 1 and 2 are sufficiently far from the turbine, we take P1 = P2 = Patm, yielding no net pressure force on the control volume. We approximate the velocities at the inlet (1) and outlet (2) to be uniform at V1 and V2, respectively; and the momentum flux correction factors are thus 𝛽1 = 𝛽2 = 1. The momentum equation reduces to FR = m ·V2 −m · V1 = m · (V2 −V1) (14–52) The smaller control volume in Fig. 14–108 encloses the turbine, but A3 = A4 = A, since this control volume is infinitesimally thin in the limit (we approximate the turbine as a disk). Since the air is considered to be incompressible, V3 = V4. However, the wind turbine extracts energy from the air, causing a pressure drop. Thus, P3 ≠ P4. When we apply the stream wise component of the control volume momentum equation on the small control volume, we get FR + P3A −P4A = 0 → FR = (P4 −P3)A (14–53) The Bernoulli equation is certainly not applicable across the turbine, since it is extracting energy from the air. However, it is a reasonable approximation between locations 1 and 3 and between locations 4 and 2: P1 ρg + V1 2 2g + z1 = P3 ρg + V3 2 2g + z3 and P4 ρg + V4 2 2g + z4 = P2 ρg + V2 2 2g + z2 In this ideal analysis, the pressure starts at atmospheric pressure far upstream (P1 = Patm), rises smoothly from P1 to P3, drops suddenly from P3 to P4 across the turbine disk, and then rises smoothly from P4 to P2, ending at atmospheric pressure far downstream (P2 = Patm) (Fig. 14–109). We add Eqs. 14–52 and 14–53, setting P1 = P2 = Patm and V3 = V4. In addition, since the wind turbine is horizontally inclined, z1 = z2 = z3 = z4 (gravitational effects are negligible in air anyway). After some algebra, this yields V1 2 −V2 2 2 = P3 −P4 ρ (14–54) Substituting m · = ρV3A into Eq. 14–52 and then combining the result with Eqs. 14–53 and 14–54 yields V3 = V1 + V2 2 (14–55) Thus, we conclude that the average velocity of the air through an ideal wind turbine is the arithmetic average of the far upstream and far downstream velocities. Of course, the validity of this result is limited by the applicability of the Bernoulli equation. For convenience, we define a new variable a as the fractional loss of velocity from far upstream to the turbine disk as a = V1 −V3 V1 (14–56) FIGURE 14–108 The large and small control volumes for analysis of ideal wind turbine performance bounded by an axisym metric diverging stream tube.
V2 V1 V3 V4 FR Patm Patm A 1 = Streamline Streamline Wind turbine 3 4 2 FIGURE 14–109 Qualitative sketch of average stream wise velocity and pressure profiles through a wind turbine.
V or P V P 0 0 Streamwise distance, x Patm 1 4 2 Turbine disk location 3 cen96537_ch14_793-884.indd 858 29/12/16 4:39 pm 859 CHAPTER 14 The velocity through the turbine thus becomes V3 = V1(1 − a), and the mass flow rate through the turbine becomes m · = 𝜌AV3 = 𝜌AV1(1 − a). Combining this expression for V3 with Eq. 14–55 yields V2 = V1(1 −2a) (14–57) For an ideal wind turbine without irreversible losses such as friction, the power generated by the turbine is simply the difference between the incom ing and outgoing kinetic energies. Performing some algebra, we get W .
ideal = m · V1 2 −V2 2 2 = ρAV1(1 −a) V1 2 −V1 2(1 −2a)2 2 = 2ρAV1 3a(1 −a)2 (14–58) Again assuming no irreversible losses in transferring power from the turbine to the turbine shaft, the efficiency of the wind turbine is expressed as the power coefficient defined in Eq. 14–51 as CP = W .
rotor shaft output 1 2 ρV 1 3A = W .
ideal 1 2 ρV 1 3A = 2 ρAV 1 3a(1 −a)2 1 2 ρV 1 3A = 4a(1 −a)2 (14–59) Finally, as any good engineer knows, we calculate the maximum possible value of CP by setting dCP /da = 0 and solving for a (Fig. 14–110). This yields a = 1 or 1/3, and the details are left as an exercise. Since a = 1 is the trivial case (no power generated), we conclude that a must equal 1/3 for maximum possible power coefficient. Substituting a = 1/3 into Eq. 14–59 gives CP, max = 41 3(1 −1 3) 2 = 16 27 ≅0.5926 (14–60) This value of CP, max represents the maximum possible power coefficient of any wind turbine and is known as the Betz limit. All real wind turbines have a maximum achievable power coefficient less than this due to irrevers ible losses which have been ignored in this ideal analysis. Figure 14–111 shows power coefficient CP as a function of the ratio of turbine blade tip speed 𝜔R to wind speed V for several types of wind turbines, where 𝜔 is the angular velocity of the wind turbine blades and R is their radius. From this plot, we see that an ideal propeller-type wind turbine approaches the Betz limit as 𝜔R/V approaches infinity. However, the power coefficient of real wind turbines reaches a maximum at some finite value of 𝜔R/V and then drops beyond that. In practice, three primary effects lead to a maximum achievable power coefficient that is lower than the Betz limit: • Rotation of the wake behind the rotor (swirl) • Finite number of rotor blades and their associated tip losses (tip vortices are generated in the wake of rotor blades for the same reason they are gen erated on finite airplane wings since both produce “lift”) (see Chap. 11) • Nonzero aerodynamic drag on the rotor blades (frictional drag as well as induced drag—see Chap. 11) Day 1, Lesson 1 To find the max or min of y(x), set dy/dx = 0 and solve for x.
FIGURE 14–110 The use of derivatives to calculate minima or maxima is one of the first things that engineers learn.
cen96537_ch14_793-884.indd 859 29/12/16 4:39 pm 860 TURBOMACHINERY See Manwell et al. (2010) for further discussion about how to account for these losses. In addition, mechanical losses due to shaft friction lead to even lower maximum achievable power coefficients. Other mechanical and electrical losses in the gearbox, generator, etc., also reduce the overall wind turbine efficiency, as previously mentioned. As seen in Fig. 14–111, the “best” wind turbine is the high-speed HAWT, and that is why you see this type of wind turbine being installed throughout the world. In summary, wind tur bines provide a “green” alternative to fossil fuels, and as the price of fossil fuels rises, wind turbines will become more commonplace.
FIGURE 14–111 Performance (power coefficient) of various types of wind turbines as a function of the ratio of turbine blade tip speed to wind speed. So far, no design has achieved better performance than the horizontal axis wind turbine (HAWT). Adapted from Robinson (1981, Ref. 10). 0.6 0.5 0.4 Power coefficient, CP Ideal, propeller type High-speed HAWT Darrieus VAWT Savonius rotor American multiblade Dutch, four arm Betz limit: CP = 0.5926 0.3 0.2 0.1 0 1 2 3 4 5 6 7 8 Turbine blade tip speed/wind speed, ωR/V EXAMPLE 14–14 Power Generation in a Wind Farm The average wind speed at a proposed HAWT wind farm site is 12.5 m/s (Fig. 14–112). The power coefficient of each wind turbine is predicted to be 0.41, and the combined efficiency of the gearbox and generator is 92 percent. Each wind turbine must produce 2.5 MW of electrical power when the wind blows at 12.5 m/s. (a) Calculate the required diameter of each turbine disk. Take the average air den sity to be 𝜌 = 1.2 kg/m3. (b) If 30 such turbines are built on the site and an average home in the area con sumes approximately 1.5 kW of electrical power, estimate how many homes can be powered by this wind farm, assuming an additional efficiency of 96 percent to account for the powerline losses.
SOLUTION We are to estimate the required disk diameter of a wind turbine and how many homes a wind farm can serve.
Assumptions 1 The power coefficient is 0.41 and the combined gearbox/ generator efficiency is 0.92. The power distribution system has an efficiency of 96 percent.
FIGURE 14–112 A wind farm site.
© T.W. van Urk/Getty Images RF cen96537_ch14_793-884.indd 860 29/12/16 4:39 pm 861 CHAPTER 14 14–5 ■ TURBINE SCALING LAWS Dimensionless Turbine Parameters We define dimensionless groups (Pi groups) for turbines in much the same way as we did in Section 14–3 for pumps. Neglecting Reynolds number and roughness effects, we deal with the same dimensional variables: gravity times net head (gH), volume flow rate (V . ), some characteristic diameter of the tur bine (D), runner rotational speed (𝜔), output brake horsepower (bhp), and fluid density (𝜌), as illustrated in Fig. 14–113. In fact, the dimensional analy sis is identical whether analyzing a pump or a turbine, except for the fact that bhp Drunner Ddischarge H = net head ω ρ V ∙ FIGURE 14–113 The main variables used for dimensional analysis of a turbine. The characteristic turbine diameter D is typically either the runner diameter Drunner or the discharge diameter Ddischarge. Properties The air density is given as 1.2 kg/m3.
Analysis (a) From the definition of power coefficient, W .
rotor shaft output = CP 1 2 ρV 3 A = CP 1 2 ρV 3(𝜋 D2/4) But the actual electrical power produced is lower than this because of gearbox and generator inefficiencies, W .
electrical output = 𝜂gearbox/generator CP𝜋 ρV 3D2 8 which we solve for diameter, D = √8 W .
electrical output 𝜂generatorCP𝜋 ρV 3 = √ 8 2.5 × 106 W( N·m/s W ) ( kg·m/s2 N ) (0.41)(0.92) 𝜋 (1.2 kg m3) (12.5 m s ) 3 = 84.86 m ≅85 m (b) For 30 machines, the total electrical power produced is 30(2.5 MW) = 75 MW However, some of that is lost (wasted – turned into heat) due to inefficiencies in the power distribution system. The electrical power that actually makes it to people’s home is thus (𝜂power distribution system)(total power) = (0.96)(75 MW) = 72 MW Since an average home consumes power at a rate of 1500 W, the number of homes served by this wind farm is calculated as Number of homes = (𝜂power distribution system)(number of turbines)(W .
electrical output per turbine) W .
electrical usage per home = (0.96)(30 turbines)(2.5 × 106 W/turbine) 1.5 × 103 W/home = 4.8 × 104 homes = 48,000 homes Discussion We give the final answers to two significant digits since we cannot expect any better than that. Wind farms of this size and larger are being constructed throughout the world.
cen96537_ch14_793-884.indd 861 29/12/16 4:39 pm 862 TURBOMACHINERY for turbines, we take bhp instead of V . as the independent variable. In addition, 𝜂turbine (Eq. 14–44) is used in place of 𝜂pump as the non dimensional efficiency. A summary of the dimensionless parameters is provided here: Dimensionless turbine parameters: CH = Head coefficient = gH 𝜔 2D2 CQ = Capacity coefficient = V .
𝜔 D3 (14–61) CP = Power coefficient = bhp ρ𝜔 3D5 𝜂turbine = Turbine efficiency = bhp ρgHV .
When plotting turbine performance curves, we use CP instead of CQ as the independent parameter. In other words, CH and CQ are functions of CP, and 𝜂turbine is thus also a function of CP, since 𝜂turbine = CP CQCH = function of CP (14–62) The affinity laws (Eqs. 14–38) can be applied to turbines as well as to pumps, allowing us to scale turbines up or down in size (Fig. 14–114). We also use the affinity laws to predict the performance of a given turbine operating at different speeds and flow rates in the same way as we did pre viously for pumps. The simple similarity laws are strictly valid only if the model and the pro totype operate at identical Reynolds numbers and are exactly geometrically similar (including relative surface roughness and tip clearance). Unfortu nately, it is not always possible to satisfy all these criteria when performing model tests, because the Reynolds number achievable in the model tests is generally much smaller than that of the prototype, and the model surfaces have larger relative roughness and tip clearances. When the full-scale pro totype is significantly larger than its model, the prototype’s performance is generally better, for the same reasons discussed previously for pumps. Some empirical equations have been developed to account for the increase in effi ciency between a small model and a full-scale prototype. One such equation was suggested by Moody (1926), and can be used as a first-order correction, Moody efficiency correction equation for turbines: 𝜂turbine, prototype ≅1 −(1 −𝜂turbine, model)( Dmodel Dprototype) 1/5 (14–63) Note that Eq. 14–63 is also used as a first-order correction when scaling model pumps to full scale (Eq. 14–34). In practice, hydroturbine engineers generally find that the actual increase in efficiency from model to prototype is only about two-thirds of the increase given by Eq. 14–63. For example, suppose the efficiency of a one-tenth scale model is 93.2 percent. Equation 14–63 predicts a full-scale efficiency of 95.7 percent, or an increase of 2.5 percent. In practice, we expect only about two-thirds of this increase, or 93.2 + 2.5(2/3) = 94.9 percent. Some more advanced correction equations are available from the International Electrotechnical Commission (IEC), a worldwide organization for standardization.
bhpB HB = net head ωB ρB DB bhpA Turbine A Turbine B HA = net head ωA ρA VA DA ∙ VB ∙ FIGURE 14–114 Dimensional analysis is useful for scaling two geometrically similar turbines. If all the dimensionless turbine parameters of turbine A are equivalent to those of turbine B, the two turbines are dynamically similar.
cen96537_ch14_793-884.indd 862 29/12/16 4:39 pm 863 CHAPTER 14 CH, A = CH, B = = 1.11 gH ω2D2 CP, A = CP, B = = 3.38 bhp ρω3D5 CQ, A = CQ, B = = 3.23 ωD3 ηturbine, A = ηturbine, B = = 94.2% bhp ρgH V ⋅ V ⋅ FIGURE 14–115 Dimensionless turbine parameters for both turbines of Example 14–15. Since the two turbines operate at homologous points, their dimensionless parameters must match.
EXAMPLE 14–15 Application of Turbine Affinity Laws A Francis turbine is being designed for a hydroelectric dam. Instead of starting from scratch, the engineers decide to geometrically scale up a previously designed hydroturbine that has an excellent performance history. The existing turbine (turbine A) has diameter DA = 2.05 m, and spins at n .
A = 120 rpm (𝜔A = 12.57 rad/s). At its best efficiency point, V .
A = 350 m3/s, HA = 75.0 m of water, and bhpA = 242 MW. The new turbine (turbine B) is for a larger facility. Its generator will spin at the same speed (120 rpm), but its net head will be higher (HB = 104 m). Calculate the diameter of the new turbine such that it operates most efficiently, and calculate V .
B, bhpB, and 𝜂turbine, B.
SOLUTION We are to design a new hydroturbine by scaling up an existing hydroturbine. Specifically we are to calculate the new turbine diameter, volume flow rate, and brake horsepower.
Assumptions 1 The new turbine is geometrically similar to the existing turbine. 2 Reynolds number effects and roughness effects are negligible. 3 The new pen stock is also geometrically similar to the existing penstock so that the flow entering the new turbine (velocity profile, turbulence intensity, etc.) is similar to that of the existing turbine.
Properties The density of water at 20°C is 𝜌 = 998.0 kg/m3.
Analysis Since the new turbine (B) is dynamically similar to the existing turbine (A), we are concerned with only one particular homologous operating point of both turbines, namely, the best efficiency point. We solve Eq. 14–38b for DB, DB = DA√ HB HA n · A n · B = (2.05 m) √ 104 m 75.0 m 120 rpm 120 rpm = 2.41 m We then solve Eq. 14–38a for V .
B, V .
B = V .
A( n · B n · A) ( DB DA) 3 = (350 m3/s)( 120 rpm 120 rpm) ( 2.41 m 2.05 m) 3 = 572 m3/s Finally, we solve Eq. 14–38c for bhpB, bhpB = bhpA( ρB ρA) ( n · B n · A) 3 ( DB DA) 5 = (242 MW)( 998.0 kg/m3 998.0 kg/m3) ( 120 rpm 120 rpm) 3 ( 2.41 m 2.05 m) 5 = 548 MW As a check, we calculate the dimensionless turbine parameters of Eq. 14–61 for both turbines to show that these two operating points are indeed homologous, and the turbine efficiency is calculated to be 0.942 for both turbines (Fig. 14–115). As discussed previously, however, total dynamic similarity may not actually be achieved between the two turbines because of scale effects (larger turbines gener ally have higher efficiency). The diameter of the new turbine is about 18 percent greater than that of the existing turbine, so the increase in efficiency due to turbine size should not be significant. We verify this by using the Moody efficiency cor rection equation (Eq. 14–63), considering turbine A as the “model” and B as the “prototype,” Efficiency correction: 𝜂turbine, B ≅1 −(1 −𝜂turbine, A)( DA DB) 1/5 = 1 −(1 −0.942)( 2.05 m 2.41 m ) 1/5 = 0.944 cen96537_ch14_793-884.indd 863 29/12/16 4:39 pm 864 TURBOMACHINERY Turbine Specific Speed In our discussion of pump scaling laws (Section 14–3), we defined another useful dimensionless parameter, pump specific speed (NSp), based on CQ and CH. We could use the same definition of specific speed for turbines, but since CP rather than CQ is the independent dimensionless parameter for tur bines, we define turbine specific speed (NSt) differently, namely, in terms of CP and CH, Turbine specific speed: NSt = CP 1/2 CH 5/4 = (bhp/ρ𝜔 3D5)1/2 (gH/𝜔 2D2)5/4 = 𝜔 (bhp)1/2 ρ1/2(gH)5/4 (14–64) Turbine specific speed is also called power specific speed in some textbooks. It is left as an exercise to compare the definitions of pump specific speed (Eq. 14–35) and turbine specific speed (Eq. 14–64) in order to show that Relationship between NSt and NSp: NSt = NSp√𝜂turbine (14–65) Note that Eq. 14–51 does not apply to a pump running backward as a tur bine or vice versa. There are applications in which the same turbomachine is used as both a pump and a turbine; these devices are appropriately called pump–turbines. For example, a coal or nuclear power plant may pump water to a higher elevation during times of low power demand, and then run that water through the same turbomachine (operating as a turbine) during times of high power demand (Fig. 14–116). Such facilities often take advan tage of natural elevation differences at mountainous sites and can achieve significant gross heads (upward of 1000 ft) without construction of a dam. A photograph of a pump–turbine is shown in Fig. 14–117. Note that there are inefficiencies in the pump–turbine when operating as a pump and also when operating as a turbine. Moreover, since one turboma chine must be designed to operate as both a pump and a turbine, neither 𝜂pump nor 𝜂turbine are as high as they would be for a dedicated pump or tur bine. Nevertheless, the overall efficiency of this type of energy storage is around 80 percent for a well-designed pump–turbine unit. In practice, the pump–turbine may operate at a different flow rate and rpm when it is acting as a turbine compared to when it is acting as a pump, since the best efficiency point of the turbine is not necessarily the same as that of the pump. However, for the simple case in which the flow rate and rpm are the same for both the pump and turbine operations, we use Eqs. 14–35 and 14–64 to compare pump specific speed and turbine specific speed. After some algebra, or 94.4 percent. Indeed, the first-order correction yields a predicted efficiency for the larger turbine that is only a fraction of a percent greater than that of the smaller turbine.
Discussion If the flow entering the new turbine from the penstock were not simi lar to that of the existing turbine (e.g., velocity profile and turbulence intensity), we could not expect exact dynamic similarity.
Motor/generator (acting as a motor) Motor/generator (acting as a generator) (a) (b) Pump–turbine (acting as a turbine) Pump–turbine (acting as a pump) FIGURE 14–116 A pump–turbine is used by some power plants for energy storage: (a) water is pumped by the pump– turbine during periods of low demand for power, and (b) electricity is generated by the pump– turbine during periods of high demand for power.
cen96537_ch14_793-884.indd 864 29/12/16 4:39 pm 865 CHAPTER 14 Pump–turbine specific speed relationship at same flow rate and rpm: NSt = NSp√𝜂turbine ( Hpump Hturbine) 3/4 = NSp(𝜂turbine)5/4(𝜂pump)3/4( bhppump bhpturbine) 3/4 (14–66) We previously discussed some problems with the units of pump specific speed. Unfortunately, these same problems also occur with turbine specific speed. Namely, although NSt is by definition a dimensionless parameter, practicing engineers have grown accustomed to using inconsistent units that transform NSt into a cumbersome dimensional quantity. In the United States, most turbine engineers write the rotational speed in units of rotations per minute (rpm), bhp in units of horsepower, and H in units of feet. Further more, they ignore gravitational constant g and density 𝜌 in the definition of NSt. (The turbine is assumed to operate on earth and the working fluid is assumed to be water.) We define Turbine specific speed, customary U.S. units: NSt, US = (n ·, rpm) (bhp, hp)1/2 (H, ft)5/4 (14–67) There is some discrepancy in the turbomachinery literature over the con versions between the two forms of turbine specific speed. To convert NSt, US to NSt we divide by g5/4 and 𝜌1/2, and then use conversion ratios to cancel all units. We set g = 32.174 ft/s2 and assume water at density 𝜌 = 62.40 lbm/ft3. When done properly by converting 𝜔 to rad/s, the conversion is NSt, US = 43.46NSt or NSt = 0.02301NSt, US. However, some authors convert n . to rotations per second, introducing a factor of 2𝜋 in the conversion, i.e., NSt, US = 273.1NSt or NSt = 0.003662NSt, US. The former conversion is more common and is summarized in Fig. 14–118. There is also a metric or SI version of turbine specific speed that is becoming more popular these days and is preferred by many hydroturbine designers. It is defined in the same way as the customary U.S. pump specific FIGURE 14–117 The runner of a pump–turbine used at the Yards Creek pumped storage sta tion in Blairstown, NJ. There are seven runner blades of outer diameter 17.3 ft (5.27 m). The turbine rotates at 240 rpm and produces 112 MW of power at a volume flow rate of 56.6 m3/s from a net head of 221 m.
© American Hydro Corporation. Used by permission.
Conversion ratios NSt = 0.02301 NSt, US NSt, US = 43.46 NSt FIGURE 14–118 Conversions between the dimensionless and the customary U.S. definitions of turbine specific speed. Numerical values are given to four significant digits. The conversions assume earth gravity and water as the working fluid.
cen96537_ch14_793-884.indd 865 29/12/16 4:39 pm 866 TURBOMACHINERY speed (Eq. 14–36), except that SI units are used (m3/s instead of gpm and m instead of ft), NSt, SI = (n ·, rpm)(V .
, m3/s)1/2 (H, m)3/4 (14–68) We may call this capacity specific speed to distinguish it from power specific speed (Eq. 14–64). One advantage is that NSt, SI can be compared more directly to pump specific speed and is thus useful for analyzing pump-turbines. It is less useful, however, to compare NSt, SI to previously published values of NSt or NSt, US because of the fundamental difference between their definitions. Technically, turbine specific speed could be applied at any operating con dition and would just be another function of CP. That is not how it is typi cally used, however. Instead, it is common to define turbine specific speed only at the best efficiency point (BEP) of the turbine. The result is a single number that characterizes the turbine.
Turbine specific speed is used to characterize the operation of a turbine at its optimum conditions (best efficiency point) and is useful for preliminary turbine selection.
As plotted in Fig. 14–119, impulse turbines perform optimally for NSt near 0.15, while Francis turbines and Kaplan or propeller turbines perform best at NSt near 1 and 2.5, respectively. It turns out that if NSt is less than about 0.3, an impulse turbine is the best choice. If NSt is between about 0.3 and 2, a Francis turbine is a better choice. When NSt is greater than about 2, a Kaplan or propeller turbine should be used. These ranges are indicated in Fig. 14–119 in terms of NSt and NSt, US.
EXAMPLE 14–16 Turbine Specific Speed Calculate and compare the turbine specific speed for both the small (A) and large (B) turbines of Example 14–15.
SOLUTION The turbine specific speed of two dynamically similar turbines is to be compared.
Properties The density of water at T = 20°C is 𝜌 = 998.0 kg/m3.
1 0.7 0.5 0.01 NSt NSt, US 10 0.8 0.9 0.6 0.02 0.05 0.1 0.2 0.5 1 1 2 5 10 20 50 100 200 2 5 ηmax Kaplan/propeller Francis Impulse FIGURE 14–119 Maximum efficiency as a function of turbine specific speed for the three main types of dynamic turbine. Horizontal scales show nondimensional turbine specific speed (NSt) and turbine specific speed in customary U.S. units (NSt, US). Sketches of the blade types are also provided on the plot for reference.
cen96537_ch14_793-884.indd 866 29/12/16 4:39 pm 867 CHAPTER 14 Analysis We calculate the dimensionless turbine specific speed for turbine A, NSt, A = 𝜔 A(bhpA)1/2 ρA 1/2(gHA)5/4 = (12.57 rad/s)(242 × 106 W)1/2 (998.0 kg/m3)1/2[(9.81 m/s2)(75.0 m)]5/4( kg·m2/s2 W·s ) 1/2 = 1.615 ≅1.62 and for turbine B, NSt, B = 𝜔 B(bhpB)1/2 ρB 1/2(gHB)5/4 = (12.57 rad/s)(548 × 106 W)1/2 (998.0 kg/m3)1/2 [(9.81 m/s2)(104 m)]5/4( kg·m2/s2 W·s ) 1/2 = 1.615 ≅1.62 We see that the turbine specific speeds of the two turbines are the same. As a check of our algebra we calculate NSt in Fig. 14–120 a different way using its definition in terms of CP and CH (Eq. 14–64). The result is the same (except for roundoff error). Finally, we calculate the turbine specific speed in customary U.S. units from the conversions of Fig. 14–118, NSt, US, A = NSt, US, B = 43.46NSt = (43.46)(1.615) = 70.2 Discussion Since turbines A and B operate at homologous points, it is no surprise that their turbine specific speeds are the same. In fact, if they weren’t the same, it would be a sure sign of an algebraic or calculation error. From Fig. 14–119, a Francis turbine is indeed the appropriate choice for a turbine specific speed of 1.6.
NSt = CP CH = (3.38)1/2 (1.11)5/4 = 1.61 1/2 5/4 Turbine Specific Speed: FIGURE 14–120 Calculation of turbine specific speed using the dimensionless parameters CP and CH for Example 14–16. (See Fig. 14–115 for values of CP and CH for turbine A and turbine B.) Guest Author: Werner J. A. Dahm, The University of Michigan The very high rotation rates at which small gas turbine engines operate, often approaching 100,000 rpm, allow rotary centrifugal atomizers to create the liquid fuel spray that is burned in the combustor. Note that a 10-cm-diameter atomizer rotating at 30,000 rpm imparts 490,000 m/s2 of acceleration (50,000 g) to the liquid fuel, which allows such fuel atomizers to potentially produce very small drop sizes. The actual drop sizes depend on the fluid properties, including the liquid and gas densities 𝜌L and 𝜌G, the viscosities 𝜇L and 𝜇G, and the liquid–gas surface tension 𝜎s. Figure 14–121 shows such a rotary atomizer rotating at rate 𝜔, with radial channels in the rim at nominal radius R ≡ (R1 + R2)/2. Fuel flows into the channels due to the acceleration R𝜔2 and forms a liquid film on the channel walls. The large acceleration leads to a typical film thickness t of only about 10 μm. The channel shape is chosen to produce desirable atomization perfor mance. For a given shape, the resulting drop sizes depend on the cross-flow velocity Vc ≡ R𝜔 into which the film issues at the channel exit, together with the liquid and gas properties. From these, there are four dimensionless groups APPLICATION SPOTLIGHT ■ Rotary Fuel Atomizers (a) (b) t d t d Fuel: ρL, μL ω R2 R1 FIGURE 14–121 Schematic diagram of (a) a rotary fuel atomizer, and (b) a close-up of the liquid fuel film along the channel walls.
Reprinted by permission of Werner J.A. Dahm, University of Michigan.
cen96537_ch14_793-884.indd 867 29/12/16 4:39 pm 868 TURBOMACHINERY that determine the atomization performance: the liquid–gas density and viscosity ratios r ≡ [𝜌L/𝜌G] and m ≡ [𝜇L/𝜇G], the film Weber number Wet ≡ [𝜌GVc 2t/𝜎s], and the Ohnesorge number Oht ≡ [𝜇L/(𝜌L𝜎st)1/2]. Note that Wet gives the characteristic ratio of the aerodynamic forces that the gas exerts on the liquid film to the surface tension forces that act on the liquid surface, while Oht gives the ratio of the viscous forces in the liquid film to the surface tension forces that act on the film. Together these express the relative importance of the three main physical effects involved in the atomization process: inertia, viscous diffusion, and surface tension. Figure 14–122 shows examples of the resulting liquid breakup process for sev eral channel shapes and rotation rates, visualized using 10-ns pulsed-laser pho tography. The drop sizes turn out to be relatively insensitive to changes in the Ohnesorge number, since the values for practical fuel atomizers are in the limit Oht ≪ 1 and thus viscous effects are relatively unimportant. The Weber number, however, remains crucial since surface tension and inertia effects dominate the atomization process. At small Wet, the liquid undergoes subcritical breakup in which surface tension pulls the thin liquid film into a single column that subse quently breaks up to form relatively large drops. At supercritical values of Wet, the thin liquid film breaks up aerodynamically into fine drop sizes on the order of the film thickness t. From results such as these, engineers can successfully develop rotary fuel atomizers for practical applications.
Reference Dahm, W. J. A., Patel, P. R., and Lerg, B. H., “Visualization and Fundamental Analysis of Liquid Atomization by Fuel Slingers in Small Gas Turbines,” AIAA Paper No. 2002-3183, AIAA, Washington, DC, 2002.
FIGURE 14–122 Visualizations of liquid breakup by rotary fuel atomizers, showing subcritical breakup at relatively low values of Wet (top), for which surface tension effects are sufficiently strong relative to inertia to pull the thin liquid film into large columns; and supercrit ical breakup at higher values of Wet (bottom), for which inertia dominates over surface tension and the thin film breaks into fine droplets.
Reprinted by permission of Werner J. A. Dahm, University of Michigan.
SUMMARY We classify turbomachinery into two broad categories, pumps and turbines. The word pump is a general term for any fluid machine that adds energy to a fluid. We explain how this energy transfer occurs for several types of pump designs— both positive-displacement pumps and dynamic pumps. The word turbine refers to a fluid machine that extracts energy from a fluid. There are also positive-displacement turbines and dynamic turbines of several varieties. The most useful equation for preliminary turbomachinery design is the Euler turbomachine equation, Tshaft = ρV .
(r2V2, t −r1V1, t) Note that for pumps, the inlet and outlet are at radii r1 and r2, respectively, while for turbines, the inlet is at radius r2 and the outlet is at radius r1. We show several examples where blade shapes for both pumps and turbines are designed based on desired flow velocities. Then, using the Euler turbomachine equation, the performance of the turbomachine is predicted. The turbomachinery scaling laws illustrate a practical application of dimensional analysis. The scaling laws are used in the design of new turbomachines that are geometri cally similar to existing turbomachines. For both pumps and turbines, the main dimensionless parameters are head coef ficient, capacity coefficient, and power coefficient, defined respectively as CH = gH 𝜔 2D2 CQ = V .
𝜔 D3 CP = bhp ρ𝜔 3D5 In addition to these, we define pump efficiency and turbine efficiency as reciprocals of each other, 𝜂pump = W .
water horsepower W .
shaft = ρgV .
H bhp 𝜂turbine = W .
shaft W .
water horsepower = bhp ρgV .
H cen96537_ch14_793-884.indd 868 29/12/16 4:39 pm 869 CHAPTER 14 Finally, two other useful dimensionless parameters called pump specific speed and turbine specific speed are defined, respectively, as NSp = CQ 1/2 CH 3/4 = 𝜔 V .
1/2 (gH)3/4 NSt = CP 1/2 CH 5/4 = 𝜔 (bhp)1/2 ρ1/2(gH)5/4 These parameters are useful for preliminary design and for selection of the type of pump or turbine that is most appro priate for a given application. We discuss the basic design features of both hydroturbines and wind turbines. For the latter we derive an upper limit to the power coefficient, namely the Betz limit, CP, max = 41 3(1 −1 3) 2 = 16 27 ≅0.5926 Turbomachinery design assimilates knowledge from sev eral key areas of fluid mechanics, including mass, energy, and momentum analysis (Chaps. 5 and 6); dimensional analysis and modeling (Chap. 7); flow in pipes (Chap. 8); differential analysis (Chaps. 9 and 10); and aerodynamics (Chap. 11). In addition, for gas turbines and other types of turbomachines that involve gases, compressible flow analysis (Chap. 12) is required. Finally, computational fluid dynamics (Chap. 15) plays an ever-increasing role in the design of highly efficient turbomachines.
REFERENCES AND SUGGESTED READING 1. ASHRAE (American Society of Heating, Refrigerating and Air Conditioning Engineers, Inc.). ASHRAE Funda mentals Handbook, ASHRAE, 1791 Tullie Circle, NE, Atlanta, GA, 30329; editions every four years: 1993, 1997, 2001, etc.
2. L. F. Moody. “The Propeller Type Turbine,” ASCE Trans., 89, p. 628, 1926.
3. Earl Logan, Jr., ed. Handbook of Turbomachinery. New York: Marcel Dekker, Inc., 1995.
4. A. J. Glassman, ed. Turbine Design and Application. NASA Sp-290, NASA Scientific and Technical Informa tion Program. Washington, DC, 1994.
5. D. Japikse and N. C. Baines. Introduction to Turboma chinery. Norwich, VT: Concepts ETI, Inc., and Oxford: Oxford University Press, 1994.
6. Earl Logan, Jr. Turbomachinery: Basic Theory and Applications, 2nd ed. New York: Marcel Dekker, Inc., 1993.
7. R. K. Turton. Principles of Turbomachinery, 2nd ed. London: Chapman & Hall, 1995.
8. Terry Wright. Fluid Machinery: Application, Selection, and Design. Boca Raton, FL: CRC Press, 2009.
9. J. F. Manwell, J. G. McGowan, and A. L. Rogers. Wind Energy Explained — Theory, Design, and Application, 2nd ed. West Sussex, England: John Wiley & Sons, LTC, 2010.
10. M. L. Robinson. “The Darrieus Wind Turbine for Electrical Power Generation,” J. Royal Aeronautical Society, Vol. 85, pp. 244–255, June 1981.
PROBLEMS General Problems 14–1C What is the more common term for an energy- producing turbomachine? How about an energy-absorbing turbomachine? Explain this terminology. In particular, from which frame of reference are these terms defined—that of the fluid or that of the surroundings?
14–2C What are the primary differences between fans, blowers, and compressors? Discuss in terms of pressure rise and volume flow rate.
14–3C List at least two common examples of fans, of blowers, and of compressors.
14–4C Discuss the primary difference between a positive-displacement turbomachine and a dynamic turbomachine. Give an example of each for both pumps and turbines.
14–5C Explain why there is an “extra” term in the Bernoulli equation in a rotating reference frame.
14–6C For a turbine, discuss the difference between brake horsepower and water horsepower, and also define turbine efficiency in terms of these quantities.
Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and SI users can ignore them. Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software.
cen96537_ch14_793-884.indd 869 29/12/16 4:39 pm 870 TURBOMACHINERY 14–7C For a pump, discuss the difference between brake horsepower and water horsepower, and also define pump efficiency in terms of these quantities.
14–8 A water pump increases the pressure of the water passing through it (Fig. P14–8). The flow is assumed to be incompressible. For each of the three cases listed below, how does average water speed change across the pump? In par ticular, is Vout less than, equal to, or greater than Vin? Show your equations, and explain.
(a) Outlet diameter is less than inlet diameter (Dout < Din) (b) Outlet and inlet diameters are equal (Dout = Din) (c) Outlet diameter is greater than inlet diameter (Dout > Din) Pump P out V out Pin Vin Din Dout FIGURE P14–8 14–9 An air compressor increases the pressure (Pout > Pin) and the density (𝜌out > 𝜌in) of the air passing through it (Fig. P14–9). For the case in which the outlet and inlet diameters are equal (Dout = Din), how does average air speed change across the compressor? In particular, is Vout less than, equal to, or greater than Vin? Explain. Answer: less than FIGURE P14–9 Compressor P out ρout, V out Pin ρin, Vin Din Dout Pumps 14–10C Define net positive suction head and required net positive suction head, and explain how these two quan tities are used to ensure that cavitation does not occur in a pump.
14–11C There are three main categories of dynamic pumps. List and define them.
14–12C For each statement about centrifugal pumps, choose whether the statement is true or false, and discuss your answer briefly: (a) A centrifugal pump with radial blades has higher efficiency than the same pump with backward-inclined blades.
(b) A centrifugal pump with radial blades produces a larger pressure rise than the same pump with backward- or forward-inclined blades over a wide range of V .
.
(c) A centrifugal pump with forward-inclined blades is a good choice when one needs to provide a large pressure rise over a wide range of volume flow rates.
(d) A centrifugal pump with forward-inclined blades would most likely have less blades than a pump of the same size with backward-inclined or radial blades.
14–13C Figure P14–13C shows two possible locations for a water pump in a piping system that pumps water from the lower tank to the upper tank. Which location is better? Why?
Pump Valve Option (a) Option (b) Valve Reservoir Reservoir Pump Valve Valve Reservoir Reservoir FIGURE P14–13C cen96537_ch14_793-884.indd 870 29/12/16 4:39 pm 871 CHAPTER 14 14–14C Consider flow through a water pump. For each statement, choose whether the statement is true or false, and discuss your answer briefly: (a) The faster the flow through the pump, the more likely that cavitation will occur.
(b) As water temperature increases, NPSHrequired also increases.
(c) As water temperature increases, the available NPSH also increases.
(d) As water temperature increases, cavitation is less likely to occur.
14–15C Write the equation that defines actual (available) net positive suction head NPSH. From this definition, dis cuss at least five ways you can decrease the likelihood of cavitation in the pump, for the same liquid, temperature, and volume flow rate.
14–16C Consider a typical centrifugal liquid pump. For each statement, choose whether the statement is true or false, and discuss your answer briefly: (a) V . at the pump’s free delivery is greater than V . at its best efficiency point.
(b) At the pump’s shutoff head, the pump efficiency is zero.
(c) At the pump’s best efficiency point, its net head is at its maximum value.
(d) At the pump’s free delivery, the pump efficiency is zero.
14–17C Explain why it is usually not wise to arrange two (or more) dissimilar pumps in series or in parallel.
14–18C Consider steady, incompressible flow through two identical pumps (pumps 1 and 2), either in series or in paral lel. For each statement, choose whether the statement is true or false, and discuss your answer briefly: (a) The volume flow rate through the two pumps in series is equal to V .
1 + V .
2.
(b) The overall net head across the two pumps in series is equal to H1 + H2.
(c) The volume flow rate through the two pumps in parallel is equal to V .
1 + V .
2.
(d) The overall net head across the two pumps in parallel is equal to H1 + H2.
14–19C In Fig. P14–19C is shown a plot of pump net head as a function of pump volume flow rate, or capacity. On the figure, label the shutoff head, the free delivery, the pump performance curve, the system curve, and the operat ing point.
H 0 0 V ⋅ FIGURE P14–19C 14–20 Suppose the pump of Fig. P14–19C is situated between two water tanks with their free surfaces open to the atmosphere. Which free surface is at a higher elevation—the one corresponding to the tank supplying water to the pump inlet, or the one corresponding to the tank connected to the pump outlet? Justify your answer through use of the energy equation between the two free surfaces.
14–21 Suppose the pump of Fig. P14–19C is situated between two large water tanks with their free surfaces open to the atmosphere. Explain qualitatively what would happen to the pump performance curve if the free surface of the out let tank were raised in elevation, all else being equal. Repeat for the system curve. What would happen to the operating point—would the volume flow rate at the operating point decrease, increase, or remain the same? Indicate the change on a qualitative plot of H versus V .
, and discuss. (Hint: Use the energy equation between the free surface of the tank upstream of the pump and the free surface of the tank down stream of the pump.) 14–22 Suppose the pump of Fig. P14–19C is situated between two large water tanks with their free surfaces open to the atmosphere. Explain qualitatively what would happen to the pump performance curve if a valve in the piping sys tem were changed from 100 percent open to 50 percent open, all else being equal. Repeat for the system curve. What would happen to the operating point—would the volume flow rate at the operating point decrease, increase, or remain the same? Indicate the change on a qualitative plot of H versus V .
, and discuss. (Hint: Use the energy equation between the free sur face of the upstream tank and the free surface of the down stream tank.) Answer: decrease 14–23E A manufacturer of small water pumps lists the per formance data for a family of its pumps as a parabolic curve fit, Havailable = H0 − aV . 2, where H0 is the pump’s shutoff head and a is a coefficient. Both H0 and a are listed in a table for the pump family, along with the pump’s free delivery. The pump head is given in units of feet of water column, and capacity is given in units of gallons per minute. (a) What are the units cen96537_ch14_793-884.indd 871 29/12/16 4:39 pm 872 TURBOMACHINERY of coefficient a? (b) Generate an expression for the pump’s free delivery V . max in terms of H0 and a. (c) Suppose one of the manufacturer’s pumps is used to pump water from one large reservoir to another at a higher elevation. The free sur faces of both reservoirs are exposed to atmospheric pressure. The system curve simplifies to Hrequired = (z2 − z1) + bV . 2. Calculate the operating point of the pump (V . operating and Hoperating) in terms of H0, a, b, and elevation difference z2 − z1.
14–24 Consider the flow system sketched in Fig. P14–24. The fluid is water, and the pump is a centrifugal pump. Gen erate a qualitative plot of the pump net head as a function of the pump capacity. On the figure, label the shutoff head, the free delivery, the pump performance curve, the system curve, and the operating point. (Hint: Carefully consider the required net head at conditions of zero flow rate.) z1 V2 Pump z2 1 2 Reservoir V1 ≅ 0 FIGURE P14–24 14–25 Suppose the pump of Fig. P14–24 is operating at free delivery conditions. The pipe, both upstream and down stream of the pump, has an inner diameter of 2.0 cm and nearly zero roughness. The minor loss coefficient associ ated with the sharp inlet is 0.50, each valve has a minor loss coefficient of 2.4, and each of the three elbows has a minor loss coefficient of 0.90. The contraction at the exit reduces the diameter by a factor of 0.60 (60% of the pipe diameter), and the minor loss coefficient of the contraction is 0.15. Note that this minor loss coefficient is based on the average exit velocity, not the average velocity through the pipe itself. The total length of pipe is 8.75 m, and the elevation difference is (z1 − z2) = 4.6 m. Estimate the volume flow rate through this piping system. Answer: 34.4 Lpm 14–26 Repeat Prob. 14–25, but with a rough pipe—pipe roughness 𝜀 = 0.12 mm. Assume that a modified pump is used, such that the new pump operates at its free delivery conditions, just as in Prob. 14–25. Assume all other dimen sions and parameters are the same as in that problem. Do your results agree with intuition? Explain.
14–27 Consider the piping system of Fig. P14–24, with all the dimensions, parameters, minor loss coef ficients, etc., of Prob. 14–25. The pump’s performance follows a parabolic curve fit, Havailable = H0 − aV . 2, where H0 = 19.8 m is the pump’s shutoff head, and a = 0.00426 m/(Lpm)2 is a coef ficient of the curve fit. Estimate the operating volume flow rate V . in Lpm (liters per minute), and compare with that of Prob. 14–25. Discuss.
14–28 Repeat Prob. 14–27, but instead of a smooth pipe, let the pipe roughness = 0.12 mm. Com pare to the smooth pipe case and discuss—does the result agree with your intuition?
14–29 The performance data for a centrifugal water pump are shown in Table P14–29 for water at 20°C (Lpm = liters per minute). (a) For each row of data, calculate the pump efficiency (percent). Show all units and unit conver sions for full credit. (b) Estimate the volume flow rate (Lpm) and net head (m) at the BEP of the pump.
TABLE P14–29 V .
, Lpm H, m bhp, W 0.0 47.5 133 6.0 46.2 142 12.0 42.5 153 18.0 36.2 164 24.0 26.2 172 30.0 15.0 174 36.0 0.0 174 14–30 For the centrifugal water pump of Prob. 14–29, plot the pump’s performance data: H (m), bhp (W), and 𝜂pump (percent) as functions of V . (Lpm), using symbols only (no lines). Perform linear least-squares polyno mial curve fits for all three parameters, and plot the fitted curves as lines (no symbols) on the same plot. For consis tency, use a first-order curve fit for H as a function of V .
2, use a second-order curve fit for bhp as a function of both V . and V .
2, and use a third-order curve fit for 𝜂pump as a function of V . , V .
2, and V .
3. List all curve-fitted equations and coefficients (with units) for full credit. Calculate the BEP of the pump based on the curve-fitted expressions.
14–31 Suppose the pump of Probs. 14–29 and 14–30 is used in a piping system that has the system requirement Hrequired = (z2 − z1) + bV .
2, where the elevation difference z2 − z1 = 13.2 m, and coefficient b = 0.0185 m/(Lpm)2. Estimate the operating point of the system, namely, V .
operating (Lpm) and Hoperating (m).
cen96537_ch14_793-884.indd 872 29/12/16 4:39 pm 873 CHAPTER 14 14–32 Suppose you are looking into purchasing a water pump with the performance data shown in Table P14–32. Your supervisor asks for some more infor mation about the pump. (a) Estimate the shutoff head H0 and the free delivery V .
max of the pump. [Hint: Perform a least-squares curve fit (regression analysis) of Havailable versus V . 2, and calculate the best-fit values of coefficients H0 and a that translate the tabulated data of Table P14–32 into the para bolic expression, Havailable = H0 − aV . 2. From these coeffi cients, estimate the free delivery of the pump.] (b) The application requires 57.0 Lpm of flow at a pressure rise across the pump of 5.8 psi. Is this pump capable of meeting the requirements? Explain.
TABLE P14–32 V .
, Lpm H, m 20 21 30 18.4 40 14 50 7.6 14–33 The performance data of a water pump follow the curve fit Havailable = H0 − aV . 2, where the pump’s shutoff head H0 = 7.46 m, coefficient a = 0.0453 m/(Lpm)2, the units of pump head H are meters, and the units of V . are liters per min ute (Lpm). The pump is used to pump water from one large res ervoir to another large reservoir at a higher elevation. The free surfaces of both reservoirs are exposed to atmospheric pres sure. The system curve simplifies to Hrequired = (z2 − z1) + bV . 2, where elevation difference z2 − z1 = 3.52 m, and coefficient b = 0.0261 m/(Lpm)2. Calculate the operating point of the pump (V .
operating and Hoperating) in appropriate units (Lpm and meters, respectively). Answers: 7.43 Lpm, 4.96 m 14–34 For the application at hand, the flow rate of Prob. 14–33 is not adequate. At least 9 Lpm is required. Repeat Prob. 14–33 for a more powerful pump with H0 = 8.13 m and a = 0.0297 m/(Lpm)2. Calculate the percentage improve ment in flow rate compared to the original pump. Is this pump able to deliver the required flow rate?
14–35E A water pump is used to pump water from one large reservoir to another large reser voir that is at a higher elevation. The free surfaces of both reservoirs are exposed to atmospheric pres sure, as sketched in Fig. P14–35E. The dimensions and minor loss coefficients are provided in the figure. The pump’s performance is approximated by the expression Havailable = H0 − aV . 2, where the shutoff head H0 = 125 ft of water column, coefficient a = 2.50 ft/(gpm)2, available pump head Havailable is in units of feet of water column, and capacity V . is in units of gallons per minute (gpm). Estimate the capacity delivered by the pump. Answer: 6.34 gpm z2 – z1 z1 V1 ≅ 0 Pump Valve 1 Valve 2 z2 – z1 = 22.0 ft (elevation difference) D = 1.20 in (pipe diameter) KL, entrance = 0.50 (pipe entrance) KL, valve 1 = 2.0 (valve 1) KL, valve 2 = 6.8 (valve 2) KL, elbow = 0.34 (each elbow—there are 3) KL, exit = 1.05 (pipe exit) L = 124 ft (total pipe length) ε = 0.0011 in (pipe roughness) z2 1 Reservoir V2 ≅ 0 2 Reservoir D FIGURE P14–35E 14–36E For the pump and piping system of Prob. 14–35E, plot the required pump head Hrequired (ft of water column) as a function of volume flow rate V . (gpm). On the same plot, com pare the available pump head Havailable versus V .
, and mark the operating point. Discuss.
14–37E Suppose that the two reservoirs in Prob. 14–35E are 1000 ft farther apart horizontally, but at the same eleva tions. All the constants and parameters are identical to those of Prob. 14–35E except that the total pipe length is 1124 ft instead of 124 ft. Calculate the volume flow rate for this case and compare with the result of Prob. 14–35E. Discuss.
14–38E Paul realizes that the pump being used in Prob. 14–35E is not well-matched for this applica tion, since its shutoff head (125 ft) is much larger than its required net head (less than 30 ft), and its capacity is fairly low. In other words, this pump is designed for high-head, low-capac ity applications, whereas the application at hand is fairly low-head, and a higher capacity is desired. Paul tries to convince his supervisor that a less expensive pump, with lower shutoff head but higher free delivery, would result in a significantly increased flow rate between the two reservoirs. Paul looks through some online brochures, and finds a pump with the performance data shown in Table P14–38E. His supervisor asks him to predict the volume flow rate between the two reservoirs if the existing pump were replaced with the new pump. (a) Perform a least-squares curve fit (regression analysis) of Havailable versus V .
2, and cen96537_ch14_793-884.indd 873 29/12/16 4:39 pm 874 TURBOMACHINERY calculate the best-fit values of coefficients H0 and a that trans late the tabulated data of Table P14–38E into the parabolic expression Havailable = H0 − aV . 2. Plot the data points as symbols and the curve fit as a line for comparison. (b) Estimate the operating volume flow rate of the new pump if it were to replace the existing pump, all else being equal. Compare to the result of Prob. 14–35E and discuss. Is Paul correct? (c) Gener ate a plot of required net head and available net head as func tions of volume flow rate and indicate the operating point on the plot.
TABLE P14–38E V .
, gpm H, ft 0 38 4 37 8 34 12 29 16 21 20 12 24 0 14–39 A water pump is used to pump water from one large reservoir to another large reservoir that is at a higher elevation. The free surfaces of both reservoirs are exposed to atmospheric pressure, as sketched in Fig. P14–39. The dimensions and minor loss coefficients are provided in the figure. The pump’s perfor mance is approximated by the expression Havailable = H0 − aV . 2, where shutoff head H0 = 24.4 m of water column, coefficient a = 0.0678 m/(Lpm)2, available pump head Havailable is in units of meters of water column, and capacity V . is in units of liters per minute (Lpm). Estimate the capacity delivered by the pump. Answer: 11.6 Lpm Pump z2 – z1 z1 V1 ≅ 0 Valve z2 – z1 = 7.85 m (elevation difference) D = 2.03 cm (pipe diameter) KL, entrance = 0.50 (pipe entrance) KL, valve = 17.5 (valve) KL, elbow = 0.92 (each elbow—there are 5) KL, exit = 1.05 (pipe exit) L = 176.5 m (total pipe length) ε = 0.25 mm (pipe roughness) z2 1 Reservoir V2 ≅ 0 2 Reservoir D FIGURE P14–39 14–40 For the pump and piping system of Prob. 14–39, plot required pump head Hrequired (m of water column) as a function of volume flow rate V . (Lpm). On the same plot, compare available pump head Havailable versus V .
, and mark the operating point. Discuss.
14–41 Suppose that the free surface of the inlet reservoir in Prob. 14–39 is 3.0 m higher in elevation, such that z2 − z1 = 4.85 m. All the constants and parameters are identical to those of Prob. 14–39 except for the elevation difference. Cal culate the volume flow rate for this case and compare with the result of Prob. 14–39. Discuss.
14–42 April’s supervisor asks her to find a replace ment pump that will increase the flow rate through the piping system of Prob. 14–39 by a factor of 2 or greater. April looks through some online brochures, and finds a pump with the performance data shown in Table P14–42. All dimensions and parameters remain the same as in Prob. 14–39—only the pump is changed. (a) Perform a least-squares curve fit (regression analysis) of Havailable versus V .
2, and calculate the best-fit values of coefficients H0 and a that translate the tabulated data of Table P14–42 into the para bolic expression Havailable = H0 − aV .
2. Plot the data points as symbols and the curve fit as a line for comparison. (b) Use the expression obtained in part (a) to estimate the operating volume flow rate of the new pump if it were to replace the existing pump, all else being equal. Compare to the result of Prob. 14–39 and discuss. Has April achieved her goal? (c) Generate a plot of required net head and available net head as functions of volume flow rate, and indicate the operating point on the plot.
TABLE P14–42 V .
, Lpm H, m 0 46.5 5 46 10 42 15 37 20 29 25 16.5 30 0 14–43 Calculate the volume flow rate between the reser voirs of Prob. 14–39 for the case in which the pipe diameter is doubled, all else remaining the same. Discuss.
14–44 Comparing the results of Probs. 14–39 and 14–43, the volume flow rate increases as expected when one dou bles the inner diameter of the pipe. One might expect that the Reynolds number increases as well. Does it? Explain.
14–45 Repeat Prob. 14–39, but neglect all minor losses. Compare the volume flow rate with that of Prob. 14–39. Are minor losses important in this problem? Discuss.
14–46 Consider the pump and piping system of Prob. 14–39. Suppose that the lower reservoir cen96537_ch14_793-884.indd 874 29/12/16 4:39 pm 875 CHAPTER 14 is huge, and its surface does not change elevation, but the upper reservoir is not so big, and its surface rises slowly as the reservoir fills. Generate a curve of volume flow rate V . (Lpm) as a function of z2 − z1 in the range 0 to the value of z2 − z1 at which the pump ceases to pump any more water. At what value of z2 − z1 does this occur? Is the curve linear? Why or why not? What would happen if z2 − z1 were greater than this value? Explain.
14–47E The performance data for a centrifugal water pump are shown in Table P14–47E for water at 77°F (gpm = gallons per minute). (a) For each row of data, calculate the pump efficiency (percent). Show all units and unit conver sions for full credit. (b) Estimate the volume flow rate (gpm) and net head (ft) at the BEP of the pump.
TABLE P14–47E V .
, gpm H, ft bhp, hp 0.0 19.0 0.06 4.0 18.5 0.064 8.0 17.0 0.069 12.0 14.5 0.074 16.0 10.5 0.079 20.0 6.0 0.08 24.0 0.0 0.078 14–48E Transform each column of the pump performance data of Prob. 14–47E to metric units: V . into Lpm (liters per minute), H into m, and bhp into W. Calculate the pump effi ciency (percent) using these metric values, and compare to that of Prob. 14–47E.
14–49E For the centrifugal water pump of Prob. 14–47E, plot the pump’s performance data: H (ft), bhp (hp), and 𝜂pump (percent) as functions of V . (gpm), using symbols only (no lines). Perform linear least-squares polynomial curve fits for all three parameters, and plot the fitted curves as lines (no symbols) on the same plot. For consistency, use a first-order curve fit for H as a function of V .
2, use a second-order curve fit for bhp as a function of both V . and V .
2, and use a third-order curve fit for 𝜂pump as a function of V .
, V .
2, and V .
3. List all curve- fitted equations and coefficients (with units) for full credit. Cal culate the BEP of the pump based on the curve-fitted expressions.
14–50E Suppose the pump of Probs. 14–47E and 14–49E is used in a piping system that has the system require ment Hrequired = (z2 − z1) + bV .
2, where elevation difference z2 − z1 = 15.5 ft, and coefficient b = 0.00986 ft/(gpm)2. Estimate the operating point of the system, namely, V .
operating (gpm) and Hoperating (ft). Answers: 9.14 gpm, 16.3 ft 14–51 A local ventilation system (a hood and duct system) is used to remove air and contaminants from a pharmaceuti cal lab (Fig. P14–51). The inner diameter (ID) of the duct is D = 150 mm, its average roughness is 0.15 mm, and its total length is L = 24.5 m. There are three elbows along the duct, each with a minor loss coefficient of 0.21. Literature from the hood manufacturer lists the hood entry loss coefficient as 3.3 based on duct velocity. When the damper is fully open, its loss coefficient is 1.8. The minor loss coefficient through the 90° tee is 0.36. Finally, a one-way valve is installed to prevent contaminants from a second hood from flowing “backward” into the room. The minor loss coefficient of the (open) one-way valve is 6.6. The performance data of the fan fit a parabolic curve of the form Havailable = H0 − aV .
2, where shutoff head H0 = 60.0 mm of water column, coefficient a = 2.50 × 10−7 mm of water column per (Lpm)2, available head Havailable is in units of mm of water column, and capacity V . is in units of Lpm of air. Estimate the volume flow rate in Lpm through this ventilation system. Answer: 7090 Lpm T = 25°C P = 1 atm Fan z2 2 One-way valve 90° Tee Branch from another hood Damper Hood z1 1 FIGURE P14–51 14–52 For the duct system of Prob. 14–51, plot required fan head Hrequired (mm of water column) as a function of volume flow rate V . (Lpm). On the same plot, compare available fan head Havailable versus V .
, and mark the operating point. Discuss.
14–53 Repeat Prob. 14–51, ignoring all minor losses. How important are the minor losses in this problem? Discuss.
14–54 Suppose the one-way valve of Fig. P14–51 malfunc tions due to corrosion and is stuck in its fully closed position (no air can get through). The fan is on, and all other condi tions are identical to those of Prob. 14–51. Calculate the gage pressure (in pascals and in mm of water column) at a point just downstream of the fan. Repeat for a point just upstream of the one-way valve.
cen96537_ch14_793-884.indd 875 29/12/16 4:39 pm 876 TURBOMACHINERY 14–55E A local ventilation system (a hood and duct system) is used to remove air and contaminants produced by a welding operation (Fig. P14–55E). The inner diameter (ID) of the duct is D = 9.06 in, its average roughness is 0.0059 in, and its total length is L = 34.0 ft. There are three elbows along the duct, each with a minor loss coefficient of 0.21. Literature from the hood manufacturer lists the hood entry loss coefficient as 4.6 based on duct velocity. When the damper is fully open, its loss coef ficient is 1.8. A squirrel cage centrifugal fan with a 9.0-in inlet is available. Its performance data fit a parabolic curve of the form H available = H0 − aV . 2, where shutoff head H0 = 2.30 inches of water column, coefficient a = 8.50 × 10−6 inches of water column per (SCFM)2, available head Havailable is in units of inches of water column, and capacity V . is in units of standard cubic feet per minute (SCFM, at 77°F). Estimate the volume flow rate in SCFM through this ventilation system. Answer: 452 SCFM Damper Fan Hood 2 z1 1 z2 FIGURE P14–55E 14–56E For the duct system and fan of Prob. 14–55E, partially closing the damper would decrease the flow rate. All else being unchanged, estimate the minor loss coefficient of the damper required to decrease the volume flow rate by a factor of 2.
14–57E Repeat Prob. 14–55E, ignoring all minor losses. How important are the minor losses in this problem? Discuss.
14–58E A centrifugal pump is used to pump water at 77°F from a reservoir whose surface is 20.0 ft above the centerline of the pump inlet (Fig. P14–58E). The piping system consists of 67.5 ft of PVC pipe with an ID of 1.2 in and negligible aver age inner roughness height. The length of pipe from the bot tom of the lower reservoir to the pump inlet is 12.0 ft. There are several minor losses in the piping system: a sharp-edged inlet (KL = 0.5), two flanged smooth 90° regular elbows (KL = 0.3 each), two fully open flanged globe valves (KL = 6.0 each), and an exit loss into the upper reservoir (KL = 1.05). The pump’s required net positive suction head is pro vided by the manufacturer as a curve fit: NPSHrequired = 1.0 ft + [0.0054 ft/(gpm)2]V . 2, where volume flow rate is in gpm. Estimate the maximum volume flow rate (in units of gpm) that can be pumped without cavitation.
Reservoir Reservoir Valve Valve Pump z3 – z1 z3 z1 z2 3 1 2 FIGURE P14–58E 14–59E Repeat Prob. 14–58E, but at a water temperature of 113°F. Discuss.
14–60 A self-priming centrifugal pump is used to pump water at 25°C from a reservoir whose surface is 2.2 m below the centerline of the pump inlet (Fig. P14–60). The pipe is PVC pipe with an ID of 24.0 mm and negligible average inner roughness height. The pipe length from the submerged pipe inlet to the pump inlet is 2.8 m. There are only two minor losses in the piping system from the pipe inlet to the pump inlet: a sharp-edged reentrant inlet (KL = 0.85), and a flanged smooth 90° regular elbow (KL = 0.3). The pump’s required net positive suction head is provided by the manufacturer as a curve fit: NPSHrequired = 2.2 m + [0.0013 m/(Lpm)2]V .
2, where volume flow rate is in Lpm. Estimate the maximum volume flow rate (in units of Lpm) that can be pumped with out cavitation.
Reservoir z2 z1 z2 – z1 1 Pump 2 FIGURE P14–60 cen96537_ch14_793-884.indd 876 29/12/16 4:39 pm 877 CHAPTER 14 14–61 Repeat Prob. 14–60, but at a water temperature of 80°C. Repeat for 90°C. Discuss.
14–62 Repeat Prob. 14–60, but with the pipe diameter increased by a factor of 2 (all else being equal). Does the volume flow rate at which cavitation occurs in the pump increase or decrease with the larger pipe? Discuss.
14–63E The two-lobe rotary pump of Fig. P14–63E moves 0.145 gal of a coal slurry in each lobe volume V lobe. Calculate the volume flow rate of the slurry (in gpm) for the case where n . = 220 rpm. Answer: 128 gpm In Out Vlobe V ⋅ V ⋅ FIGURE P14–63E 14–64E Repeat Prob. 14–63E for the case in which the pump has three lobes on each rotor instead of two, and V lobe = 0.103 gal.
14–65 Consider the gear pump of Fig. 14–26c. Suppose the volume of fluid confined between two gear teeth is 0.350 cm3. How much fluid volume is pumped per rotation? Answer: 9.80 cm3 14–66 A centrifugal pump rotates at n . = 750 rpm. Water enters the impeller normal to the blades (𝛼1 = 0°) and exits at an angle of 35° from radial (𝛼2 = 35°). The inlet radius is r1 = 12.0 cm, at which the blade width b1 = 18.0 cm. The outlet radius is r2 = 24.0 cm, at which the blade width b2 = 16.2 cm. The volume flow rate is 0.573 m3/s. Assuming 100 percent efficiency, calculate the net head produced by this pump in cm of water column height. Also calculate the required brake horsepower in W.
14–67 Suppose the pump of Prob. 14–66 has some swirl at the inlet such that 𝛼1 = 10° instead of 0°. Calculate the net head and required horsepower and compare to Prob. 14–66. Discuss. In particular, is the angle at which the fluid impinges on the impeller blade a critical parameter in the design of centrifugal pumps?
14–68 Suppose the pump of Prob. 14–66 has some reverse swirl at the inlet such that 𝛼1 = −10° instead of 0°. Calcu late the net head and required horsepower and compare to Prob. 14–66. Discuss. In particular, is the angle at which the fluid impinges on the impeller blade a critical parameter in the design of centrifugal pumps? Does a small amount of reverse swirl increase or decrease the net head of the pump— in other words, is it desirable? Note: Keep in mind that we are neglecting losses here.
14–69 A vane-axial flow fan is being designed with the stator blades upstream of the rotor blades (Fig. P14–69). To reduce expenses, both the stator and rotor blades are to be constructed of sheet metal. The stator blade is a simple cir cular arc with its leading edge aligned axially and its trail ing edge at angle 𝛽st = 26.6° from the axis as shown in the sketch. (The subscript notation indicates stator trailing edge.) There are 18 stator blades. At design conditions, the axial-flow speed through the blades is 31.4 m/s, and the impel ler rotates at 1800 rpm. At a radius of 0.50 m, calculate the leading and trailing edge angles of the rotor blade, and sketch the shape of the blade. How many rotor blades should there be?
Rotor ω Stator Vin βst Hub and motor r ωr ? ? ?
Vout FIGURE P14–69 14–70 Two water pumps are arranged in series. The per formance data for both pumps follow the parabolic curve fit Havailable = H0 − aV . 2. For pump 1, H0 = 6.33 m and coef ficient a = 0.0633 m/(Lpm)2; for pump 2, H0 = 9.25 m and coefficient a = 0.0472 m/(Lpm)2. In either case, the units of net pump head H are m, and the units of capacity V . are Lpm. Calculate the combined shutoff head and free deliv ery of the two pumps working together in series. At what volume flow rate should pump 1 be shut off and bypassed? Explain. Answers: 15.6 m, 14.0 Lpm, 10.0 Lpm 14–71 The same two water pumps of Prob. 14–70 are arranged in parallel. Calculate the shutoff head and free delivery of the two pumps working together in parallel. At what combined net head should pump 1 be shut off and bypassed? Explain.
Turbines 14–72C Give at least two reasons why turbines often have greater efficiencies than do pumps.
cen96537_ch14_793-884.indd 877 29/12/16 4:39 pm 878 TURBOMACHINERY 14–73C Name and briefly describe the differences between the two basic types of dynamic turbine.
14–74C Discuss the meaning of reverse swirl in reaction hydroturbines, and explain why some reverse swirl may be desirable. Use an equation to support your answer. Why is it not wise to have too much reverse swirl?
14–75C What is a draft tube, and what is its purpose? Describe what would happen if turbomachinery designers did not pay attention to the design of the draft tube.
14–76C Briefly discuss the main difference in the way that dynamic pumps and reaction turbines are classified as cen trifugal (radial), mixed flow, or axial.
14–77 A hydroelectric plant has 14 identical Francis turbines, a gross head of 245 m, and a volume flow rate of 11.5 m3/s through each turbine. The water is at 25°C. The efficien cies are 𝜂turbine = 95.9%, 𝜂generator = 94.2%, and 𝜂other = 95.6%, where 𝜂other accounts for all other mechanical energy losses. Estimate the electrical power production from this plant in MW.
14–78 A Pelton wheel is used to produce hydroelectric power. The average radius of the wheel is 1.83 m, and the jet velocity is 102 m/s from a nozzle of exit diameter equal to 10.0 cm. The turning angle of the buckets is 𝛽 = 165°. (a) Calculate the volume flow rate through the turbine in m3/s. (b) What is the optimum rotation rate (in rpm) of the wheel (for maximum power)? (c) Calculate the output shaft power in MW if the efficiency of the turbine is 82 percent. Answers: (a) 0.801 m3/s, (b) 266 rpm, (c) 3.35 MW 14–79 Some engineers are evaluating potential sites for a small hydroelectric dam. At one such site, the gross head is 340 m, and they estimate that the volume flow rate of water through each turbine would be 0.95 m3/s. Estimate the ideal power production per turbine in MW.
14–80 Prove that for a given jet speed, volume flow rate, turning angle, and wheel radius, the maximum shaft power produced by a Pelton wheel occurs when the turbine bucket moves at half the jet speed.
14–81 Wind (𝜌 = 1.204 kg/m3) blows through a HAWT wind turbine. The turbine diameter is 60.0 m. The combined efficiency of the gearbox and generator is 88 percent. (a) For a realistic power coefficient of 0.42, estimate the electrical power production when the wind blows at 9.5 m/s. (b) Repeat and compare using the Betz limit, assuming the same gear box and generator. 14–82 A Francis radial-flow hydroturbine is being designed with the following dimensions: r2 = 2.00 m, r1 = 1.42 m, b2 = 0.731 m, and b1 = 2.20 m. The runner rotates at n . = 180 rpm. The wicket gates turn the flow by angle 𝛼2 = 30° from radial at the runner inlet, and the flow at the runner outlet is at angle 𝛼1 = 10° from radial (Fig. P14–82). The volume flow rate at design conditions is 340 m3/s, and the gross head provided by the dam is Hgross = 90.0 m. For the preliminary design, irre versible losses are neglected. Calculate the inlet and outlet run ner blade angles 𝛽2 and 𝛽1, respectively, and predict the power output (MW) and required net head (m). Is the design feasible?
α2 α1 r2 r1 Control volume V 1 ω V1, n V1, t V2, t V2, n V 2 FIGURE P14–82 14–83 Reconsider Prob. 14–82. Using appropriate software, investigate the effect of the runner outlet angle 𝛼1 on the required net head and the output power. Let the outlet angle vary from −20° to 20° in increments of 1°, and plot your results. Determine the minimum possible value of 𝛼1 such that the flow does not violate the laws of thermodynamics.
14–84E A hydroelectric power plant is being designed. The gross head from the reservoir to the tailrace is 735 ft, and the volume flow rate of water through each turbine is 189,400 gpm at 50°F. There are 8 identical parallel turbines, each with an efficiency of 96.3 percent, and all other mechanical energy losses (through the penstock, etc.) are estimated to reduce the output by 3.6 percent. The generator itself has an efficiency of 93.9 percent. Estimate the electric power production from the plant in MW.
14–85E A Francis radial-flow hydroturbine has the following dimensions, where location 2 is the inlet and location 1 is the outlet: r2 = 6.60 ft, r1 = 4.40 ft, b2 = 2.60 ft, and b1 = 7.20 ft. The runner blade angles are 𝛽2 = 82° and 𝛽1 = 46° at the turbine inlet and outlet, respectively. The runner rotates at n . = 120 rpm. The volume flow rate at design conditions is 4.70 × 106 gpm. Irreversible losses are neglected in this preliminary analysis. Calculate the angle 𝛼2 through which the wicket gates should turn the flow, where 𝛼2 is measured from the radial direction at the runner inlet (Fig. P14–82). Calculate the swirl angle 𝛼1, where 𝛼1 is measured from the radial direction at the runner outlet (Fig. P14–82). Does this turbine have forward or reverse swirl? Predict the power output (hp) and required net head (ft).
cen96537_ch14_793-884.indd 878 29/12/16 4:39 pm 879 CHAPTER 14 14–86E Using appropriate software, adjust the runner blade trailing edge angle 𝛽1 of Prob. 14–85E, keeping all other parameters the same, such that there is no swirl at the turbine outlet. Report 𝛽1 and the corresponding shaft power.
14–87 A simple single-stage axial turbine is being designed to produce power from water flowing through a tube as in Fig. P14–87. We approximate both the stator and rotor as thin (bent sheet metal). The 16 stator (upstream) blades have 𝛽sl = 0° and 𝛽st = 50.3°, where subscripts “sl” and “st” mean stator leading edge and stator trailing edge, respectively. At design conditions, the axial flow speed is 8.31 m/s, the rotor turns at 360 rpm, and it is desired that there be no swirl downstream of the turbine. At a radius of 0.324 m, calcu late angles 𝛽rl and 𝛽rt (rotor leading and trailing edge angles), sketch what the rotor vanes should look like, and specify how many rotor vanes there should be.
Rotor ω Stator Vout Hub and Generator r ωr ? ? ?
βst Vin FIGURE P14–87 14–88 In the section on wind turbines, an expression was derived for the ideal power coefficient of a wind turbine, CP = 4a(1 − a)2. Prove that the maximum possible power coefficient occurs when a = 1/3.
14–89 A Francis radial-flow hydroturbine has the following dimensions, where location 2 is the inlet and location 1 is the outlet: r2 = 2.00 m, r1 = 1.30 m, b2 = 0.85 m, and b1 = 2.10 m.
The runner blade angles are 𝛽2 = 71.4° and 𝛽1 = 15.3° at the turbine inlet and outlet, respectively. The runner rotates at n . = 160 rpm. The volume flow rate at design conditions is 80.0 m3/s. Irreversible losses are neglected in this pre liminary analysis. Calculate the angle 𝛼2 through which the wicket gates should turn the flow, where 𝛼2 is measured from the radial direction at the runner inlet (Fig. P14–82). Calcu late the swirl angle 𝛼1, where 𝛼1 is measured from the radial direction at the runner outlet (Fig. P14–82). Does this tur bine have forward or reverse swirl? Predict the power output (MW) and required net head (m).
Pump and Turbine Scaling Laws 14–90C Pump specific speed and turbine specific speed are “extra” parameters that are not necessary in the scaling laws for pumps and turbines. Explain, then, their purpose.
14–91C For each statement, choose whether the statement is true or false, and discuss your answer briefly: (a) If the rpm of a pump is doubled, all else staying the same, the capacity of the pump goes up by a factor of about 2.
(b) If the rpm of a pump is doubled, all else staying the same, the net head of the pump goes up by a factor of about 2.
(c) If the rpm of a pump is doubled, all else staying the same, the required shaft power goes up by a factor of about 4.
(d) If the rpm of a turbine is doubled, all else staying the same, the output shaft power of the turbine goes up by a fac tor of about 8.
14–92C Discuss which dimensionless pump performance parameter is typically used as the independent parameter. Repeat for turbines instead of pumps. Explain.
14–93C Look up the word affinity in a dictionary. Why do you suppose some engineers refer to the turbomachinery scal ing laws as affinity laws?
14–94 Apply the necessary conversion factors to prove the relationship between dimensionless turbine specific speed and customary U.S. turbine specific speed, NSt = 43.46NSt, US. Note that we assume water as the fluid and standard earth gravity.
14–95 Consider the fan of Prob. 14–51. The fan diameter is 30.0 cm, and it operates at n . = 600 rpm. Nondimensional ize the fan performance curve, i.e., plot CH versus CQ. Show sample calculations of CH and CQ at V . = 13,600 Lpm.
14–96 Calculate the fan specific speed of the fan of Probs. 14–51 and 14–95 at the best efficiency point for the case in which the BEP occurs at 13,600 Lpm. Provide answers in both dimensionless form and in customary U.S. units. What kind of fan is it?
14–97 Calculate the pump specific speed of the pump of Example 14–11 at its best efficiency point. Provide answers in both dimensionless form and in customary U.S. units. What kind of pump is it?
14–98 Len is asked to design a small water pump for an aquarium. The pump should deliver 18.0 Lpm of water at a net head of 1.6 m at its best efficiency point. A motor that spins at 1200 rpm is available. What kind of pump (cen trifugal, mixed, or axial) should Len design? Show all your calculations and justify your choice. Estimate the maximum pump efficiency Len can hope for with this pump. Answers: centrifugal, 75% 14–99 Consider the pump of Prob. 14–98. Suppose the pump is modified by attaching a different motor, for which cen96537_ch14_793-884.indd 879 29/12/16 4:39 pm 880 TURBOMACHINERY the rpm is 1800 rpm. If the pumps operate at homologous points (namely, at the BEP) for both cases, predict the vol ume flow rate and net head of the modified pump. Calculate the pump specific speed of the modified pump, and compare to that of the original pump. Discuss.
14–100E A large water pump is being designed for a nuclear reactor. The pump should deliver 2500 gpm of water at a net head of 45 ft at its best efficiency point. A motor that spins at 300 rpm is available. What kind of pump (centrifugal, mixed, or axial) should be designed? Show all your calculations and justify your choice. Estimate the maximum pump efficiency that can be hoped for with this pump. Estimate the power (brake horsepower) required to run the pump.
14–101 Consider the pump of Prob. 14–39. The pump diameter is 1.80 cm, and it operates at n . = 4200 rpm. Nondimensionalize the pump performance curve, i.e., plot CH versus CQ. Show sample calculations of CH and CQ at V . = 14.0 Lpm.
14–102 Calculate the pump specific speed of the pump of Prob. 14–101 at the best efficiency point for the case in which the BEP occurs at 14.0 Lpm. Provide answers in both dimensionless form and in customary U.S. units. What kind of pump is it? Answers: 0.199, 545, centrifugal 14–103 Calculate the turbine specific speed of the turbine in Prob. 14–82. Provide answers in both dimensionless form and in customary U.S. units. Is it in the normal range for a Francis turbine? If not, what type of turbine would be more appropriate?
14–104 Calculate the turbine specific speed of the Warwick hydroturbine of Fig 14–91. Does it fall within the range of NSt appropriate for that type of turbine?
14–105 Calculate the turbine specific speed of the turbine of Example 14–13 for the case where 𝛼1 = 10°. Provide answers in both dimensionless form and in customary U.S. units. Is it in the normal range for a Francis turbine? If not, what type of turbine would be more appropriate?
14–106 Calculate the turbine specific speed of the turbine in Prob. 14–89. Provide answers in both dimensionless form and in customary U.S. units. Is it in the normal range for a Francis turbine? If not, what type of turbine would be more appropriate?
14–107E Calculate the turbine specific speed of the turbine in Prob. 14–85E using customary U.S. units. Is it in the nor mal range for a Francis turbine? If not, what type of turbine would be more appropriate?
14–108 Calculate the turbine specific speed of the Round Butte hydroturbine of Fig 14–89. Does it fall within the range of NSt appropriate for that type of turbine?
14–109 A one-fifth scale model of a water turbine is tested in a laboratory at T = 20°C. The diameter of the model is 8.0 cm, its volume flow rate is 17.0 m3/h, it spins at 1500 rpm, and it operates with a net head of 15.0 m. At its best efficiency point, it delivers 450 W of shaft power. Calculate the efficiency of the model turbine. What is the most likely kind of turbine being tested? Answers: 64.9%, impulse 14–110 The prototype turbine corresponding to the one-fifth scale model turbine discussed in Prob. 14–109 is to operate across a net head of 50 m. Determine the appropriate rpm and volume flow rate for best efficiency. Predict the brake horsepower output of the prototype turbine, assuming exact geometric similarity.
14–111 Prove that the model turbine (Prob. 14–109) and the prototype turbine (Prob. 14–110) operate at homologous points by comparing turbine efficiency and turbine specific speed for both cases.
14–112 In Prob. 14–111, we scaled up the model turbine test results to the full-scale prototype assuming exact dynamic similarity. However, as discussed in the text, a large prototype typically yields higher efficiency than does the model. Esti mate the actual efficiency of the prototype turbine. Briefly explain the higher efficiency.
14–113 Verify that turbine specific speed and pump spe cific speed are related as follows: NSt = NSp√𝜂turbine.
14–114 Consider a pump–turbine that operates both as a pump and as a turbine. Under conditions in which the rota tional speed 𝜔 and the volume flow rate V . are the same for the pump and the turbine, verify that turbine specific speed and pump specific speed are related as NSt = NSp√𝜂turbine ( Hpump Hturbine) 3/4 = NSp(𝜂turbine)5/4(𝜂pump)3/4( bhppump bhpturbine) 3/4 Review Problems 14–115C What is a pump–turbine? Discuss an application where a pump–turbine is useful.
14–116C The common water meter found in most homes can be thought of as a type of turbine, since it extracts energy from the flowing water to rotate the shaft connected to the volume-counting mechanism (Fig. P14–116C). From the point of view of a piping system, however (Chap. 8), what kind of device is a water meter? Explain.
Water meter FIGURE P14–116C cen96537_ch14_793-884.indd 880 29/12/16 4:39 pm 881 CHAPTER 14 14–117C For each statement, choose whether the statement is true or false, and discuss your answer briefly: (a) A gear pump is a type of positive-displacement pump.
(b) A rotary pump is a type of positive-displacement pump.
(c) The pump performance curve (net head versus capacity) of a positive-displacement pump is nearly vertical through‑ out its recommended operating range at a given rotational speed.
(d) At a given rotational speed, the net head of a positive-displacement pump decreases with fluid viscosity.
14–118 A two-lobe rotary positive-displacement pump, similar to that of Fig. 14–30, moves 3.64 cm3 of tomato paste in each lobe volume V lobe. Calculate the volume flow rate of tomato paste for the case where n . = 336 rpm.
14–119 For two dynamically similar pumps, manipulate the dimensionless pump parameters to show that DB = DA (HA/HB)1/4(V .
B/V .
A)1/2. Does the same relationship apply to two dynamically similar turbines?
14–120 For two dynamically similar turbines, manipu late the dimensionless turbine parameters to show that DB = DA(HA/HB)3/4(𝜌A/𝜌B)1/2(bhpB/bhpA)1/2. Does the same relationship apply to two dynamically similar pumps?
14–121 A group of engineers is designing a new hydro turbine by scaling up an existing one. The existing tur bine (turbine A) has diameter DA = 1.50 m, and spins at n .
A = 150 rpm. At its best efficiency point, V . A = 162 m3/s, HA = 90.0 m of water, and bhpA = 132 MW. The new turbine (turbine B) will spin at 120 rpm, and its net head will be HB = 110 m. Calculate the diameter of the new turbine such that it operates most efficiently, and calculate V . B and bhpB. Answers: 2.07 m, 342 m3/s, 341 MW 14–122 Calculate and compare the efficiency of the two turbines of Prob. 14–121. They should be the same since we are assuming dynamic similarity. However, the larger turbine will actually be slightly more efficient than the smaller tur bine. Use the Moody efficiency correction equation to predict the actual expected efficiency of the new turbine. Discuss.
14–123 Calculate and compare the turbine specific speed for both the small (A) and large (B) turbines of Prob. 14–121. What kind of turbine are these most likely to be?
14–124 Calculate the turbine specific speed of the Smith Mountain hydroturbine of Fig 14–90. Does it fall within the range of NSt appropriate for that type of turbine?
Fundamentals of Engineering (FE) Exam Problems 14–125 Which turbomachine is designed to deliver a very high pressure rise, typically at low to moderate flow rates?
(a) Compressor (b) Blower (c) Turbine (d ) Pump (e) Fan 14–126 In the turbomachinery industry, capacity refers to (a) Power (b) Mass flow rate (c) Volume flow rate (d) Net head (e) Energy grade line 14–127 In the pump performance curve, the point at which the net head is zero is called (a) Best efficiency point (b) Free delivery (c) Shutoff head (d ) Operating point (e) Duty point 14–128 A pump increases the pressure of water from 100 kPa to 1.2 MPa at a rate of 0.5 m3/min. The inlet and outlet diameters are identical and there is no change in eleva tion across the pump. If the efficiency of the pump is 77 per cent, the power supplied to the pump is (a) 11.9 kW (b) 12.6 kW (c) 13.3 kW (d ) 14.1 kW (e) 15.5 kW 14–129 A pump increases the pressure of water from 100 kPa to 900 kPa to an elevation of 35 m. The inlet and outlet diameters are identical. The net head of the pump is (a) 143 m (b) 117 m (c) 91 m (d) 70 m (e) 35 m 14–130 The brake horsepower and water horsepower of a pump are determined to be 15 kW and 12 kW, respectively. If the flow rate of water to the pump under these conditions is 0.05 m3/s, the total head loss of the pump is (a) 11.5 m (b) 9.3 m (c) 7.7 m (d ) 6.1 m (e) 4.9 m 14–131 A power plant requires 940 L/min of water. The required net head is 5 m at this flow rate. An examination of pump performance curves indicates that two centrifugal pumps with different impeller diameters can deliver this flow rate. The pump with an impeller diameter of 203 mm has a pump efficiency of 73 percent and delivers 10 m of net head. The pump with an impeller diameter of 111 mm has a lower pump efficiency of 67 percent and delivers 5 m of net head. What is the ratio of the required brake horse power (bhp) of the pump with 203-mm-diameter impeller to that of the pump with 111-mm-diameter impeller?
(a) 0.45 (b) 0.68 (c) 0.86 (d) 1.84 (e) 2.11 14–132 Water enters the pump of a steam power plant at 15 kPa and 50°C at a rate of 0.15 m3/s. The diameter of the pipe at the pump inlet is 0.25 m. What is the net positive suction head (NPSH) at the pump inlet? (a) 1.70 m (b) 1.49 m (c) 1.26 m (d ) 0.893 m (e) 0.746 m 14–133 Which quantities are added when two pumps are connected in series and parallel?
(a) Series: Pressure change. Parallel: Net head (b) Series: Net head. Parallel: Pressure change (c) Series: Net head. Parallel: Flow rate (d ) Series: Flow rate. Parallel: Net head (e) Series: Flow rate. Parallel: Pressure change 14–134 Three pumps are connected in series. According to pump performance curves, the free delivery of each pump is as follows: Pump 1: 1600 L/min Pump 2: 2200 L/min Pump 3: 2800 L/min If the flow rate for this pump system is 2500 L/min, which pump(s) should be shut off?
cen96537_ch14_793-884.indd 881 29/12/16 4:39 pm 882 TURBOMACHINERY (a) Pump 1 (b) Pump 2 (c) Pump 3 (d ) Pumps 1 and 2 (e) Pumps 2 and 3 14–135 Three pumps are connected in parallel. According to pump performance curves, the shutoff head of each pump is as follows: Pump 1: 7 m Pump 2: 10 m Pump 3: 15 m If the net head for this pump system is 9 m, which pump(s) should be shut off?
(a) Pump 1 (b) Pump 2 (c) Pump 3 (d ) Pumps 1 and 2 (e) Pumps 2 and 3 14–136 A two-lobe rotary positive-displacement pump moves 0.60 cm3 of motor oil in each lobe volume. For every 90° of rotation of the shaft, one lobe volume is pumped. If the rotation rate is 550 rpm, the volume flow rate of oil is (a) 330 cm3/min (b) 660 cm3/min (c) 1320 cm3/min (d) 2640 cm3/min (e) 3550 cm3/min 14–137 A centrifugal blower rotates at 1400 rpm. Air enters the impeller normal to the blades (𝛼1 = 0°) and exits at an angle of 25° (𝛼2 = 25°). The inlet radius is r1 = 6.5 cm, and the inlet blade width b1 = 8.5 cm. The outlet radius and blade width are r2 = 12 cm and b2 = 4.5 cm, respectively. The volume flow rate is 0.22 m3/s. What is the net head produced by this blower in meters of air?
(a) 12.3 m (b) 3.9 m (c) 8.8 m (d) 5.4 m (e) 16.4 m 14–138 A pump is designed to deliver 9500 L/min of water at a required head of 8 m. The pump shaft rotates at 1100 rpm. The pump specific speed in nondimensional form is (a) 0.277 (b) 0.515 (c) 1.17 (d ) 1.42 (e) 1.88 14–139 The net head delivered by a pump at a rotational speed of 1000 rpm is 10 m. If the rotational speed is doubled, the net head delivered will be (a) 5 m (b) 10 m (c) 20 m (d ) 40 m (e) 80 m 14–140 The rotating part of a turbine is called (a) Propeller (b) Scroll (c) Blade row (d ) Impeller (e) Runner 14–141 Which choice is correct for the comparison of the operation of impulse and reaction turbines?
(a) Impulse: Higher flow rate (b) Impulse: Higher head (c) Reaction: Higher head (d) Reaction: Smaller flow rate (e) None of these 14–142 Which turbine type is an impulse turbine?
(a) Kaplan (b) Francis (c) Pelton (d) Propeller (e) Centrifugal 14–143 In a hydroelectric power plant, water flows through a large tube through the dam. This tube is called a (a) Tailrace (b) Draft tube (c) Runner (d ) Penstock (e) Propeller 14–144 A turbine is placed at the bottom of a 20-m-high water body. Water flows through the turbine at a rate of 30 m3/s. If the shaft power delivered by the turbine is 5 MW, the turbine efficiency is (a) 85% (b) 79% (c) 88% (d ) 74% (e) 82% 14–145 A hydroelectric power plant is to be built at a dam with a gross head of 240 m. The head losses in the head gate and penstock are estimated to be 6 m. The flow rate through the turbine is 18,000 L/min. The efficiencies of the turbine and the generator are 88 percent and 96 percent, respectively. The electricity production from this turbine is (a) 6930 kW (b) 5750 kW (c) 6440 kW (d ) 5820 kW (e) 7060 kW 14–146 In wind turbines, the minimum wind speed at which useful power can be generated is called the (a) Rated speed (b) Cut-in speed (c) Cut-out speed (d) Available speed (e) Betz speed 14–147 A wind turbine is installed in a location where the wind blows at 8 m/s. The air temperature is 10°C and the diameter of turbine blade is 30 m. If the overall turbine-generator efficiency is 35 percent, the electrical power pro duction is (a) 79 kW (b) 109 kW (c) 142 kW (d ) 154 kW (e) 225 kW 14–148 The available power from a wind turbine is calcu lated to be 100 kW when the wind speed is 5 m/s. If the wind velocity is doubled, the available wind power becomes (a) 100 kW (b) 200 kW (c) 400 kW (d ) 800 kW (e) 1600 kW 14–149 A new hydraulic turbine is to be designed to be similar to an existing turbine with following parameters at its best efficiency point: DA = 3 m, n .
A = 90 rpm, V .
A = 200 m3/s, HA = 55 m, bhpA = 100 MW. The new turbine will have a speed of 110 rpm and the net head will be 40 m. What is the bhp of the new turbine such that it operates most efficiently?
(a) 17.6 MW (b) 23.5 MW (c) 30.2 MW (d ) 40.0 MW (e) 53.7 MW 14–150 A hydraulic turbine operates at the following param eters at its best efficiency point: n . = 110 rpm, V . = 200 m3/s, H = 55 m, bhp = 100 MW. The turbine specific speed of this turbine is (a) 0.74 (b) 0.38 (c) 1.40 (d ) 2.20 (e) 1.15 Design and Essay Problems 14–151 This problem is useful for the preliminary design of a hydroturbine. From the material learned in this chapter, it is fairly simple to estimate how much power a hydrotur bine can generate, given only the flow rate of water and the elevation difference upstream and downstream of the dam. A cen96537_ch14_793-884.indd 882 29/12/16 4:39 pm 883 CHAPTER 14 dam has a gross head of 15.5 m and a flow rate of 0.22 m3/s. Approximating the overall efficiency of the turbine/generator to be 75%, estimate the electrical power (in kW) that could be produced.
14–152 A hydroturbine is proposed for a dam with an estimated gross head of 20.5 m and a flow rate of 4.52 m3/s. The generator is to turn at 120 rpm. You are asked to identify the type of turbine best suited for this dam. For estimation purposes, set 𝜂turbine = 85% and 𝜂other = 95%. Note that a reasonable approximation is that net head H = 𝜂turbine 𝜂other Hgross. The electrical generator efficiency is 𝜂generator = 95%.
(a) If all the flow is to go through one turbine, calculate the turbine specific speed and then select the most appropriate type of turbine for this situation.
(b) If the flow is split into 8 identical turbines, calculate the turbine specific speed and then select the most appropriate type of turbine for this situation.
(c) Which case would generate the most overall power? Explain.
14–153 Develop a general-purpose computer appli cation (using appropriate software) that employs the affinity laws to design a new pump (B) that is dynamically similar to a given pump (A). The inputs for pump A are diameter, net head, capacity, density, rotational speed, and pump efficiency. The inputs for pump B are density (𝜌B may differ from 𝜌A), desired net head, and desired capacity. The outputs for pump B are diameter, rota tional speed, and required shaft power. Test your program using the following inputs: DA = 5.0 cm, HA = 120 cm, V . A = 400 cm3/s, 𝜌A = 998.0 kg/m3, n .
A = 1725 rpm, 𝜂pump, A = 81 percent, 𝜌B = 1226 kg/m3, HB = 450 cm, and V . B = 2400 cm3/s. Verify your results manually. Answers: DB = 8.80 cm, n .
B = 1898 rpm, and bhpB = 160 W 14–154 Experiments on an existing pump (A) yield the following BEP data: DA = 10.0 cm, HA = 210 cm, V . A = 1350 cm3/s, 𝜌A = 998.0 kg/m3, n .
A = 1500 rpm, 𝜂pump, A = 87 percent. You are to design a new pump (B) that has the following requirements: 𝜌B = 998.0 kg/m3, HB = 570 cm, and V . B = 3670 cm3/s. Apply the computer program you developed in Prob. 14–153 to calculate DB (cm), n .
B (rpm), and bhpB (W). Also calculate the pump specific speed. What kind of pump is this (most likely)?
14–155 Develop a general-purpose computer application (using appropriate software) that employs the affinity laws to design a new turbine (B) that is dynamically similar to a given turbine (A). The inputs for turbine A are diameter, net head, capacity, density, rotational speed, and brake horsepower. The inputs for turbine B are density (𝜌B may differ from 𝜌A), available net head, and rotational speed. The outputs for turbine B are diameter, capacity, and brake horsepower. Test your program using the following inputs: DA = 1.40 m, HA = 80.0 m, V . A = 162 m3/s, 𝜌A = 998.0 kg/m3, n .
A = 150 rpm, bhpA = 118 MW, 𝜌B = 998.0 kg/m3, HB = 95.0 m, and n . B = 120 rpm. Verify your results manually. Answers: DB = 1.91 m, V . B = 328 m3/s, and bhpB = 283 MW 14–156 Experiments on an existing turbine (A) yield the following data: DA = 86.0 cm, HA = 22.0 m, V . A = 69.5 m3/s, 𝜌A = 998.0 kg/m3, n .
A = 240 rpm, bhpA = 11.4 MW. You are to design a new turbine (B) that has the following requirements: 𝜌B = 998.0 kg/m3, HB = 95.0 m, and n .
B = 210 rpm. Apply the computer program you developed in Prob. 14–155 to calculate DB (m), V . B (m3/s), and bhpB (MW). Also calculate the turbine specific speed. What kind of turbine is this (most likely)?
14–157 Calculate and compare the efficiency of the two turbines of Prob. 14–156. They should be the same since we are assuming dynamic similarity. How ever, the larger turbine will actually be slightly more efficient than the smaller turbine. Use the Moody efficiency correc tion equation to predict the actual expected efficiency of the new turbine. Discuss.
cen96537_ch14_793-884.indd 883 29/12/16 4:39 pm This page intentionally left blank 15 CHAPTER 885 OBJECTIVES When you finish reading this chapter, you should be able to ■ ■ Understand the importance of a high-quality, good resolution mesh ■ ■ Apply appropriate boundary conditions to computational domains ■ ■ Understand how to apply CFD to basic engineering problems and how to determine whether the output is physically meaningful ■ ■ Realize that you need much further study and practice to use CFD successfully CFD simulation of airflow pathlines around a hovering Harrier aircraft near the ground in a headwind. The visualization shows the horseshoe-shaped vortex that forms along the ground from the interaction of the downward pointing jet nozzles. Under certain combinations of wind and jet speeds, hot air and debris from the exhaust jets rebound ing off the ground may be ingested by the engines, which could cause the engines to stall and the aircraft to crash.
© Science Source I NT R O D UC T I O N TO C O M P UTAT I O N AL F LUI D DY N AM I CS A brief introduction to computational fluid dynamics (CFD) is pre sented in this chapter. While any intelligent, computer-literate per son can run a CFD code, the results he or she obtains may not be physically correct. In fact, if the grid is not properly generated, or if the boundary conditions or flow parameters are improperly applied, the results may even be completely erroneous. Therefore, the goal of this chapter is to present guidelines about how to generate a grid, how to specify boundary conditions, and how to determine if the computer output is meaningful. We stress the application of CFD to engineering problems, rather than details about grid generation techniques, discretization schemes, CFD algorithms, or numerical stability.
The examples presented here have been obtained with the commercial com putational fluid dynamics code ANSYS-FLUENT. Other CFD codes would yield similar, but not identical results. Sample CFD solutions are shown for incompressible and compressible laminar and turbulent flows, flows with heat transfer, and flows with free surfaces. As always, one learns best by hands-on practice. For this reason, we provide several homework problems that utilize many additional CFD problems are provided or the books website at www.mhhe.com/cengel.
cen96537_ch15_885-946.indd 885 14/01/17 2:10 pm 886 COMPUTATIONAL FLUID DYNAMICS 15–1 ■ INTRODUCTION AND FUNDAMENTALS Motivation There are two fundamental approaches to design and analysis of engineering systems that involve fluid flow: experimentation and calculation. The former typically involves construction of models that are tested in wind tunnels or other facilities (Chap. 7), while the latter involves solution of differential equa tions, either analytically (Chaps. 9 and 10) or computationally. In the present chapter, we provide a brief introduction to computational fluid dynamics (CFD), the field of study devoted to solution of the equations of fluid flow through use of a computer (or, more recently, several computers working in parallel). Modern engineers apply both experimental and CFD analyses, and the two complement each other. For example, engineers may obtain global properties, such as lift, drag, pressure drop, or power, experimentally, but use CFD to obtain details about the flow field, such as shear stresses, velocity and pressure profiles (Fig. 15–1), and flow streamlines. In addition, experimental data are often used to validate CFD solutions by matching the computation ally and experimentally determined global quantities. CFD is then employed to shorten the design cycle through carefully controlled parametric studies, thereby reducing the required amount of experimental testing.
The current state of computational fluid dynamics is that CFD can handle laminar flows with ease, but turbulent flows of practical engineering interest are impossible to solve without invoking turbulence models. Unfortunately, no turbulence model is universal, and a turbulent CFD solution is only as good as the appropriateness of the turbulence model. In spite of this limi tation, the standard turbulence models yield reasonable results for many practical engineering problems.
There are several aspects of CFD that are not covered in this chapter— grid generation techniques, numerical algorithms, finite difference and finite volume schemes, stability issues, turbulence modeling, etc. You need to study these topics in order to fully understand both the capabilities and limi tations of computational fluid dynamics. In this chapter, we merely scratch the surface of this exciting field. Our goal is to present the fundamentals of CFD from a user’s point of view, providing guidelines about how to gener ate a grid, how to specify boundary conditions, and how to determine if the computer output is physically meaningful.
We begin this section by presenting the differential equations of fluid flow that are to be solved, and then we outline a solution procedure. Sub sequent sections of this chapter are devoted to example CFD solutions for laminar flow, turbulent flow, flows with heat transfer, compressible flow, and open-channel flow.
Equations of Motion For steady laminar flow of a viscous, incompressible, Newtonian fluid with out free-surface effects, the equations of motion are the continuity equation ∇ → ·V → = 0 (15–1) and the Navier–Stokes equation (V → ·∇ → )V → = −1 ρ ∇ → Pʹ + 𝜈∇2V → (15–2) FIGURE 15–1 CFD calculations of the ascent of the space shuttle launch vehicle (SSLV). The grid consists of more than 16 million points, and filled pressure contours are shown. Free-stream conditions are Ma = 1.25, and the angle of attack is −3.3°.
NASA Photo/Photo by Ray J. Gomez, NASA Johnson Space Center, Houston, Texas.
cen96537_ch15_885-946.indd 886 14/01/17 2:10 pm 887 CHAPTER 15 Strictly speaking, Eq. 15–1 is a conservation equation, while Eq. 15–2 is a transport equation that represents transport of linear momentum through out the computational domain. In Eqs. 15–1 and 15–2, V › is the velocity of the fluid, 𝜌 is its density, and ν is its kinematic viscosity (ν = 𝜇/𝜌). The lack of free-surface effects enables us to use the modified pressure Pʹ, thereby eliminating the gravity term from Eq. 15–2 (see Chap. 10). Note that Eq. 15–1 is a scalar equation, while Eq. 15–2 is a vector equation. Equa tions 15–1 and 15–2 apply only to incompressible flows in which we also assume that both 𝜌 and ν are constants. Thus, for three-dimensional flow in Cartesian coordinates, there are four coupled differential equations for four unknowns, u, 𝜐, w, and Pʹ (Fig. 15–2). If the flow were compressible, Eqs. 15–1 and 15–2 would need to be modified appropriately, as discussed in Section 15–5. Liquid flows can almost always be treated as incompressible, and for many gas flows, the gas is at a low enough Mach number that it behaves as a nearly incompressible fluid.
Solution Procedure To solve Eqs. 15–1 and 15–2 numerically, the following steps are performed. Note that the order of some of the steps (particularly steps 2 through 5) is interchangeable.
1. A computational domain is chosen, and a grid (also called a mesh) is generated; the domain is divided into many small elements called cells. For two-dimensional (2-D) domains, the cells are areas, while for three-dimensional (3-D) domains the cells are volumes (Fig. 15–3). You can think of each cell as a tiny control volume in which discretized versions of the conservation equations are solved. Note that we limit our discussion here to cell-centered finite volume CFD codes. The quality of a CFD solution is highly dependent on the quality of the grid. Therefore, you are advised to make sure that your grid is of high quality before proceeding to the next step (Fig. 15–4).
2. Boundary conditions are specified on each edge of the computational domain (2-D flows) or on each face of the domain (3-D flows).
3. The type of fluid (water, air, gasoline, etc.) is specified, along with fluid properties (temperature, density, viscosity, etc.). Many CFD codes FIGURE 15–2 The equations of motion to be solved by CFD for the case of steady, incompressible, laminar flow of a Newtonian fluid with constant properties and without free-surface effects. A Cartesian coordinate system is used. There are four equations and four unknowns: u, 𝜐, w, and Pʹ.
Continuity: x-momentum: y-momentum: z-momentum: 휕u 휕 x 휕 x 휕 x 휕 υ 휕 y 휕 w 0 = + + 휕z 1 – = 휕u + u 휕 2u 휕 2u 휕 2u 휕 x2 v 휕u 휕 y 휕 P΄ υ + + + 휕y2 휕z2 + 휕u 휕z w = 휕 υ 휕x + u 휕 υ 휕 y υ + 휕 υ 휕z w ) ) ρ 1 – 휕 2 υ 휕 2 υ 휕 2 υ 휕 x2 v 휕P΄ 휕 y + + 휕 y2 휕z2 + ) ) ρ = 휕 w 휕x + u 휕 w 휕 y υ + 휕 w 휕z w 1 – 휕 2w 휕 2w 휕 2w 휕x2 v 휕P΄ 휕z + + 휕y2 휕z2 + ) ) ρ FIGURE 15–3 A computational domain is the region in space in which the equations of motion are solved by CFD. A cell is a small subset of the computational domain. Shown are (a) a two-dimensional domain and quadrilateral cell, and (b) a three-dimensional domain and hexahedral cell. The boundaries of a 2-D domain are called edges, while those of a 3-D domain are called faces.
Computational domain Cell Cell Boundaries Boundaries Computational domain (b) (a) cen96537_ch15_885-946.indd 887 14/01/17 2:10 pm 888 COMPUTATIONAL FLUID DYNAMICS have built-in property databases for common fluids, making this step relatively painless.
4. Numerical parameters and solution algorithms are selected. These are specific to each CFD code and are not discussed here. The default settings of most modern CFD codes are appropriate for the simple problems discussed in this chapter.
5. Starting values for all flow field variables are specified for each cell. These are initial conditions, which may or may not be correct, but are necessary as a starting point, so that the iteration process may proceed (step 6). We note that for proper unsteady-flow calculations, the initial conditions must be correct.
6. Beginning with the initial guesses, discretized forms of Eqs. 15–1 and 15–2 are solved iteratively, usually at the center of each cell. If one were to put all the terms of Eq. 15–2 on one side of the equation, the solution would be “exact” when the sum of these terms, defined as the residual, is zero for every cell in the domain. In a CFD solution, however, the sum is never identically zero, but (hopefully) decreases with progressive iterations. A residual can be thought of as a measure of how much the solution to a given transport equation deviates from exact, and you monitor the average residual associated with each transport equation to help determine when the solution has converged. Sometimes hundreds or even thousands of iterations are required to converge on a final solution, and the residuals may decrease by several orders of magnitude.
7. Once the solution has converged, flow field variables such as velocity and pressure are plotted and analyzed graphically. You can also define and analyze additional custom functions that are formed by algebraic combinations of flow field variables. Most commercial CFD codes have built-in postprocessors, designed for quick graphical analysis of the flow field. There are also stand-alone postprocessor software packages available for this purpose. Since the graphics output is often displayed in vivid colors, CFD has earned the nickname colorful fluid dynamics.
8. Global properties of the flow field, such as pressure drop, and integral properties, such as forces (lift and drag) and moments acting on a body, are calculated from the converged solution (Fig. 15–5). With most CFD codes, this can also be done “on the fly” as the iterations proceed. In many cases, in fact, it is wise to monitor these quantities along with the residuals during the iteration process; when a solution has converged, the global and integral properties should settle down to constant values as well.
For unsteady flow, a physical time step is specified, appropriate initial conditions are assigned, and an iteration loop is carried out to solve the transport equations to simulate changes in the flow field over this small span of time. Since the changes between time steps are small, a relatively small number of iterations (on the order of tens) is usually required between each time step. Upon convergence of this “inner loop,” the code marches to the next time step. If a flow has a steady-state solution, that solution is sometimes easier to find by marching in time—after enough time has past, the flow field variables settle down to their steady-state values. Most CFD codes take advantage of this fact by internally specifying a pseudo-time step (artificial time) and marching toward a steady-state solution. In such cases, FL FD M FIGURE 15–5 Global and integral properties of a flow, such as forces and moments on an object, are calculated after a CFD solution has converged. They can also be calculated during the iteration process to monitor convergence.
FIGURE 15–4 A quality grid is essential to a quality CFD simulation.
NOTICE Do not proceed with CFD calculations until you have generated a high-quality grid.
cen96537_ch15_885-946.indd 888 14/01/17 2:10 pm 889 CHAPTER 15 the pseudo-time step can even be different for different cells in the computa tional domain and can be tuned appropriately to decrease convergence time.
Other “tricks” are often used to reduce computation time, such as multi gridding, in which the flow field variables are updated first on a coarse grid so that gross features of the flow are quickly established. That solution is then interpolated to finer and finer grids, the final grid being the one speci fied by the user (Fig. 15–6). In some commercial CFD codes, several layers of multigridding may occur “behind the scenes” during the iteration process, without user input (or awareness). You can learn more about computational algorithms and other numerical techniques that improve convergence by read ing books devoted to computational methods, such as Tannehill, Anderson, and Pletcher (2012).
Additional Equations of Motion If energy conversion or heat transfer is important in the problem, another transport equation, the energy equation, must also be solved. If tempera ture differences lead to significant changes in density, an equation of state (such as the ideal-gas law) is used. If buoyancy is important, the effect of temperature on density is reflected in the gravity term (which must then be separated from the modified pressure term in Eq. 15–2).
For a given set of boundary conditions, a laminar flow CFD solution approaches an “exact” solution, limited only by the accuracy of the discret ization scheme used for the equations of motion, the level of convergence, and the degree to which the grid is resolved. The same would be true of a turbulent flow simulation if the grid could be fine enough to resolve all the unsteady, three-dimensional turbulent eddies. Unfortunately, this kind of direct simulation of turbulent flow is usually not possible for practical engineering applications due to computer limitations. Instead, additional approximations are made in the form of turbulence models so that turbulent flow solutions are possible. The turbulence models generate additional transport equations that model the enhanced mixing and diffusion of turbulence; these additional transport equations must be solved along with those of mass and momentum. Turbulence modeling is discussed in more detail in Section 15–3.
Modern CFD codes include options for calculation of particle trajecto ries, species transport, heat transfer, and turbulence. The codes are easy to use, and solutions can be obtained without knowledge about the equations or their limitations. Herein lies the danger of CFD: When in the hands of someone without knowledge of fluid mechanics, erroneous results are likely to occur (Fig. 15–7). It is critical that users of CFD possess some funda mental knowledge of fluid mechanics so that they can discern whether a CFD solution makes physical sense or not.
Grid Generation and Grid Independence The first step (and arguably the most important step) in a CFD solution is generation of a grid that defines the cells on which flow variables (velocity, pressure, etc.) are calculated throughout the computational domain. Modern commercial CFD codes come with their own grid generators, and third-party grid generation programs are also available. The grids used in this chapter are generated with ANSYS-FLUENT’s grid generation package.
FIGURE 15–7 CFD solutions are easy to obtain, and the graphical outputs can be beautiful; but correct answers depend on correct inputs and knowledge about the flow field.
FIGURE 15–6 With multigridding, solutions of the equations of motion are obtained on a coarse grid first, followed by successively finer grids. This speeds up convergence.
cen96537_ch15_885-946.indd 889 14/01/17 2:10 pm 890 COMPUTATIONAL FLUID DYNAMICS Many CFD codes can run with either structured or unstructured grids. A structured grid consists of planar cells with four edges (2-D) or volu metric cells with six faces (3-D). Although the cells may be distorted from rectangular, each cell is numbered according to indices (i, j, k) that do not necessarily correspond to coordinates x, y, and z. An illustration of a 2-D structured grid is shown in Fig. 15–8. To construct this grid, nine nodes are specified on the top and bottom edges; these nodes correspond to eight intervals along these edges. Similarly, five nodes are specified on the left and right edges, corresponding to four intervals along these edges. The intervals correspond to i = 1 through 8 and j = 1 through 4, and are num bered and marked in Fig. 15–8. An internal grid is then generated by con necting nodes one-for-one across the domain such that rows ( j = constant) and columns (i = constant) are clearly defined, even though the cells them selves may be distorted (not necessarily rectangular). In a 2-D structured grid, each cell is uniquely specified by an index pair (i, j). For example, the shaded cell in Fig. 15–8 is at (i = 4, j = 3). You should be aware that some CFD codes number nodes rather than intervals.
An unstructured grid consists of cells of various shapes, but typically triangles or quadrilaterals (2-D) and tetrahedrons or hexahedrons (3-D) are used. Two unstructured grids for the same domain as that of Fig. 15–8 are generated, using the same interval distribution on the edges; these grids are shown in Fig. 15–9. Unlike the structured grid, one cannot uniquely identify cells in the unstructured grid by indices i and j; instead, cells are numbered in some other fashion internally in the CFD code.
For complex geometries, an unstructured grid is usually much easier for the user of the grid generation code to create. However, there are some advan tages to structured grids. For example, some (usually older) CFD codes are written specifically for structured grids; these codes converge more rapidly, and often more accurately, by utilizing the index feature of structured grids. For modern general-purpose CFD codes that can handle both structured and unstructured grids, however, this is no longer an issue. More impor tantly, fewer cells are usually generated with a structured grid than with an unstructured grid. In Fig. 15–8, for example, the structured grid has 8 × 4 = 32 cells, while the unstructured triangular grid of Fig. 15–9a has 76 cells, and the unstructured quadrilateral grid has 38 cells, even though the identi cal node distribution is applied at the edges in all three cases. In boundary y j = 4 i = 1 2 2 3 3 1 4 5 6 7 8 x FIGURE 15–8 Sample structured 2-D grid with nine nodes and eight intervals on the top and bottom edges, and five nodes and four intervals on the left and right edges. Indices i and j are shown. The red cell is at (i = 4, j = 3).
y Unstructured triangular grid (a) (b) x y Unstructured quadrilateral grid x FIGURE 15–9 Sample 2-D unstructured grids with nine nodes and eight intervals on the top and bottom edges, and five nodes and four intervals on the left and right edges. These grids use the same node distribution as that of Fig. 15–8: (a) unstructured triangular grid, and (b) unstructured quadrilateral grid. The red cell in the upper right corner of (a) is moderately skewed.
cen96537_ch15_885-946.indd 890 14/01/17 2:10 pm 891 CHAPTER 15 layers, where flow variables change rapidly normal to the wall and highly resolved grids are required close to the wall, structured grids enable much finer resolution than do unstructured grids for the same number of cells. This can be seen by comparing the grids of Figs. 15–8 and 15–9 near the far right edge. The cells of the structured grid are thin and tightly packed near the right edge, while those of the unstructured grids are not.
We must emphasize that regardless of the type of grid you choose (struc tured or unstructured, quadrilateral or triangular, etc.), it is the quality of the grid that is most critical for reliable CFD solutions. In particular, you must always be careful that individual cells are not highly skewed, as this can lead to convergence difficulties and inaccuracies in the numerical solution. The shaded cell in Fig. 15–9a is an example of a cell with moderately high skewness, defined as the departure from symmetry. There are various kinds of skewness, for both two- and three-dimensional cells. Three-dimensional cell skewness is beyond the scope of the present textbook—the type of skewness most appropriate for two-dimensional cells is equiangle skewness, defined as Equiangle skewness: QEAS = MAX( 𝜃max −𝜃equal 180° −𝜃equal , 𝜃equal −𝜃min 𝜃equal ) (15–3) where 𝜃min and 𝜃max are the minimum and maximum angles (in degrees) between any two edges of the cell, and 𝜃equal is the angle between any two edges of an ideal equilateral cell with the same number of edges. For trian gular cells 𝜃equal = 60° and for quadrilateral cells 𝜃equal = 90°. You can show by Eq. 15–3 that 0 < QEAS < 1 for any 2-D cell. By definition, an equilateral triangle has zero skewness. In the same way, a square or rectangle has zero skewness. A grossly distorted triangular or quadrilateral element may have unacceptably high skewness (Fig. 15–10). Some grid generation codes use numerical schemes to smooth the grid so as to minimize skewness.
Other factors affect the quality of the grid as well. For example, abrupt changes in cell size can lead to numerical or convergence difficulties in the CFD code. Also, cells with a very large aspect ratio can sometimes cause problems. While you can often minimize the cell count by using a struc tured grid instead of an unstructured grid, a structured grid is not always the best choice, depending on the shape of the computational domain. You must always be cognizant of grid quality. Keep in mind that a high-quality unstructured grid is better than a poor-quality structured grid. An example is shown in Fig. 15–11 for the case of a computational domain with a small (a) Triangular cells Zero skewness High skewness High skewness Zero skewness (b) Quadrilateral cells FIGURE 15–10 Skewness is shown in two dimensions: (a) an equilateral triangle has zero skewness, but a highly distorted triangle has high skewness. (b) Similarly, a rectangle has zero skewness, but a highly distorted quadrilateral cell has high skewness.
(a) (b) (c) (d) FIGURE 15–11 Comparison of four 2-D grids for a highly distorted computational domain: (a) structured 8 × 8 grid with 64 cells and (QEAS)max = 0.83, (b) unstructured triangular grid with 70 cells and (QEAS)max = 0.76, (c) unstructured quadrilateral grid with 67 cells and (QEAS)max = 0.87, and (d) hybrid grid with 62 cells and (QEAS)max = 0.76.
cen96537_ch15_885-946.indd 891 14/01/17 2:10 pm 892 COMPUTATIONAL FLUID DYNAMICS acute angle at the upper-right corner. For this example we have adjusted the node distribution so that the grid in any case contains between 60 and 70 cells for direct comparison. The structured grid (Fig. 15–11a) has 8 × 8 = 64 cells; but even after smoothing, the maximum equiangle skewness is 0.83—cells near the upper right corner are highly skewed. The unstructured triangu lar grid (Fig. 15–11b) has 70 cells, but the maximum skewness is reduced to 0.76. More importantly, the overall skewness is lower throughout the entire computational domain. The unstructured quad grid (Fig. 15–11c) has 67 cells. Although the overall skewness is better than that of the structured mesh, the maximum skewness is 0.87—higher than that of the structured mesh. The hybrid grid shown in Fig. 15–11d is discussed shortly.
Situations arise in which a structured grid is preferred (e.g., the CFD code requires structured grids, boundary layer zones need high resolu tion, or the simulation is pushing the limits of available computer mem ory). Generation of a structured grid is straightforward for geometries with straight edges. All we need to do is divide the computational domain into four-sided (2-D) or six-sided (3-D) blocks or zones. Inside each block, we generate a structured grid (Fig. 15–12a). Such an analysis is called multi block analysis. For more complicated geometries with curved surfaces, we need to determine how the computational domain can be divided into indi vidual blocks that may or may not have flat edges (2-D) or faces (3-D). A two-dimensional example involving circular arcs is shown in Fig. 15–12b. Most CFD codes require that the nodes match on the common edges and faces between blocks.
Many commercial CFD codes allow you to split the edges or faces of a block and assign different boundary conditions to each segment of the edge or face. In Fig. 15–12a for example, the left edge of block 2 is split about two-thirds of the way up to accommodate the junction with block 1. The lower segment of this edge is a wall, and the upper segment of this edge is an interior edge. (These and other boundary conditions are discussed shortly.) A similar situation occurs on the right edge of block 2 and on the top edge of block 3. Some CFD codes accept only elementary blocks, namely, blocks whose edges or faces cannot be split. For example, the four-block grid of Fig. 15–12a requires seven elementary blocks under this limitation (Fig. 15–13). The total number of cells is the same, which you can verify. Finally, for CFD codes that allow blocks with split edges or faces, we can sometimes combine two or more blocks into one. For example, it is left as an exercise to show how the structured grid of Fig. 5–11b can be simpli fied to just three nonelementary blocks.
When developing the block topology with complicated geometries as in Fig. 15–12b, the goal is to create blocks in such a way that no cells in the grid are highly skewed. In addition, cell size should not change abruptly in any direction, and the blocking topology should lend itself to clustering cells near solid walls so that boundary layers can be resolved. With practice you can master the art of creating sophisticated multiblock structured grids. Multiblock grids are necessary for structured grids of complex geometry. They may also be used with unstructured grids, but are not necessary since the cells can accommodate complex geometries.
Finally, a hybrid grid is one that combines regions or blocks of struc tured and unstructured grids. For example, you can mate a structured grid FIGURE 15–12 Examples of structured grids generated for multiblock CFD analysis: (a) a simple 2-D computational domain composed of rectangular four-sided blocks, and (b) a more complicated 2-D domain with curved surfaces, but again composed of four-sided blocks and quadrilateral cells. The number of i- and j-intervals is shown in parentheses for each block. There are, of course, acceptable alternative ways to divide these computational domains into blocks.
(a) Block 1 (12 × 8) Block 2 (10 × 21) Block 3 (9 × 5) Block 4 (3 × 5) Block 4 (5 × 16) (b) Block 6 (8 × 16) Block 5 (5 × 8) Block 1 (12 × 8) Block 2 (5 × 16) Block 3 (5 × 8) cen96537_ch15_885-946.indd 892 14/01/17 2:10 pm 893 CHAPTER 15 block close to a wall with an unstructured grid block outside of the region of influence of the boundary layer. A hybrid grid is often used to enable high resolution near a wall without requiring high resolution away from the wall (Fig. 15–14). When generating any type of grid (structured, unstruc tured, or hybrid), you must always be careful that individual cells are not highly skewed. For example, none of the cells in Fig. 15–14 has any signifi cant skewness. Another example of a hybrid grid is shown in Fig. 15–11d. Here we have split the computational domain into two blocks. The four-sided block on the left is meshed with a structured grid, while the three-sided block on the right is meshed with an unstructured triangular grid. The maximum skewness is 0.76, the same as that of the unstructured tri angular grid of Fig. 15–11b, but the total number of cells is reduced from 70 to 62.
Computational domains with very small angles like the one shown in Fig. 15–11 are difficult to mesh at the sharp corner, regardless of the type of cells used. One way to avoid large values of skewness at a sharp corner is to simply chop off or round off the sharp corner. This can be done very close to the corner so that the geometric modification is imperceptible from an over all view and has little if any effect on the flow, yet greatly improves the per formance of the CFD code by reducing the skewness. For example, the trou blesome sharp corner of the computational domain of Fig. 15–11 is chopped off and replotted in Fig. 15–15. Through use of multiblocking and hybrid grids, the grid shown in Fig. 15–15 has 62 cells and a maximum skewness of only 0.53—a vast improvement over any of the grids in Fig. 15–11.
The examples shown here are for two dimensions. In three dimensions, you can still choose between structured, unstructured, and hybrid grids. If a four-sided 2-D face with structured cells is swept in the third dimension, a fully structured 3-D mesh is produced, consisting of hexahedral cells (n = 6 faces per cell). When a 2-D face with unstructured triangular cells is swept in the third direction, the 3-D mesh can consist of prism cells (n = 5 faces per cell) or tetrahedral cells (n = 4 faces per cell—like a pyramid). These are illustrated in Fig. 15–16. When a hexahedral mesh is impracti cal to apply (e.g., complex geometry), a tetrahedral mesh (also called a tet mesh) is a common alternative approach. Automatic grid generation codes often generate a tet mesh by default. However, just as in the 2-D case, a 3-D unstructured tet mesh results in greater overall cell count than a structured hexahedral mesh with the same resolution along boundaries.
The most recent enhancement in grid generation is the use of polyhedral meshes. As the name implies, such a mesh consists of cells of many faces, called polyhedral cells. Some modern grid generators can create unstruc tured three-dimensional meshes with a mixture of n-sided cells, where n Block 7 (3 × 5) Block 2 (10 × 8) Block 1 (12 × 8) Block 3 (10 × 8) Block 4 (10 × 5) Block 5 (6 × 5) Block 6 (3 × 5) FIGURE 15–13 The multiblock grid of Fig. 15–12a modified for a CFD code that can handle only elementary blocks.
Structured Structured Unstructured FIGURE 15–14 Sample two-dimensional hybrid grid near a curved surface; two structured regions and one unstructured region are labeled.
(a) (b) FIGURE 15–15 Hybrid grid for the computational domain of Fig. 15–11 with the sharp corner chopped off: (a) overall view—the grid contains 62 cells with (QEAS)max = 0.53, (b) magnified view of the chopped off corner.
cen96537_ch15_885-946.indd 893 14/01/17 2:10 pm 894 COMPUTATIONAL FLUID DYNAMICS can be any integer greater than 3. An example polyhedral mesh is shown in Fig. 15–17. In some codes, the polyhedral cells are formed by merg ing tetrahedral cells, reducing total cell count. This saves a significant amount of computer memory and speeds up the CFD calculations. Overall cell-count reductions (and corresponding CPU time savings) by a factor of as much as 5 have been reported without compromising solution accu racy. Another advantage of polyhedral meshes is that cell skewness can be reduced, improving the overall mesh quality and also speeding up conver gence. Finally, polyhedral cells with large n have many more neighbor cells than do simple tetrahedral or prism cells. This is advantageous for tasks such as calculating gradients (derivatives) of flow parameters—details are beyond the level of the present text.
Generation of a good grid is often tedious and time consuming; engineers who use CFD on a regular basis will agree that grid generation usually takes more of their time than does the CFD solution itself (engineer’s time, not CPU time). However, time spent generating a good grid is time well spent, since the CFD results will be more reliable and may converge more rapidly (Fig. 15–18). A high-quality grid is critical to an accurate CFD solution; a poorly resolved or low-quality grid may even lead to an incorrect solu tion. It is important, therefore, for users of CFD to test if their solution is grid independent. The standard method to test for grid independence is to increase the resolution (by a factor of 2 in all directions if feasible) and repeat the simulation. If the results do not change appreciably, the original grid is probably adequate. If, on the other hand, there are significant differ ences between the two solutions, the original grid is likely of inadequate resolution. In such a case, an even finer grid should be tried until the grid is adequately resolved. This method of testing for grid independence is time consuming, and unfortunately, not always feasible, especially for large engi neering problems in which the solution pushes computer resources to their limits. In a 2-D simulation, if one doubles the number of intervals on each edge, the number of cells increases by a factor of 22 = 4; the required com putation time for the CFD solution also increases by approximately a factor of 4. For three-dimensional flows, doubling the number of intervals in each direction increases the cell count by a factor of 23 = 8. You can see how grid independence studies can easily get beyond the range of a computer’s memory capacity and/or CPU availability. If you cannot double the number of intervals because of computer limitations, a good rule of thumb is that you should increase the number of intervals by at least 20 percent in all directions to test for grid independence.
On a final note about grid generation, the trend in CFD today is auto mated grid generation, coupled with automated grid refinement based on error estimates. Yet even in the face of these emerging trends, it is critical that you understand how the grid impacts the CFD solution.
Boundary Conditions While the equations of motion, the computational domain, and even the grid may be the same for two CFD calculations, the type of flow that is modeled is determined by the imposed boundary conditions. Appropriate boundary conditions are required in order to obtain an accurate CFD solution FIGURE 15–16 Examples of three-dimensional cells: (a) hexahedral, (b) prism, and (c) tetrahedral, along with the number of faces n for each case.
(a) n = 6 (b) n = 5 (c) n = 4 FIGURE 15–17 Example hybrid structured/polyhedral mesh with a structured mesh along the walls of the rocket where there is a thin boundary layer, and a polyhedral mesh expanding outward from the boundary layer mesh.
© Matthew Erdman. Reprinted by permission.
cen96537_ch15_885-946.indd 894 14/01/17 2:10 pm 895 CHAPTER 15 (Fig. 15–19). There are several types of boundary conditions available; the most relevant ones are listed and briefly described in the following. The names are those used by ANSYS-FLUENT; other CFD codes may use somewhat different terminology, and the details of their boundary conditions may differ. In the descriptions given, the words face or plane are used, implying three-dimensional flow. For a two-dimensional flow, the word edge or line should be substituted for face or plane.
Wall Boundary Conditions The simplest boundary condition is that of a wall. Since fluid cannot pass through a wall, the normal component of velocity is set to zero relative to the wall along a face on which the wall boundary condition is prescribed. In addition, because of the no-slip condition, we usually set the tangential component of velocity at a stationary wall to zero as well. In Fig. 15–19, for example, the upper and lower edges of this simple domain are spec ified as wall boundary conditions with no slip. If the energy equation is being solved, either wall temperature or wall heat flux must also be speci fied (but not both; see Section 15–4). If a turbulence model is being used, turbulence transport equations are solved, and wall roughness may need to be specified, since turbulent boundary layers are influenced greatly by the roughness of the wall. In addition, you must choose among various kinds of turbulence wall treatments (wall functions, etc.). These turbulence options are beyond the scope of the present text (see Wilcox, 2006); fortunately the default options of most modern CFD codes are sufficient for many applica tions involving turbulent flow.
Moving walls and walls with specified shear stresses can also be simulated in many CFD codes. There are situations where we desire to let the fluid slip along the wall (we call this an “inviscid wall”). For example, we can specify a zero-shear-stress wall boundary condition along the free surface of a swimming pool or hot tub when simulating such a flow (Fig. 15–20). Note that with this simplification, the fluid is allowed to “slip” along the surface, since the viscous shear stress caused by the air above it is negligibly small (Chap. 9). When making this approximation, however, surface waves and their associated pressure fluctuations cannot be taken into account.
Inflow/Outflow Boundary Conditions There are several options at the boundaries through which fluid enters the computational domain (inflow) or leaves the domain (outflow). They are generally categorized as either velocity-specified conditions or pressure-specified conditions. At a velocity inlet, we specify the velocity of the incoming flow along the inlet face. If energy and/or turbulence equations are being solved, the temperature and/or turbulence properties of the incoming flow need to be specified as well.
At a pressure inlet, we specify the total pressure along the inlet face (for example, flow coming into the computational domain from a pressurized tank of known pressure or from the far field where the ambient pressure is known). At a pressure outlet, fluid flows out of the computational domain. We specify the static pressure along the outlet face; in many cases this is atmospheric pressure (zero gage pressure). For example, the pressure is FIGURE 15–18 Time spent generating a good grid is time well spent.
Wall Computational domain Outlet Inlet Wall FIGURE 15–19 Boundary conditions must be carefully applied at all boundaries of the computational domain. Appropriate boundary conditions are required in order to obtain an accurate CFD solution.
cen96537_ch15_885-946.indd 895 14/01/17 2:10 pm 896 COMPUTATIONAL FLUID DYNAMICS atmospheric at the outlet of a subsonic exhaust pipe open to ambient air (Fig. 15–21). Flow properties, such as temperature, and turbulence proper ties are also specified at pressure inlets and pressure outlets. For outlets, however, these properties are not used unless the solution demands reverse flow across the outlet. Reverse flow at a pressure outlet is usually an indi cation that the computational domain is not large enough. If reverse flow warnings persist as the CFD solution iterates, the computational domain should be extended.
Pressure is not specified at a velocity inlet, as this would lead to math ematical overspecification, since pressure and velocity are coupled in the equations of motion. Rather, pressure at a velocity inlet adjusts itself to match the rest of the flow field. In similar fashion, velocity is not specified at a pressure inlet or outlet, as this would also lead to mathematical over specification. Rather, velocity at a pressure-specified boundary condition adjusts itself to match the rest of the flow field (Fig. 15–22).
Another option at an outlet of the computational domain is the outflow boundary condition. At an outflow boundary, no flow properties are speci fied; instead, flow properties such as velocity, turbulence quantities, and temperature are forced to have zero gradients normal to the outflow face (Fig. 15–23). For example, if a duct is sufficiently long so that the flow is fully developed at the outlet, the outflow boundary condition would be appropriate, since velocity does not change in the direction normal to the outlet face. Note that the flow direction is not constrained to be perpendic ular to the outflow boundary, as also illustrated in Fig. 15–23. If the flow is still developing, but the pressure at the outlet is known, a pressure outlet boundary condition would be more appropriate than an outflow boundary condition. The outflow boundary condition is often preferred over the pres sure outlet in rotating flows since the swirling motion leads to radial pressure gradients that are not easily handled by a pressure outlet.
A common situation in a simple CFD application is to specify one or more velocity inlets along portions of the boundary of the computational domain, and one or more pressure outlets or outflows at other portions of Pout Pin Pressure inlet; Pin specified Pressure outlet; Pout specified Outlet velocity calculated, not specified Inlet velocity calculated, not specified Computational domain FIGURE 15–22 At a pressure inlet or a pressure outlet, we specify the pressure on the face, but we cannot specify the velocity through the face. As the CFD solution converges, the velocity adjusts itself such that the prescribed pressure boundary conditions are satisfied.
The free surface is approximated as a wall boundary condition with slip (zero shear stress).
Velocity inlet Standard no-slip wall boundary condition Pressure outlet Pout Vin Computational domain FIGURE 15–20 The standard wall boundary condition is imposed on stationary solid boundaries, where we also impose either a wall temperature or a wall heat flux. The shear stress along the wall can be set to zero to simulate the free surface of a liquid, as shown here for the case of a swimming pool. There is slip along this “wall” that simulates the free surface (in contact with air).
Pressure outlet Pout = Patm FIGURE 15–21 When modeling an incompressible flow field, with the outlet of a pipe or duct exposed to ambient air, the proper boundary condition is a pressure outlet with Pout = Patm. Shown here is the tail pipe of an automobile. Photo by Po-Ya Abel Chuang. Used by permission.
cen96537_ch15_885-946.indd 896 14/01/17 2:10 pm 897 CHAPTER 15 the boundary, with walls defining the geometry of the rest of the computa tional domain. For example, in our swimming pool (Fig. 15–20), we set the left-most face of the computational domain as a velocity inlet and the bottom-most face as a pressure outlet. The rest of the faces are walls, with the free surface modeled as a wall with zero shear stress.
Finally, for compressible flow simulations, the inlet and outlet boundary conditions are further complicated by introduction of Riemann invariants and characteristic variables related to incoming and outgoing waves, discussion of which is beyond the scope of the present text. Fortunately, many CFD codes have a pressure far field boundary condition for compressible flows. This boundary condition is used to specify the Mach number, pressure, and temperature at an inlet. The same boundary condition can be applied at an outlet; when flow exits the computational domain, flow variables at the out let are extrapolated from the interior of the domain. Again, you must ensure that there is no reverse flow at an outlet.
Miscellaneous Boundary Conditions Some boundaries of a computational domain are neither walls nor inlets or outlets, but rather enforce some kind of symmetry or periodicity. For exam ple, the periodic boundary condition is useful when the geometry involves repetition. Flow field variables along one face of a periodic boundary are numerically linked to a second face of identical shape (and in most CFD codes, also identical face mesh). Thus, flow leaving (crossing) the first periodic boundary can be imagined as entering (crossing) the second peri odic boundary with identical properties (velocity, pressure, temperature, etc.). Periodic boundary conditions always occur in pairs and are useful for flows with repetitive geometries, such as flow between the blades of a turbomachine or through an array of heat exchanger tubes (Fig. 15–24). The periodic boundary condition enables us to work with a computational domain that is much smaller than the full flow field, thereby conserving computer resources. In Fig. 15–24, you can imagine an infinite number of repeated domains (dashed lines) above and below the actual computational domain (the light blue shaded region). Periodic boundary conditions must be specified as either translational (periodicity applied to two parallel faces, as in Fig. 15–24) or rotational (periodicity applied to two radially oriented faces). The region of flow between two neighboring blades of a fan (a flow passage) is an example of a rotationally periodic domain (see Fig. 15–58).
The symmetry boundary condition forces flow field variables to be mirror-imaged across a symmetry plane. Mathematically, gradients of most flow field variables in the direction normal to the symmetry plane are set to zero across the plane of symmetry, although some variables are specified as even functions and some as odd functions across a symmetry bound ary condition. For physical flows with one or more symmetry planes, this boundary condition enables us to model a portion of the physical flow domain, thereby conserving computer resources. The symmetry bound ary differs from the periodic boundary in that no “partner” boundary is required for the symmetry case. In addition, fluid can flow parallel to a symmetry boundary, but not through a symmetry boundary, whereas flow u x Outflow boundary FIGURE 15–23 At an outflow boundary condition, the gradient or slope of velocity normal to the outflow face is zero, as illustrated here for u as a function of x along a horizontal line. Note that neither pressure nor velocity are specified at an outflow boundary.
Computational domain Periodic Periodic Out In FIGURE 15–24 The periodic boundary condition is imposed on two identical faces. Whatever happens at one of the faces must also happen at its periodic partner face, as illustrated by the velocity vectors crossing the periodic faces.
cen96537_ch15_885-946.indd 897 14/01/17 2:10 pm 898 COMPUTATIONAL FLUID DYNAMICS can cross a periodic boundary. Consider, for example, flow across an array of heat exchanger tubes (Fig. 15–24). If we assume that no flow crosses the periodic boundaries of that computational domain, we can use symmetry boundary conditions instead. Alert readers will notice that we can even cut the size of the computational domain in half by wise choice of symmetry planes (Fig. 15–25).
For axisymmetric flows, the axis boundary condition is applied to a straight edge that represents the axis of symmetry (Fig. 15–26a). Fluid can flow parallel to the axis, but cannot flow through the axis. The axisymmet ric option enables us to solve the flow in only two dimensions, as sketched in Fig. 15–26b. The computational domain is simply a rectangle in the xy-plane; you can imagine rotating this plane about the x-axis to generate the axisymmetry. In the case of swirling axisymmetric flows, fluid may also flow tangentially in a circular path around the axis of symmetry. Swirling axisymmetric flows are sometimes called rotationally symmetric.
Internal Boundary Conditions The final classification of boundary conditions is imposed on faces or edges that do not define a boundary of the computational domain, but rather exist inside the domain. When an interior boundary condition is specified on a face, flow crosses through the face without any user-forced changes, just as it would cross from one interior cell to another (Fig. 15–27). This boundary condition is necessary for situations in which the computational domain is divided into separate blocks or zones, and enables communication between blocks. We have found this boundary condition to be useful for postprocess ing as well, since a predefined face is present in the flow field, on whose surface we can plot velocity vectors, pressure contours, etc. In more sophis ticated CFD applications in which there is a sliding or rotating mesh, the interface between the two blocks is called upon to smoothly transfer infor mation from one block to another.
Out Symmetry Symmetry Computational domain In FIGURE 15–25 The symmetry boundary condition is imposed on a face so that the flow across that face is a mirror image of the calculated flow. We sketch imaginary domains (dashed lines) above and below the computational domain (the light blue shaded region) in which the velocity vectors are mirror images of those in the computational domain. In this heat exchanger example, the left face of the domain is a velocity inlet, the right face is a pressure outlet or outflow outlet, the cylinders are walls, and both the top and bottom faces are symmetry planes.
FIGURE 15–26 The axis boundary condition is applied to the axis of symmetry (here the x-axis) in an axisymmetric flow, since there is rotational symmetry about that axis. (a) A slice defining the xy- or r𝜃-plane is shown, and the velocity components can be either (u, 𝜐) or (ur, u𝜃). (b) The computational domain (light blue shaded region) for this problem is reduced to a plane in two dimensions (x and y). In many CFD codes, x and y are used as axisymmetric coordinates, with y being understood as the distance from the x-axis.
Rotational symmetry (a) Axisymmetric body Axis y uθ ur υ u r x θ V (b) In Out y x Axis x y Computational domain Wall υ u V cen96537_ch15_885-946.indd 898 14/01/17 2:10 pm 899 CHAPTER 15 The fan boundary condition is specified on a plane across which a sudden pressure increase (or decrease) is to be assigned. This boundary condition is similar to an interior boundary condition except for the forced pressure rise. The CFD code does not solve the detailed, unsteady flow field through indi vidual fan blades, but simply models the plane as an infinitesimally thin fan that changes the pressure across the plane. The fan boundary condition is useful, for example, as a simple model of a fan inside a duct (Fig. 15–27), a ceiling fan in a room, or the propeller or jet engine that provides thrust to an airplane. If the pressure rise across the fan is specified as zero, this bound ary condition behaves the same as an interior boundary condition.
Practice Makes Perfect The best way to learn computational fluid dynamics is through examples and practice. You are encouraged to experiment with various grids, bound ary conditions, numerical parameters, etc., in order to get a feel for CFD. Before tackling a complicated problem, it is best to solve simpler problems, especially ones for which analytical or empirical solutions are known (for comparison and verification). For this reason, dozens of practice problems are provided on the book’s website.
In the following sections, we solve several example problems of general engineering interest to illustrate many of the capabilities and limitations of CFD. We start with laminar flows, and then provide some introductory turbulent flow examples. Finally we provide examples of flows with heat transfer, compressible flows, and liquid flows with free surfaces.
15–2 ■ LAMINAR CFD CALCULATIONS Computational fluid dynamics does an excellent job at computing incom pressible, steady or unsteady, laminar flow, provided that the grid is well resolved and the boundary conditions are properly specified. We show several simple examples of laminar flow solutions, paying particular attention to grid resolution and appropriate application of boundary conditions. In all examples in this section, the flows are incompressible and two-dimensional (or axisymmetric).
Pipe Flow Entrance Region at Re = 500 Consider flow of room-temperature water inside a smooth round pipe of length L = 40.0 cm and diameter D = 1.00 cm. We assume that the water enters at a uniform speed equal to V = 0.05024 m/s. The kinematic viscos ity of the water is ν = 1.005 × 10−6 m2/s, producing a Reynolds number of Re = VD/ν = 500. We assume incompressible, steady, laminar flow. We are interested in the entrance region in which the flow gradually becomes fully developed. Because of the axisymmetry, we set up a computational domain that is a two-dimensional slice from the axis to the wall, rather than a three-dimensional cylindrical volume (Fig. 15–28). We generate six structured grids for this computational domain: very coarse (40 intervals in the axial direction × 8 intervals in the radial direction), coarse (80 × 16), medium (160 × 32), fine (320 × 64), very fine (640 × 128), and ultrafine (1280 × 256). (Note that the number of intervals is doubled in both directions for In Fan Interior Out P P + ΔP FIGURE 15–27 The fan boundary condition imposes an abrupt change in pressure across the fan face to simulate an axial-flow fan in a duct. When the specified pressure rise is zero, the fan boundary condition degenerates to an interior boundary condition.
x Computational domain Axis Velocity inlet Wall Pressure outlet V D L r FIGURE 15–28 Because of axisymmetry about the x-axis, flow through a round pipe can be solved computationally with a two-dimensional slice through the pipe from r = 0 to D/2. The computational domain is the light blue shaded region, and the drawing is not to scale. Boundary conditions are indicated.
cen96537_ch15_885-946.indd 899 14/01/17 2:10 pm 900 COMPUTATIONAL FLUID DYNAMICS each successive grid; the number of computational cells increases by a factor of 4 for each grid.) In all cases the nodes are evenly distributed axially, but are concentrated near the wall radially, since we expect larger velocity gra dients near the pipe wall. Close-up views of the first three of these grids are shown in Fig. 15–29.
We run the CFD program ANSYS-FLUENT in double precision for all six cases. (Double precision arithmetic is not always necessary for engineer ing calculations—we use it here to obtain the best possible precision in our comparisons.) Since the flow is laminar, incompressible, and axisymmet ric, only three transport equations are solved—continuity, x-momentum, and y-momentum. Note that coordinate y is used in the CFD code instead of r as the distance from the axis of rotation (Fig. 15–26). The CFD code is run until convergence (all the residuals level off). Recall that a residual is a measure of how much the solution to a given transport equation deviates from exact; the lower the residual, the better the convergence. For the very coarse grid case, this occurs in about 500 iterations, and the residuals level off to less than 10−12 (relative to their initial values). The decay of the resid uals is plotted in Fig. 15–30 for the very coarse case. Note that for more complicated flow problems with finer grids, you cannot always expect such low residuals; in some CFD solutions, the residuals level off at much higher values, like 10−3.
We define P1 as the average pressure at an axial location one pipe diam eter downstream of the inlet. Similarly we define P20 at 20 pipe diameters. The average axial pressure drop from 1 to 20 diameters is thus ∆P = P1 − P20, and is equal to 4.404 Pa (to four significant digits of preci sion) for the very coarse grid case. Centerline pressure and axial velocity are plotted in Fig. 15–31a as functions of downstream distance. The solution 100 10–6 10–14 10–16 0 200 400 Iteration number 600 10–12 10–4 10–2 10–10 10–8 Continuity x-momentum y-momentum FIGURE 15–30 Decay of the residuals with iteration number for the very coarse grid laminar pipe flow solution (double precision arithmetic).
(a) (b) (c) FIGURE 15–29 Portions of the three coarsest structured grids generated for the laminar pipe flow example: (a) very coarse (40 × 8), (b) coarse (80 × 16), and (c) medium (160 × 32). The number of computational cells is 320, 1280, and 5120, respectively. In each view, the pipe wall is at the top and the pipe axis is at the bottom, as in Fig. 15–28.
cen96537_ch15_885-946.indd 900 14/01/17 2:10 pm 901 CHAPTER 15 appears to be physically reasonable. We see the increase of centerline axial velocity to conserve mass as the boundary layer on the pipe wall grows downstream. We see a sharp drop in pressure near the pipe entrance where viscous shear stresses on the pipe wall are highest. The pressure drop approaches linear toward the end of the entrance region where the flow is nearly fully developed, as expected. Finally, we compare in Fig. 15–31b the axial velocity profile at the end of the pipe to the known analytical solution for fully developed laminar pipe flow (see Chap. 8). The agreement is excellent, especially considering that there are only eight intervals in the radial direction.
Is this CFD solution grid independent? To find out, we repeat the calcu lations using the coarse, medium, fine, very fine, and ultrafine grids. The convergence of the residuals is qualitatively similar to that of Fig. 15–30 for all cases, but CPU time increases significantly as grid resolution improves, and the levels of the final residuals are not as low as those of the coarse resolution case. The number of iterations required until convergence also increases with improved grid resolution. The pressure drop from x/D = 1 to 20 is listed in Table 15–1 for all six cases. ∆P is also plotted as a func tion of number of cells in Fig. 15–32. We see that even the very coarse 8 5 0 2 1.5 (a) (b) 0 0 10 20 30 40 6 7 1 3 4 2 P gage, Pa 0.5 1 x/D V uCL 0.5 0.2 0 0 0.5 1 1.5 2 0.3 0.4 0.1 u/V r D CFD uCL/V P gage Analytical FIGURE 15–31 CFD results for the very coarse grid laminar pipe flow simulation: (a) development of centerline pressure and centerline axial velocity with downstream distance, and (b) axial velocity profile at pipe outlet compared to analytical prediction.
4.5 4.2 3.8 102 103 104 Number of cells ΔP, Pa 106 3.9 4.3 4.4 4 4.1 105 FIGURE 15–32 Calculated pressure drop from x /D = 1 to 20 in the entrance flow region of axisymmetric pipe flow as a function of number of cells.
TABLE 15–1 Pressure drop from x /D = 1 to 20 for the various grid resolution cases in the entrance flow region of axisymmetric pipe flow Case Number of Cells ∆P, Pa Very coarse 320 4.404 Coarse 1280 3.983 Medium 5120 3.998 Fine 20,480 4.016 Very fine 81,920 4.033 Ultrafine 327,680 4.035 cen96537_ch15_885-946.indd 901 14/01/17 2:10 pm 902 COMPUTATIONAL FLUID DYNAMICS grid does a reasonable job at predicting ∆P. The difference in pressure drop from the very coarse grid to the ultrafine grid is less than 10 percent. Thus, the very coarse grid may be adequate for some engineering calculations. If greater precision is needed, however, we must use a finer grid. We see grid independence to three significant digits by the very fine case. The change in ∆P from the very fine grid to the ultrafine grid is less than 0.07 percent—a grid as finely resolved as the ultrafine grid is unnecessary in any practical engineering analysis.
The most significant differences between the six cases occur very close to the pipe entrance, where pressure gradients and velocity gradients are largest. In fact, there is a singularity at the inlet, where the axial velocity changes suddenly from V to zero at the wall because of the no-slip con dition. We plot in Fig. 15–33 contour plots of normalized axial velocity, u/V near the pipe entrance. We see that although global properties of the flow field (like overall pressure drop) vary by only a few percent as the grid is refined, details of the flow field (like the velocity contours shown here) change considerably with grid resolution. You can see that as the grid is continually refined, the axial velocity contour shapes become smoother and more well defined. The greatest differences in the contour shapes occur near the pipe wall.
r x r x r x r x 1.1 1.2 1.3 (a) (b) (c) (d) 1.4 1.1 1.2 1.3 1.4 1.1 1.2 1.3 1.4 1.1 1.2 1.3 1.4 FIGURE 15–33 Normalized axial velocity contours (u/V) for the laminar pipe flow example. Shown is a close-up view of the entrance region of the pipe for each of the first four grids: (a) very coarse (40 × 8), (b) coarse (80 × 16), (c) medium (160 × 32), and (d ) fine (320 × 64).
cen96537_ch15_885-946.indd 902 14/01/17 2:10 pm 903 CHAPTER 15 Flow around a Circular Cylinder at Re = 150 To illustrate that reliable CFD results require correct problem formulation, consider the seemingly simple problem of steady, incompressible, two-dimensional flow of air over a circular cylinder of diameter D = 2.0 cm (Fig. 15–34). The two-dimensional computational domain used for this sim ulation is sketched in Fig. 15–35. Only the upper half of the flow field is solved, due to symmetry along the bottom edge of the computational domain; a symmetry boundary condition is specified along this edge to ensure that no flow crosses the plane of symmetry. With this boundary condition imposed, the required computational domain size is reduced by a factor of 2. A stationary, no-slip wall boundary condition is applied at the cylinder sur face. The left half of the far field outer edge of the domain has a velocity inlet boundary condition, on which is specified the velocity components u = V and 𝜐 = 0. A pressure outlet boundary condition is specified along the right half of the outer edge of the domain. (The gage pressure there is set to zero, but since the velocity field in an incompressible CFD code depends only on pressure differences, not absolute value of pressure, the value of pressure specified for the pressure outlet boundary condition is irrelevant.) Three two-dimensional structured grids are generated for comparison: coarse (30 radial intervals × 60 intervals along the cylinder surface = 1800 cells), medium (60 × 120 = 7200 cells), and fine (120 × 240 = 28,800 cells), as seen in Fig. 15–36. Note that only a small portion of the computational domain is shown here; the full domain extends 15 cylinder diameters outward from the origin, and the cells get progressively larger further away from the cylinder.
We apply a free-stream flow of air at a temperature of 25°C, at standard atmospheric pressure, and at velocity V = 0.1096 m/s from left to right around this circular cylinder. The Reynolds number of the flow, based on cylinder diameter (D = 2.0 cm), is thus Re = 𝜌VD/𝜇 = 150. Experiments at this Reynolds number reveal that the boundary layer is laminar and sepa rates almost 10° before the top of the cylinder, at 𝛼 ≅ 82° from the front stagnation point. The wake also remains laminar. Experimentally measured values of drag coefficient at this Reynolds number show much discrepancy in the literature; the range is from CD ≅ 1.1 to 1.4, and the differences are most likely due to the quality of the free-stream and three-dimensional effects (oblique vortex shedding, etc.). (Recall that CD = 2F D /𝜌V 2A, where A is the frontal area of the cylinder, and A = D times the span of the cylinder, taken as unit length in a two-dimensional CFD calculation.) CFD solutions are obtained for each of the three grids shown in Fig. 15–36, assuming steady laminar flow. All three cases converge without problems, but y x Cylinder D V FIGURE 15–34 Flow of fluid at free-stream speed V over a two-dimensional circular cylinder of diameter D.
Far field inflow (velocity inlet) Far field outflow (pressure outlet) Computational domain Cylinder surface (wall) Symmetry line (symmetry) V 0.3 m 0.02 m y α x FIGURE 15–35 Computational domain (light blue shaded region) used to simulate steady two-dimensional flow over a circular cylinder (not to scale). It is assumed that the flow is symmetric about the x-axis. Applied boundary conditions are shown for each edge in parentheses. We also define 𝛼, the angle measured along the cylinder sur face from the front stagnation point.
cen96537_ch15_885-946.indd 903 14/01/17 2:10 pm 904 COMPUTATIONAL FLUID DYNAMICS (a) (b) (c) FIGURE 15–36 Structured two-dimensional grids around the upper half of a circular cylinder: (a) coarse (30 × 60), (b) medium (60 × 120), and (c) fine (120 × 240). The bottom edge is a line of symmetry. Only a portion of each computational domain is shown—the domain extends well beyond the portion shown here.
cen96537_ch15_885-946.indd 904 14/01/17 2:10 pm 905 CHAPTER 15 the results do not necessarily agree with physical intuition or with experimen tal data. Streamlines are shown in Fig. 15–37 for the three grid resolutions. In all cases, the image is mirrored about the symmetry line so that even though only the top half of the flow field is solved, the full flow field is displayed.
For the coarse resolution case (Fig. 15–37a), the boundary layer separates at 𝛼 = 120°, well past the top of the cylinder, and CD is 1.00. The boundary layer is not well enough resolved to yield the proper boundary layer separation point, and the drag is somewhat smaller than it should be. Two large counter-rotating separation bubbles are seen in the wake; they stretch several cylin der diameters downstream. For the medium resolution case (Fig. 15–37b), the flow field is significantly different. The boundary layer separates a little further upstream at 𝛼 = 110°, which is more in line with the experimental results, but CD has decreased to about 0.982—further away from the experi mental value. The separation bubbles in the cylinder’s wake have grown much longer than those of the coarse grid case. Does refining the grid even further improve the numerical results? Figure 15–37c shows streamlines for the fine resolution case. The results look qualitatively similar to those of the medium resolution case, with 𝛼 = 109°, but the drag coefficient is even smaller (CD = 0.977), and the separation bubbles are even longer. A fourth calculation (not shown) at even finer grid resolution shows the same trend—the separation bubbles stretch downstream and the drag coefficient decreases somewhat.
Shown in Fig. 15–38 is a contour plot of tangential velocity component (u𝜃) for the medium resolution case. We plot values of u𝜃 over a very small range around zero, so that we can clearly see where along the cylinder the flow changes direction. This is thus a clever way to locate the separation point along a cylinder wall. Note that this works only for a circular cylinder because of its unique geometry. A more general way to determine the separa tion point is to identify the point along the wall where the wall shear stress 𝜏w is zero; this technique works for bodies of any shape. From Fig. 15–38, we see that the boundary layer separates at an angle of 𝛼 = 110° from the front stagnation point, much further downstream than the experimentally obtained value of 82°. In fact, all our CFD results predict boundary layer separation on the rear side rather than the front side of the cylinder.
These CFD results are unphysical—such elongated separation bubbles could not remain stable in a real flow situation, the separation point is too far downstream, and the drag coefficient is too low compared to experi mental data. Furthermore, repeated grid refinement does not lead to better results as we would hope; on the contrary, the results get worse with grid refinement. Why do these CFD simulations yield such poor agreement with experiment? The answer is twofold: 1. We have forced the CFD solution to be steady, when in fact flow over a circular cylinder at this Reynolds number is not steady. Experiments show that a periodic Kármán vortex street forms behind the cylinder (Tritton, 1977; see also Fig. 4–25 of this text).
2. All three cases in Fig. 15–37 are solved for the upper half-plane only, and symmetry is enforced about the x-axis. In reality, flow over a circular cylinder is highly nonsymmetric; vortices are shed alternately from the top and the bottom of the cylinder, forming the Kármán vortex street.
To correct both of these problems, we need to run an unsteady CFD simu lation with a full grid (top and bottom)—without imposing the symmetry (a) (b) (c) FIGURE 15–37 Streamlines produced by steady-state CFD calculations of flow over a circular cylinder at Re = 150: (a) coarse grid (30 × 60), (b) medium grid (60 × 120), and (c) fine grid (120 × 240). Note that only the top half of the flow is calculated—the bottom half is displayed as a mirror image of the top.
Separation point uθ > 0 uθ < 0 θ y x α FIGURE 15–38 Contour plot of tangential velocity component u𝜃 for flow over a circular cylinder at Re = 150 and for the medium grid resolution case (60 × 120). Values in the range −10−4 < u𝜃 < 10−4 m/s are plotted, so as to reveal the precise location of boundary layer separation, i.e., where u𝜃 changes sign just outside the cylinder wall, as sketched. For this case, the flow separates at 𝛼 = 110°.
cen96537_ch15_885-946.indd 905 14/01/17 2:10 pm 906 COMPUTATIONAL FLUID DYNAMICS condition. We run the simulation as an unsteady two-dimensional laminar flow, using the computational domain sketched in Fig. 15–39. The top and bottom (far field) edges are specified as a periodic boundary condition pair so that nonsymmetric oscillations in the wake are not suppressed (flow can cross these boundaries as necessary). The far field boundaries are also very far away (75 to 200 cylinder diameters), so that their effect on the calcula tions is insignificant.
The mesh is very fine near the cylinder to resolve the boundary layer. The grid is also fine in the wake region to resolve the shed vortices as they travel downstream. For this particular simulation, we use a hybrid grid somewhat like that shown in Fig. 15–14. The fluid is air, the cylinder diam eter is 1.0 m, and the free-stream air speed is set to 0.00219 m/s. These values produce a Reynolds number of 150 based on cylinder diameter. Note that the Reynolds number is the important parameter in this problem— the choices of D, V, and type of fluid are not critical, so long as they pro duce the desired Reynolds number (Fig. 15–40).
As we march in time, small nonuniformities in the flow field amplify, and the flow becomes unsteady and antisymmetric with respect to the x-axis. A Kármán vortex street forms naturally. After sufficient CPU time, the simulated flow settles into a periodic vortex shedding pattern, much like the real flow. A contour plot of vorticity at one instant in time is shown in Fig. 15–41, along with a photograph showing streaklines of the same flow obtained experimentally in a wind tunnel. It is clear from the CFD simulation that the Kármán vortices decay downstream, since the magnitude of vortic ity in the vortices decreases with downstream distance. This decay is partly physical (viscous), and partly artificial (numerical dissipation). Nevertheless, physical experiments verify the decay of the Kármán vortices. The decay is not so obvious in the streakline photograph (Fig. 15–41b); this is due to the time-integrating property of streaklines, as was pointed out in Chap. 4. A close-up view of vortices shedding from the cylinder at a particular instant in time is shown in Fig. 15–42, again with a comparison between CFD results Far field inflow (velocity inlet) Far field edge (periodic) Far field edge (periodic) Far field outflow (pressure outlet) Cylinder surface (wall) 75D 200D V D y x FIGURE 15–39 Computational domain (light blue shaded region) used to simulate unsteady, two-dimensional, laminar flow over a circular cylinder (not to scale). Applied boundary conditions are in parentheses.
= Re = ρVD μ VD v Reynolds number is defined as for flow at free-stream speed V over a circular cylinder of diameter D in a fluid of density ρ and dynamic viscosity μ (kinematic viscosity v).
FIGURE 15–40 In an incompressible CFD simulation of flow around a cylinder, the choice of free-stream speed, cylinder diameter, or even type of fluid is not critical, provided that the desired Reynolds number is achieved.
cen96537_ch15_885-946.indd 906 14/01/17 2:10 pm 907 CHAPTER 15 x/D y/D y/D D D x/D 10 0 20 30 40 50 60 70 80 90 100 (a) 10 0 20 30 40 50 60 70 80 90 100 (b) FIGURE 15–41 Laminar flow in the wake of a circular cylinder at Re ≅ 150: (a) an instantaneous snapshot of vorticity contours produced by CFD, and (b) time-integrated streaklines produced by a smoke wire located at x/D = 5. The vorticity contours show that Kármán vortices decay rapidly in the wake, whereas the streaklines retain a “memory” of their history from upstream, making it appear that the vortices continue for a great distance downstream.
From Cimbala, et al., 1988.
1 0 2 3 4 5 6 7 8 9 10 11 (a) x/D (b) D D 1 0 2 3 4 5 6 7 8 9 10 11 x/D y/D y/D FIGURE 15–42 Close-up view of vortices shedding from a circular cylinder: (a) instantaneous vorticity contour plot produced by CFD at Re = 150, and (b) dye streaklines produced by dye introduced at the cylinder surface at Re = 140. An animated version of this CFD picture is available on the book’s website.
(b) Reprinted by permission of Sadatoshi Taneda.
cen96537_ch15_885-946.indd 907 14/01/17 2:10 pm 908 COMPUTATIONAL FLUID DYNAMICS and experimental results—this time from experiments in a water channel. An animated version of Fig. 15–42 is provided on the book’s website so that you can watch the dynamic process of vortex shedding.
We compare the CFD results to experimental results in Table 15–2. The calculated time-averaged drag coefficient on the cylinder is 1.14. As men tioned previously, experimental values of CD at this Reynolds number vary from about 1.1 to 1.4, so the agreement is within the experimental scatter. Note that the present simulation is two-dimensional, inhibiting any kind of oblique vortex shedding or other three-dimensional nonuniformities. This may be why our calculated drag coefficient is on the lower end of the reported experimental range. The Strouhal number of the Kármán vortex street is defined as Strouhal number: St = fsheddingD V (15–4) where fshedding is the shedding frequency of the vortex street. From our CFD simulation, we calculate St = 0.16. The experimentally obtained value of Strouhal number at this Reynolds number is about 0.18 (Williamson, 1989), so again the agreement is reasonable, although the CFD results are a bit low compared to experiment. Perhaps a finer grid would help somewhat, but the major reason for the discrepancy is more likely due to unavoidable three-dimensional effects in the experiments, which are not present in these two-dimensional simulations. Overall this CFD simulation is a success, as it captures all the major physical phenomena in the flow field.
This exercise with “simple” laminar flow over a circular cylinder has dem onstrated some of the capabilities of CFD, but has also revealed several aspects of CFD about which one must be cautious. Poor grid resolution can lead to incorrect solutions, particularly with respect to boundary layer separation, but continued refinement of the grid does not necessarily lead to more physically correct results, particularly if the boundary conditions are not set appropriately (Fig. 15–43). For example, forced numerical flow symmetry is not always wise, even for cases in which the physical geometry is entirely symmetric.
Symmetric geometry does not guarantee symmetric flow.
In addition, forced steady flow may yield incorrect results when the flow is inherently unstable and/or oscillatory. Likewise, forced two-dimensionality may yield incorrect results when the flow is inherently three-dimensional.
How then can we ensure that a laminar CFD calculation is correct? Only by systematic study of the effects of computational domain size, grid resolu tion, boundary conditions, flow regime (steady or unsteady, 2-D or 3-D, etc.), along with experimental validation. As with most other areas of engineering, experience is of paramount importance.
15–3 ■ TURBULENT CFD CALCULATIONS CFD simulations of turbulent flow are much more difficult than those of laminar flow, even for cases in which the flow field is steady in the mean (statisticians refer to this condition as stationary). The reason is that the finer features of the turbulent flow field are always unsteady and three-dimensional—random, swirling, vortical structures called turbulent eddies FIGURE 15–43 Poor grid resolution can lead to incorrect CFD results, but a finer grid does not guarantee a more physically correct solution. If the boundary conditions are not specified properly, the results may be unphysical, regardless of how fine the grid.
The main cause of the disagreement is most likely due to three-dimensional effects rather than grid resolution or numerical issues.
TABLE 15–2 Comparison of CFD results and experimental results for unsteady laminar flow over a circular cylinder at Re = 150 CD St Experiment 1.1 to 1.4 0.18 CFD 1.14 0.16 cen96537_ch15_885-946.indd 908 14/01/17 2:10 pm 909 CHAPTER 15 of all orientations arise in a turbulent flow (Fig. 15–44). Some CFD cal culations use a technique called direct numerical simulation (DNS), in which an attempt is made to resolve the unsteady motion of all the scales of the turbulent flow. However, the size difference and the time scale differ ence between the largest and smallest eddies can be several orders of magni tude (L ≫ 𝜂 in Fig. 15–44). Furthermore, these differences increase with the Reynolds number (Tennekes and Lumley, 1972), making DNS calculations of turbulent flows even more difficult as the Reynolds number increases. DNS solutions require extremely fine, fully three-dimensional grids, large computers, and an enormous amount of CPU time. With today’s computers, DNS results are not yet feasible for practical high Reynolds number turbu lent flows of engineering interest such as flow over a full-scale airplane. This situation is not expected to change for several more decades, even if the fantastic rate of computer improvement continues at today’s pace.
Thus, we find it necessary to make some simplifying assumptions in order to simulate complex, high Reynolds number, turbulent flow fields. The next level below DNS is large eddy simulation (LES). With this technique, large unsteady features of the turbulent eddies are resolved, while small-scale dis sipative turbulent eddies are modeled (Fig. 15–45). The basic assumption is that the smaller turbulent eddies are isotropic; i.e., it is assumed that the small eddies are independent of coordinate system orientation and always behave in a statistically similar and predictable way, regardless of the tur bulent flow field. LES requires significantly less computer resources than does DNS, because we eliminate the need to resolve the smallest eddies in the flow field. In spite of this, the computer requirements for practical engi neering analysis and design are nevertheless still formidable using today’s technology. Further discussion about DNS and LES is beyond the scope of the present text, but these are areas of much current research.
The next lower level of sophistication is to model all the unsteady tur bulent eddies with some kind of turbulence model. No attempt is made to resolve the unsteady features of any of the turbulent eddies, not even the largest ones (Fig. 15–46). Instead, mathematical models are employed to take into account the enhanced mixing and diffusion caused by turbulent eddies. For simplicity, we consider only steady (that is, stationary), incom pressible flow. When using a turbulence model, the steady Navier–Stokes equation (Eq. 15–2) is replaced by what is called the Reynolds-averaged Navier–Stokes (RANS) equation, shown here for steady (stationary), incompressible, turbulent flow, Steady RANS equation: (V → ·∇ → )V → = −1 ρ ∇ → Pʹ + 𝜈∇2V → + ∇ → · (𝜏 ij, turbulent) (15–5) Compared to Eq. 15–2, there is an additional term on the right side of Eq. 15–5 that accounts for the turbulent fluctuations. 𝜏ij, turbulent is a tensor known as the specific Reynolds stress tensor, so named because it acts in a similar fashion as the viscous stress tensor 𝜏ij (Chap. 9). In Cartesian coordinates, 𝜏ij, turbulent is 𝜏 ij, turbulent = − ( uʹ2 uʹ𝜐ʹ uʹwʹ uʹ𝜐ʹ 𝜐ʹ2 𝜐ʹwʹ uʹwʹ 𝜐ʹwʹ wʹ2 ) (15–6) η L FIGURE 15–44 All turbulent flows, even those that are steady in the mean (stationary), contain unsteady, three-dimensional turbulent eddies of various sizes. Shown is the average velocity profile and some of the eddies; the smallest turbulent eddies (size 𝜂) are orders of magnitude smaller than the largest turbulent eddies (size L). Direct numerical simulation (DNS) is a CFD technique that simulates all relevant turbulent eddies in the flow.
FIGURE 15–45 Large eddy simulation (LES) is a simplification of direct numerical simulation in which only the large turbulent eddies are resolved—the small eddies are modeled, significantly reducing computer requirements. Shown is the average velocity profile and the resolved eddies.
cen96537_ch15_885-946.indd 909 14/01/17 2:10 pm 910 COMPUTATIONAL FLUID DYNAMICS where the overbar indicates the time average of the product of two fluc tuating velocity components and primes denote fluctuating velocity com ponents. Since the Reynolds stress is symmetric, six (rather than nine) additional unknowns are introduced into the problem. These new unknowns are modeled in various ways by turbulence models. A detailed description of turbulence models is beyond the scope of this text; you are referred to Wilcox (2006) or Chen and Jaw (1998) for further discussion.
There are many turbulence models in use today, including algebraic, one-equation, two-equation, and Reynolds stress models. Three of the most pop ular turbulence models are the k-𝜀 model, the k-𝜔 model, and the q-𝜔 model. These so-called two-equation turbulence models add two more transport equations, which must be solved simultaneously with the equations of mass and linear momentum (and also energy if this equation is being used). Along with the two additional transport equations that must be solved when using a two-equation turbulence model, two additional boundary conditions must be specified for the turbulence properties at inlets and at outlets. (Note that the properties specified at outlets are not used unless reverse flow is encoun tered at the outlet.) For example, in the k-𝜀 model you may specify both k (turbulent kinetic energy) and 𝜀 (turbulent dissipation rate). However, appropriate values of these variables are not always known. A more useful option is to specify turbulence intensity I (ratio of characteristic turbulent eddy velocity to free-stream velocity or some other characteristic or aver age velocity) and turbulent length scale ℓ (characteristic length scale of the energy-containing turbulent eddies). If detailed turbulence data are not available, a good rule of thumb at inlets is to set I to 10 percent and to set ℓ to one-half of some characteristic length scale in the flow field (Fig. 15–47).
We emphasize that turbulence models are approximations that rely heavily on empirical constants for mathematical closure of the equations. The models are calibrated with the aid of direct numerical simulation and exper imental data obtained from simple flow fields like flat plate boundary lay ers, shear layers, and isotropic decaying turbulence downstream of screens. Unfortunately, no turbulence model is universal, meaning that although the model works well for flows similar to those used for calibration, it is not guaranteed to yield a physically correct solution when applied to general turbulent flow fields, especially those involving flow separation and reattachment and/or large-scale unsteadiness.
Turbulent flow CFD solutions are only as good as the appropriateness and validity of the turbulence model used in the calculations.
We emphasize also that this statement remains true regardless of how fine we make the computational grid. When applying CFD to laminar flows, we can usually improve the physical accuracy of the simulation by refining the grid (provided that the boundary conditions are properly specified, of course). This is not always the case for turbulent flow CFD analyses using turbulence models, even when the boundary conditions are correct. While a refined grid produces better numerical accuracy, the physical accuracy of the solution is always limited by the physical accuracy of the turbulence model itself.
With these cautions in mind, we now present some practical examples of CFD calculations of turbulent flow fields. In all the turbulent flow examples FIGURE 15–46 When a turbulence model is used in a CFD calculation, all the turbulent eddies are modeled, and only Reynolds-averaged flow properties are calculated. Shown is the average velocity profile. There are no resolved turbulent eddies.
• V • I • 퓁 Velocity inlet: D D FIGURE 15–47 A useful rule of thumb for turbulence properties at a pressure inlet or velocity inlet boundary condition is to specify a turbulence intensity of 10 percent and a turbulent length scale of one-half of some characteristic length scale in the problem (ℓ = D/2).
cen96537_ch15_885-946.indd 910 14/01/17 2:10 pm 911 CHAPTER 15 discussed in this chapter, we employ the k-𝜀 turbulence model with wall functions. This model is the default turbulence model in many commercial CFD codes such as ANSYS-FLUENT. In all cases we assume stationary flow; no attempt is made to model unsteady features of the flow, such as vortex shedding in the wake of a bluff body. It is assumed that the tur bulence model accounts for all the inherent unsteadiness due to turbu lent eddies in the flow field. Note that unsteady (nonstationary) turbulent flows are also solvable with turbulence models, through the use of time-marching schemes (unsteady RANS calculations), but only when the time scale of the unsteadiness is much longer than that of individual turbulent eddies. For example, suppose you are calculating the forces and moments on a blimp during a gust of wind (Fig. 15–48). At the inlet boundary, you would impose the time-varying wind velocity and turbulence levels, and an unsteady turbulent flow solution could then be calculated using turbulence models. The large-scale, overall features of the flow (flow separation, forces and moments on the body, etc.) would be unsteady, but the fine-scale fea tures of the turbulent boundary layer, for example, would be modeled by the quasi-steady turbulence model.
Flow around a Circular Cylinder at Re = 10,000 As our first example of a turbulent flow CFD solution, we calculate flow over a circular cylinder at Re = 10,000. For illustration, we use the same two-dimensional computational domain that was used for the laminar cylin der flow calculations, as sketched in Fig. 15–35. As with the laminar flow calculation, only the upper half of the flow field is solved here, due to sym metry along the bottom edge of the computational domain. We use the same three grids used for the laminar flow case as well—coarse, medium, and fine resolution (Fig. 15–36). We point out, however, that grids designed for turbulent flow calculations (especially those employing turbulence models with wall functions) are generally not the same as those designed for lami nar flow of the same geometry, particularly near walls.
We apply a free-stream flow of air at 25°C and at velocity V = 7.304 m/s from left to right around this circular cylinder. The Reynolds number of the flow, based on cylinder diameter (D = 2.0 cm), is approximately 10,000. Experiments at this Reynolds number reveal that the boundary layer is lam inar and separates several degrees upstream of the top of the cylinder (at 𝛼 ≅ 82°). The wake, however, is turbulent; such a mixture of laminar and turbulent flow is particularly difficult for CFD codes. The measured drag coefficient at this Reynolds number is CD ≅ 1.15 (Tritton, 1977). CFD solu tions are obtained for each of the three grids, assuming stationary (steady in the mean) turbulent flow. We employ the k-𝜀 turbulence model with wall functions. The inlet turbulence level is set to 10 percent with a length scale of 0.01 m (half of the cylinder diameter). All three cases converge nicely. Streamlines are plotted in Fig. 15–49 for the three grid resolution cases. In each plot, the image is mirrored about the symmetry line so that even though only the top half of the flow field is solved, the full flow field is visualized.
For the coarse resolution case (Fig. 15–49a), the boundary layer separates well past the top of the cylinder, at 𝛼 ≊ 140°. Furthermore, the drag coef ficient CD is only 0.647, almost a factor of 2 smaller than it should be. Let’s FL FD V(t) FIGURE 15–48 While most CFD calculations with turbulence models are stationary (steady in the mean), it is also possible to calculate unsteady turbulent flow fields using turbulence models. In the case of flow over a body, we may impose unsteady boundary conditions and march in time to predict gross features of the unsteady flow field.
cen96537_ch15_885-946.indd 911 14/01/17 2:10 pm 912 COMPUTATIONAL FLUID DYNAMICS see if a finer mesh improves the agreement with experimental data. For the medium resolution case (Fig. 15–49b), the flow field is significantly different. The boundary layer separates nearer to the top of the cylinder, at 𝛼 = 104°, and CD has increased to about 0.742—closer, but still significantly less than the experimental value. We also notice that the recirculating eddies in the cylinder’s wake have grown in length by nearly a factor of 2 compared to those of the coarse grid case. Figure 15–49c shows streamlines for the fine resolution case. The results look very similar to those of the medium resolu tion case, and the drag coefficient has increased only slightly (CD = 0.753). The boundary layer separation point for this case is at 𝛼 = 102°.
Further grid refinement (not shown) does not change the results signifi cantly from those of the fine grid case. In other words, the fine grid appears to be sufficiently resolved, yet the results do not agree with experiment. Why? There are several problems with our calculations: we are modeling a steady flow, even though the actual physical flow is unsteady; we are enforc ing symmetry about the x-axis, even though the physical flow is unsymmetric (a Kármán vortex street can be observed in experiments at this Reynolds number); and we are using a turbulence model instead of resolving all the small eddies of the turbulent flow. Another significant source of error in our calculations is that the CFD code is run with turbulence turned on in order to reasonably model the wake region, which is turbulent; however, the boundary layer on the cylinder surface is actually still laminar. The pre dicted location of the separation point downstream of the top of the cylinder is more in line with turbulent boundary layer separation, which does not occur until much higher values of Reynolds number (after the “drag crisis” at Re greater than 2 × 105).
The bottom line is that CFD codes have a hard time in the transitional regime between laminar and turbulent flow, and when there is a mixture of laminar and turbulent flow in the same computational domain. In fact, most commer cial CFD codes give the user a choice between laminar and turbulent—there is no “middle ground.” In the present calculations, we model the boundary layer as turbulent, even though the physical boundary layer is laminar; it is not surprising, then, that the results of our calculations do not agree well with experiment. If we would have instead specified laminar flow over the entire computational domain, the CFD results would have been even worse (less physical).
Is there any way around this problem of poor physical accuracy for the case of mixed laminar and turbulent flow? Perhaps. In some CFD codes you can specify the flow to be laminar or turbulent in different regions of the flow. But even then, the transitional process from laminar to turbulent flow is somewhat abrupt, again not physically correct. Furthermore, you would need to know where the transition takes place in advance—this defeats the purpose of a stand-alone CFD calculation for fluid flow prediction. Advanced wall treatment models are being generated that may some day do a better job in the transitional region. In addition, some new turbulence models are being developed that are better tuned to low Reynolds number turbulence. Transition is an area of active research in CFD.
In summary, we cannot accurately model the mixed laminar/turbulent flow problem of flow over a circular cylinder at Re ∽ 10,000 using standard turbulence models and the steady RANS equation. It appears that accurate (a) (b) (c) FIGURE 15–49 Streamlines produced by CFD calculations of stationary turbulent flow over a circular cylinder at Re = 10,000: (a) coarse grid (30 × 60), (b) medium grid (60 × 120), and (c) fine grid (120 × 240). Note that only the top half of the flow is calculated—the bottom half is displayed as a mirror image of the top.
cen96537_ch15_885-946.indd 912 14/01/17 2:10 pm 913 CHAPTER 15 results can be obtained only through use of time-accurate (unsteady RANS), LES, or DNS solutions that are orders of magnitude more computationally demanding.
Flow around a Circular Cylinder at Re = 107 As a final cylinder example, we use CFD to calculate flow over a circular cylinder at Re = 107—well beyond the drag crisis. The cylinder for this case is of 1.0 m diameter, and the fluid is water. The free-stream velocity is 10.05 m/s. At this value of Reynolds number the experimentally mea sured value of drag coefficient is around 0.7 (Tritton, 1977). The bound ary layer is turbulent at the separation point, which occurs at around 120°. Thus we do not have the mixed laminar/turbulent boundary layer problem as in the lower Reynolds number example—the boundary layer is turbulent everywhere except near the nose of the cylinder, and we should expect bet ter results from the CFD predictions. We use a two-dimensional half-grid similar to that of the fine resolution case of the previous examples, but the mesh near the cylinder wall is adapted appropriately for this high Reynolds number. As previously, we use the k-𝜀 turbulence model with wall functions. The inlet turbulence level is set to 10 percent with a length scale of 0.5 m. Unfortunately, the drag coefficient is calculated to be 0.262—less than half of the experimental value at this Reynolds number. Streamlines are shown in Fig. 15–50. The boundary layer separates a bit too far downstream, at 𝛼 = 129°. There are several possible reasons for the discrepancy. We are forcing the simulated flow to be steady and symmetric, whereas the actual flow is neither, due to vortex shedding. (Vortices are shed even at high Reynolds numbers.) In addition, the turbulence model and its near wall treatment (wall functions) may not be capturing the proper physics of the flow field. Again we must conclude that accurate results for flow over a cir cular cylinder can be obtained only through use of a full grid rather than a half grid, and with time-accurate (unsteady RANS), LES, or DNS solutions that are orders of magnitude more computationally demanding.
Design of the Stator for a Vane-Axial Flow Fan The next turbulent flow CFD example involves design of the stator for a vane-axial flow fan that is to be used to drive a wind tunnel. The overall fan diameter is D = 1.0 m, and the design point of the fan is at an axial-flow speed of V = 50 m/s. The stator vanes span from radius r = rhub = 0.25 m at the hub to r = rtip = 0.50 m at the tip. The stator vanes are upstream of the rotor blades in this design (Fig. 15–51). A preliminary stator vane shape is chosen that has a trailing edge angle of 𝛽st = 63° and a chord length of 20 cm. At any value of radius r, the actual amount of turning depends on the number of stator vanes—we expect that the fewer the number of vanes, the smaller the average angle at which the flow is turned by the stator vanes because of the greater spacing between vanes. It is our goal to determine the minimum number of stator vanes required so that the flow impinging on the leading edges of the rotor blades (located one chord length downstream of the stator vane trailing edges) is turned at an average angle of at least 45°. We also require there to be no significant flow separation from the stator vane surface.
FIGURE 15–50 Streamlines produced by CFD calculations of stationary turbulent flow over a circular cylinder at Re = 107. Unfortunately, the predicted drag coefficient is still not accurate for this case.
Rotor 휔 Stator V 훽st Hub and motor Vane tip Vane hub r D 휔r FIGURE 15–51 Schematic diagram of the vane-axial flow fan being designed. The stator precedes the rotor, and the flow through the stator vanes is to be modeled with CFD.
cen96537_ch15_885-946.indd 913 14/01/17 2:10 pm 914 COMPUTATIONAL FLUID DYNAMICS As a first approximation, we model the stator vanes at any desired value of r as a two-dimensional cascade of vanes (see Chap. 14). Each vane is separated by blade spacing s at this radius, as defined in Fig. 15–52. We use CFD to predict the maximum allowable value of s, from which we esti mate the minimum number of stator vanes that meet the given requirements of the design.
Since the flow through the two-dimensional cascade of stator vanes is infi nitely periodic in the y-direction, we need to model only one flow passage through the vanes, specifying two pairs of periodic boundary conditions on the top and bottom edges of the computational domain (Fig. 15–53). We run six cases, each with a different value of blade spacing. We choose s = 10, 20, 30, 40, 50, and 60 cm, and generate a structured grid for each of these values of blade spacing. The grid for the case with s = 20 cm is shown in Fig. 15–54; the other grids are similar, but more intervals are specified in the y-direction as s increases. Notice how we have made the grid spacing fine near the pressure and suction surfaces so that the boundary layer on these surfaces can be better resolved. We specify V = 50 m/s at the velocity inlet, zero gage pressure at the pressure outlet, and a smooth wall boundary condition with no slip at both the pressure and suction surfaces. Since we are modeling the flow with a turbulence model (k-𝜀 with wall functions), we must also specify turbulence properties at the velocity inlet. For these simulations we specify a turbulence intensity of 10 percent and a turbulence length scale of 0.01 m (1.0 cm).
We run the CFD calculations long enough to converge as far as possible for all six cases, and we plot streamlines in Fig. 15–55 for six blade spac ings: s = 10, 20, 30, 40, 50, and 60 cm. Although we solve for flow through only one flow passage, we plot several duplicate flow passages, stacked one on top of the other, in order to visualize the flow field as a periodic cascade. The streamlines for the first three cases look very similar at first glance, but closer inspection reveals that the average angle of flow down stream of the trailing edge of the stator vane decreases with s. (We define (b) (a) y r D s at r = rtip rhub rtip x s s c s FIGURE 15–52 Definition of blade spacing s: (a) frontal view of the stator, and (b) the stator modeled as a two-dimensional cascade in edge view. Twelve radial stator vanes are shown in the frontal view, but the actual number of vanes is to be determined. Three stator vanes are shown in the cascade, but the actual cascade consists of an infinite number of vanes, each displaced by blade spacing s, which increases with radius r. The two-dimensional cascade is an approximation of the three-dimensional flow at one value of radius r and blade spacing s. Chord length c is defined as the horizontal length of the stator vane.
V Translationally periodic 2 Velocity inlet Translationally periodic 1 Pressure surface Pressure outlet Suction surface s y x FIGURE 15–53 Computational domain (light blue shaded region) defined by one flow passage through two stator vanes. The top wall of the passage is the pressure surface, and the bottom wall is the suction surface. Two translationally periodic pairs are defined: periodic 1 upstream and periodic 2 downstream.
cen96537_ch15_885-946.indd 914 14/01/17 2:10 pm 915 CHAPTER 15 flow angle 𝛽 relative to horizontal as sketched in Fig. 15–55a.) Also, the gap (white space) between the wall and the closest streamline to the suction surface increases in size as s increases, indicating that the flow speed in that region decreases. In fact, it turns out that the boundary layer on the suction surface of the stator vane must resist an ever-increasingly adverse pressure gradient (decelerating flow speed and positive pressure gradient) as blade spacing is increased. At large enough s, the boundary layer on the suction surface cannot withstand the severely adverse pressure gradient and sepa rates off the wall. For s = 40, 50, and 60 cm (Fig. 15–55d through f ), flow separation off the suction surface is clearly seen in these streamline plots. Furthermore, the severity of the flow separation increases with s. This is not unexpected if we imagine the limit as s →∞. In that case, the stator vane is isolated from its neighbors, and we surely expect massive flow separation since the vane has such a high degree of camber.
We list average outlet flow angle 𝛽avg, average outlet flow speed Vavg, and predicted drag force on a stator vane per unit depth FD /b in Table 15–3 as functions of blade spacing s. (Depth b is into the page of Fig. 15–55 and is assumed to be 1 m in two-dimensional calculations such as these.) While 𝛽avg and Vavg decrease continuously with s, FD /b first rises to a maximum for the s = 20 cm case, and then decreases from there on. You may recall from the previously stated design criteria for this exam ple that the average outlet flow angle must be greater than 45°, and there must be no significant flow separation. From our CFD results, it appears that both of these criteria break down somewhere between s = 30 and 40 cm. We obtain a better picture of flow separation by plotting vorticity con tours (Fig. 15–56). In these color contour plots, blue represents large nega tive vorticity (clockwise rotation), red represents large positive vorticity (counterclockwise rotation), and green is zero vorticity. When the boundary layer remains attached, we expect the vorticity to be concentrated within thin FIGURE 15–54 Structured grid for the two-dimensional stator vane cascade at blade spacing s = 20 cm. The outflow region in the wake of the vanes is inten tionally longer than that at the inlet to avoid backflow at the pressure outlet in case of flow separation on the suction surface of the stator vane. The outlet is one chord length downstream of the stator vane trailing edges; the outlet is also the location of the leading edges of the rotor blades (not shown).
All calculated values are reported to three significant digits. The CFD calculations are performed using the k-𝜀 turbulence model with wall functions.
TABLE 15–3 Variation of average outlet flow angle 𝛽avg, average outlet flow speed Vavg, and predicted drag force per unit depth FD /b as functions of blade spacing s 𝛽avg, Vavg, FD /b, s, cm degrees m/s N/m 1 0 60.8 103 554 20 56.1 89.6 722 30 49.7 77.4 694 40 43.2 68.6 612 50 37.2 62.7 538 60 32.3 59.1 489 cen96537_ch15_885-946.indd 915 14/01/17 2:10 pm 916 COMPUTATIONAL FLUID DYNAMICS (a) (b) FIGURE 15–56 Vorticity contour plots produced by CFD calculations of stationary turbulent flow through a stator vane flow passage: blade spacing (a) s = 30 cm and (b) s = 40 cm. The flow field is largely irrotational (zero vorticity) except in the thin boundary layer along the walls and in the wake region. However, when the boundary layer separates, as in case (b), the vorticity spreads throughout the separated flow region.
(a) (b) (c) (d) (e) ( f ) β FIGURE 15–55 Streamlines produced by CFD calculations of stationary turbulent flow through a stator vane flow passage: (a) blade spacing s = 10 cm, (b) 20 cm, (c) 30 cm, (d ) 40 cm, (e) 50 cm, and ( f ) 60 cm. The CFD cal culations are performed using the k-𝜀 turbulence model with wall functions. Flow angle 𝛽 is defined in image (a) as the average angle of flow, relative to horizontal, just downstream of the trailing edge of the stator vane.
cen96537_ch15_885-946.indd 916 14/01/17 2:10 pm 917 CHAPTER 15 boundary layers along the stator vane surfaces, as is the case in Fig. 15–56a for s = 30 cm. When the boundary layer separates, however, the vorticity suddenly spreads out away from the suction surface, as seen in Fig. 15–56b for s = 40 cm. These results verify that significant flow separation occurs somewhere between s = 30 and 40 cm. As a side note, notice how the vortic ity is concentrated not only in the boundary layer, but also in the wake for both cases shown in Fig. 15–56.
Finally, we compare velocity vector plots in Fig. 15–57 for three cases: s = 20, 40, and 60 cm. We generate several equally spaced parallel lines in the computational domain; each line is tilted at 45° from the horizontal. Velocity vectors are then plotted along each of these parallel lines. When s = 20 cm (Fig. 15–57a), the boundary layer remains attached on both the suction and pressure surfaces of the stator vane all the way to its trailing edge. When s = 40 cm (Fig. 15–57b), flow separation and reverse flow along the suction surface appears. When s = 60 cm (Fig. 15–57c), the separa tion bubble and the reverse flow region have grown—this is a “dead” flow region, in which the air speeds are very small. In all cases, the flow on the pressure surface (lower left side) of the stator vane remains attached. How many vanes (N) does a blade spacing of s = 30 cm represent? We can easily calculate N by noting that at the vane tip (r = rtip = D/2 = 50 cm), where s is largest, the total available circumference (C) is Available circumference: C = 2𝜋rtip = 𝜋D (15–7) The number of vanes that can be placed within this circumference with a blade spacing of s = 30 cm is thus Maximum number of vanes: N = C s = 𝜋D s = 𝜋(100 cm) 30 cm = 10.5 (15–8) Obviously we can have only an integer value of N, so we conclude from our preliminary analysis that we should have at least 10 or 11 stator vanes. How good is our approximation of the stator as a two-dimensional cas cade of vanes? To answer this question, we perform a full three-dimensional CFD analysis of the stator. Again, we take advantage of the periodicity by modeling only one flow passage—a three-dimensional passage between two radial stator vanes (Fig. 15–58). We choose N = 10 stator vanes by specify ing an angle of periodicity of 360/10 = 36°. From Eq. 15–8, this represents s = 31.4 at the vane tips and s = 15.7 at the hub, for an average value of savg = 23.6. We generate a hexagonal structured grid in a computational domain bounded by a velocity inlet, an outflow outlet, a section of cylindri cal wall at the hub and another at the tip, the pressure surface of the vane, the suction surface of the vane, and two pairs of periodic boundary condi tions. In this three-dimensional case, the periodic boundaries are rotationally periodic instead of translationally periodic. Note that we use an outflow boundary condition rather than a pressure outlet boundary condition, because we expect the swirling motion to produce a radial pressure distribu tion on the outlet face. The grid is finer near the walls than elsewhere (as usual), to better resolve the boundary layer. The incoming velocity, turbu lence level, turbulence model, etc., are all the same as those used for the (a) (b) (c) FIGURE 15–57 Velocity vectors produced by CFD calculations of stationary turbulent flow through a stator vane flow passage: blade spacing s = (a) 20 cm, (b) 40 cm, and (c) 60 cm.
cen96537_ch15_885-946.indd 917 14/01/17 2:10 pm 918 COMPUTATIONAL FLUID DYNAMICS two-dimensional approximation. The total number of computational cells is almost 800,000.
Pressure contours on the stator vane surfaces and on the inner cylindri cal wall are plotted in Fig. 15–59. This view is from the same angle as that of Fig. 15–60, but we have zoomed out and duplicated the computational domain nine times circumferentially about the axis of rotation (the x-axis) for a total of 10 flow passages to aid in visualization of the flow field. You can see that the pressure is higher (red) on the pressure surface than on the suction surface (blue). You can also see an overall drop in pressure along the hub surface from upstream to downstream of the stator. The change in average pressure from the inlet to the outlet is calculated to be 3.29 kPa.
Outflow outlet Inner cylinder wall Rotationally periodic 2 Rotationally periodic 1 Suction surface Pressure surface Velocity inlet Outer cylinder wall V x z y FIGURE 15–58 Three-dimensional computational domain defined by one flow passage through two stator vanes for N = 10 (angle between vanes = 36°). The computational domain volume is defined between the pressure and suction surfaces of the stator vanes, between the inner and outer cylinder walls, and from the inlet to the outlet. Two pairs of rotationally periodic boundary conditions are defined as shown.
Pressure surface Inlet V –6.30e + 03 –5.60e + 03 –4.89e + 03 –4.19e + 03 3.58e + 03 2.88e + 03 2.17e + 03 1.46e + 03 7.58e + 02 5.16e + 01 –6.55e + 02 –1.36e + 03 –2.07e + 03 –2.77e + 03 –3.48e + 03 Suction surface Outlet x z y FIGURE 15–59 Pressure contour plot produced by three-dimensional CFD calculations of stationary turbulent flow through a stator vane flow passage. Pressure is shown in N/m2 on the vane surfaces and the inner cylinder wall (the hub). Outlines of the inlet and outlet are also shown for clarity. Although only one flow passage is modeled in the CFD calculations, we duplicate the image circumferentially around the x-axis nine times to visualize the entire stator flow field. In this image, high pressures (as on the pressure surfaces of the vanes) are red, while low pressures (as on the suction surfaces of the vanes, especially near the hub) are blue.
cen96537_ch15_885-946.indd 918 14/01/17 2:10 pm 919 CHAPTER 15 To compare our three-dimensional results directly with the two-dimensional approximation, we run one additional two-dimensional case at the average blade spacing, s = savg = 23.6 cm. A comparison between the two- and three-dimensional cases is shown in Table 15–4. From the three-dimensional calcu lation, the net axial force on one stator vane is FD = 183 N. We compare this to the two-dimensional value by converting to force per unit depth (force per unit span of the stator vane). Since the stator vane spans 0.25 m, FD /b = (183 N)/(0.25 m) = 732 N/m. The corresponding two-dimensional value from Table 15–4 is FD /b = 724 N/m, so the agreement is very good (≅ 1 percent difference). The average speed at the outlet of the three-dimensional domain is Vavg = 84.7 m/s, almost identical to the two-dimensional value of 84.8 m/s in Table 15–4. The two-dimensional approximation differs by less than 1 percent. Finally, the average outlet flow angle 𝛽avg obtained from our full three-dimensional calculation is 53.3°, which easily meets the design cri terion of 45°. We compare this to the two-dimensional approximation of 53.9° in Table 15–4; the agreement is again around 1 percent.
Contours of tangential velocity component at the outlet of the computational domain are plotted in Fig. 15–60. We see that the tangential velocity distribution is not uniform; it decreases as we move radially outward from hub to tip as we should expect, since blade spacing s increases from hub to tip. We also find (not shown here) that the outflow pressure increases radially from hub to tip. This also agrees with our intuition, since we know that a radial pressure gradient is required to sustain a tangential flow—the pressure rise with increasing radius provides the centripetal acceleration necessary to turn the flow about the x-axis.
Another comparison can be made between the three-dimensional and two-dimensional calculations by plotting vorticity contours in a slice through the x z y 9.00e + 01 0.00e + 00 6.00e + 00 1.20e + 01 1.80e + 01 2.40e + 01 3.00e + 01 3.60e + 01 4.20e + 01 4.80e + 01 5.40e + 01 6.00e + 01 6.60e + 01 7.20e + 01 7.80e + 01 8.40e + 01 Pressure surface Inlet V Outlet Suction surface FIGURE 15–60 Tangential velocity contour plot produced by three-dimensional CFD calculations of stationary turbulent flow through a stator vane flow passage. The tangential velocity component is shown in m/s at the outlet of the computational domain (and also on the vane surfaces, where the velocity is zero). An outline of the inlet to the computational domain is also shown for clarity. Although only one flow passage is modeled, we duplicate the image circumferentially around the x-axis nine times to visualize the entire stator flow field. In this image, the tangential velocity ranges from 0 (blue) to 90 m/s (red).
Values are shown to three significant digits.
TABLE 15–4 CFD results for flow through a stator vane flow passage: the two-dimensional cascade approximation at the average blade spacing, (s = savg = 23.6 cm) is compared to the full three-dimensional calculation 2-D, s = 23.6 cm Full 3-D 𝛽avg 53.9° 53.3° Vavg, m/s 84.8 84.7 FD/b, N/m 724 732 cen96537_ch15_885-946.indd 919 14/01/17 2:10 pm 920 COMPUTATIONAL FLUID DYNAMICS computational domain within the flow passage between vanes. Two such slices are created—a slice close to the hub and a slice close to the tip, and vorticity contours are plotted in Fig. 15–61. In both slices, the vorticity is confined to the thin boundary layer and wake. There is no flow separation near the hub, but we see that near the tip, the flow has just begun to sepa rate on the suction surface near the trailing edge of the stator vane. Notice that the air leaves the trailing edge of the vane at a steeper angle at the hub than at the tip. This also agrees with our two-dimensional approximation (and our intuition), since blade spacing s at the hub (15.7 cm) is smaller than s at the tip (31.4 cm).
In conclusion, the approximation of this three-dimensional stator as a two-dimensional cascade of stator vanes turns out to be quite good overall, particularly for preliminary analysis. The discrepancy between the two- and three-dimensional calculations for gross flow features, such as force on the vane, outlet flow angle, etc., is around 1 percent or less for all reported quan tities. It is therefore no wonder that the two-dimensional cascade approach is such a popular approximation in turbomachinery design. The more detailed three-dimensional analysis gives us confidence that a stator with 10 vanes is Outlet Outlet Inlet Inlet Flow separation Suction surface Suction surface Pressure surface Pressure surface V V (a) (b) β β y z x y z x FIGURE 15–61 Vorticity contour plots produced by three-dimensional stationary turbulent CFD calculations of flow through a stator vane flow passage: (a) a slice near the hub or root of the vanes and (b) a slice near the tip of the vanes. Contours of z-vorticity are plotted, since the faces are nearly perpendicular to the z-axis. In these images, blue regions (as in the upper half of the wake and in the flow separation zone) represent negative (clockwise) z-vorticity, while red regions (as in the lower half of the wake) represent positive (counterclockwise) z-vorticity. Near the hub, there is no sign of flow separation, but near the tip, there is some indication of flow separation near the trailing edge of the suction side of the vane. Also shown are arrows indicating how the periodic boundary condition works. Flow leaving the bottom of the periodic boundary enters at the same speed and direction into the top of the periodic boundary. Outflow angle 𝛽 is larger near the hub than near the tip of the stator vanes, because blade spacing s is smaller at the hub than at the tip, and also because of the mild flow separation near the tip.
cen96537_ch15_885-946.indd 920 14/01/17 2:10 pm 921 CHAPTER 15 sufficient to meet the imposed design criteria for this axial-flow fan. How ever, our three-dimensional calculations have revealed a small separated region near the tip of the stator vane. It may be wise to apply some twist to the stator vanes (reduce the pitch angle or angle of attack toward the tip) in order to avoid this separation. (Twist is discussed in more detail in Chap. 14.) Alternatively, we can increase the number of stator vanes to 11 or 12 to hopefully eliminate flow separation at the vane tips.
As a final comment on this example flow field, all the calculations were performed in a fixed coordinate system. Modern CFD codes contain options for modeling zones in the flow field with rotating coordinate systems so that similar analyses can be performed on rotor blades as well as on stator vanes.
15–4 ■ CFD WITH HEAT TRANSFER By coupling the differential form of the energy equation with the equa tions of fluid motion, we can use a computational fluid dynamics code to calculate properties associated with heat transfer (e.g., temperature distri butions or rate of heat transfer from a solid surface to a fluid). Since the energy equation is a scalar equation, only one extra transport equation (typi cally for either temperature or enthalpy) is required, and the computational expense (CPU time and RAM requirements) is not increased significantly. Heat transfer capability is built into most commercially available CFD codes, since many practical problems in engineering involve both fluid flow and heat transfer. As mentioned previously, additional boundary conditions related to heat transfer need to be specified. At solid wall boundaries, we may specify either wall temperature Twall (K) or the wall heat flux q .
wall (W/m2), defined as the rate of heat transfer per unit area from the wall to the fluid (but not both at the same time, as illustrated in Fig. 15–62). When we model a zone in a computational domain as a solid body that involves the gen eration of thermal energy via electric heating (as in electronic components) or chemical or nuclear reactions (as in nuclear fuel rods), we may instead specify the heat generation rate per unit volume g . (W/m3) within the solid since the ratio of the total heat generation rate to the exposed surface area must equal the average wall heat flux. In that case, neither Twall nor q .
wall is specified; both converge to values that match the specified heat generation rate. In addition, the temperature distribution inside the solid object itself can then be calculated. Other boundary conditions (such as those associated with radiation heat transfer) may also be applied in CFD codes.
In this section we do not go into details about the equations of motion or the numerical techniques used to solve them. Rather, we show some basic examples that illustrate the capability of CFD to calculate practical flows of engineering interest that involve heat transfer.
Temperature Rise through a Cross-Flow Heat Exchanger Consider flow of cool air through a series of hot tubes as sketched in Fig. 15–63. In heat exchanger terminology, this geometrical configuration is called a cross-flow heat exchanger. If the airflow were to enter hori zontally (𝛼 = 0) at all times, we could cut the computational domain in half Fluid Solid (a) Fluid Solid Twall specified Twall computed qwall computed (b) .
qwall specified .
FIGURE 15–62 At a wall boundary, we may specify either (a) the wall temperature or (b) the wall heat flux, but not both, as this would be mathematically overspecified.
Computational domain α 3D D Translationally periodic Translationally periodic Out In 3D FIGURE 15–63 The computational domain (light blue shaded region) used to model turbulent flow through a cross-flow heat exchanger. Flow enters from the left at angle 𝛼 from the horizontal.
cen96537_ch15_885-946.indd 921 14/01/17 2:10 pm 922 COMPUTATIONAL FLUID DYNAMICS and apply symmetry boundary conditions on the top and bottom edges of the domain (see Fig. 15–25). In the case under consideration, however, we allow the airflow to enter the computational domain at some angle (𝛼 ≠ 0). Thus, we impose translationally periodic boundary conditions on the top and bottom edges of the domain as sketched in Fig. 15–63. We set the inlet air temperature to 300 K and the surface temperature of each tube to 500 K. The diameter of the tubes and the speed of the air are chosen such that the Reynolds number is approximately 1 × 105 based on tube diameter. The tube surfaces are assumed to be hydrodynamically smooth (zero roughness) in this first set of calculations. The hot tubes are staggered as sketched in Fig. 15–63 and are spaced three diameters apart both horizontally and verti cally. We assume two-dimensional stationary turbulent flow without gravity effects and set the turbulence intensity of the inlet air to 10 percent. We run two cases for comparison: 𝛼 = 0 and 10°. Our goal is to see whether the heat transfer to the air is enhanced or inhibited by a nonzero value of 𝛼. Which case do you think will provide greater heat transfer?
We generate a two-dimensional, multiblock, structured grid with very fine resolution near the tube walls as shown in Fig. 15–64, and we run the CFD code to convergence for both cases. Temperature contours are shown for the 𝛼 = 0° case in Fig. 15–65, and for the 𝛼 = 10° case in Fig. 15–66. The aver age rise of air temperature leaving the outlet of the control volume for the case with 𝛼 = 0° is 5.51 K, while that for 𝛼 = 10° is 5.65 K. Thus we con clude that the off-axis inlet flow leads to more effective heating of the air, although the improvement is only about 2.5 percent. We compute a third case (not shown) in which 𝛼 = 0° but the turbulence intensity of the incoming air is increased to 25 percent. This leads to improved mixing, and the average air temperature rise from inlet to outlet increases by about 6.5 percent to 5.87 K.
Finally, we study the effect of rough tube surfaces. We model the tube walls as rough surfaces with a characteristic roughness height of 0.01 m (1 percent of cylinder diameter). Note that we had to coarsen the grid some what near each tube so that the distance from the center of the closest com putational cell to the wall is greater than the roughness height; otherwise the roughness model in the CFD code is unphysical. The flow inlet angle is set to 𝛼 = 0° for this case, and flow conditions are identical to those of Fig. 15–65. FIGURE 15–64 Close-up view of the structured grid near one of the cross-flow heat exchanger tubes. The grid is fine near the tube walls so that the wall boundary layer can be better resolved.
FIGURE 15–65 Temperature contour plots produced by CFD calculations of stationary turbulent flow through a cross-flow heat exchanger at 𝛼 = 0° with smooth tubes. Contours range from 300 K (blue) to 315 K (red) or higher (white). The average air temperature at the outlet increases by 5.51 K compared to the inlet air temperature. Note that although the calculations are performed in the computational domain of Fig. 15–63, the image is duplicated here three times for purposes of illustration.
cen96537_ch15_885-946.indd 922 14/01/17 2:10 pm 923 CHAPTER 15 Temperature contours are plotted in Fig. 15–67. Pure white regions in the con tour plot represent locations where the air temperature is greater than 315 K. The average air temperature rise from inlet to outlet is 14.48 K, a 163 percent increase over the smooth wall case at 𝛼 = 0°. Thus we see that wall roughness is a critical parameter in turbulent flows. This example provides some insight as to why the tubes in heat exchangers are often purposely roughened.
Cooling of an Array of Integrated Circuit Chips In electronics equipment, instrumentation, and computers, electronic com ponents, such as integrated circuits (ICs or “chips”), resistors, transistors, diodes, and capacitors, are soldered onto printed circuit boards (PCBs). The PCBs are often stacked in rows as sketched in Fig. 15–68. Because many of these electronic components must dissipate heat, cooling air is often blown through the air gap between each pair of PCBs to keep the components from getting too hot. Consider the design of a PCB for an outer space application. Several identical PCBs are to be stacked as in Fig. 15–68. Each PCB is 10 cm high and 30 cm long, and the spacing between boards is FIGURE 15–66 Temperature contour plots produced by CFD calculations of stationary turbulent flow through a cross-flow heat exchanger at 𝛼 = 10° with smooth tubes. Contours range from 300 K (blue) to 315 K (red) or higher (white). The average air temperature at the outlet increases by 5.65 K compared to the inlet air temperature. Thus, off-axis inlet flow (𝛼 = 10°) yields a ∆T that is 2.5 percent higher than that for the on-axis inlet flow (𝛼 = 0°).
FIGURE 15–67 Temperature contour plots produced by CFD calculations of stationary turbulent flow through a cross-flow heat exchanger at 𝛼 = 0° with rough tubes (average wall roughness equal to 1 percent of tube diameter; wall functions utilized in the CFD calculations). Contours range from 300 K (blue) to 315 K (red) or higher (white). The average air temperature at the outlet increases by 14.48 K compared to the inlet air temperature. Thus, even this small amount of surface roughness yields a ∆T that is 163 percent higher than that for the case with smooth tubes.
cen96537_ch15_885-946.indd 923 14/01/17 2:10 pm 924 COMPUTATIONAL FLUID DYNAMICS 2.0 cm. Cooling air enters the gap between the PCBs at a speed of 2.60 m/s and a temperature of 30°C. The electrical engineers must fit eight identical ICs on a 10 cm × 15 cm portion of each board. Each of the ICs dissipates 6.24 W of heat: 5.40 W from its top surface and 0.84 W from its sides. (There is assumed to be no heat transfer from the bottom of the chip to the PCB.) The rest of the components on the board have negligible heat trans fer compared to that from the eight ICs. To ensure adequate performance, the average temperature on the chip surface should not exceed 150°C, and the maximum temperature anywhere on the surface of the chip should not exceed 180°C. Each chip is 2.5 cm wide, 4.5 cm long, and 0.50 cm thick. The electrical engineers come up with two possible configurations of the eight chips on the PCB as sketched in Fig. 15–69: in the long configuration, the chips are aligned with their long dimension parallel to the flow, and in the short configuration, the chips are aligned with their short dimension parallel to the flow. The chips are staggered in both cases to enhance cool ing. We are to determine which arrangement leads to the lower maximum surface temperature on the chips, and whether the electrical engineers will meet the surface temperature requirements.
For each configuration, we define a three-dimensional computational domain consisting of a single flow passage through the air gap between two PCBs (Fig. 15–70). We generate a structured hexagonal grid with 267,520 cells for each configuration. The Reynolds number based on the 2.0 cm gap between boards is about 3600. If this were a simple two-dimensional chan nel flow, this Reynolds number would be barely high enough to establish turbulent flow. However, since the surfaces leading up to the velocity inlet are very rough, the flow is most likely turbulent. We note that low Reyn olds number turbulent flows are challenging for most turbulence models, since the models are calibrated at high Reynolds numbers. Nevertheless, we assume stationary turbulent flow and employ the k-𝜀 turbulence model with wall functions. While the absolute accuracy of these calculations may be suspect because of the low Reynolds number, comparisons between the long and short configurations should be reasonable. We ignore buoyancy effects in the calculations since this is a space application. The inlet is specified PCB IC Cooling air at V = 2.60 m/s and T = 30°C ∞ FIGURE 15–68 Four printed circuit boards (PCBs) stacked in rows, with air blown in between each PCB to provide cooling.
Long configuration Short configuration FIGURE 15–69 Two possible configurations of the eight ICs on the PCB: long configuration and short configuration. Without peeking ahead, which configuration do you think will offer the best cooling to the chips?
cen96537_ch15_885-946.indd 924 14/01/17 2:10 pm 925 CHAPTER 15 as a velocity inlet (air) with V = 2.60 m/s and T∞ = 30°C; we set the inlet turbulence intensity to 20 percent and the turbulent length scale to 1.0 mm. The outlet is a pressure outlet at zero gage pressure. The PCB is modeled as a smooth adiabatic wall (zero heat transfer from the wall to the air). The top and sides of the computational domain are also approximated as smooth adiabatic walls.
Based on the given chip dimensions, the surface area of the top of a chip is 4.5 cm × 2.5 cm = 11.25 cm2. The total surface area of the four sides of the chip is 7.0 cm2. From the given heat transfer rates, we calculate the rate of heat transfer per unit area from the top surface of each chip, q · top = 5.4 W 11.25 cm2 = 0.48 W/cm2 So, we model the top surface of each chip as a smooth wall with a surface heat flux of 4800 W/m2 from the wall to the air. Similarly, the rate of heat transfer per unit area from the sides of each chip is q · sides = 0.84 W 7.0 cm2 = 0.12 W/cm2 Since the sides of the chip have electrical leads, we model each side sur face of each chip as a rough wall with an equivalent roughness height of 0.50 mm and a surface heat flux of 1200 W/m2 from the wall to the air.
The CFD code ANSYS-FLUENT is run for each case to convergence. Results are summarized in Table 15–5, and temperature contours are plot ted in Figs. 15–71 and 15–72. The average temperature on the top surfaces of the chips is about the same for either configuration (144.4°C for the long case and 144.7°C for the short case) and is below the recommended limit of 150°C. There is more of a difference in average temperature on the side sur faces of the chips, however (84.2°C for the long case and 91.4°C for the short case), although these values are well below the limit. Of greatest concern are the maximum temperatures. For the long configuration, Tmax = 187.5°C and occurs on the top surface of chip 7 (the middle chip of the last row). For the Velocity inlet Adiabatic walls Pressure outlet Long configuration 1 2 3 6 7 8 4 5 1 2 3 6 7 8 4 5 Short configuration y z x y z x Velocity inlet Adiabatic walls Pressure outlet FIGURE 15–70 Computational domains for the chip cooling example. Air flowing through the gap between two PCBs is modeled. Two separate grids are generated, one for the long configuration and one for the short configuration. Chips 1 through 8 are labeled for reference. The surfaces of these chips transfer heat to the air; all other walls are adiabatic.
cen96537_ch15_885-946.indd 925 14/01/17 2:10 pm 926 COMPUTATIONAL FLUID DYNAMICS short configuration, Tmax = 182.1°C and occurs close to midboard on the top surfaces of chips 7 and 8 (the two chips in the last row). For both configura tions these values exceed the recommended limit of 180°C, although not by much. The short configuration does a better job at cooling the top surfaces of the chips, but at the expense of a slightly larger pressure drop and poorer cooling along the side surfaces of the chips.
Notice from Table 15–5 that the average change in air temperature from inlet to outlet is identical for both configurations (7.83°C). This should not be surprising, because the total rate of heat transferred from the chips to the air is the same regardless of chip configuration. In fact, in a CFD anal ysis it is wise to check values like this—if average ∆T were not the same between the two configurations, we would suspect some kind of error in our calculations.
4.60e + 02 4.50e + 02 4.40e + 02 4.30e + 02 4.20e + 02 4.10e + 02 4.00e + 02 3.90e + 02 3.80e + 02 3.70e + 02 3.60e + 02 3.50e + 02 3.40e + 02 3.30e + 02 3.20e + 02 3.10e + 02 3.00e + 02 Long configuration Air flow 3 2 1 5 4 8 7 6 y z x Tmax = 460.7 K FIGURE 15–71 CFD results for the chip cooling example, long configuration: temperature contours as viewed from directly above the chip surfaces, with T values in K on the legend. The location of maximum surface temperature is indicated; it occurs near the end of chip 7. Red regions near the leading edges of chips 1, 2, and 3 are also seen, indicating high surface temperatures at those locations.
TABLE 15–5 Comparison of CFD results for the chip cooling example, long and short configurations Long Short Tmax, top surfaces of chips 187.5°C 182.1°C Tavg, top surfaces of chips 144.5°C 144.7°C Tmax, side surfaces of chips 154.0°C 170.6°C Tavg, side surfaces of chips 84.2°C 91.4°C Average ∆T, inlet to outlet 7.83°C 7.83°C Average ∆P, inlet to outlet −5.14 Pa −5.58 Pa cen96537_ch15_885-946.indd 926 14/01/17 2:10 pm 927 CHAPTER 15 We point out many other interesting features of these flow fields. For either configuration, the average surface temperature on the downstream chips is greater than that on the upstream chips. This makes sense physi cally, since the first chips receive the coolest air, while those downstream are cooled by air that has already been warmed up somewhat. We notice that the front chips (1, 2, and 3 in the long configuration and 1 and 2 in the short configuration) have regions of high temperature just downstream of their leading edges. A close-up view of the temperature distribution on one of these chips is shown in Fig. 15–73a. Why is the temperature so high there? It turns out that the flow separates off the sharp corner at the front of the chip and forms a recirculating eddy called a separation bubble on the top of the chip (Fig. 15–73b). The air speed is slow in that region, especially along the reattachment line where the flow reattaches to the surface. The slow air speed leads to a local “hot spot” in that region of the chip surface since convective cooling is minimal there. Finally, we notice in Fig. 15–73a that downstream of the separation bubble, T increases down the chip surface. There are two reasons for this: (1) the air warms up as it travels down the chip, and (2) the boundary layer on the chip surface grows downstream. The larger the boundary layer thickness, the lower the air speed near the surface, and thus the lower the amount of convective cooling at the surface.
In summary, our CFD calculations have predicted that the short configu ration leads to a lower value of maximum temperature on the chip surfaces and appears at first glance to be the preferred configuration for heat trans fer. However, the short configuration demands a higher pressure drop at the same volume flow rate (Table 15–5). For a given cooling fan, this additional pressure drop would shift the operating point of the fan to a lower volume 4.60e + 02 Short configuration Air flow 2 1 4 5 6 7 8 3 Tmax = 455.3 K y z x 3.00e + 02 3.10e + 02 3.20e + 02 3.30e + 02 3.40e + 02 3.50e + 02 3.60e + 02 3.70e + 02 3.80e + 02 3.90e + 02 4.00e + 02 4.10e + 02 4.20e + 02 4.30e + 02 4.40e + 02 4.50e + 02 FIGURE 15–72 CFD results for the chip cooling example, short configuration: temperature contours as viewed from directly above the chip surfaces, with T values in K on the legend. The same temperature scale is used here as in Fig. 15–71. The locations of maximum surface temperature are indicated; they occur near the end of chips 7 and 8 near the center of the PCB. Red regions near the leading edges of chips 1 and 2 are also seen, indicating high surface temperatures at those locations.
cen96537_ch15_885-946.indd 927 14/01/17 2:10 pm 928 COMPUTATIONAL FLUID DYNAMICS flow rate (Chap. 14), decreasing the cooling effect. It is not known whether this shift would be enough to favor the long configuration—more informa tion about the fan and more analysis would be required. The bottom line in either case is that there is not sufficient cooling to keep the chip surface temperature below 180°C everywhere on every chip. To rectify the situa tion, we recommend that the designers spread the eight hot chips over the entire PCB rather than in the limited 10 cm × 15 cm area of the board. The increased space between chips should result in sufficient cooling for the given flow rate. Another option is to install a more powerful fan that would increase the speed of the inlet air.
15–5 ■ COMPRESSIBLE FLOW CFD CALCULATIONS All the examples discussed in this chapter so far have been for incompress ible flow (𝜌 = constant). When the flow is compressible, density is no lon ger a constant, but becomes an additional variable in the equation set. We limit our discussion here to ideal gases. When we apply the ideal-gas law, we introduce yet another unknown, namely, temperature T. Hence, the energy equation must be solved along with the compressible forms of the equations of conservation of mass and conservation of momentum (Fig. 15–74). In addition, fluid properties, such as viscosity and thermal conductivity, are no longer necessarily treated as constants, since they are functions of tem perature; thus, they appear inside the derivative operators in the differen tial equations of Fig. 15–74. While the equation set looks ominous, many commercially available CFD codes are able to handle compressible flow problems, including shock waves.
When solving compressible flow problems with CFD, the boundary con ditions are somewhat different than those of incompressible flow. For exam ple, at a pressure inlet we need to specify both stagnation pressure and static pressure, along with stagnation temperature. A special boundary condition (called pressure far field in ANSYS-FLUENT) is also available for com pressible flows. With this boundary condition, we specify the Mach number, the static pressure, and the temperature; it can be applied to both inlets and outlets and is well suited for supersonic external flows.
The equations of Fig. 15–74 are for laminar flow, whereas many compress ible flow problems occur at high flow speeds in which the flow is turbulent. Therefore, the equations of Fig. 15–74 must be modified accordingly (into the RANS equation set) to include a turbulence model, and more transport equa tions must be added, as discussed previously. The equations then get quite long and complicated and are not included here. Fortunately, in many situa tions, we can approximate the flow as inviscid, eliminating the viscous terms from the equations of Fig. 15–74 (the Navier–Stokes equation reduces to the Euler equation). As we shall see, the inviscid flow approximation turns out to be quite good for many practical high-speed flows, since the boundary lay ers along walls are very thin at high Reynolds numbers. In fact, compress ible CFD calculations can predict flow features that are often quite difficult to obtain experimentally. For example, many experimental measurement tech niques require optical access, which is limited in three-dimensional flows, and even in some axisymmetric flows. CFD is not limited in this way.
Chip 2, long configuration Cooling air Region of high T Approximate location of reattachment line Separation bubble (a) (b) FIGURE 15–73 (a) Close-up top view of temperature contours on the surface of chip 2 of the long configuration. The region of high temperature is outlined. Temperature contour levels are the same as in Fig. 15–71. (b) An even closer view (an edge view) of streamlines outlining the separation bubble in that region. The approximate location of the reattachment line on the chip surface is also shown.
cen96537_ch15_885-946.indd 928 14/01/17 2:10 pm 929 CHAPTER 15 Compressible Flow through a Converging–Diverging Nozzle For our first example, we consider compressible flow of air through an axisymmetric converging–diverging nozzle. The computational domain is shown in Fig. 15–75. The inlet radius is 0.10 m, the throat radius is 0.075 m, and the outlet radius is 0.12 m. The axial distance from the inlet to the throat is 0.30 m—the same as the axial distance from the throat to the out let. A structured grid with approximately 12,000 quadrilateral cells is used in the calculations. At the pressure inlet boundary, the stagnation pressure P0, inlet is set to 220 kPa (absolute), the static pressure Pinlet is set to 210 kPa, and the stagnation temperature T0, inlet is set to 300 K. For the first case, we set the static pressure Pb at the pressure outlet boundary to 50.0 kPa (Pb/P0, inlet = 0.227)—low enough that the flow is supersonic through the entire diverging section of the nozzle, without any normal shocks in the nozzle. This back pressure ratio corresponds to a value between cases E and F in Fig. 12–22, in which a complex shock pattern occurs downstream of the nozzle exit; these shock waves do not influence the flow in the nozzle itself, since the flow exiting the nozzle is supersonic. We do not attempt to model the flow downstream of the nozzle exit.
Δ Δ Δ Δ Δ 휕(ρu) 휕 x 휕(ρυ) = ρgx 휕 y 휕(ρw) 휕z P = ρRT + + 휕u 휕 x + = 0 u 휕u 휕 y 휕P 휕 x 휕 휕 x υ + + – 휕u 휕 y w Ideal gas law: Continuity: x-momentum: ) ) ρ 휕u 휕y + μ 휕υ 휕x ] ] ) ) 휕u 휕 x + + 휕 휕z 휕w 휕x + μ 휕u 휕z ] ] ) ) + 2μ ) ) λ . V = ρgy 휕υ 휕x + u 휕υ 휕y 휕P 휕y 휕 휕x 휕 휕z υ + + – 휕υ 휕z w y-momentum: ) ) ρ 휕υ 휕z + μ 휕w 휕y ] ] ) ) 휕υ 휕x + μ 휕u 휕y ] ] ) ) + 휕 휕y + 휕υ 휕y + 2μ ) ) λ . V = ρgz 휕w 휕x + u 휕w 휕y 휕P 휕z 휕 휕x υ + + – 휕w 휕z w z-momentum: ) ) ρ 휕w 휕x + μ 휕u 휕z ] ] ) ) + 휕 휕y 휕υ 휕z + μ 휕w 휕y ] ] ) ) + 휕 휕z 휕w 휕z + 2μ ) ) λ . V = 휕P 휕x + u 휕P 휕y υ + 휕P 휕z w ) ) 훽 T 휕T 휕x + u 휕T 휕y υ + 휕T 휕z w Energy: ) ) ρcp + (k T) + ɸ .
휕 휕y FIGURE 15–74 The equations of motion for the case of steady, compressible, laminar flow of a Newtonian fluid in Cartesian coordinates. There are six equations and six unknowns: 𝜌, u, 𝜐, w, T, and P. Five of the equations are nonlinear partial differential equations, while the ideal-gas law is an algebraic equation. R is the specific ideal-gas constant, 𝜆 is the second coefficient of viscosity, often set equal to −2𝜇/3; cp is the specific heat at constant pressure; k is the thermal conductivity; 𝛽 is the coefficient of thermal expansion; and Φ is the dissipation function, given by White (2005) as Φ = 2𝜇( ∂u ∂x) 2 + 2𝜇( ∂𝜐 ∂y) 2 + 2𝜇( ∂w ∂z ) 2 + 𝜇( ∂𝜐 ∂x + ∂u ∂y) 2 + 𝜇( ∂w ∂y + ∂𝜐 ∂z) 2 + 𝜇( ∂u ∂z + ∂w ∂x ) 2 + 𝜆( ∂u ∂x + ∂𝜐 ∂y + ∂w ∂z ) 2 Pressure inlet Pressure outlet Wall Axis FIGURE 15–75 Computational domain for compressible flow through a converging–diverging nozzle. Since the flow is axisymmetric, only one two-dimensional slice is needed for the CFD solution.
cen96537_ch15_885-946.indd 929 14/01/17 2:10 pm 930 COMPUTATIONAL FLUID DYNAMICS The CFD code is run to convergence in its steady, inviscid, compressible flow mode. The average values of the Mach number Ma and pressure ratio P/P0, inlet are calculated at 25 axial locations along the converging–diverging nozzle (every 0.025 m) and are plotted in Fig. 15–76a. The results match almost perfectly with the predictions of one-dimensional isentropic flow (Chap. 12). At the throat (x = 0.30 m), the average Mach number is 0.997, and the average value of P/P0, inlet is 0.530. One-dimensional isentropic flow theory predicts Ma = 1 and P/P0, inlet = 0.528 at the throat. The small dis crepancies between CFD and theory are due to the fact that the computed flow is not one-dimensional, since there is a radial velocity component and, therefore, a radial variation of the Mach number and static pressure. Care ful examination of the Mach number contour lines of Fig. 15–76b reveal that they are curved, not straight as would be predicted by one-dimensional isentropic theory. The sonic line (Ma = 1) is identified for clarity in the figure. Although Ma = 1 right at the wall of the throat, sonic conditions along the axis of the nozzle are not reached until somewhat downstream of the throat.
Next, we run a series of cases in which back pressure Pb is varied, while keeping all other boundary conditions fixed. Results for three cases are Ma Ma P P0, inlet P/P0, inlet 0.6 0.5 0.3 0.4 0.2 0.1 Throat Sonic line 0 1 0.9 0.8 0.7 0.0 0.5 1.0 1.5 2.0 2.5 0 0.1 0.2 0.3 x, m (a) (b) 0.4 0.5 0.6 FIGURE 15–76 CFD results for steady, adiabatic, inviscid compressible flow through an axisymmetric converging–diverging nozzle: (a) calculated average Mach number and pressure ratio at 25 axial locations (circles), compared to predictions from isentropic, one-dimensional compressible flow theory (solid lines); (b) Mach number contours, ranging from Ma = 0.3 (blue) to 2.7 (red). Although only the top half is calculated, a mirror image about the x-axis is shown for clarity. The sonic line (Ma = 1) is also highlighted. It is parabolic rather than straight in this axisymmetric flow due to the radial component of velocity, as discussed in Schreier (1982).
cen96537_ch15_885-946.indd 930 14/01/17 2:10 pm 931 CHAPTER 15 shown in Fig. 15–77: Pb = (a) 100, (b) 150, and (c) 200 kPa, i.e., Pb/P0, inlet = (a) 0.455, (b) 0.682, and (c) 0.909, respectively. For all three cases, a nor mal shock occurs in the diverging portion of the nozzle. Furthermore, as back pressure increases, the shock moves upstream toward the throat, and decreases in strength. Since the flow is choked at the throat, the mass flow rate is identical in all three cases (and also in the previous case shown in Fig. 15–76). We notice that the normal shock is not straight, but rather is curved due to the radial component of velocity, as previously mentioned.
For case (b), in which Pb/P0, inlet = 0.682, the average values of the Mach number and pressure ratio P/P0, inlet are calculated at 25 axial loca tions along the converging–diverging nozzle (every 0.025 m), and are plotted in Fig. 15–78. For comparison with theory, the one-dimensional isentropic flow relations are used upstream and downstream of the shock, and the normal shock relations are used to calculate the pressure jump across Shock (a) Shock (b) (c) Shock FIGURE 15–77 CFD results for steady, adiabatic, inviscid, compressible flow through a converging-diverging nozzle: contours of stagnation pressure ratio P0/P0, inlet are shown for Pb/P0, inlet = (a) 0.455; (b) 0.682; and (c) 0.909. Since stagnation pressure is constant upstream of the shock and decreases suddenly across the shock, it serves as a convenient indicator of the location and strength of the normal shock in the nozzle. In these contour plots, P0/P0, inlet ranges from 0.5 (blue) to 1.01 (red). It is clear from the colors downstream of the shock that the farther downstream the shock, the stronger the shock (larger magnitude of stagnation pressure drop across the shock). We also note the shape of the shocks—curved rather than straight, because of the radial component of velocity.
cen96537_ch15_885-946.indd 931 14/01/17 2:10 pm 932 COMPUTATIONAL FLUID DYNAMICS the shock (Chap. 12). To match the specified back pressure, one-dimensional analysis requires that the normal shock be located at x = 0.4436 m, account ing for the change in both P0 and A across the shock. The agreement between CFD calculations and one-dimensional theory is again excellent. The small discrepancy in both the pressure and the Mach number just downstream of the shock is attributed to the curved shape of the shock (Fig. 15–77b), as discussed previously. In addition, the shock in the CFD calculations is not infinitesimally thin, as predicted by one-dimensional theory, but is spread out over a few computational cells. The latter inaccuracy can be reduced somewhat by refining the grid in the area of the shock wave (not shown).
The previous CFD calculations are for steady, inviscid, adiabatic flow. When there are no shock waves (Fig. 15–76), the flow is also isentropic, since it is both adiabatic and reversible (no irreversible losses). However, when a shock wave exists in the flow field (Fig. 15–77), the flow is no lon ger isentropic since there are irreversible losses across the shock, although it is still adiabatic.
One final CFD case is run in which two additional irreversibilities are included, namely, friction and turbulence. We modify case (b) of Fig. 15–77 by running a steady, adiabatic, turbulent case using the k-𝜀 turbulence model with wall functions. The turbulence intensity at the inlet is set to 10 percent with a turbulence length scale of 0.050 m. A contour plot of P/P0, inlet is shown in Fig. 15–79, using the same color contour scale as in Fig. 15–77. Comparison of Figs. 15–77b and 15–79 reveals that the shock wave for the turbulent case occurs further upstream and is therefore somewhat weaker. In addition, the stagnation pressure is small in a very thin region along the channel walls. This is due to frictional losses in the thin boundary layer. Turbulent and viscous irreversibilities in the boundary layer region are responsible for this decrease in stagnation pressure. Furthermore, the boundary layer separates just downstream of the shock, leading to more irreversibilities. A close-up view of velocity vectors in the vicinity of the Ma P P0, inlet 0.6 0.5 0.3 0.4 0.2 0.1 Throat 0 1 0.9 0.8 0.7 0.0 0.5 1.0 1.5 2.0 2.5 0 0.1 0.2 0.3 x, m 0.4 0.5 0.6 Shock Ma P/P0, inlet FIGURE 15–78 Mach number and pressure ratio as functions of axial distance along a converging–diverging nozzle for the case in which Pb/P0, inlet = 0.682. Averaged CFD results at 25 axial locations (circles) for steady, inviscid, adiabatic, compressible flow are compared to predictions from one-dimensional compressible flow theory (solid lines).
cen96537_ch15_885-946.indd 932 14/01/17 2:10 pm 933 CHAPTER 15 separation point along the wall is shown in Fig. 15–80. We note that this case does not converge well and is inherently unsteady; the interaction between shock waves and boundary layers is a very difficult task for CFD. Because we use wall functions, flow details within the turbulent boundary layer are not resolved in this CFD calculation. Experiments reveal, however, that the shock wave interacts much more significantly with the boundary layer, pro ducing “𝜆-feet,” as discussed in the Application Spotlight of Chap. 12.
Finally, we compare the mass flow rate for this viscous, turbulent case to that of the inviscid case, and find that m . has decreased by about 0.7 percent. Why? As discussed in Chap. 10, a boundary layer along a wall impacts the outer flow such that the wall appears to be thicker by an amount equal to the displacement thickness 𝛿. The effective throat area is thus reduced somewhat by the presence of the boundary layer, leading to a reduction in mass flow rate through the converging–diverging nozzle. The effect is small in this example since the boundary layers are so thin relative to the dimen sions of the nozzle, and it turns out that the inviscid approximation is quite good (less than one percent error).
Oblique Shocks over a Wedge For our final compressible flow example, we model steady, adiabatic, two-dimensional, inviscid, compressible flow of air over a wedge of half-angle 𝜃 (Fig. 15–81). Since the flow has top–bottom symmetry, we model only the upper half of the flow and use a symmetry boundary condition along the bottom edge. We run three cases: 𝜃 = 10, 20, and 30°, at an inlet Mach number of 2.0. CFD results are shown in Fig. 15–82 for all three cases. In the CFD plots, a mirror image of the computational domain is projected across the line of symmetry for clarity.
For the 10° case (Fig. 15–82a), a straight oblique shock originating at the apex of the wedge is observed, as also predicted by inviscid theory. The flow turns across the oblique shock by 10° so that it is parallel to the wedge wall. The shock angle 𝛽 predicted by inviscid theory is 39.31°, and the pre dicted Mach number downstream of the shock is 1.64. Measurements with a protractor on Fig. 15–82a yield 𝛽 ≅ 40°, and the CFD calculation of the Mach number downstream of the shock is 1.64; the agreement with theory is thus excellent.
Shock Boundary layer irreversibilities Flow separation FIGURE 15–79 CFD results for stationary, adiabatic, turbulent, compressible flow through a converging–diverging nozzle. Contours of stagnation pressure ratio P0/P0, inlet are shown for the case with Pb/P0, inlet = 0.682, the same back pressure and color scale as that of Fig. 15–77b. Flow separation and irreversibilities in the boundary layer are identified.
Shock FIGURE 15–80 Close-up view of velocity vectors and stagnation pressure contours in the vicinity of the separated flow region of Fig. 15–79. The sudden decrease in velocity magnitude across the shock is seen, as is the reverse flow region downstream of the shock.
y x θ Pressure far field Wedge (wall) Symmetry FIGURE 15–81 Computational domain and boundary conditions for compressible flow over a wedge of half-angle 𝜃. Since the flow is symmetric about the x-axis, only the upper half is modeled in the CFD analysis.
cen96537_ch15_885-946.indd 933 14/01/17 2:10 pm 934 COMPUTATIONAL FLUID DYNAMICS For the 20° case (Fig. 15–82b), the CFD calculations yield a Mach number of 1.21 downstream of the shock. The shock angle measured from the CFD calculations is about 54°. Inviscid theory predicts a Mach number of 1.21 and a shock angle of 53.4°, so again the agreement between theory and CFD is excellent. Since the shock for the 20° case is at a steeper angle (closer to a normal shock), it is stronger than the shock for the 10° case, as indicated by the redder coloring in the Mach contours downstream of the shock for the 20° case.
At Mach number 2.0 in air, inviscid theory predicts that a straight oblique shock can form up to a maximum wedge half-angle of about 23° (Chap. 12). At wedge half-angles greater than this, the shock must move upstream of the wedge (become detached), forming a detached shock, which takes the shape of a bow wave (Chap. 12). The CFD results at 𝜃 = 30° (Fig. 15–82c) show that this is indeed the case. The portion of the detached shock just upstream of the leading edge is a normal shock, and thus the flow down stream of that portion of the shock is subsonic. As the shock curves back ward, it becomes progressively weaker, and the Mach number downstream of the shock increases, as indicated by the coloring.
CFD Methods for Two-Phase Flows Two-phase flows, such as those with liquid and gas, pose significant modeling challenges due to the discontinuities in material properties between the phases, localized interfacial forces, and potential for con tinuously changing spatial distribution of phases. As an example, in a water boiler operating at atmospheric pressure, the liquid density is more than 1500 times greater than that of the vapor, and the volume fraction occupied by the vapor phase increases along the device length as heat is added. Due to such complexities, few analytical solutions are available for two-phase flows, and computational modeling approaches are invaluable. θ =10° β (a) Ma2 Ma1 Oblique shock (b) θ = 20° Ma2 Ma1 β Oblique shock (c) θ = 30° Ma1 Ma2 Detached shock FIGURE 15–82 CFD results (Mach number contours) for steady, adiabatic, inviscid, compressible flow at Ma1 = 2.0 over a wedge of half-angle 𝜃 = (a) 10°, (b) 20°, and (c) 30°. The Mach number contours range from Ma = 0.2 (blue) to 2.0 (red) in all cases. For the two smaller wedge half-angles, an attached weak oblique shock forms at the leading edge of the wedge, but for the 30° case, a detached shock (bow wave) forms ahead of the wedge. Shock strength increases with 𝜃, as indicated by the color change downstream of the shock as 𝜃 increases.
This section was contributed by Professor Alex Rattner of Penn State University. cen96537_ch15_885-946.indd 934 14/01/17 2:10 pm 935 CHAPTER 15 A number of computational frameworks have been developed to describe different classes of two-phase flow. In dispersed two-phase flows, a field of small particles, bubbles, or droplets is distributed in a continuous medium. An example could be the mist of liquid droplets generated by an aerosol spray can. In Lagrangian CFD approaches, such droplets would be modeled as point features (parcels), each advanced at each time step by forces from gravity, drag, and interactions with other particles. The net mass and momentum transport from parcels in each mesh cell are then applied as source terms in the governing equations for the continuous phase. Eulerian approaches incorporate closure models that predict the average interactions between the continuous and dispersed phases in each mesh cell. These inter-phase closure models are often referred to as sub-grid scale (SGS) models because they describe the average effect of features smaller than the resolved mesh scales. Eulerian formulations can have reduced computational costs as only average quantities need to be tracked in each mesh cell rather than the states of each parcel.
Eulerian and Lagrangian formulations are well suited for dispersed two-phase flows, but do not predict local interfacial dynamics. Interface resolv ing methods are needed for flows where such phenomena are important. One representative technique is the volume of fluid (VOF) method (Hirt and Nichols, 1981), which solves a transport equation for the volume fraction (𝛼) of one phase in each mesh cell. 𝛼 is bounded between 0 and 1, and is advected by the velocity field (Fig. 15–83). An example application of interface resolving (VOF) CFD is presented for two-phase Taylor flow in a vertical tube (Rattner and Garimella, 2015). Here, gas rises in continuous elongated bullet-shaped Taylor bub bles (Davies and Taylor, 1950), surrounded by thin downward flowing liquid films on the tube wall, and separated by upward flowing liquid regions (slugs). Side-by-side axisymmetric simulation results and experi mental photographs are presented in Fig. 15–84 for an air-water flow case in an 8.0-mm diameter tube at Reexp = 885 and ReCFD = 880. Close agreement was found between experiment and simulation for this case, with bubble rise velocities of Ub, exp = 0.16 ± 0.01 m/s and Ub, CFD = 0.159 m/s. While experiments are needed for validating CFD methods, 2 1 2 1 (a) (b) (c) (d) (e) FIGURE 15–83 Schematic of the volume of fluid (VOF) method. (a) A droplet of fluid 2 is contained in fluid 1. (b) The volume fraction of fluid 2 (𝛼) in each mesh cell is determined. (c–d) The velocity field advects the 𝛼 field. (e) The modified 𝛼 field represents the new distribution of the phases.
Courtesy of Alex Rattner, Penn State University.
Re = 880 Re = 885 FIGURE 15–84 Comparison of axisymmetric VOF simulation (left) and experimental measurement (right) of vertical upward air-water Taylor flow in an 8.0 mm diameter tube at nearly identi cal Reynolds numbers. Courtesy of Alex Rattner, Penn State University.
cen96537_ch15_885-946.indd 935 14/01/17 2:10 pm 936 COMPUTATIONAL FLUID DYNAMICS this case also suggests ways that experiments and simulations can be complementary. Here, CFD provides quantities that are challenging to measure, such as the pressure field in the bubble wake. Experiments pro vide statistical properties from many more distinct bubbles than can be feasibly simulated due to computational costs.
15–6 ■ OPEN-CHANNEL FLOW CFD CALCULATIONS So far, all our examples have been for one single-phase fluid (air or water). However, many commercially available CFD codes can handle flow of a mixture of gases (e.g., carbon monoxide in air), flow with two phases of the same fluid (e.g., steam and liquid water), and even flow of two fluids of dif ferent phase (e.g., liquid water and gaseous air). The latter case is of interest in this section, namely, the flow of water with a free surface, above which is gaseous air, i.e., open-channel flow. We present here some simple examples of CFD solutions of open-channel flows.
Flow over a Bump on the Bottom of a Channel Consider a two-dimensional channel with a flat, horizontal bottom. At a cer tain location along the bottom of the channel, there is a smooth bump, 1.0 m long and 0.10 m high at its center (Fig. 15–85). The velocity inlet is split into two parts—the lower part for liquid water and the upper part for air. In the CFD calculations, the inlet velocity of both the air and the water is specified as Vinlet. The water depth at the inlet of the computational domain is specified as yinlet, but the location of the water surface in the rest of the domain is calculated. The flow is modeled as inviscid.
We consider cases with both subcritical and supercritical inlets (Chap. 13). Results from the CFD calculations are shown in Fig. 15–86 for three cases for comparison. For the first case (Fig. 15–86a), yinlet is specified as 0.30 m, and Vinlet is specified as 0.50 m/s. The corresponding Froude number is cal culated to be Froude number: Fr = Vinlet √gyinlet = 0.50 m/s √(9.81 m/s2)(0.30 m) = 0.291 Velocity inlet Air Water Top of domain (inviscid wall) Pressure outlet Channel bottom (wall) Channel bottom (wall) Bump (wall) Vinlet Vinlet yinlet FIGURE 15–85 Computational domain for steady, incompressible, two-dimensional flow of water over a bump along the bottom of a channel, with boundary conditions identified. Two fluids are modeled in the flow field—liquid water and air above the free surface of the water. Liquid depth yinlet and inlet speed Vinlet are specified.
(a) (b) (c) FIGURE 15–86 CFD results for incompressible, two-dimensional flow of water over a bump along the channel bottom. Phase contours are plotted, where blue indicates liquid water and white indicates gaseous air: (a) subcritical-to-subcritical, (b) supercritical-to-supercritical, and (c) subcritical-to-supercritical.
cen96537_ch15_885-946.indd 936 14/01/17 2:10 pm 937 CHAPTER 15 Since Fr < 1, the flow at the inlet is subcritical, and the liquid surface dips slightly above the bump (Fig. 15–86a). The flow remains subcritical down stream of the bump, and the liquid surface height slowly rises back to its prebump level. The flow is thus subcritical everywhere.
For the second case (Fig. 15–86b), yinlet is specified as 0.50 m, and Vinlet is specified as 4.0 m/s. The corresponding Froude number is calculated to be 1.81. Since Fr > 1, the flow at the inlet is supercritical, and the liquid sur face rises above the bump (Fig. 15–86b). Far downstream, the liquid depth returns to 0.50 m, and the average velocity returns to 4.0 m/s, yielding Fr = 1.81—the same as at the inlet. Thus, this flow is supercritical everywhere.
Finally, we show results for a third case (Fig. 15–86c) in which the flow entering the channel is subcritical (yinlet = 0.50 m, Vinlet = 1.0 m/s, and Fr = 0.452). In this case, the water surface dips downward over the bump, as expected for subcritical flow. However, on the downstream side of the bump, youtlet = 0.25 m, Voutlet = 2.0 m/s, and Fr = 1.28. Thus, this flow starts subcritical, but changes to supercritical downstream of the bump. If the domain had extended further downstream, we would likely see a hydraulic jump that would bring the Froude number back below unity (subcritical).
Flow through a Sluice Gate (Hydraulic Jump) As a final example, we consider a two-dimensional channel with a flat, horizontal bottom, but this time with a sluice gate (Fig. 15–87). The water depth at the inlet of the computational domain is specified as yinlet, and the inlet flow velocity is specified as Vinlet. The bottom of the sluice gate is at distance a from the channel bottom. The flow is modeled as inviscid.
We run the CFD code with yinlet = 12.0 m and Vinlet = 0.833 m/s, yield ing an inlet Froude number of Frinlet = 0.0768 (subcritical). The bottom of the sluice gate is at a = 0.125 m from the channel bottom. Results from the CFD calculations are shown in Fig. 15–88. After the water passes under the sluice gate, its average velocity increases to 12.8 m/s, and its depth decreases to y = 0.78 m. Thus, Fr = 4.63 (supercritical) downstream of the sluice gate and upstream of the hydraulic jump. Some distance downstream, we see a hydraulic jump in which the average water depth increases to y = 3.54 m, and the average water velocity decreases to 2.82 m/s. The Froude number downstream of the hydraulic jump is thus Fr = 0.478 (subcritical). We notice that the downstream water depth is significantly lower than that upstream of the sluice gate, indicating relatively large dissipation through Velocity inlet Sluice gate (wall) Pressure outlet Top of domain (inviscid wall) Channel bottom (wall) Air Water Vinlet Vinlet yinlet a FIGURE 15–87 Computational domain for steady, incompressible, two-dimensional flow of water through a sluice gate, with boundary conditions identified. Two fluids are modeled in the flow field— liquid water, and air above the free surface of the water. Liquid depth yinlet and inlet speed Vinlet are specified.
cen96537_ch15_885-946.indd 937 14/01/17 2:10 pm 938 COMPUTATIONAL FLUID DYNAMICS SUMMARY Although neither as ubiquitous as spreadsheets, nor as easy to use as mathematical solvers, computational fluid dynam ics codes are continually improving and are becoming more commonplace. Once the realm of specialized scientists who wrote their own codes and used supercomputers, commer cial CFD codes with numerous features and user-friendly interfaces can now be obtained for personal computers at a reasonable cost and are available to engineers of all dis ciplines. As shown in this chapter, however, a poor grid, improper choice of laminar versus turbulent flow, inap propriate boundary conditions, and/or any of a number of other miscues can lead to CFD solutions that are physically incorrect, even though the colorful graphical output always looks pretty. Therefore, it is imperative that CFD users be well grounded in the fundamentals of fluid mechanics in order to avoid erroneous answers from a CFD simulation. In addition, appropriate comparisons should be made to experi mental data whenever possible to validate CFD predictions. Bearing these cautions in mind, CFD has enormous potential for diverse applications involving fluid flows. We show examples of both laminar and turbulent CFD solu tions. For incompressible laminar flow, computational fluid dynamics does an excellent job, even for unsteady flows with separation. In fact, laminar CFD solutions are “exact” to the extent that they are limited by grid resolution and boundary conditions. Unfortunately, many flows of practical engineering interest are turbulent, not laminar. Direct numerical simu lation (DNS) has great potential for simulation of complex turbulent flow fields, and algorithms for solving the equations of motion (the three-dimensional continuity and Navier– Stokes equations) are well established. However, resolution of all the fine scales of a high Reynolds number complex tur bulent flow requires computers that are orders of magnitude faster than today’s fastest machines. It will be decades before computers advance to the point where DNS is useful for prac tical engineering problems. In the meantime, the best we can do is employ turbulence models, which are semi-empirical transport equations that model (rather than solve) the increased mixing and diffusion caused by turbulent eddies. When run ning CFD codes that utilize turbulence models, we must be careful that we have a fine-enough mesh and that all boundary conditions are properly applied. In the end, however, regard less of how fine the mesh, or how valid the boundary condi tions, turbulent CFD results are only as good as the turbu lence model used. Nevertheless, while no turbulence model is universal (applicable to all turbulent flows), we obtain reason able performance for many practical flow simulations. We also demonstrate in this chapter that CFD can yield useful results for flows with heat transfer, compressible flows, and open-channel flows. In all cases, however, users of CFD must be careful that they choose an appropriate computational domain, apply proper boundary conditions, generate a good grid, and use the proper models and approx imations. As computers continue to become faster and more powerful, CFD will take on an ever-increasing role in design and analysis of complex engineering systems. We have only scratched the surface of computational fluid dynamics in this brief chapter. In order to become proficient and competent at CFD, you must take advanced courses of study in numerical methods, fluid mechanics, turbulence, and heat transfer. We hope that, if nothing else, this chapter has spurred you on to further study of this exciting topic.
the hydraulic jump and a corresponding decrease in the specific energy of the flow (Chap. 13). The analogy between specific energy loss through a hydraulic jump in open-channel flow and stagnation pressure loss through a shock wave in compressible flow is reinforced.
Sluice gate Hydraulic jump (b) (a) FIGURE 15–88 CFD results for incompressible, two-dimensional flow of water through a sluice gate in an open channel. Phase contours are plotted, where blue indi cates liquid water and white indicates gaseous air: (a) overall view of the sluice gate and hydraulic jump, and (b) close-up view of the hydraulic jump. The flow is highly unsteady, and these are instan taneous snapshots at an arbitrary time.
cen96537_ch15_885-946.indd 938 14/01/17 2:10 pm 939 CHAPTER 15 Guest Authors: James G. Brasseur and Anupam Pal, The Pennsylvania State University The mechanical function of the stomach (called gastric “motility”) is central to proper nutrition, reliable drug delivery, and many gastric dysfunctions such as gastroparesis. Figure 15–89 shows a magnetic resonance image (MRI) of the stomach. The stomach is a mixer, a grinder, a storage chamber, and a sophisti cated pump that controls the release of liquid and solid gastric content into the small intestines where nutrient uptake occurs. Nutrient release is controlled by the opening and closing of a valve at the end of the stomach (the pylorus) and the time variations in pressure difference between the stomach and duode num. Gastric pressure is controlled by muscle tension over the stomach wall and peristaltic contraction waves that pass through the antrum (Fig. 15–89). These antral peristaltic contraction waves also break down food particles and mix material within the stomach, both food and drugs. It is currently impossi ble, however, to measure the mixing fluid motions in the human stomach. The MRI, for example, gives only an outline of special magnetized fluid within the stomach. In order to study these invisible fluid motions and their effects, we have developed a computer model of the stomach using computational fluid dynamics.
The mathematics underlying our computational model is derived from the laws of fluid mechanics. The model is a way of extending MRI mea surements of time-evolving stomach geometry to the fluid motions within. Whereas computer models cannot describe the full complexity of gastric physiology, they have the great advantage of allowing controlled systematic variation of parameters, so sensitivities that cannot be measured experimen tally can be studied computationally. Our virtual stomach applies a numerical method called the “lattice Boltzmann” algorithm that is well suited to fluid flows in complex geometries, and the boundary conditions are obtained from MRI data. In Fig. 15–90 we predict the motions, breakdown, and mixing of 1-cm-size extended-release drug tablets in the stomach. In this numeri cal experiment the drug tablet is denser than the surrounding highly vis cous meal. We predict that the antral peristaltic waves generate recirculat ing eddies and retropulsive “jets” within the stomach, which in turn generate high shear stresses that wear away the tablet surface and release the drug. The drug then mixes by the same fluid motions that release the drug. We find that gastric fluid motions and mixing depend on the details of the time variations in stomach geometry and pylorus.
References Indireshkumar, K., Brasseur, J. G., Faas, H., Hebbard, G. S., Kunz, P., Dent, J., Boesinger, P., Feinle, C., Fried, M., Li, M., and Schwizer, W., “Relative Contribution of ‘Pressure Pump’ and ‘Peristaltic Pump’ to Slowed Gastric Emptying,” Amer J Physiol, 278, pp. G604–616, 2000.
Pal, A., Indireshkumar, K., Schwizer, W., Abrahamsson, B., Fried, M., Brasseur, J. G., “2004 Gastric Flow and Mixing Studied Using Computer Simulation,” Proc. Royal Soc. London, Biological Sciences, October 2004.
APPLICATION SPOTLIGHT ■ A Virtual Stomach FIGURE 15–89 Magnetic resonance image of the human stomach in vivo at one instant in time showing peristaltic (i.e., propagating) contraction waves (CW) in the end region of the stomach (the antrum). The pylorus is a sphincter, or valve, that allows nutrients into the duodenum (small intestines).
Developed by Anupam Pal and James Brasseur. Used by permission.
Antral CW Antrum Pylorus 15 y (cm) 10 5 0 0 5 10 x (cm) 15 20 FIGURE 15–90 Computer simulation of fluid motions within the stomach (velocity vectors) from peristaltic antral contraction waves (Fig. 15–89), and the release of a drug (red trail) from an extended release tablet (red circle).
Developed by Anupam Pal and James Brasseur. Used by permission.
cen96537_ch15_885-946.indd 939 14/01/17 2:10 pm 940 COMPUTATIONAL FLUID DYNAMICS REFERENCES AND SUGGESTED READING 1. C-J. Chen and S-Y. Jaw. Fundamentals of Turbulence Modeling. Washington, DC: Taylor & Francis, 1998.
2. J. M. Cimbala, H. Nagib, and A. Roshko. “Large Structure in the Far Wakes of Two-Dimensional Bluff Bodies,” Fluid Mech., 190, pp. 265–298, 1988.
3. R. M. Davies and G. I. Taylor. “The Mechanics of Large Bubbles Rising through Extended Liquids and Through Liquids in Tubes,” Proc. R. Soc. A, 200(1062), pp. 375–390, 1950.
4. C. W. Hirt and B. D. Nichols. “Volume of Fluid (VOF) Method for the Dynamics of Free Boundaries,” J. Comput. Phys., 39(1), pp. 201–225, 1981.
5. A. S. Rattner and S. Garimella. “Vertical Upward Inter mediate Scale Taylor Flow: Experiments and Kinematic Closure,” Int. J. Multiph. Flow, 75, pp. 107–123, 2015.
6. S. Schreier. Compressible Flow. New York: Wiley-Interscience, Chap. 6 (Transonic Flow), pp. 285–293, 1982.
7. J. C. Tannehill, D. A. Anderson, and R. H. Pletcher. Computational Fluid Mechanics and Heat Transfer, 3rd ed. Washington, DC: Taylor & Francis, 2012.
8. H. Tennekes and J. L. Lumley. A First Course in Turbulence. Cambridge, MA: The MIT Press, 1972.
9. D. J. Tritton. Physical Fluid Dynamics. New York: Van Nostrand Reinhold Co., 1977.
10. M. Van Dyke. An Album of Fluid Motion. Stanford, CA: The Parabolic Press, 1982.
11. F. M. White. Viscous Fluid Flow, 3rd ed. New York: McGraw-Hill, 2005.
12. D. C. Wilcox. Turbulence Modeling for CFD, 3rd ed. La Cañada, CA: DCW Industries, Inc., 2006.
13. C. H. K. Williamson. “Oblique and Parallel Modes of Vortex Shedding in the Wake of a Circular Cylinder at Low Reynolds Numbers,” J. Fluid Mech., 206, pp. 579–627, 1989.
14. J. Tu, G. H. Yeoh, and C. Liu. Computational Fluid Dynamics: A Practical Approach. Burlington, MA: Elsevier, 2008.
PROBLEMS Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and the SI users can ignore them. Additional CFD problems are posted on the text website.
Fundamentals, Grid Generation, and Boundary Conditions 15–1C A CFD code is used to solve a two-dimensional (x and y), incompressible, laminar flow without free surfaces. The fluid is Newtonian. Appropriate boundary conditions are used. List the variables (unknowns) in the problem, and list the corresponding equations to be solved by the computer.
15–2C Write a brief (a few sentences) definition and description of each of the following, and provide example(s) if helpful: (a) computational domain, (b) mesh, (c) transport equation, (d) coupled equations.
15–3C What is the difference between a node and an interval and how are they related to cells? In Fig. P15–3C, how many nodes and how many intervals are on each edge?
FIGURE P15–3C 15–4C For the two-dimensional computational domain of Fig. P15–3C, with the given node distribution, sketch a sim ple structured grid using four-sided cells and sketch a simple unstructured grid using three-sided cells. How many cells are in each? Discuss.
15–5C For the two-dimensional computational domain of Fig. P15–3C, with the given node distribution, sketch a simple structured grid using four-sided cells and sketch a simple unstructured polyhedral grid using at least one of each: 3-sided, 4-sided, and 5-sided cells. Try to avoid large skewness. Compare the cell count for each case and discuss your results.
15–6C Summarize the eight steps involved in a typical CFD analysis of a steady, laminar flow field.
cen96537_ch15_885-946.indd 940 14/01/17 2:10 pm 941 CHAPTER 15 15–7C Suppose you are using CFD to simulate flow through a duct in which there is a circular cylinder as in Fig. P15–7C. The duct is long, but to save computer resources you choose a computational domain in the vicinity of the cylinder only. Explain why the downstream edge of the computational domain should be further from the cylin der than the upstream edge.
Computational domain In Out FIGURE P15–7C 15–8C Write a brief (a few sentences) discussion about the significance of each of the following in regards to an iterative CFD solution: (a) initial conditions, (b) residual, (c) iteration, and (d) postprocessing.
15–9C Briefly discuss how each of the following is used by CFD codes to speed up the iteration process: (a) multigrid ding and (b) artificial time.
15–10C Of the boundary conditions discussed in this chap ter, list all the boundary conditions that may be applied to the right edge of the two-dimensional computational domain sketched in Fig. P15–10C. Why can’t the other boundary conditions be applied to this edge?
BC to be specified on this edge FIGURE P15–10C 15–11C What is the standard method to test for adequate grid resolution when using CFD?
15–12C What is the difference between a pressure inlet and a velocity inlet boundary condition? Explain why you cannot specify both pressure and velocity at a velocity inlet bound ary condition or at a pressure inlet boundary condition.
15–13C An incompressible CFD code is used to simulate the flow of air through a two-dimensional rectangular chan nel (Fig. P15–13C). The computational domain consists of four blocks, as indicated. Flow enters block 4 from the upper right and exits block 1 to the left as shown. Inlet velocity V is known and outlet pressure Pout is also known. Label the boundary conditions that should be applied to every edge of every block of this computational domain.
Block 1 Block 2 Block 3 Block 4 In Out V Pout FIGURE P15–13C 15–14C Consider Prob. 15–13C again, except let the boundary condition on the common edge between blocks 1 and 2 be a fan with a specified pressure rise from right to left across the fan. Suppose an incompressible CFD code is run for both cases (with and without the fan). All else being equal, will the pressure at the inlet increase or decrease? Why? What will happen to the velocity at the outlet? Explain.
15–15C List six boundary conditions that are used with CFD to solve incompressible fluid flow problems. For each one, provide a brief description and give an example of how that boundary condition is used.
15–16 A CFD code is used to simulate flow over a two-dimensional airfoil at an angle of attack. A portion of the com putational domain near the airfoil is outlined in Fig. P15–16 (the computational domain extends well beyond the region outlined by the dashed line). Sketch a coarse structured grid using four-sided cells and sketch a coarse unstructured grid using three-sided cells in the region shown. Be sure to cluster the cells where appropriate. Discuss the advantages and dis advantages of each grid type.
FIGURE P15–16 15–17 For the airfoil of Prob. 15–16, sketch a coarse hybrid grid and explain the advantages of such a grid.
15–18 An incompressible CFD code is used to simulate the flow of water through a two-dimensional rectangular channel in which there is a circular cylinder (Fig. P15–18). A time-averaged turbulent flow solution is generated using a turbu lence model. Top–bottom symmetry about the cylinder is assumed. Flow enters from the left and exits to the right as shown. Inlet velocity V is known, and outlet pressure Pout is also known. Generate the blocking for a structured grid using four-sided blocks, and sketch a coarse grid using four-sided cen96537_ch15_885-946.indd 941 14/01/17 2:10 pm 942 COMPUTATIONAL FLUID DYNAMICS cells, being sure to cluster cells near walls. Also be careful to avoid highly skewed cells. Label the boundary conditions that should be applied to every edge of every block of your com putational domain. (Hint: Six to seven blocks are sufficient.) V In Out Pout FIGURE P15–18 15–19 An incompressible CFD code is used to simulate the flow of gasoline through a two-dimensional rectangular channel in which there is a large circular settling chamber (Fig. P15–19). Flow enters from the left and exits to the right as shown. A time-averaged turbulent flow solution is generated using a tur bulence model. Top–bottom symmetry is assumed. Inlet veloc ity V is known, and outlet pressure Pout is also known. Generate the blocking for a structured grid using four-sided blocks, and sketch a coarse grid using four-sided cells, being sure to cluster cells near walls. Also be careful to avoid highly skewed cells. Label the boundary conditions that should be applied to every edge of every block of your computational domain.
Pout In V Out FIGURE P15–19 15–20 Redraw the structured multiblock grid of Fig. 15–12b for the case in which your CFD code can handle only elemen tary blocks. Renumber all the blocks and indicate how many i- and j-intervals are contained in each block. How many ele mentary blocks do you end up with? Add up all the cells, and verify that the total number of cells does not change.
15–21 Suppose your CFD code can handle nonelementary blocks. Combine as many blocks of Fig. 15–12b as you can. The only restriction is that in any one block, the number of i-intervals and the number of j-intervals must be constants. Show that you can create a structured grid with only three nonelementary blocks. Renumber all the blocks and indicate how many i- and j-intervals are contained in each block. Add up all the cells and verify that the total number of cells does not change.
15–22 A new heat exchanger is being designed with the goal of mixing the fluid downstream of each stage as thoroughly as possible. Anita comes up with a design whose cross section for one stage is sketched in Fig. P15–22. The geometry extends periodically up and down beyond the region shown here. She uses several dozen rectangular tubes inclined at a high angle of attack to ensure that the flow separates and mixes in the wakes. The performance of this geometry is to be tested using two-dimensional time-averaged CFD simulations with a turbulence model, and the results will be compared to those of competing geometries. Sketch the simplest possible computational domain that can be used to simulate this flow. Label and indicate all boundary conditions on your diagram. Discuss.
FIGURE P15–22 15–23 Sketch a coarse structured multiblock grid with four-sided elementary blocks and four-sided cells for the computa tional domain of Prob. 15–22.
15–24 Anita runs a CFD code using the computational domain and grid developed in Probs. 15–22 and 15–23. Unfortunately, the CFD code has a difficult time converging and Anita real izes that there is reverse flow at the outlet (far right edge of the computational domain). Explain why there is reverse flow, and discuss what Anita should do to correct the problem.
15–25 As a follow-up to the heat exchanger design of Prob. 15–22, suppose Anita’s design is chosen based on the results of a preliminary single-stage CFD analysis. Now she is asked to simulate two stages of the heat exchanger. The second row of rectangular tubes is staggered and inclined oppositely to that of the first row to promote mixing (Fig. P15–25). The geometry extends periodically up and down beyond the region shown here. Sketch a computational domain that can be used to simulate this flow. Label and indicate all boundary conditions on your diagram. Discuss.
FIGURE P15–25 cen96537_ch15_885-946.indd 942 14/01/17 2:10 pm 943 CHAPTER 15 15–26 Sketch a structured multiblock grid with four-sided elementary blocks for the computational domain of Prob. 15–25. Each block is to have four-sided structured cells, but you do not have to sketch the grid, just the block topology. Try to make all the blocks as rectangular as possible to avoid highly skewed cells in the corners. Assume that the CFD code requires that the node distribution on periodic pairs of edges be identical (the two edges of a periodic pair are “linked” in the grid generation process). Also assume that the CFD code does not allow a block’s edges to be split for application of boundary conditions.
General CFD Problems 15–27 Consider the two-dimensional wye of Fig. P15–27. Dimensions are in meters, and the drawing is not to scale. Incompressible flow enters from the left, and splits into two parts. Generate three coarse grids, with identical node distri butions on all edges of the computational domain: (a) struc tured multiblock grid, (b) unstructured triangular grid, and (c) unstructured quadrilateral grid. Compare the number of cells in each case and comment about the quality of the grid in each case.
(2, 1) (0, 1) (0, 0) (5, 0) (5, 0.5) (4.5, 3.5) (2.5, 0.5) (5, 3) FIGURE P15–27 15–28 Choose one of the grids generated in Prob. 15–27, and run a CFD solution for laminar flow of air with a uniform inlet velocity of 0.02 m/s. Set the outlet pressure at both outlets to the same value, and calculate the pressure drop through the wye. Also calculate the percentage of the inlet flow that goes out of each branch. Generate a plot of streamlines.
15–29 Repeat Prob. 15–28, except for turbulent flow of air with a uniform inlet velocity of 10.0 m/s. In addition, set the turbulence intensity at the inlet to 10 percent with a turbulent length scale of 0.5 m. Use the k-𝜀 turbulence model with wall functions. Set the outlet pressure at both outlets to the same value, and calculate the pressure drop through the wye. Also calculate the percentage of the inlet flow that goes out of each branch. Generate a plot of streamlines. Compare results with those of laminar flow (Prob. 15–28).
15–30 Generate a computational domain to study the lami nar boundary layer growing on a flat plate at Re = 10,000. Generate a very coarse mesh, and then continually refine the mesh until the solution becomes grid independent. Discuss.
15–31 Repeat Prob. 15–30, except for a turbulent boundary layer at Re = 106. Discuss.
15–32 Generate a computational domain to study ventila tion in a room (Fig. P15–32). Specifically, generate a rectan gular room with a velocity inlet in the ceiling to model the supply air, and a pressure outlet in the ceiling to model the return air. You may make a two-dimensional approximation for simplicity (the room is infinitely long in the direction normal to the page in Fig. P15–32). Use a structured rect angular grid. Plot streamlines and velocity vectors. Discuss.
Air supply Air return FIGURE P15–32 15–33 Repeat Prob. 15–32, except use an unstructured triangular grid, keeping everything else the same. Do you get the same results as those of Prob. 15–32? Compare and discuss.
15–34 Repeat Prob. 15–32, except move the supply and/or return vents to various locations in the ceiling. Compare and discuss.
15–35 Choose one of the room geometries of Probs. 15–32 and 15–34, and add the energy equation to the calculations. In particular, model a room with air-conditioning, by spec ifying the supply air as cool (T = 18°C), while the walls, floor, and ceiling are warm (T = 26°C). Adjust the supply air speed until the average temperature in the room is as close as possible to 22°C. How much ventilation (in terms of number of room air volume changes per hour) is required to cool this room to an average temperature of 22°C? Discuss.
15–36 Repeat Prob. 15–35, except create a three-dimensional room, with an air supply and an air return in the ceiling. Compare the two-dimensional results of Prob. 15–35 with the more realistic three-dimensional results of this problem. Discuss.
15–37 Generate a computational domain to study compress ible flow of air through a converging nozzle with atmospheric pressure at the nozzle exit (Fig. P15–37). The nozzle walls may be approximated as inviscid (zero shear stress). Run sev eral cases with various values of inlet pressure. How much These problems require CFD software, although not any particular brand. Students must do the following problems “from scratch,” including generation of an appropriate mesh.
cen96537_ch15_885-946.indd 943 14/01/17 2:10 pm 944 COMPUTATIONAL FLUID DYNAMICS inlet pressure is required to choke the flow? What happens if the inlet pressure is higher than this value? Discuss.
Pressure outlet Pressure inlet FIGURE P15–37 15–38 Repeat Prob. 15–37, except remove the inviscid flow approximation. Instead, let the flow be turbulent, with smooth, no-slip walls. Compare your results to those of Prob. 15–37. What is the major effect of friction in this problem? Discuss.
15–39 Generate a computational domain to study incom pressible, laminar flow over a two-dimensional streamlined body (Fig. P15–39). Generate various body shapes, and cal culate the drag coefficient for each shape. What is the small est value of CD that you can achieve? (Note: For fun, this problem can be turned into a contest between students. Who can generate the lowest-drag body shape?) Body FD V FIGURE P15–39 15–40 Repeat Prob. 15–39, except for an axisymmetric, rather than a two-dimensional, body. Compare to the two-dimensional case. For the same sectional slice shape, which has the lower drag coefficient? Discuss.
15–41 Repeat Prob. 15–40, except for turbulent, rather than laminar, flow. Compare to the laminar case. Which has the lower drag coefficient? Discuss.
15–42 Generate a computational domain to study Mach waves in a two-dimensional supersonic channel (Fig. P15–42). Specifically, the domain should consist of a simple rectan gular channel with a supersonic inlet (Ma = 2.0), and with a very small bump on the lower wall. Using air with the invis cid flow approximation, generate a Mach wave, as sketched. Measure the Mach angle, and compare with theory (Chap. 12). Also discuss what happens when the Mach wave hits the opposite wall. Does it disappear, or does it reflect, and if so, what is the reflection angle? Discuss.
Ma Bump ?
FIGURE P15–42 15–43 Repeat Prob. 15–42, except for several values of the Mach number, ranging from 1.10 to 3.0. Plot the calculated Mach angle as a function of Mach number and compare to the theoretical Mach angle (Chap. 12). Discuss.
Review Problems 15–44C For each statement, choose whether the statement is true or false, and discuss your answer briefly: (a) The physical validity of a CFD solution always improves as the grid is refined.
(b) The x-component of the Navier–Stokes equation is an example of a transport equation.
(c) For the same number of nodes in a two-dimensional mesh, a structured grid typically has fewer cells than an unstructured triangular grid.
(d ) A time-averaged turbulent flow CFD solution is only as good as the turbulence model used in the calculations.
15–45C In Prob. 15–19 we take advantage of top–bottom symmetry when constructing our computational domain and grid. Why can’t we also take advantage of the right–left sym metry in this exercise? Repeat the discussion for the case of potential flow.
15–46C Gerry creates the computational domain sketched in Fig. P15–46C to simulate flow through a sudden contrac tion in a two-dimensional duct. He is interested in the time-averaged pressure drop and the minor loss coefficient created by the sudden contraction. Gerry generates a grid and calcu lates the flow with a CFD code, assuming steady, turbulent, incompressible flow (with a turbulence model).
(a) Discuss one way that Gerry could improve his computa tional domain and grid so that he would get the same results in approximately half the computer time.
(b) There may be a fundamental flaw in how Gerry has set up his computational domain. What is it? Discuss what should be different about Gerry’s setup.
Out In FIGURE P15–46C cen96537_ch15_885-946.indd 944 14/01/17 2:10 pm 945 CHAPTER 15 15–47C Think about modern high-speed, large-memory computer systems. What feature of such computers lends itself nicely to the solution of CFD problems using a multi block grid with approximately equal numbers of cells in each individual block? Discuss.
15–48C What is the difference between multigridding and multiblocking? Discuss how each may be used to speed up a CFD calculation. Can these two be applied together?
15–49C Suppose you have a fairly complex geometry and a CFD code that can handle unstructured grids with triangular cells. Your grid generation code can create an unstructured grid very quickly. Give some reasons why it might be wiser to take the time to create a multiblock structured grid instead. In other words, is it worth the effort? Discuss.
15–50 Generate a computational domain and grid, and calculate flow through the single-stage heat exchanger of Prob. 15–22, with the heating elements set at a 45° angle of attack with respect to horizontal. Set the inlet air temperature to 20°C, and the wall temperature of the heating elements to 120°C. Calculate the average air temperature at the outlet.
15–51 Repeat the calculations of Prob. 15–50 for several angles of attack of the heating elements, from 0 (horizon tal) to 90° (vertical). Use identical inlet conditions and wall conditions for each case. Which angle of attack provides the most heat transfer to the air? Specifically, which angle of attack yields the highest average outlet temperature?
15–52 Generate a computational domain and grid, and calcu late flow through the two-stage heat exchanger of Prob. 15–25, with the heating elements of the first stage set at a 45° angle of attack with respect to horizontal, and those of the second stage set to an angle of attack of −45°. Set the inlet air temperature to 20°C, and the wall temperature of the heating elements to 120°C. Calculate the average air temperature at the outlet.
15–53 Repeat the calculations of Prob. 15–52 for several angles of attack of the heating elements, from 0 (horizontal) to 90° (vertical). Use identical inlet conditions and wall con ditions for each case. Note that the second stage of heating elements should always be set to an angle of attack that is the negative of that of the first stage. Which angle of attack provides the most heat transfer to the air? Specifically, which angle of attack yields the highest average outlet temperature? Is this the same angle as calculated for the single-stage heat exchanger of Prob. 15–51? Discuss.
15–54 Generate a computational domain and grid, and cal culate stationary turbulent flow over a spinning circular cyl inder (Fig. P15–54). In which direction is the side force on the body—up or down? Explain. Plot streamlines in the flow. Where is the upstream stagnation point?
V ω D FIGURE P15–54 15–55 For the spinning cylinder of Fig. P15–54, generate a dimensionless parameter for rotational speed relative to free-stream speed (combine variables 𝜔, D, and V into a nondi mensional Pi group). Repeat the calculations of Prob. 15–54 for several values of angular velocity 𝜔. Use identical inlet conditions for each case. Plot lift and drag coefficients as functions of your dimensionless parameter. Discuss.
15–56 Consider the flow of air into a two-dimensional slot along the floor of a large room, where the floor is coincident with the x-axis (Fig. P15–56). Generate an appropriate com putational domain and grid. Using the inviscid flow approxi mation in the CFD code, calculate vertical velocity compo nent 𝜐 as a function of distance away from the slot along the y-axis. Compare with the potential flow results of Chap. 10 for flow into a line sink. Discuss.
⋅ y x Room Floor V FIGURE P15–56 15–57 For the slot flow of Prob. 15–56, change to laminar flow instead of inviscid flow, and recompute the flow field. Compare your results to the inviscid flow case and to the poten tial flow case of Chap. 10. Plot contours of vorticity. Where is the irrotational flow approximation appropriate? Discuss.
15–58 Generate a computational domain and grid, and cal culate the flow of air into a two-dimensional vacuum cleaner inlet (Fig. P15–58), using the inviscid flow approximation in the CFD code. Compare your results with those predicted in Chap. 10 for potential flow. Discuss.
y x Stagnation point Maximum speed, potential flow Vacuum nozzle b b b w V .
FIGURE P15–58 15–59 For the vacuum cleaner of Prob. 15–58, change to laminar flow instead of inviscid flow, and recompute the flow field. Compare your results to the inviscid flow case and to the potential flow case of Chap. 10. Discuss.
cen96537_ch15_885-946.indd 945 14/01/17 2:10 pm This page intentionally left blank 947 APPENDIX 1 P RO P E RT Y TABLE S AN D C H A RTS ( S I U N I TS) TABLE A–1 Molar Mass, Gas Constant, and Ideal-Gas Specific Heats of Some Substances 948 TABLE A–2 Boiling and Freezing Point Properties 949 TABLE A–3 Properties of Saturated Water 950 TABLE A–4 Properties of Saturated Refrigerant-134a 951 TABLE A–5 Properties of Saturated Ammonia 952 TABLE A–6 Properties of Saturated Propane 953 TABLE A–7 Properties of Liquids 954 TABLE A–8 Properties of Liquid Metals 955 TABLE A–9 Properties of Air at 1 atm Pressure 956 TABLE A–10 Properties of Gases at 1 atm Pressure 957 TABLE A–11 Properties of the Atmosphere at High Altitude 959 FIGURE A–12 The Moody Chart for the Friction Factor for Fully Developed Flow in Circular Pipes 960 TABLE A–13 One-Dimensional Isentropic Compressible Flow Functions for an Ideal Gas with k = 1.4 961 TABLE A–14 One-Dimensional Normal Shock Functions for an Ideal Gas with k = 1.4 962 TABLE A–15 Rayleigh Flow Functions for an Ideal Gas with k = 1.4 963 TABLE A–16 Fanno Flow Functions for an Ideal Gas with k = 1.4 964 Most properties in the tables are obtained from the property database of EES, and the original sources are listed under the tables. Properties are often listed to more significant digits than the claimed accuracy for the purpose of minimizing accumulated round-off error in hand calculations and ensuring a close match with the results obtained with EES.
cen96537_app1_947_964.indd 947 16/01/17 12:01 pm 948 PROPERTY TABLES AND CHARTS The unit kJ/kg·K is equivalent to kPa·m3/kg·K. The gas constant is calculated from R = Ru /M, where Ru = 8.31447 kJ/kmol·K is the universal gas constant and M is the molar mass.
Source: Specific heat values are obtained primarily from the property routines prepared by The National Institute of Standards and Technology (NIST), Gaithersburg, MD.
TABLE A–1 Molar mass, gas constant, and ideal-gas specific heats of some substances Molar Mass M, kg/kmol Gas Constant R, kJ/kg·K Specific Heat Data at 25°C Substance cp, kJ/kg·K cv, kJ/kg·K k = cp /cv Air 28.97 0.2870 1.005 0.7180 1.400 Ammonia, NH3 17.03 0.4882 2.093 1.605 1.304 Argon, Ar 39.95 0.2081 0.5203 0.3122 1.667 Bromine, Br2 159.81 0.05202 0.2253 0.1732 1.300 Isobutane, C4H10 58.12 0.1430 1.663 1.520 1.094 n-Butane, C4H10 58.12 0.1430 1.694 1.551 1.092 Carbon dioxide, CO2 44.01 0.1889 0.8439 0.6550 1.288 Carbon monoxide, CO 28.01 0.2968 1.039 0.7417 1.400 Chlorine, Cl2 70.905 0.1173 0.4781 0.3608 1.325 Chlorodifluoromethane (R-22), CHClF2 86.47 0.09615 0.6496 0.5535 1.174 Ethane, C2H6 30.070 0.2765 1.744 1.468 1.188 Ethylene, C2H4 28.054 0.2964 1.527 1.231 1.241 Fluorine, F2 38.00 0.2187 0.8237 0.6050 1.362 Helium, He 4.003 2.077 5.193 3.116 1.667 n-Heptane, C7H16 100.20 0.08297 1.649 1.566 1.053 n-Hexane, C6H14 86.18 0.09647 1.654 1.558 1.062 Hydrogen, H2 2.016 4.124 14.30 10.18 1.405 Krypton, Kr 83.80 0.09921 0.2480 0.1488 1.667 Methane, CH4 16.04 0.5182 2.226 1.708 1.303 Neon, Ne 20.183 0.4119 1.030 0.6180 1.667 Nitrogen, N2 28.01 0.2968 1.040 0.7429 1.400 Nitric oxide, NO 30.006 0.2771 0.9992 0.7221 1.384 Nitrogen dioxide, NO2 46.006 0.1889 0.8060 0.6171 1.306 Oxygen, O2 32.00 0.2598 0.9180 0.6582 1.395 n-Pentane, C5H12 72.15 0.1152 1.664 1.549 1.074 Propane, C3H8 44.097 0.1885 1.669 1.480 1.127 Propylene, C3H6 42.08 0.1976 1.531 1.333 1.148 Steam, H2O 18.015 0.4615 1.865 1.403 1.329 Sulfur dioxide, SO2 64.06 0.1298 0.6228 0.4930 1.263 Tetrachloromethane, CCl4 153.82 0.05405 0.5415 0.4875 1.111 Tetrafluoroethane (R-134a), C2H2F4 102.03 0.08149 0.8334 0.7519 1.108 Trifluoroethane (R-143a), C2H3F3 84.04 0.09893 0.9291 0.8302 1.119 Xenon, Xe 131.30 0.06332 0.1583 0.09499 1.667 cen96537_app1_947_964.indd 948 16/01/17 12:01 pm 949 APPENDIX 1 TABLE A–2 Boiling and freezing point properties Boiling Data at 1 atm Freezing Data Liquid Properties Normal Latent Heat of Latent Heat Specific Boiling Vaporization Freezing of Fusion Temperature, Density Heat Substance Point, °C hfg, kJ/kg Point, °C hif, kJ/kg °C 𝜌, kg/m3 cp, kJ/kg·K Ammonia −33.3 1357 −77.7 322.4 −33.3 682 4.43 −20 665 4.52 0 639 4.60 25 602 4.80 Argon −185.9 161.6 −189.3 28 −185.6 1394 1.14 Benzene 80.2 394 5.5 126 20 879 1.72 Brine (20% sodium chloride by mass) 103.9 — −17.4 — 20 1150 3.11 n-Butane −0.5 385.2 −138.5 80.3 −0.5 601 2.31 Carbon dioxide −78.4 230.5 (at 0°C) −56.6 0 298 0.59 Ethanol 78.2 838.3 −114.2 109 25 783 2.46 Ethyl alcohol 78.6 855 −156 108 20 789 2.84 Ethylene glycol 198.1 800.1 −10.8 181.1 20 1109 2.84 Glycerine 179.9 974 18.9 200.6 20 1261 2.32 Helium −268.9 22.8 — — −268.9 146.2 22.8 Hydrogen −252.8 445.7 −259.2 59.5 −252.8 70.7 10.0 Isobutane −11.7 367.1 −160 105.7 −11.7 593.8 2.28 Kerosene 204–293 251 −24.9 — 20 820 2.00 Mercury 356.7 294.7 −38.9 11.4 25 13,560 0.139 Methane −161.5 510.4 −182.2 58.4 −161.5 423 3.49 −100 301 5.79 Methanol 64.5 1100 −97.7 99.2 25 787 2.55 Nitrogen −195.8 198.6 −210 25.3 −195.8 809 2.06 −160 596 2.97 Octane 124.8 306.3 −57.5 180.7 20 703 2.10 Oil (light) 25 910 1.80 Oxygen −183 212.7 −218.8 13.7 −183 1141 1.71 Petroleum — 230–384 20 640 2.0 Propane −42.1 427.8 −187.7 80.0 −42.1 581 2.25 0 529 2.53 50 449 3.13 Refrigerant-134a −26.1 216.8 −96.6 — −50 1443 1.23 −26.1 1374 1.27 0 1295 1.34 25 1207 1.43 Water 100 2257 0.0 333.7 0 1000 4.22 25 997 4.18 50 988 4.18 75 975 4.19 100 958 4.22 Sublimation temperature. (At pressures below the triple-point pressure of 518 kPa, carbon dioxide exists as a solid or gas. Also, the freezing-point temperature of carbon dioxide is the triple-point temperature of −56.5°C.) cen96537_app1_947_964.indd 949 16/01/17 12:01 pm 950 PROPERTY TABLES AND CHARTS TABLE A–3 Properties of saturated water Volume Enthalpy Specific Thermal Prandtl Expansion Surface Saturation Density of Heat Conductivity Dynamic Viscosity Number Coefficient Tension, Temp. Pressure 𝜌, kg/m3 Vaporization cp, J/kg·K k, W/m·K 𝜇, kg/m·s Pr 𝛽, 1/K N/m T, °C Psat, kPa Liquid Vapor hfg, kJ/kg Liquid Vapor Liquid Vapor Liquid Vapor Liquid Vapor Liquid Liquid 0.01 0.6113 999.8 0.0048 2501 4217 1854 0.561 0.0171 1.792 × 10−3 0.922 × 10−5 13.5 1.00 −0.068 × 10−3 0.0756 5 0.8721 999.9 0.0068 2490 4205 1857 0.571 0.0173 1.519 × 10−3 0.934 × 10−5 11.2 1.00 0.015 × 10−3 0.0749 10 1.2276 999.7 0.0094 2478 4194 1862 0.580 0.0176 1.307 × 10−3 0.946 × 10−5 9.45 1.00 0.733 × 10−3 0.0742 15 1.7051 999.1 0.0128 2466 4186 1863 0.589 0.0179 1.138 × 10−3 0.959 × 10−5 8.09 1.00 0.138 × 10−3 0.0735 20 2.339 998.0 0.0173 2454 4182 1867 0.598 0.0182 1.002 × 10−3 0.973 × 10−5 7.01 1.00 0.195 × 10−3 0.0727 25 3.169 997.0 0.0231 2442 4180 1870 0.607 0.0186 0.891 × 10−3 0.987 × 10−5 6.14 1.00 0.247 × 10−3 0.0720 30 4.246 996.0 0.0304 2431 4178 1875 0.615 0.0189 0.798 × 10−3 1.001 × 10−5 5.42 1.00 0.294 × 10−3 0.0712 35 5.628 994.0 0.0397 2419 4178 1880 0.623 0.0192 0.720 × 10−3 1.016 × 10−5 4.83 1.00 0.337 × 10−3 0.0704 40 7.384 992.1 0.0512 2407 4179 1885 0.631 0.0196 0.653 × 10−3 1.031 × 10−5 4.32 1.00 0.377 × 10−3 0.0696 45 9.593 990.1 0.0655 2395 4180 1892 0.637 0.0200 0.596 × 10−3 1.046 × 10−5 3.91 1.00 0.415 × 10−3 0.0688 50 12.35 988.1 0.0831 2383 4181 1900 0.644 0.0204 0.547 × 10−3 1.062 × 10−5 3.55 1.00 0.451 × 10−3 0.0679 55 15.76 985.2 0.1045 2371 4183 1908 0.649 0.0208 0.504 × 10−3 1.077 × 10−5 3.25 1.00 0.484 × 10−3 0.0671 60 19.94 983.3 0.1304 2359 4185 1916 0.654 0.0212 0.467 × 10−3 1.093 × 10−5 2.99 1.00 0.517 × 10−3 0.0662 65 25.03 980.4 0.1614 2346 4187 1926 0.659 0.0216 0.433 × 10−3 1.110 × 10−5 2.75 1.00 0.548 × 10−3 0.0654 70 31.19 977.5 0.1983 2334 4190 1936 0.663 0.0221 0.404 × 10−3 1.126 × 10−5 2.55 1.00 0.578 × 10−3 0.0645 75 38.58 974.7 0.2421 2321 4193 1948 0.667 0.0225 0.378 × 10−3 1.142 × 10−5 2.38 1.00 0.607 × 10−3 0.0636 80 47.39 971.8 0.2935 2309 4197 1962 0.670 0.0230 0.355 × 10−3 1.159 × 10−5 2.22 1.00 0.653 × 10−3 0.0627 85 57.83 968.1 0.3536 2296 4201 1977 0.673 0.0235 0.333 × 10−3 1.176 × 10−5 2.08 1.00 0.670 × 10−3 0.0617 90 70.14 965.3 0.4235 2283 4206 1993 0.675 0.0240 0.315 × 10−3 1.193 × 10−5 1.96 1.00 0.702 × 10−3 0.0608 95 84.55 961.5 0.5045 2270 4212 2010 0.677 0.0246 0.297 × 10−3 1.210 × 10−5 1.85 1.00 0.716 × 10−3 0.0599 100 101.33 957.9 0.5978 2257 4217 2029 0.679 0.0251 0.282 × 10−3 1.227 × 10−5 1.75 1.00 0.750 × 10−3 0.0589 110 143.27 950.6 0.8263 2230 4229 2071 0.682 0.0262 0.255 × 10−3 1.261 × 10−5 1.58 1.00 0.798 × 10−3 0.0570 120 198.53 943.4 1.121 2203 4244 2120 0.683 0.0275 0.232 × 10−3 1.296 × 10−5 1.44 1.00 0.858 × 10−3 0.0550 130 270.1 934.6 1.496 2174 4263 2177 0.684 0.0288 0.213 × 10−3 1.330 × 10−5 1.33 1.01 0.913 × 10−3 0.0529 140 361.3 921.7 1.965 2145 4286 2244 0.683 0.0301 0.197 × 10−3 1.365 × 10−5 1.24 1.02 0.970 × 10−3 0.0509 150 475.8 916.6 2.546 2114 4311 2314 0.682 0.0316 0.183 × 10−3 1.399 × 10−5 1.16 1.02 1.025 × 10−3 0.0487 160 617.8 907.4 3.256 2083 4340 2420 0.680 0.0331 0.170 × 10−3 1.434 × 10−5 1.09 1.05 1.145 × 10−3 0.0466 170 791.7 897.7 4.119 2050 4370 2490 0.677 0.0347 0.160 × 10−3 1.468 × 10−5 1.03 1.05 1.178 × 10−3 0.0444 180 1,002.1 887.3 5.153 2015 4410 2590 0.673 0.0364 0.150 × 10−3 1.502 × 10−5 0.983 1.07 1.210 × 10−3 0.0422 190 1,254.4 876.4 6.388 1979 4460 2710 0.669 0.0382 0.142 × 10−3 1.537 × 10−5 0.947 1.09 1.280 × 10−3 0.0399 200 1,553.8 864.3 7.852 1941 4500 2840 0.663 0.0401 0.134 × 10−3 1.571 × 10−5 0.910 1.11 1.350 × 10−3 0.0377 220 2,318 840.3 11.60 1859 4610 3110 0.650 0.0442 0.122 × 10−3 1.641 × 10−5 0.865 1.15 1.520 × 10−3 0.0331 240 3,344 813.7 16.73 1767 4760 3520 0.632 0.0487 0.111 × 10−3 1.712 × 10−5 0.836 1.24 1.720 × 10−3 0.0284 260 4,688 783.7 23.69 1663 4970 4070 0.609 0.0540 0.102 × 10−3 1.788 × 10−5 0.832 1.35 2.000 × 10−3 0.0237 280 6,412 750.8 33.15 1544 5280 4835 0.581 0.0605 0.094 × 10−3 1.870 × 10−5 0.854 1.49 2.380 × 10−3 0.0190 300 8,581 713.8 46.15 1405 5750 5980 0.548 0.0695 0.086 × 10−3 1.965 × 10−5 0.902 1.69 2.950 × 10−3 0.0144 320 11,274 667.1 64.57 1239 6540 7900 0.509 0.0836 0.078 × 10−3 2.084 × 10−5 1.00 1.97 0.0099 340 14,586 610.5 92.62 1028 8240 11,870 0.469 0.110 0.070 × 10−3 2.255 × 10−5 1.23 2.43 0.0056 360 18,651 528.3 144.0 720 14,690 25,800 0.427 0.178 0.060 × 10−3 2.571 × 10−5 2.06 3.73 0.0019 374.14 22,090 317.0 317.0 0 — — — — 0.043 × 10−3 4.313 × 10−5 0 Note 1: Kinematic viscosity 𝜈 and thermal diffusivity 𝛼 can be calculated from their definitions, 𝜈 = 𝜇/𝜌 and 𝛼 = k/𝜌cp = 𝜈/Pr. The temperatures 0.01°C, 100°C, and 374.14°C are the triple-, boiling-, and critical-point temperatures of water, respectively. The properties listed above (except the vapor density) can be used at any pressure with negligible error except at temperatures near the critical-point value.
Note 2: The unit kJ/kg·°C for specific heat is equivalent to kJ/kg·K, and the unit W/m·°C for thermal conductivity is equivalent to W/m·K.
Source: Viscosity and thermal conductivity data are from J. V. Sengers and J. T. R. Watson, Journal of Physical and Chemical Reference Data 15 (1986), pp. 1291–1322. Other data are obtained from various sources or calculated.
cen96537_app1_947_964.indd 950 16/01/17 12:01 pm 951 APPENDIX 1 Note 1: Kinematic viscosity 𝜈 and thermal diffusivity 𝛼 can be calculated from their definitions, 𝜈 = 𝜇/𝜌 and 𝛼 = k/𝜌cp = 𝜈/Pr. The properties listed here (except the vapor density) can be used at any pressures with negligible error except at temperatures near the critical-point value.
Note 2: The unit kJ/kg·°C for specific heat is equivalent to kJ/kg·K, and the unit W/m·°C for thermal conductivity is equivalent to W/m·K.
Source: Data generated from the EES software developed by S. A. Klein and F. L. Alvarado. Original sources: R. Tillner-Roth and H. D. Baehr, “An International Standard Formulation for the Thermodynamic Properties of 1,1,1,2-Tetrafluoroethane (HFC-134a) for Temperatures from 170 K to 455 K and Pressures up to 70 MPa,” J. Phys. Chem, Ref. Data, Vol. 23, No. 5, 1994; M. J. Assael, N. K. Dalaouti, A. A. Griva, and J. H. Dymond, “Viscosity and Thermal Conductivity of Halogenated Methane and Ethane Refrigerants,” IJR, Vol. 22, pp. 525–535, 1999; NIST REFPROP 6 program (M. O. McLinden, S. A. Klein, E. W. Lemmon, and A. P. Peskin, Physical and Chemical Properties Division, National Institute of Standards and Technology, Boulder, CO 80303, 1995).
TABLE A–4 Properties of saturated refrigerant-134a Volume Enthalpy Specific Thermal Prandtl Expansion Surface Saturation Density of Heat Conductivity Dynamic Viscosity Number Coefficient Tension, Temp. Pressure 𝜌, kg/m3 Vaporization cp, J/kg·K k, W/m·K 𝜇, kg/m·s Pr 𝛽, 1/K N/m T, °C P, kPa Liquid Vapor hfg, kJ/kg Liquid Vapor Liquid Vapor Liquid Vapor Liquid Vapor Liquid Liquid −40 51.2 1418 2.773 225.9 1254 748.6 0.1101 0.00811 4.878 × 10−4 2.550 × 10−6 5.558 0.235 0.00205 0.01760 −35 66.2 1403 3.524 222.7 1264 764.1 0.1084 0.00862 4.509 × 10−4 3.003 × 10−6 5.257 0.266 0.00209 0.01682 −30 84.4 1389 4.429 219.5 1273 780.2 0.1066 0.00913 4.178 × 10−4 3.504 × 10−6 4.992 0.299 0.00215 0.01604 −25 106.5 1374 5.509 216.3 1283 797.2 0.1047 0.00963 3.882 × 10−4 4.054 × 10−6 4.757 0.335 0.00220 0.01527 −20 132.8 1359 6.787 213.0 1294 814.9 0.1028 0.01013 3.614 × 10−4 4.651 × 10−6 4.548 0.374 0.00227 0.01451 −15 164.0 1343 8.288 209.5 1306 833.5 0.1009 0.01063 3.371 × 10−4 5.295 × 10−6 4.363 0.415 0.00233 0.01376 −10 200.7 1327 10.04 206.0 1318 853.1 0.0989 0.01112 3.150 × 10−4 5.982 × 10−6 4.198 0.459 0.00241 0.01302 −5 243.5 1311 12.07 202.4 1330 873.8 0.0968 0.01161 2.947 × 10−4 6.709 × 10−6 4.051 0.505 0.00249 0.01229 0 293.0 1295 14.42 198.7 1344 895.6 0.0947 0.01210 2.761 × 10−4 7.471 × 10−6 3.919 0.553 0.00258 0.01156 5 349.9 1278 17.12 194.8 1358 918.7 0.0925 0.01259 2.589 × 10−4 8.264 × 10−6 3.802 0.603 0.00269 0.01084 10 414.9 1261 20.22 190.8 1374 943.2 0.0903 0.01308 2.430 × 10−4 9.081 × 10−6 3.697 0.655 0.00280 0.01014 15 488.7 1244 23.75 186.6 1390 969.4 0.0880 0.01357 2.281 × 10−4 9.915 × 10−6 3.604 0.708 0.00293 0.00944 20 572.1 1226 27.77 182.3 1408 997.6 0.0856 0.01406 2.142 × 10−4 1.075 × 10−5 3.521 0.763 0.00307 0.00876 25 665.8 1207 32.34 177.8 1427 1028 0.0833 0.01456 2.012 × 10−4 1.160 × 10−5 3.448 0.819 0.00324 0.00808 30 770.6 1188 37.53 173.1 1448 1061 0.0808 0.01507 1.888 × 10−4 1.244 × 10−5 3.383 0.877 0.00342 0.00742 35 887.5 1168 43.41 168.2 1471 1098 0.0783 0.01558 1.772 × 10−4 1.327 × 10−5 3.328 0.935 0.00364 0.00677 40 1017.1 1147 50.08 163.0 1498 1138 0.0757 0.01610 1.660 × 10−4 1.408 × 10−5 3.285 0.995 0.00390 0.00613 45 1160.5 1125 57.66 157.6 1529 1184 0.0731 0.01664 1.554 × 10−4 1.486 × 10−5 3.253 1.058 0.00420 0.00550 50 1318.6 1102 66.27 151.8 1566 1237 0.0704 0.01720 1.453 × 10−4 1.562 × 10−5 3.231 1.123 0.00456 0.00489 55 1492.3 1078 76.11 145.7 1608 1298 0.0676 0.01777 1.355 × 10−4 1.634 × 10−5 3.223 1.193 0.00500 0.00429 60 1682.8 1053 87.38 139.1 1659 1372 0.0647 0.01838 1.260 × 10−4 1.704 × 10−5 3.229 1.272 0.00554 0.00372 65 1891.0 1026 100.4 132.1 1722 1462 0.0618 0.01902 1.167 × 10−4 1.771 × 10−5 3.255 1.362 0.00624 0.00315 70 2118.2 996.2 115.6 124.4 1801 1577 0.0587 0.01972 1.077 × 10−4 1.839 × 10−5 3.307 1.471 0.00716 0.00261 75 2365.8 964 133.6 115.9 1907 1731 0.0555 0.02048 9.891 × 10−5 1.908 × 10−5 3.400 1.612 0.00843 0.00209 80 2635.2 928.2 155.3 106.4 2056 1948 0.0521 0.02133 9.011 × 10−5 1.982 × 10−5 3.558 1.810 0.01031 0.00160 85 2928.2 887.1 182.3 95.4 2287 2281 0.0484 0.02233 8.124 × 10−5 2.071 × 10−5 3.837 2.116 0.01336 0.00114 90 3246.9 837.7 217.8 82.2 2701 2865 0.0444 0.02357 7.203 × 10−5 2.187 × 10−5 4.385 2.658 0.01911 0.00071 95 3594.1 772.5 269.3 64.9 3675 4144 0.0396 0.02544 6.190 × 10−5 2.370 × 10−5 5.746 3.862 0.03343 0.00033 100 3975.1 651.7 376.3 33.9 7959 8785 0.0322 0.02989 4.765 × 10−5 2.833 × 10−5 11.77 8.326 0.10047 0.00004 cen96537_app1_947_964.indd 951 16/01/17 12:01 pm 952 PROPERTY TABLES AND CHARTS Note 1: Kinematic viscosity 𝜈 and thermal diffusivity 𝛼 can be calculated from their definitions, 𝜈 = 𝜇/𝜌 and 𝛼 = k/𝜌cp = 𝜈/Pr. The properties listed here (except the vapor density) can be used at any pressures with negligible error except at temperatures near the critical-point value.
Note 2: The unit kJ/kg·°C for specific heat is equivalent to kJ/kg·K, and the unit W/m·°C for thermal conductivity is equivalent to W/m·K.
Source: Data generated from the EES software developed by S. A. Klein and F. L. Alvarado. Original sources: Tillner-Roth, Harms-Watzenberg, and Baehr, “Eine neue Fundamentalgleichung fur Ammoniak,” DKV-Tagungsbericht 20:167–181, 1993; Liley and Desai, “Thermophysical Properties of Refrigerants,” ASHRAE, 1993, ISBN 1-1883413-10-9.
TABLE A–5 Properties of saturated ammonia Volume Enthalpy Specific Thermal Prandtl Expansion Surface Saturation Density of Heat Conductivity Dynamic Viscosity Number Coefficient Tension, Temp. Pressure 𝜌, kg/m3 Vaporization cp, J/kg·K k, W/m·K 𝜇, kg/m·s Pr 𝛽, 1/K N/m T, °C P, kPa Liquid Vapor hfg, kJ/kg Liquid Vapor Liquid Vapor Liquid Vapor Liquid Vapor Liquid Liquid −40 71.66 690.2 0.6435 1389 4414 2242 — 0.01792 2.926 × 10−4 7.957 × 10−6 — 0.9955 0.00176 0.03565 −30 119.4 677.8 1.037 1360 4465 2322 — 0.01898 2.630 × 10−4 8.311 × 10−6 — 1.017 0.00185 0.03341 −25 151.5 671.5 1.296 1345 4489 2369 0.5968 0.01957 2.492 × 10−4 8.490 × 10−6 1.875 1.028 0.00190 0.03229 −20 190.1 665.1 1.603 1329 4514 2420 0.5853 0.02015 2.361 × 10−4 8.669 × 10−6 1.821 1.041 0.00194 0.03118 −15 236.2 658.6 1.966 1313 4538 2476 0.5737 0.02075 2.236 × 10−4 8.851 × 10−6 1.769 1.056 0.00199 0.03007 −10 290.8 652.1 2.391 1297 4564 2536 0.5621 0.02138 2.117 × 10−4 9.034 × 10−6 1.718 1.072 0.00205 0.02896 −5 354.9 645.4 2.886 1280 4589 2601 0.5505 0.02203 2.003 × 10−4 9.218 × 10−6 1.670 1.089 0.00210 0.02786 0 429.6 638.6 3.458 1262 4617 2672 0.5390 0.02270 1.896 × 10−4 9.405 × 10−6 1.624 1.107 0.00216 0.02676 5 516 631.7 4.116 1244 4645 2749 0.5274 0.02341 1.794 × 10−4 9.593 × 10−6 1.580 1.126 0.00223 0.02566 10 615.3 624.6 4.870 1226 4676 2831 0.5158 0.02415 1.697 × 10−4 9.784 × 10−6 1.539 1.147 0.00230 0.02457 15 728.8 617.5 5.729 1206 4709 2920 0.5042 0.02492 1.606 × 10−4 9.978 × 10−6 1.500 1.169 0.00237 0.02348 20 857.8 610.2 6.705 1186 4745 3016 0.4927 0.02573 1.519 × 10−4 1.017 × 10−5 1.463 1.193 0.00245 0.02240 25 1003 602.8 7.809 1166 4784 3120 0.4811 0.02658 1.438 × 10−4 1.037 × 10−5 1.430 1.218 0.00254 0.02132 30 1167 595.2 9.055 1144 4828 3232 0.4695 0.02748 1.361 × 10−4 1.057 × 10−5 1.399 1.244 0.00264 0.02024 35 1351 587.4 10.46 1122 4877 3354 0.4579 0.02843 1.288 × 10−4 1.078 × 10−5 1.372 1.272 0.00275 0.01917 40 1555 579.4 12.03 1099 4932 3486 0.4464 0.02943 1.219 × 10−4 1.099 × 10−5 1.347 1.303 0.00287 0.01810 45 1782 571.3 13.8 1075 4993 3631 0.4348 0.03049 1.155 × 10−4 1.121 × 10−5 1.327 1.335 0.00301 0.01704 50 2033 562.9 15.78 1051 5063 3790 0.4232 0.03162 1.094 × 10−4 1.143 × 10−5 1.310 1.371 0.00316 0.01598 55 2310 554.2 18.00 1025 5143 3967 0.4116 0.03283 1.037 × 10−4 1.166 × 10−5 1.297 1.409 0.00334 0.01493 60 2614 545.2 20.48 997.4 5234 4163 0.4001 0.03412 9.846 × 10−5 1.189 × 10−5 1.288 1.452 0.00354 0.01389 65 2948 536.0 23.26 968.9 5340 4384 0.3885 0.03550 9.347 × 10−5 1.213 × 10−5 1.285 1.499 0.00377 0.01285 70 3312 526.3 26.39 939.0 5463 4634 0.3769 0.03700 8.879 × 10−5 1.238 × 10−5 1.287 1.551 0.00404 0.01181 75 3709 516.2 29.90 907.5 5608 4923 0.3653 0.03862 8.440 × 10−5 1.264 × 10−5 1.296 1.612 0.00436 0.01079 80 4141 505.7 33.87 874.1 5780 5260 0.3538 0.04038 8.030 × 10−5 1.292 × 10−5 1.312 1.683 0.00474 0.00977 85 4609 494.5 38.36 838.6 5988 5659 0.3422 0.04232 7.645 × 10−5 1.322 × 10−5 1.338 1.768 0.00521 0.00876 90 5116 482.8 43.48 800.6 6242 6142 0.3306 0.04447 7.284 × 10−5 1.354 × 10−5 1.375 1.871 0.00579 0.00776 95 5665 470.2 49.35 759.8 6561 6740 0.3190 0.04687 6.946 × 10−5 1.389 × 10−5 1.429 1.999 0.00652 0.00677 100 6257 456.6 56.15 715.5 6972 7503 0.3075 0.04958 6.628 × 10−5 1.429 × 10−5 1.503 2.163 0.00749 0.00579 cen96537_app1_947_964.indd 952 16/01/17 12:01 pm 953 APPENDIX 1 TABLE A–6 Properties of saturated propane Volume Enthalpy Specific Thermal Prandtl Expansion Surface Saturation Density of Heat Conductivity Dynamic Viscosity Number Coefficient Tension, Temp. Pressure 𝜌, kg/m3 Vaporization cp, J/kg·K k, W/m·K 𝜇, kg/m·s Pr 𝛽, 1/K N/m T, °C P, kPa Liquid Vapor hfg, kJ/kg Liquid Vapor Liquid Vapor Liquid Vapor Liquid Vapor Liquid Liquid −120 0.4053 664.7 0.01408 498.3 2003 1115 0.1802 0.00589 6.136 × 10−4 4.372 × 10−6 6.820 0.827 0.00153 0.02630 −110 1.157 654.5 0.03776 489.3 2021 1148 0.1738 0.00645 5.054 × 10−4 4.625 × 10−6 5.878 0.822 0.00157 0.02486 −100 2.881 644.2 0.08872 480.4 2044 1183 0.1672 0.00705 4.252 × 10−4 4.881 × 10−6 5.195 0.819 0.00161 0.02344 −90 6.406 633.8 0.1870 471.5 2070 1221 0.1606 0.00769 3.635 × 10−4 5.143 × 10−6 4.686 0.817 0.00166 0.02202 −80 12.97 623.2 0.3602 462.4 2100 1263 0.1539 0.00836 3.149 × 10−4 5.409 × 10−6 4.297 0.817 0.00171 0.02062 −70 24.26 612.5 0.6439 453.1 2134 1308 0.1472 0.00908 2.755 × 10−4 5.680 × 10−6 3.994 0.818 0.00177 0.01923 −60 42.46 601.5 1.081 443.5 2173 1358 0.1407 0.00985 2.430 × 10−4 5.956 × 10−6 3.755 0.821 0.00184 0.01785 −50 70.24 590.3 1.724 433.6 2217 1412 0.1343 0.01067 2.158 × 10−4 6.239 × 10−6 3.563 0.825 0.00192 0.01649 −40 110.7 578.8 2.629 423.1 2258 1471 0.1281 0.01155 1.926 × 10−4 6.529 × 10−6 3.395 0.831 0.00201 0.01515 −30 167.3 567.0 3.864 412.1 2310 1535 0.1221 0.01250 1.726 × 10−4 6.827 × 10−6 3.266 0.839 0.00213 0.01382 −20 243.8 554.7 5.503 400.3 2368 1605 0.1163 0.01351 1.551 × 10−4 7.136 × 10−6 3.158 0.848 0.00226 0.01251 −10 344.4 542.0 7.635 387.8 2433 1682 0.1107 0.01459 1.397 × 10−4 7.457 × 10−6 3.069 0.860 0.00242 0.01122 0 473.3 528.7 10.36 374.2 2507 1768 0.1054 0.01576 1.259 × 10−4 7.794 × 10−6 2.996 0.875 0.00262 0.00996 5 549.8 521.8 11.99 367.0 2547 1814 0.1028 0.01637 1.195 × 10−4 7.970 × 10−6 2.964 0.883 0.00273 0.00934 10 635.1 514.7 13.81 359.5 2590 1864 0.1002 0.01701 1.135 × 10−4 8.151 × 10−6 2.935 0.893 0.00286 0.00872 15 729.8 507.5 15.85 351.7 2637 1917 0.0977 0.01767 1.077 × 10−4 8.339 × 10−6 2.909 0.905 0.00301 0.00811 20 834.4 500.0 18.13 343.4 2688 1974 0.0952 0.01836 1.022 × 10−4 8.534 × 10−6 2.886 0.918 0.00318 0.00751 25 949.7 492.2 20.68 334.8 2742 2036 0.0928 0.01908 9.702 × 10−5 8.738 × 10−6 2.866 0.933 0.00337 0.00691 30 1076 484.2 23.53 325.8 2802 2104 0.0904 0.01982 9.197 × 10−5 8.952 × 10−6 2.850 0.950 0.00358 0.00633 35 1215 475.8 26.72 316.2 2869 2179 0.0881 0.02061 8.710 × 10−5 9.178 × 10−6 2.837 0.971 0.00384 0.00575 40 1366 467.1 30.29 306.1 2943 2264 0.0857 0.02142 8.240 × 10−5 9.417 × 10−6 2.828 0.995 0.00413 0.00518 45 1530 458.0 34.29 295.3 3026 2361 0.0834 0.02228 7.785 × 10−5 9.674 × 10−6 2.824 1.025 0.00448 0.00463 50 1708 448.5 38.79 283.9 3122 2473 0.0811 0.02319 7.343 × 10−5 9.950 × 10−6 2.826 1.061 0.00491 0.00408 60 2110 427.5 49.66 258.4 3283 2769 0.0765 0.02517 6.487 × 10−5 1.058 × 10−5 2.784 1.164 0.00609 0.00303 70 2580 403.2 64.02 228.0 3595 3241 0.0717 0.02746 5.649 × 10−5 1.138 × 10−5 2.834 1.343 0.00811 0.00204 80 3127 373.0 84.28 189.7 4501 4173 0.0663 0.03029 4.790 × 10−5 1.249 × 10−5 3.251 1.722 0.01248 0.00114 90 3769 329.1 118.6 133.2 6977 7239 0.0595 0.03441 3.807 × 10−5 1.448 × 10−5 4.465 3.047 0.02847 0.00037 Note 1: Kinematic viscosity 𝜈 and thermal diffusivity 𝛼 can be calculated from their definitions, 𝜈 = 𝜇/𝜌 and 𝛼 = k/𝜌cp = 𝜈/Pr. The properties listed here (except the vapor density) can be used at any pressures with negligible error except at temperatures near the critical-point value.
Note 2: The unit kJ/kg·°C for specific heat is equivalent to kJ/kg·K, and the unit W/m·°C for thermal conductivity is equivalent to W/m·K.
Source: Data generated from the EES software developed by S. A. Klein and F. L. Alvarado. Original sources: Reiner Tillner-Roth, “Fundamental Equations of State,” Shaker, Verlag, Aachan, 1998; B. A. Younglove and J. F. Ely, “Thermophysical Properties of Fluids. II Methane, Ethane, Propane, Isobutane, and Normal Butane,” J. Phys. Chem. Ref. Data, Vol. 16, No. 4, 1987; G.R. Somayajulu, “A Generalized Equation for Surface Tension from the Triple-Point to the Critical-Point,” International Journal of Thermophysics, Vol. 9, No. 4, 1988.
cen96537_app1_947_964.indd 953 16/01/17 12:01 pm 954 PROPERTY TABLES AND CHARTS TABLE A–7 Properties of liquids Volume Specific Thermal Thermal Dynamic Kinematic Prandtl Expansion Temp. Density Heat cp, Conductivity Diffusivity Viscosity Viscosity Number Coeff. T, °C 𝜌, kg/m3 J/kg·K k, W/m·K 𝛼, m2/s 𝜇, kg/m·s 𝜈, m2/s Pr 𝛽, 1/K Methane (CH4 ) −160 420.2 3492 0.1863 1.270 × 10−7 1.133 × 10−4 2.699 × 10−7 2.126 0.00352 −150 405.0 3580 0.1703 1.174 × 10−7 9.169 × 10−5 2.264 × 10−7 1.927 0.00391 −140 388.8 3700 0.1550 1.077 × 10−7 7.551 × 10−5 1.942 × 10−7 1.803 0.00444 −130 371.1 3875 0.1402 9.749 × 10−8 6.288 × 10−5 1.694 × 10−7 1.738 0.00520 −120 351.4 4146 0.1258 8.634 × 10−8 5.257 × 10−5 1.496 × 10−7 1.732 0.00637 −110 328.8 4611 0.1115 7.356 × 10−8 4.377 × 10−5 1.331 × 10−7 1.810 0.00841 −100 301.0 5578 0.0967 5.761 × 10−8 3.577 × 10−5 1.188 × 10−7 2.063 0.01282 −90 261.7 8902 0.0797 3.423 × 10−8 2.761 × 10−5 1.055 × 10−7 3.082 0.02922 Methanol [CH3 (OH)] 20 788.4 2515 0.1987 1.002 × 10−7 5.857 × 10−4 7.429 × 10−7 7.414 0.00118 30 779.1 2577 0.1980 9.862 × 10−8 5.088 × 10−4 6.531 × 10−7 6.622 0.00120 40 769.6 2644 0.1972 9.690 × 10−8 4.460 × 10−4 5.795 × 10−7 5.980 0.00123 50 760.1 2718 0.1965 9.509 × 10−8 3.942 × 10−4 5.185 × 10−7 5.453 0.00127 60 750.4 2798 0.1957 9.320 × 10−8 3.510 × 10−4 4.677 × 10−7 5.018 0.00132 70 740.4 2885 0.1950 9.128 × 10−8 3.146 × 10−4 4.250 × 10−7 4.655 0.00137 Isobutane (R600a) −100 683.8 1881 0.1383 1.075 × 10−7 9.305 × 10−4 1.360 × 10−6 12.65 0.00142 −75 659.3 1970 0.1357 1.044 × 10−7 5.624 × 10−4 8.531 × 10−7 8.167 0.00150 −50 634.3 2069 0.1283 9.773 × 10−8 3.769 × 10−4 5.942 × 10−7 6.079 0.00161 −25 608.2 2180 0.1181 8.906 × 10−8 2.688 × 10−4 4.420 × 10−7 4.963 0.00177 0 580.6 2306 0.1068 7.974 × 10−8 1.993 × 10−4 3.432 × 10−7 4.304 0.00199 25 550.7 2455 0.0956 7.069 × 10−8 1.510 × 10−4 2.743 × 10−7 3.880 0.00232 50 517.3 2640 0.0851 6.233 × 10−8 1.155 × 10−4 2.233 × 10−7 3.582 0.00286 75 478.5 2896 0.0757 5.460 × 10−8 8.785 × 10−5 1.836 × 10−7 3.363 0.00385 100 429.6 3361 0.0669 4.634 × 10−8 6.483 × 10−5 1.509 × 10−7 3.256 0.00628 Glycerin 0 1276 2262 0.2820 9.773 × 10−8 10.49 8.219 × 10−3 84,101 5 1273 2288 0.2835 9.732 × 10−8 6.730 5.287 × 10−3 54,327 10 1270 2320 0.2846 9.662 × 10−8 4.241 3.339 × 10−3 34,561 15 1267 2354 0.2856 9.576 × 10−8 2.496 1.970 × 10−3 20,570 20 1264 2386 0.2860 9.484 × 10−8 1.519 1.201 × 10−3 12,671 25 1261 2416 0.2860 9.388 × 10−8 0.9934 7.878 × 10−4 8,392 30 1258 2447 0.2860 9.291 × 10−8 0.6582 5.232 × 10−4 5,631 35 1255 2478 0.2860 9.195 × 10−8 0.4347 3.464 × 10−4 3,767 40 1252 2513 0.2863 9.101 × 10−8 0.3073 2.455 × 10−4 2,697 Engine Oil (unused) 0 899.0 1797 0.1469 9.097 × 10−8 3.814 4.242 × 10−3 46,636 0.00070 20 888.1 1881 0.1450 8.680 × 10−8 0.8374 9.429 × 10−4 10,863 0.00070 40 876.0 1964 0.1444 8.391 × 10−8 0.2177 2.485 × 10−4 2,962 0.00070 60 863.9 2048 0.1404 7.934 × 10−8 0.07399 8.565 × 10−5 1,080 0.00070 80 852.0 2132 0.1380 7.599 × 10−8 0.03232 3.794 × 10−5 499.3 0.00070 100 840.0 2220 0.1367 7.330 × 10−8 0.01718 2.046 × 10−5 279.1 0.00070 120 828.9 2308 0.1347 7.042 × 10−8 0.01029 1.241 × 10−5 176.3 0.00070 140 816.8 2395 0.1330 6.798 × 10−8 0.006558 8.029 × 10−6 118.1 0.00070 150 810.3 2441 0.1327 6.708 × 10−8 0.005344 6.595 × 10−6 98.31 0.00070 Source: Data generated from the EES software developed by S. A. Klein and F. L. Alvarado. Originally based on various sources.
cen96537_app1_947_964.indd 954 16/01/17 12:01 pm 955 APPENDIX 1 TABLE A–8 Properties of liquid metals Volume Specific Thermal Thermal Dynamic Kinematic Prandtl Expansion Temp. Density Heat cp, Conductivity Diffusivity Viscosity Viscosity Number Coeff.
T, °C 𝜌, kg/m3 J/kg·K k, W/m·K 𝛼, m2/s 𝜇, kg/m·s 𝜈, m2/s Pr 𝛽, 1/K Mercury (Hg) Melting Point: −39°C 0 13595 140.4 8.18200 4.287 × 10−6 1.687 × 10−3 1.241 × 10−7 0.0289 1.810 × 10−4 25 13534 139.4 8.51533 4.514 × 10−6 1.534 × 10−3 1.133 × 10−7 0.0251 1.810 × 10−4 50 13473 138.6 8.83632 4.734 × 10−6 1.423 × 10−3 1.056 × 10−7 0.0223 1.810 × 10−4 75 13412 137.8 9.15632 4.956 × 10−6 1.316 × 10−3 9.819 × 10−8 0.0198 1.810 × 10−4 100 13351 137.1 9.46706 5.170 × 10−6 1.245 × 10−3 9.326 × 10−8 0.0180 1.810 × 10−4 150 13231 136.1 10.07780 5.595 × 10−6 1.126 × 10−3 8.514 × 10−8 0.0152 1.810 × 10−4 200 13112 135.5 10.65465 5.996 × 10−6 1.043 × 10−3 7.959 × 10−8 0.0133 1.815 × 10−4 250 12993 135.3 11.18150 6.363 × 10−6 9.820 × 10−4 7.558 × 10−8 0.0119 1.829 × 10−4 300 12873 135.3 11.68150 6.705 × 10−6 9.336 × 10−4 7.252 × 10−8 0.0108 1.854 × 10−4 Bismuth (Bi) Melting Point: 271°C 350 9969 146.0 16.28 1.118 × 10−5 1.540 × 10−3 1.545 × 10−7 0.01381 400 9908 148.2 16.10 1.096 × 10−5 1.422 × 10−3 1.436 × 10−7 0.01310 500 9785 152.8 15.74 1.052 × 10−5 1.188 × 10−3 1.215 × 10−7 0.01154 600 9663 157.3 15.60 1.026 × 10−5 1.013 × 10−3 1.048 × 10−7 0.01022 700 9540 161.8 15.60 1.010 × 10−5 8.736 × 10−4 9.157 × 10−8 0.00906 Lead (Pb) Melting Point: 327 °C 400 10506 158 15.97 9.623 × 10−6 2.277 × 10−3 2.167 × 10−7 0.02252 450 10449 156 15.74 9.649 × 10−6 2.065 × 10−3 1.976 × 10−7 0.02048 500 10390 155 15.54 9.651 × 10−6 1.884 × 10−3 1.814 × 10−7 0.01879 550 10329 155 15.39 9.610 × 10−6 1.758 × 10−3 1.702 × 10−7 0.01771 600 10267 155 15.23 9.568 × 10−6 1.632 × 10−3 1.589 × 10−7 0.01661 650 10206 155 15.07 9.526 × 10−6 1.505 × 10−3 1.475 × 10−7 0.01549 700 10145 155 14.91 9.483 × 10−6 1.379 × 10−3 1.360 × 10−7 0.01434 Sodium (Na) Melting Point: 98 °C 100 927.3 1378 85.84 6.718 × 10−5 6.892 × 10−4 7.432 × 10−7 0.01106 200 902.5 1349 80.84 6.639 × 10−5 5.385 × 10−4 5.967 × 10−7 0.008987 300 877.8 1320 75.84 6.544 × 10−5 3.878 × 10−4 4.418 × 10−7 0.006751 400 853.0 1296 71.20 6.437 × 10−5 2.720 × 10−4 3.188 × 10−7 0.004953 500 828.5 1284 67.41 6.335 × 10−5 2.411 × 10−4 2.909 × 10−7 0.004593 600 804.0 1272 63.63 6.220 × 10−5 2.101 × 10−4 2.614 × 10−7 0.004202 Potassium (K) Melting Point: 64 °C 200 795.2 790.8 43.99 6.995 × 10−5 3.350 × 10−4 4.213 × 10−7 0.006023 300 771.6 772.8 42.01 7.045 × 10−5 2.667 × 10−4 3.456 × 10−7 0.004906 400 748.0 754.8 40.03 7.090 × 10−5 1.984 × 10−4 2.652 × 10−7 0.00374 500 723.9 750.0 37.81 6.964 × 10−5 1.668 × 10−4 2.304 × 10−7 0.003309 600 699.6 750.0 35.50 6.765 × 10−5 1.487 × 10−4 2.126 × 10−7 0.003143 Sodium–Potassium (%22Na-%78K) Melting Point: −11°C 100 847.3 944.4 25.64 3.205 × 10−5 5.707 × 10−4 6.736 × 10−7 0.02102 200 823.2 922.5 26.27 3.459 × 10−5 4.587 × 10−4 5.572 × 10−7 0.01611 300 799.1 900.6 26.89 3.736 × 10−5 3.467 × 10−4 4.339 × 10−7 0.01161 400 775.0 879.0 27.50 4.037 × 10−5 2.357 × 10−4 3.041 × 10−7 0.00753 500 751.5 880.1 27.89 4.217 × 10−5 2.108 × 10−4 2.805 × 10−7 0.00665 600 728.0 881.2 28.28 4.408 × 10−5 1.859 × 10−4 2.553 × 10−7 0.00579 Source: Data generated from the EES software developed by S. A. Klein and F. L. Alvarado. Originally based on various sources.
cen96537_app1_947_964.indd 955 16/01/17 12:01 pm 956 PROPERTY TABLES AND CHARTS TABLE A–9 Properties of air at 1 atm pressure Specific Thermal Thermal Dynamic Kinematic Prandtl Temp. Density Heat cp , Conductivity Diffusivity Viscosity Viscosity Number T, °C 𝜌, kg/m3 J/kg·K k, W/m·K 𝛼, m2/s 𝜇, kg/m·s 𝜈, m2/s Pr −150 2.866 983 0.01171 4.158 × 10−6 8.636 × 10−6 3.013 × 10−6 0.7246 −100 2.038 966 0.01582 8.036 × 10−6 1.189 × 10−6 5.837 × 10−6 0.7263 −50 1.582 999 0.01979 1.252 × 10−5 1.474 × 10−5 9.319 × 10−6 0.7440 −40 1.514 1002 0.02057 1.356 × 10−5 1.527 × 10−5 1.008 × 10−5 0.7436 −30 1.451 1004 0.02134 1.465 × 10−5 1.579 × 10−5 1.087 × 10−5 0.7425 −20 1.394 1005 0.02211 1.578 × 10−5 1.630 × 10−5 1.169 × 10−5 0.7408 −10 1.341 1006 0.02288 1.696 × 10−5 1.680 × 10−5 1.252 × 10−5 0.7387 0 1.292 1006 0.02364 1.818 × 10−5 1.729 × 10−5 1.338 × 10−5 0.7362 5 1.269 1006 0.02401 1.880 × 10−5 1.754 × 10−5 1.382 × 10−5 0.7350 10 1.246 1006 0.02439 1.944 × 10−5 1.778 × 10−5 1.426 × 10−5 0.7336 15 1.225 1007 0.02476 2.009 × 10−5 1.802 × 10−5 1.470 × 10−5 0.7323 20 1.204 1007 0.02514 2.074 × 10−5 1.825 × 10−5 1.516 × 10−5 0.7309 25 1.184 1007 0.02551 2.141 × 10−5 1.849 × 10−5 1.562 × 10−5 0.7296 30 1.164 1007 0.02588 2.208 × 10−5 1.872 × 10−5 1.608 × 10−5 0.7282 35 1.145 1007 0.02625 2.277 × 10−5 1.895 × 10−5 1.655 × 10−5 0.7268 40 1.127 1007 0.02662 2.346 × 10−5 1.918 × 10−5 1.702 × 10−5 0.7255 45 1.109 1007 0.02699 2.416 × 10−5 1.941 × 10−5 1.750 × 10−5 0.7241 50 1.092 1007 0.02735 2.487 × 10−5 1.963 × 10−5 1.798 × 10−5 0.7228 60 1.059 1007 0.02808 2.632 × 10−5 2.008 × 10−5 1.896 × 10−5 0.7202 70 1.028 1007 0.02881 2.780 × 10−5 2.052 × 10−5 1.995 × 10−5 0.7177 80 0.9994 1008 0.02953 2.931 × 10−5 2.096 × 10−5 2.097 × 10−5 0.7154 90 0.9718 1008 0.03024 3.086 × 10−5 2.139 × 10−5 2.201 × 10−5 0.7132 100 0.9458 1009 0.03095 3.243 × 10−5 2.181 × 10−5 2.306 × 10−5 0.7111 120 0.8977 1011 0.03235 3.565 × 10−5 2.264 × 10−5 2.522 × 10−5 0.7073 140 0.8542 1013 0.03374 3.898 × 10−5 2.345 × 10−5 2.745 × 10−5 0.7041 160 0.8148 1016 0.03511 4.241 × 10−5 2.420 × 10−5 2.975 × 10−5 0.7014 180 0.7788 1019 0.03646 4.593 × 10−5 2.504 × 10−5 3.212 × 10−5 0.6992 200 0.7459 1023 0.03779 4.954 × 10−5 2.577 × 10−5 3.455 × 10−5 0.6974 250 0.6746 1033 0.04104 5.890 × 10−5 2.760 × 10−5 4.091 × 10−5 0.6946 300 0.6158 1044 0.04418 6.871 × 10−5 2.934 × 10−5 4.765 × 10−5 0.6935 350 0.5664 1056 0.04721 7.892 × 10−5 3.101 × 10−5 5.475 × 10−5 0.6937 400 0.5243 1069 0.05015 8.951 × 10−5 3.261 × 10−5 6.219 × 10−5 0.6948 450 0.4880 1081 0.05298 1.004 × 10−4 3.415 × 10−5 6.997 × 10−5 0.6965 500 0.4565 1093 0.05572 1.117 × 10−4 3.563 × 10−5 7.806 × 10−5 0.6986 600 0.4042 1115 0.06093 1.352 × 10−4 3.846 × 10−5 9.515 × 10−5 0.7037 700 0.3627 1135 0.06581 1.598 × 10−4 4.111 × 10−5 1.133 × 10−4 0.7092 800 0.3289 1153 0.07037 1.855 × 10−4 4.362 × 10−5 1.326 × 10−4 0.7149 900 0.3008 1169 0.07465 2.122 × 10−4 4.600 × 10−5 1.529 × 10−4 0.7206 1000 0.2772 1184 0.07868 2.398 × 10−4 4.826 × 10−5 1.741 × 10−4 0.7260 1500 0.1990 1234 0.09599 3.908 × 10−4 5.817 × 10−5 2.922 × 10−4 0.7478 2000 0.1553 1264 0.11113 5.664 × 10−4 6.630 × 10−5 4.270 × 10−4 0.7539 Note: For ideal gases, the properties cp, k, 𝜇, and Pr are independent of pressure. The properties 𝜌, 𝜈, and 𝛼 at a pressure P (in atm) other than 1 atm are determined by multiplying the values of 𝜌 at the given temperature by P and by dividing 𝜈 and 𝛼 by P.
Source: Data generated from the EES software developed by S. A. Klein and F. L. Alvarado. Original sources: Keenan, Chao, Kaye, Gas Tables, Wiley, 1983; and Thermo physical Properties of Matter, Vol. 3: Thermal Conductivity, Y. S. Touloukian, P. E. Liley, S. C. Saxena, Vol. 11: Viscosity, Y. S. Touloukian, S. C. Saxena, and P. Hestermans, IFI/Plenun, NY, 1970, ISBN 0-306067020-8.
cen96537_app1_947_964.indd 956 16/01/17 12:01 pm 957 APPENDIX 1 TABLE A–10 Properties of gases at 1 atm pressure Specific Thermal Thermal Dynamic Kinematic Prandtl Temp. Density Heat cp, Conductivity Diffusivity Viscosity Viscosity Number T, °C 𝜌, kg/m3 J/kg·K k, W/m·K 𝛼, m2/s 𝜇, kg/m·s 𝜈, m2/s Pr Carbon Dioxide, CO2 −50 2.4035 746 0.01051 5.860 × 10−6 1.129 × 10−5 4.699 × 10−6 0.8019 0 1.9635 811 0.01456 9.141 × 10−6 1.375 × 10−5 7.003 × 10−6 0.7661 50 1.6597 866.6 0.01858 1.291 × 10−5 1.612 × 10−5 9.714 × 10−6 0.7520 100 1.4373 914.8 0.02257 1.716 × 10−5 1.841 × 10−5 1.281 × 10−5 0.7464 150 1.2675 957.4 0.02652 2.186 × 10−5 2.063 × 10−5 1.627 × 10−5 0.7445 200 1.1336 995.2 0.03044 2.698 × 10−5 2.276 × 10−5 2.008 × 10−5 0.7442 300 0.9358 1060 0.03814 3.847 × 10−5 2.682 × 10−5 2.866 × 10−5 0.7450 400 0.7968 1112 0.04565 5.151 × 10−5 3.061 × 10−5 3.842 × 10−5 0.7458 500 0.6937 1156 0.05293 6.600 × 10−5 3.416 × 10−5 4.924 × 10−5 0.7460 1000 0.4213 1292 0.08491 1.560 × 10−4 4.898 × 10−5 1.162 × 10−4 0.7455 1500 0.3025 1356 0.10688 2.606 × 10−4 6.106 × 10−5 2.019 × 10−4 0.7745 2000 0.2359 1387 0.11522 3.521 × 10−4 7.322 × 10−5 3.103 × 10−4 0.8815 Carbon Monoxide, CO −50 1.5297 1081 0.01901 1.149 × 10−5 1.378 × 10−5 9.012 × 10−6 0.7840 0 1.2497 1048 0.02278 1.739 × 10−5 1.629 × 10−5 1.303 × 10−5 0.7499 50 1.0563 1039 0.02641 2.407 × 10−5 1.863 × 10−5 1.764 × 10−5 0.7328 100 0.9148 1041 0.02992 3.142 × 10−5 2.080 × 10−5 2.274 × 10−5 0.7239 150 0.8067 1049 0.03330 3.936 × 10−5 2.283 × 10−5 2.830 × 10−5 0.7191 200 0.7214 1060 0.03656 4.782 × 10−5 2.472 × 10−5 3.426 × 10−5 0.7164 300 0.5956 1085 0.04277 6.619 × 10−5 2.812 × 10−5 4.722 × 10−5 0.7134 400 0.5071 1111 0.04860 8.628 × 10−5 3.111 × 10−5 6.136 × 10−5 0.7111 500 0.4415 1135 0.05412 1.079 × 10−4 3.379 × 10−5 7.653 × 10−5 0.7087 1000 0.2681 1226 0.07894 2.401 × 10−4 4.557 × 10−5 1.700 × 10−4 0.7080 1500 0.1925 1279 0.10458 4.246 × 10−4 6.321 × 10−5 3.284 × 10−4 0.7733 2000 0.1502 1309 0.13833 7.034 × 10−4 9.826 × 10−5 6.543 × 10−4 0.9302 Methane, CH4 −50 0.8761 2243 0.02367 1.204 × 10−5 8.564 × 10−6 9.774 × 10−6 0.8116 0 0.7158 2217 0.03042 1.917 × 10−5 1.028 × 10−5 1.436 × 10−5 0.7494 50 0.6050 2302 0.03766 2.704 × 10−5 1.191 × 10−5 1.969 × 10−5 0.7282 100 0.5240 2443 0.04534 3.543 × 10−5 1.345 × 10−5 2.567 × 10−5 0.7247 150 0.4620 2611 0.05344 4.431 × 10−5 1.491 × 10−5 3.227 × 10−5 0.7284 200 0.4132 2791 0.06194 5.370 × 10−5 1.630 × 10−5 3.944 × 10−5 0.7344 300 0.3411 3158 0.07996 7.422 × 10−5 1.886 × 10−5 5.529 × 10−5 0.7450 400 0.2904 3510 0.09918 9.727 × 10−5 2.119 × 10−5 7.297 × 10−5 0.7501 500 0.2529 3836 0.11933 1.230 × 10−4 2.334 × 10−5 9.228 × 10−5 0.7502 1000 0.1536 5042 0.22562 2.914 × 10−4 3.281 × 10−5 2.136 × 10−4 0.7331 1500 0.1103 5701 0.31857 5.068 × 10−4 4.434 × 10−5 4.022 × 10−4 0.7936 2000 0.0860 6001 0.36750 7.120 × 10−4 6.360 × 10−5 7.395 × 10−4 1.0386 Hydrogen, H2 −50 0.11010 12635 0.1404 1.009 × 10−4 7.293 × 10−6 6.624 × 10−5 0.6562 0 0.08995 13920 0.1652 1.319 × 10−4 8.391 × 10−6 9.329 × 10−5 0.7071 50 0.07603 14349 0.1881 1.724 × 10−4 9.427 × 10−6 1.240 × 10−4 0.7191 100 0.06584 14473 0.2095 2.199 × 10−4 1.041 × 10−5 1.582 × 10−4 0.7196 150 0.05806 14492 0.2296 2.729 × 10−4 1.136 × 10−5 1.957 × 10−4 0.7174 200 0.05193 14482 0.2486 3.306 × 10−4 1.228 × 10−5 2.365 × 10−4 0.7155 300 0.04287 14481 0.2843 4.580 × 10−4 1.403 × 10−5 3.274 × 10−4 0.7149 400 0.03650 14540 0.3180 5.992 × 10−4 1.570 × 10−5 4.302 × 10−4 0.7179 500 0.03178 14653 0.3509 7.535 × 10−4 1.730 × 10−5 5.443 × 10−4 0.7224 1000 0.01930 15577 0.5206 1.732 × 10−3 2.455 × 10−5 1.272 × 10−3 0.7345 1500 0.01386 16553 0.6581 2.869 × 10−3 3.099 × 10−5 2.237 × 10−3 0.7795 2000 0.01081 17400 0.5480 2.914 × 10−3 3.690 × 10−5 3.414 × 10−3 1.1717 (continued) cen96537_app1_947_964.indd 957 16/01/17 12:01 pm 958 PROPERTY TABLES AND CHARTS TABLE A–10 (Continued) Properties of gases at 1 atm pressure Specific Thermal Thermal Dynamic Kinematic Prandtl Temp. Density Heat cp, Conductivity Diffusivity Viscosity Viscosity Number T, °C 𝜌, kg/m3 J/kg·K k, W/m·K 𝛼, m2/s 𝜇, kg/m·s 𝜈, m2/s Pr Nitrogen, N2 −50 1.5299 957.3 0.02001 1.366 × 10−5 1.390 × 10−5 9.091 × 10−6 0.6655 0 1.2498 1035 0.02384 1.843 × 10−5 1.640 × 10−5 1.312 × 10−5 0.7121 50 1.0564 1042 0.02746 2.494 × 10−5 1.874 × 10−5 1.774 × 10−5 0.7114 100 0.9149 1041 0.03090 3.244 × 10−5 2.094 × 10−5 2.289 × 10−5 0.7056 150 0.8068 1043 0.03416 4.058 × 10−5 2.300 × 10−5 2.851 × 10−5 0.7025 200 0.7215 1050 0.03727 4.921 × 10−5 2.494 × 10−5 3.457 × 10−5 0.7025 300 0.5956 1070 0.04309 6.758 × 10−5 2.849 × 10−5 4.783 × 10−5 0.7078 400 0.5072 1095 0.04848 8.727 × 10−5 3.166 × 10−5 6.242 × 10−5 0.7153 500 0.4416 1120 0.05358 1.083 × 10−4 3.451 × 10−5 7.816 × 10−5 0.7215 1000 0.2681 1213 0.07938 2.440 × 10−4 4.594 × 10−5 1.713 × 10−4 0.7022 1500 0.1925 1266 0.11793 4.839 × 10−4 5.562 × 10−5 2.889 × 10−4 0.5969 2000 0.1502 1297 0.18590 9.543 × 10−4 6.426 × 10−5 4.278 × 10−4 0.4483 Oxygen, O2 −50 1.7475 984.4 0.02067 1.201 × 10−5 1.616 × 10−5 9.246 × 10−6 0.7694 0 1.4277 928.7 0.02472 1.865 × 10−5 1.916 × 10−5 1.342 × 10−5 0.7198 50 1.2068 921.7 0.02867 2.577 × 10−5 2.194 × 10−5 1.818 × 10−5 0.7053 100 1.0451 931.8 0.03254 3.342 × 10−5 2.451 × 10−5 2.346 × 10−5 0.7019 150 0.9216 947.6 0.03637 4.164 × 10−5 2.694 × 10−5 2.923 × 10−5 0.7019 200 0.8242 964.7 0.04014 5.048 × 10−5 2.923 × 10−5 3.546 × 10−5 0.7025 300 0.6804 997.1 0.04751 7.003 × 10−5 3.350 × 10−5 4.923 × 10−5 0.7030 400 0.5793 1025 0.05463 9.204 × 10−5 3.744 × 10−5 6.463 × 10−5 0.7023 500 0.5044 1048 0.06148 1.163 × 10−4 4.114 × 10−5 8.156 × 10−5 0.7010 1000 0.3063 1121 0.09198 2.678 × 10−4 5.732 × 10−5 1.871 × 10−4 0.6986 1500 0.2199 1165 0.11901 4.643 × 10−4 7.133 × 10−5 3.243 × 10−4 0.6985 2000 0.1716 1201 0.14705 7.139 × 10−4 8.417 × 10−5 4.907 × 10−4 0.6873 Water Vapor, H2O −50 0.9839 1892 0.01353 7.271 × 10−6 7.187 × 10−6 7.305 × 10−6 1.0047 0 0.8038 1874 0.01673 1.110 × 10−5 8.956 × 10−6 1.114 × 10−5 1.0033 50 0.6794 1874 0.02032 1.596 × 10−5 1.078 × 10−5 1.587 × 10−5 0.9944 100 0.5884 1887 0.02429 2.187 × 10−5 1.265 × 10−5 2.150 × 10−5 0.9830 150 0.5189 1908 0.02861 2.890 × 10−5 1.456 × 10−5 2.806 × 10−5 0.9712 200 0.4640 1935 0.03326 3.705 × 10−5 1.650 × 10−5 3.556 × 10−5 0.9599 300 0.3831 1997 0.04345 5.680 × 10−5 2.045 × 10−5 5.340 × 10−5 0.9401 400 0.3262 2066 0.05467 8.114 × 10−5 2.446 × 10−5 7.498 × 10−5 0.9240 500 0.2840 2137 0.06677 1.100 × 10−4 2.847 × 10−5 1.002 × 10−4 0.9108 1000 0.1725 2471 0.13623 3.196 × 10−4 4.762 × 10−5 2.761 × 10−4 0.8639 1500 0.1238 2736 0.21301 6.288 × 10−4 6.411 × 10−5 5.177 × 10−4 0.8233 2000 0.0966 2928 0.29183 1.032 × 10−3 7.808 × 10−5 8.084 × 10−4 0.7833 Note: For ideal gases, the properties cp, k, 𝜇, and Pr are independent of pressure. The properties 𝜌, 𝜈, and 𝛼 at a pressure P (in atm) other than 1 atm are determined by multiplying the values of 𝜌 at the given temperature by P and by dividing 𝜈 and 𝛼 by P.
Source: Data generated from the EES software developed by S. A. Klein and F. L. Alvarado. Originally based on various sources.
cen96537_app1_947_964.indd 958 16/01/17 12:01 pm 959 APPENDIX 1 TABLE A–11 Properties of the atmosphere at high altitude Speed of Thermal Altitude, Temperature, Pressure, Gravity Sound, Density Viscosity Conductivity, m °C kPa g, m/s2 m/s 𝜌, kg/m3 𝜇, kg/m·s W/m·K 0 15.00 101.33 9.807 340.3 1.225 1.789 × 10−5 0.0253 200 13.70 98.95 9.806 339.5 1.202 1.783 × 10−5 0.0252 400 12.40 96.61 9.805 338.8 1.179 1.777 × 10−5 0.0252 600 11.10 94.32 9.805 338.0 1.156 1.771 × 10−5 0.0251 800 9.80 92.08 9.804 337.2 1.134 1.764 × 10−5 0.0250 1000 8.50 89.88 9.804 336.4 1.112 1.758 × 10−5 0.0249 1200 7.20 87.72 9.803 335.7 1.090 1.752 × 10−5 0.0248 1400 5.90 85.60 9.802 334.9 1.069 1.745 × 10−5 0.0247 1600 4.60 83.53 9.802 334.1 1.048 1.739 × 10−5 0.0245 1800 3.30 81.49 9.801 333.3 1.027 1.732 × 10−5 0.0244 2000 2.00 79.50 9.800 332.5 1.007 1.726 × 10−5 0.0243 2200 0.70 77.55 9.800 331.7 0.987 1.720 × 10−5 0.0242 2400 −0.59 75.63 9.799 331.0 0.967 1.713 × 10−5 0.0241 2600 −1.89 73.76 9.799 330.2 0.947 1.707 × 10−5 0.0240 2800 −3.19 71.92 9.798 329.4 0.928 1.700 × 10−5 0.0239 3000 −4.49 70.12 9.797 328.6 0.909 1.694 × 10−5 0.0238 3200 −5.79 68.36 9.797 327.8 0.891 1.687 × 10−5 0.0237 3400 −7.09 66.63 9.796 327.0 0.872 1.681 × 10−5 0.0236 3600 −8.39 64.94 9.796 326.2 0.854 1.674 × 10−5 0.0235 3800 −9.69 63.28 9.795 325.4 0.837 1.668 × 10−5 0.0234 4000 −10.98 61.66 9.794 324.6 0.819 1.661 × 10−5 0.0233 4200 −12.3 60.07 9.794 323.8 0.802 1.655 × 10−5 0.0232 4400 −13.6 58.52 9.793 323.0 0.785 1.648 × 10−5 0.0231 4600 −14.9 57.00 9.793 322.2 0.769 1.642 × 10−5 0.0230 4800 −16.2 55.51 9.792 321.4 0.752 1.635 × 10−5 0.0229 5000 −17.5 54.05 9.791 320.5 0.736 1.628 × 10−5 0.0228 5200 −18.8 52.62 9.791 319.7 0.721 1.622 × 10−5 0.0227 5400 −20.1 51.23 9.790 318.9 0.705 1.615 × 10−5 0.0226 5600 −21.4 49.86 9.789 318.1 0.690 1.608 × 10−5 0.0224 5800 −22.7 48.52 9.785 317.3 0.675 1.602 × 10−5 0.0223 6000 −24.0 47.22 9.788 316.5 0.660 1.595 × 10−5 0.0222 6200 −25.3 45.94 9.788 315.6 0.646 1.588 × 10−5 0.0221 6400 −26.6 44.69 9.787 314.8 0.631 1.582 × 10−5 0.0220 6600 −27.9 43.47 9.786 314.0 0.617 1.575 × 10−5 0.0219 6800 −29.2 42.27 9.785 313.1 0.604 1.568 × 10−5 0.0218 7000 −30.5 41.11 9.785 312.3 0.590 1.561 × 10−5 0.0217 8000 −36.9 35.65 9.782 308.1 0.526 1.527 × 10−5 0.0212 9000 −43.4 30.80 9.779 303.8 0.467 1.493 × 10−5 0.0206 10,000 −49.9 26.50 9.776 299.5 0.414 1.458 × 10−5 0.0201 12,000 −56.5 19.40 9.770 295.1 0.312 1.422 × 10−5 0.0195 14,000 −56.5 14.17 9.764 295.1 0.228 1.422 × 10−5 0.0195 16,000 −56.5 10.53 9.758 295.1 0.166 1.422 × 10−5 0.0195 18,000 −56.5 7.57 9.751 295.1 0.122 1.422 × 10−5 0.0195 Source: U.S. Standard Atmosphere Supplements, U.S. Government Printing Office, 1966. Based on year-round mean conditions at 45° latitude and varies with the time of the year and the weather patterns. The conditions at sea level (z = 0) are taken to be P = 101.325 kPa, T = 15°C, 𝜌 = 1.2250 kg/m3, g = 9.80665 m2/s.
cen96537_app1_947_964.indd 959 16/01/17 12:01 pm 960 PROPERTY TABLES AND CHARTS 0.1 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.025 0.02 0.01 0.015 0.009 0.008 103 2(103) 3 4 5 6 8 104 2(104) 3 4 5 6 8 105 Reynolds number, Re 2(105) 3 4 5 6 8 106 2(106) 3 4 5 6 8 107 108 2(107) 3 4 5 6 8 0.05 0.04 0.03 0.02 0.015 0.01 0.008 0.006 0.004 0.002 0.001 0.0008 0.0006 0.0004 0.0002 0.0001 0.00005 0.00001 Relative roughness, ε/D Darcy friction factor, f ε/D = 0.000001 ε/D = 0.000005 Smooth pipes ε/D = 0 Laminar flow, f = 64/Re Glass, plastic Concrete Wood stave Rubber, smoothed Copper or brass tubing Cast iron Galvanized iron Wrought iron Stainless steel Commercial steel Material 0 0.003–0.03 0.0016 0.000033 0.000005 0.00085 0.0005 0.00015 0.000007 0.00015 0 0.9–9 0.5 0.01 0.0015 0.26 0.15 0.046 0.002 0.045 ft mm Laminar flow Transitional flow Fully rough turbulent flow ( f levels off) Turbulent flow Roughness, ε FIGURE A–12 The Moody chart for the friction factor for fully developed flow in circular pipes for use in the head loss relation hL = f L D V 2 2g. Friction factors in the turbulent flow are evaluated from the Colebrook equation 1 √f = −2 log10 ( 𝜀/D 3.7 + 2.51 Re √f).
cen96537_app1_947_964.indd 960 16/01/17 12:01 pm 961 APPENDIX 1 TABLE A–13 One-dimensional isentropic compressible flow functions for an ideal gas with k = 1.4 Ma Ma A/A P/P0 𝜌/𝜌0 T/T0 0 0 ∞ 1.0000 1.0000 1.0000 0.1 0.1094 5.8218 0.9930 0.9950 0.9980 0.2 0.2182 2.9635 0.9725 0.9803 0.9921 0.3 0.3257 2.0351 0.9395 0.9564 0.9823 0.4 0.4313 1.5901 0.8956 0.9243 0.9690 0.5 0.5345 1.3398 0.8430 0.8852 0.9524 0.6 0.6348 1.1882 0.7840 0.8405 0.9328 0.7 0.7318 1.0944 0.7209 0.7916 0.9107 0.8 0.8251 1.0382 0.6560 0.7400 0.8865 0.9 0.9146 1.0089 0.5913 0.6870 0.8606 1.0 1.0000 1.0000 0.5283 0.6339 0.8333 1.2 1.1583 1.0304 0.4124 0.5311 0.7764 1.4 1.2999 1.1149 0.3142 0.4374 0.7184 1.6 1.4254 1.2502 0.2353 0.3557 0.6614 1.8 1.5360 1.4390 0.1740 0.2868 0.6068 2.0 1.6330 1.6875 0.1278 0.2300 0.5556 2.2 1.7179 2.0050 0.0935 0.1841 0.5081 2.4 1.7922 2.4031 0.0684 0.1472 0.4647 2.6 1.8571 2.8960 0.0501 0.1179 0.4252 2.8 1.9140 3.5001 0.0368 0.0946 0.3894 3.0 1.9640 4.2346 0.0272 0.0760 0.3571 5.0 2.2361 25.000 0.0019 0.0113 0.1667 ∝ 2.2495 ∝ 0 0 0 Ma = Ma√ k + 1 2 + (k −1)Ma2 A A = 1 Ma[( 2 k + 1) (1 + k −1 2 Ma2) ] 0.5(k+1)/(k−1) P P0 = (1 + k −1 2 Ma2) −k/(k−1) ρ ρ0 = (1 + k −1 2 Ma2) −1/(k−1) T T0 = (1 + k −1 2 Ma2) −1 A/A Ma T/T0 ρ/ρ0 P/P0 2.5 3.0 2.0 1.5 Compressible flow functions 1.0 0.5 0 0 0.5 1.0 1.5 Ma 2.0 2.5 3.0 cen96537_app1_947_964.indd 961 16/01/17 12:01 pm 962 PROPERTY TABLES AND CHARTS TABLE A–14 One-dimensional normal shock functions for an ideal gas with k = 1.4 Ma1 Ma2 P2/P1 𝜌2/𝜌1 T2/T1 P02/P01 P02/P1 1.0 1.0000 1.0000 1.0000 1.0000 1.0000 1.8929 1.1 0.9118 1.2450 1.1691 1.0649 0.9989 2.1328 1.2 0.8422 1.5133 1.3416 1.1280 0.9928 2.4075 1.3 0.7860 1.8050 1.5157 1.1909 0.9794 2.7136 1.4 0.7397 2.1200 1.6897 1.2547 0.9582 3.0492 1.5 0.7011 2.4583 1.8621 1.3202 0.9298 3.4133 1.6 0.6684 2.8200 2.0317 1.3880 0.8952 3.8050 1.7 0.6405 3.2050 2.1977 1.4583 0.8557 4.2238 1.8 0.6165 3.6133 2.3592 1.5316 0.8127 4.6695 1.9 0.5956 4.0450 2.5157 1.6079 0.7674 5.1418 2.0 0.5774 4.5000 2.6667 1.6875 0.7209 5.6404 2.1 0.5613 4.9783 2.8119 1.7705 0.6742 6.1654 2.2 0.5471 5.4800 2.9512 1.8569 0.6281 6.7165 2.3 0.5344 6.0050 3.0845 1.9468 0.5833 7.2937 2.4 0.5231 6.5533 3.2119 2.0403 0.5401 7.8969 2.5 0.5130 7.1250 3.3333 2.1375 0.4990 8.5261 2.6 0.5039 7.7200 3.4490 2.2383 0.4601 9.1813 2.7 0.4956 8.3383 3.5590 2.3429 0.4236 9.8624 2.8 0.4882 8.9800 3.6636 2.4512 0.3895 10.5694 2.9 0.4814 9.6450 3.7629 2.5632 0.3577 11.3022 3.0 0.4752 10.3333 3.8571 2.6790 0.3283 12.0610 4.0 0.4350 18.5000 4.5714 4.0469 0.1388 21.0681 5.0 0.4152 29.000 5.0000 5.8000 0.0617 32.6335 ∞ 0.3780 ∞ 6.0000 ∞ 0 ∞ T01 = T02 Ma2 = √ (k −1)Ma 2 1 + 2 2kMa 2 1 −k + 1 P2 P1 = 1 + kMa 2 1 1 + kMa 2 2 = 2kMa 2 1 −k + 1 k + 1 ρ2 ρ1 = P2/P1 T2/T1 = (k + 1)Ma 2 1 2 + (k −1)Ma 2 1 = V1 V2 T2 T1 = 2 + Ma 2 1 (k −1) 2 + Ma 2 2 (k −1) P02 P01 = Ma1 Ma2[ 1 + Ma 2 1 (k −1)/2 1 + Ma 2 1 (k −1)/2] (k+1)/[2(k−1)] P02 P1 = (1 + kMa 2 1 )[1 + Ma 2 2 (k −1)/2]k/(k−1) 1 + kMa 2 2 Ma2 T2/T1 ρ2/ρ1 P02/P1 P02/P01 P2/P1 3.0 4.0 5.0 2.0 Normal shock functions 1.0 0 1.0 1.5 2.0 2.5 Ma1 3.0 cen96537_app1_947_964.indd 962 16/01/17 12:01 pm 963 APPENDIX 1 TABLE A–15 Rayleigh flow functions for an ideal gas with k = 1.4 Ma T0/T0 P0/P0 T/T P/P V/V 0.0 0.0000 1.2679 0.0000 2.4000 0.0000 0.1 0.0468 1.2591 0.0560 2.3669 0.0237 0.2 0.1736 1.2346 0.2066 2.2727 0.0909 0.3 0.3469 1.1985 0.4089 2.1314 0.1918 0.4 0.5290 1.1566 0.6151 1.9608 0.3137 0.5 0.6914 1.1141 0.7901 1.7778 0.4444 0.6 0.8189 1.0753 0.9167 1.5957 0.5745 0.7 0.9085 1.0431 0.9929 1.4235 0.6975 0.8 0.9639 1.0193 1.0255 1.2658 0.8101 0.9 0.9921 1.0049 1.0245 1.1246 0.9110 1.0 1.0000 1.0000 1.0000 1.0000 1.0000 1.2 0.9787 1.0194 0.9118 0.7958 1.1459 1.4 0.9343 1.0777 0.8054 0.6410 1.2564 1.6 0.8842 1.1756 0.7017 0.5236 1.3403 1.8 0.8363 1.3159 0.6089 0.4335 1.4046 2.0 0.7934 1.5031 0.5289 0.3636 1.4545 2.2 0.7561 1.7434 0.4611 0.3086 1.4938 2.4 0.7242 2.0451 0.4038 0.2648 1.5252 2.6 0.6970 2.4177 0.3556 0.2294 1.5505 2.8 0.6738 2.8731 0.3149 0.2004 1.5711 3.0 0.6540 3.4245 0.2803 0.1765 1.5882 P0/P0 T0/T0 T/T P/P V/V 3.5 3.0 2.5 1.5 2.0 Rayleigh flow functions 1.0 0.5 0 0 0.5 1.0 1.5 Ma 2.0 2.5 3.0 T0 T 0 = (k + 1)Ma2[2 + (k −1)Ma2] (1 + kMa2)2 P0 P0 = k + 1 1 + kMa2 ( 2 + (k −1)Ma2 k + 1 ) k/(k−1 T T = ( Ma(1 + k) 1 + kMa2 ) 2 P P = 1 + k 1 + kMa2 V V = ρ ρ = (1 + k)Ma2 1 + kMa2 cen96537_app1_947_964.indd 963 16/01/17 12:01 pm 964 PROPERTY TABLES AND CHARTS TABLE A–16 Fanno flow functions for an ideal gas with k = 1.4 Ma P0/P0 T/T P/P V/V fL/D 0.0 ∞ 1.2000 ∞ 0.0000 ∞ 0.1 5.8218 1.1976 10.9435 0.1094 66.9216 0.2 2.9635 1.1905 5.4554 0.2182 14.5333 0.3 2.0351 1.1788 3.6191 0.3257 5.2993 0.4 1.5901 1.1628 2.6958 0.4313 2.3085 0.5 1.3398 1.1429 2.1381 0.5345 1.0691 0.6 1.1882 1.1194 1.7634 0.6348 0.4908 0.7 1.0944 1.0929 1.4935 0.7318 0.2081 0.8 1.0382 1.0638 1.2893 0.8251 0.0723 0.9 1.0089 1.0327 1.1291 0.9146 0.0145 1.0 1.0000 1.0000 1.0000 1.0000 0.0000 1.2 1.0304 0.9317 0.8044 1.1583 0.0336 1.4 1.1149 0.8621 0.6632 1.2999 0.0997 1.6 1.2502 0.7937 0.5568 1.4254 0.1724 1.8 1.4390 0.7282 0.4741 1.5360 0.2419 2.0 1.6875 0.6667 0.4082 1.6330 0.3050 2.2 2.0050 0.6098 0.3549 1.7179 0.3609 2.4 2.4031 0.5576 0.3111 1.7922 0.4099 2.6 2.8960 0.5102 0.2747 1.8571 0.4526 2.8 3.5001 0.4673 0.2441 1.9140 0.4898 3.0 4.2346 0.4286 0.2182 1.9640 0.5222 T0 = T 0 P0 P 0 = ρ0 ρ 0 = 1 Ma( 2 + (k −1)Ma2 k + 1 ) (k+1)/2(k−1) T T = k + 1 2 + (k −1)Ma2 P P = 1 Ma( k + 1 2 + (k −1)Ma2) 1/2 V V = ρ ρ = Ma( k + 1 2 + (k −1)Ma2) 1/2 fL D = 1 −Ma2 kMa2 + k + 1 2k ln (k + 1)Ma2 2 + (k −1)Ma2 2.5 3.0 2.0 1.5 Fanno flow functions 1.0 0.5 0 0 0.5 1.0 1.5 Ma 2.0 2.5 3.0 T/T V/V P/P fL/D P0/P0 cen96537_app1_947_964.indd 964 16/01/17 12:01 pm 965 APPENDIX 2 P RO P E R T Y T A B LE S AN D C H A R T S ( E N G L I S H UNI T S) TABLE A–1E Molar Mass, Gas Constant, and Ideal-Gas Specific Heats of Some Substances 966 TABLE A–2E Boiling and Freezing Point Properties 967 TABLE A–3E Properties of Saturated Water 968 TABLE A–4E Properties of Saturated Refrigerant-134a 969 TABLE A–5E Properties of Saturated Ammonia 970 TABLE A–6E Properties of Saturated Propane 971 TABLE A–7E Properties of Liquids 972 TABLE A–8E Properties of Liquid Metals 973 TABLE A–9E Properties of Air at 1 atm Pressure 974 TABLE A–10E Properties of Gases at 1 atm Pressure 975 TABLE A–11E Properties of the Atmosphere at High Altitude 977 Most properties in the tables are obtained from the property database of EES, and the original sources are listed under the tables. Properties are often listed to more significant digits than the claimed accuracy for the purpose of minimizing accumulated round-off error in hand calculations and ensuring a close match with the results obtained with EES.
cen96537_app2_965_978.indd 965 16/01/17 12:03 pm 966 PROPERTY TABLES AND CHARTS TABLE A–1E Molar mass, gas constant, and ideal-gas specific heats of some substances Gas Constant R Specific Heat Data at 77°F Molar Mass, Btu/ psia·ft3/ cp, cv, Substance M, lbm/lbmol lbm·R lbm·R Btu/lbm·R Btu/lbm·R k = cp/cv Air 28.97 0.06855 0.3704 0.2400 0.1715 1.400 Ammonia, NH3 17.03 0.1166 0.6301 0.4999 0.3834 1.304 Argon, Ar 39.95 0.04970 0.2686 0.1243 0.07457 1.667 Bromine, Br2 159.81 0.01242 0.06714 0.0538 0.04137 1.300 Isobutane, C4H10 58.12 0.03415 0.1846 0.3972 0.3631 1.094 n-Butane, C4H10 58.12 0.03415 0.1846 0.4046 0.3705 1.092 Carbon dioxide, CO2 44.01 0.04512 0.2438 0.2016 0.1564 1.288 Carbon monoxide, CO 28.01 0.07089 0.3831 0.2482 0.1772 1.400 Chlorine, Cl2 70.905 0.02802 0.1514 0.1142 0.08618 1.325 Chlorodifluoromethane (R-22), CHClF2 86.47 0.02297 0.1241 0.1552 0.1322 1.174 Ethane, C2H6 30.070 0.06604 0.3569 0.4166 0.3506 1.188 Ethylene, C2H4 28.054 0.07079 0.3826 0.3647 0.2940 1.241 Fluorine, F2 38.00 0.05224 0.2823 0.1967 0.1445 1.362 Helium, He 4.003 0.4961 2.681 1.2403 0.7442 1.667 n-Heptane, C7H16 100.20 0.01982 0.1071 0.3939 0.3740 1.053 n-Hexane, C6H14 86.18 0.02304 0.1245 0.3951 0.3721 1.062 Hydrogen, H2 2.016 0.9850 5.323 3.416 2.431 1.405 Krypton, Kr 83.80 0.02370 0.1281 0.05923 0.03554 1.667 Methane, CH4 16.04 0.1238 0.6688 0.5317 0.4080 1.303 Neon, Ne 20.183 0.09838 0.5316 0.2460 0.1476 1.667 Nitrogen, N2 28.01 0.07089 0.3831 0.2484 0.1774 1.400 Nitric oxide, NO 30.006 0.06618 0.3577 0.2387 0.1725 1.384 Nitrogen dioxide, NO2 46.006 0.04512 0.2438 0.1925 0.1474 1.306 Oxygen, O2 32.00 0.06205 0.3353 0.2193 0.1572 1.395 n-Pentane, C5H12 72.15 0.02752 0.1487 0.3974 0.3700 1.074 Propane, C3H8 44.097 0.04502 0.2433 0.3986 0.3535 1.127 Propylene, C3H6 42.08 0.04720 0.2550 0.3657 0.3184 1.148 Steam, H2O 18.015 0.1102 0.5957 0.4455 0.3351 1.329 Sulfur dioxide, SO2 64.06 0.03100 0.1675 0.1488 0.1178 1.263 Tetrachloromethane, CCl4 153.82 0.01291 0.06976 0.1293 0.1164 1.111 Tetrafluoroethane (R-134a), C2H2F4 102.03 0.01946 0.1052 0.1991 0.1796 1.108 Trifluoroethane (R-143a), C2H3F3 84.04 0.02363 0.1277 0.2219 0.1983 1.119 Xenon, Xe 131.30 0.01512 0.08173 0.03781 0.02269 1.667 The gas constant is calculated from R = Ru/M, where Ru = 1.9859 Btu/lbmol·R = 10.732 psia·ft3/lbmol·R is the universal gas constant and M is the molar mass.
Source: Specific heat values are mostly obtained from the property routines prepared by The National Institute of Standards and Technology (NIST), Gaithersburg, MD.
cen96537_app2_965_978.indd 966 16/01/17 12:03 pm 967 APPENDIX 2 TABLE A–2E Boiling and freezing point properties Boiling Data at 1 atm Freezing Data Liquid Properties Normal Latent Heat of Latent Heat Specific Boiling Vaporization Freezing of Fusion Tempera- Density Heat cp, Substance Point, °F hfg, Btu/lbm Point, °F hif, Btu/lbm ture, °F 𝜌, lbm/ft3 Btu/lbm·R Ammonia −27.9 24.54 −107.9 138.6 −27.9 42.6 1.06 0 41.3 1.083 40 39.5 1.103 80 37.5 1.135 Argon −302.6 69.5 −308.7 12.0 −302.6 87.0 0.272 Benzene 176.4 169.4 41.9 54.2 68 54.9 0.411 Brine (20% sodium chloride by mass) 219.0 — 0.7 — 68 71.8 0.743 n-Butane 31.1 165.6 −217.3 34.5 31.1 37.5 0.552 Carbon dioxide −109.2 99.6 (at 32°F) −69.8 — 32 57.8 0.583 Ethanol 172.8 360.5 −173.6 46.9 77 48.9 0.588 Ethyl alcohol 173.5 368 −248.8 46.4 68 49.3 0.678 Ethylene glycol 388.6 344.0 12.6 77.9 68 69.2 0.678 Glycerine 355.8 419 66.0 86.3 68 78.7 0.554 Helium −452.1 9.80 — — −452.1 9.13 5.45 Hydrogen −423.0 191.7 −434.5 25.6 −423.0 4.41 2.39 Isobutane 10.9 157.8 −255.5 45.5 10.9 37.1 0.545 Kerosene 399–559 108 −12.8 — 68 51.2 0.478 Mercury 674.1 126.7 −38.0 4.90 77 847 0.033 Methane −258.7 219.6 296.0 25.1 −258.7 26.4 0.834 −160 20.0 1.074 Methanol 148.1 473 −143.9 42.7 77 49.1 0.609 Nitrogen −320.4 85.4 −346.0 10.9 −320.4 50.5 0.492 −260 38.2 0.643 Octane 256.6 131.7 −71.5 77.9 68 43.9 0.502 Oil (light) — — 77 56.8 0.430 Oxygen −297.3 91.5 −361.8 5.9 −297.3 71.2 0.408 Petroleum — 99–165 68 40.0 0.478 Propane −43.7 184.0 −305.8 34.4 −43.7 36.3 0.538 32 33.0 0.604 100 29.4 0.673 Refrigerant-134a −15.0 93.2 −141.9 — −40 88.5 0.283 −15 86.0 0.294 32 80.9 0.318 90 73.6 0.348 Water 212 970.5 32 143.5 32 62.4 1.01 90 62.1 1.00 150 61.2 1.00 212 59.8 1.01 Sublimation temperature. (At pressures below the triple-point pressure of 75.1 psia, carbon dioxide exists as a solid or gas. Also, the freezing-point temperature of carbon dioxide is the triple-point temperature of −69.8°F.) cen96537_app2_965_978.indd 967 16/01/17 12:03 pm 968 PROPERTY TABLES AND CHARTS TABLE A–3E Properties of saturated water Volume Enthalpy Specific Thermal Prandtl Expansion Surface Saturation Density of Heat Conductivity Dynamic Viscosity Number Coefficient Tension, Temp. Pressure 𝜌, lbm/ft3 Vaporization cp, Btu/lbm·R k, Btu/h·ft·R 𝜇, lbm/ft·s Pr 𝛽, 1/R lbf/ft T, °F Psat, psia Liquid Vapor hfg, Btu/lbm Liquid Vapor Liquid Vapor Liquid Vapor Liquid Vapor Liquid Liquid 32.02 0.0887 62.41 0.00030 1075 1.010 0.446 0.324 0.0099 1.204 × 10−3 6.194 × 10−6 13.5 1.00 −0.038 × 10−3 0.00518 40 0.1217 62.42 0.00034 1071 1.004 0.447 0.329 0.0100 1.038 × 10−3 6.278 × 10−6 11.4 1.01 0.003 × 10−3 0.00514 50 0.1780 62.41 0.00059 1065 1.000 0.448 0.335 0.0102 8.781 × 10−4 6.361 × 10−6 9.44 1.01 0.047 × 10−3 0.00509 60 0.2563 62.36 0.00083 1060 0.999 0.449 0.341 0.0104 7.536 × 10−4 6.444 × 10−6 7.95 1.00 0.080 × 10−3 0.00503 70 0.3632 62.30 0.00115 1054 0.999 0.450 0.347 0.0106 6.556 × 10−4 6.556 × 10−6 6.79 1.00 0.115 × 10−3 0.00497 80 0.5073 62.22 0.00158 1048 0.999 0.451 0.352 0.0108 5.764 × 10−4 6.667 × 10−6 5.89 1.00 0.145 × 10−3 0.00491 90 0.6988 62.12 0.00214 1043 0.999 0.453 0.358 0.0110 5.117 × 10−4 6.778 × 10−6 5.14 1.00 0.174 × 10−3 0.00485 100 0.9503 62.00 0.00286 1037 0.999 0.454 0.363 0.0112 4.578 × 10−4 6.889 × 10−6 4.54 1.01 0.200 × 10−3 0.00479 110 1.2763 61.86 0.00377 1031 0.999 0.456 0.367 0.0115 4.128 × 10−4 7.000 × 10−6 4.05 1.00 0.224 × 10−3 0.00473 120 1.6945 61.71 0.00493 1026 0.999 0.458 0.371 0.0117 3.744 × 10−4 7.111 × 10−6 3.63 1.00 0.246 × 10−3 0.00467 130 2.225 61.55 0.00636 1020 0.999 0.460 0.375 0.0120 3.417 × 10−4 7.222 × 10−6 3.28 1.00 0.267 × 10−3 0.00460 140 2.892 61.38 0.00814 1014 0.999 0.463 0.378 0.0122 3.136 × 10−4 7.333 × 10−6 2.98 1.00 0.287 × 10−3 0.00454 150 3.722 61.19 0.0103 1008 1.000 0.465 0.381 0.0125 2.889 × 10−4 7.472 × 10−6 2.73 1.00 0.306 × 10−3 0.00447 160 4.745 60.99 0.0129 1002 1.000 0.468 0.384 0.0128 2.675 × 10−4 7.583 × 10−6 2.51 1.00 0.325 × 10−3 0.00440 170 5.996 60.79 0.0161 996 1.001 0.472 0.386 0.0131 2.483 × 10−4 7.722 × 10−6 2.90 1.00 0.346 × 10−3 0.00434 180 7.515 60.57 0.0199 990 1.002 0.475 0.388 0.0134 2.317 × 10−4 7.833 × 10−6 2.15 1.00 0.367 × 10−3 0.00427 190 9.343 60.35 0.0244 984 1.004 0.479 0.390 0.0137 2.169 × 10−4 7.972 × 10−6 2.01 1.00 0.382 × 10−3 0.00420 200 11.53 60.12 0.0297 978 1.005 0.483 0.391 0.0141 2.036 × 10−4 8.083 × 10−6 1.88 1.00 0.395 × 10−3 0.00412 210 14.125 59.87 0.0359 972 1.007 0.487 0.392 0.0144 1.917 × 10−4 8.222 × 10−6 1.77 1.00 0.412 × 10−3 0.00405 212 14.698 59.82 0.0373 970 1.007 0.488 0.392 0.0145 1.894 × 10−4 8.250 × 10−6 1.75 1.00 0.417 × 10−3 0.00404 220 17.19 59.62 0.0432 965 1.009 0.492 0.393 0.0148 1.808 ×10−4 8.333 × 10−6 1.67 1.00 0.429 × 10−3 0.00398 230 20.78 59.36 0.0516 959 1.011 0.497 0.394 0.0152 1.711 × 10−4 8.472 × 10−6 1.58 1.00 0.443 × 10−3 0.00390 240 24.97 59.09 0.0612 952 1.013 0.503 0.394 0.0156 1.625 × 10−4 8.611 × 10−6 1.50 1.00 0.462 × 10−3 0.00383 250 29.82 58.82 0.0723 946 1.015 0.509 0.395 0.0160 1.544 × 10−4 8.611 × 10−6 1.43 1.00 0.480 × 10−3 0.00375 260 35.42 58.53 0.0850 939 1.018 0.516 0.395 0.0164 1.472 × 10−4 8.861 × 10−6 1.37 1.00 0.497 × 10−3 0.00367 270 41.85 58.24 0.0993 932 1.020 0.523 0.395 0.0168 1.406 × 10−4 9.000 × 10−6 1.31 1.01 0.514 × 10−3 0.00360 280 49.18 57.94 0.1156 925 1.023 0.530 0.395 0.0172 1.344 × 10−4 9.111 × 10−6 1.25 1.01 0.532 × 10−3 0.00352 290 57.53 57.63 0.3390 918 1.026 0.538 0.395 0.0177 1.289 × 10−4 9.250 × 10−6 1.21 1.01 0.549 × 10−3 0.00344 300 66.98 57.31 0.1545 910 1.029 0.547 0.394 0.0182 1.236 × 10−4 9.389 × 10−6 1.16 1.02 0.566 × 10−3 0.00336 320 89.60 56.65 0.2033 895 1.036 0.567 0.393 0.0191 1.144 × 10−4 9.639 × 10−6 1.09 1.03 0.636 × 10−3 0.00319 340 117.93 55.95 0.2637 880 1.044 0.590 0.391 0.0202 1.063 × 10−4 9.889 × 10−6 1.02 1.04 0.656 × 10−3 0.00303 360 152.92 55.22 0.3377 863 1.054 0.617 0.389 0.0213 9.972 × 10−5 1.013 × 10−5 0.973 1.06 0.681 × 10−3 0.00286 380 195.60 54.46 0.4275 845 1.065 0.647 0.385 0.0224 9.361 × 10−5 1.041 × 10−5 0.932 1.08 0.720 × 10−3 0.00269 400 241.1 53.65 0.5359 827 1.078 0.683 0.382 0.0237 8.833 × 10−5 1.066 × 10−5 0.893 1.11 0.771 × 10−3 0.00251 450 422.1 51.46 0.9082 775 1.121 0.799 0.370 0.0271 7.722 × 10−5 1.130 × 10−5 0.842 1.20 0.912 × 10−3 0.00207 500 680.0 48.95 1.479 715 1.188 0.972 0.352 0.0312 6.833 × 10−5 1.200 × 10−5 0.830 1.35 1.111 × 10−3 0.00162 550 1046.7 45.96 4.268 641 1.298 1.247 0.329 0.0368 6.083 × 10−5 1.280 × 10−5 0.864 1.56 1.445 × 10−3 0.00118 600 1541 42.32 3.736 550 1.509 1.759 0.299 0.0461 5.389 × 10−5 1.380 × 10−5 0.979 1.90 1.885 × 10−3 0.00074 650 2210 37.31 6.152 422 2.086 3.103 0.267 0.0677 4.639 × 10−5 1.542 × 10−5 1.30 2.54 0.00034 700 3090 27.28 13.44 168 13.80 25.90 0.254 0.1964 3.417 × 10−5 2.044 × 10−5 6.68 9.71 0.00002 705.44 3204 19.79 19.79 0 ∞ ∞ ∞ ∞ 2.897 × 10−5 2.897 × 10−5 0 Note 1: Kinematic viscosity 𝜈 and thermal diffusivity 𝛼 can be calculated from their definitions, 𝜈 = 𝜇/𝜌 and 𝛼 = k/𝜌cp = 𝜈/Pr. The temperatures 32.02°F, 212°F, and 705.44°F are the triple-, boiling-, and critical-point temperatures of water, respectively. All properties listed above (except the vapor density) can be used at any pressure with negligible error except at temperatures near the critical-point value.
Note 2: The unit Btu/lbm·°F for specific heat is equivalent to Btu/lbm·R, and the unit Btu/h·ft·°F for thermal conductivity is equivalent to Btu/h·ft·R.
Source: Viscosity and thermal conductivity data are from J. V. Sengers and J. T. R. Watson, Journal of Physical and Chemical Reference Data 15 (1986), pp. 1291–1322. Other data are obtained from various sources or calculated.
cen96537_app2_965_978.indd 968 16/01/17 12:03 pm 969 APPENDIX 2 TABLE A–4E Properties of saturated refrigerant-134a Volume Enthalpy Specific Thermal Prandtl Expansion Surface Saturation Density of Heat Conductivity Dynamic Viscosity Number Coefficient Tension, Temp. Pressure 𝜌, lbm/ft3 Vaporization cp, Btu/lbm·R k, Btu/h·ft·R 𝜇, lbm/ft·s Pr 𝛽, 1/R lbf/ft T, °F P, psia Liquid Vapor hfg, Btu/lbm Liquid Vapor Liquid Vapor Liquid Vapor Liquid Vapor Liquid Liquid −40 7.4 88.51 0.1731 97.1 0.2996 0.1788 0.0636 0.00466 3.278 × 10−4 1.714 × 10−6 5.558 0.237 0.00114 0.001206 −30 9.9 87.5 0.2258 95.6 0.3021 0.1829 0.0626 0.00497 3.004 × 10−4 2.053 × 10−6 5.226 0.272 0.00117 0.001146 −20 12.9 86.48 0.2905 94.1 0.3046 0.1872 0.0613 0.00529 2.762 × 10−4 2.433 × 10−6 4.937 0.310 0.00120 0.001087 −10 16.6 85.44 0.3691 92.5 0.3074 0.1918 0.0602 0.00559 2.546 × 10−4 2.856 × 10−6 4.684 0.352 0.00124 0.001029 0 21.2 84.38 0.4635 90.9 0.3103 0.1966 0.0589 0.00589 2.354 × 10−4 3.314 × 10−6 4.463 0.398 0.00128 0.000972 10 26.6 83.31 0.5761 89.3 0.3134 0.2017 0.0576 0.00619 2.181 × 10−4 3.811 × 10−6 4.269 0.447 0.00132 0.000915 20 33.1 82.2 0.7094 87.5 0.3167 0.2070 0.0563 0.00648 2.024 × 10−4 4.342 × 10−6 4.098 0.500 0.00137 0.000859 30 40.8 81.08 0.866 85.8 0.3203 0.2127 0.0550 0.00676 1.883 × 10−4 4.906 × 10−6 3.947 0.555 0.00142 0.000803 40 49.8 79.92 1.049 83.9 0.3240 0.2188 0.0536 0.00704 1.752 × 10−4 5.494 × 10−6 3.814 0.614 0.00149 0.000749 50 60.2 78.73 1.262 82.0 0.3281 0.2253 0.0522 0.00732 1.633 × 10−4 6.103 × 10−6 3.697 0.677 0.00156 0.000695 60 72.2 77.51 1.509 80.0 0.3325 0.2323 0.0507 0.00758 1.522 × 10−4 6.725 × 10−6 3.594 0.742 0.00163 0.000642 70 85.9 76.25 1.794 78.0 0.3372 0.2398 0.0492 0.00785 1.420 × 10−4 7.356 × 10−6 3.504 0.810 0.00173 0.000590 80 101.4 74.94 2.122 75.8 0.3424 0.2481 0.0476 0.00810 1.324 × 10−4 7.986 × 10−6 3.425 0.880 0.00183 0.000538 90 119.1 73.59 2.5 73.5 0.3481 0.2572 0.0460 0.00835 1.234 × 10−4 8.611 × 10−6 3.357 0.955 0.00195 0.000488 100 138.9 72.17 2.935 71.1 0.3548 0.2674 0.0444 0.00860 1.149 × 10−4 9.222 × 10−6 3.303 1.032 0.00210 0.000439 110 161.2 70.69 3.435 68.5 0.3627 0.2790 0.0427 0.00884 1.068 × 10−4 9.814 × 10−6 3.262 1.115 0.00227 0.000391 120 186.0 69.13 4.012 65.8 0.3719 0.2925 0.0410 0.00908 9.911 × 10−5 1.038 × 10−5 3.235 1.204 0.00248 0.000344 130 213.5 67.48 4.679 62.9 0.3829 0.3083 0.0392 0.00931 9.175 × 10−5 1.092 × 10−5 3.223 1.303 0.00275 0.000299 140 244.1 65.72 5.455 59.8 0.3963 0.3276 0.0374 0.00954 8.464 × 10−5 1.144 × 10−5 3.229 1.416 0.00308 0.000255 150 277.8 63.83 6.367 56.4 0.4131 0.3520 0.0355 0.00976 7.778 × 10−5 1.195 × 10−5 3.259 1.551 0.00351 0.000212 160 314.9 61.76 7.45 52.7 0.4352 0.3839 0.0335 0.00998 7.108 × 10−5 1.245 × 10−5 3.324 1.725 0.00411 0.000171 170 355.8 59.47 8.762 48.5 0.4659 0.4286 0.0314 0.01020 6.450 × 10−5 1.298 × 10−5 3.443 1.963 0.00498 0.000132 180 400.7 56.85 10.4 43.7 0.5123 0.4960 0.0292 0.01041 5.792 × 10−5 1.356 × 10−5 3.661 2.327 0.00637 0.000095 190 449.9 53.75 12.53 38.0 0.5929 0.6112 0.0267 0.01063 5.119 × 10−5 1.431 × 10−5 4.090 2.964 0.00891 0.000061 200 504.0 49.75 15.57 30.7 0.7717 0.8544 0.0239 0.01085 4.397 × 10−5 1.544 × 10−5 5.119 4.376 0.01490 0.000031 210 563.8 43.19 21.18 18.9 1.4786 1.6683 0.0199 0.01110 3.483 × 10−5 1.787 × 10−5 9.311 9.669 0.04021 0.000006 Note 1: Kinematic viscosity 𝜈 and thermal diffusivity 𝛼 can be calculated from their definitions, 𝜈 = 𝜇/𝜌 and 𝛼 = k/𝜌cp = 𝜈/Pr. The properties listed here (except the vapor density) can be used at any pressures with negligible error except at temperatures near the critical-point value.
Note 2: The unit Btu/lbm·°F for specific heat is equivalent to Btu/lbm·R, and the unit Btu/h·ft·°F for thermal conductivity is equivalent to Btu/h·ft·R.
Source: Data generated from the EES software developed by S. A. Klein and F. L. Alvarado. Original sources: R. Tillner-Roth and H. D. Baehr, “An International Standard Formulation for the Thermodynamic Properties of 1,1,1,2-Tetrafluoroethane (HFC-134a) for Temperatures from 170 K to 455 K and Pressures up to 70 MPa,” J. Phys. Chem. Ref. Data, Vol. 23, No. 5, 1994; M. J. Assael, N. K. Dalaouti, A. A. Griva, and J. H. Dymond, “Viscosity and Thermal Conductivity of Halogenated Methane and Ethane Refrigerants,” IJR, Vol. 22, pp. 525–535, 1999; NIST REFPROP 6 program (M. O. McLinden, S. A. Klein, E. W. Lemmon, and A. P. Peskin, Physicial and Chemical Properties Division, National Institute of Standards and Technology, Boulder, CO 80303, 1995).
cen96537_app2_965_978.indd 969 16/01/17 12:03 pm 970 PROPERTY TABLES AND CHARTS TABLE A–5E Properties of saturated ammonia Volume Enthalpy Specific Thermal Prandtl Expansion Surface Saturation Density of Heat Conductivity Dynamic Viscosity Number Coefficient Tension, Temp. Pressure 𝜌, lbm/ft3 Vaporization cp, Btu/lbm·R k, Btu/h·ft·R 𝜇, lbm/ft·s Pr 𝛽, 1/R lbf/ft T, °F P, psia Liquid Vapor hfg, Btu/lbm Liquid Vapor Liquid Vapor Liquid Vapor Liquid Vapor Liquid Liquid −40 10.4 43.08 0.0402 597.0 1.0542 0.5354 — 0.01026 1.966 × 10−4 5.342 × 10−6 — 1.003 0.00098 0.002443 −30 13.9 42.66 0.0527 590.2 1.0610 0.5457 — 0.01057 1.853 × 10−4 5.472 × 10−6 — 1.017 0.00101 0.002357 −20 18.3 42.33 0.0681 583.2 1.0677 0.5571 0.3501 0.01089 1.746 × 10−4 5.600 × 10−6 1.917 1.031 0.00103 0.002272 −10 23.7 41.79 0.0869 575.9 1.0742 0.5698 0.3426 0.01121 1.645 × 10−4 5.731 × 10−6 1.856 1.048 0.00106 0.002187 0 30.4 41.34 0.1097 568.4 1.0807 0.5838 0.3352 0.01154 1.549 × 10−4 5.861 × 10−6 1.797 1.068 0.00109 0.002103 10 38.5 40.89 0.1370 560.7 1.0873 0.5992 0.3278 0.01187 1.458 × 10−4 5.994 ×10−6 1.740 1.089 0.00112 0.002018 20 48.2 40.43 0.1694 552.6 1.0941 0.6160 0.3203 0.01220 1.371 × 10−4 6.125 ×10−6 1.686 1.113 0.00116 0.001934 30 59.8 39.96 0.2075 544.4 1.1012 0.6344 0.3129 0.01254 1.290 × 10−4 6.256 ×10−6 1.634 1.140 0.00119 0.001850 40 73.4 39.48 0.2521 535.8 1.1087 0.6544 0.3055 0.01288 1.213 × 10−4 6.389 ×10−6 1.585 1.168 0.00123 0.001767 50 89.2 38.99 0.3040 526.9 1.1168 0.6762 0.2980 0.01323 1.140 × 10−4 6.522 ×10−6 1.539 1.200 0.00128 0.001684 60 107.7 38.50 0.3641 517.7 1.1256 0.6999 0.2906 0.01358 1.072 × 10−4 6.656 ×10−6 1.495 1.234 0.00132 0.001601 70 128.9 37.99 0.4332 508.1 1.1353 0.7257 0.2832 0.01394 1.008 × 10−4 6.786 ×10−6 1.456 1.272 0.00137 0.001518 80 153.2 37.47 0.5124 498.2 1.1461 0.7539 0.2757 0.01431 9.486 × 10−5 6.922 ×10−6 1.419 1.313 0.00143 0.001436 90 180.8 36.94 0.6029 487.8 1.1582 0.7846 0.2683 0.01468 8.922 × 10−5 7.056 ×10−6 1.387 1.358 0.00149 0.001354 100 212.0 36.40 0.7060 477.0 1.1719 0.8183 0.2609 0.01505 8.397 × 10−5 7.189 ×10−6 1.358 1.407 0.00156 0.001273 110 247.2 35.83 0.8233 465.8 1.1875 0.8554 0.2535 0.01543 7.903 × 10−5 7.325 × 10−6 1.333 1.461 0.00164 0.001192 120 286.5 35.26 0.9564 454.1 1.2054 0.8965 0.2460 0.01582 7.444 × 10−5 7.458 × 10−6 1.313 1.522 0.00174 0.001111 130 330.4 34.66 1.1074 441.7 1.2261 0.9425 0.2386 0.01621 7.017 × 10−5 7.594 × 10−6 1.298 1.589 0.00184 0.001031 140 379.2 34.04 1.2786 428.8 1.2502 0.9943 0.2312 0.01661 6.617 × 10−5 7.731 × 10−6 1.288 1.666 0.00196 0.000951 150 433.2 33.39 1.4730 415.2 1.2785 1.0533 0.2237 0.01702 6.244 × 10−5 7.867 × 10−6 1.285 1.753 0.00211 0.000872 160 492.7 32.72 1.6940 400.8 1.3120 1.1214 0.2163 0.01744 5.900 × 10−5 8.006 × 10−6 1.288 1.853 0.00228 0.000794 170 558.2 32.01 1.9460 385.4 1.3523 1.2012 0.2089 0.01786 5.578 × 10−5 8.142 × 10−6 1.300 1.971 0.00249 0.000716 180 630.1 31.26 2.2346 369.1 1.4015 1.2965 0.2014 0.01829 5.278 × 10−5 8.281 × 10−6 1.322 2.113 0.00274 0.000638 190 708.6 30.47 2.5670 351.6 1.4624 1.4128 0.1940 0.01874 5.000 × 10−5 8.419 × 10−6 1.357 2.286 0.00306 0.000562 200 794.4 29.62 2.9527 332.7 1.5397 1.5586 0.1866 0.01919 4.742 × 10−5 8.561 × 10−6 1.409 2.503 0.00348 0.000486 210 887.9 28.70 3.4053 312.0 1.6411 1.7473 0.1791 0.01966 4.500 × 10−5 8.703 × 10−6 1.484 2.784 0.00403 0.000411 220 989.5 27.69 3.9440 289.2 1.7798 2.0022 0.1717 0.02015 4.275 × 10−5 8.844 × 10−6 1.595 3.164 0.00480 0.000338 230 1099.8 25.57 4.5987 263.5 1.9824 2.3659 0.1643 0.02065 4.064 × 10−5 8.989 × 10−6 1.765 3.707 0.00594 0.000265 240 1219.4 25.28 5.4197 234.0 2.3100 2.9264 0.1568 0.02119 3.864 × 10−5 9.136 × 10−6 2.049 4.542 0.00784 0.000194 Note 1: Kinematic viscosity 𝜈 and thermal diffusivity 𝛼 can be calculated from their definitions, 𝜈 = 𝜇/𝜌 and 𝛼 = k/𝜌cp = 𝜈/Pr. The properties listed here (except the vapor density) can be used at any pressures with negligible error except at temperatures near the critical-point value.
Note 2: The unit Btu/lbm·°F for specific heat is equivalent to Btu/lbm·R, and the unit Btu/h·ft·°F for thermal conductivity is equivalent to Btu/h·ft·R.
Source: Data generated from the EES software developed by S. A. Klein and F. L. Alvarado. Original sources: Tillner-Roth, Harms-Watzenterg, and Baehr, “Eine neue Fundamentalgleichung fur Ammoniak,” DKV-Tagungsbericht 20: 167–181, 1993; Liley and Desai, “Thermophysical Properties of Refrigerants,” ASHRAE, 1993, ISBN 1-1883413-10-9.
cen96537_app2_965_978.indd 970 16/01/17 12:03 pm 971 APPENDIX 2 TABLE A–6E Properties of saturated propane Volume Enthalpy Specific Thermal Prandtl Expansion Surface Saturation Density of Heat Conductivity Dynamic Viscosity Number Coefficient Tension, Temp. Pressure 𝜌, lbm/ft3 Vaporization cp, Btu/lbm·R k, Btu/h·ft·R 𝜇, lbm/ft·s Pr 𝛽, 1/R lbf/ft T, °F P, psia Liquid Vapor hfg, Btu/lbm Liquid Vapor Liquid Vapor Liquid Vapor Liquid Vapor Liquid Liquid −200 0.0201 42.06 0.0003 217.7 0.4750 0.2595 0.1073 0.00313 5.012 × 10−4 2.789 × 10−6 7.991 0.833 0.00083 0.001890 −180 0.0752 41.36 0.0011 213.4 0.4793 0.2680 0.1033 0.00347 3.941 × 10−4 2.975 × 10−6 6.582 0.826 0.00086 0.001780 −160 0.2307 40.65 0.0032 209.1 0.4845 0.2769 0.0992 0.00384 3.199 × 10−4 3.164 × 10−6 5.626 0.821 0.00088 0.001671 −140 0.6037 39.93 0.0078 204.8 0.4907 0.2866 0.0949 0.00423 2.660 × 10−4 3.358 × 10−6 4.951 0.818 0.00091 0.001563 −120 1.389 39.20 0.0170 200.5 0.4982 0.2971 0.0906 0.00465 2.252 × 10−4 3.556 × 10−6 4.457 0.817 0.00094 0.001455 −100 2.878 38.46 0.0334 196.1 0.5069 0.3087 0.0863 0.00511 1.934 × 10−4 3.756 × 10−6 4.087 0.817 0.00097 0.001349 −90 4.006 38.08 0.0453 193.9 0.5117 0.3150 0.0842 0.00534 1.799 × 10−4 3.858 × 10−6 3.936 0.819 0.00099 0.001297 −80 5.467 37.70 0.0605 191.6 0.5169 0.3215 0.0821 0.00559 1.678 × 10−4 3.961 × 10−6 3.803 0.820 0.00101 0.001244 −70 7.327 37.32 0.0793 189.3 0.5224 0.3284 0.0800 0.00585 1.569 × 10−4 4.067 × 10−6 3.686 0.822 0.00104 0.001192 −60 9.657 36.93 0.1024 186.9 0.5283 0.3357 0.0780 0.00611 1.469 × 10−4 4.172 × 10−6 3.582 0.825 0.00106 0.001140 −50 12.54 36.54 0.1305 184.4 0.5345 0.3433 0.0760 0.00639 1.378 × 10−4 4.278 × 10−6 3.490 0.828 0.00109 0.001089 −40 16.05 36.13 0.1641 181.9 0.5392 0.3513 0.0740 0.00668 1.294 × 10−4 4.386 × 10−6 3.395 0.831 0.00112 0.001038 −30 20.29 35.73 0.2041 179.3 0.5460 0.3596 0.0721 0.00697 1.217 × 10−4 4.497 × 10−6 3.320 0.835 0.00115 0.000987 −20 25.34 35.31 0.2512 176.6 0.5531 0.3684 0.0702 0.00728 1.146 × 10−4 4.611 × 10−6 3.253 0.840 0.00119 0.000937 −10 31.3 34.89 0.3063 173.8 0.5607 0.3776 0.0683 0.00761 1.079 × 10−4 4.725 × 10−6 3.192 0.845 0.00123 0.000887 0 38.28 34.46 0.3703 170.9 0.5689 0.3874 0.0665 0.00794 1.018 × 10−4 4.842 × 10−6 3.137 0.850 0.00127 0.000838 10 46.38 34.02 0.4441 167.9 0.5775 0.3976 0.0647 0.00829 9.606 × 10−5 4.961 × 10−6 3.088 0.857 0.00132 0.000789 20 55.7 33.56 0.5289 164.8 0.5867 0.4084 0.0629 0.00865 9.067 × 10−5 5.086 × 10−6 3.043 0.864 0.00138 0.000740 30 66.35 33.10 0.6259 161.6 0.5966 0.4199 0.0612 0.00903 8.561 × 10−5 5.211 × 10−6 3.003 0.873 0.00144 0.000692 40 78.45 32.62 0.7365 158.1 0.6072 0.4321 0.0595 0.00942 8.081 × 10−5 5.342 × 10−6 2.967 0.882 0.00151 0.000644 50 92.12 32.13 0.8621 154.6 0.6187 0.4452 0.0579 0.00983 7.631 × 10−5 5.478 × 10−6 2.935 0.893 0.00159 0.000597 60 107.5 31.63 1.0046 150.8 0.6311 0.4593 0.0563 0.01025 7.200 × 10−5 5.617 × 10−6 2.906 0.906 0.00168 0.000551 70 124.6 31.11 1.1659 146.8 0.6447 0.4746 0.0547 0.01070 6.794 × 10−5 5.764 × 10−6 2.881 0.921 0.00179 0.000505 80 143.7 30.56 1.3484 142.7 0.6596 0.4915 0.0532 0.01116 6.406 × 10−5 5.919 × 10−6 2.860 0.938 0.00191 0.000460 90 164.8 30.00 1.5549 138.2 0.6762 0.5103 0.0517 0.01165 6.033 × 10−5 6.081 × 10−6 2.843 0.959 0.00205 0.000416 100 188.1 29.41 1.7887 133.6 0.6947 0.5315 0.0501 0.01217 5.675 × 10−5 6.256 × 10−6 2.831 0.984 0.00222 0.000372 120 241.8 28.13 2.3562 123.2 0.7403 0.5844 0.0472 0.01328 5.000 × 10−5 6.644 × 10−6 2.825 1.052 0.00267 0.000288 140 306.1 26.69 3.1003 111.1 0.7841 0.6613 0.0442 0.01454 4.358 × 10−5 7.111 × 10−6 2.784 1.164 0.00338 0.000208 160 382.4 24.98 4.1145 96.4 0.8696 0.7911 0.0411 0.01603 3.733 × 10−5 7.719 × 10−6 2.845 1.371 0.00459 0.000133 180 472.9 22.79 5.6265 77.1 1.1436 1.0813 0.0376 0.01793 3.083 × 10−5 8.617 × 10−6 3.380 1.870 0.00791 0.000065 Note 1: Kinematic viscosity 𝜈 and thermal diffusivity 𝛼 can be calculated from their definitions, 𝜈 = 𝜇/𝜌 and 𝛼 = k/𝜌cp = 𝜈/Pr. The properties listed here (except the vapor density) can be used at any pressures with negligible error except at temperatures near the critical-point value.
Note 2: The unit Btu/lbm·°F for specific heat is equivalent to Btu/lbm·R, and the unit Btu/h·ft·°F for thermal conductivity is equivalent to Btu/h·ft·R.
Source: Data generated from the EES software developed by S. A. Klein and F. L. Alvarado. Original sources: Reiner Tillner-Roth, “Fundamental Equations of State,” Shaker, Verlag, Aachan, 1998; B. A. Younglove and J. F. Ely, “Thermophysical Properties of Fluids. II Methane, Ethane, Propane, Isobutane, and Normal Butane,” J. Phys. Chem. Ref. Data, Vol. 16, No. 4, 1987; G. R. Somayajulu, “A Generalized Equation for Surface Tension from the Triple-Point to the Critical-Point,” International Journal of Thermophysics, Vol. 9, No. 4, 1988.
cen96537_app2_965_978.indd 971 16/01/17 12:03 pm 972 PROPERTY TABLES AND CHARTS TABLE A–7E Properties of liquids Volume Specific Thermal Thermal Dynamic Kinematic Prandtl Expansion Temp. Density Heat cp, Conductivity Diffusivity Viscosity Viscosity Number Coeff. 𝛽, T, °F 𝜌, lbm/ft3 Btu/lbm·R k, Btu/h·ft·R 𝛼, ft2/s 𝜇, lbm/ft·s 𝜈, ft2/s Pr 1/R Methane (CH4) −280 27.41 0.8152 0.1205 1.497 × 10−6 1.057 × 10−4 3.857 × 10−6 2.575 0.00175 −260 26.43 0.8301 0.1097 1.389 × 10−6 8.014 × 10−5 3.032 × 10−6 2.183 0.00192 −240 25.39 0.8523 0.0994 1.276 × 10−6 6.303 × 10−5 2.482 × 10−6 1.945 0.00215 −220 24.27 0.8838 0.0896 1.159 × 10−6 5.075 × 10−5 2.091 × 10−6 1.803 0.00247 −200 23.04 0.9314 0.0801 1.036 × 10−6 4.142 × 10−5 1.798 × 10−6 1.734 0.00295 −180 21.64 1.010 0.0709 9.008 × 10−7 3.394 × 10−5 1.568 × 10−6 1.741 0.00374 −160 19.99 1.158 0.0616 7.397 × 10−7 2.758 × 10−5 1.379 × 10−6 1.865 0.00526 −140 17.84 1.542 0.0518 5.234 × 10−7 2.168 × 10−5 1.215 × 10−6 2.322 0.00943 Methanol [CH3(OH)] 70 49.15 0.6024 0.1148 1.076 × 10−6 3.872 × 10−4 7.879 × 10−6 7.317 0.000656 90 48.50 0.6189 0.1143 1.057 × 10−6 3.317 × 10−4 6.840 × 10−6 6.468 0.000671 110 47.85 0.6373 0.1138 1.036 × 10−6 2.872 × 10−4 6.005 × 10−6 5.793 0.000691 130 47.18 0.6576 0.1133 1.014 × 10−6 2.513 × 10−4 5.326 × 10−6 5.250 0.000716 150 46.50 0.6796 0.1128 9.918 × 10−7 2.218 × 10−4 4.769 × 10−6 4.808 0.000749 170 45.80 0.7035 0.1124 9.687 × 10−7 1.973 × 10−4 4.308 × 10−6 4.447 0.000789 Isobutane (R600a) −150 42.75 0.4483 0.0799 1.157 × 10−6 6.417 × 10−4 1.500 × 10−5 12.96 0.000785 −100 41.06 0.4721 0.0782 1.120 × 10−6 3.669 × 10−4 8.939 × 10−6 7.977 0.000836 −50 39.31 0.4986 0.0731 1.036 × 10−6 2.376 × 10−4 6.043 × 10−6 5.830 0.000908 0 37.48 0.5289 0.0664 9.299 × 10−7 1.651 × 10−4 4.406 × 10−6 4.738 0.001012 50 35.52 0.5643 0.0591 8.187 × 10−7 1.196 × 10−4 3.368 × 10−6 4.114 0.001169 100 33.35 0.6075 0.0521 7.139 × 10−7 8.847 × 10−5 2.653 × 10−6 3.716 0.001421 150 30.84 0.6656 0.0457 6.188 × 10−7 6.558 × 10−5 2.127 × 10−6 3.437 0.001883 200 27.73 0.7635 0.0400 5.249 × 10−7 4.750 × 10−5 1.713 × 10−6 3.264 0.002970 Glycerin 32 79.65 0.5402 0.163 1.052 × 10−6 7.047 0.08847 84101 40 79.49 0.5458 0.1637 1.048 × 10−6 4.803 0.06042 57655 50 79.28 0.5541 0.1645 1.040 × 10−6 2.850 0.03594 34561 60 79.07 0.5632 0.1651 1.029 × 10−6 1.547 0.01956 18995 70 78.86 0.5715 0.1652 1.018 × 10−6 0.9422 0.01195 11730 80 78.66 0.5794 0.1652 1.007 × 10−6 0.5497 0.00699 6941 90 78.45 0.5878 0.1652 9.955 × 10−7 0.3756 0.004787 4809 100 78.24 0.5964 0.1653 9.841 × 10−7 0.2277 0.00291 2957 Engine Oil (unused) 32 56.12 0.4291 0.0849 9.792 × 10−7 2.563 4.566 × 10−2 46636 0.000389 50 55.79 0.4395 0.08338 9.448 × 10−7 1.210 2.169 × 10−2 22963 0.000389 75 55.3 0.4531 0.08378 9.288 × 10−7 0.4286 7.751 × 10−3 8345 0.000389 100 54.77 0.4669 0.08367 9.089 × 10−7 0.1630 2.977 × 10−3 3275 0.000389 125 54.24 0.4809 0.08207 8.740 × 10−7 7.617 × 10−2 1.404 × 10−3 1607 0.000389 150 53.73 0.4946 0.08046 8.411 × 10−7 3.833 × 10−2 7.135 × 10−4 848.3 0.000389 200 52.68 0.5231 0.07936 7.999 × 10−7 1.405 × 10−2 2.668 × 10−4 333.6 0.000389 250 51.71 0.5523 0.07776 7.563 × 10−7 6.744 × 10−3 1.304 × 10−4 172.5 0.000389 300 50.63 0.5818 0.07673 7.236 × 10−7 3.661 × 10−3 7.232 × 10−5 99.94 0.000389 Source: Data generated from the EES software developed by S. A. Klein and F. L. Alvarado. Originally based on various sources.
cen96537_app2_965_978.indd 972 16/01/17 12:03 pm 973 APPENDIX 2 TABLE A–8E Properties of liquid metals Volume Specific Thermal Thermal Dynamic Kinematic Prandtl Expansion Temp. Density Heat cp, Conductivity Diffusivity Viscosity Viscosity Number Coeff. 𝛽, T, °F 𝜌, lbm/ft3 Btu/lbm·R k, Btu/h·ft·R 𝛼, ft2/s 𝜇, lbm/ft·s 𝜈, ft2/s Pr 1/R Mercury (Hg) Melting Point: −38°F 32 848.7 0.03353 4.727 4.614 × 10−5 1.133 × 10−3 1.335 × 10−6 0.02895 1.005 × 10−4 50 847.2 0.03344 4.805 4.712 × 10−5 1.092 × 10−3 1.289 × 10−6 0.02737 1.005 × 10−4 100 842.9 0.03319 5.015 4.980 × 10−5 9.919 × 10−4 1.176 × 10−6 0.02363 1.005 × 10−4 150 838.7 0.03298 5.221 5.244 × 10−5 9.122 × 10−4 1.087 × 10−6 0.02074 1.005 × 10−4 200 834.5 0.03279 5.422 5.504 × 10−5 8.492 × 10−4 1.017 × 10−6 0.01849 1.005 × 10−4 300 826.2 0.03252 5.815 6.013 × 10−5 7.583 × 10−4 9.180 × 10−7 0.01527 1.005 × 10−4 400 817.9 0.03236 6.184 6.491 × 10−5 6.972 × 10−4 8.524 × 10−7 0.01313 1.008 × 10−4 500 809.6 0.03230 6.518 6.924 × 10−5 6.525 × 10−4 8.061 × 10−7 0.01164 1.018 × 10−4 600 801.3 0.03235 6.839 7.329 × 10−5 6.186 × 10−4 7.719 × 10−7 0.01053 1.035 × 10−4 Bismuth (Bi) Melting Point: 520°F 700 620.7 0.03509 9.361 1.193 × 10−4 1.001 × 10−3 1.614 × 10−6 0.01352 800 616.5 0.03569 9.245 1.167 × 10−4 9.142 × 10−4 1.482 × 10−6 0.01271 900 612.2 0.0363 9.129 1.141 × 10−4 8.267 × 10−4 1.350 × 10−6 0.01183 1000 608.0 0.0369 9.014 1.116 × 10−4 7.392 × 10−4 1.215 × 10−6 0.0109 1100 603.7 0.0375 9.014 1.105 × 10−4 6.872 × 10−4 1.138 × 10−6 0.01029 Lead (Pb) Melting Point: 621°F 700 658 0.03797 9.302 1.034 × 10−4 1.612 × 10−3 2.450 × 10−6 0.02369 800 654 0.03750 9.157 1.037 × 10−4 1.453 × 10−3 2.223 × 10−6 0.02143 900 650 0.03702 9.013 1.040 × 10−4 1.296 × 10−3 1.994 × 10−6 0.01917 1000 645.7 0.03702 8.912 1.035 × 10−4 1.202 × 10−3 1.862 × 10−6 0.01798 1100 641.5 0.03702 8.810 1.030 × 10−4 1.108 × 10−3 1.727 × 10−6 0.01676 1200 637.2 0.03702 8.709 1.025 × 10−4 1.013 × 10−3 1.590 × 10−6 0.01551 Sodium (Na) Melting Point: 208°F 300 57.13 0.3258 48.19 7.192 × 10−4 4.136 × 10−4 7.239 × 10−6 0.01007 400 56.28 0.3219 46.58 7.142 × 10−4 3.572 × 10−4 6.350 × 10−6 0.008891 500 55.42 0.3181 44.98 7.087 × 10−4 3.011 × 10−4 5.433 × 10−6 0.007667 600 54.56 0.3143 43.37 7.026 × 10−4 2.448 × 10−4 4.488 × 10−6 0.006387 800 52.85 0.3089 40.55 6.901 × 10−4 1.772 × 10−4 3.354 × 10−6 0.004860 1000 51.14 0.3057 38.12 6.773 × 10−4 1.541 × 10−4 3.014 × 10−6 0.004449 Potassium (K) Melting Point: 147°F 300 50.40 0.1911 26.00 7.500 × 10−4 2.486 × 10−4 4.933 × 10−6 0.006577 400 49.58 0.1887 25.37 7.532 × 10−4 2.231 × 10−4 4.500 × 10−6 0.005975 500 48.76 0.1863 24.73 7.562 × 10−4 1.976 × 10−4 4.052 × 10−6 0.005359 600 47.94 0.1839 24.09 7.591 × 10−4 1.721 × 10−4 3.589 × 10−6 0.004728 800 46.31 0.1791 22.82 7.643 × 10−4 1.210 × 10−4 2.614 × 10−6 0.003420 1000 44.62 0.1791 21.34 7.417 × 10−4 1.075 × 10−4 2.409 × 10−6 0.003248 Sodium–Potassium (%22Na–%78K) Melting Point: 12°F 200 52.99 0.2259 14.79 3.432 × 10−4 3.886 × 10−4 7.331 × 10−6 0.02136 300 52.16 0.2230 14.99 3.580 × 10−4 3.467 × 10−4 6.647 × 10−6 0.01857 400 51.32 0.2201 15.19 3.735 × 10−4 3.050 × 10−4 5.940 × 10−6 0.0159 600 49.65 0.2143 15.59 4.070 × 10−4 2.213 × 10−4 4.456 × 10−6 0.01095 800 47.99 0.2100 15.95 4.396 × 10−4 1.539 × 10−4 3.207 × 10−6 0.007296 1000 46.36 0.2103 16.20 4.615 × 10−4 1.353 × 10−4 2.919 × 10−6 0.006324 Source: Data generated from the EES software developed by S. A. Klein and F. L. Alvarado. Originally based on various sources.
cen96537_app2_965_978.indd 973 16/01/17 12:03 pm 974 PROPERTY TABLES AND CHARTS TABLE A–9E Properties of air at 1 atm pressure Specific Thermal Thermal Dynamic Kinematic Prandtl Temp. Density Heat cp, Conductivity Diffusivity Viscosity Viscosity Number T, °F 𝜌, lbm/ft3 Btu/lbm·R k, Btu/h·ft·R 𝛼, ft2/s 𝜇, lbm/ft·s 𝜈, ft2/s Pr −300 0.24844 0.5072 0.00508 1.119 × 10−5 4.039 × 10−6 1.625 × 10−5 1.4501 −200 0.15276 0.2247 0.00778 6.294 × 10−5 6.772 × 10−6 4.433 × 10−5 0.7042 −100 0.11029 0.2360 0.01037 1.106 × 10−4 9.042 × 10−6 8.197 × 10−5 0.7404 −50 0.09683 0.2389 0.01164 1.397 × 10−4 1.006 × 10−5 1.039 × 10−4 0.7439 0 0.08630 0.2401 0.01288 1.726 × 10−4 1.102 × 10−5 1.278 × 10−4 0.7403 10 0.08446 0.2402 0.01312 1.797 × 10−4 1.121 × 10−5 1.328 × 10−4 0.7391 20 0.08270 0.2403 0.01336 1.868 × 10−4 1.140 × 10−5 1.379 × 10−4 0.7378 30 0.08101 0.2403 0.01361 1.942 × 10−4 1.158 × 10−5 1.430 × 10−4 0.7365 40 0.07939 0.2404 0.01385 2.016 × 10−4 1.176 × 10−5 1.482 × 10−4 0.7350 50 0.07783 0.2404 0.01409 2.092 × 10−4 1.194 × 10−5 1.535 × 10−4 0.7336 60 0.07633 0.2404 0.01433 2.169 × 10−4 1.212 × 10−5 1.588 × 10−4 0.7321 70 0.07489 0.2404 0.01457 2.248 × 10−4 1.230 × 10−5 1.643 × 10−4 0.7306 80 0.07350 0.2404 0.01481 2.328 × 10−4 1.247 × 10−5 1.697 × 10−4 0.7290 90 0.07217 0.2404 0.01505 2.409 × 10−4 1.265 × 10−5 1.753 × 10−4 0.7275 100 0.07088 0.2405 0.01529 2.491 × 10−4 1.281 × 10−5 1.809 × 10−4 0.7260 110 0.06963 0.2405 0.01552 2.575 × 10−4 1.299 × 10−5 1.866 × 10−4 0.7245 120 0.06843 0.2405 0.01576 2.660 × 10−4 1.316 × 10−5 1.923 × 10−4 0.7230 130 0.06727 0.2405 0.01599 2.746 × 10−4 1.332 × 10−5 1.981 × 10−4 0.7216 140 0.06615 0.2406 0.01623 2.833 × 10−4 1.349 × 10−5 2.040 × 10−4 0.7202 150 0.06507 0.2406 0.01646 2.921 × 10−4 1.365 × 10−5 2.099 × 10−4 0.7188 160 0.06402 0.2406 0.01669 3.010 × 10−4 1.382 × 10−5 2.159 × 10−4 0.7174 170 0.06300 0.2407 0.01692 3.100 × 10−4 1.398 × 10−5 2.220 × 10−4 0.7161 180 0.06201 0.2408 0.01715 3.191 × 10−4 1.414 × 10−5 2.281 × 10−4 0.7148 190 0.06106 0.2408 0.01738 3.284 × 10−4 1.430 × 10−5 2.343 × 10−4 0.7136 200 0.06013 0.2409 0.01761 3.377 × 10−4 1.446 × 10−5 2.406 × 10−4 0.7124 250 0.05590 0.2415 0.01874 3.857 × 10−4 1.524 × 10−5 2.727 × 10−4 0.7071 300 0.05222 0.2423 0.01985 4.358 × 10−4 1.599 × 10−5 3.063 × 10−4 0.7028 350 0.04899 0.2433 0.02094 4.879 × 10−4 1.672 × 10−5 3.413 × 10−4 0.6995 400 0.04614 0.2445 0.02200 5.419 × 10−4 1.743 × 10−5 3.777 × 10−4 0.6971 450 0.04361 0.2458 0.02305 5.974 × 10−4 1.812 × 10−5 4.154 × 10−4 0.6953 500 0.04134 0.2472 0.02408 6.546 × 10−4 1.878 × 10−5 4.544 × 10−4 0.6942 600 0.03743 0.2503 0.02608 7.732 × 10−4 2.007 × 10−5 5.361 × 10−4 0.6934 700 0.03421 0.2535 0.02800 8.970 × 10−4 2.129 × 10−5 6.225 × 10−4 0.6940 800 0.03149 0.2568 0.02986 1.025 × 10−3 2.247 × 10−5 7.134 × 10−4 0.6956 900 0.02917 0.2599 0.03164 1.158 × 10−3 2.359 × 10−5 8.087 × 10−4 0.6978 1000 0.02718 0.2630 0.03336 1.296 × 10−3 2.467 × 10−5 9.080 × 10−4 0.7004 1500 0.02024 0.2761 0.04106 2.041 × 10−3 2.957 × 10−5 1.460 × 10−3 0.7158 2000 0.01613 0.2855 0.04752 2.867 × 10−3 3.379 × 10−5 2.095 × 10−3 0.7308 2500 0.01340 0.2922 0.05309 3.765 × 10−3 3.750 × 10−5 2.798 × 10−3 0.7432 3000 0.01147 0.2972 0.05811 4.737 × 10−3 4.082 × 10−5 3.560 × 10−3 0.7516 3500 0.01002 0.3010 0.06293 5.797 × 10−3 4.381 × 10−5 4.373 × 10−3 0.7543 4000 0.00889 0.3040 0.06789 6.975 × 10−3 4.651 × 10−5 5.229 × 10−3 0.7497 Note: For ideal gases, the properties cp, k, 𝜇, and Pr are independent of pressure. The properties 𝜌, 𝜈, and 𝛼 at a pressure P (in atm) other than 1 atm are determined by multi plying the values of 𝜌 at the given temperature by P and by dividing 𝜈 and 𝛼 by P.
Source: Data generated from the EES software developed by S. A. Klein and F. L. Alvarado. Original sources: Keenan, Chao, Kaye, Gas Tables, Wiley, 1983; and Thermo physical Properties of Matter, Vol. 3: Thermal Conductivity, Y. S. Touloukian, P. E. Liley, S. C. Saxena, Vol. 11: Viscosity, Y. S. Touloukian, S. C. Saxena, and P. Hester mans, IFI/Plenun, NY, 1970, ISBN 0-306067020-8.
cen96537_app2_965_978.indd 974 16/01/17 12:03 pm 975 APPENDIX 2 TABLE A–10E Properties of gases at 1 atm pressure Specific Thermal Thermal Dynamic Kinematic Prandtl Temp. Density Heat cp, Conductivity Diffusivity Viscosity Viscosity Number T, °F 𝜌, lbm/ft3 Btu/lbm·R k, Btu/h·ft·R 𝛼, ft2/s 𝜇, lbm/ft·s 𝜈, ft2/s Pr Carbon Dioxide, CO2 −50 0.14712 0.1797 0.00628 6.600 × 10−5 7.739 × 10−6 5.261 × 10−5 0.7970 0 0.13111 0.1885 0.00758 8.522 × 10−5 8.661 × 10−6 6.606 × 10−5 0.7751 50 0.11825 0.1965 0.00888 1.061 × 10−4 9.564 × 10−6 8.086 × 10−5 0.7621 100 0.10769 0.2039 0.01017 1.286 × 10−4 1.045 × 10−5 9.703 × 10−5 0.7543 200 0.09136 0.2171 0.01273 1.784 × 10−4 1.217 × 10−5 1.332 × 10−4 0.7469 300 0.07934 0.2284 0.01528 2.341 × 10−4 1.382 × 10−5 1.743 × 10−4 0.7445 500 0.06280 0.2473 0.02027 3.626 × 10−4 1.696 × 10−5 2.700 × 10−4 0.7446 1000 0.04129 0.2796 0.03213 7.733 × 10−4 2.381 × 10−5 5.767 × 10−4 0.7458 1500 0.03075 0.2995 0.04281 1.290 × 10−3 2.956 × 10−5 9.610 × 10−4 0.7445 2000 0.02450 0.3124 0.05193 1.885 × 10−3 3.451 × 10−5 1.408 × 10−3 0.7474 Carbon Monoxide, CO −50 0.09363 0.2571 0.01118 1.290 × 10−4 9.419 × 10−6 1.005 × 10−4 0.7798 0 0.08345 0.2523 0.01240 1.636 × 10−4 1.036 × 10−5 1.242 × 10−4 0.7593 50 0.07526 0.2496 0.01359 2.009 × 10−4 1.127 × 10−5 1.498 × 10−4 0.7454 100 0.06854 0.2484 0.01476 2.408 × 10−4 1.214 × 10−5 1.772 × 10−4 0.7359 200 0.05815 0.2485 0.01702 3.273 × 10−4 1.379 × 10−5 2.372 × 10−4 0.7247 300 0.05049 0.2505 0.01920 4.217 × 10−4 1.531 × 10−5 3.032 × 10−4 0.7191 500 0.03997 0.2567 0.02331 6.311 × 10−4 1.802 × 10−5 4.508 × 10−4 0.7143 1000 0.02628 0.2732 0.03243 1.254 × 10−3 2.334 × 10−5 8.881 × 10−4 0.7078 1500 0.01957 0.2862 0.04049 2.008 × 10−3 2.766 × 10−5 1.413 × 10−3 0.7038 2000 0.01559 0.2958 0.04822 2.903 × 10−3 3.231 × 10−5 2.072 × 10−3 0.7136 Methane, CH4 −50 0.05363 0.5335 0.01401 1.360 × 10−4 5.861 × 10−6 1.092 × 10−4 0.8033 0 0.04779 0.5277 0.01616 1.780 × 10−4 6.506 × 10−6 1.361 × 10−4 0.7649 50 0.04311 0.5320 0.01839 2.228 × 10−4 7.133 × 10−6 1.655 × 10−4 0.7428 100 0.03925 0.5433 0.02071 2.698 × 10−4 7.742 × 10−6 1.972 × 10−4 0.7311 200 0.03330 0.5784 0.02559 3.690 × 10−4 8.906 × 10−6 2.674 × 10−4 0.7245 300 0.02892 0.6226 0.03077 4.748 × 10−4 1.000 × 10−5 3.457 × 10−4 0.7283 500 0.02289 0.7194 0.04195 7.075 × 10−4 1.200 × 10−5 5.244 × 10−4 0.7412 1000 0.01505 0.9438 0.07346 1.436 × 10−3 1.620 × 10−5 1.076 × 10−3 0.7491 1500 0.01121 1.1162 0.10766 2.390 × 10−3 1.974 × 10−5 1.760 × 10−3 0.7366 2000 0.00893 1.2419 0.14151 3.544 × 10−3 2.327 × 10−5 2.605 × 10−3 0.7353 Hydrogen, H2 −50 0.00674 3.0603 0.08246 1.110 × 10−3 4.969 × 10−6 7.373 × 10−4 0.6638 0 0.00601 3.2508 0.09049 1.287 × 10−3 5.381 × 10−6 8.960 × 10−4 0.6960 50 0.00542 3.3553 0.09818 1.500 × 10−3 5.781 × 10−6 1.067 × 10−3 0.7112 100 0.00493 3.4118 0.10555 1.742 × 10−3 6.167 × 10−6 1.250 × 10−3 0.7177 200 0.00419 3.4549 0.11946 2.295 × 10−3 6.911 × 10−6 1.652 × 10−3 0.7197 300 0.00363 3.4613 0.13241 2.924 × 10−3 7.622 × 10−6 2.098 × 10−3 0.7174 500 0.00288 3.4572 0.15620 4.363 × 10−3 8.967 × 10−6 3.117 × 10−3 0.7146 1000 0.00189 3.5127 0.20989 8.776 × 10−3 1.201 × 10−5 6.354 × 10−3 0.7241 1500 0.00141 3.6317 0.26381 1.432 × 10−2 1.477 × 10−5 1.048 × 10−2 0.7323 2000 0.00112 3.7656 0.31923 2.098 × 10−2 1.734 × 10−5 1.544 × 10−2 0.7362 (continued) cen96537_app2_965_978.indd 975 16/01/17 12:03 pm 976 PROPERTY TABLES AND CHARTS TABLE A–10E (Continued) Properties of gases at 1 atm pressure Specific Thermal Thermal Dynamic Kinematic Prandtl Temp. Density Heat cp, Conductivity Diffusivity Viscosity Viscosity Number T, °F 𝜌, lbm/ft3 Btu/lbm·R k, Btu/h·ft·R 𝛼, ft2/s 𝜇, lbm/ft·s 𝜈, ft2/s Pr Nitrogen, N2 −50 0.09364 0.2320 0.01176 1.504 × 10−4 9.500 × 10−6 1.014 × 10−4 0.6746 0 0.08346 0.2441 0.01300 1.773 × 10−4 1.043 × 10−5 1.251 × 10−4 0.7056 50 0.07527 0.2480 0.01420 2.113 × 10−4 1.134 × 10−5 1.507 × 10−4 0.7133 100 0.06854 0.2489 0.01537 2.502 × 10−4 1.221 × 10−5 1.783 × 10−4 0.7126 200 0.05815 0.2487 0.01760 3.379 × 10−4 1.388 × 10−5 2.387 × 10−4 0.7062 300 0.05050 0.2492 0.01970 4.349 × 10−4 1.543 × 10−5 3.055 × 10−4 0.7025 500 0.03997 0.2535 0.02359 6.466 × 10−4 1.823 × 10−5 4.559 × 10−4 0.7051 1000 0.02628 0.2697 0.03204 1.255 × 10−3 2.387 × 10−5 9.083 × 10−4 0.7232 1500 0.01958 0.2831 0.04002 2.006 × 10−3 2.829 × 10−5 1.445 × 10−3 0.7202 2000 0.01560 0.2927 0.04918 2.992 × 10−3 3.212 × 10−5 2.059 × 10−3 0.6882 Oxygen, O2 −50 0.10697 0.2331 0.01216 1.355 × 10−4 1.104 × 10−5 1.032 × 10−4 0.7622 0 0.09533 0.2245 0.01346 1.747 × 10−4 1.218 × 10−5 1.277 × 10−4 0.7312 50 0.08598 0.2209 0.01475 2.157 × 10−4 1.326 × 10−5 1.543 × 10−4 0.7152 100 0.07830 0.2200 0.01601 2.582 × 10−4 1.429 × 10−5 1.826 × 10−4 0.7072 200 0.06643 0.2221 0.01851 3.484 × 10−4 1.625 × 10−5 2.446 × 10−4 0.7020 300 0.05768 0.2262 0.02096 4.463 × 10−4 1.806 × 10−5 3.132 × 10−4 0.7018 500 0.04566 0.2352 0.02577 6.665 × 10−4 2.139 × 10−5 4.685 × 10−4 0.7029 1000 0.03002 0.2520 0.03698 1.357 × 10−3 2.855 × 10−5 9.509 × 10−4 0.7005 1500 0.02236 0.2626 0.04701 2.224 × 10−3 3.474 × 10−5 1.553 × 10−3 0.6985 2000 0.01782 0.2701 0.05614 3.241 × 10−3 4.035 × 10−5 2.265 × 10−3 0.6988 Water Vapor, H2O −50 0.06022 0.4512 0.00797 8.153 × 10−5 4.933 × 10−6 8.192 × 10−5 1.0050 0 0.05367 0.4484 0.00898 1.036 × 10−4 5.592 × 10−6 1.041 × 10−4 1.0049 50 0.04841 0.4472 0.01006 1.291 × 10−4 6.261 × 10−6 1.293 × 10−4 1.0018 100 0.04408 0.4473 0.01121 1.579 × 10−4 6.942 × 10−6 1.574 × 10−4 0.9969 200 0.03740 0.4503 0.01372 2.263 × 10−4 8.333 × 10−6 2.228 × 10−4 0.9845 300 0.03248 0.4557 0.01648 3.093 × 10−4 9.756 × 10−6 3.004 × 10−4 0.9713 500 0.02571 0.4707 0.02267 5.204 × 10−4 1.267 × 10−5 4.931 × 10−4 0.9475 1000 0.01690 0.5167 0.04134 1.314 × 10−3 2.014 × 10−5 1.191 × 10−3 0.9063 1500 0.01259 0.5625 0.06315 2.477 × 10−3 2.742 × 10−5 2.178 × 10−3 0.8793 2000 0.01003 0.6034 0.08681 3.984 × 10−3 3.422 × 10−5 3.411 × 10−3 0.8563 Note: For ideal gases, the properties cp, k, 𝜇, and Pr are independent of pressure. The properties 𝜌, 𝜈, and 𝛼 at a pressure P (in atm) other than 1 atm are determined by multiplying the values of 𝜌 at the given temperature by P and by dividing 𝜈 and 𝛼 by P.
Source: Data generated from the EES software developed by S. A. Klein and F. L. Alvarado. Originally based on various sources.
cen96537_app2_965_978.indd 976 16/01/17 12:03 pm 977 APPENDIX 2 TABLE A–11E Properties of the atmosphere at high altitude Speed of Thermal Altitude, Temperature, Pressure, Gravity, Sound, Density Viscosity Conductivity, ft °F psia g, ft/s2 ft/s 𝜌, lbm/ft3 𝜇, lbm/ft·s Btu/h·ft·R 0 59.00 14.7 32.174 1116 0.07647 1.202 × 10−5 0.0146 500 57.22 14.4 32.173 1115 0.07536 1.199 × 10−5 0.0146 1000 55.43 14.2 32.171 1113 0.07426 1.196 × 10−5 0.0146 1500 53.65 13.9 32.169 1111 0.07317 1.193 × 10−5 0.0145 2000 51.87 13.7 32.168 1109 0.07210 1.190 × 10−5 0.0145 2500 50.09 13.4 32.166 1107 0.07104 1.186 × 10−5 0.0144 3000 48.30 13.2 32.165 1105 0.06998 1.183 × 10−5 0.0144 3500 46.52 12.9 32.163 1103 0.06985 1.180 × 10−5 0.0143 4000 44.74 12.7 32.162 1101 0.06792 1.177 × 10−5 0.0143 4500 42.96 12.5 32.160 1099 0.06690 1.173 × 10−5 0.0142 5000 41.17 12.2 32.159 1097 0.06590 1.170 × 10−5 0.0142 5500 39.39 12.0 32.157 1095 0.06491 1.167 × 10−5 0.0141 6000 37.61 11.8 32.156 1093 0.06393 1.164 × 10−5 0.0141 6500 35.83 11.6 32.154 1091 0.06296 1.160 × 10−5 0.0141 7000 34.05 11.3 32.152 1089 0.06200 1.157 × 10−5 0.0140 7500 32.26 11.1 32.151 1087 0.06105 1.154 × 10−5 0.0140 8000 30.48 10.9 32.149 1085 0.06012 1.150 × 10−5 0.0139 8500 28.70 10.7 32.148 1083 0.05919 1.147 × 10−5 0.0139 9000 26.92 10.5 32.146 1081 0.05828 1.144 × 10−5 0.0138 9500 25.14 10.3 32.145 1079 0.05738 1.140 × 10−5 0.0138 10,000 23.36 10.1 32.145 1077 0.05648 1.137 × 10−5 0.0137 11,000 19.79 9.72 32.140 1073 0.05473 1.130 × 10−5 0.0136 12,000 16.23 9.34 32.137 1069 0.05302 1.124 × 10−5 0.0136 13,000 12.67 8.99 32.134 1065 0.05135 1.117 × 10−5 0.0135 14,000 9.12 8.63 32.131 1061 0.04973 1.110 × 10−5 0.0134 15,000 5.55 8.29 32.128 1057 0.04814 1.104 × 10−5 0.0133 16,000 +1.99 7.97 32.125 1053 0.04659 1.097 × 10−5 0.0132 17,000 −1.58 7.65 32.122 1049 0.04508 1.090 × 10−5 0.0132 18,000 −5.14 7.34 32.119 1045 0.04361 1.083 × 10−5 0.0130 19,000 −8.70 7.05 32.115 1041 0.04217 1.076 × 10−5 0.0129 20,000 −12.2 6.76 32.112 1037 0.04077 1.070 × 10−5 0.0128 22,000 −19.4 6.21 32.106 1029 0.03808 1.056 × 10−5 0.0126 24,000 −26.5 5.70 32.100 1020 0.03553 1.042 × 10−5 0.0124 26,000 −33.6 5.22 32.094 1012 0.03311 1.028 × 10−5 0.0122 28,000 −40.7 4.78 32.088 1003 0.03082 1.014 × 10−5 0.0121 30,000 −47.8 4.37 32.082 995 0.02866 1.000 × 10−5 0.0119 32,000 −54.9 3.99 32.08 987 0.02661 0.986 × 10−5 0.0117 34,000 −62.0 3.63 32.07 978 0.02468 0.971 × 10−5 0.0115 36,000 −69.2 3.30 32.06 969 0.02285 0.956 × 10−5 0.0113 38,000 −69.7 3.05 32.06 968 0.02079 0.955 × 10−5 0.0113 40,000 −69.7 2.73 32.05 968 0.01890 0.955 × 10−5 0.0113 45,000 −69.7 2.148 32.04 968 0.01487 0.955 × 10−5 0.0113 50,000 −69.7 1.691 32.02 968 0.01171 0.955 × 10−5 0.0113 55,000 −69.7 1.332 32.00 968 0.00922 0.955 × 10−5 0.0113 60,000 −69.7 1.048 31.99 968 0.00726 0.955 × 10−5 0.0113 Source: U.S. Standard Atmosphere Supplements, U.S. Government Printing Office, 1966. Based on year-round mean conditions at 45° latitude and varies with the time of the year and the weather patterns. The conditions at sea level (z = 0) are taken to be P = 14.696 psia, T = 59°F, 𝜌 = 0.076474 lbm/ft3, g = 32.1741 ft2/s.
cen96537_app2_965_978.indd 977 16/01/17 12:03 pm This page intentionally left blank G l o s s a r y 979 Note: Boldface color glossary terms correspond to boldface color terms in the text. Italics indicates a term defined elsewhere in the glossary.
Boldface terms without page numbers are concepts that are not defined in the text but are defined or cross-referenced in the glossary for students to review.
absolute pressure: See stress, pressure stress. Contrast with gage pressure.
absolute viscosity: See viscosity.
acceleration field: See field.
adiabatic process: A process with no heat transfer.
advective acceleration: In order to reduce confusion of terminology in flows where buoyancy forces generate convec tive fluid motions, the term “convective acceleration” is often replaced with the term “advective acceleration.” aerodynamics: The application of fluid dynamics to air, land, and water-going vehicles. Often the term is specifically applied to the flow surrounding, and forces and moments on, flight vehicles in air, as opposed to vehicles in water or other liquids (hydrodynamics).
angle of attack: The angle between an airfoil or wing and the free-stream flow velocity vector.
average: An area/volume/time average of a fluid property is the integral of the property over an area/volume/time period divided by the corresponding area/volume/time period. Also called mean.
axisymmetric flow: A flow that when specified appropri ately using cylindrical coordinates (r, 𝜃, x) does not vary in the azimuthal (𝜃) direction. Thus, all partial derivatives in 𝜃 are zero. The flow is therefore either one-dimensional or two-dimensional (see also dimensionality and planar flow).
barometer: A device that measures atmospheric pressure.
basic dimensions: See dimensions.
Bernoulli equation: A useful reduction of conservation of momentum (and conservation of energy) that describes a balance between pressure (flow work), velocity (kinetic energy), and position of fluid particles relative to the gravity vector (potential energy) in regions of a fluid flow where frictional force on fluid particles is negligible compared to pressure force in that region of the flow (see inviscid flow). There are multiple forms of the Bernoulli equation for incompressible vs. compressible, steady vs. nonsteady, and derivations through Newton’s law vs. the first law of thermodynamics. The most commonly used forms are for steady incompressible fluid flow derived through conserva tion of momentum.
bluff (or blunt) body: A moving object with a blunt rear por tion. Bluff bodies have wakes resulting from massive flow separation over the rear of the body.
boundary condition: In solving for flow field variables (velocity, temperature) from governing equations, it is neces sary to mathematically specify a function of the variable at the surface. These mathematical statements are called bound ary conditions. The no-slip condition that the flow velocity must equal the surface velocity at the surface is an example of a boundary condition that is used with the Navier–Stokes equation to solve for the velocity field.
boundary layer: At high Reynolds numbers relatively thin “boundary layers” exist in the flow adjacent to surfaces where the flow is brought to rest (see no-slip condition). Boundary layers are characterized by high shear with the highest velocities away from the surface. Frictional force, viscous stress, and vorticity are significant in boundary layers. The approximate form of the two components of the Navier–Stokes equation, simplified by neglecting the terms that are small within the boundary layer, are called the boundary layer equations. The associated approximation based on the existence of thin boundary layers surrounded by irrotational or inviscid flow is called the boundary layer approximation.
boundary layer approximation: See boundary layer.
boundary layer equations: See boundary layer.
boundary layer thickness measures: Different measures of the thickness of a boundary layer as a function of downstream distance are used in fluid flow analyses. These are: boundary layer thickness: The full thickness of the viscous layer that defines the boundary layer, from the surface to the edge. Defining the edge is difficult to do precisely, so the “edge” of the boundary layer is often defined as the point where the boundary layer velocity is a large fraction of the free-stream velocity (e.g., 𝛿99 is the distance from the surface to the point where the streamwise velocity component is 99 percent of the free-stream velocity).
displacement thickness: A boundary layer thickness measure that quantifies the deflection of fluid streamlines in the direction away from the surface as a result Guest Author: James G. Brasseur, The Pennsylvania State University Note: This glossary covers boldface color terms found in Chapters 1 to 11.
cen96534_glos_979_992.indd 979 01/12/16 10:56 am 980 FLUID MECHANICS of friction-induced reduction in mass flow adjacent to the surface. Displacement thickness (𝛿) is a measure of the thickness of this mass flow rate deficit layer. In all boundary layers, 𝛿 < 𝛿.
momentum thickness: A measure of the layer of highest deficit in momentum flow rate adjacent to the surface as a result of frictional resisting force (shear stress). Because Newton’s second law states that force equals time rate of momentum change, momentum thickness 𝜃 is proportional to surface shear stress. In all boundary layers, 𝜃 < 𝛿.
Buckingham Pi theorem: A mathematical theorem used in dimensional analysis that predicts the number of nondimen sional groups that must be functionally related from a set of dimensional parameters that are thought to be functionally related.
buffer layer: The part of a turbulent boundary layer, close to the wall, lying between the viscous and inertial sublayers. This thin layer is a transition from the friction-dominated layer adjacent to the wall where viscous stresses are large, to the inertial layer where turbulent stresses are large compared to viscous stresses.
bulk modulus of elasticity: See compressibility.
buoyant force: The net upward hydrostatic pressure force acting on an object submerged, or partially submerged, in a fluid.
cavitation: The formation of vapor bubbles in a liquid as a result of pressure going below the vapor pressure.
center of pressure: The effective point of application of pressure distributed over a surface. This is the point where a counteracting force (equal to integrated pressure) must be placed for the net moment from pressure about that point to be zero.
centripetal acceleration: Acceleration associated with the change in the direction of the velocity (vector) of a material particle.
closed system: See system.
coefficient of compressibility: See compressibility.
compressibility: The extent to which a fluid particle changes volume when subjected to either a change in pressure or a change in temperature.
bulk modulus of elasticity: Synonymous with coefficient of compressibility.
coefficient of compressibility: The ratio of pressure change to relative change in volume of a fluid particle. This coefficient quantifies compressibility in response to pressure change, an important effect in high Mach number flows.
coefficient of volume expansion: The ratio of relative density change to change in temperature of a fluid particle. This coefficient quantifies compressibility in response to temperature change.
computational fluid dynamics (CFD): The application of the conservation laws with boundary and initial conditions in mathematical discretized form to estimate field variables quantitatively on a discretized grid (or mesh) spanning part of the flow field.
conservation laws: The fundamental principles upon which all engineering analysis is based, whereby the material properties of mass, momentum, energy, and entropy can change only in balance with other physical properties involving forces, work, and heat transfer. These laws are predictive when written in mathematical form and appropriately combined with boundary conditions, initial conditions, and constitutive relationships.
conservation of energy principle: This is the first law of thermodynamics, a fundamental law of physics stating that the time rate of change of total energy of a fixed mass (system) is balanced by the net rate at which work is done on the mass and heat energy is transferred to the mass. Note: To mathematically convert the time derivative of mass, momentum, and energy of fluid mass in a system to that in a control volume, one applies the Reynolds transport theorem.
conservation of mass principle: A fundamental law of physics stating that a volume always containing the same atoms and molecules (system) must always contain the same mass. Thus the time rate of change of mass of a system is zero. This law of physics must be revised when matter moves at speeds approaching the speed of light so that mass and energy can be exchanged as per Einstein’s laws of relativity.
conservation of momentum: This is Newton’s second law of motion, a fundamental law of physics stating that the time rate of change of momentum of a fixed mass (system) is balanced by the net sum of all forces applied to the mass.
constitutive equations: An empirical relationship between a physical variable in a conservation law of physics and other physical variables in the equation that are to be predicted. For example, the energy equation written for temperature includes the heat flux vector. It is known from experiments that heat flux for most common materials is accurately approximated as proportional to the gradient in temperature (this is called Fourier’s law). In Newton’s law written for a fluid particle, the viscous stress tensor (see stress) must be written as a func tion of velocity to solve the equation. The most common con stitutive relationship for viscous stress is that for a Newtonian fluid. See also rheology.
continuity equation: Mathematical form of conservation of mass applied to a fluid particle in a flow.
continuum: Treatment of matter as a continuous (without holes) distribution of finite mass differential volume elements. Each volume element must contain huge numbers of molecules so that the macroscopic effect of the molecules can be modeled without considering individual molecules.
cen96534_glos_979_992.indd 980 01/12/16 10:56 am GLOSSARY 981 contour plot: Also called an isocontour plot, this is a way of plotting data as lines of constant variable through a flow field. Streamlines, for example, may be identified as lines of constant stream function in two-dimensional incompressible steady flows.
control mass: See system.
control volume: A volume specified for analysis where flow enters and/or exits through some portion(s) of the volume surface. Also called an open system (see system).
convective acceleration: Synonymous with advective acceleration, this term must be added to the partial time derivative of velocity to properly quantify the acceleration of a fluid particle within an Eulerian frame of reference. For example, a fluid particle moving through a contraction in a steady flow speeds up as it moves, yet the time derivative is zero. The additional convective acceleration term required to quantify fluid acceleration (e.g., in Newton’s second law) is called the convective derivative. See also Eulerian description, Lagrangian description, material derivative, and steady flow.
convective derivative: See material derivative and convective acceleration.
creeping flow: Fluid flow in which frictional forces domi nate fluid accelerations to the point that the flow can be well modeled with the acceleration term in Newton’s second law set to zero. Such flows are characterized by Reynolds num bers that are small compared to 1 (Re ≪ 1). Since Reynolds number typically can be written as characteristic velocity times characteristic length divided by kinematic viscosity (VL /𝜈), creeping flows are often slow-moving flows around very small objects (e.g., sedimentation of dust particles in air or motion of spermatozoa in water), or with very viscous fluids (e.g., glacier and tar flows). Also called Stokes flow.
deformation rate: See strain rate.
derived dimensions: See dimensions.
deviatoric stress tensor: Another term for viscous stress tensor. See stress.
differential analysis: Analysis at a point in the flow (as opposed to over a control volume).
differential volume/area/length: A small volume 𝛿V, area 𝛿A, or length 𝛿x in the limit of the volume/area/length shrink ing to a point. Derivatives are often produced in this limit. (Note that 𝛿 is sometimes written as Δ or d.) dimensional analysis: A process of analysis based solely on the variables of relevance to the flow system under study, the dimensions of the variables, and dimensional homogene ity. After determining the other variables on which a variable of interest depends (e.g., drag on a car depends on the speed and size of the car, fluid viscosity, fluid density, and surface roughness), one applies the principle of dimensional homoge neity with the Buckingham Pi theorem to relate an appropri ately nondimensionalized variable of interest (e.g., drag) with the other variables appropriately nondimen sionalized (e.g., Reynolds numbers, roughness ratio, and Mach number).
dimensional homogeneity: The requirement that summed terms must have the same dimensions (e.g., 𝜌V2, pressure P, and shear stress 𝜏xy are dimensionally homogeneous while power, specific enthalpy h, and Pm . are not). Dimensional homogeneity is the basis of dimensional analysis.
dimensionality: The number of spatial coordinates in whose direction velocity components and/or other variables vary for a specified coordinate system. For example, fully developed flow in a tube is one-dimensional (1-D) in the radial direction r since the only nonzero velocity component (the axial, or x-, component) is constant in the x- and 𝜃-directions, but varies in the r-direction. Planar flows are two-dimensional (2-D). Flows over bluff bodies such as cars, airplanes, and buildings are three-dimensional (3-D). Spatial derivatives are nonzero only in the directions of dimensionality.
dimensions: The required specification of a physical quantity beyond its numerical value. See also units.
derived (or secondary) dimensions: Combinations of fundamental dimensions. Examples of derived dimen sions are: velocity (L/t), stress or pressure (F/L2 = m/Lt2), energy or work (mL2/t2 = FL), density (m/L3), specific weight (F/L3), and specific gravity (unitless).
fundamental (primary, basic) dimensions: Mass (m), length (L), time (t), temperature (T), electrical current (I), amount of light (C), and amount of matter (N) without reference to a specific system of units. Note that the force dimension is obtained through Newton’s law as F = mL/t2 (thus, the mass dimension can be replaced with a force dimension by replacing m with Ft2/L).
drag coefficient: Nondimensional drag given by the drag force on an object nondimensionalized by dynamic pressure of the free-stream flow times frontal area of the object: CD = FD 1 2 ρV 2A Note that at high Reynolds numbers (Re ≫ 1), CD is a normalized variable, whereas at Re ≪ 1, CD is nondimensional but is not normalized (see normalization). See also lift coefficient.
drag force: The force on an object opposing the motion of the object. In a frame of reference moving with the object, this is the force on the object in the direction of flow. There are multiple components to drag force: friction drag: The part of the drag on an object resulting from integrated surface shear stress in the direction of flow relative to the object.
induced drag: The component of the drag force on a finite-span wing that is “induced” by lift and associated with the tip vortices that form at the tips of the wing and “downwash” behind the wing.
cen96534_glos_979_992.indd 981 01/12/16 10:56 am 982 FLUID MECHANICS pressure (or form) drag: The part of the drag on an object resulting from integrated surface pressure in the direction of flow relative to the object. Larger pressure on the front of a moving bluff body (such as a car) relative to the rear results from massive flow separation and wake formation at the rear.
dynamic pressure: When the Bernoulli equation in incom pressible steady flow and/or the conservation of energy equa tion along a streamline are written in forms where each term in the equations has the dimensions force/area, dynamic pres sure is the kinetic energy (per unit volume) term (i.e., 1 2ρV 2).
dynamic similarity: See similarity.
dynamic viscosity: See viscosity.
dynamics: When contrasted with statics the term refers to the application of Newton’s second law of motion to moving matter. When contrasted with kinematics the term refers to forces or accelerations through Newton’s law force balances.
eddy viscosity: See turbulence models.
efficiency: A ratio that describes levels of losses of useful power obtained from a device. Efficiency of 1 implies no losses in the particular function of the device for which a particular definition of efficiency is designed. For example, mechanical efficiency of a pump is defined as the ratio of useful mechanical power transferred to the flow by the pump to the mechanical energy, or shaft work, required to drive the pump. Pump-motor efficiency of a pump is defined as the ratio of useful mechanical power transferred to the flow over the electrical power required to drive the pump. Pump-motor efficiency, therefore, includes additional losses and is thus lower than mechanical pump efficiency.
energy: A state of matter described by the first law of thermodynamics that can be altered at the macroscopic level by work, and at the microscopic level through adjustments in thermal energy.
flow energy: Synonymous with flow work. The work associated with pressure acting on a flowing fluid.
heat (transfer): The term “heat” is generally used synon ymously with thermal energy. Heat transfer is the transfer of thermal energy from one physical location to another.
internal energy: Forms of energy arising from the microscopic motions of molecules and atoms, and from the structure and motions of the subatomic particles com prising the atoms and molecules, within matter.
kinetic energy: Macroscopic (or mechanical) form of energy arising from the speed of matter relative to an inertial frame of reference.
mechanical energy: The nonthermal components of energy; examples include kinetic and potential energy.
potential energy: A mechanical form of energy that changes as a result of macroscopic displacement of matter relative to the gravitational vector.
thermal energy: Internal energy associated with micro scopic motions of molecules and atoms. For single-phase systems, it is the energy represented by temperature.
total energy: Sum of all forms of energy. Total energy is the sum of kinetic, potential, and internal energies. Equivalently, total energy is the sum of mechanical and thermal energies.
work energy: The integral of force over the distance in which a mass is moved by the force. Work is energy associated with the movement of matter by a force.
energy grade line: See grade lines.
English system: See units.
entry length: The entry flow region in a pipe or duct flow where the wall boundary layers are thickening toward the center with axial distance x of the duct, so that axial deriva tives are nonzero. As with the fully developed region, the hydrodynamic entry length involves growth of a velocity boundary layer, and the thermal entry length involves growth of a temperature boundary layer.
Eulerian derivative: See material derivative.
Eulerian description: In contrast with a Lagrangian description, an Eulerian analysis of fluid flow is developed from a frame of reference through which the fluid particles move. In this frame the acceleration of fluid particles is not simply the time derivative of fluid velocity, and must include another term, called convective acceleration, to describe the change in velocity of fluid particles as they move through a velocity field. Note that velocity fields are always defined in an Eulerian frame of reference.
extensional strain rate: See strain rate.
extensive property: A fluid property that depends on total volume or total mass (e.g., total internal energy). See inten sive property.
field: The representation of a flow variable as a function of Eulerian coordinates (x, y, z). For example, the velocity and acceleration fields are the fluid velocity and acceleration vectors (V ›, a ›) as functions of position (x, y, z) in the Eulerian description at a specified time t.
flow field: The field of flow variables. Generally, this term refers to the velocity field, but it may also mean all field variables in a fluid flow.
first law of thermodynamics: See conservation laws, con servation of energy.
flow separation: A phenomenon where a boundary layer adjacent to a surface is forced to leave, or “separate” from, the surface due to “adverse” pressure forces (i.e., increas ing pressure) in the flow direction. Flow separation occurs in regions of high surface curvature, for example, at the rear of an automobile and other bluff bodies.
flow work: The work term in first law of thermodynamics applied to fluid flow associated with pressure forces on the flow. See energy, flow energy.
cen96534_glos_979_992.indd 982 01/12/16 10:56 am GLOSSARY 983 fluid: A material that when sheared deforms continuously in time during the period that shear forces are applied. By con trast, shear forces applied to a solid cause the material either to deform to a fixed static position (after which deforma tion stops), or cause the material to fracture. Consequently, whereas solid deformations are generally analyzed using strain and shear, fluid flows are analyzed using rates of strain and shear (see strain rate).
fluid mechanics/dynamics: The study and analysis of fluids through the macroscopic conservation laws of physics, i.e., conservation of mass, momentum (Newton’s second law), and energy (first law of thermodynamics), and the second law of thermodynamics.
fluid particle/element: A differential particle, or element, embedded in a fluid flow containing always the same atoms and molecules. Thus a fluid particle has fixed mass 𝛿m and moves with the flow with local flow velocity V ›, accel eration a › particle = DV ›/Dt and trajectory (xparticle(t), yparticle(t), tparticle(t)). See also material derivative, material particle, material position vector, and pathline.
forced flow: Flow resulting from an externally applied force. Examples include liquid flow through tubes driven by a pump and fan-driven airflow for cooling computer components. Natural flows, in contrast, result from internal buoyancy forces driven by temperature (i.e., density) varia tions within a fluid in the presence of a gravitational field. Examples include buoyant plumes around a human body or in the atmosphere.
friction/frictional: See Newtonian fluid, viscosity, and viscous force.
friction factor: It can be shown from dimensional analysis and conservation of momentum applied to a steady fully developed pipe flow that the frictional contribution to the pressure drop along the pipe, nondimensionalized by flow dynamic pressure (1 2ρV2 avg), is proportional to the length-to-diameter ratio (L /D) of the pipe. The proportionality factor f is called the friction factor. The friction factor is quantified from experiment (turbulent flow) and theory (laminar flow) in empirical relationships, and in the Moody chart, as a function of the Reynolds number and nondimensional roughness. Conservation of momentum shows that the friction factor is proportional to the nondimensional wall shear stress (i.e., the skin friction).
frictionless flow: Mathematical treatments of fluid flows sometimes use conservation of momentum and energy equa tions without the frictional terms. Such mathematical treat ments “assume” that the flow is “frictionless,” implying no viscous force (Newton’s second law), nor viscous dissipation ( first law of thermodynamics). However, no real fluid flow of engineering interest can exist without viscous forces, dissipation, and/or head losses in regions of practical importance. The engineer should always identify the flow regions where frictional effects are concentrated. When developing models for prediction, the engineer should consider the role of these viscous regions in the prediction of variables of interest and should estimate levels of error in simplified treatments of the viscous regions. In high Reynolds number flows, frictional regions include boundary layers, wakes, jets, shear layers, and flow regions surround ing vortices.
Froude number: An order-of-magnitude estimate of the ratio of the inertial term in Newton’s law of motion to the gravity force term. The Froude number is an important nondimensional group in free-surface flows, as is generally the case in channels, rivers, surface flows, etc.
fully developed: Used by itself, the term is generally un derstood to imply hydrodynamically fully developed, a flow region where the velocity field is constant along a specified direction in the flow. In the fully developed region of pipe or duct flow, the velocity field is constant in the axial direction, x (i.e., it is independent of x), so that x-derivatives of velocity are zero in the fully developed region. There also exists the concept of “thermally fully developed” for the temperature field; however, unlike hydrodynamically fully developed regions where both the magnitude and shape of the velocity profile are constant in x, in thermally fully developed regions only the shape of the temperature profile is constant in x. See also entry length.
fundamental dimensions: See dimensions.
gage pressure: Pressure (P) relative to atmospheric pres sure (Patm). That is, Pgage = P − Patm. See also stress, pressure stress. Thus Pgage > 0 or Pgage < 0 is simply the pressure above or below atmospheric pressure.
gas dynamics: The study and analysis of gases and vapors through the macroscopic conservation laws of physics (see fluid mechanics/dynamics).
geometric similarity: See similarity.
grade lines: Lines of head summations.
energy grade line: Line describing the sum of pressure head, velocity head, and elevation head. See head.
hydraulic grade line: Line describing the sum of pressure head and elevation head. See head.
Hagen–Poiseuille flow: See Poiseuille flow.
head: A quantity (pressure, kinetic energy, etc.) expressed as an equivalent column height of a fluid. Conservation of energy for steady flow written for a control volume surrounding a central streamline with one inlet and one outlet, or shrunk to a streamline, can be written such that each term has the dimensions of length. Each of these terms is called a head term: elevation head: The term in the head form of conser vation of energy (see head) involving distance in the direction opposite to the gravitational vector relative to a predefined datum (z).
cen96534_glos_979_992.indd 983 01/12/16 10:56 am 984 FLUID MECHANICS head loss: The term in the head form of conservation of energy (see head) that contains frictional losses and other irreversibilities. Without this term, the energy equation for streamlines becomes the Bernoulli equation in head form.
pressure head: The term in the head form of conserva tion of energy (see head) involving pressure (P/𝜌g).
velocity head: The (kinetic energy) term in the head form of conservation of energy (see head) involving velocity (V2/2g).
heat: See energy.
hot-film anemometer: Similar to a hot-wire anemometer except using a metallic film rather than a wire; used primarily for liquid flows. The measurement portion of a hot-film probe is generally larger and more rugged than that of a hot-wire probe.
hot-wire anemometer: A device used to measure a veloc ity component locally in a gas flow based on the relationship between the flow around a thin heated wire (the hot wire), temperature of the wire, and heating of the wire resulting from a current. See also hot-film anemometer.
hydraulic grade line: See grade lines.
hydraulics: The hydrodynamics of liquid and vapor flow in pipes, ducts, and open channels. Examples include water piping systems and ventilation systems.
hydrodynamic entry length: See entry length.
hydrodynamically fully developed: See fully developed.
hydrodynamics: The study and analysis of liquids through the macroscopic conservation laws of physics (see fluid mechanics/dynamics). The term is sometimes applied to incompressible vapor and gas flows, but when the fluid is air, the term aerodynamics is generally used instead.
hydrostatic pressure: The component of pressure variation in a fluid flow that would exist in the absence of flow as a result of gravitational body force. This term appears in the hydrostatic equation and in the Bernoulli equation. See also dynamic and static pressure.
hypersonic: An order of magnitude or more above the speed of sound (Mach number ≫ 1).
ideal fluid: See perfect fluid.
ideal gas: A gas at low enough density and/or high enough temperature that (a) density, pressure, and temperature are related by the ideal-gas equation of state, P = 𝜌RT, and (b) specific internal energy and enthalpy are functions only of temperature.
incompressible flow: A fluid flow where variations in den sity are sufficiently small to be negligible. Flows are generally incompressible either because the fluid is incompressible (liquids) or because the Mach number is low (roughly < 0.3).
induced drag: See drag force.
inertia/inertial: The acceleration term in Newton’s sec ond law, or effects related to this term. Thus, a flow with higher inertia requires larger deceleration to be brought to rest.
inertial sublayer: A highly turbulent part of a turbulent boundary layer, close to the wall but just outside the viscous sublayer and buffer layer, where turbulent stresses are large compared to viscous stresses.
intensive property: A fluid property that is independent of total volume or total mass (i.e., an extensive property per unit mass or sometimes per unit volume).
internal energy: See energy.
inviscid (region of) flow: Region of a fluid flow where viscous forces are sufficiently small relative to other forces (typically, pressure force) on fluid particles in that region of the flow to be neglected in Newton’s second law of motion to a good level of approximation (compare with viscous flow). See also frictionless flow. An inviscid region of flow is not necessarily irrotational.
irrotational (region of) flow: A region of a flow with negli gible vorticity (i.e., fluid particle rotation). Also called potential flow. An irrotational region of flow is also inviscid.
isocontour plot: See contour plot.
jet: A friction-dominated region issuing from a tube or orifice and formed by surface boundary layers that have been swept behind by the mean velocity. Jets are characterized by high shear with the highest velocities in the center of the jet and lowest velocities at the edges. Frictional force, viscous stress, and vorticity are significant in jets.
Kármán vortex street: The two-dimensional alternating unsteady pattern of vortices that is commonly observed behind circular cylinders in a flow (e.g., the vortex street behind wires in the wind is responsible for the distinct tone sometimes heard).
kinematic similarity: See similarity.
kinematic viscosity: Fluid viscosity divided by density.
kinematics: In contrast with dynamics, the kinematic aspects of a fluid flow are those that do not directly involve Newton’s second law force balance. Kinematics refers to descriptions and mathematical derivations based only on conservation of mass (continuity) and definitions related to flow and deformation.
kinetic energy: See energy.
kinetic energy correction factor: Control volume analy sis of the conservation of energy equation applied to tubes contains area integrals of kinetic energy flux. The integrals are often approximated as proportional to kinetic energy formed with area-averaged velocity, Vavg. The inaccuracy in this approximation can be significant, so a kinetic energy correction factor, 𝛼, multiplies the term to improve the cen96534_glos_979_992.indd 984 01/12/16 10:56 am GLOSSARY 985 approximation. The correction 𝛼 depends on the shape of the velocity profile, is largest for laminar profiles (Poiseuille flow), and is closest to 1 in turbulent pipe flows at very high Reynolds numbers.
Lagrangian derivative: See material derivative.
Lagrangian description: In contrast with the Eulerian description, a Lagrangian analysis is developed from a frame of reference attached to moving material particles. For example, solid particle acceleration in the standard Newton’s second law form, F › = ma ›, is in a coordinate system that moves with the particle so that acceleration a › is given by the time derivative of particle velocity. This is the typical analytical approach used for analysis of the motion of solid objects.
laminar flow: A stable well-ordered state of fluid flow in which all pairs of adjacent fluid particles move alongside one another forming laminates. A flow that is not laminar is either turbulent or transitional to turbulence, which occurs above a critical Reynolds number.
laser Doppler velocimetry (LDV): Also called laser Doppler anemometry (LDA). A technique for measuring a velocity component locally in a flow based on the Doppler shift associated with the passage of small particles in the flow through the small target volume formed by the crossing of two laser beams. Unlike hot-wire and hot-film anemometry and like particle image velocimetry, there is no interference to the flow.
lift coefficient: Nondimensional lift given by the lift force on a lifting object (such as an airfoil or wing) nondimension alized by dynamic pressure of the free-stream flow times planform area of the object: CL = FL 1 2ρV 2A Note that at high Reynolds numbers (Re ≫ 1), CL is a normalized variable, whereas at Re ≪ 1, CL is nondimensional but is not normalized (see normalization). See also drag coefficient.
lift force: The net aerodynamic force on an object perpen dicular to the motion of the object.
linear strain rate: Synonymous with extensional strain rate. See strain rate.
losses: Frictional head losses in pipe flows are separated into those losses in the fully developed pipe flow regions of a pip ing network, the major losses, plus head losses in other flow regions of the network, the minor losses. Minor loss regions include entry lengths, pipe couplings, bends, valves, etc. It is not unusual for minor losses to be larger than major losses.
Mach number: Nondimensional ratio of the characteristic speed of the flow to the speed of sound. Mach number characterizes the level of compressibility in response to pres sure variations in the flow.
major losses: See losses.
manometer: A device that measures pressure based on hy drostatic pressure principles in liquids.
material acceleration: The acceleration of a fluid particle at the point (x, y, z) in a flow at time t. This is given by the mate rial derivative of fluid velocity: DV ›(x, y, z, t)/Dt.
material derivative: Synonymous terms are total derivative, substantial derivative, and particle derivative. These terms mean the time rate of change of fluid variables (temperature, velocity, etc.) moving with a fluid particle. Thus, the mate rial derivative of temperature at a point (x, y, z) at time t is the time derivative of temperature attached to a moving fluid particle at the point (x, y, z) in the flow at the time t. In a Lagrangian frame of reference (i.e., a frame attached to the moving particle), particle temperature Tparticle depends only on time, so a time derivative is a total derivative dTparticle(t)/dt. In an Eulerian frame, the temperature field T(x, y, z, t) depends on both position (x, y, z) and time t, so the material derivative must include both a partial derivative in time and a convective derivative: dTparticle(t)/dt = DT(x, y, z, t)/Dt = ∂T/∂t + V ›·∇ ›T. See also field.
material particle: A differential particle, or element, that contains always the same atoms and molecules. Thus a mate rial particle has fixed mass 𝛿m. In a fluid flow, this is the same as a fluid particle.
material position vector: A vector [xparticle(t), yparticle(t), zparticle(t)] that defines the location of a material particle as a function of time. Thus the material position vector in a fluid flow defines the trajectory of a fluid particle in time.
mean: Synonymous with average.
mechanical energy: See energy.
mechanics: The study and analysis of matter through the macroscopic conservation laws of physics (mass, momentum, energy, second law).
minor losses: See losses.
mixing length: See turbulence models.
momentum: The momentum of a material particle (or fluid particle) is the mass of the material particle times its velocity. The momentum of a macroscopic volume of material particles is the integrated momentum per unit volume over the volume, where momentum per unit volume is the density of the material particle times its velocity. Note that momentum is a vector.
momentum flux correction factor: A correction factor added to correct for approximations made in the simplifica tion of the area integrals for the momentum flux terms in the control volume form of conservation of momentum.
Moody chart: A commonly used plot of the friction factor as a function of the Reynolds number and roughness param eter for fully developed pipe flow. The chart is a combination cen96534_glos_979_992.indd 985 01/12/16 10:56 am 986 FLUID MECHANICS of flow theory for laminar flow with a graphical representa tion of an empirical formula by Colebrook to a large set of experimental data for turbulent pipe flow of various values of “sandpaper” roughness.
natural flow: Contrast with forced flow.
Navier–Stokes equation: Newton’s second law of fluid motion (or conservation of momentum) written for a fluid particle (the differential form) with the viscous stress tensor replaced by the constitutive relationship between stress and strain rate for Newtonian fluids. Thus the Navier–Stokes equation is simply Newton’s law written for Newtonian fluids.
Newtonian fluid: When a fluid is subjected to a shear stress, the fluid continuously changes shape (deformation). If the fluid is Newtonian, the rate of deformation (i.e., strain rate) is proportional to the applied shear stress and the constant of proportionality is called viscosity. In general flows, the rate of deformation of a fluid particle is described mathematically by a strain rate tensor and the stress by a stress tensor. In flows of Newtonian fluids, the stress tensor is proportional to the strain rate tensor, and the constant of proportionality is called viscosity. Most common fluids (water, oil, gasoline, air, most gases, and vapors) without particles or large molecules in suspension are Newtonian.
Newton’s second law: See conservation of momentum.
nondimensionalization: The process of making a dimen sional variable dimensionless by dividing the variable by a scaling parameter (a single variable or a combination of variables) that has the same dimensions. For example, the surface pressure on a moving ball might be nondimensionalized by dividing it by 𝜌V 2, where 𝜌 is fluid density and V is free-stream velocity. See also normalization.
non-Newtonian fluid: A non-Newtonian fluid is one that deforms at a rate that is not linearly proportional to the stress causing the deformation. Depending on the manner in which viscosity varies with strain rate, non-Newtonian fluids can be labeled shear thinning (viscosity decreases with increasing strain rate), shear thickening (viscosity increases with increas ing strain rate), and viscoelastic (when the shearing forces are removed, the fluid particles return partially to an earlier shape). Suspensions and liquids with long-chain molecules are generally non-Newtonian. See also Newtonian fluid and viscosity.
normal stress: See stress.
normalization: A particular nondimensionalization where the scaling parameter is chosen so that the nondimensional ized variable attains a maximum value that is of order 1 (say, within roughly 0.5 to 2). Normalization is more restrictive (and more difficult to do properly) than nondimensionalization. For example, P/(𝜌V2) discussed under nondimensionalization is also normalized pressure on a flying baseball (where Reynolds number Re ≫1), but is simply nondimensionalization of surface pressure on a small glass bead dropping slowly through honey (where Re ≪ 1).
no-slip condition: The requirement that at the interface between a fluid and a solid surface, the fluid velocity and surface velocity are equal. Thus if the surface is fixed, the fluid must obey the boundary condition that fluid velocity = 0 at the surface.
one-dimensional: See dimensionality.
open system: Same as control volume.
particle derivative: See material derivative.
particle image velocimetry (PIV): A technique for measur ing a velocity component locally in a flow based on tracking the movement of small particles in the flow over a short time using pulsed lasers. Unlike hot-wire and hot-film anemometry and like laser Doppler velocimetry, there is no interference to the flow.
pathline: A curve mapping the trajectory of a fluid particle as it travels through a flow over a period of time. Mathemati cally, this is the curve through the points mapped out by the material position vector [xparticle(t), yparticle(t), zparticle(t)] over a defined period of time. Thus, pathlines are formed over time, and each fluid particle has its own pathline. In a steady flow, fluid particles move along streamlines, so pathlines and streamlines coincide. In a nonsteady flow, however, pathlines and streamlines are generally very different. Contrast with streamline.
perfect fluid: Also called an ideal fluid, the concept of a fictitious fluid that can flow in the absence of all frictional effects. There is no such thing as a perfect fluid, even as an approximation, so the engineer need not consider the concept further.
periodic: An unsteady flow in which the flow oscillates about a steady mean.
Pitot-static probe: A device used to measure fluid velocity through the application of the Bernoulli equation with simul taneous measurement of static and stagnation pressures. Also called a Pitot-Darcy probe.
planar flow: A two-dimensional flow with two nonzero components of velocity in Cartesian coordinates that vary only in the two coordinate directions of the flow. Thus, all partial derivatives perpendicular to the plane of the flow are zero. See also axisymmetric flow and dimensionality.
Poiseuille flow: Fully developed laminar flow in a pipe or duct. Also called Hagen–Poiseuille flow. The mathematical model relationships for Poiseuille flow relating the flow rate and/or velocity profile to the pressure drop along the pipe/ duct, fluid viscosity and geometry are sometimes referred to as Poiseuille’s law (although strictly not a “law” of mechan ics). The velocity profile of all Poiseuille flows is parabolic, and the rate of axial pressure drop is constant.
cen96534_glos_979_992.indd 986 01/12/16 10:56 am GLOSSARY 987 Poiseuille’s law: See Poiseuille flow.
potential energy: See energy.
potential flow: Synonymous with irrotational flow. This is a region of a flow with negligible vorticity (i.e., fluid particle rotation). In such regions, a velocity potential function exists (thus the name).
potential function: If a region of a flow has zero vorticity (fluid particle spin), the velocity vector in that region can be written as the gradient of a scalar function called the velocity potential function, or simply the potential function. In prac tice, potential functions are often used to model flow regions where vorticity levels are small but not necessarily zero.
power: Work per unit time; time rate at which work is done.
pressure: See stress.
pressure force: As applicable to Newton’s second law, this is the force acting on a fluid particle that arises from spatial gradients in pressure within the flow. See also stress, pressure stress.
pressure work: See flow work.
primary dimension: See dimensions.
profile plot: A graphical representation of the spatial varia tion of a fluid property (temperature, pressure, strain rate, etc.) through a region of a fluid flow. A profile plot defines property variations in part of a field (e.g., a temperature profile might define the variation of temperature along a line within the temperature field).
velocity profile: The spatial variation in a velocity component or vector through a region of a fluid flow. For example, in a pipe flow the velocity profile generally de fines the variation in axial velocity with radius across the pipe cross section, while a boundary layer velocity profile generally defines variation in axial velocity normal to the surface. The velocity profile is part of a velocity field.
quasi-steady flow: See steady flow.
Reynolds number: An order-of-magnitude estimate of the ratio of the following two terms in Newton’s second law of motion over a region of the flow: the inertial (or acceleration) term over the viscous force term. Most but not all Reynolds numbers can be written as an appropriate characteristic veloc ity V times a characteristic length scale L consistent with the velocity V, divided by the kinematic viscosity 𝜈 of the fluid: Re = VL/𝜈. The Reynolds number is arguably the most important nondimensional similarity parameter in fluid flow analysis since it gives a rough estimate of the importance of frictional force in the overall flow.
Reynolds stress: Velocity components (and other variables) in turbulent flows are separated into mean plus fluctuat ing components. When the equation for mean streamwise velocity component is derived from the Navier–Stokes equation, six new terms appear given by fluid density times the averaged product of two velocity components. Because these terms have the same units as stress (force/ area), they are called turbulent stresses or Reynolds stresses (in memory of Osborne Reynolds who first quantified turbulent variables as mean + fluctuation). Just as viscous stresses can be written as a tensor (or matrix), we define a Reynolds stress tensor with Reynolds normal stress components and Reynolds shear stress components. Although Reynolds stresses are not true stresses, they have qualitatively similar effects as do viscous stresses, but as a result of the large chaotic vortical motions of turbulence rather than the microscopic molecular motions that underlie viscous stresses.
Reynolds transport theorem: The mathematical relationship between the time rate of change of a fluid property in a system (volume of fixed mass moving with the flow) and the time rate of change of a fluid property in a control volume (volume, usually fixed in space, with fluid mass moving across its surface). This finite volume expression is closely related to the material (time) derivative of a fluid property attached to a moving fluid particle. See also conservation laws.
rheology: The study and mathematical representation of the deformation of different fluids in response to surface forces, or stress. The mathematical relationships between stress and deformation rate (or strain rate) are called constitutive equa tions. The Newtonian relationship between stress and strain rate is the simplest example of a rheological constitutive equation. See also Newtonian and non-Newtonian fluid.
rotation rate: The angular velocity, or rate of spin, of a fluid particle (a vector, with units rad/s, given by 1/2 the curl of the velocity vector). See also vorticity.
rotational flow: Synonymous with vortical flow, this term describes a flow field, or a region of a flow field, with signifi cant levels of vorticity.
saturation pressure: The pressure at which the phase of a simple compressible substance changes between liquid and vapor at fixed temperature.
saturation temperature: The temperature at which the phase of a simple compressible substance changes between liquid and vapor at fixed pressure.
scaling parameter: A single variable, or a combination of variables, that is chosen to nondimensionalize a variable of interest. See also nondimensionalization and normalization.
schlieren technique: An experimental technique to visualize flows based on the refraction of light from varying fluid density. The illuminance level in a schlieren image responds to the first spatial derivative of density.
secondary dimensions: See dimensions.
cen96534_glos_979_992.indd 987 01/12/16 10:56 am 988 FLUID MECHANICS shadowgraph technique: An experimental technique to visualize flows based on the refraction of light from varying fluid density. The illuminance level in a shadowgraph image responds to the second spatial derivative of density.
shear: Refers to gradients (derivatives) in velocity components in directions normal to the velocity component.
shear force: See stress, shear stress.
shear layer: A quasi two-dimensional flow region with a high gradient in streamwise velocity component in the transverse flow direction. Shear layers are inherently viscous and vortical in nature.
shear rate: The gradient in streamwise velocity in the direction perpendicular to the velocity. Thus, if streamwise (x) velocity u varies in y, the shear rate is du/dy. The term is applied to shear flows, where shear rate is twice the shear strain rate. See also strain rate.
shear strain: See strain rate.
shear stress: See stress, shear stress.
shear thickening fluid: See non-Newtonian fluid.
shear thinning fluid: See non-Newtonian fluid.
SI system: See units.
similarity: The principle that allows one to quantitatively relate one flow to another when certain conditions are met. Geometric similarity, for example, must be true before one can hope for kinematic or dynamic similarity. The quantitative relationship that relates one flow to another is developed us ing a combination of dimensional analysis and data (generally, experimental, but also numerical or theoretical).
dynamic similarity: If two objects are geometrically and kinematically similar, then if the ratios of all forces (pressure, viscous stress, gravity force, etc.) between a point in the flow surrounding one object, and the same point scaled appropriately in the flow surrounding the other object, are all the same at all corresponding pairs of points, the flow is dynamically similar.
geometric similarity: Two objects of different size are geometrically similar if they have the same geometrical shape (i.e., if all dimensions of one are a constant multiple of the corresponding dimensions of the other).
kinematic similarity: If two objects are geometrically similar, then if the ratios of all velocity components be tween a point in the flow surrounding one object, and the same point scaled appropriately in the flow surrounding the other object, are all the same at all corresponding pairs of points, the flow is kinematically similar.
skin friction: Surface shear stress 𝜏w nondimensionalized by an appropriate dynamic pressure 1 2ρV 2. Also called the skin friction coefficient, Cf.
solid: A material that when sheared either deforms to a fixed static position (after which deformation stops) or fractures. See also fluid.
sonic: At the speed of sound (Mach number = 1).
specific gravity: Fluid density nondimensionalized by the density of liquid water at 4°C and atmospheric pressure (1 g/cm3 or 1000 kg/m3). Thus, specific gravity, SG = 𝜌/𝜌water.
specific weight: The weight of a fluid per unit volume, i.e., fluid density times acceleration due to gravity (specific weight, 𝛾 = 𝜌g).
spin: See rotation rate and vorticity.
stability: A general term that refers to the tendency of a material particle or object (fluid or solid) to move away from or return when displaced slightly from its original position.
neutrally stable: See stability. When displaced slightly, the particle or object will remain in its displaced position.
stable: See stability. When displaced slightly, the particle or object will return to its original position.
unstable: See stability. When displaced slightly, the particle or object will continue to move from its original position.
stagnation point: A point in a fluid flow where the velocity goes to zero. For example, the point on the streamline that intersects the nose of a moving projectile is a stagnation point.
stall: The phenomenon of massive flow separation from the surface of a wing when angle of attack exceeds a criti cal value, and consequent dramatic loss of lift and increase in drag. A plane in stall drops rapidly and must have its nose brought down to reestablish attached boundary layer flow and regenerate lift and reduce drag.
static pressure: Another term for pressure, used in context with the Bernoulli equation to distinguish it from dynamic pressure.
statics: The mechanical study and analysis of material that is fully at rest in a specific frame of reference.
steady flow: A flow in which all fluid variables (velocity, pressure, density, temperature, etc.) at all fixed points in the flow are constant in time (but generally vary from place to place). Thus, in steady flows all partial derivatives in time are zero. Flows that are not precisely steady but that change suf ficiently slowly in time to neglect time derivative terms with relatively little error are called quasi-steady.
Stokes flow: See creeping flow.
strain: See strain rate.
strain rate: Strain rate can also be called deformation rate. This is the rate at which a fluid particle deforms (i.e., changes shape) at a given position and time in a fluid flow. To fully quantify all possible changes in shape of a three-dimensional fluid particle require six numbers. Mathemat ically, these are the six independent components of a second-rank symmetric strain rate tensor, generally written as a symmetric 3 × 3 matrix. Strain is time-integrated strain rate and describes de formation of a fluid particle after a period of time. See stress.
cen96534_glos_979_992.indd 988 01/12/16 10:56 am GLOSSARY 989 extensional strain rate: The components of strain rate that describe elongation or compression of a fluid particle in one of the three coordinate directions. These are the three diagonal elements of the strain rate tensor. The definition of extensional strain depends on one’s choice of coordinate axes. Also called linear strain rate.
shear strain rate: The components of strain rate that de scribe deformation of a fluid particle in response to shear changing an angle between planes mutually perpendicular to the three coordinate axes. These are the off-diagonal elements of the strain rate tensor. The definition of shear strain depends on one’s choice of coordinate axes.
volumetric strain rate: Rate of change of volume of a fluid particle per unit volume. Also called bulk strain rate and rate of volumetric dilatation.
streakline: Used in flow visualization of fluid flows, this is a curve defined over time by the release of a marker (dye or smoke) from a fixed point in the flow. Contrast with pathline and streamline. In a steady flow, streamlines, pathlines, and streaklines all coincide. In a nonsteady flow, however, these sets of curves are each different from one another.
stream function: The two velocity components in a two-dimensional steady incompressible flow can be defined in terms of a single two-dimensional function 𝜓 that automati cally satisfies conservation of mass (the continuity equation), reducing the solution of the two-component velocity field to the solution of this single stream function. This is done by writing the two velocity components as spatial derivatives of the stream function. A wonderful property of the stream function is that (iso)contours of constant 𝜓 define streamlines in the flow.
streamline: A curve that is everywhere tangent to a velocity vector in a fluid velocity field at a fixed instant in time. Thus, the streamlines indicate the direction of the fluid motions at each point. In a steady flow, streamlines are constant in time and fluid particles move along streamlines. In a nonsteady flow the streamlines change with time and fluid particles do not move along streamlines. Contrast with pathline.
streamtube: A bundle of streamlines. A streamtube is usu ally envisioned as a surface formed by an infinite number of streamlines initiated within the flow on a circular circuit and tending to form a tubelike surface in some region of the flow.
stress: A component of a force distributed over an area is written as the integral of a stress over that area. Thus, stress is the force component dFi on a differential area element divided by the area of the element dAj (in the limit dAj → 0), where i and j indicate a coordinate direction x, y, or z. Stress 𝜎ij = dFi/dAj is therefore a force component per unit area in the i-direction on surface j. To obtain the surface force from stress, one integrates stress over the corresponding surface area. Mathematically, there are six independent components of a second-rank symmetric stress tensor, generally written as a symmetric 3 × 3 matrix.
normal stress: A stress (force component per unit area) that acts perpendicular to the area. Therefore, 𝜎xx, 𝜎yy, and 𝜎zz are normal stresses. The normal force over a surface is the net force from shear stress, given by integrating the shear stress over the surface area. The normal stresses are the diagonal elements of the stress tensor.
pressure stress: In a fluid at rest all stresses are normal stresses and all stresses act inward on a surface. At a fixed point, the three normal stresses are equal and the magnitude of these equal normal stresses is called pressure. Thus, in a static fluid 𝜎xx = 𝜎yy = 𝜎zz = −P, where P is pressure. In a moving fluid, stresses in addition to pressure are viscous stresses. A pressure force on a surface is the pressure stress integrated over the surface. The pressure force per unit volume on a fluid particle for Newton’s second law, however, is the negative of the gradient (spatial derivatives) of pressure at that point.
Reynolds stress: See Reynolds stress.
shear stress: A stress (force component per unit area) that acts tangent to the area. Therefore, 𝜎xy, 𝜎yx, 𝜎xz, 𝜎zx, 𝜎yz, and 𝜎zy are shear stresses. The shear force over a surface is the net force from shear stress, given by integrating the shear stress over the surface area. The shear stresses are the off-diagonal elements of the stress tensor.
turbulent stress: See Reynolds stress.
viscous stress: Flow creates stresses in the fluid that are in addition to hydrostatic pressure stresses. These additional stresses are viscous since they arise from friction-induced fluid deformations within the flow. For example, 𝜎xx = −P + 𝜏xx, 𝜎yy = −P + 𝜏yy, and 𝜎zz = −P + 𝜏zz, where 𝜏xx, 𝜏yy, and 𝜏zz are viscous normal stresses. All shear stresses result from friction in a flow and are therefore viscous stresses. A viscous force on a surface is a viscous stress integrated over the surface. The viscous force per unit volume on a fluid particle for Newton’s second law, however, is the divergence (spatial derivatives) of the viscous stress tensor at that point.
stress tensor: See stress.
subsonic: Below the speed of sound (Mach number < 1).
substantial derivative: See material derivative.
supersonic: Above the speed of sound (Mach number > 1).
surface tension: The force per unit length at a liquid–vapor or liquid–liquid interface resulting from the imbalance in attractive forces among like liquid molecules at the interface.
system: Usually when the word system is used by itself, closed system is implied, in contrast with a control volume or open system.
closed system: A volume specified for analysis that en closes always the same fluid particles. Therefore, no flow crosses any part of the volume’s surface and a closed system must move with the flow. Note that Newton’s law analysis of solid particles is generally a closed system analysis, sometimes referred to as a free body.
cen96534_glos_979_992.indd 989 01/12/16 10:56 am 990 FLUID MECHANICS open system: A volume specified for analysis where flow crosses at least part of the volume’s surface. Also called a control volume.
thermal energy: See energy.
three-dimensional: See dimensionality.
timeline: Used for visualization of fluid flows, this is a curve defined at some instant in time by the release of a marker from a line in the flow at some earlier instant in time. The timeline, often used to approximate a velocity profile in a laboratory flow, is very different from streaklines, pathlines, and streamlines.
tip vortex: Vortex formed off each tip of an airplane wing as a byproduct of lift. Synonymous with trailing vortex. See also induced drag.
total derivative: See material derivative.
total energy: See energy.
trailing vortex: See tip vortex.
trajectory: See pathline.
transient period: A time-dependent period of flow evolution leading to a new equilibrium period that is generally, but not necessarily, steady. An example is the start-up period after a jet engine is switched on, leading to a steady (equilibrium) jet flow.
transitional flow: An unstable vortical fluid flow at a Reyn olds number higher than a critical value that is large relative to 1, but is not sufficiently high that the flow has reached a fully turbulent flow state. Transitional flows often oscillate randomly between laminar and turbulent states.
turbulence models: Constitutive model relationships be tween Reynolds stresses and the mean velocity field in turbulent flows. Such model equations are necessary to solve the equation for mean velocity. A simple and widely used modeled form for the Reynolds stresses is to write them like the Newtonian relationship for viscous stresses, as proportional to the mean strain rate, with the proportionality being a turbulent viscosity or eddy viscosity. However, unlike Newtonian fluids, the eddy viscosity is a strong function of the flow itself, and the different ways in which eddy viscosity is modeled as a function of other calculated flow field variables constitute different eddy viscosity models. One traditional approach to modeling eddy viscosity is in terms of a mixing length, which is made proportional to a length set by the flow.
turbulent flow: An unstable disordered state of vortical fluid flow that is inherently unsteady and that contains eddying mo tions over a wide range of sizes (or scales). Turbulent flows are always at Reynolds numbers above a critical value that is large relative to 1. Mixing is hugely enhanced, surface shear stresses are much higher, and head loss is greatly increased in turbulent flows as compared to corresponding laminar flows.
turbulent stress: See Reynolds stress.
turbulent viscosity: See turbulence models.
two-dimensional: See dimensionality.
uniform flow: A flow in which all fluid properties, such as velocity, pressure, temperature, etc., do not vary with position.
units: A specific system to quantify numerically the dimen sions of a physical quantity. The most common systems of units are SI (kg, N, m, s), English (lbm, lbf, ft, s), BGS (slug, lb, ft, s), and cgs (g, dyne, cm, s). See also dimensions.
unsteady flow: A flow in which at least one variable at a fixed point in the flow changes with time. Thus, in unsteady flows a partial derivative in time is nonzero for at least one point in the flow.
vapor pressure: The pressure below which a fluid, at a given temperature, will exist in the vapor state. See also cavitation and saturation pressure.
velocity: A vector that quantifies the rate of change in position and the direction of motion of a material particle.
velocity field: See field.
velocity profile: See profile plots.
viscoelastic fluid: See non-Newtonian fluid.
viscosity: See Newtonian fluid. Viscosity is a property of a fluid that quantifies the ratio of shear stress to rate of defor mation (strain rate) of a fluid particle. (Therefore, viscosity has the dimensions of stress/strain rate, or Ft/L2 = m/Lt.) Qualitatively, viscosity quantifies the level by which a par ticular fluid resists deformation when subjected to shear stress (frictional resistance or friction). Viscosity is a measured property of a fluid and is a function of temperature. For Newtonian fluids, viscosity is independent of the rate of applied stress and strain rate. The viscous nature of non-Newtonian fluids is more difficult to quantify in part because viscosity varies with strain rate. The terms absolute viscosity, dynamic viscosity, and viscosity are synonymous. See also kinematic viscosity.
viscous (regions of) flow: Regions of a fluid flow where viscous forces are significant relative to other forces (typically, pressure force) on fluid particles in that region of the flow, and therefore cannot be neglected in Newton’s second law of motion (compare with inviscid flow).
viscous (or frictional) force: As applicable to Newton’s second law, this is the force acting on a fluid particle that arises from spatial gradients in viscous (or frictional) stresses within the flow. The viscous force on a surface is the viscous stress integrated over the surface. See also stress, viscous stress.
viscous stress tensor: See stress. Also called the deviatoric stress tensor.
viscous sublayer: The part of a turbulent boundary layer adjacent to the surface that contains the highest viscous cen96534_glos_979_992.indd 990 01/12/16 10:56 am GLOSSARY 991 stresses. The velocity gradient in this layer adjacent to the wall is exceptionally high. See also inertial layer and buffer layer.
vortex: A local structure in a fluid flow characterized by a concentration of vorticity (i.e., fluid particle spin or rotation) in a tubular core with circular streamlines around the core axis. A tornado, hurricane, and bathtub vortex are common examples of vortices. Turbulent flow is filled with small vortices of various sizes, strengths, and orientations.
vortical flow: Synonymous with rotational flow, this term describes a flow field, or a region of a flow field, with significant levels of vorticity.
vorticity: Twice the angular velocity, or rate of spin, of a fluid particle (a vector, with units rad/s, given by the curl of the velocity vector). See also rotation rate.
wake: The friction-dominated region behind a body formed by surface boundary layers that are swept to the rear by the free-stream velocity. Wakes are characterized by high shear with the lowest velocities in the center of the wake and high est velocities at the edges. Frictional force, viscous stress, and vorticity are significant in wakes.
work: See energy.
cen96534_glos_979_992.indd 991 01/12/16 10:56 am This page intentionally left blank 993 A Absolute pressure, 78 transducers, 90 of vacuum chamber, 79 Absolute velocity, 167, 195, 252 fluid velocity, 171 Absolute viscosity, 52 Acceleration advective, 141, 142 convective, 141–142 field, 139, 140 fluid particle, 141 Newton’s second law, 141 of fluid particle, 140–141, 204 through nozzle, 142–143 on straight path, 108–110 Accuracy, 28–30 error, 28 Actual peak, 154 Adhesive forces, 59 Adiabatic duct flow with friction, 710 choked Fanno flow, 716–719 control volume, 711 effects of friction on properties of Fanno flow, 713 friction causes mach number, 712 property relations for Fanno flow, 713–716 T-s diagram for adiabatic frictional flow in constant-area duct, 712 Adiabatic process, 220 Adiabatic reversible flow, 227 Adsorption coefficient, 335–336 Advective acceleration, 141, 142 Adverse pressure gradient, 583 Aerodynamic(s), 2 drag force, 308–309 shoulder of body, 553 Aerostatics, 91 Affinity laws, 835 effects of doubling pump speed, 836 geometrically similar pump, 837–839 Air at 1 atm pressure properties, 956, 974 Airflow through converging–diverging nozzle, 684–685 Airfoils, 638 Air–fuel mixture, compression of, 448–449 Algebraic form, 257 Alternate depths, 742, 743–744 Alternate derivation of RTT, 169–171 Ampere, 16 Analytical approach, 22 Analytical solution, 146 Angle, 59 Angular displacement, 2 Angular momentum, 269–272. See also Linear momentum equations, 190, 249, 250–251, 271–273 flow with no external moments, 275 radial-flow devices, 275–280 special cases, 274–275 primary dimensions, 298–299 Angular momentum equation, 190 Angular velocity, 155 vector, 156 Apparent viscosity of fluid, 52 Approximate solution, 520 approximation for inviscid regions of flow, 529 Bernoulli equation in inviscid regions of flow, 530–533 Euler equation, 529–530 Archimedes’ principle, 101 Area moment of inertia. See Second moment of area Aspect ratio, 643 Atmosphere, 84 air, 42 at high altitude properties, 959, 977 pressure, 42, 255, 733 Atomic weight, 334 Available wind power, 856 Average rotation rate, 155 Average velocity, 192 Average wind power density, 857 Axial acceleration, 143 Axial pumps, 822–830 analysis of vane-axial flow fan, 827 axial-flow fans, 823 blade pitch angle, 825 calculation of twist in airplane propeller, 824 design of vane-axial flow fan for wind tunnel, 829–830 fan performance curves, 825 tube-axial fan, 826 well-designed rotor blade, 823 Axial-flow, 812 devices, 275 pump, 812 turbine, 845 I n d e x cen96534_ind_993_1018.indd 993 01/12/16 10:57 am 994 INDEX Axis boundary condition, 897 Axisymmetric flow, 14, 462, 538, 540, 695 Axisymmetric irrotational regions of flow, 540–541 B Back pressure, 677 Backward-inclined blades, 813 Balance equations, 190 Band, 844 Barometer, 84. See also Manometer density of air, 84–85 gravity driven flow, 85–86 hydrostatic pressure in solar pond, 86–87 measuring atmospheric pressure with, 85 Barometric pressure, 84 Basic dimensions, 298 Beam splitter, 410 Bellows flowmeters, 402 Bends, 384 BEP. See Best efficiency point (BEP) Bernoulli equations, 189, 203, 226, 559, 638, 674. See also Energy equations acceleration of fluid particle, 204 applications, 212 bernoulli equation for compressible flow, 217–218 conservation of energy, 190–191 conservation of mass, 190–198 derivation, 204–206 dimensional homogeneity, 300 energy analysis of steady flows, 223–233 force balance across streamlines, 206 HGLand EGL, 210–212 in inviscid regions of flow, 530–533 in irrotational regions of flow, 535–536 limitations on use, 208–209, 210 linear momentum equation, 190 mechanical energy and efficiency, 198–203 rise of ocean due to Hurricane, 216–217 in rotating reference frame, 818 siphoning out gasoline from fuel tank, 213–215 spraying water into air, 212 static, dynamic, and stagnation pressures, 207–208 unsteady, compressible flow, 207 velocity measurement by pitot tube, 215 water discharge from large tank, 213 Bernoulli head, 796 Best efficiency point (BEP), 797, 834, 866 Best estimate, 663 Best hydraulic cross sections, 751 rectangular channels, 753 trapezoidal channels, 753–755 variation of flow rate, 752 variation of hydraulic radius, 752 Betz limit, 859 Bias error, 29 Bingham plastic fluids, 472 Bingham plastics, 53 Biofluid mechanics, 414 blood flow through aortic bifurcation, 418–420 cardiovascular system, 414, 415 differential analysis of biofluid mechanics flows, 499–502 flow measurement techniques, 416 MRI-PCV, 421 overall flow structure, 418 particle traces for BSM valve, 417 PSV, 422 three-dimensional flow structures, 419 Blade row, 827 Blade spacing, 914 Blast wave. See Shock wave Blockage, 327 Blood flow through aortic bifurcation, 418–420 Blower, 794 Body forces, 106, 252, 274 Boiling and freezing point properties, 949, 967 Boundary, 15 Boundary layer, 9, 355, 529, 559 approximation, 558–559. See also Navier–Stokes equation displacement thickness, 572–575 Euler equation, 559–560 free shear layers, 561 laminar boundary layer on flat plate, 569–572 laminar/turbulent boundary layer, 563 momentum integral technique for boundary layers, 587–595 momentum thickness, 575–576 transition of laminar boundary, 562 trip wire, 562 turbulent flat plate boundary layer, 576–582 conditions, 444, 894 inflow/outflow boundary conditions, 895–897 internal boundary conditions, 898–899 miscellaneous boundary conditions, 897–898 wall boundary conditions, 895 coordinate system, 563 equations, 559, 563 centripetal acceleration, 565 order of magnitude, 564–565, 566 outer flow speed, 567 pressure in irrotational region of flow, 566 set, 567 with pressure gradients, 582–587 procedure, 568 region, 355 thickness, 560 Bourdon tube, 90 Bow wave, 695, 934 cen96534_ind_993_1018.indd 994 01/12/16 10:57 am 995 INDEX Brake horsepower, 797 Bridge scour, 779 British thermal unit (Btu), 19, 43 Broad-crested weir, 774–775 Btu. See British thermal unit (Btu) Buckets, 796, 853 Buckingham PI theorem, 309 comparison of wall shear stress and nondimensionalized wall shear stress, 325 friction in pipe, 322–324 guidelines for manipulation, 314 lift on wing, 319–322 from method of repeating variables, 313 method of repeating variables indicating zero Π’s, 318 nondimensional parameters, 315–316 pressure in soap bubble, 318–319 steps, 310–311, 314 Buffer layer, 368 Bulk modulus of compressibility, 45 Bulk modulus of elasticity, 45 Bulk strain rate, 157 Buoyancy, 100–102 effect, 11 force, 7 height of ice block below water surface, 103–104 measuring specific gravity by hydrometer, 102–103 Buoyant force, 100 C Calorie (cal), 19, 43 Candela (cd), 16 Capacity, 796 specific speed, 866 Capillaries, 59 Capillary effect, 59 capillary rise of water, 60 capillary rise to generating power in hydraulic turbine, 61 meniscus of colored water, 59 Cardiovascular system, 414, 415 Cartesian coordinates, 141–142, 155. See also Cylindrical coordinates calculating pressure field, 476–478 Navier–Stokes equation and continuity equations in, 474 stream function in, 456 calculation of stream function for known velocity field, 458–459 calculation of velocity field from stream function, 457–458 curves of constant stream function, 457 relative velocity deduced from streamlines, 461 two-dimensional streamlines, 459 volume flow rate deduced from streamlines, 461–462 Cartesian tensor notation, 158 Cascade, 827 Cauchy’s equation, 465, 466, 470. See also Continuity equation alternative form of Cauchy’s equation, 469 derivation using divergence theorem, 465–466 derivation using infinitesimal control volume, 466–469 derivation using Newton’s second law, 469–470 Cavitation, 41–43, 63, 211 bubbles, 42, 803 nuclei, 63 pump, 803–806 Center of buoyancy (B), 104–105 Center of gravity (G), 104–105 Center of pressure, 92, 93 Centipoise, 52 Centrifugal flow, 812 Centrifugal forces, 110, 822 Centrifugal pumps, 276, 812 centrifugal-flow pump, 812 circulatory flow loss, 821 close-up side view, 815 idealized blower performance, 816–817 law of cosines, 817 net head and brake horsepower performance curves, 814 net head as function of volume flow rate, 822 preliminary design of, 818–820 typical centrifugal blower, 813 Centripetal acceleration, 144, 272 Centripetal forces, 272, 813 CFD. See Computational fluid dynamics (CFD) CGPM. See General Conference of Weights and Measures (CGPM) Characteristic curve, 389 Chezy coefficient, 746 Choked Fanno flow, 716 in duct, 717–718 exit conditions in duct, 718–719 Choked flow, 678, 773 Choked Rayleigh flow, 708 in tubular combustor, 709–710 Churchill equation, 374 Circular pipe, laminar, 360 Circular pipes, 352 Circulatory flow loss, 821 Closed system, 15, 140, 164, 190, 191 Closed volume, 811 Coefficient of compressibility, 45 discharge, 398, 770 of surface tension, 56 of viscosity, 52 of volume expansion, 46–47 cen96534_ind_993_1018.indd 995 01/12/16 10:57 am 996 INDEX Cohesive forces, 59 Colebrook equation, 371–374, 716 Color Schlieren imaging techniques, 152 Combined efficiency, 200 Compressibility, 45 coefficient of compressibility, 45 coefficient of volume expansion, 46–47 density variation with temperature and pressure, 48 speed of sound and mach number, 49–51 vapor cloud arounds F/A-18F Super Hornet, 48 Compressible converging duct design, 451–453 Compressible flow, 10–11, 192, 667, 696. See also Incompressible flow; Internal flow; Open-channel flow adiabatic duct flow with friction, 710–719 Bernoulli equation for, 217–218 CFD calculations, 928 compressible flow through converging–diverging nozzle, 929–933 equations of motion, 929 oblique shocks over wedge, 933–934 for two-phase flows, 934–936 duct flow with heat transfer and negligible friction, 701–710 isentropic flow through nozzles, 677–685 one-dimensional isentropic flow, 671–677 shock waves and expansion waves, 685–701 shock-wave/boundary-layer interactions, 720 stagnation properties, 668–671 Compressible fluid element, 157 flow, 158 Compressible stream function, 464–465 Compression of high-speed air in aircraft, 670–671 Compressor, 794 Computational domain, 887 Computational fluid dynamics (CFD), 145, 324, 412, 478, 524, 584, 829, 885 additional equations of motion, 889 boundary conditions, 894–899 calculations, 886 for compressible flow, 928–936 for open-channel flow, 936–938 codes, 154 equations of motion, 886–887 grid generation and grid independence, 889–894 with heat transfer, 921–928 laminar CFD calculations, 899–908 motivation, 886 practice makes perfect, 899 software, 28, 559 solution procedure, 887–889 turbulent CFD calculations, 908–921 Conservation of angular momentum principle, 251 of energy, 190–191, 686 for oscillating steel ball, 202–203 principle, 219 equation, 887 laws, 190, 251 of linear momentum principle, 203 of mass, 190, 191, 387, 444, 675, 686, 744–745 alternative form of continuity equation, 449–450 conservation of mass principle, 193–195 continuity equation in cylindrical coordinates, 450 derivation using divergence theorem, 445–446 derivation using infinitesimal control volume, 446–449 incompressible flow, 196–198 incompressible flow, 451–456 mass balance for steady-flow processes, 195 moving or deforming control volumes, 195 principle, 193–195, 353 special cases of continuity equation, 450 steady compressible flow, 450–451 and volume flow rates, 191–193 of mechanical energy principle, 206 of momentum principle, 190, 250 principles, 189 of volume, 196 Constant error, 29 Constitutive equations, 470 Contact angle of water, 59, 60 Continuity and volumetric strain rate comparison, 455 Continuity equation, 190, 443, 444, 446, 476, 533–534, 702, 744, 886. See also Cauchy’s equation adiabatic duct flow with friction, 711 alternative form of continuity equation, 449–450 boundary conditions, 482 in Cartesian coordinates, 474 Couette flow with applied pressure gradient, 487–492 in cylindrical coordinates, 450, 475 derivation using divergence theorem, 445–446 derivation using infinitesimal control volume, 446–449 free-surface boundary conditions, 483 fully developed Couette flow, 483–486 fully developed flow in round pipe, 492–496 incompressible flow, 451–456 procedure, 481–482 property relations for Fanno flow, 713 rotational viscometer, 486–487 solutions of, 481–499 special cases of, 450 steady compressible flow, 450–451 sudden motion of infinite flat plate, 496–499 Continuum, 138 Contour lines, 801 Contour plots, 154–155 cen96534_ind_993_1018.indd 996 01/12/16 10:57 am 997 INDEX Contracted product, 254 Control mass. See Closed system Control points, 762 Control surface, 165 Control volume (CV), 15–16, 137, 138, 140, 164, 190, 191, 193, 222, 251–252, 255, 693 analysis, 252 forces acting on, 252–255 Convective acceleration, 141–142 Converging, 667 nozzles, 678 air loss from flat tire, 681 effect of back pressure, 679 isentropic flow of air in nozzle, 680–681 variation of mass flow rate, 679 Converging–diverging nozzles, 667, 673, 677, 682 airflow through converging–diverging nozzle, 684–685 compressible flow through, 929 CFD results, 930–931, 933 computational domain for, 929 mach number and pressure ratio, 932 velocity vectors and stagnation pressure contours, 933 effects of back pressure, 683 Conversion efficiency, 199 Core flow region. See Irrotational flow—region Couette flow, 183, 340, 483–486 with applied pressure gradient, 487–492 Counter-rotating axial-flow fan, 825 Coupled equations, 444 Creeping flow, 340, 524. See also Doublet flow approximation, 524–525 drag on a sphere in, 527–529 drag on object in, 526–527 Reynolds number, 524, 525–526 Critical depth, 738, 742 Critical energy, 742 Critical flow, 737 Critical length, 716 Critical point, 676, 742 Critical properties, 676 Critical ratios, 676 Critical Reynolds number, 354, 561 Critical slope, 747–748 Critical temperature and pressure in gas flow, 677 Critical uniform flow, 747–748 Critical velocity, 742 Cross-flow heat exchanger, temperature rising through, 921 close-up view of structured grid, 922 computational domain, 921 temperature contour plots, 922–923 Cross-flow over a cylinder, 633 Cunningham correction factor, 663 Curved surfaces. See also Plane surfaces gravity-controlled cylindrical gate, 99–100 Hydrostatic forces on submerged, 97–98 multilayered fluid, 98–99 Cut-in speed, 856 Cut-out speed, 856 CV. See Control volume (CV) Cylinders and spheres, flow over, 633 average drag coefficient, 635 flow visualization, 635 laminar boundary layer, 634 effect of surface roughness, 636–637 Cylindrical coordinates. See also Cartesian coordinates calculating pressure field in cylindrical coordinates, 479–481 continuity equation in, 450 Navier–Stokes equation and continuity equations in, 475 stream function in, 463–464 Cylindrical polar coordinates, 450 D D’Alembert’s paradox, 553 Darcy friction factor, 323, 359 Darcy–Weisbach friction factor, 359 Deadweight tester, 91 Deflection angle, 692 Deformation rates, 155 Deforming control volumes, 195, 252 Degree Kelvin (°K), 16 Del operator, 141 Demand curve, 388 Density, 39 density, specific gravity, and mass of air in room, 41 density of ideal gases, 40–41 specific weight, 40 variation with temperature and pressure, 48 Dependent Π, 306 Derived dimensions, 16 Design condition, 818 Detached oblique shock, 695 Detached shock, 934 Deviatoric stress tensor, 471 Different index of refraction, 151 Differential analysis of fluid flow of biofluid mechanics flows, 499–502 conservation of mass, 444–456 differential equations application of, 444 of fluid motion, 443 differential linear momentum equation, 465–470 Navier–Stokes equation, 470–475 exact solutions of continuity and, 481–499 no-slip boundary condition, 503 pressure field calculation for known velocity field, 476–481 problems, 476 stream function, 456–465 cen96534_ind_993_1018.indd 997 01/12/16 10:57 am 998 INDEX Differential approaches, 249 Differential equations, 22 application of, 444 of fluid motion, 443 Differential linear momentum equation, 465 alternative form of Cauchy’s equation, 469 derivation using divergence theorem, 465–466 derivation using infinitesimal control volume, 466–469 derivation using Newton’s second law, 469–470 Differential mass flow rate, 194 Differential pressure transducers, 90 Dilatant fluids, 53, 472 Dimensional analysis, 297, 830–833 Buckingham PI theorem, 309–325 dimensional homogeneity, 299–305 dimensions, 297–299 experiment and correlation of experimental data, 325–326 flows with free surfaces, 329–332 incomplete similarity, 326 method of repeating variables, 309–325 and similarity, 305–309 units, 297–299 wind tunnel testing, 326–329 Dimensional constant, 301 Dimensional homogeneity, 19, 299 of Bernoulli equation, 300 electric power generation by wind turbine, 19–20 nondimensionalization of equations, 300–305 obtaining formulas from unit consideration, 20–21 Dimensional variables, 301 Dimensionless group, 302 pump parameters, 831 turbine parameters, 861–864 application of turbine affinity laws, 863–864 for scaling two geometrically similar turbines, 862 Dimensions, 16, 297–299 dimensional homogeneity, 19–21 primary dimensions and units, 17 SI and English units, 17–19 standard prefixes in SI units, 17 unity conversion ratios, 21–22 weight of one pound-mass, 21 Direct numerical simulation (DNS), 908 Discharge coefficient, 398, 770 Discharge of water from tank example, 197–198 Disk area, 856 Dispersed two-phase flows, 935 Displacement thickness, 572–575 Distorted model, 329 Divergence theorem, 445–446, 465–466 Diverging flow sections, 209 DNS. See Direct numerical simulation (DNS) Doppler effect, 405 flowmeters, 404 ultrasonic flowmeters, 405 Double regulated turbines, 845 Doublet flow, 548–549. See also Creeping flow flow into vacuum cleaner attachment, 554–558 pressure coefficient, 552–553 pressure distribution on circular cylinder, 551 SP, 553–554 superposition of, 550 symmetric flow, 550–551 Doublet strength (K), 548 Downwash, 822 Draft tube, 846 Drag balance, 308 characteristics, 640 crisis, 635 Drag force, 51 acting on pipe in river, 637 Drift velocity, 348 Droplet formation, 597 Drowned outflow, 770 sluice gate with, 771–772 Drum gate, 770 Duct flow with heat transfer and negligible friction, 701 choked Rayleigh flow, 708–710 effects of heating and cooling on properties of Rayleigh flow, 704 extrema of Rayleigh line, 704–706 effect of heat transfer on flow velocity, 706 ideal gas with constant specific heats, 702 property relations for Rayleigh flow, 706–708 T-s diagram for flow, 703 Ducted axial-flow fan, 822 Ducted pumps, 796 Duty point, 798 Dynamic(s), 2 machines, 795, 796 pressures, 207–208, 798 pumps, 796, 812 similarity, 328 stall, 649 temperature, 669 turbines, 796, 840–841 viscosity, 52 E Eddies, 365 Eddy diffusivity of momentum. See Kinematic turbulent viscosity cen96534_ind_993_1018.indd 998 01/12/16 10:57 am 999 INDEX Eddy viscosity, 367 EES. See Engineering Equation Solver (EES) EGL. See Energy grade line (EGL) Elbows, 384 Electrostatic precipitator (ESP), 348 Elementary planar irrotational flows, 542 doublet flow, 548–549 line source/line sink flow, 544–546 line vortex flow, 546–548 superposition technique, 542–543 uniform stream flow, 543 Elevation head, 210, 741 Enclosed pumps, 796 Enclosed turbines, 796 Endplates, 643 Energy, 43–44 absorbing devices, 794 analysis of steady flows, 223 control volume, 224 incompressible flow with no mechanical work devices, 226 kinetic energy correction factor, 226–233 lost mechanical energy, 224 mechanical energy flow chart, 226 typical power plant, 225 dissipation ratio, 767 pattern factor, 857 principles, 675 producing devices, 794 transfer by heat, 220 by work, 220–223 Energy equations, 189, 219, 702, 744–745, 889. See also Bernoulli equations adiabatic duct flow with friction, 711 conservation of energy, 190–191 conservation of mass, 190–198 energy analysis of steady flows, 223–233 linear momentum equation, 190 mechanical energy and efficiency, 198–203 property relations for Fanno flow, 714 Energy grade line (EGL), 210–212, 796 Engineering Equation Solver (EES), 27 Engineering software packages, 26 CFD software, 28 equation solvers, 27 solving system of equations numerically, 27–28 English system, 16, 17–19 English units boiling and freezing point properties, 967 molar mass, gas constant, and ideal-gas specific heats of some substances, 966 properties of air at 1 atm pressure, 974 of atmosphere at high altitude, 977 of gases at 1 atm pressure, 975–976 of liquid metals, 973 of liquids, 972 of saturated ammonia, 970 of saturated propane, 971 of saturated refrigerant-134a, 969 of saturated water, 968 Enthalpy, 44, 668 Entrance region, 351, 355–357 Entropy, 702 change, 702, 711 increase of, 686 Entry lengths, 356–357 Equation of motion, 107 Equation of state, 40, 470, 702, 711 Equiangle skewness, 891 Equipotential lines, 539 Equivalent length, 380 ESP. See Electrostatic precipitator (ESP) Euler equation, 529 Euler number (Eu), 523 Euler turbomachine equation, 816, 841, 849 Euler’s turbine equation, 276 formula. See Euler turbomachine equation Eulerian approaches, 935 Eulerian derivative, 143 Eulerian descriptions, 138 acceleration field, 140–143 of fluid flow, 137 material derivative, 143–144 steady two-dimensional velocity field, 139–140 Eulerian formulations, 935 Expanding flow, 696 Expansion fan, 696 Expansion waves, 685 Prandtl–Meyer, 696–701 Experimental approaches,, 22 249 Extensional strain, 155 Extensive properties, 38, 166 External flow, 10. See also Internal flow over cylinders and spheres, 633–637 fluid flow, 351 lift, 638–650 F Family of curves, 145 Fan, 794 selection for air cooling of computer, 230–232 Fanning friction factor, 360 cen96534_ind_993_1018.indd 999 01/12/16 10:57 am 1000 INDEX Fanno flow, 710–719 choked, 716–719 effects of friction on properties, 713 functions, 964 property relations for, 713–716 Fanno line, 686, 687, 712 point of maximum entropy on, 689–690 Favorable pressure gradient, 583 Field variables, 138, 141 Filled contour plot, 154 Finite control volume approach, 249 Finite-span wings and induced drag, 642–643 First law of thermodynamics, 140, 219 First-order finite difference approximation, 143 Fixed control volume, 167, 251, 258 Floatmeter. See Variable-area flowmeters Flow over bump with negligible friction, 772–773 contraction, 383 control and measurement, 769 overflow gates, 772–778 underflow gates, 770–772 depth, 734 domain, 138, 444 energy, 199, 668 field, 139 with free surfaces, 329–332 measurement techniques, 416 with no external forces, 260 deceleration of spacecraft, 266–268 force to hold deflector elbow in place, 261–262 force to hold reversing elbow in place, 263 net force on flange, 268–269 power generation and wind loading of wind turbine, 265–266 water jet striking moving cart, 263–265 with no external moments, 275 separation process, 9 along streamline, 209 work, 44, 198, 223 Flow around circular cylinder, 903, 911–913 CFD simulation of flow, 906 computational domain, 903, 906 contour plot, 905 flow of fluid at free-stream speed, 903 grid resolution, 908 laminar flow, 907 streamlines, 905 two-dimensional grids, 904 vortices shedding from circular cylinder, 907 Flow patterns, 145 pathlines, 146–148 refractive flow visualization techniques, 151–152 spinning baseball, 145 streaklines, 148–150 streamlines and streamtubes, 145–146 surface flow visualization techniques, 152 timelines, 150–151 in unsteady flow, 150 Flow rate, 251, 373, 396. See also Internal flow biofluid mechanics, 414–422 electromagnetic flowmeters, 406–407 in laminar flow, gravity effect on, 361–362 LDV, 410–411 obstruction flowmeters, 398–401 in open channel in uniform flow, 748–749 pitot probes, 396–398 pitot-static probes, 396–398 PIV, 411–414 positive displacement flowmeters, 401–402 thermal anemometers, 408–409 turbine flowmeters, 402–403 ultrasonic flowmeters, 404–405 variable-area flowmeters, 403–404 vortex flowmeters, 407–408 Flow visualization, 145, 635 pathlines, 146–148 refractive flow visualization techniques, 151–152 spinning baseball, 145 streaklines, 148–150 streamlines and streamtubes, 145–146, 147 surface flow visualization techniques, 152 timelines, 150–151 Fluid elements compressed by piston in cylinder, 157 deformation of, 155 fundamental types of fluid element motion, 156 kinematic properties in two-dimensional flow, 158–160 per unit volume, 157 rate of rotation, 156 shear strain rate, 158 Fluid flow, 8, 249, 351 axisymmetric flow, 14 classification, 9 compressible vs. incompressible flow, 10–11 internal vs. external flow, 10 laminar vs. turbulent flow, 11 natural vs. forced flow, 11 one-, two-, and three-dimensional flows, 13–14 problems, 372 diameter of an air duct, 374–375 flow rate of air in duct, 377–378 pressure drop in water pipe, 375–376 turbulent draining from pool, 378–379 steady vs. unsteady flow, 12–13 three-dimensional Leibniz theorem applied to, 170 cen96534_ind_993_1018.indd 1000 01/12/16 10:57 am 1001 INDEX uniform vs. nonuniform flow, 14–15 viscous flows vs. inviscid flow regions, 10 Fluid kinematics, 137, 138 flow patterns and flow visualization, 145–152 fluidic actuators, 173 kinematic descriptions, 155–160 Lagrangian and Eulerian descriptions, 138–144 plots of fluid flow data, 152–155 RTT, 164–172 smelling food, 174 vorticity and rotationality, 160–164 Fluid mechanics, 2, 155, 189 accuracy, precision, and significant digits, 28–30 application areas of, 4–5 classification of fluid flows, 9–15 development of fluid systems, 6–7 dimensions and units, 16–22 engineering problems, 6 engineering software packages, 26–28 fluid, 2–4 fluid flow analysis, 1 fluid theory, 7–8 modeling in engineering, 22–24 no-slip condition, 8–9 nuclear blasts and raindrops, 32 problem-solving technique, 24–25 research and work, 8 significant digits and volume flow rate, 31 system and control volume, 15–16 Fluid(s), 138 disk element, 358 dynamics, 2, 4, 145 fluidic actuators, 173 friction effect on fluid temperature, 227–228 machines, 794, 795 particle, 138, 140, 141 acceleration of, 204 in rigid-body motion, 106 acceleration on straight path, 108–110 fluids at rest, 108 free fall of fluid body, 108 paraboloids of revolution, 111 pressure at point, 112 rigid-body motion of fluids, 107–108 rising of liquid during rotation, 113 rotation in cylindrical container, 110 statics, 91–92, 520 systems, 198 velocity, 151, 678, 682 Focal volume, 410 Foot (ft), 17 Force balance across streamlines, 206 Forced flow, 11 Forced vortex motion, 110 Forces acting on control volume, 252–255 Forward-inclined blades, 814 Fourier’s law of heat conduction, 334–335 Fractional factorial test matrices, 325 Francis mixed-flow turbine, 844 Francis radial-flow turbine, 844 Francis turbine, 844 Free delivery, 389, 797 Free outflow, 770 Free surface, 734 Frequency shift flowmeters. See Doppler effect—flowmeters Friction, 352, 932 coefficient, 360 drag, 634 effect on fluid temperature and head loss, 227–228 effects, 208 heating in pump, 228–229 lines, 152 slope, 745 velocity, 368, 580 Frictionless flow, 529 Froude number (Fr2), 302, 523, 737–741, 936 Full dynamic pressure, 552 Full factorial test matrix, 325 Full-flow electromagnetic flowmeter, 406 Full-scale prototype, 305 Fully developed region, 351 Fully rough flow, 372 Fully rough turbulent flow, 372, 746 Fundamental dimensions, 16, 298 G Gage pressure, 78 head, 741 transducers, 90 Gas(es), 4 at 1 atm pressure properties, 957–958, 975–976 dynamics, 2 flow, 352 turbines, 839, 853 Gauckler–Manning equations, 747 Gauss’s theorem, 445 General Conference of Weights and Measures (CGPM), 16 Generator efficiency, 200 Geometric similarity, 305, 306, 307 Geometrically scaled model, 305 Gliders, 648 Gliding, 648 Gradient operator, 141 Gradually varied flow (GVF), 735, 755. See also Rapidly varied flows (RVF) liquid surface profiles in open channels, 757–760 numerical solution of surface profile, 762–764 cen96534_ind_993_1018.indd 1001 01/12/16 10:57 am 1002 INDEX Gradually varied flow (GVF) (continued) representative surface profiles, 760–762 slow-moving river, 757 variation of properties over differential flow section, 756 Gravitational constant (gc), 21 Gravitational force acting on fluid element, 253 Gravitational vector in Cartesian coordinates, 253 Gravity, 253 effect on velocity and flow rate in laminar flow, 361–362 forces, 18, 274 gravity-controlled cylindrical gate, 99–100 Grid generation and grid independence, 889–894 commercial CFD codes, 892 computational domains, 893 hybrid grid, 893 sample two-dimensional hybrid grid, 893 skewness, 891 structured 2-D grid, 890 three-dimensional cells, 894 2-D unstructured grids, 890 unstructured grid, 890 Gross head, 845 Ground vortex, 177 GVF. See Gradually varied flow (GVF) H Harmonic conjugates, 539 Harmonic functions, 539 HAWTs. See Horizontal axis wind turbines (HAWTs) Head gate, 846 Head loss, 189, 359–361, 380, 383, 385–386 friction effect on, 227–228 Heat energy, 43 Heat transfer (Q), 219, 220 CFD with, 921 cooling of array of integrated circuit chips, 923–928 temperature rising through cross-flow heat exchanger, 921–923 duct flow with, 701–710 effect on flow velocity, 706 rate, 220 Hele–Shaw flow, 461 HGL. See Hydraulic grade line (HGL) Higher-viscosity fluid, 54 Homologous pump, 832 Horizontal axis wind turbines (HAWTs), 853 Horizontal pipe, 360 “Horse-shoe” vortex, 779 Horsepower (hp), 19 Hot-film anemometer. See Thermal anemometers Hot-wire anemometer. See Thermal anemometers hp. See Horsepower (hp) Hybrid grid, 892–893 Hydraulic grade line (HGL), 210–212, 733 Hydraulic(s), 2 depth, 740 diameter, 354, 736 Jack operation, 83 jump, 739, 765–769, 937 radius, 736 for best cross section, 754 turbines, 839 Hydrodynamic(s), 2 developing flow, 355 entrance region, 355 entry length, 355 fully developed region, 355 Hydroelectric power generation from dam, 229–230 Hydrogen bubble wire, 151 Hydrostatic(s), 91 forces on submerged curved surfaces, 97–100 on submerged plane surfaces, 92–97 pressure, 86 207 Hydroturbines, 839, 851–852 Hypersonic flow, 11 I ICs chips. See Integrated circuits chips (ICs chips) ID. See Inner diameter (ID) Ideal flow, 224 Ideal gas(es), 40, 669 density of, 40–41 property relations for Fanno flow, 714–716 property relations for isentropic flow, 675–677 Ideal power production, 845 Ideal-gas equation of state, 40, 687 Idealized Bernoulli-type flow, 211 IEC. See International Electrotechnical Commission (IEC) Image sink, 555 Impact pressure. See Full dynamic pressure Impeller, 813, 825, 839 Impeller blades, 796, 812 Impulse turbines, 841–843 Inclined manometers, 88 Incomplete similarity, 326 Incompressibility of unsteady two-dimensional flow example, 453 Incompressible continuity equation, 451 in Cartesian coordinates, 451 in cylindrical coordinates, 451 Incompressible flow, 10–11, 146, 147, 196–198, 209, 450, 451, 454–455. See also Compressible flow; Internal flow cen96534_ind_993_1018.indd 1002 01/12/16 10:57 am 1003 INDEX comparison of continuity and volumetric strain rate, 455 conditions for, 455–456 design of compressible converging duct, 451–453 incompressibility of unsteady two-dimensional flow, 453 missing velocity component, 453–454 Navier–Stokes equation derivation for, 472–474 with no mechanical work devices and negligible friction, 226 two-dimensional, incompressible, vortical flow, 454–455 Incompressible stream function in Cartesian coordinates, 456 Incompressible substances, 10, 44 Independent Π, 306 Induced drag, 642 Inertial coordinate system, 256 Inertial reference frame, 256 Inertial sublayer, 368 Infinitesimal control volume, 446–449, 466–469 Infinitesimal surface waves, 740 Inflow boundary conditions, 895–897 Inlet boundary conditions, 483 Inner diameter (ID), 875 Inner product, 254 Insertion electromagnetic flowmeters, 406 Inspectional analysis, 300 Instantaneous flow pattern, 148 Integral form, 257 Integrated circuits chips (ICs chips), 923 CFD results for chip cooling example, 926–927 comparison of CFD results, 926 computational domains for chip cooling example, 925 configurations of eight ICs on PCB, 924 cooling of array of, 923 PCBs, 924 temperature contours, 928 Integrated tracer particle location, 149 Intense mixing of fluid, 353–354 Intensive properties, 38, 166 Interferometry, 151 Internal boundary conditions, 898–899 Internal energy, 43, 224, 668 Internal flow, 10, 351, 633. See also Compressible flow; External flow; Incompressible flow circular pipes, 352 entrance region, 355–357 fluid flow, 351 laminar and turbulent flows, 353–355 laminar flow in pipes, 357–365 minor losses, 379–386 through pipes, 351 piping networks and pump selection, 386–396 turbulent flow in pipes, 365–379 International Electrotechnical Commission (IEC), 862 International System of Units (SI Units), 16, 17–19 boiling and freezing point properties, 949 Fanno flow functions, 964 molar mass, gas constant, and ideal-gas specfic heats of some substances, 948 Moody chart for friction factor, 960 one-dimensional isentropic compressible flow functions, 961 one-dimensional normal shock functions, 962 properties of air at 1 atm pressure, 956 of atmosphere at high altitude, 959 of gases at 1 atm pressure, 957–958 of liquid metals, 955 of liquids, 954 of saturated ammonia, 952 of saturated propane, 953 of saturated refrigerant-134a, 951 of saturated water, 950 Rayleigh flow functions, 963 Interrogation regions, 412–413 Inviscid flow, 529 Inviscid regions of flow, 10, 203 approximation for, 529 Bernoulli equation in, 530–533 Euler equation, 529–530 Irreversible head losses, 225, 798 Irrotational flow, 160, 163 region, 355 Irrotational flow approximation, 521, 533. See also Navier–Stokes equation Bernoulli equation in irrotational regions of flow, 535–536 continuity equation, 533–534 elementary planar irrotational flows, 542–549 irrotational flows forming by superposition technique, 549–558 momentum equation, 535 region model of Tornado, 536–538 superposition in irrotational regions of flow, 542 two-dimensional irrotational regions of flow, 538–541 Isentropic flow, 218, 674, 701, 932 of air in nozzle, 680–681 of ideal gases, 218 through nozzles, 677 converging nozzles, 678–681 converging–diverging nozzles, 682–685 property relations for isentropic flow of ideal gases, 675–677 Isentropic stagnation state, 669 Isocontour plots. See Contour plots Isolated system, 15 cen96534_ind_993_1018.indd 1003 01/12/16 10:57 am 1004 INDEX Isothermal compressibility (α), 46 Isothermal flow, Navier–Stokes equation derivation for, 472–474 Isothermal process, 218 J Joule (J), 19, 43 K Kaplan turbine, 844 Kármán integral equation, 589 flat plate boundary layer analysis using, 590–591 Kármán vortex street, 149, 905 Kelvin (K), 40 Kilogram (kg), 16, 17 Kilogram-force (kgf), 18 Kilojoule (kJ), 19, 43 Kilopascal (kPa), 78 Kilowatt-hour (kWh), 19 Kinematic(s), 138 eddy viscosity, 367 properties in two-dimensional flow, 158–160 similarity, 305, 306 turbulent viscosity, 367 viscosity, 53 Kinetic energy, 43, 198, 199, 224 correction factor, 226 fan selection for air cooling of computer, 230–232 friction effect on fluid temperature and head loss, 227–228 hydroelectric power generation from dam, 229–230 pumping power and frictional heating in pump, 228–229 pumping water from lake to pool, 232–233 of flow, 384 nergy flux correction factor, 274 King’s law, 409 Knudsen number, 39, 663 L Lagrangian CFD approaches, 935 Lagrangian derivative, 143 Lagrangian descriptions, 138 acceleration field, 140–143 of fluid flow, 137 material derivative, 143–144 steady two-dimensional velocity field, 139–140 Lagrangian formulations, 935 Laminar boundary layer on flat plate, 569–572 Laminar component, 366 Laminar flow, 11, 259, 353–355. See also Turbulent flow CFD calculations, 899 flow around circular cylinder, 903–908 pipe flow entrance region, 899–902 in channels, 735–737 in noncircular pipes, 362–365 in pipes, 357. See also Turbulent flow—in pipes fluid disk element, 358 effect of gravity on velocity and flow rate in, 361–362 laminar flow in noncircular pipes, 362–365 pressure drop and head loss, 359–361 Laminar pipe flow, momentum-flux correction factor for, 258–259 Laminar regime, 832 Laminar sublayer, 368 Langley full-scale wind tunnel (LFST), 327 Laplace equation, 534 Laplacian operator (∇2), 473, 534 Large eddy simulation (LES), 909 Laser Doppler anemometry (LDA), 410 Laser Doppler velocimetry (LDV), 410–411 Laser velocimetry (LV), 410 Laval nozzles, 673 Law of cosines, 817 Law of dimensional homogeneity, 299 Law of wake, 580 LDA. See Laser Doppler anemometry (LDA) LDV. See Laser Doppler velocimetry (LDV) Le Système International d’ Unités. See International System of units (SI units) Leading edge angle, 815 Leading edge vortex (LEV), 648, 649 Left-side convention, 460 Leibniz theorem, 168, 169 LES. See Large eddy simulation (LES) LEV. See Leading edge vortex (LEV) LFST. See Langley full-scale wind tunnel (LFST) Lift, 638 airfoils, 638 birds/insects using flapping wing motion to fly, 649–650 coefficient, 320 and drag characteristics, 640 and drag of commercial airplane, 644–645 finite-span wings and induced drag, 642–643 effect of flaps on, 641 flying in nature, 647 generation by spinning, 643, 645, 647 gliders, 648 irrotational and actual flow, 639 large airplanes with flapping wings, 650 large birds, 648 medium birds, 649 small birds and insects, 649 cen96534_ind_993_1018.indd 1004 01/12/16 10:57 am 1005 INDEX effect of spin on tennis ball, 646 variation of lift coefficient, 642 variation of lift-to-drag ratio, 640 Lift-to-drag ratio, 639, 640 Line sink flow, 544–546 rotationality of, 164 superposition of, 549 Line source flow, 544–546, 546–548 Line vortex, 163, 454 superposition of, 549 Linear momentum, 249, 250, 255. See also Angular momentum equation, 190, 204, 250, 255, 686 flow systems, 256 flow with no external forces, 260–269 momentum-flux correction factor, 257–259 RTT, 256 special cases, 257 steady flow, 259–260 Linear strain, 155 Linear strain rate, 137, 155, 156, 157 in Cartesian coordinates, 156 Linear sublayer, 368 Lines rotate counterclockwise, 155 Liquid flow, 352 Liquid layer of thickness, 736–737 Liquid metals properties English units, 973 SI units, 955 Liquid surface profiles in open channels, 757–760 Liquid–gas interface, 734 Liquids properties English units, 972 SI units, 954 Load, 797 Local acceleration, 141 Log law, 580 Logarithmic law, 369–370 Loss coefficient, 379, 382–383 Low Reynolds number flow, 524 Lower-viscosity fluid, 54 LV. See Laser velocimetry (LV) M Mach angle, 696 Mach number (Ma), 11, 667, 673, 675, 696 of air entering diffuser, 51 and pressure ratio, 932 property relations for Fanno flow, 714 speed of sound and, 49–51 Mach wave, 696 Macroscopic energy, 43 Magnitude of angular momentum, 271 of moment of force, 272 of torque, 270 of velocity, 153 Magnus effect, 643 Major losses, 379. See also Minor losses Manning coefficient, 747 Manning equations, 747 Manometer, 87, 88. See also Barometer measuring pressure with, 88 measuring pressure with multifluid, 89–90 Manta ray swimming, 280–281 Markers, 413 Mass, 38 of air in room, 41 balance, 193 for steady-flow processes, 195 concentration, 335 equation, 189 flow rate, 191–193, 195, 257, 796 Material acceleration, 143 of steady velocity field, 144 Material derivative, 143, 172 of pressure, 143–144 Material element, 449–450 Material particle, 140 Material position vector, 140 Material volume, 170 of fluid element, 157 Mathematical singularity, 163 MAVs. See Micro Air Vehicles (MAVs) Maximum length, 716 Mean pressure, 471 Measurement volume, 410 Mechanical energy, 189, 224 balance, 189, 206 conservation of energy for oscillating steel ball, 202–203 and efficiency, 198 ideal hydraulic turbine coupled, 199 loss, 189 pumping water from lake to storage tank, 201–202 of water at bottom of container, 200 Mechanics, 2 efficiency, 198, 199 power generation, 853 pressure, 471 Megapascal (MPa), 78 MEKA arrays. See Microelectrokinetic actuator arrays (MEKA arrays) MEMS. See Microelectromechanical systems (MEMS) Meniscus, 59 Metacentric height (GM), 105 Meter (m), 16, 17 cen96534_ind_993_1018.indd 1005 01/12/16 10:57 am 1006 INDEX Method of repeating variables, 297, 309 common parameters as repeating parameters, 313 comparison of wall shear stress and nondimensionalized wall shear stress, 325 friction in pipe example, 322–324 guidelines for choosing repeating parameters, 312 guidelines for manipulation of Π’s resulting from, 314 indicating zero Π’s, 318 lift on wing example, 319–322 pressure in soap bubble example, 318–319 steps, 310–311, 314 Micro Air Vehicles (MAVs), 649 Microelectrokinetic actuator arrays (MEKA arrays), 652 Microelectromechanical systems (MEMS), 652 Microscopic energy, 43 Microsoft Excel, 27 Minor losses, 379 constant-diameter section of pipe with, 379 flow contraction and head loss, 383 head loss, 380 head loss and pressure rise during gradual expansion example, 385–386 kinetic energy of flow, 384 loss coefficient, 382–383 losses during changes of direction, 384 effect of rounding of pipe inlet, 384 total head loss, 380–381 Mixed flow, 812. See also Turbulent flow pump, 812 turbine, 844–845 Mixing length, 367 Modified pressure (P′), 524 Molar mass, 40, 334 gas constant, and ideal-gas specific heats of some substances, 948, 966 Mole (mol), 16 Molecular weight. See Molar mass Moment, 270 Moment arm, 270 Moment of force, 272 Moment of inertia, 270 Moment of momentum. See Angular momentum Momentum, 250 of body, 190 equation, 535 flow rate, 257 thickness, 575–576 Momentum analysis of flow systems angular momentum equation, 272–280 control volume, 251–252 forces acting on control volume, 252–255 linear momentum equation, 255–269 manta ray swimming, 280–281 Newton’s laws, 250–251 rotational motion and angular momentum, 269–272 Momentum integral technique for boundary layers, 587 conservation of x-momentum, 588–589 control volume, 587–588 creeping flow approximation, 592–594 drag on wall of wind tunnel test section, 591–592 flat plate boundary layer analysis, 590–591 velocity overshoot, 594–595 Momentum-flux correction factor, 257 for laminar pipe flow, 258–259 Moody chart, 372 and equations, 370–372 for friction factor, 960 Moody efficiency correction equation for pumps, 833 correction equation for turbines, 862 Motion, 138 compressed by piston in cylinder, 157 kinematic properties in two-dimensional flow, 158–160 rate of rotation, 156 shear strain rate, 158 types of, 155, 156 Motor efficiency, 200 Moving control volumes, 195 MRI-phase contrast velocimetry (MRI-PCV), 421 Multigridding, 889 Multistage axial-flow compressor, 828 pump, 828 turbine, 828 Mutual orthogonality, 539 N NACA. See National Advisory Committee for Aeronautics (NACA) Nappe, 775 National Advisory Committee for Aeronautics (NACA), 641 Natural flow, 11 Navier–Stokes equation, 443, 470, 476, 520, 886–887, 928. See also Boundary layer—approximation; Irrotational flow approximation approximate solution, 520–521 approximation for inviscid regions of flow, 529–533 boundary conditions, 482 and continuity equations in Cartesian coordinates, 474 in cylindrical coordinates, 475 Couette flow with applied pressure gradient, 487–492 creeping flow approximation, 524–529 derivation for incompressible, isothermal flow, 472–474 cen96534_ind_993_1018.indd 1006 01/12/16 10:57 am 1007 INDEX droplet formation, 597 free-surface boundary conditions, 483 fully developed Couette flow, 483–486 fully developed flow in round pipe, 492–496 Newtonian fluids vs. non-Newtonian fluids, 471–472 nondimensionalized equations of motion, 521–524 procedure, 481–482 rotational viscometer, 486–487 solutions of, 481 sudden motion of infinite flat plate, 496–499 Negative shear strain, 157 Negligible friction duct flow with, 701–710 flow over bump with, 772–773 incompressible flow with, 226 Negligible heat transfer, 209 Negligible viscous effects, 208 Net force on flange, 268–269 Net head, 796 for hydraulic turbine, 847 for liquid pump, 796 Net mass flow rate, 194 transfer, 193 Net positive suction head (NPSH), 803–806 Neutrally stable situation, 104–105 Newton (N), 17 Newton-meter (N.m), 19 Newton’s law(s), 138, 250–251 first law, 250 laws of motion, 255 second law, 17, 140, 141, 189, 190, 250, 255, 469–470 second law of motion, 106 third law, 250 Newtonian fluids, 52, 471–472 Newtons per square meter (N/m2), 78 No external moments, 275 No shaft work, 209 No-slip condition, 8–9 boundary condition, 503 No-temperature-jump condition, 9 Non-Newtonian fluids, 471–472 Noncircular pipes, 352 Nondimensional, 300 parameters, 297, 300, 302 pump performance curves, 832 roughness parameter, 831 variables, 301 Nondimensionalization of equation(s), 300, 305 advantages, 302, 303 extrapolation of nondimensionalized data, 304 Froude number, 302 nondimensionalized form of Bernoulli equation, 300 scaling parameters, 301 throwing baseball on Moon, 304–305 unsteady fluid flow problem, 305 vertical velocity, 301 Nondimensionalized equations of motion, 521 modified pressure, 524 nondimensional grouping of parameters, 522–523 scaling parameters, 521–522 Nondimensionalized Navier–Stokes equation, 523 Nonuniform flow, 14–15, 734, 735 perimeters, superposition method for, 748–751 roughness, channels with, 750–751 Nonzero even for steady flow, 142 Normal acceleration, 204 Normal components, 253 of velocity, 194 Normal depth, 735, 746, 762 Normal shock(s), 683, 686. See also Oblique shocks air inlet of supersonic fighter jet, 688 h-s diagram for flow, 687 point of maximum entropy on Fanno line, 689–690 Schlieren image of blast wave, 688 shock wave in converging–diverging nozzle, 690–691 variation of flow properties, 687 waves, 686 Normal stress, 3, 78, 91, 253 Normalized equation, 300 Nozzles, 853 meters, 398–401 NPSH. See Net positive suction head (NPSH) Nuclear blasts and raindrops, 32 Numerical solution(s), 145 of surface profile, 762 classification of channel slope, 764 gradually varied flow with M1 surface profile, 762–764 Nutating disc, 840 Nutating disk flowmeters, 402 O Oblique shocks, 684, 691, 695 color Schlieren image, 696 dependence of straight oblique shock deflection angle, 694 detached oblique shock, 695 normal mach numbers, 693 relationships across oblique shock for ideal gas, 694 Schlieren image of small model, 692 shock angle 𝛽 formed by slender, 692 still frames from Schlieren videography, 695 velocity vectors, 693 over wedge, 933–934 cen96534_ind_993_1018.indd 1007 01/12/16 10:57 am 1008 INDEX Ohm’s law, 334 One inlet, steady flow with, 260 One outlet, steady flow with, 260 One-dimensional diffusion equation, 497 One-dimensional flows, 13–14 One-dimensional isentropic compressible flow functions, 961 One-dimensional isentropic flow, 671 gas flow through converging–diverging duct, 672–673 property relations for isentropic flow of ideal gases, 675–677 variation of fluid properties in flow direction, 673–675 One-dimensional Leibniz theorem, 169, 170 One-dimensional normal shock functions, 962 One-dimensional variable, 734 One-seventh-power law, 577 Open axial-flow fans, 822 Open pumps, 796 Open system, 15, 140 Open turbines, 796 Open-channel flow, 10, 211, 351, 733. See also Compressible flow best hydraulic cross sections, 751–755 bridge scour, 779 CFD calculations, 936 flow over bump on bottom of channel, 936–937 flow through sluice gate, 937–938 classification, 734 conservation of mass and energy equations, 744–745 flow control and measurement, 769–778 Froude number and wave speed, 737–741 GVF, 755–764 laminar and turbulent flows in channels, 735–737 RVF and hydraulic jump, 765–769 specific energy, 741–744 uniform and varied flows, 734–735 uniform flow in channels, 745–751 Operating point, 389, 798 of fan in ventilation system, 799–801 Orifice meters, 398–401 Outer flow region, 559 Outer layer, 368, 580 Outer product of velocity vector, 466 Outflow boundary conditions, 895–897 Outlet boundary conditions, 483 Output shaft power, 843 Overall efficiency, 200 Overflow gates, 772 broad-crested weir, 774–775 flow over bump with negligible friction, 772–773 sharp-crested weirs, 775–778 Overlap layer, 368 P Paddlewheel flowmeters, 403 Parallel, pumps in, 806–809 Parallel axis theorem, 94 Parallel pipes, 386–387 Parameters, 301 Partial derivative operator, 141 Partial pressure, 42 Particle derivative, 143 Particle image velocimetry (PIV), 147, 173, 411–414 Particle relaxation time, 346 Particle shadow velocimetry (PSV), 422 Pascal (Pa), 78 Pascal’s law, 82, 89 Passage losses, 821 Path functions, 191 Pathlines, 146–148 PCBs. See Printed circuit boards (PCBs) Pediatric ventricular assist device (PVAD), 416 Pelton wheel, 841 Penstock, 846 Performance curve, 389 Periodic boundary conditions, 897 Periodic flow, 12 Peristaltic pump, 809 Piezoelectric transducers, 91 Piezometer, 208 tube, 208 Pipe diameter, 373 Pipe flow entrance region axisymmetry, 899 CFD results, 901 coarsest structured grids, 900 laminar CFD calculations, 899 normalized axial velocity contours, 902 pressure drop, 901 residuals with iteration number, 900 Pipes, 352 Piping networks, 386, 387 Piping system pump to, 797 example of manufacturer’s performance plot, 802 impeller size, 802–803 operating point of fan in ventilation system, 799–801 operating point of piping system, 798 with pumps and turbines, 388 characteristic pump curves, 389 efficiency of pump–motor combination, 388 effect of flushing on flow rate from shower example, 393–396 gravity-driven water flow in pipe example, 392–393 pumping water through two parallel pipes example, 389–391 cen96534_ind_993_1018.indd 1008 01/12/16 10:57 am 1009 INDEX Pitch angle, 822 Pitot formula, 397–398 Pitot tube, 208, 215 Pitot-static probe, 208 Pitot–Darcy probes, 208, 397 PIV. See Particle image velocimetry (PIV) Planar flow, 463, 538 Planar irrotational regions of flow, 539–540 Plane surfaces. See also Curved surfaces center of pressure, 92, 93 centroid and centroidal moments of inertia, 94 hydrostatic forces on submerged, 92 pressure at centroid of, 93 pressure prism, 95 second moment of area, 93–94 submerged rectangular plate, 95–97 Planform area, 319, 638 Plastic fluids, 472 Plot flow data, 152 Plots of fluid flow data, 152 contour plots, 154–155 profile plots, 153 vector plots, 153–154 Point function, 143, 192 Poise, viscosity unit, 52 Poiseuille flow, 185, 345, 492–496 Poiseuille’s law, 360 Position vector, 138 Positive shear strain, 157 Positive-displacement machines, 795 pump, 795, 809–812 turbine, 795, 840 Postprocessors, 154 Potential energy, 43, 198, 199, 224 Potential function, 533 Pound (lb), 18 Pound-force (lbf), 17 Pound-mass (lbm), 17 Power, 220 coefficient, 857 generation from sprinkler system, 278–280 specific speed, 864 Power-law velocity profile, 370 Prandtl equation, 371 Prandtl–Meyer expansion waves, 696 axisymmetric cone, 697 calculations, 700–701 complex interactions between shock waves and expansion waves, 698 estimation of Mach number from mach lines, 698–699 oblique shock calculations, 699–700 Pratt & Whitney PW4000 turbofan engine, 828 Precision, 28–30 error, 28 Pressure (P), 3, 38, 78 Barometer, 84–87 coefficient, 337, 551 drag, 634 drop, 359–361, 373, 387 far field in ANSYS-FLUENT, 928 in fluid, 81–83, 794 forces, work done by, 221–223 head, 210 hydraulic Jack operation, 83 inlet, 895 intensifiers, 83 manometer, 87–90 material derivative, 143–144 measurement devices, 84 other pressure measurement devices, 90–91 outlet, 895 at point, 79–80 prism, 95 transducers, 90 variation with depth, 80 Pressure correction algorithm, 479 Pressure field, 138 calculation, 476 in Cartesian coordinates, 476–478 compressible flow fields, 478–479 in cylindrical coordinates, 479–481 Pressure gradient(s), 107 boundary layer separation, 583–584 boundary layers with, 582 CFD, 584–586 calculation of turbulent flow, 587 flat plate boundary layers, 582–583 outer flow, 586–587 Preswirl, 826 Primary dimensions, 16, 298 of angular momentum, 298–299 and primary SI and English units, 298 Principle of dimensional homogeneity, 297 Printed circuit boards (PCBs), 923 Problem-solving technique, 1, 24–25 Profile plots, 153 Propeller, 822 flowmeters, 402 mixed-flow turbine, 844 turbine, 845 Property of system, 38 of fluids capillary effect, 59–61 cavitation, 63 compressibility and speed of sound, 45–51 continuum, 38–39 cen96534_ind_993_1018.indd 1009 01/12/16 10:57 am 1010 INDEX Property of system (continued) density and specific gravity, 39–41 energy and specific heats, 43–44 surface tension, 56–58 vapor pressure and cavitation, 41–43 viscosity, 51–56 relations for Fanno flow, 713–716 relations for isentropic flow of ideal gases, 675 critical temperature and pressure in gas flow, 677 critical-pressure, critical-temperature, and critical-density ratios, 677 properties at nozzle throat critical properties, 676 relations for Rayleigh flow, 706–708 Pseudoplastic fluids, 53, 472 PSV. See Particle shadow velocimetry (PSV) Pump scaling laws affinity laws, 835–839 dimensional analysis, 830–833 pump specific speed, 833–835 Pump(s), 793, 794, 796. See also Piping system; Turbomachinery axial pumps, 822–830 centrifugal pumps, 812–821, 822 dynamic pumps, 812 efficiency, 200, 797 head, 225, 226 net head of pump, 797 net positive suction head, 803–806 performance curves, 797 performance curves, 797–803 positive-displacement pumps, 809–812 for preliminary pump design, 834–835 pump cavitation, 803–806 pump to piping system, 797–803 pumping power and frictional heating in, 228–229 in series and parallel, 806–809 specific speed, 833 Pumping power, in pump, 228–229 Pumping water from lake to pool, 232–233 from lake to storage tank, 201–202 Pump–turbine(s), 245, 864 Pure constants, 301 PVAD. See Pediatric ventricular assist device (PVAD) Q Quantity of matter of fixed identity, 164 R Radial blades, 814 Radial-flow. See also Turbulent flow devices, 275 annular control volume, 276 bending moment acting at base of water pipe, 277–278 power generation from sprinkler system, 278–280 side and frontal views of typical centrifugal pump, 276 pumps, 812 Raindrops, nuclear blasts and, 32 Rake of streaklines, 149 Rankine scale (R), 40 RANS equation. See Reynolds-averaged Navier–Stokes equation (RANS equation) Rapid flow, 738 Rapidly varied flow (RVF), 735, 755, 765–769. See also Gradually varied flow (GVF) Rarefied gas flow theory, 39 Rate of change of angular velocity, 270 Rate of rotation, 137, 155, 156, 159 vector, 156, 160 in Cartesian coordinates, 156 Rate of translation, 137, 159 vector, 155 Rate of volumetric dilatation, 157 Rated speed, 856 Ratio of required shaft power, 836 Rayleigh flows, 701 functions, 963 property relations for, 706–708 Rayleigh line, 686, 688, 703 extrema of, 704–706 Reaction turbines, 843. See also Turbine(s) aerial view of Hoover dam, 848 effect of component efficiencies on plant efficiency, 850–851 differs significantly from impulse turbine, 844 five-bladed propeller turbine, 846 hydroturbine design, 851–852 relative and absolute velocity vectors, 849 runner of Francis radial-flow turbine, 845 typical setup and terminology for hydroelectric plant, 847 visualization from CFD analysis, 850 Real flow, 224 Reattachment line, 927 Rectangular channel, 736, 753 Rectangular weir, 775 Refracted rays, 151, 152 Refractive flow visualization techniques, 151–152 Refractive property, 151 Region in space chosen for study, 165 Regions of potential flow, 533 Regression analysis, 325–326 Relative density, 39 Relative roughness, 370–371 Relative velocity, 168, 195, 251 RTT in terms of, 171 Repeating variables method, 830 cen96534_ind_993_1018.indd 1010 01/12/16 10:57 am 1011 INDEX Required net head, 798 Required net positive suction head (NPSHrequired), 804 Residence time, 142, 143 Resistance coefficient, 379 Retronasal olfaction, 174 Reverse flow, 583, 896 Reverse swirl, 849 Reverse thrust, 825 Reverse-flow cyclone, 348 Reversible adiabatic flow, 701 Revolutions per minute (rpm), 54 Reynolds number (Re), 11, 307, 351, 354–355, 523, 524, 633, 692, 735, 800, 831, 903 independence, 327, 328, 832 Reynolds stresses, 366 Reynolds transport theorem (RTT), 137, 164, 194, 249, 256, 335 alternate derivation, 169–171 control surface gives net amount, 167 moving system, 166 one well-defined inlet, 169 relationship between material derivative and, 172 relative velocity, 168 two methods of analyzing spraying of deodorant, 165 Reynolds-averaged Navier–Stokes equation (RANS equation), 720, 909 Rheology, 471 Richardson number, 339 Rigid-body motion, fluids in, 106 acceleration on straight path, 108–110 fluids at rest, 108 free fall of fluid body, 108 paraboloids of revolution, 111 pressure at point, 112 rigid-body motion of fluids, 107–108 rising of liquid during rotation, 113 rotation in cylindrical container, 110 Rotameter. See Variable-area flowmeters Rotary fuel atomizers, 867–868 Rotary pump, 809, 812 Rotational/rotation, 155 kinetic energy, 271–272 motion, 269–272 periodic boundary conditions, 897 rate of, 137 viscometer, 486–487 Rotationality, 160 comparison of circular flows, 163–164 rotation of fluid elements, 160–161 rotationality of line sink, 164 rotationally symmetric flows, 897 in two-dimensional flow, 162 two-dimensional flow in Cartesian coordinates, 161 vorticity contours in two-dimensional flow, 161–162 Rotodynamic pumps, 812 Rotor, 813, 825 blades, 812 rpm. See Revolutions per minute (rpm) RTT. See Reynolds transport theorem (RTT) Runge–Kutta numerical integration technique, 150 Runner, 839 blades, 796, 843 leading edge, 849 trailing edge, 849 RVF. See Rapidly varied flow (RVF) S Saturated ammonia properties English units, 970 SI units, 952 Saturated propane properties English units, 971 SI units, 953 Saturated refrigerant-134a properties, English units, 969 properties, SI units, 951 Saturated water properties, English units, 968 properties, SI units, 950 Saturation pressure, 41, 803 temperature, 41 Scalar field variable, 138 Scalar quantity, 156 Scalar variable, 153 Scaling parameters, 301, 521–522 Schlieren image, 152 Schlieren technique, 151 Scroll, 812 Second (s), 16, 17 Second moment of area, 93 Second-order tensor, 253 Secondary dimensions, 16 Seeding particles, 411 Seeds, 411 Self-priming pump, 811 Separation bubble, 583, 927 Series pipes in, 386–387 pumps in, 806–809 SGS models. See Sub-grid scale models (SGS models) Shadowgram, 152 Shadowgraph technique, 151 Shaft power, 271 Shaft work, 199, 221 cen96534_ind_993_1018.indd 1011 01/12/16 10:57 am 1012 INDEX Sharp-crested rectangular weir, 776 Sharp-crested triangular weir, 777 Sharp-crested weirs, 775 measuring flow rate by weir, 778 subcritical flow over bump, 777–778 triangular weir, 776 Shear force, 52 Shear strain, 2, 155 Shear strain rate, 137, 155, 157, 158 in Cartesian coordinates, 158 Shear stress, 3, 253 Shear thickening fluids, 53, 472 Shear thinning fluids, 53, 472 Shock angle, 692 Shock wave(s), 344, 451, 667, 685, 686 normal shocks, 686–691 oblique shocks, 691–696 Shock-wave/boundary-layer interactions, 720 Shockless entry condition, 815 Shrink, 159 Shroud, 813 Shutoff head, 389, 797 SI Units. See International System of Units (SI Units) Significant digits, 28–30 volume flow rate and, 31 Similarity dimensional analysis, 305–309 Similarity rules, 835 Simple blood viscosity model, 501–502 Single regulated Kaplan turbines, 845 Single-stream devices, 224 steady-flow systems, 195 systems, 260 Singular point, 544 Singularity, 544 Siphoning out gasoline from fuel tank, 213215 Skewness, 891 Skin friction drag, 9 Slip length, 503 Sluice gate, 770 with drowned outflow, 771–772 Smoke wire, 148 Snail-shaped casing, 812 diffuser, 813 Solid-body rotation, 163 Solid-state pressure transducers, 91 Sonar system. See Sound navigation and ranging system (Sonar system) Sonic flow, 11 Sonic length, 716 Sonic speed, 49 Sonic velocity, 674 Sonoluminescence, 63 Sound intensity, 348 Sound navigation and ranging system (Sonar system), 63 Sound waves, 685 SP. See Stagnation point (SP) Spalding’s law of wall, 580 Span, 638 Specific energy, 741 character of flow and alternate depth, 743–744 liquid in open channel, 742 sluice gate illustrates alternate depths, 742 variation of specific energy Es, 742 Specific gravity, 39 density, specific gravity, and mass of air in room, 41 density of ideal gases, 40–41 specific weight, 40 of substance, 39 Specific heats, 43–44 Specific properties, 38 Specific Reynolds stress tensor, 909 Specific volume, 39 Specific weight, 18, 40 Speed of sound, 45, 667, 671 coefficient of compressibility, 45 coefficient of volume expansion, 46–47 density variation with temperature and pressure, 48 mach number and, 49–51 vapor cloud around F/A-18F Super Hornet, 48 Speed of surface waves, 739–741 Spin effect on tennis ball, 646 Spinning, lift generation by, 643, 645, 647 Spiral casing, 843 Spraying water into air, 212 Stability of immersed and floating bodies, 104–106 Stable situation, 104–105 Stagnation enthalpy, 668 pressure, 207–208, 669, 707–708 properties, 668–671 state, 667 streamline, 208 temperature, 669, 678, 707–708 Stagnation point (SP), 208, 553 Stall, 641 condition, 584 Starting vortex, 639 State postulate, 38 Static enthalpy, 668 Static pressure, 207–208, 798 Static pressure tap, 207, 551–552 Static-to-stagnation pressure ratio, 679 Statics, 2 Stationary bodies, 211 Statistical approach, 4 cen96534_ind_993_1018.indd 1012 01/12/16 10:57 am 1013 INDEX Stator blades, 825 Stator design for vane-axial flow fan, 913 blade spacing, 914 CFD results for flow, 919 computational domain, 914 grid for two-dimensional stator vane cascade, 915 pressure contour plot, 918 streamlines producing by CFD calculations, 916 tangential velocity contour plot, 919 three-dimensional computational domain, 918 vane-axial flow fan, 913 variation of average outlet flow, 915 velocity vectors producing by CFD calculations, 917 vorticity contour plots, 920 vorticity contour plots producing by CFD calculations, 916 Stay vanes, 843 Steady compressible flow, 450–451 Steady flow(s), 12–13, 142, 148, 195, 204, 205, 208, 257, 259, 274, 734 devices, 12 energy analysis of, 223–233 with one inlet and one outlet, 260 process, 673 processes, mass balance for, 195 system, 195 Steady incompressible flow, 227 Steady linear momentum equation, 259 Steady two-dimensional velocity field, 139–140 Steady velocity field, material acceleration of, 144 Steam turbines, 839, 853 Stoke, 53 Stokes approximation with Cunningham correction, 663 Stokes flow, 524 Stored potential energy per unit volume, 198 Straight blades. See Radial blades Strain rate tensor in Cartesian coordinates, 158 Strain-gage pressure transducers, 91 Streaklines, 148 flow patterns in unsteady flow, 150 Kármán vortices, 149 smoke streaklines, 149 Stream function, 336, 443, 539 in Cartesian coordinates, 456–462 compressible, 464–465 in cylindrical coordinates, 463–464 Streamline(s), 145, 147, 204, 586 relative velocity deduced from, 461 two-dimensional, 459 volume flow rate deduced from, 461–462 in xy-plane, 146 Streamtubes, 145–146, 147 Streamwise acceleration, 204 Stress, 3 Stress tensor, 253, 465 in Cartesian coordinates, 253 Strouhal number (St), 407, 523, 908 Sub-grid scale models (SGS models), 935 Subcritical flow, 737, 738 over bump, 777–778 Submerged outflow, 770 Subsonic diffuser, 675 Subsonic flow, 11, 674, 695 Substantial derivative, 143 Sudden expansion, 384 Supercritical flow, 737, 738 Superposition method irrotational flows forming by, 549 in irrotational regions of flow, 542 for nonuniform perimeters, 748 channels with nonuniform roughness, 750–751 flow rate in open channel in uniform flow, 748–749 height of rectangular channel, 749 superposition of line sink and line vortex, 549 superposition of uniform stream and doublet flow, 550–558 Superposition of uniform stream and doublet, 550–558 Supersonic flow, 11, 695 Supply curve, 389 Surface drag, 9 Surface energy, 56 Surface flow visualization techniques, 152 Surface forces, 106, 252, 274 on differential surface element, 254 Surface oil visualization, 152 Surface profile, 757, 760–762 Surface roughness, 636–637 Surface tension, 56–58 Surfactants, 56 Surge tower, 46 Surroundings, 15 Swirl, 823 Swirling axisymmetric flows, 897 Symmetry boundary condition, 897 System, 15–16 analysis, 138 curve, 388 system-to-control-volume transformation, 167 Systematic error, 29 T Tailrace, 846 Tangential components, 253 Tangential momentum equation, 693 Tangential velocity, 842 Taylor series expansion, 446 cen96534_ind_993_1018.indd 1013 01/12/16 10:57 am 1014 INDEX Temperature (T), 38 Tensor notation, 253 Thermal conductivity, 334 Thermal energy, 43, 189, 220 Thermodynamics, 140 thermodynamic pressure, 470 Thermosiphoning effect, 11 Three-dimensional flows, 13–14 Three-dimensional Leibniz theorem, 171 Throat, 673, 680 Thrust, 260, 261 Time plots, 152 Time rate of change, 191 Time-exposed flow path, 148 Time-integrated flow pattern, 148 Timelines, 150–151 Tip vortex, 642 Torque, 270 Torr unit, 84 Torricelli equation, 213 Total angular momentum, 251 Total body force acting on control volume, 253 Total derivative, 143 operator, 141 Total energy (E), 43 of fluid, 668 Total enthalpy, 668 Total force, 254 acting on control volume, 252 Total head, 210 Total irreversible head loss, 800 Total pressure, 207 Total surface force acting on control surface, 254 Total temperature, 669 Tracer particle location at time, 148 Tracer particles, 147 Trailing edge angle, 815 Trailing vortices, 642 Tranquil flow, 738 Transient flow, 12 Transit time flowmeters, 404 Transition from laminar to turbulent flow, 353 Transition layer, 368 Transition Reynolds number, 561 Transitional flow, 11 Transitions, 760 Translation, 155 rate of, 137 translational, periodic boundary conditions, 897, 922 Transport equation, 887 Trapezoidal channels, 753, 754–755 Triangular weir, 776 Truncated Taylor series expansion, 107 Tube-axial fan, 825, 826 Tubes, 352 Tufts, 152 Turbine scaling laws dimensionless turbine parameters, 861–864 turbine specific speed, 864–867 Turbine specific speed, 864 conversions between dimensionless and customary U. S., 865 example, 866–867 maximum efficiency as function of turbine, 866 runner of pump–turbine, 865 Turbine(s), 793, 794, 839. See also Turbomachinery affinity laws, application of, 863–864 dynamic, 840–841 efficiency, 200, 847, 848 gas, 853 head, 225, 226 impulse, 841–843 piping systems with, 388 characteristic pump curves, 389 efficiency of pump–motor combination, 388 effect of flushing on flow rate from shower example, 393–396 gravity-driven water flow in pipe example, 392–393 pumping water through two parallel pipes example, 389–391 positive-displacement turbines, 840 reaction, 843–852 steam, 853 wind, 853–861 Turbofan engine, 828 Turbomachinery, 793. See also Pumps; Turbine(s) classifications and terminology, 794 human heart, 795 pump scaling laws, 830–839 rotary fuel atomizers, 867–868 rotating shaft, 795 scaling laws, 793 turbine scaling laws, 861–867 Turbomachines, 272, 795 Turbulent flat plate boundary layer, 576–577 expression for laminar and turbulent boundary layer, 578 laminar and turbulent boundary layers comparison, 578–580 one-seventh-power law, 577 turbulent boundary layer profile equations comparison, 581–582 wall-wake law, 580–581 Turbulent flow, 11, 353–355, 357. See also Laminar flow; Radial-flow in channels, 735–737 in pipes, 365 intense mixing in turbulent flow, 366 cen96534_ind_993_1018.indd 1014 01/12/16 10:57 am 1015 INDEX moody chart and equations, 370–372 turbulent shear stress, 366–367 turbulent velocity profile, 368–370 types of fluid flow problems, 372–379 water exiting tube, 365 Turbulent/turbulence, 932 boundary layer separation, 912 CFD calculations, 908 design of stator for vane-axial flow fan, 913–921 flow around circular cylinder, 911–913 LES, 909 turbulence model, 910 turbulent flows, 909 component, 366 dissipation rate, 910 eddies, 908 intensity, 910 kinetic energy, 910 layer, 368, 580 length scale, 910 model, 367, 909, 910 shear stress, 366–367 stresses, 366 viscosity, 367, 368–370 Turning angle, 692 across expansion fan, 697 Twist in blade, 822 Two circular flows, 163–164 Two-dimensional flow(s), 13–14, 454–455 in Cartesian coordinates, 161 kinematic properties in, 158–160 rotationality in, 162 vorticity contours in, 161–162 Two-dimensional irrotational regions of flow, 538 axisymmetric irrotational regions of flow, 540–541 planar irrotational regions of flow, 539–540 velocity components, 541 Two-dimensional shear flow, 181 Two-dimensional stream function in Cartesian coordinates, 456 Two-equation turbulence models, 910 Two-phase flows, CFD methods for, 934 axisymmetric VOF simulation, 935 pressure field in bubble wake, 936 VOF method, 935 Two-phase Taylor flow, 935 U U-tube manometer, 87 Underflow gate(s), 769, 770 sluice gate with drowned outflow, 771–772 Unfavorable pressure gradient, 583 Unforced flow, 11 Uniform critical flow, 747 Uniform flow, 14–15, 258, 734–735, 755 in channels, 745 critical uniform flow, 747–748 flow depth y, average flow velocity V, 746 superposition method for nonuniform perimeters, 748–751 Uniform stream flow, 543 superposition of uniform stream and doublet flow, 550 Uniform-flow velocity, 746 Unit outer normal, 166–167 United States Customary System (USCS). See English system Units, 16–22, 297–299 dimensional homogeneity, 19–21 primary dimensions and units, 17 SI and English units, 17–19 standard prefixes in SI units, 17 unity conversion ratios, 21–22 weight of one pound-mass, 21 Unity conversion ratios, 21–22 Universal gas constant, 40 Unstable situation, 104–105 Unsteady, compressible flow, 207 Unsteady flow, 12–13, 146 Useful pump head, 798 Useful pumping power, 200 V V-notch weir. See Triangular weir Vacuum pressures, 78 Valves, 385 Vane-axial flow fan blade spacing, 914 CFD results for flow, 919 computational domain, 914 design for wind tunnel, 829–830 grid for two-dimensional stator vane cascade, 915 pressure contour plot, 918 stator design for, 913 streamlines producing by CFD calculations, 916 tangential velocity contour plot, 919 three-dimensional computational domain, 918 vane-axial fan, 825 vane-axial flow fan, 913 variation of average outlet flow, 915 velocity vectors producing by CFD calculations, 917 vorticity contour plots, 920 vorticity contour plots producing by CFD calculations, 916 Vanes, 825 Vapor, 4 Vapor pressure, 41–43, 803 Variable control volume, 197 cen96534_ind_993_1018.indd 1015 01/12/16 10:57 am 1016 INDEX Variable pitch, 825 Variation of fluid properties in flow direction, 673–675 Varied flows, 734–735 VAWTs. See Vertical axis wind turbines (VAWTs) Vector, 156 equation, 256 field variable, 138 plots, 153–154 product, 466 quantity, 156 Velocity, 153, 814 boundary layer, 355 components, 541 defect, 370 defect law, 370 in flow, 547–548 head, 210 inlet, 895 in laminar flow, gravity effect on, 361–362 potential function, 533 profile plot, 153 vector, 138, 147, 153, 154, 155, 204 Velocity field, 138 calculation from stream function example, 457–458 calculation of pressure field for known, 476–481 calculation of stream function for, 458–459 Velocity measurement, 396. See also Internal flow biofluid mechanics, 414–422 electromagnetic flowmeters, 406–407 LDV, 410–411 obstruction flowmeters, 398–401 Pitot probes, 396–398 by Pitot tube, 215 Pitot-static probes, 396–398 PIV, 411–414 positive displacement flowmeters, 401–402 thermal anemometers, 408–409 turbine flowmeters, 402–403 ultrasonic flowmeters, 404–405 variable-area flowmeters, 403–404 vortex flowmeters, 407–408 Vena contracta, 381 Ventilation system, operating point of fan in, 799–801 Venturi meters, 398–401 Venturi nozzles, 673 Vertical axis wind turbines (VAWTs), 853 Viscoelastic fluid, 471 Viscometer, 55 Viscosity, 10, 51 of fluid, 53–54, 55–56 fluid behavior in laminar flow, 52 Newtonian fluids, 52 variation of dynamic viscosity, 55 Viscous effects, 203 Viscous flows, 10 Viscous length, 369 Viscous stress tensor, 471 Viscous sublayer, 368 Viscous–inviscid interaction, 720 VOCs. See Volatile organic compounds (VOCs) VOF method. See Volume of fluid method (VOF method) Volatile organic compounds (VOCs), 335 Volume (V), 38 Volume flow rate, 191–193, 796, 811, 830 through positive-displacement pump, 812 significant digits and, 31 Volume of fluid method (VOF method), 935 Volumetric and continuity strain rate comparison, 455 Volumetric strain, 158 Volumetric strain rate, 157 in Cartesian coordinates, 157 comparison with continuity strain rate, 455 Volute, 812, 843 von Kármán equation, 372 Vortex ring, 32 Vortex shedding, 280 Vortex strength, 546 Vortical flow, 454–455 Vortices shed, 149 Vorticity, 160–164 Vorticity contours in two-dimensional flow, 161–162 Vorticity vector, 160 in Cartesian coordinates, 161 in cylindrical coordinates, 162–163 W Wake function, 580 Wall boundary conditions, 895 Wall functions, 895 Wall sublayer, 368 Wall-wake law, 580 Water discharge from large tank, 213 Water flow through garden hose nozzle example, 196–197 Water hammer arrestor, 45–46 Water horsepower, 797 Watt (W), 19 Wave angle, 692 Wave speed, 737–741 Weak oblique shocks, 695 Weber number, 319 Weight density, 40 Weir(s), 769, 774 discharge coefficient, 774 head, 774 measuring flow rate by, 778 Wicket gates, 843 cen96534_ind_993_1018.indd 1016 01/12/16 10:57 am 1017 INDEX Wind farms, 856 Wind power density, 856 Wind tunnel testing, 326–329 Wind turbine(s), 839, 853 derivatives to calculating minima or maxima, 859 disk area of, 857 large and small control volumes, 858 power generation in wind farm, 860–861 qualitative sketch of average streamwise velocity, 858 typical qualitative wind-turbine power performance curve, 857 various wind turbine designs and categorization, 854–855 wind farms popping up, 856 Windmill, 839, 853 Wing loading, 638 Winglets, 643 Wingspan, 638 With-rotation swirl, 849 Work, 19 energy transfer by, 220 done by pressure forces, 221–223 shaft work, 221 transfer, 219 X x-momentum equation, 701–702 adiabatic duct flow with friction, 711 property relations for Fanno flow, 713–714 Y Yield stress, 472 Z Zero pressure point, 553 cen96534_ind_993_1018.indd 1017 01/12/16 10:57 am This page intentionally left blank Conversion Factors DIMENSION METRIC METRIC/ENGLISH Acceleration 1 m/s2 = 100 cm/s2 1 m/s2 = 3.2808 ft/s2 1 ft/s2 = 0.3048 m/s2 Area 1 m2 = 104 cm2 = 106 mm2 = 10−6 km2 1 m2 = 1550 in2 = 10.764 ft2 1 ft2 = 144 in2 = 0.09290304 m2 Density 1 g/cm3 = 1 kg/L = 1000 kg/m3 1 g/cm3 = 62.428 lbm/ft3 = 0.036127 lbm/in3 1 lbm/in3 = 1728 lbm/ft3 1 kg/m3 = 0.062428 lbm/ft3 Energy, heat, work, 1 kJ = 1000 J = 1000 N ⋅ m = 1 kPa ⋅ m3 1 kJ = 0.94782 Btu and specific energy 1 kJ/kg = 1000 m2/s2 1 Btu = 1.055056 kJ 1 kWh = 3600 kJ = 5.40395 psia ⋅ ft3 = 778.169 lbf ⋅ ft 1 Btu/lbm = 25,037 ft2/s2 = 2.326 kJ/kg 1 kWh = 3412.14 Btu Force 1 N = 1 kg ⋅ m/s2 = 105 dyne 1 N = 0.22481 lbf 1 kgf = 9.80665 N 1 lbf = 32.174 lbm ⋅ ft/s2 = 4.44822 N 1 lbf = 1 slug ⋅ ft/s2 Length 1 m = 100 cm = 1000 mm = 106 µm 1 m = 39.370 in = 3.2808 ft = 1.0926 yd 1 km = 1000 m 1 ft = 12 in = 0.3048 m 1 mile = 5280 ft = 1.6093 km 1 in = 2.54 cm Mass 1 kg = 1000 g 1 kg = 2.2046226 lbm 1 metric ton = 1000 kg 1 lbm = 0.45359237 kg 1 ounce = 28.3495 g 1 slug = 32.174 lbm = 14.5939 kg 1 short ton = 2000 lbm = 907.1847 kg Power 1 W = 1 J/s 1 kW = 3412.14 Btu/h = 1.341 hp 1 kW = 1000 W = 1 kJ/s = 737.56 lbf ⋅ ft/s 1 hp‡ = 745.7 W 1 hp = 550 lbf ⋅ ft/s = 0.7068 Btu/s = 42.41 Btu/min = 2544.5 Btu/h = 0.74570 kW 1 Btu/h = 1.055056 kJ/h Pressure or stress, 1 Pa = 1 N/m2 1 Pa = 1.4504 × 10−4 psi and pressure 1 kPa = 103 Pa = 10−3 MPa = 0.020886 lbf/ft2 expressed as 1 atm = 101.325 kPa = 1.01325 bar 1 psi = 144 lbf/ft2 = 6.894757 kPa a head = 760 mm Hg at 0°C 1 atm = 14.696 psi = 1.03323 kgf/cm2 = 29.92 inches Hg at 30°F 1 mm Hg = 0.1333 kPa 1 inch Hg = 13.60 inches H2O = 3.387 kPa Specific heat 1 kJ/kg ⋅ °C = 1 kJ/kg ⋅ K 1 Btu/lbm ⋅ °F = 4.1868 kJ/kg ⋅ °C = 1 J/g ⋅ °C 1 Btu/lbmol ⋅ R = 4.1868 kJ/kmol ⋅ K 1 kJ/kg ⋅ °C = 0.23885 Btu/lbm ⋅ °F = 0.23885 Btu/lbm ⋅ R Specific volume 1 m3/kg = 1000 L/kg 1 m3/kg = 16.02 ft3/lbm = 1000 cm3/g 1 ft3/lbm = 0.062428 m3/kg Temperature T(K) = T(°C) + 273.15 T(R) = T(°F) + 459.67 = 1.8T(K) ΔT(K) = ΔT(°C) T(°F) = 1.8 T(°C) + 32 ΔT(°F) = ΔT(R) = 1.8 ΔT(K) Velocity 1 m/s = 3.60 km/h 1 m/s = 3.2808 ft/s = 2.237 mi/h 1 mi/h = 1.46667 ft/s 1 mi/h = 1.6093 km/h Viscosity, dynamic 1 kg/m ⋅ s = 1 N ⋅ s/m2 = 1 Pa ⋅ s = 10 poise 1 kg/m ⋅ s = 2419.1 lbm/ft ⋅ h = 0.020886 lbf ⋅ s/ft2 = 0.67197 lbm/ft ⋅ s Exact conversion factor between metric and English units.
‡ Mechanical horsepower. The electrical horsepower is taken to be exactly 746 W.
1019 cen96534_cf_1019-1020.indd 1019 01/12/16 10:55 am 1020 CONVERSION FACTORS DIMENSION METRIC METRIC/ENGLISH Viscosity, kinematic 1 m2/s = 104 cm2/s 1 m2/s = 10.764 ft2/s = 3.875 × 104 ft2/h 1 stoke = 1 cm2/s = 10−4 m2/s 1 m2/s = 10.764 ft2/s Volume 1 m3 = 1000 L = 106 cm3 (cc) 1 m3 = 6.1024 × 104 in3 = 35.315 ft3 = 264.17 gal (U.S.) 1 U.S. gallon = 231 in3 = 3.7854 L 1 fl ounce = 29.5735 cm3 = 0.0295735 L 1 U.S. gallon = 128 fl ounces Volume flow rate 1 m3/s = 60,000 L/min = 106 cm3/s 1 m3/s = 15,850 gal/min = 35.315 ft3/s = 2118.9 ft3/min (CFM) Exact conversion factor between metric and English units.
Some Physical Constants PHYSICAL CONSTANT METRIC ENGLISH Standard acceleration of gravity g = 9.80665 m/s2 g = 32.174 ft/s2 Standard atmospheric pressure Patm = 1 atm = 101.325 kPa Patm = 1 atm = 14.696 psia = 1.01325 bar = 2116.2 lbf/ft2 = 760 mm Hg (0°C) = 29.9213 inches Hg (32°F) = 10.3323 m H2O (4°C) = 406.78 inches H2O (39.2°F) Universal gas constant Ru = 8.31447 kJ/kmol ⋅ K Ru = 1.9859 Btu/lbmol ⋅ R = 8.31447 kN ⋅ m/kmol ⋅ K = 1545.37 ft ⋅ lbf/lbmol ⋅ R Commonly Used Properties PROPERTY METRIC ENGLISH Air at 20°C (68°F) and 1 atm Specific gas constant Rair = 0.2870 kJ/kg ⋅ K Rair = 0.06855 Btu/lbm ⋅ R = 287.0 m2/s2 ⋅ K = 53.34 ft ⋅ lbf/lbm ⋅ R = 1716 ft2/s2 ⋅ R Specific heat ratio k = cp/cv = 1.40 k = cp/cv = 1.40 Specific heats cp = 1.005 kJ/kg ⋅ K cp = 0.2400 Btu/lbm ⋅ R = 1005 m2/s2 ⋅ K = 186.8 ft ⋅ lbf/lbm ⋅ R cv = 0.7180 kJ/kg ⋅ K = 6009 ft2/s2 ⋅ R = 718.0 m2/s2 ⋅ K cv = 0.1715 Btu/lbm ⋅ R = 133.5 ft ⋅ lbf/lbm ⋅ R = 4294 ft2/s2 ⋅ R Speed of sound c = 343.2 m/s = 1236 km/h c = 1126 ft/s = 767.7 mi/h Density 𝜌 = 1.204 kg/m3 𝜌 = 0.07518 lbm/ft3 Viscosity 𝜇 = 1.825 × 10−5 kg/m ⋅ s 𝜇 = 1.227 × 10−5 lbm/ft ⋅ s Kinematic viscosity 𝜈 = 1.516 × 10−5 m2/s 𝜈 = 1.632 × 10−4 ft2/s Liquid water at 20°C (68°F) and 1 atm Specific heat (c = cp = cv) c = 4.182 kJ/kg ⋅ K c = 0.9989 Btu/lbm ⋅ R = 4182 m2/s2 ⋅ K = 777.3 ft ⋅ lbf/lbm ⋅ R = 25,009 ft2/s2 ⋅ R Density 𝜌 = 998.0 kg/m3 𝜌 = 62.30 lbm/ft3 Viscosity 𝜇 = 1.002 × 10−3 kg/m ⋅ s 𝜇 = 6.733 × 10−4 lbm/ft ⋅ s Kinematic viscosity 𝜈 = 1.004 × 10−6 m2/s 𝜈 = 1.081 × 10−5 ft2/s Independent of pressure or temperature.
cen96534_cf_1019-1020.indd 1020 01/12/16 10:55 am a Manning constant, m1/3/s; height from channel bottom to bottom of sluice gate, m a ›, a Acceleration and its magnitude, m/s2 A, Ac Area, m2; cross-sectional area, m2 Ar Archimedes number AR Aspect ratio b Width or other distance, m; intensive property in RTT analysis; turbomachinery blade width, m bhp Brake horsepower, hp or kW B Center of buoyancy; extensive property in RTT analysis Bi Biot number Bo Bond number c Specific heat for incompressible substance, kJ/kg·K; speed of sound, m/s; speed of light in a vacuum, m/s; chord length of an airfoil, m c0 Wave speed, m/s cp Constant-pressure specific heat, kJ/kg·K cv Constant-volume specific heat, kJ/kg·K C Dimension of the amount of light C Bernoulli constant, m2/s2 or m/t2·L, depending on the form of Bernoulli equation; Chezy coefficient, m1/2/s; circumference, m Ca Cavitation number CD, CD,x Drag coefficient; local drag coefficient Cd Discharge coefficient Cf, Cf,x Fanning friction factor or skin friction coefficient; local skin friction coefficient CH Head coefficient CL, CL,x Lift coefficient; local lift coefficient CNPSH Suction head coefficient CP Center of pressure Cp Pressure coefficient CP Power coefficient CQ Capacity coefficient CS Control surface CV Control volume Cwd Weir discharge coefficient D or d Diameter, m (d typically for a smaller diameter than D) DAB Species diffusion coefficient, m2/s Dh Hydraulic diameter, m Dp Particle diameter, m e Specific total energy, kJ/kg e → r, e → 𝜃 Unit vector in r- and 𝜃-direction, respectively E Voltage, V E, E . Total energy, kJ; and rate of energy, kJ/s Ec Eckert number EGL Energy grade line, m Es Specific energy in open-channel flows, m Eu Euler number f Frequency, cycles/s; Blasius boundary layer dependent similarity variable f, fx Darcy friction factor; and local Darcy friction factor F → , F Force and its magnitude, N FB Magnitude of buoyancy force, N FD Magnitude of drag force, N Ff Magnitude of drag force due to friction, N FL Magnitude of lift force, N Fo Fourier number Fr Froude number FT Magnitude of tension force, N g →, g Gravitational acceleration and its magnitude, m/s2 g . Heat generation rate per unit volume, W/m3 G Center of gravity GM Metacentric height, m Gr Grashof number h Specific enthalpy, kJ/kg; height, m; head, m; convective heat transfer coefficient, W/m2·K hfg Latent heat of vaporization, kJ/kg hL Head loss, m H Boundary layer shape factor; height, m; net head of a pump or turbine, m; total energy of a liquid in open-channel flow, expressed as a head, m; weir head, m H → , H Moment of momentum and its magnitude, N·m·s HGL Hydraulic grade line, m Hgross Gross head acting on a turbine, m i Index of intervals in a CFD grid (typically in x-direction) i → Unit vector in x-direction 1021 N o m e n c l a t u r e cen96534_nc_1021-1030.indd 1021 01/12/16 10:56 am 1022 NOMENCLATURE I Dimension of electric current I Moment of inertia, N·m·s2; current, A; turbulence intensity Ixx Second moment of inertia, m4 j Reduction in Buckingham Pi theorem; index of intervals in a CFD grid (typically in y-direction) j › Unit vector in y-direction Ja Jakob number k Specific heat ratio; expected number of Πs in Buckingham Pi theorem; thermal conductivity, W/m·K; turbulent kinetic energy per unit mass, m2/s2; index of intervals in a CFD grid (typically in z-direction) k › Unit vector in z-direction ke Specific kinetic energy, kJ/kg K Doublet strength, m3/s KE Kinetic energy, kJ KL Minor loss coefficient Kn Knudsen number ℓ Length or distance, m; turbulent length scale, m L Dimension of length L Length or distance, m Le Lewis number Lc Chord length of an airfoil, m; characteristic length, m Lh Hydrodynamic entry length, m Lw Weir length, m m Dimension of mass m, m . Mass, kg; and mass flow rate, kg/s M Molar mass, kg/kmol M ›, M Moment of force and its magnitude, N·m Ma Mach number n Number of parameters in Buckingham Pi theorem; Manning coefficient n, n . Number of rotations; and rate of rotation, rpm n → Unit normal vector N Dimension of the amount of matter N Number of moles, mol or kmol; number of blades in a turbomachine NP Power number NPSH Net positive suction head, m NSp Pump specific speed NSt Turbine specific speed Nu Nusselt number p Wetted perimeter, m pe Specific potential energy, kJ/kg P, P′ Pressure and modified pressure, N/m2 or Pa PE Potential energy, kJ Pe Peclet number Pgage Gage pressure, N/m2 or Pa Pm Mechanical pressure, N/m2 or Pa Pr Prandtl number Psat or Pv Saturation pressure or vapor pressure, kPa Pvac Vacuum pressure, N/m2 or Pa Pw Weir height, m q Heat transfer per unit mass, kJ/kg q . Heat flux (rate of heat transfer per unit area), W/m2 Q, Q . Total heat transfer, kJ; and rate of heat transfer, W or kW QEAS Equiangle skewness in a CFD grid r› , r Moment arm and its magnitude, m; radial coordinate, m; radius, m R Gas constant, kJ/kg·K; radius, m; electrical resistance, Ω Ra Rayleigh number Re Reynolds number Rh Hydraulic radius, m Ri Richardson number Ru Universal gas constant, kJ/kmol·K s Submerged distance along the plane of a plate, m; distance along a surface or streamline, m; specific entropy, kJ/kg·K; fringe spacing in LDV, m; turbomachinery blade spacing, m S0 Slope of the bottom of a channel in open-channel flow Sc Schmidt number Sc Critical slope in open-channel flow Sf Friction slope in open-channel flow SG Specific gravity Sh Sherwood number SP Property at a stagnation point St Stanton number; Strouhal number Stk Stokes number t Dimension of time t Time, s T Dimension of temperature T Temperature, °C or K cen96534_nc_1021-1030.indd 1022 01/12/16 10:56 am 1023 NOMENCLATURE T →, T Torque and its magnitude, N·m u Specific internal energy, kJ/kg; Cartesian velocity component in x-direction, m/s u Friction velocity in turbulent boundary layer, m/s ur Cylindrical velocity component in r-direction, m/s u𝜃 Cylindrical velocity component in 𝜃-direction, m/s uz Cylindrical velocity component in z-direction, m/s U Internal energy, kJ; x-component of velocity outside a boundary layer (parallel to the wall), m/s 𝜐 Cartesian velocity component in y-direction, m/s v Specific volume, m3/kg V, V · Volume, m3; and volume flow rate, m3/s V →, V Velocity and its magnitude (speed), m/s; average velocity, m/s V0 Uniform-flow velocity in open-channel flow, m/s w Work per unit mass, kJ/kg; Cartesian velocity component in z-direction, m/s; width, m W Weight, N; width, m W, W . Work transfer, kJ; and rate of work (power), W or kW We Weber number x Cartesian coordinate (usually to the right), m x → Position vector, m y Cartesian coordinate (usually up or into the page), m; depth of liquid in open-channel flow, m yn Normal depth in open-channel flow, m z Cartesian coordinate (usually up), m Greek Letters 𝛼 Angle; angle of attack; kinetic energy correction factor; thermal diffusivity, m2/s; isothermal compressibility, kPa−1 or atm−1 𝛼 →, 𝛼 Angular acceleration and its magnitude, s−2 𝛽 Coefficient of volume expansion, K−1; momentum-flux correction factor; angle; diameter ratio in obstruction flowmeters; oblique shock angle; turbomachinery blade angle 𝛿 Boundary layer thickness, m; distance between streamlines, m; angle; small change in a quantity 𝛿 Boundary layer displacement thickness, m 𝜀 Mean surface roughness, m; turbulent dissipation rate, m2/s3 𝜀ij Strain rate tensor, s−1 Φ Dissipation function, kg/m·s3 𝜙 Angle; velocity potential function, m2/s 𝛾s Specific weight, N/m3 Г Circulation or vortex strength, m2/s 𝜂 Efficiency; Blasius boundary layer independent similarity variable 𝜅 Bulk modulus of compressibility, kPa or atm; log law constant in turbulent boundary layer 𝜆 Mean free path length, m; wavelength, m; second coefficient of viscosity, kg/m·s 𝜇 Viscosity, kg/m·s; Mach angle 𝜈 Kinematic viscosity m2/s 𝜈(Ma) Prandtl–Meyer function for expansion waves, degrees or rad Π Nondimensional parameter in dimensional analysis 𝜃 Angle or angular coordinate; boundary layer momentum thickness, m; pitch angle of a turbomachinery blade; turning or deflection angle of oblique shock 𝜌 Density, kg/m3 𝜎 Normal stress, N/m2 𝜎ij Stress tensor, N/m2 𝜎s Surface tension, N/m 𝜏 Shear stress, N/m2 𝜏ij Viscous stress tensor (also called shear stress tensor), N/m2 𝜏ij, turbulent Specific Reynolds stress tensor, m2/s2 𝜔 →, 𝜔 Angular velocity vector and its magnitude, rad/s; angular frequency, rad/s 𝜓 Stream function, m2/s 𝜁 → , 𝜁 Vorticity vector and its magnitude, s−1 Subscripts ∞ Property of the far field 0 Stagnation property; property at the origin or at a reference point abs Absolute atm Atmospheric avg Average quantity cen96534_nc_1021-1030.indd 1023 01/12/16 10:56 am 1024 NOMENCLATURE b Property of the back or exit of a nozzle, e.g., back pressure Pb C Acting at the centroid c Pertaining to a cross section cr Critical property CL Pertaining to the centerline CS Pertaining to a control surface CV Pertaining to a control volume e Property at an exit; extracted portion eff Effective property f Property of a fluid, usually of a liquid H Acting horizontally lam Property of a laminar flow L Portion lost by irreversibilities m Property of a model max Maximum value mech Mechanical property min Minimum value n Normal component P Acting at the center of pressure p Property of a prototype; property of a particle; property of a piston R Resultant r Relative (moving frame of reference) rec Rectangular property rl Property of the rotor leading edge rt Property of the rotor trailing edge S Acting on a surface s Property of a solid sat Saturation property; property of a satellite sl Property of the stator leading edge st Property of the stator trailing edge sub Submerged portion sys Pertaining to a system t Tangential component tri Trianglular property turb Property of a turbulent flow u Useful portion V Acting vertically v Property of a vapor vac Vacuum w Property at a wall Superscripts _ (overbar) Averaged quantity · (overdot) Quantity per unit time; time derivative ′ (prime) Fluctuating quantity; derivative of a variable; modified variable Nondimensional property; sonic property + Law of the wall variable in turbulent boundary layer → (over arrow) Vector quantity cen96534_nc_1021-1030.indd 1024 01/12/16 10:56 am |
14390 | https://www.youtube.com/shorts/xd0zh1SYeoA | Visual Explanation of Counting 90 Degree Angles Around the Unit Circle - YouTube
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Visual Explanation of Counting 90 Degree Angles Around the Unit Circle
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2024 Feb 18
Degree angles around the unit circle represent the measurement of angles in degrees within a circle with a radius of 1 unit. In trigonometry, the unit circle serves as a fundamental tool for understanding the relationships between angles and trigonometric functions such as sine, cosine, and tangent. Each degree corresponds to a specific point on the unit circle, facilitating calculations and visualizations of trigonometric properties. By mapping angles onto the unit circle, mathematicians and scientists can solve complex problems involving angles, rotations, and periodic phenomena. Understanding degree angles around the unit circle is crucial for mastering trigonometry and related fields. More Lessons: Twitter:
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14392 | https://www.youtube.com/watch?v=R_bgo8j6jxc | Analyzing an author's purpose | Reading | Khan Academy
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Keep going! Check out the next lesson and practice what you’re learning:
When it comes to nonfiction texts (and especially opinion pieces), nearly everybody's selling something. Stay sharp, stay skeptical, and watch out for manipulative tactics—what an author doesn't say can be as important as what they do say.
Khan Academy is a nonprofit organization with the mission of providing a free, world-class education for anyone, anywhere. We offer quizzes, questions, instructional videos, and articles on a range of academic subjects, including math, biology, chemistry, physics, history, economics, finance, grammar, preschool learning, and more. We provide teachers with tools and data so they can help their students develop the skills, habits, and mindsets for success in school and beyond. Khan Academy has been translated into dozens of languages, and 15 million people around the globe learn on Khan Academy every month. As a 501(c)(3) nonprofit organization, we would love your help!
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Uma Pândega de Paranóias, Fobias e Mania da Perseguição by Stealing Orchestra is licensed under a Attribution-NonCommercial-NoDerivatives (aka Music Sharing) 3.0 International License.
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Introduction hello readers today we are going on a dangerous journey inside the mind of the author every piece of text is written for a purpose and especially in informational texts every author structures their texts words and their ideas with that purpose in mind and sometimes that purpose will be harder to see as readers our job is to consider the author's purpose as we read how is it influencing the information the author shares how is it influencing my understanding of the topic is what i think the same as what the author thinks do i agree with them do i disagree you may be familiar with the memory device of pi persuade inform entertain as three categories of purpose but i'd like to go deeper than that Lobbyists i live in washington dc which is home to an entire professional class of what are called lobbyists people whose job it is to advocate to congress on behalf of a special interest for example the oil industry or the cheese industry this isn't always bad you can learn more about lobbyists and advocates and khan academy's government and politics course but frequently it takes the form of someone from an industry trying to convince congress to give them a competitive advantage over other industries now let's pull real world politics out of this and pretend for the purposes of this lesson that there are two warring lobby groups in dc one that represents the cake industry and another that represents the pie industry an age-old conflict so hold that conflict in your mind for a second cake versus pie now when we read informational text we should be learning new information but the author's opinions can shape the text to the point where the information becomes biased or misleading when you read informational text maintain an air of healthy skepticism Questions ask yourself the following questions constantly what's the author's opinion what information did they include or what information did they leave out and if so was that on purpose what's the connotation of the author's word choice you know how do the words feel and and what's the overall tone of the piece so now we have these questions to ask ourselves let's return to our pie versus cake lobby fight in washington so imagine you're reading the washington post and you see this opinion column cake connected to cavities and poor dental health study fines a recent study by the american dental association connected the consumption of cake or similar foods with a 30 increased risk of tooth decay cavities and gingivitis people who reported having consumed cake in the previous 60 days were significantly more likely to experience problems related to dental health than those people who did not cake is dangerous cake which rarely contains fruit will rot your teeth and then if we squint down at the very bottom of the column it'll say in small text the author is the ceo of circular solutions a pie advocacy network and now let's ask ourselves those same questions again while acknowledging that this is a fake story that i made up for the purposes of this video there is nobody named wendell apricot jam although i wish there were that'd be a great name so what's the author's opinion well it seems to me that the author really thinks cake is dangerous and poses a threat to the nation's dental health we know this because they literally say cake is dangerous they want people to buy and eat fewer cakes what information did the author choose to include well so this is interesting the author talks about the results of this study and the big takeaways about the danger of cake or similar foods and then they go on to mention incidentally that cake rarely contains fruit that's a curious thing to say it's almost like they're saying pie which has fruit in it is better for you than cake without actually saying it what information did the author choose to leave out well notice that the piece says cake or similar foods i would be hard-pressed to come up with a definition of foods similar to cake that does not also include desserts like pie but the way that this piece is written it swerves out of its way to avoid pinning the same tooth decay risks on pie the way that it blames cake i think it would be reasonable to assume that if we read the underlying study that this opinion piece is based on it would include pies and cakes in the same category of sugary desserts that are associated with bad dental outcomes why would the author do this well they want you to buy pie instead of cake but they don't want you to think too hard about it because if you did you would buy and eat both less cake and less pie and the pie lobby doesn't want that what's the connotation or the feeling of the author's word choices well they're citing a scientific study by dentists so they want you to take their warning seriously and they're using words like risk and danger rot and problems they want you to make an association between eating cake and bad tooth health they want you to associate scientific language with trustworthiness so i'm going to put down scientific authority here what's the overall tone it's negative the author does not want you to eat cake they want you to think it's bad and they've planted a little hint here that fruity foods are better what's a fruity food pie can they say pie is safer no not without lying outright now i want to be clear about two things sometimes an author's purpose really will just be to inform straightforwardly not everything has to be a tug-of-war between you and the author over the truth sometimes a writer just wants you to know the life cycle of a butterfly or how to make egg salad or the order the u.s presidents came in but most of the time there's a purpose and opinion behind the text now the second thing i want to be super clear about is that i i made up this whole thing i made up the article the statistics wendell you know uh all of that's fake um i like both pie and cake very much and now that i've said that i'm sure you're wondering oh is david in the pocket of big dessert and the answer is no i'm in the pocket of big free education for everyone everywhere you can learn anything dave it out |
14393 | https://www1.udel.edu/leipzig/depende/preterito_imperfecto.htm | Pretérito vs. imperfecto
Pretérito vs. imperfecto: usos
Ambos, el pretérito y el imperfecto, se usan para hablar sobre eventos que ocurrieron en el pasado, pero cada uno de ellos nos presenta estos eventos de una perspectiva diferente.
El pretérito:
El pretérito sirve para presentar acciones como terminadas o cumplidas. Las presentamos como “históricas”. Se presentan como si no tuvieran ninguna conexión con el presente.
Ayer fue un día horrible. No pude bañarme porque no hubo agua caliente. Por un minuto pensé ir al gimnasio para hacer ejercicios y tomar una ducha pero no encontré mi carné de identidad. …
¿Cuál es tu reacción después de escuchar esto?
El imperfecto:
El imperfecto no tiene este rasgo semántico. Solo expresa que un evento era en el pasado. No contiene ningún límite. Pretendemos que no sabemos ni cuando empezó ni cuando terminó.
Aunque lógicamente sabemos que lo que se cuenta ya está pasado, emocionalmente nos situamos en medio de alguna acción inacabada que todavía no se había acabado cuando hablábamos de ella.
El lunes era un día horrible. Después de levantarme quería bañarme pero no había agua caliente. Pensaba ir al gimnasio para hacer ejercicios y tomar una ducha después pero no encontraba mi carné de identidad. …
¿Cuál es tu reacción después de escuchar esto?
Indicadores temporales
Muchas veces el pretérito se usa con indicadores temporales como ayer, la semana pasada, el año pasado, tres horas, 10 veces, toda la noche, hasta la madrugada, etc., que subrayan o destacan el límite temporal.
Cambio de significado del verbo
Algunas veces la combinación del significado propio del verbo con el pretérito cambia el significado tanto que parecen significados diferentes que hasta se traducen con verbos diferentes en inglés y que le dan a la frase un significado diferente:
SABER
Yo sabía que Juan estaba enfermo. (I knew that Juan was sick.)
Yo supe que Juan estaba enfermo por casualidad. (I found out that Juan was sick by accident)
Saber + pretérito = to find out, to realize, to begin to know
CONOCER
Nos conocíamos desde siempre. (We knew each other forever.)
Nos conocimos ayer en la universidad. (We got to know each other yesterday at the university.)
Conocer + pretérito = to get to know, to meet, to get acquainted
QUERER
Yo quería salir con Juana. (I wanted to go out with Juana.)
Yo quise salir con Juana pero sabía que no era su tipo. (I wanted to go out with Juana but I knew that I was not her type.)
Querer + pretérito = To try to do something (but then give up the initial idea)
TENER
Juan tenía 9 años cuando lo conocí. (Juan was 9 when I (first) met him. )
Juan tuvo 20 años ayer. (Juan turned 20 yesterday.)
Tener + pretérito = to get, to turn
SER
Juan era mi hermano mayor. (Juan was my older brother)
Juan fue mi hermano mayor. (Juan was my older brother.) (but something happened to him)
Ser + pretérito = no longer existing
Usos del pretérito y del imperfecto
Verbos durativos normalmente se usan con el imperfecto y los demás verbos con el pretérito.
Buenos escritores y usuarios de la lengua española usan la existencia de los dos tiempos del pasado para manipular la mente de los lectores.
La fuerza del imperfecto
Usamos el imperfecto para hacer sentir al lector en el medio de una situación, como si fuera parte de la situación o del evento cuyo desenlace no sabemos todavía, o pretendemos no saber.
También se usa para descripciones (decoración, escenografía, sentimientos de los actores, paisaje, tiempo, …)
Normalmente combinamos el pretérito y el imperfecto. Creamos el trasfondo con el imperfecto y las acciones concretas con el pretérito.
Leía el periódico. Eran las cinco de la tarde cuando sonó el telefono. Era mi amigo Juan. Me invitó al cine. Colgué el teléfono y me fui.
Descripciones y narraciones
También podemos decir que el imperfecto es el tiempo para describir (cómo eran o estaban o cómo solían ser/ estar las cosas y personas) y el pretérito para narrar (lo que hicieron).
Pretérito o imperfecto
A veces el imperfecto, por ejemplo hablaba, se traduce con ‘was talking’ o ‘used to talk’ en inglés. Esto te puede servir de indicador cuando no sabes si es el pretérito o el imperfecto que tienes que usar en cierta situación, pero ten cuidado porque las formas con –ing en inglés son mucho más populares y tienen una variedad de significados:
I was talking to her till about 8:00.
She used to be an actress before she retired. Hablé con ella hasta las 8:00.
Ella fue actriz antes de jubilarse. |
14394 | https://study.com/academy/lesson/solving-the-derivative-of-cos-x.html | Derivative of Cos(x) | Definition, Proof & Functions - Lesson | Study.com
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Math Courses
AP Calculus AB & BC: Exam Prep
Derivative of Cos(x) | Definition, Proof & Functions
Contributors: Katherine Kaylegian-Starkey, Laura Pennington
Author
KK
Author:
Katherine Kaylegian-Starkey
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Instructor
Instructor:
Laura Pennington
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Calculate the derivative of cos(x) & understand the proof of the derivation of the derivative of cos(x). Learn the derivatives of the other trigonometric functions. Updated: 11/21/2023
Table of Contents
What is the Derivative of Cos(x) ?
Derivative of Cos(x) Proof
Derivative of Trigonometric Functions
Lesson Summary
Show
FAQ
What is the derivative of cos(x)tan(x)?
The derivative of cos(x)tan(x) can be found by writing tan(x) as sin(x)/cos(x). Writing tan(x) in this way causes the cosines to cancel, and the expression reduces to sin(x). The derivative of sin(x) is cos(x).
What is the derivative of cos(x)/x?
The derivative of cos(x)/x can be found by using the quotient rule. The quotient rule is used to calculate derivatives of fractions, and d/dx cos(x)/x = -(xsin(x) - cos(x))/x^2.
What is the derivative of sin(x) and cos(x)?
The derivative of cos(x) is -sin(x). The derivative of sin(x) is cos(x). Both derivatives can be derived using Euler's complex representation of sine and cosine or by rewriting sin(x) as 1/csc(x) and cos(x) as 1/sec(x).
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What is the Derivative of Cos(x) ? ----------------------------------
A derivative calculates the instantaneous rate of change of a function, and while there are formulas to calculate the derivatives of polynomials, many other derivatives, such as trigonometric derivatives, are often memorized. It is perfectly fine to know a derivative and to use it, but it is important to understand where that derivative comes from.
So, what is the derivative of cos(x)? The derivative of cos(x) is simply -sin(x), and there are two ways to differentiate cos(x). The first method of differentiation uses Euler's form of cos(x). Euler is a famous mathematician who discovered that sine and cosine together create a complex representation of the exponential, and his famous formula, e i x=c o s(x)+i s i n(x) where i stands for an imaginary number, can be rearranged to form a complex representation of cos(x), c o s(x)=1 2(e i x+e−i x), and a complex representation of sin(x), s i n(x)=i 2(e i x−e−i x).
The second method of differentiation requires the quotient rule. Recall that the quotient rule is an algorithm that takes the derivative of a fraction. Mathematically, the quotient rule states that, for a function of the form h(x)=f(x)g(x), the derivative h'(x) is h′(x)=f′(x)g(x)−g′(x)f(x)g 2(x). Using the quotient rule, the derivative of cos(x) can be calculated in terms of other trigonometric functions by noticing that s e c(x)=1 c o s(x) so 1 s e c(x)=c o s(x). Regardless of the choice of method for differentiation, the derivative of cos(x) is written as d d x c o s(x).
Figure 1: cos(x) is in green, and the derivative of cos(x), -sin(x) is in blue.
Derivative of Cos(x) Proof --------------------------
The derivative of cos(x) proof is a step-by-step example explaining how to take the derivative of cos(x). Both methods for differentiating cos(x) will be proved here.
Steps to Solve the Derivative of Cos(x)
Method 1: Using Euler's form of cos(x)
Euler's form of cos(x) is c o s(x)=1 2(e i x+e−i x). This expression contains two exponential terms and a constant term. Since both exponential terms have a constant term, i, the chain rule will be needed to take this derivative. Recall that the chain rule is an algorithm that calculates the derivative of a compound function, such as h(x) = f(g(x)). Mathematically, the chain rule says that h′(x)=f′(g(x)g′(x).
1) Write the derivative.
d d x c o s(x)=d d x ℜ[1 2(e i x+e−i x)]
The ℜ means that the final answer must be real-valued.
2) Pull the constant out of the derivative on the right side of the equal sign.
1 2 d d x(e i x+e−i x)
3) Use the chain rule.
−d d x e i x=i e i x
−d d x e−i x=−i e−i x
4) Put it all together.
1 2 d d x(e i x+e−i x)=1 2(i e i x−i e−i x)
5) Simplify.
1 2(i e i x−i e−i x)=i 2(e i x−i e−i x)
The right side of this expression is Euler's formula for sin(x).
6) Take the real part of the right side of this expression.
ℜ[i 2(e i x−i e−i x)]=−i 2(e i x−i e−i x)
−i 2(e i x−i e−i x)=−s i n(x)
Method 2: Using the trigonometric expansion of cos(x)
The trigonometric expansion of cos(x) is 1 s e c(x)=c o s(x). This method requires knowing the derivative of a constant, the quotient rule for derivatives, and what the derivative of sec(x) is. The derivative of sec(x) is sec'(x) = sec(x)tan(x).
1) Write the derivative.
d d x c o s(x)=d d x 1 s e c(x)
2) Define one function to be f(x) and one function to be g(x). Take the derivative of each function.
− f(x) = 1
−f'(x) = 0
− g(x) = sec(x)
− g'(x) = sec(x)tan(x)
3) Put it all together using the quotient rule, h′(x)=f′(x)g(x)−g′(x)f(x)g 2(x).
d d x 1 s e c(x)=0∗s e c(x)−s e c(x)t a n(x)∗1 s e c 2(x)
4) Simplify.
0∗s e c(x)−s e c(x)t a n(x)∗1 s e c 2(x)=−t a n(x)s e c(x)
5) Expand the remaining trig functions and simplify.
−t a n(x)s e c(x)=−s i n(x)c o s(x)1 c o s(x)
−s i n(x)c o s(x)1 c o s(x)=−s i n(x)c o s(x)c o s(x)1
−s i n(x)c o s(x)c o s(x)1=−s i n(x)
Derivative of Trigonometric Functions -------------------------------------
The six trigonometric functions appear frequently in applied mathematics, physics, and chemistry because their oscillatory nature makes it possible to describe many natural phenomena, such as the movement of sound waves or the spinning of electrons. Each trigonometric function has a derivative that can be obtained using a trigonometric expansion or by using Euler's formula. Figure 2 lists the six trigonometric functions with their corresponding derivatives.
Figure 2:
| Function | Derivative |
--- |
| C | 0 |
| cos(x) | -sin(x) |
| sin(x) | cos(x) |
| tan(x) | sec^2 |
| |
| sec(x) | sec(x)tan(x) |
| |
| csc(x) | -csc(x)cot(x) |
| |
| cot(x) | -csc^2(x) |
| |
Lesson Summary --------------
What is the derivative of cos(x)? The derivative of cos(x) is -sin(x), and there are two different methods to differentiate cos(x). The first method for differentiating cos(x) is to use Euler's formula and the complex representation of cosine, c o s(x)=1 2(e i x+e−i x), and sine, s i n(x)=1 2(e i x−e−i x). Since the arguments of the exponentials in the complex representation of cosine involve a constant and a variable, the chain rule must be used to evaluate this derivative.
The second method to differentiate cos(x) is to notice that s e c(x)=1 c o s(x). This expression can be rearranged so that 1 s e c(x)=c o s(x). This method for finding the derivative of cos(x) requires using the quotient rule. Regardless of which method is used, the derivative of cos(x) proof derives the same result. It is possible to use a complex representation or a trigonometric expansion to obtain the derivatives for all six trig functions: sin(x), cos(x), tan(x), sec(x), csc(x), and cot(x).
Video Transcript
Steps to Solve
The derivative of cos(x) is often one that people know by memory, but they don't know how to show it. Let's take a look at how we can find the derivative of cos(x) if we can't remember it off the top of our heads. To do this, we're going to make use of the fact that cos(x) = 1 / sec(x). We're going to find the derivative of 1 / sec(x), and in doing so, we will also find the derivative of cos(x), since they're equal.
Notice that 1 / sec(x) is a quotient. Therefore, we are going to use the quotient rule for derivatives to find the derivative. The quotient rule for derivatives is as follows.
The other facts that we will need to know to find this derivative are as follows:
The derivative of sec(x) is sec(x)tan(x)
The derivative of a constant is 0
tan(x) = sin(x) / cos(x)
(a/b) / (c/d) = (a/b) (d/c) = (ad / bc)
Okay, now that we have a game plan and all the facts that we're going to need, let's dive in! The first thing we want to do is use the quotient rule on 1 / sec(x). In 1 / sec(x), the function in the numerator is f(x) = 1, and the function in the denominator is g(x) = sec(x). We plug these into the quotient rule, which you can see play out here:
Now we're going to simplify. This is where our facts come into play. During the first part of simplification, we use facts 1 and 2, which you can see here:
And now, we can and will simplify further using facts 3 and 4, like this right here:
We can put all this work in a nice organized and compact form, as follows:
So, with all said and done, we see that the derivative of cos(x) is -sin(x).
If you want to make sure that you get all of the different steps that we're showing, feel free to pause and take note of them. If this is just a handy review for you, let's move on.
Trigonometric Function Derivatives
As was said, the derivative of cos(x) is one that is well known. Though it's always useful to know how to find a derivative, knowing some key derivatives by memory is quite useful. Suppose you were solving a problem and along the way, you had to find the derivative of cos(x). If you had it memorized that the derivative of cos(x) is -sin(x), you wouldn't have to go through all the steps we just went through. We see that knowing certain derivatives can save us a lot of time and work.
Derivatives of the trigonometric functions show up in physics, engineering, building, astronomy, trigonometry, and many other areas. On top of that, the derivatives of the basic trigonometric functions (sine, cosine, tangent, cosecant, secant, and cotangent) show up often in the solving process of more complex derivatives. We even saw this in our example. The derivative of sec(x) came up when we were finding the derivative of cos(x).
Because of this, it's an extremely good idea to put the derivatives of the basic trigonometric functions into memory. The following chart appearing on your screen gives the derivatives for each of these functions. Again, feel free to pause so you can make sure you catch all of the different variations.
| Trigonometric Function | Derivative |
--- |
| sin(x) | cos(x) |
| cos(x) | -sin(x) |
| tan(x) | sec 2 (x) |
| csc(x) | -csc(x)cot(x) |
| sec(x) | sec(x)tan(x) |
| cot(x) | -csc 2 (x) |
It's really important to know how to find these derivatives. However, once you've found them, it's a good idea to put them into memory so you don't have to find them every time they come up. Especially because the derivatives of these basic trigonometric functions show up really often.
Lesson Summary
Let's take a brief moment to review what we've learned. The most important thing we learned was that the derivative of cos(x) = -sin(x). We also showed how it was solved, starting with the fact that cos(x) = 1 / sec(x), followed by the following facts:
The derivative of sec(x) is sec(x)tan(x)
The derivative of a constant is 0.
tan(x) = sin(x) / cos(x)
(a/b) / (c/d) = (a/b) (d/c) = (ad / bc)
Then, as you recall, we followed these steps:
Use the quotient rule on 1 / sec(x)
Plug the appropriate information from 1 / sec(x) into the quotient rule
Simplify and use facts 1 and 2 from before
Simplify further using facts 3 and 4 from before
Organize and you'll find the answer -sin(x)
Simple as that!
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AP Calculus AB & BC: Exam Prep
24 chapters
173 lessons
Chapter 1
Graph Basics
Parts of a Graph | Labels & Examples 6:21 min
Function Graphs | Types, Equations & Examples 5:06 min
Chapter 2
The Basics of Functions
Function in Math | Definition & Examples 7:57 min
Functions: Identification, Notation & Practice Problems 9:24 min
Transformations: How to Shift Graphs on a Plane 7:12 min
Domain & Range of a Function | Definition, Equation & Examples 8:32 min
How to Add, Subtract, Multiply and Divide Functions 6:43 min
How to Compose Functions 6:52 min
Inverse Functions | Definition, Methods & Calculation 6:05 min
Applying Function Operations Practice Problems 5:17 min
Chapter 3
How to Graph Functions
Graphing Basic Functions 8:01 min
Compounding Functions and Graphing Functions of Functions 7:47 min
Inverse Function | Graph & Examples 7:31 min
Polynomial Functions: Properties and Factoring 7:45 min
Polynomial Functions: Exponentials and Simplifying 7:45 min
Exponentials, Logarithms & the Natural Log 8:36 min
Horizontal and Vertical Asymptotes 7:47 min
Implicit Function Overview, Formula & Examples 4:30 min
Chapter 4
Overview of Limits of Functions
Understanding Limits: Using Notation 3:43 min
Symbolic Logic Overview, List & Examples 6:30 min
Using a Graph to Define Limits 5:24 min
One-Sided Limits and Continuity 4:33 min
Infinite Limits Definition, Determination & Examples
Limit of a Function | Definition, Rules & Examples 5:15 min
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Graphs and Limits: Defining Asymptotes and Infinity 3:29 min
Limit of a Function | Overview & Existence 5:26 min
Vertical, Horizontal & Slant Asymptotes | Functions & Limits 8:00 min
Comparing Relative Magnitudes of Functions
Chapter 5
Overview of Function Continuity
Continuity in a Function 5:37 min
Points of Discontinuity | Overview, Types & Examples 6:26 min
Regions of Continuity in a Function 5:22 min
Intermediate Value Theorem: Definition 4:50 min
Intermediate Value Theorem | Definition, Proof & Examples 6:30 min
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Chapter 6
Understanding Exponentials & Logarithms
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Exponential Growth & Decay | Formula, Function & Graphs 8:41 min
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Chapter 7
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Optimization and Differentiation 5:49 min
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The Derivative at a Point
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Anti-Derivatives: Calculating Indefinite Integrals of Polynomials 11:55 min
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Power Series | Definition, Operations & Examples 6:17 min
Chapter 23
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Chapter 24
AP Calculus AB & BC Flashcards
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14395 | https://mathoverflow.net/questions/453649/elegant-proof-for-xy-yx-leftrightarrow-x-mathbbn-y-mathbbn | co.combinatorics - Elegant proof for $xy < yx \Leftrightarrow x^\mathbb{N} < y^\mathbb{N}$ - MathOverflow
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Elegant proof for x y<y x⇔x N<y N x y<y x⇔x N<y N
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Let x,y x,y be finite words over totally ordered alphabet and << denote the lexicographical order, i.e for two not necessarily finite words we say x<y x<y iff one of the following holds
There are words u,x′,y′u,x′,y′ and letters a<b a<b such that x=u a x′,y=u b y′x=u a x′,y=u b y′
y=x u y=x u for some non-empty word u u
Then the following are equivalent
x y<y x x y<y x
x N<y N x N<y N, where x N x N denotes infinite word x x x x…x x x x…
x<∗y x<∗y iff x y<y x x y<y x is a total order(which we're trying to prove by equivalence)
This new total order is another extension of "weak lexicographical order" where only first condition considered, i.e x x and x u x u are incomparable for non-empty u u(note that x y x y and y x y x always have the same length so we're not using "strong" order). It is naturally arising from the following problem:
Let x 1,…,x n x 1,…,x n be finite words on totally ordered alphabet, find σ∈S n σ∈S n minimizing x σ(1)x σ(2)…x σ(n)x σ(1)x σ(2)…x σ(n)
The solution is to sort x i x i with respect to order <∗<∗. If the sequence is not sorted according to this order, then there are adjacent elements x,y x,y such that x y>y x x y>y x, by swapping them we'll decrease word which we are minimizing.
But it might be tedious to prove that <∗<∗ is transitive, so instead I suggest equivalent definition of <∗<∗, that is x<∗y x<∗y iff x N<y N x N<y N. Now transitivity is obvious.
But is there elegant and simple proof of equivalence not involving much of casework on |x|<>|y||x|<>|y|, etc?
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asked Aug 29, 2023 at 18:11
thematdevthematdev
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4 Note sure if this elegant enough, but observe that for all n∈N n∈N the inequality x y<y x x y<y x implies that x n y n<y n x n x n y n<y n x n by repeatedly swapping pairs of x x and y y. It follows that x N≤y N x N≤y N. Note that inequality can only hold if the basic period of the infinite word divides both x x and y y, in which case we would have x y=y x x y=y x.1001 –1001 2023-08-29 19:42:08 +00:00 Commented Aug 29, 2023 at 19:42
@1001 yes, I think it's elegant enough, in some sense your claim is taking limit of both sides lim n→∞x n y n=x N lim n→∞x n y n=x N. You can post an answer if you prove 2 -> 1 too, your comment just proves 1 -> 2 thematdev –thematdev 2023-08-29 20:11:11 +00:00 Commented Aug 29, 2023 at 20:11
3 But 2->1 follows from this: if 1 does not hold, then x y≥y x x y≥y x, hence x n y n≥y n x n x n y n≥y n x n and taking the limit in n n we see that 2 also does not hold Fedor Petrov –Fedor Petrov 2023-08-29 20:48:31 +00:00 Commented Aug 29, 2023 at 20:48
1 Technically you didn't define the << order on infinite words.Sam Hopkins –Sam Hopkins♦ 2023-08-29 22:27:02 +00:00 Commented Aug 29, 2023 at 22:27
3 1 and 2 are only equivalent if x x is nonempty.Emil Jeřábek –Emil Jeřábek 2023-08-30 08:57:29 +00:00 Commented Aug 30, 2023 at 8:57
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Let x,y∈Σ+x,y∈Σ+.
Observe that for all n∈N n∈N the inequality x y<y x x y<y x implies that
x n y n<y n x n x n y n<y n x n
by repeatedly swapping pairs of x x and y y. It follows that x N≤y N x N≤y N (i.e. the two infinite sequences are equal, or y N y N is greater at the first position they differ). Note that equality can only hold if the basic period of the infinite word divides both |x||x| and |y||y|, in which case we would have x y=y x x y=y x.
Conversely, x y≥y x x y≥y x implies x N≥y N x N≥y N.
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edited Sep 1, 2023 at 7:25
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2 revs1001
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3 You need x x to be nonempty in the converse argument.Emil Jeřábek –Emil Jeřábek 2023-08-30 09:00:42 +00:00 Commented Aug 30, 2023 at 9:00
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14396 | https://www.lix.polytechnique.fr/Labo/Claire.Kenyon/Publis/rectiles.ps | %!PS-Adobe-2.0 %%Creator: dvipsk 5.55a Copyright 1986, 1994 Radical Eye Software %%Title: rectiles.dvi %%Pages: 10 %%PageOrder: Ascend %%BoundingBox: 0 0 596 842 %%EndComments %DVIPSCommandLine: dvips -o rectiles.ps rectiles.dvi %DVIPSParameters: dpi=300, compressed, comments removed %DVIPSSource: TeX output 1997.05.12:1745 %%BeginProcSet: texc.pro /TeXDict 250 dict def TeXDict begin /N{def}def /B{bind def}N /S{exch}N /X{S N}B /TR{translate}N /isls false N /vsize 11 72 mul N /hsize 8.5 72 mul N /landplus90{false}def /@rigin{isls{[0 landplus90{1 -1}{-1 1} ifelse 0 0 0]concat}if 72 Resolution div 72 VResolution div neg scale isls{landplus90{VResolution 72 div vsize mul 0 exch}{Resolution -72 div hsize mul 0}ifelse TR}if Resolution VResolution vsize -72 div 1 add mul TR matrix currentmatrix dup dup 4 get round 4 exch put dup dup 5 get round 5 exch put setmatrix}N /@landscape{/isls true N}B /@manualfeed{ 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ch-height true[1 0 0 -1 -.1 ch-xoff sub ch-yoff .1 add]/id ch-image N /rw ch-width 7 add 8 idiv string N /rc 0 N /gp 0 N /cp 0 N{ rc 0 ne{rc 1 sub /rc X rw}{G}ifelse}imagemask restore}B /G{{id gp get /gp gp 1 add N dup 18 mod S 18 idiv pl S get exec}loop}B /adv{cp add /cp X}B /chg{rw cp id gp 4 index getinterval putinterval dup gp add /gp X adv}B /nd{/cp 0 N rw exit}B /lsh{rw cp 2 copy get dup 0 eq{pop 1}{dup 255 eq{pop 254}{dup dup add 255 and S 1 and or}ifelse}ifelse put 1 adv} B /rsh{rw cp 2 copy get dup 0 eq{pop 128}{dup 255 eq{pop 127}{dup 2 idiv S 128 and or}ifelse}ifelse put 1 adv}B /clr{rw cp 2 index string putinterval adv}B /set{rw cp fillstr 0 4 index getinterval putinterval adv}B /fillstr 18 string 0 1 17{2 copy 255 put pop}for N /pl[{adv 1 chg} {adv 1 chg nd}{1 add chg}{1 add chg nd}{adv lsh}{adv lsh nd}{adv rsh}{ adv rsh nd}{1 add adv}{/rc X nd}{1 add set}{1 add clr}{adv 2 chg}{adv 2 chg nd}{pop nd}]dup{bind pop}forall N /D{/cc X dup type /stringtype ne{] }if nn /base get cc ctr put nn /BitMaps get S ctr S sf 1 ne{dup dup length 1 sub dup 2 index S get sf div put}if put /ctr ctr 1 add N}B /I{ cc 1 add D}B /bop{userdict /bop-hook known{bop-hook}if /SI save N @rigin 0 0 moveto /V matrix currentmatrix dup 1 get dup mul exch 0 get dup mul add .99 lt{/QV}{/RV}ifelse load def pop pop}N /eop{SI restore showpage userdict /eop-hook known{eop-hook}if}N /@start{userdict /start-hook known{start-hook}if pop /VResolution X /Resolution X 1000 div /DVImag X /IE 256 array N 0 1 255{IE S 1 string dup 0 3 index put cvn put}for 65781.76 div /vsize X 65781.76 div /hsize X}N /p{show}N /RMat[1 0 0 -1 0 0]N /BDot 260 string N /rulex 0 N /ruley 0 N /v{/ruley X /rulex X V}B /V {}B /RV statusdict begin /product where{pop product dup length 7 ge{0 7 getinterval dup(Display)eq exch 0 4 getinterval(NeXT)eq or}{pop false} ifelse}{false}ifelse end{{gsave TR -.1 -.1 TR 1 1 scale rulex ruley false RMat{BDot}imagemask grestore}}{{gsave TR -.1 -.1 TR rulex ruley scale 1 1 false RMat{BDot}imagemask grestore}}ifelse B /QV{gsave transform round exch round exch itransform moveto rulex 0 rlineto 0 ruley neg rlineto rulex neg 0 rlineto fill grestore}B /a{moveto}B /delta 0 N /tail{dup /delta X 0 rmoveto}B /M{S p delta add tail}B /b{S p tail} B /c{-4 M}B /d{-3 M}B /e{-2 M}B /f{-1 M}B /g{0 M}B /h{1 M}B /i{2 M}B /j{ 3 M}B /k{4 M}B /w{0 rmoveto}B /l{p -4 w}B /m{p -3 w}B /n{p -2 w}B /o{p -1 w}B /q{p 1 w}B /r{p 2 w}B /s{p 3 w}B /t{p 4 w}B /x{0 S rmoveto}B /y{ 3 2 roll p a}B /bos{/SS save N}B /eos{SS restore}B end %%EndProcSet TeXDict begin 39158280 55380996 1000 300 300 (rectiles.dvi) @start /Fa 1 27 df<1470EB01F0EB03C0EB0780EB0E005B5B5BA213F05BB3AC485AA3 485A48C7FC1206120E12385A12C012707E120E120612076C7E6C7EA36C7EB3AC7F1370A2 7F7F7FEB0780EB03C0EB01F0EB007014637B811F>26 D E /Fb 4 117 df103 DI<13E0EA01F0A213E0EA00C013 00A7120FEA1380EA23C01263EA478012C71207EA0F00A3121E1320EA3C60A21340EA38C0 EA1980EA0F000C1E7D9D0F>I116 D E /Fc 6 111 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02EA1C01003813E0EA7800A25AA71278A2EA3801121CEA0C02EA070CEA01F0C7FCA9EB0F FE171F7E941A>113 DII<1202A41206A3120E 121E123EEAFFFCEA0E00AB1304A6EA07081203EA01F00E1F7F9E13>I<000E137038FE07 F0EA1E00000E1370AD14F0A238060170380382783800FC7F18157F941B>I<38FF83FE38 1F00F0000E13C06C1380EB8100EA0383EA01C2EA00E41378A21338133C134E138FEA0187 EB0380380201C0000413E0EA0C00383E01F038FF03FE17157F941A>120 D<38FF80FE381E00781430000E1320A26C1340A2EB80C000031380A23801C100A2EA00E2 A31374A21338A31310A25BA35B12F05B12F10043C7FC123C171F7F941A>I E /Fs 16 122 df80 D82 D<007FB71280A39039807F807FD87C00140F007815 07A20070150300F016C0A2481501A5C791C7FCB3A490B612C0A32A287EA72F>84 D<3803FF80000F13F0381F01FC383F80FE147F801580EA1F00C7FCA4EB3FFF3801FC3FEA 0FE0EA1F80EA3F00127E5AA4145F007E13DF393F839FFC381FFE0F3803FC031E1B7E9A21 >97 D99 D101 D<9038FF80F00003EBE3F8390FC1FE1C391F007C7C4813 7E003EEB3E10007EEB3F00A6003E133E003F137E6C137C380FC1F8380BFFE00018138090 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Fe(0)834 2534 y Fl(;)7 b(x)877 2519 y Fe(00)897 2534 y Fn())-57 2576 y Fo(is)15 b(bije)n(ctive.)-57 2654 y Fn(No)o(w,)22 b(the)g(closed)g(path)g Fl(w)g Fn(in)f(\000()p Fl(G)p Fn())g(asso)q(ciated)h(to)g(the)-57 2700 y(b)q(oundary)14 b(of)f Fl(P)19 b Fn(maps)13 b(to)h(a)f(closed)i(path)f Fl(w)668 2685 y Fe(0)693 2700 y Fn(in)f(\000()p Fl(H)s Fn().)1063 117 y(Ro)q(ot)f(\000()p Fl(H)s Fn())g(at)h(some)f (arbitrary)g(v)o(ertex)i Fl(r)q Fn(,)e(far)g(from)f Fl(w)1960 101 y Fe(0)1971 117 y Fn(,)1013 158 y(and)g(de\014ne)h(the)g(heigh)o(t) f(of)g Fl(x)g Fm(2)g Fl(H)k Fn(to)c(b)q(e)h(the)g(distance)g(from)1013 200 y Fl(x)k Fn(to)h Fl(r)h Fn(in)e(the)i(graph)f(\000()p Fl(H)s Fn().)27 b((F)m(or)17 b(example,)f(if)g Fl(n)h Fn(is)f(the)1013 248 y(surface)d(area)g(of)f Fl(P)6 b Fn(,)12 b(w)o(e)h(can)f(de\014ne)i Fl(r)f Fn(as)g(the)g(p)q(oin)o(t)f (()p Fl(ab)p Fn())1927 233 y Fj(10)p Fc(n)1013 289 y Fn(in)h Fl(H)s Fn().)1013 369 y Fd(Lemma)i(1)21 b Fo(If)15 b Fl(P)20 b Fo(is)15 b(tile)n(able,)g(then)g(ther)n(e)g(is)g (a)h(tiling)e(of)i Fl(P)1013 410 y Fo(whose)f(highest)g(p)n(oint)g(is)f (on)i(the)f(b)n(oundary)g Fl(@)r(P)6 b Fo(.)1013 488 y Fd(Pro)q(of:)17 b Fn(Let)d Fm(T)24 b Fn(b)q(e)14 b(a)f(tiling)f(of)h (the)h(p)q(olygon.)i(Ev)o(ery)e(p)q(oin)o(t)1013 530 y(inside)i Fl(P)22 b Fn(is)17 b(on)f(the)i(b)q(oundary)e(of)g(a)h (tile,)f(and)h(so)f(has)h(an)1013 571 y(asso)q(ciated)f(elemen)o(t)g (of)f Fl(G)g Fn((and)h(so)g(a)g(heigh)o(t)f(in)h Fl(H)s Fn().)24 b(As-)1013 613 y(sume)11 b(that)h(a)f(highest)h(p)q(oin)o(t)f Fl(x)g Fn(is)h(in)f(the)h(in)o(terior)g(of)f Fl(P)6 b Fn(.)16 b(W)m(e)1013 654 y(claim)f(that)j Fl(x)f Fn(cannot)h(b)q(e)g (at)g(the)g(corner)h(of)e(a)g(tile.)29 b(F)m(or,)1013 696 y(if)13 b Fl(x)h Fn(is,)g(for)g(example,)e(the)j(lo)o(w)o(er)f (left)g(corner)h(of)f(some)f(tile,)1013 737 y(then)i Fl(x)g Fn(has)g(neigh)o(b)q(ors)g(lab)q(elled)f Fl(xa)g Fn(and)h Fl(xb)p Fn(.)20 b(In)15 b(\000()p Fl(H)s Fn(),)f(it)1013 779 y(is)k(easy)g(to)g(see)i(that)e(no)g(matter)f(ho)o(w)h Fl(x)g Fn(is)g(with)g(resp)q(ect)1013 820 y(to)e Fl(r)q Fn(,)h(either)g Fl(xa)f Fn(or)g Fl(xb)g Fn(is)h(farther)g(from)d Fl(r)k Fn(than)e Fl(x)p Fn(,)h(hence)1013 862 y(higher:)h(con)o (tradiction.)1063 904 y(Without)10 b(loss)h(of)f(generalit)o(y)m(,)g(w) o(e)h(can)g(assume)f(then)i(that)1013 946 y Fl(x)h Fn(is)g(on)g(the)h (left)f(side)h(of)f(a)g(v)o(ertical)g(tile.)k(Then)d(the)g(tile)f(on) 1013 987 y(the)k(left)g(side)g(of)f Fl(x)h Fn(m)o(ust)f(also)g(b)q(e)h (v)o(ertical)g((b)q(ecause)i Fl(x)d Fn(is)1013 1029 y(not)g(a)g(corner)i(and)e(not)h(on)f(the)h(b)q(oundary).)26 b(Assume)16 b(for)1013 1077 y(example)8 b(that)i Fl(x)f Fn(is)h(as)g(in)f(\014gure)h(3.)16 b(Then)10 b Fl(xb)1725 1062 y Fj(2)1744 1077 y Fn(,)g(whic)o(h)f(lab)q(els)1013 1119 y(a)i(neigh)o(b)q(or)g(of)g Fl(x)p Fn(,)g(is)g(also)f(highest.)18 b(Since)12 b(it)e(is)i(not)f(a)g(corner)1013 1160 y(either,)17 b(this)g(determines)g(the)g(p)q(osition)e(of)h(the)h(tile)f(left)h(of) 1013 1202 y Fl(x)p Fn(:)g(the)e(t)o(w)o(o)e(tiles)h(surrounding)g Fl(x)f Fn(m)o(ust)g(form)f(a)i(rectangle.)1063 1249 y(Th)o(us)g Fl(y)g Fn(=)f Fl(xbab)1329 1234 y Fe(\000)p Fj(1)1387 1249 y Fn(and)i Fl(z)f Fn(=)f Fl(xba)1611 1234 y Fe(\000)p Fj(1)1655 1249 y Fl(b)1673 1234 y Fe(\000)p Fj(1)1731 1249 y Fn(are)i(t)o(w)o(o)f(lab)q(els)1013 1290 y(of)i(the)h(tiling.)24 b(W)m(e)16 b(observ)o(e)h(that)g(no)f(matter)g(where)h Fl(r)h Fn(is,)1013 1332 y(one)g(of)f(the)i(t)o(w)o(o)e(p)q(oin)o(ts)h Fl(y)i Fn(and)e Fl(z)i Fn(m)o(ust)d(b)q(e)h(at)g(the)h(same)1013 1373 y(distance)d(as)f Fl(x)g Fn(from)f Fl(r)q Fn(,)h(hence)i(also)e (of)f(maxim)o(um)d(heigh)o(t.)1013 1415 y(F)m(or)18 b(example,)f Fl(z)j Fn(is)e(also)g(a)f(maxim)o(um)c(in)18 b(\014gure)h(4.)30 b(If)18 b Fl(z)1013 1456 y Fn(is)h(on)f(the)i(b)q(order)g(of)e Fl(P)6 b Fn(,)20 b(w)o(e)f(are)g(done.)34 b(Otherwise,)21 b(b)o(y)1013 1498 y(the)13 b(same)f(argumen)o(t)f(as)i(ab)q(o)o(v)o(e,) f(the)h(tile)f(left)g(of)g Fl(z)j Fn(m)o(ust)c(b)q(e)1013 1539 y(v)o(ertical)18 b(and)g(form)f(a)h(rectangle)h(with)f(the)h (previous)f(t)o(w)o(o)1013 1581 y(tiles.)27 b(W)m(e)17 b(can)g(no)o(w)g(replace)g(the)h(three)g(v)o(ertical)f(tiles)g(b)o(y) 1013 1622 y(three)g(horizon)o(tal)e(tiles,)h(forming)d(a)j(new,)g(lo)o (w)o(er)g(tiling)e((in)1013 1664 y(the)g(sense)h(that)f(ev)o(ery)h(p)q (oin)o(t)e(of)g Fl(P)19 b Fn(is)14 b(lo)o(w)o(er;)e(see)j(\014gure)g (5).)1013 1705 y(Therefore)e(a)f(lo)o(w)o(est)h(tiling)d(m)o(ust)i(ha) o(v)o(e)g(a)g(highest)g(p)q(oin)o(t)g(on)1013 1747 y(its)i(b)q(oundary) g(.)1063 1790 y(Giv)o(en)i Fl(P)6 b Fn(,)17 b(w)o(e)g(no)o(w)f (consider)i(the)g(highest)f(p)q(oin)o(t)g Fl(x)f Fn(on)1013 1831 y(the)i(b)q(oundary)f Fl(@)r(P)24 b Fn(of)17 b Fl(P)6 b Fn(.)28 b(F)m(rom)16 b(the)i(ab)q(o)o(v)o(e)f(lemma,)e(w)o(e)1013 1873 y(kno)o(w)i(that)h Fl(x)g Fn(is)g(also)f(the)h(highest)h(p)q(oin)o (t)e(of)g(some)g(tiling)1013 1914 y(of)d Fl(P)20 b Fn((if)14 b Fl(P)20 b Fn(is)14 b(tileable).)20 b(It)15 b(is)f(easy)h(to)g(see)h (that)e Fl(x)h Fn(cannot)1013 1956 y(b)q(e)h(at)f(a)g(corner)h(of)e Fl(@)r(P)6 b Fn(.)22 b(Assume)15 b(that)g Fl(x)g Fn(is)g(on)g(a)g(v)o (ertical)1013 1997 y(side)g(of)g Fl(@)r(P)6 b Fn(,)15 b(as)g(in)g(\014gure)h(6.)22 b(The)16 b(neigh)o(b)q(ors)f(of)g Fl(x)g Fn(on)g Fl(@)r(P)1013 2043 y Fn(are)e(lab)q(elled)f Fl(xb)f Fn(and)i Fl(xb)1408 2028 y Fe(\000)p Fj(1)1452 2043 y Fn(.)k(One)d(of)d(them)h(m)o(ust)f(also)h(ha)o(v)o(e)1013 2085 y(maxim)n(um)d(heigh)o(t,)j(for)h(example)f Fl(xb)p Fn(.)17 b(Since)d(neither)g Fl(x)f Fn(nor)1013 2126 y Fl(xb)i Fn(can)g(b)q(e)h(at)f(a)g(corner)h(of)f(a)g(tile,)g(the)h(only) e(p)q(ossibilit)o(y)g(is)1013 2168 y(a)f(v)o(ertical)h(tile)g(co)o(v)o (ering)f(b)q(oth)h Fl(x)g Fn(and)g Fl(xb)p Fn(.)1063 2210 y(Remo)o(ving)g(that)i(tile,)h(w)o(e)g(get)g(a)f(region)h(whic)o (h)f(can)h(b)q(e)1013 2252 y(tiled)e(i\013)g Fl(P)20 b Fn(can)c(b)q(e)f(tiled:)21 b(iterating)14 b(the)i(pro)q(cess,)h(w)o (e)e(will)1013 2293 y(obtain)i(an)h(algorithm)e(for)i(tiling)e Fl(P)24 b Fn(or)18 b(deciding)g(that)h Fl(P)1013 2335 y Fn(cannot)14 b(b)q(e)g(tiled)g(b)o(y)g(1)9 b Fm(\002)g Fn(3)14 b(and)f(3)c Fm(\002)h Fn(1)j(tiles.)1013 2394 y Fp(2.3)53 b(The)18 b(algorithm)1063 2446 y Fd(Input:)f Fn(a)c(p)q(olygonal)f(region)i Fl(P)19 b Fn(of)13 b Fl(Z)1683 2431 y Fj(2)1716 2446 y Fn(of)g(area)i Fl(n)p Fn(.)1013 2488 y Fd(Initiali)o(zati)o(on:)31 b Fn(Lab)q(el)21 b(some)g(p)q(oin)o (t)g(of)g Fl(P)27 b Fn(b)o(y)21 b Fl(e)p Fn(.)41 b(Let)1013 2534 y Fl(r)21 b Fn(=)f(()p Fl(ab)p Fn())1177 2519 y Fj(10)p Fc(n)1252 2534 y Fm(2)g Fl(H)j Fn(=)d Fm(h)p Fl(a;)7 b(b)p Fm(j)p Fl(a)1519 2519 y Fj(3)1536 2534 y Fl(;)g(b)1573 2519 y Fj(3)1591 2534 y Fm(i)p Fn(.)33 b(Going)18 b(around)g(the)1013 2575 y(b)q(oundary)m(,)f(lab)q(el)f(eac) o(h)i(p)q(oin)o(t)f(of)f Fl(@)r(P)23 b Fn(b)o(y)17 b(a)g(w)o(ord)g Fl(w)h Fm(2)e Fl(H)1013 2617 y Fn(and)e(compute)f(its)h(heigh)o(t.)1013 2658 y Fd(Rep)q(eat:)k Fn(Let)c Fl(x)g Fn(b)q(e)g(a)g(highest)g(p)q (oin)o(t)f(of)h Fl(@)r(P)6 b Fn(.)1013 2700 y(One)15 b(of)f Fl(x)p Fn('s)g(neigh)o(b)q(ors,)h(sa)o(y)f Fl(y)q Fn(,)h(is)f(also)g(a)h(highest)g(p)q(oin)o(t)f(of)p eop %%Page: 3 3 3 2 bop 402 642 a Fn(Figure)14 b(1:)k(A)c(necessary)i(but)e(not)g (su\016cien)o(t)g(condition)f(for)h(tiling)e Fl(P)477 1281 y Fn(Figure)i(2:)k(The)c(Ca)o(yley)f(graph)h(of)g Fl(H)g Fn(=)e Fm(h)p Fl(a;)7 b(b)p Fm(j)p Fl(a)1249 1265 y Fj(3)1278 1281 y Fn(=)12 b Fl(b)1340 1265 y Fj(3)1369 1281 y Fn(=)g Fl(e)p Fm(i)243 2034 y Fn(Figure)i(3:)k(Pro)q(of)13 b(of)g(lemma)e(1:)18 b(p)q(ositioning)13 b(the)h(tiles)g(surrounding)g (the)g(highest)h(p)q(oin)o(t)380 2667 y(Figure)f(4:)k(Pro)q(of)c(of)f (lemma)e(1:)17 b(building)c(a)g(square)i(with)f(highest)g(tiles)p eop %%Page: 4 4 4 3 bop 145 1286 a Fn(Figure)14 b(5:)j(Pro)q(of)d(of)f(lemma)e(1:)18 b(Remo)o(ving)11 b(in)o(terior)j(maxim)o(a)d(b)o(y)j(c)o(hanging)f(the) h(tiling)f(of)g(a)g(square)544 2378 y(Figure)h(6:)k(Tiling)12 b Fl(P)6 b Fn(:)17 b(putting)d(do)o(wn)f(the)i(\014rst)f(tile)p eop %%Page: 5 5 5 4 bop -57 117 a Fl(@)r(P)6 b Fn(.)26 b(T)m(ak)o(e)16 b(the)h(tile)f Fl(T)23 b Fn(in)16 b Fl(P)22 b Fn(co)o(v)o(ering)16 b Fl(x)g Fn(and)h Fl(y)h Fn((v)o(ertical)-57 158 y(if)13 b Fl(xy)j Fn(is)e(a)f(v)o(ertical)h(segmen)o(t,)g(horizon)o(tal)f (otherwise).)19 b(Re-)-57 200 y(mo)o(v)o(e)12 b Fl(T)20 b Fn(from)12 b Fl(P)6 b Fn(.)-57 241 y(Up)q(date)15 b Fl(@)r(P)6 b Fn(,)13 b(the)h(lab)q(els)g(and)g(heigh)o(ts.)-57 283 y Fd(Un)o(til:)i Fl(P)k Fn(is)13 b(tiled)h(or)g(some)f(p)q(oin)o(t) h(of)g Fl(P)19 b Fn(is)14 b(giv)o(en)f(t)o(w)o(o)h(dif-)-57 324 y(feren)o(t)h(lab)q(els)e((then)i Fl(P)k Fn(cannot)14 b(b)q(e)h(tiled).)-7 366 y(It)d(is)g(easy)h(to)f(see)i(that,)e(since)h (all)e(the)i(up)q(dates)g(are)g(lo)q(cal)-57 407 y(op)q(erations,)e (the)h(algorithm)d(can)i(b)q(e)h(implemen)o(ted)d(so)i(as)g(to)-57 449 y(ha)o(v)o(e)j(linear)f(running)h(time.)-7 490 y(If)9 b(the)h(tiles)g(are)f Fl(k)r Fm(\002)p Fn(1)h(and)f(1)p Fm(\002)p Fl(l)j Fn(instead)e(of)e(1)p Fm(\002)p Fn(3)j(and)e(3)p Fm(\002)p Fn(1,)-57 532 y(the)k(algorithm)d(can)j(easily)g(b)q(e)g (generalized.)18 b(The)c(group)e(to)-57 579 y(consider)k(is)f Fl(H)i Fn(=)d Fm(h)p Fl(a;)7 b(b)p Fm(j)p Fl(a)356 564 y Fc(k)375 579 y Fl(;)g(b)412 564 y Fc(l)424 579 y Fm(i)p Fn(,)16 b(whose)g(Ca)o(yley)e(graph)h(is)g(a)-57 621 y(tree)k(of)d(alternating)h Fl(k)q Fn(-cycles)i(and)e Fl(l)q Fn(-cycles.)29 b(The)18 b(heigh)o(t)-57 662 y(is)c(similarly)d (de\014ned,)k(as)f(the)h(distance)g(to)f(some)f(arbitrary)-57 704 y(ro)q(ot)f((su\016cien)o(tly)g(far)f(a)o(w)o(a)o(y).)16 b(Giv)o(en)c(a)f(highest)h(b)q(oundary)-57 745 y(p)q(oin)o(t,)f(the)g (lo)o(w)o(est)g(tiling)f(m)o(ust)f(con)o(tain)i(the)h(tile)e(whic)o(h)h (has)-57 787 y Fl(x)j Fn(in)f(the)h(middle)f(or)g(co-middle)g(of)g(its) h(long)f(side.)-57 860 y Fd(Theorem)i(1)20 b Fo(A)15 b(p)n(olygon)g(of)f(ar)n(e)n(a)g Fl(n)g Fo(c)n(an)h(b)n(e)g(tile)n(d)e (by)i Fn(1)8 b Fm(\002)g Fl(l)-57 901 y Fo(and)19 b Fl(k)13 b Fm(\002)e Fn(1)18 b Fo(tiles,)g(or)g(pr)n(ove)n(d)h(not)f(to)g(b)n(e) h(tile)n(able,)e(in)i(time)-57 943 y(line)n(ar)14 b(in)h Fl(n)p Fo(.)-57 1033 y Fk(3)56 b(Tiling)18 b(a)h(p)r(olygon)f(with)h Ff(k)14 b Fg(\002)d Ff(l)20 b Fk(and)f Ff(l)12 b Fg(\002)f Ff(k)27 1091 y Fk(rectangles)-7 1136 y Fn(W)m(e)f(will)e(only)i(presen) o(t)i(the)f(algorithm)c(in)j(the)h(case)g Fl(k)i Fn(=)e(2)-57 1177 y(and)e Fl(l)k Fn(=)f(3,)e(since)g(generalizing)f(the)h(algorithm) d(is)j(easy)g(from)-57 1219 y(that)k(sp)q(ecial)g(case.)-57 1272 y Fp(3.1)53 b(The)18 b(tiling)h(group)g(and)f(its)f(quotien)o(ts) -7 1323 y Fn(The)10 b(t)o(w)o(o)f(rectangular)g(tiles)h(de\014ne)g(the) g(relations)f Fl(a)814 1308 y Fj(2)833 1323 y Fl(b)851 1308 y Fj(3)881 1323 y Fn(=)-57 1369 y Fl(b)-39 1354 y Fj(3)-21 1369 y Fl(a)1 1354 y Fj(2)36 1369 y Fn(and)17 b Fl(a)142 1354 y Fj(3)160 1369 y Fl(b)178 1354 y Fj(2)212 1369 y Fn(=)f Fl(b)278 1354 y Fj(2)297 1369 y Fl(a)319 1354 y Fj(3)337 1369 y Fn(,)h(hence)g(the)g(tiling)e(group)h(is)h Fl(G)e Fn(=)-57 1415 y Fm(h)p Fl(a;)7 b(b)p Fm(j)p Fn([)p Fl(a)64 1400 y Fj(2)81 1415 y Fl(;)g(b)118 1400 y Fj(3)136 1415 y Fn(])p Fl(;)g Fn([)p Fl(a)201 1400 y Fj(3)218 1415 y Fl(;)g(b)255 1400 y Fj(2)273 1415 y Fn(])p Fm(i)p Fn(.)40 b(Let)22 b Fl(H)j Fn(b)q(e)d(the)g(quotien)o(t)f(group)-57 1463 y Fl(H)26 b Fn(=)e Fl(G=a)136 1448 y Fj(3)154 1463 y Fl(;)7 b(b)191 1448 y Fj(3)232 1463 y Fn(=)24 b Fm(h)p Fl(a;)7 b(b)p Fm(j)p Fl(a)397 1448 y Fj(3)415 1463 y Fl(;)g(b)452 1448 y Fj(3)469 1463 y Fm(i)p Fn(,)23 b(and)e Fl(K)j Fn(b)q(e)e(the)f(group)-57 1511 y Fl(K)16 b Fn(=)d Fl(G=a)115 1496 y Fj(2)133 1511 y Fl(;)7 b(b)170 1496 y Fj(2)200 1511 y Fn(=)13 b Fm(h)p Fl(a;)7 b(b)p Fm(j)p Fl(a)354 1496 y Fj(2)371 1511 y Fl(;)g(b)408 1496 y Fj(2)426 1511 y Fm(i)p Fn(.)20 b(W)m(e)14 b(ha)o(v)o(e:)20 b Fl(Z)691 1496 y Fj(2)722 1511 y Fn(=)13 b Fl(G=)p Fn([)p Fl(a;)7 b(b)p Fn(].)-57 1557 y(Giv)o(en)15 b Fl(w)h Fm(2)f Fl(G)p Fn(,)g(let)h Fl(w)307 1542 y Fe(0)319 1557 y Fl(;)7 b(w)369 1542 y Fe(00)389 1557 y Fl(;)g(w)439 1542 y Fe(000)485 1557 y Fn(b)q(e)17 b(the)f(pro)r(jections)h(of)e Fl(w)-57 1603 y Fn(on)f Fl(H)s Fn(,)f Fl(K)k Fn(and)d Fl(Z)228 1588 y Fj(2)247 1603 y Fn(.)-57 1676 y Fd(F)l(act)i(4)21 b Fo(The)14 b(map)i Fl(w)c Fm(7!)f Fn(()p Fl(w)411 1661 y Fe(0)423 1676 y Fl(;)c(w)473 1661 y Fe(00)493 1676 y Fl(;)g(w)543 1661 y Fe(000)573 1676 y Fn())15 b Fo(is)g(bije)n (ctive.)-57 1749 y Fn(De\014ning)22 b(a)g(heigh)o(t)g(function)g(is)g (no)o(w)f(a)h(more)f(complex)-57 1791 y(op)q(eration,)c(since)h(it)e (apparen)o(tly)h(in)o(v)o(olv)o(es)e(b)q(oth)i(quotien)o(t)-57 1832 y(groups)12 b Fl(H)i Fn(and)d Fl(K)s Fn(.)17 b(Ho)o(w)o(ev)o(er,) 12 b(it)f(turns)h(out)f(that)h(using)f(the)-57 1874 y(lab)q(els)g(in)g Fl(H)k Fn(already)c(giv)o(es)g(us)h(\most")e(of)h(the)h(information) -57 1915 y(that)i(w)o(e)g(need)h(to)f(construct)h(a)f(tiling.)-57 1969 y Fp(3.2)53 b(De\014ning)18 b(a)g(heigh)o(t)g(function)-7 2014 y Fn(W)m(e)e(lab)q(el)f(some)h(arbitrary)g(p)q(oin)o(t)g(of)f Fl(@)r(P)22 b Fn(b)o(y)16 b Fl(e)p Fn(.)26 b(Going)-57 2055 y(around)12 b(the)g(b)q(oundary)m(,)g(w)o(e)g(lab)q(el)f(eac)o(h)h (p)q(oin)o(t)g(of)f Fl(@)r(P)18 b Fn(b)o(y)11 b(an)-57 2097 y(elemen)o(t)i(of)h Fl(H)s Fn(.)k(Ro)q(oting)12 b Fl(H)17 b Fn(at)d(some)f(p)q(oin)o(t)g Fl(r)i Fn(su\016cien)o(tly)-57 2138 y(far)e(a)o(w)o(a)o(y)m(,)e(w)o(e)j(can)f(de\014ne)i(a)e(heigh)o (t)g(function)g(as)g(b)q(eing)h(the)-57 2180 y(distance)21 b(from)e Fl(r)i Fn(in)g(the)g(graph)f(\000()p Fl(H)s Fn().)38 b(A)21 b(tiling)e(of)h Fl(P)-57 2222 y Fn(extends)c(the)e (lab)q(ellings)f(and)h(heigh)o(ts)h(to)f(all)f(the)i(p)q(oin)o(ts)f(of) -57 2263 y Fl(P)19 b Fn(whic)o(h)14 b(are)h(on)e(the)i(b)q(order)g(of)e (a)h(tile)f((eac)o(h)i(tile)f(has)g(t)o(w)o(o)-57 2305 y(in)o(terior)g(p)q(oin)o(ts)f(whose)i(heigh)o(t)f(is)f(unde\014ned).) -57 2378 y Fd(Lemma)i(2)21 b Fo(If)15 b Fl(P)21 b Fo(is)15 b(tile)n(able,)f(then)i(ther)n(e)e(is)h(a)h(tiling)f(of)g Fl(P)-57 2419 y Fo(whose)g(maximum)h(height)f Fl(h)g Fo(o)n(c)n(curs)h(either)e(on)i(the)f(b)n(ound-)-57 2461 y(ary)c(or)g(within)g(distanc)n(e)h(1)f(of)h(the)f(b)n(oundary)h((in)g (which)f(c)n(ase)-57 2502 y(ther)n(e)j(is)h(a)g(b)n(oundary)h(p)n(oint) f(of)g(height)g Fl(h)9 b Fm(\000)g Fn(1)p Fo().)-57 2575 y Fd(Pro)q(of)15 b(sk)o(etc)o(h:)i Fn(Let)d Fl(x)f Fn(b)q(e)i(a)e(highest)h(in)o(terior)f(p)q(oin)o(t.)18 b(Us-)-57 2617 y(ing)g(the)i(same)e(elemen)o(tary)g(argumen)o(ts)h(as)g (in)f(section)i(2,)-57 2658 y(it)15 b(is)h(easy)g(to)f(see)i(that)f Fl(x)f Fn(cannot)h(b)q(e)g(at)g(the)g(corner)h(or)f(in)-57 2700 y(the)j(middle)e(of)h(the)h(short)g(side)g(of)f(a)h(rectangle:)28 b Fl(x)18 b Fn(m)o(ust)1013 117 y(b)q(e)j(on)e(the)i(long)e(side)i(of)e (a)h(rectangle.)38 b(Doing)18 b(case-b)o(y-)1013 158 y(case)g(analysis,)f(w)o(e)g(\014nd)h(that)f(the)h(tiling)d(m)o(ust)h (con)o(tain)h(a)1013 200 y(\blo)q(c)o(k")11 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y(on)f Fl(@)r(P)20 b Fn(is)14 b(also)f(highest,)h(sa)o(y)g Fl(y)q Fn(.)1063 1544 y(If)d(the)i(tile)f(at)g Fl(x)g Fn(is)f(a)h(v)o(ertical)g(rectangle,)h(then)g(it)f(is)g(easy)1013 1586 y(to)19 b(see)i(that)e(it)g(has)h(to)f(b)q(e)h(a)f(rectangle)h(co) o(v)o(ering)f Fl(x)g Fn(and)1013 1627 y Fl(y)e Fn((see)g(\014gure)f(9) g({)f(otherwise)i(some)d(in)o(terior)i(p)q(oin)o(t)f(w)o(ould)1013 1669 y(ha)o(v)o(e)f(heigh)o(t)f(2)h(more)f(than)h(the)g(heigh)o(t)g(of) f Fl(x)p Fn().)1063 1712 y(If)h(it)g(is)g(horizon)o(tal,)g(then)h(the) g(p)q(oin)o(ts)f(along)g(one)g(of)g(the)1013 1753 y(long)d(sides)h(of)f (the)h(rectangle)h(ha)o(v)o(e)e(heigh)o(t)h(equal)f(to)h(1)f(more)1013 1795 y(than)16 b(the)g(heigh)o(t)f(of)h Fl(x)p Fn(:)21 b(they)16 b(m)o(ust)f(therefore)i(b)q(e)g(highest)1013 1836 y(p)q(oin)o(ts)c(of)g Fl(P)6 b Fn(.)17 b(As)d(in)e(the)i(previous) g(section,)g(b)o(y)f(doing)f(case)1013 1878 y(analysis)i(to)h(study)g (the)h(p)q(osition)e(of)g(the)i(tiles)f(adjacen)o(t)g(to)1013 1919 y(that)g(\014rst)h(tile,)f(w)o(e)g(see)h(that)f(there)i(has)e(to)g (b)q(e)h(a)f(blo)q(c)o(k,)f(in)1013 1961 y(t)o(w)o(o)f(p)q(ossible)h(p) q(ositions.)1013 2040 y Fd(Lemma)h(3)21 b Fo(L)n(et)16 b Fl(B)j Fo(b)n(e)d(a)h(vertic)n(al)e(blo)n(ck)i(of)f(a)h(tiling)e(of)i Fl(P)6 b Fo(.)1013 2081 y(The)15 b(ar)n(e)n(a)f(to)h(the)g(right)f(of)h Fl(B)j Fo(must)1013 2123 y(1))d(c)n(ontain)g(some)g(p)n(oints)g(in)g (the)g(exterior)f(of)h Fl(P)20 b Fo(or)1013 2164 y(2))13 b(b)n(e)g(c)n(over)n(e)n(d)f(by)h(another)g(blo)n(ck)g(at)g(the)g(same) g(level)f((\014gur)n(e)1013 2206 y(10a))j(or)1013 2247 y(3))f(b)n(e)g(c)n(over)n(e)n(d)f(by)h(two)g(vertic)n(al)f(r)n(e)n (ctangles)g((\014gur)n(e)h(10b))g(or)1013 2289 y(4))j(b)n(e)h(c)n (over)n(e)n(d)f(by)h(one)g(vertic)n(al)f(r)n(e)n(ctangle)g(and)h(one)g (blo)n(ck)1013 2330 y(shifte)n(d)c(by)h(3)g((\014gur)n(e)g(10c,c'))g (or)1013 2372 y(5))g(b)n(e)g(c)n(over)n(e)n(d)g(by)g(two)f(blo)n(cks)h (shifte)n(d)f(by)i(3)f((\014gur)n(e)g(10d).)1013 2450 y Fn(The)f(pro)q(of)g(is)f(b)o(y)h(insp)q(ection.)1063 2492 y(Up)e(to)h(rotating)f(the)h(6)7 b Fm(\002)g Fn(6)12 b(\014gure)h(to)g(c)o(hange)g(the)g(tiling,)1013 2534 y(w)o(e)i(can)g(assume)g(that)g(case)h Fl(a)f Fn(nev)o(er)g(o)q(ccurs)i (in)d(the)i(tiling.)1013 2575 y(W)m(e)f(will)f(no)o(w)h(apply)g(the)h (lemma)c(rep)q(eatedly)m(.)24 b(Let)16 b Fl(B)1906 2581 y Fj(0)1940 2575 y Fn(b)q(e)1013 2617 y(the)10 b(blo)q(c)o(k)g(at)g Fl(x)p Fn(.)16 b(W)m(e)10 b(sa)o(y)g(that)g(a)g(blo)q(c)o(k)f Fl(B)k Fn(is)d Fo(r)n(e)n(achable)g Fn(from)1013 2658 y Fl(B)1044 2664 y Fj(0)1077 2658 y Fn(if)i(there)j(is)f(a)f(sequence)j (of)d(blo)q(c)o(ks)h Fl(B)1672 2664 y Fj(1)1691 2658 y Fl(;)7 b(B)1741 2664 y Fj(2)1760 2658 y Fl(;)g(:::;)g(B)1865 2664 y Fc(m)1906 2658 y Fn(=)k Fl(B)1013 2700 y Fn(suc)o(h)k(that)f Fl(B)1228 2706 y Fc(i)1256 2700 y Fn(is)g(in)g(the)h(righ)o(t)e(neigh)o (b)q(orho)q(o)q(d)h(of)g Fl(B)1859 2706 y Fc(i)p Fe(\000)p Fj(1)1930 2700 y Fn((as)p eop %%Page: 6 6 6 5 bop 575 710 a Fn(Figure)14 b(7:)k(In)o(terior)c(highest)g(p)q(oin)o (ts)g(of)f(a)h(tiling)431 1832 y(Figure)g(8:)k(Pushing)c(the)g(highest) g(p)q(oin)o(t)g(to)o(w)o(ards)g(the)g(b)q(oundary)634 2600 y(Figure)g(9:)j(Putting)d(do)o(wn)g(the)g(\014rst)h(tile)p eop %%Page: 7 7 7 6 bop 550 1433 a Fn(Figure)14 b(10:)j(The)e(righ)o(t)e(neigh)o(b)q (orho)q(o)q(d)h(of)f(a)h(blo)q(c)o(k)573 2467 y(Figure)g(11:)k(Pushing) c(v)o(ertical)g(tiles)f(to)h(the)h(left)p eop %%Page: 8 8 8 7 bop -57 117 a Fn(in)13 b(cases)h Fl(c;)7 b(c)149 101 y Fe(0)160 117 y Fl(;)g(e;)g(e)236 101 y Fe(0)260 117 y Fn(and)13 b Fl(d)g Fn(of)g(\014gure)g(10).)18 b(W)m(e)13 b(consider)h(the)-57 158 y(set)h Fl(S)h Fn(of)e(all)e(blo)q (c)o(ks)i(reac)o(hable)h(from)d Fl(B)593 164 y Fj(0)626 158 y Fn(in)h(the)i(tiling.)-7 201 y(If)g(these)j(blo)q(c)o(ks)e(nev)o (er)h(touc)o(h)f(the)h(b)q(order)g Fl(@)r(P)22 b Fn(outside)-57 243 y Fl(B)-26 249 y Fj(0)-7 243 y Fn(,)16 b(the)h(only)e(cases)i(whic) o(h)f(o)q(ccur)h(are)f Fl(b;)7 b(c;)g(c)692 228 y Fe(0)718 243 y Fn(and)16 b Fl(d)p Fn(.)24 b(All)-57 284 y(the)14 b(blo)q(c)o(ks)g(farthest)g(from)e Fl(B)422 290 y Fj(0)454 284 y Fn(ha)o(v)o(e)i(righ)o(t)f(neigh)o(b)q(orho)q(o)q(ds)-57 326 y(as)20 b(in)g(\014gure)h(10b.)36 b(Then)21 b(w)o(e)f(can)g (\push")h(the)f(v)o(ertical)-57 367 y(blo)q(c)o(ks)13 b(to)o(w)o(ards)g(the)h(left)f((as)g(in)g(\014gure)h(11))f(and)f (construct)-57 409 y(a)i(new)g(tiling)e(of)i Fl(P)19 b Fn(suc)o(h)14 b(that)g(the)h(tile)e(at)h Fl(x)g Fn(is)f(v)o(ertical.) -7 452 y(Otherwise,)h(there)g(m)o(ust)d(b)q(e)i(a)g(part)f Fl(S)k Fn(of)c(the)h(b)q(oundary)-57 494 y Fl(@)r(P)24 b Fn(reac)o(hable)19 b(from)e Fl(x)h Fn(b)o(y)g(a)g(sequence)i(of)e (blo)q(c)o(ks,)h(as)g(in)-57 535 y(\014gure)f(12.)29 b(W)m(e)18 b(observ)o(e)g(that)g(in)f(that)h(tiling,)f(there)i(is)e(a) -57 577 y(p)q(oin)o(t)c Fl(x)75 562 y Fe(0)101 577 y Fn(of)g Fl(S)k Fn(with)c(the)i(same)e(lab)q(el)g(as)h Fl(x)f Fn(in)h Fl(H)s Fn(.)-57 659 y Fd(De\014nitio)o(n)f(2)21 b Fo(Given)g(a)g(tiling)f(of)h Fl(P)6 b Fo(,)21 b(a)g(p)n(air)g Fn(()p Fl(x;)7 b(x)832 644 y Fe(0)842 659 y Fn())21 b Fo(of)-57 705 y(p)n(oints)c(of)g Fl(@)r(P)22 b Fo(is)17 b Fb(tight)f Fo(if)g Fl(x)h Fo(and)g Fl(x)537 690 y Fe(0)565 705 y Fo(ar)n(e)g(b)n(oth)g(highest)g(on)-57 746 y Fl(@)r(P)6 b Fo(,)14 b(with)h(the)g(same)g(lab)n(el,)f(and)i(ar)n(e)e(linke)n(d)h (by)g(a)g(se)n(quenc)n(e)-57 788 y(of)g(blo)n(cks)g(in)g(the)f(tiling.) -7 868 y Fn(T)m(o)e(ha)o(v)o(e)h(an)f(algorithm)f(similar)f(to)j(the)g (one)h(in)e(the)i(pre-)-57 910 y(vious)f(section,)i(w)o(e)f(w)o(ould)f (lik)o(e)g(to)g(b)q(e)i(able)f(to)f(sa)o(y)h(that)g(w)o(e)-57 951 y(can)h(alw)o(a)o(ys)e(place)i(a)f(v)o(ertical)g(tile)h(at)f Fl(x)p Fn(,)g(whic)o(h)g(w)o(ould)g(en-)-57 993 y(able)d(us)g(to)h(put) f(do)o(wn)g(the)h(\014rst)f(tile)g(using)g(lo)q(cal)f(conditions)-57 1034 y(only)m(.)24 b(Unfortunately)16 b(tigh)o(t)f(pairs)h(of)g(p)q (oin)o(ts)g(are)g(a)g(prob-)-57 1076 y(lem)11 b(in)h(this)g(approac)o (h.)18 b(Ho)o(w)o(ev)o(er)13 b(w)o(e)g(will)d(sho)o(w)j(that)f(suc)o(h) -57 1117 y(pairs)j(are)h(actually)f(indep)q(enden)o(t)i(of)e(the)h (tiling.)21 b(In)15 b(order)-57 1159 y(to)f(c)o(haracterize)h(them,)e (w)o(e)h(will)e(use)j(a)f(di\013eren)o(t)h(notion)e(of)-57 1205 y(heigh)o(t.)23 b(Let)16 b Fl(K)i Fn(=)d Fm(h)p Fl(a;)7 b(b)p Fm(j)p Fl(a)375 1190 y Fj(2)392 1205 y Fl(;)g(b)429 1190 y Fj(2)447 1205 y Fm(i)16 b Fn(b)q(e)g(ro)q(oted)g (at)g(some)f(arbi-)-57 1250 y(trary)g(far)f(a)o(w)o(a)o(y)g(p)q(oin)o (t)g Fl(r)347 1235 y Fe(0)358 1250 y Fn(.)21 b(As)15 b(w)o(e)f(go)h(around)f Fl(@)r(P)20 b Fn(from)13 b Fl(e)p Fn(,)-57 1292 y(p)q(oin)o(ts)j(can)h(also)f(b)q(e)h(lab)q(elled)f(with) g(a)g(w)o(ord)g Fl(w)h Fm(2)f Fl(K)s Fn(.)26 b(W)m(e)-57 1333 y(de\014ne)16 b(their)f Fl(K)s Fn(-heigh)o(t)f(to)h(b)q(e)g(the)g (distance)h(from)d Fl(w)i Fn(to)g Fl(r)902 1318 y Fe(0)-57 1375 y Fn(in)j(the)g(graph)g(\000()p Fl(K)s Fn())h((whic)o(h)f(in)g (this)g(case)h(is)f(just)h(a)f(line)-57 1416 y(graph).)-57 1499 y Fd(Lemma)d(4)21 b Fo(Given)12 b(a)f(tile)n(able)g(p)n(olygon)h Fl(P)6 b Fo(,)11 b(a)h(p)n(air)f(of)g(p)n(oints)-57 1540 y Fl(x;)c(x)10 1525 y Fe(0)32 1540 y Fm(2)k Fl(@)r(P)18 b Fo(is)12 b(tight)f(for)h(some)g(tiling)f(of)h Fl(P)18 b Fo(i\013)12 b(the)g(fol)r(lowing)-57 1582 y(3)j(c)n(onditions)g(ar)n (e)g(satis\014e)n(d:)-57 1623 y(1))g Fl(x)g Fo(and)g Fl(x)139 1608 y Fe(0)165 1623 y Fo(ar)n(e)g(highest)g(p)n(oints)g(of)g Fl(@)r(P)6 b Fo(.)-57 1665 y(2))12 b(It)g(is)g(p)n(ossible)g(to)h(c)n (onstruct)f(a)g(se)n(quenc)n(e)h(of)f Fn(3)t Fm(\002)t Fn(6)g Fo(r)n(e)n(ctan-)-57 1706 y(gles)i((r)n(esp)n(e)n(ctively,)e (se)n(quenc)n(e)j(of)f Fn(6)7 b Fm(\002)g Fn(3)14 b Fo(r)n(e)n (ctangles))f(linking)-57 1752 y Fl(x)i Fo(to)g Fl(x)56 1737 y Fe(0)68 1752 y Fo(,)g(wher)n(e)f(every)i(r)n(e)n(ctangle)e(is)h (to)h(the)f(right)g(of)g((r)n(esp.)-57 1793 y(b)n(elow))i(the)h(pr)n (evious)g(one)h(and)f(shifte)n(d)g(by)g(thr)n(e)n(e,)g(and)h(al)r(l)-57 1835 y(r)n(e)n(ctangles)14 b(ar)n(e)h(inside)g Fl(P)6 b Fo(.)-57 1876 y(3))147 1930 y Fm(j)p Fl(K)s Fo(-height)p Fn(()p Fl(x)p Fn())j Fm(\000)h Fl(K)s Fo(-height)p Fn(()p Fl(x)626 1913 y Fe(0)637 1930 y Fn())p Fm(j)i Fn(=)58 1982 y Fa(\032)110 2018 y Fn(2)p Fm(j)p Fl(z)e Fm(\000)g Fl(z)235 2003 y Fe(0)247 2018 y Fm(j)p Fl(=)p Fn(3)126 b Fo(in)15 b(c)n(ase)g(of)g(\014gur)n(e)g(12)110 2061 y Fn(2)p Fm(j)p Fl(z)10 b Fm(\000)g Fl(z)235 2046 y Fe(0)247 2061 y Fm(j)p Fl(=)p Fn(3)e Fm(\000)h Fn(1)56 b Fo(in)15 b(c)n(ase)g(of)g(\014gur)n(e)g(13)-57 2146 y(wher)n(e)28 b Fl(z)j Fo(and)e Fl(z)239 2131 y Fe(0)279 2146 y Fo(ar)n(e)f(the)h Fl(x)p Fo(-c)n(o)n(or)n(dinates)f((r)n(esp.) 59 b Fl(y)q Fo(-)-57 2194 y(c)n(o)n(or)n(dinates))15 b(of)g Fl(x)f Fo(and)i Fl(x)368 2179 y Fe(0)394 2194 y Fo(in)f Fl(Z)476 2179 y Fj(2)495 2194 y Fo(.)-57 2279 y Fd(Pro)q(of)d(idea:)k Fn(If)11 b(()p Fl(x;)c(x)309 2264 y Fe(0)320 2279 y Fn())k(is)g(tigh)o(t)g(for)g(some)f(tiling,)g (then)i(the)-57 2320 y(three)j(conditions)f(of)f(the)i(lemma)10 b(are)15 b(clearly)e(satis\014ed.)-7 2364 y(Con)o(v)o(ersely)m(,)25 b(giv)o(en)d(()p Fl(x;)7 b(x)423 2349 y Fe(0)434 2364 y Fn())23 b(whic)o(h)g(satisfy)g(the)h(three)-57 2405 y(conditions)13 b(of)g(the)h(lemma)o(,)c(and)j(giv)o(en)g(a)g(tiling)f (of)h Fl(P)6 b Fn(,)12 b(con-)-57 2447 y(sider)g(a)f(shortest)i(path)f (from)e Fl(x)h Fn(to)g Fl(x)525 2432 y Fe(0)548 2447 y Fn(in)g Fl(P)6 b Fn(.)17 b(One)12 b(can)g(pro)o(v)o(e)-57 2488 y(that)g(the)h(path)f(m)o(ust)e(only)i(tra)o(v)o(erse)h(horizon)o (tal)e(rectangles)-57 2530 y(of)k(the)i(tiling)d((otherwise)j(the)g Fl(K)s Fn(-heigh)o(t)e(w)o(ould)h(not)f(ha)o(v)o(e)-57 2575 y(time)h(to)h(c)o(hange)h(b)o(y)f(2)p Fm(j)p Fl(z)c Fm(\000)e Fl(z)426 2560 y Fe(0)438 2575 y Fm(j)p Fl(=)p Fn(3)g Fm(\000)g Fn(1).)28 b(It)18 b(is)f(p)q(ossible)g(to)-57 2617 y(sho)o(w)c(that)h(those)g(horizon)o(tal)f(rectangles)i(m)o(ust)d (b)q(e)i(part)g(of)-57 2658 y(blo)q(c)o(ks,)21 b(whic)o(h)e(form)f(a)h (sequence.)38 b(Therefore)21 b(()p Fl(x;)7 b(x)839 2643 y 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y(edges)f(of)e(in)o(tegral)g(lengths.)1013 960 y Fd(Initiali)o(zati)o(on:)20 b Fn(Lab)q(el)c(the)h(v)o(ertices)g(of)f Fl(@)r(P)22 b Fn(with)16 b(w)o(ords)1013 1002 y(of)d Fl(H)k Fn(and)d(of)f Fl(K)s Fn(;)g(calculate)h(their)h(heigh)o(t)e(and) h Fl(K)s Fn(-heigh)o(t.)1013 1043 y Fd(Rep)q(eat:)22 b Fn(T)m(ak)o(e)15 b Fl(x)g Fm(2)f Fl(@)r(P)22 b Fn(of)15 b(maxim)o(um)c(heigh)o(t.)24 b(If)16 b(there)1013 1085 y(exists)i Fl(y)h Fm(2)e Fl(@)r(P)23 b Fn(suc)o(h)18 b(that)g(()p Fl(x;)7 b(y)q Fn())17 b(is)g(tigh)o(t,)h(then)g(place)f (a)1013 1126 y(horizon)o(tal)f(tile)h(at)h Fl(x)f Fn(accordingly;)h (else)g(place)g(a)f(v)o(ertical)1013 1168 y(tile)c(at)h Fl(x)g Fn(appropriately)m(.)1013 1209 y(Up)q(date)d Fl(P)6 b Fn(,)11 b Fl(@)r(P)6 b Fn(,)11 b Fl(H)s Fn(-lab)q(els)f(and)h(heigh)o (ts,)g Fl(K)s Fn(-lab)q(els)g(and)g Fl(K)s Fn(-)1013 1251 y(heigh)o(ts.)1013 1292 y Fd(Un)o(til:)k Fl(P)j Fn(is)12 b(tiled)g(or)g(there)i(is)e(a)g(con\015ict)h(in)f(the)h(lab)q (els)f((in)1013 1334 y(whic)o(h)i(case)g Fl(P)20 b Fn(is)13 b(not)h(tileable).)1063 1375 y(The)j(exact)h(running)e(time)g(of)g (the)i(algorithm)c(dep)q(ends)1013 1417 y(on)f(the)i(implemen)o(tatio)o (n.)h(With)d(a)g(little)h(bit)f(of)h(w)o(ork,)f(one)1013 1458 y(can)g(see)i(that)e(there)i(is)e(an)g(implemen)o(tation)d(for)j (whic)o(h)g(the)1013 1500 y(algorithm)e(tak)o(es)j(quadratic)g(time.) 1063 1542 y(F)m(or)d(general)i Fl(k)7 b Fm(\002)f Fl(l)13 b Fn(rectangles)h((with)e(the)g(t)o(w)o(o)g(tile)g(t)o(yp)q(es)1013 1583 y Fl(k)f Fm(\002)f Fl(l)17 b Fn(and)e Fl(l)c Fm(\002)g Fl(k)q Fn(),)k(the)h(algorithm)d(can)i(b)q(e)h(generalized)g(to)1013 1625 y(yield)d(a)h(quadratic-time)e(algorithm.)1013 1709 y Fd(Theorem)j(2)20 b Fo(Given)f(a)g(p)n(olygon)h(of)f(ar)n(e)n(a)f Fl(n)p Fo(,)i(ther)n(e)e(is)g(an)1013 1750 y(algorithm)f(for)g(tiling)g (the)h(p)n(olygon)h(with)e Fl(k)12 b Fm(\002)g Fl(l)19 b Fo(and)g Fl(l)12 b Fm(\002)g Fl(k)1013 1792 y Fo(tiles)20 b(or)g(pr)n(oving)h(that)g(it)f(is)g(not)h(tile)n(able,)g(with)f (running)1013 1838 y(time)14 b Fl(O)q Fn(()p Fl(n)1181 1822 y Fj(2)1200 1838 y Fn())p Fo(.)1063 1921 y Fn(W)m(e)19 b(conjecture)i(that)f(the)g(running)g(time)e(can)i(b)q(e)g(im-)1013 1963 y(pro)o(v)o(ed.)1013 2040 y Fk(4)56 b(Prop)r(erties)17 b(of)i(rectangle)e(tilings)1063 2085 y Fn(As)d(a)g(corollary)f(of)g (the)h(ab)q(o)o(v)o(e)g(constructions)h(w)o(e)f(\014nd:)1013 2160 y Fd(Theorem)h(3)20 b Fo(A)o(ny)14 b(two)g(tilings)f(of)h Fl(P)19 b Fo(by)14 b Fl(m)7 b Fm(\002)h Fn(1)13 b Fo(and)i Fn(1)7 b Fm(\002)g Fl(n)1013 2202 y Fo(tiles)15 b(ar)n(e)i(obtaine)n(d) g(fr)n(om)e(one)j(another)e(by)h(\r)n(otations")g(of)1013 2243 y(the)12 b(form)g(il)r(lustr)n(ate)n(d)g(\014gur)n(e)g(14.)19 b(That)12 b(is,)h(m)f(vertic)n(al)g Fn(1)t Fm(\002)t Fl(n)1013 2285 y Fo(tiles)17 b(c)n(an)h(b)n(e)g(r)n(eplac)n(e)n(d)g(by) g Fl(n)g Fo(horizontal)g Fl(m)12 b Fm(\002)f Fn(1)18 b Fo(tiles)f(and)1013 2326 y(c)n(onversely.)1063 2368 y(A)o(ny)h(two)g(tilings)f(of)h Fl(P)24 b Fo(by)18 b Fn(2)11 b Fm(\002)h Fn(3)18 b Fo(and)h Fn(3)11 b Fm(\002)h Fn(2)18 b Fo(tiles)f(ar)n(e)1013 2410 y(obtainable)g(fr)n(om)f(one)h (another)g(by)f(r)n(otations)h(of)f(the)h(form)1013 2451 y(il)r(lustr)n(ate)n(d)c(\014gur)n(e)i(15.)1013 2526 y Fn(It)i(is)h(not)f(hard)g(to)h(sho)o(w)f(that)g(the)h(distance)g(in)f (terms)h(of)1013 2568 y(the)f(n)o(um)o(b)q(er)e(of)h(rotations)g(b)q (et)o(w)o(een)h(t)o(w)o(o)f(tilings)f(of)h Fl(P)6 b Fn(,)15 b(in)1013 2617 y(either)h(of)e(the)h(ab)q(o)o(v)o(e)g(cases,)h(is)f(at) f(most)g Fl(O)q Fn(()p Fl(n)1759 2602 y Fj(3)p Fc(=)p Fj(2)1811 2617 y Fn())h(for)g(area)1013 2658 y Fl(n)p Fn(.)j(F)m(urthermore,)c(\014gure)h(16)e(sho)o(ws)i(t)o(w)o(o)e (tilings)g(of)h(a)g(2)p Fl(k)c Fm(\002)1013 2700 y Fn(3)p Fl(k)20 b Fn(square)h(b)o(y)e(2)13 b Fm(\002)g Fn(1)20 b(and)f(1)13 b Fm(\002)h Fn(3)19 b(tiles)h(ha)o(ving)e(distance)p eop %%Page: 9 9 9 8 bop 413 946 a Fn(Figure)14 b(12:)j(Reac)o(hing)c(the)i(b)q(oundary) f(with)f(a)h(sequence)i(of)d(blo)q(c)o(ks)714 1832 y(Figure)h(13:)k (Chain)13 b(of)g(blo)q(c)o(ks)439 2600 y Fo(Figur)n(e)i(14:)k(r)n (otations)14 b(for)h(tr)n(ansforming)f Fl(m)c Fm(\002)f Fn(1)p Fo(,)14 b Fn(1)9 b Fm(\002)h Fl(n)15 b Fo(tilings)p eop %%Page: 10 10 10 9 bop 509 618 a Fo(Figur)n(e)14 b(15:)20 b(r)n(otations)14 b(for)h(tr)n(ansforming)f Fn(3)9 b Fm(\002)g Fn(2)15 b Fo(tilings)566 1337 y Fn(Figure)f(16:)j(Tw)o(o)d(tiling)e(whic)o(h)i (are)g(\far)f(apart")-57 1469 y(\002()p Fl(k)14 1454 y Fj(3)33 1469 y Fn().)18 b(This)13 b(distance)g(can)h(b)q(e)f (calculated)g(b)o(y)g(in)o(tegrating)-57 1511 y(the)g(di\013erence)g (in)f(heigh)o(t)f(function)h(o)o(v)o(er)g(the)h(tilings.)j(Eac)o(h)-57 1552 y(rotation)f(decreases)j(the)e(in)o(tegral)f(b)o(y)g(a)g(constan)o (t)h(amoun)o(t)-57 1594 y(only)m(.)-57 1670 y Fk(References)-57 1715 y Fn()k(William)12 b(P)m(.)j(Th)o(urston,)i(\Con)o(w)o(a)o (y's)e(Tiling)g(Groups",)8 1756 y Fo(A)o(m.)f(Math.)h(Monthly)p Fn(,)f(Oct.)g(1990,)e(pp.)i(757-773.)-57 1831 y()20 b(J.M.Robson,)i(\Sur)h(le)f(pa)o(v)n(age)f(de)i(\014gures)g(du)g(plan) 8 1873 y(par)c(des)h(barres",)h Fo(A)n(ctes)e(des)i(Journ)o(\023)-20 b(ees)20 b(Polyominos)8 1914 y(et)14 b(Pavages)p Fn(,)h(June)f(1991,)f (pp.)g(95-103.)1013 1469 y()19 b(D.Beauquier,)14 b(M.Niv)n(at,)d (\P)o(a)o(v)n(ages)i(de)h(pi)o(\022)-20 b(eces)15 b(par)f Fl(h)1952 1475 y Fc(m)1077 1511 y Fn(et)g Fl(v)1145 1517 y Fc(n)1167 1511 y Fn(",)p Fo(A)n(ctes)f(des)h(Journ)o(\023)-20 b(ees)14 b(Polyominos)g(et)g(Pavages)p Fn(,)1077 1552 y(June)h(1991,)e(pp.)g(45-60.)1013 1627 y()19 b(Eric)11 b(Remila,)d(\T)m(rois)i(algorithmes)f(de)i(pa)o(v)n(ages)f(par)h Fl(h)1952 1633 y Fc(m)1077 1668 y Fn(et)h Fl(v)1143 1674 y Fc(n)1165 1668 y Fn(",)f Fo(A)n(ctes)g(des)h(Journ)o(\023)-20 b(ees)13 b(Polyominos)f(et)g(Pavages)p Fn(,)1077 1710 y(June)j(1991,)e(pp.)g(83-94.)1013 1785 y()19 b(F)m(ran)c(Berman,)f (Da)o(vid)f(Johnson,)i(T)m(om)e(Leigh)o(ton,)h(P)o(e-)1077 1826 y(ter)20 b(W.)e(Shor,)h(Larry)g(Sn)o(yder,)h(\Generalized)f (Planar)1077 1868 y(Matc)o(hing",)c Fo(Journal)h(of)g(A)o(lgorithms)f Fn(11,)f(no)h(2,)g(1990,)1077 1909 y(pp.)f(153-184.)p eop %%Trailer end userdict /end-hook known{end-hook}if %%EOF |
14397 | https://discuss.elastic.co/t/count-based-on-condition-not-just-simple-aggregate/113792 | Count based on condition (not just simple aggregate)
Hello, I need to produce a line graph which counts the number of patients who are active in care in a given month. A patient is deemed active within a given month if they have a "return to clinic" date < 90 days from the end of the month. For example, say a patient's last encounter datetime was June, 1, 2017 and return to clinic date is July 31, 2017, This patient is considered to be active in June, July, August, Sept, October (even if the patient has not yet returned to clinic). Each document has the encounter_datetime, rtc_date and next_encounter_datetime. In sql, we could have a condition like:
count(if(timestampdiff(day, rtc_date, reporting_month_date) < 90 and (next_encounter_datetime is null or next_encounter_datetime > reporting_month_date),1,0))
I should add that we create a dataset (via sql) that for each month in the analysis, we join the most recent patient encounter relative to that month.
Is there a way to create a similar count in Elastic and visualize this in kibana?
Thanks for your help.
Elasticsearch works a little differently.
In this case, you would probably use datemath and range queries to get the matching documents, and then just look at the number of hits to get the count.
I don't quite follow how all your date fields work, but the short of it is you'll craft a query and apply it as a filter in Kibana. Click on "Add a filter" at the top, and then "Edit Query DSL", and in here you can type a raw Elasticsearch query.
You can get the and operation by applying two filters.
and
I'm unfortunately not super well versed in Elasticsearch queries though, so I'm not sure how you would compare the dates across two fields. I think you could fall back on scripted fields, do the comparison there and then filter where the value is < 90. You'll want to use the expression syntax to get date operators.
You'll also want to use a scripted field to compare next_encounter_datetime and reporting_month_date, and maybe use that comparison to return a boolean.
next_encounter_datetime
reporting_month_date
So you'll have two filters, one that uses a scripted field to get the difference of reporting_month_date and rtc_date and filters on the value being less than 90, and another that uses a scripted field to compare next_encounter_datetime and reporting_month_date and filters where result is true.
reporting_month_date
rtc_date
next_encounter_datetime
reporting_month_date
That'll give you all the matching documents, which you can create as a saved search. Then you can use that saved search to make a metric visualization that simply shows the count.
Thanks @Joe_Fleming. After re-evaluating this, I think the easiest solution is probably to change the datasets we are using to report on rather than come with more complexing querying of the existing dataset. Thanks again for your help!
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14398 | https://phdeconomics.unisi.it/wp-content/uploads/sites/127/2023/11/ThesisGuilhermeSpinatoMorlin.pdf | ESSAYS ON OPEN ECONOMY MACROECONOMICS GUILHERME SPINATO MORLIN SUPERVISOR: RICCARDO PARIBONI COORDINATOR: SIMONE D’ALESSANDRO PHD PROGRAMME IN ECONOMICS OF THE TUSCAN UNIVERSITIES ESSAYS ON OPEN ECONOMY MACROECONOMICS GUILHERME SPINATO MORLIN A THESIS PRESENTED FOR THE DEGREE OF DOCTOR OF PHILOSOPHY IN ECONOMICS ACADEMIC YEAR 2021/2022 XXXIV CYCLE PHD PROGRAMME IN ECONOMICS OF THE TUSCAN UNIVERSITIES SUPERVISOR: RICCARDO PARIBONI CO-SUPERVISOR: SIMONE D’ALESSANDRO PROGRAM COORDINATOR: SIMONE D’ALESSANDRO UNIVERSITIES OF FLORENCE, PISA, AND SIENA Contents Introduction 1 1 Inflation and Distributive Conflict 6 1.1 Conflict inflation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 1.2 Endogenous money . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10 1.3 Demand-pull inflation in New Keynesian economics . . . . . . . . . . . . .
13 1.4 Kaleckian conflicting claims models . . . . . . . . . . . . . . . . . . . . . .
17 1.5 The Classical-Keynesian approach to inflation . . . . . . . . . . . . . . . .
22 1.5.1 Critique on the Kaleckian distributive closure . . . . . . . . . . . .
22 1.5.2 Inflation in the Classical-Keynesian approach . . . . . . . . . . . .
23 1.6 Final Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
28 2 Inflation and conflicting claims in the open economy 30 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
30 2.2 Inflation in the open economy . . . . . . . . . . . . . . . . . . . . . . . . . .
33 2.3 A formal model of inflation in an open economy . . . . . . . . . . . . . . .
40 2.3.1 Distributive closure in the open economy . . . . . . . . . . . . . . .
40 2.3.2 Inflation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
43 2.3.3 Exchange rate devaluation . . . . . . . . . . . . . . . . . . . . . . .
45 2.3.4 Price of non-tradable commodity . . . . . . . . . . . . . . . . . . . .
48 2.3.5 Real exchange rate . . . . . . . . . . . . . . . . . . . . . . . . . . . .
51 2.4 Quantities, Unemployment and Conflict . . . . . . . . . . . . . . . . . . . .
55 2.4.1 Output and employment . . . . . . . . . . . . . . . . . . . . . . . .
56 2.4.2 Inflation, wages and exchange rate . . . . . . . . . . . . . . . . . . .
62 2.4.3 Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
64 2.5 Conflict inflation and distribution in the open economy . . . . . . . . . . .
69 2.6 Final Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
74 i CONTENTS Appendices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
75 A Mathematical Appendix . . . . . . . . . . . . . . . . . . . . . . . . .
75 B Real exchange rate . . . . . . . . . . . . . . . . . . . . . . . . . . . .
78 3 International inflation and trade linkages in Brazil under inflation targeting 80 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
80 3.2 Inflation and International Shocks . . . . . . . . . . . . . . . . . . . . . . .
83 3.2.1 The globalization of inflation . . . . . . . . . . . . . . . . . . . . . .
83 3.2.2 Inflation targeting and international shocks in Brazil . . . . . . . .
85 3.3 Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
87 3.3.1 Stationarity Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . .
89 3.4 Methodology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
91 3.4.1 Foreign PPI . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
91 3.4.2 VARS and SVARS . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
94 3.5 Foreign PPI Shocks and Inflation in Brazil . . . . . . . . . . . . . . . . . . .
96 3.6 Robustness Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
100 3.6.1 Industrial Production . . . . . . . . . . . . . . . . . . . . . . . . . .
100 3.6.2 COVID Shocks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
101 3.6.3 Monetary Policy . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
103 3.6.4 Local Projections . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
104 3.7 Final Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
106 Appendices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
107 A Unit Root tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
107 B Cointegration Analysis . . . . . . . . . . . . . . . . . . . . . . . . . .
107 C Baseline Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
108 D Impulse Response Functions of the Robustness Tests . . . . . . . .
111 4 Growth and debt stability in a supermultiplier model with public expenditures and foreign trade 117 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
117 4.2 The Sraffian supermultiplier . . . . . . . . . . . . . . . . . . . . . . . . . . .
119 4.2.1 Sraffian superrmultiplier model . . . . . . . . . . . . . . . . . . . .
121 4.3 The Sraffian supermultiplier in an open economy with government . . . .
127 4.3.1 Fully adjusted position . . . . . . . . . . . . . . . . . . . . . . . . . .
131 4.3.2 Stability analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
131 ii CONTENTS 4.4 Public Expenditure, Exports and Debt stability . . . . . . . . . . . . . . . .
132 4.4.1 Public debt . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
137 4.4.2 Foreign debt . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
138 4.4.3 Fully adjusted position and debt stability . . . . . . . . . . . . . . .
141 4.4.4 Sensitivity analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . .
143 4.5 The external constraint and Thirlwall’s Law . . . . . . . . . . . . . . . . . .
144 4.5.1 Thirlwall’s Law in the supermultiplier . . . . . . . . . . . . . . . . .
145 4.6 Fiscal Policy and the External Debt . . . . . . . . . . . . . . . . . . . . . . .
149 4.7 Simulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
153 4.8 Final Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
160 Appendices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
161 A Equilibrium growth and the share of each autonomous expenditure 161 B Stability of the supermultiplier in an open economy with public expenditure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
163 C Sensitivity Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . .
166 D Discrete time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
171 E Simulation Parameters and Equilibrium values . . . . . . . . . . .
174 Bibliography 191 iii Introduction Capitalist economies are open economies. Historically, capitalist economies were born as open economies. Trade and finance played fundamental role in the emergence and worldwide generalization of this economic system.
From a different angle, modern capitalism is inseparable from the constitution of national states in a competitive international order, shaped by war, colonization and imperialism.
Yet, capitalist economies are often theorized as closed economies. Open economy issues are often extensions or particular cases of closed economy analyses. Economics textbooks would argue that, after all, the global economy is a closed economy, and what works for the global economy as a whole should work for its composing parts. It may not be the case that such an assumption improves our understanding of reality. On the contrary, capitalist economies may be analogous to the Möbius strip, whose non-orientable surface blurs the distinction between exterior and interior. If this is the case, macroeconomics should not study closed and open economy issues separately.
Rather, macroeconomics should be studied primarily from an open economy perspective.
I take this premise seriously in this thesis. I discuss inflation, growth, and distribution with special attention to exchange rate, trade, and external finance. These topics are approached with theoretical and empirical methods. Theoretical inquiry pays attention to the history of ideas to discuss modern economic models. The theoretical research in this thesis also develops economic models and simulations. Empirical methods rely on state of-art time series econometrics techniques.
In the course of this thesis, I adress open economy macroeconomic theory and policy from a Classical-Keynesian perspective. The theoretical and empirical ideas presented in the next pages are therefore in line with two fundamental notions: the Classical theory of distribution and the Principle of Effective Demand.
The classical theory of value and distribution determines relative prices and income 1 INTRODUCTION distribution in the same system of long period price equations.
Classical theory emphasizes institutional and political factors in explaining distribution.
Alternative distributive closures explore such causes in ways that are coherent with the classical analytical method.
There is no rigid relation between the theory of output and the theory of value in this approach. Rather, the Classical theory analytically distinguishes the theory of value and distribution from the theory of output and accumulation.
Therefore, it is possible speak of a Classical-Keynesian approach, in which classical approach to distribution can be integrated with a Keynesian view on output and accumulation. A Classical-Keynesian perspective considers that demand determines the output level according to the Principle of Effective Demand, as proposed by Keynes and Kalecki. In the long period, in which productive capacity changes, demand also drives output growth and capital accumulation. As productive capacity adjusts to effective demand, an independent level of investment determines savings also in the long run.
While open economy concerns matter for macroeconomics as a whole, this thesis focus on economic concerns of peripheral economies. The division between core and peripheral economies was a fundamental premise in the Latin American Structuralist School.
Peripheral economies were included in the international division of labor through its relation with core economies, thereby becoming a constitutive element of the global economy. The peripheral condition meant that the external sector provided the main source of dynamism for these economies, but also imposed crucial constraints to growth and development.
Structuralist scholars showed that domestic prices and distribution in peripheral economies were directly affected by events in the international markets. They realized that movements in international prices affected the way peripheral countries disputed their share of surplus with other countries. In a closed economy, the Classical theory of value can explain both relative prices and distribution among social classes. In an open economy setup, relative prices are also associated with the distribution of the surplus among nations.
The declining trend of the terms of trade of peripheral economies identified in the post-war thus had fundamental implications for peripheral economies.
As a relative price itself, exchange rates also affects the distribution of the surplus across nations. As we will discuss, this effects have crucial implications for distributive conflict and inflation in peripheral economies.
Owing to their vulnerability to international shocks, peripheral economies were particularly affected by the COVID-19 pandemic crisis.
The outbreak of COVID-19 2 INTRODUCTION engendered the deepest recession of the last decades. Social distancing measures shut down economic activities to slow the spread of the disease. The supply shock in these activities resulted in a sharp fall in output and employment. The rise in unemployment, and the fear related to the consumption of goods and services involving higher risk of contagion reduced consumers’ spending.
This combination of supply and demand shocks caused a crisis of global dimensions.
After two years, economic recovery still faces important challenges in advanced and developing economies. Such challenges are directly related to the impact of international variables on inflation and distribution, trade and growth, debt and financial stability. I address these topics in this thesis, hoping that it provides a small contribution to the understanding of the current situation. The thesis is structured in four chapters.
The first essay reviews conflict inflation models, contrasting alternative theoretical perspectives underlying conflicting claims models.
Conflicting claims models have stressed the race between prices and money wages, in the struggle among capitalists and workers, as the main inflationary pressure. The essay shows how these models describe conflict inflation and the related outcome for income distribution. The chapter also explores criticism to the New-Keynesian Phillips curve and the relation between inflation and endogenous money theory.
A deeper understanding of distributive conflict requires an analytical exposition of the relation between prices and distribution.
In general, conflicting claims models rely on Kaleckian explanation of distribution, based on the notion of mark-up pricing according to the degree of monopoly. Conflict inflation allows wage bargaining to affect income distribution and, thus, the real mark-up level.
However, this theory contains unsolved theoretical shortcomings, lacking an ultimate explanation for profits and overlooking input-output relations. An alternative theory of distribution can be found in modern appraisals of the Classical surplus approach, which has been extended to the study of inflation, providing a consistent relation between inflation and conflict.
The second essay develops a conflicting claims models for a price-taker open economy. We start from the view that the evolution of prices and income distribution in open economies cannot be studied independently from international prices and exchange rates, especially in small open economies. Exchange rates and international prices are thus fundamental to explaining inflation in open economies. Conflict inflation models usually account for these variables by including imported inputs and, in some cases, a distributive impact of exchange rates. A different viewpoint emerges from the 3 INTRODUCTION Classical-Keynesian distribution theory for a price-taker open economy.
Thus, we explore this alternative by developing a conflict inflation model along the lines of the Classical-Keynesian approach. The chapter contributes to the literature by combining the conflicting claims approach with the Classical-Keynesian open economy framework.
Including tradable prices, the model considers their direct impact on distribution.
Therefore, it addresses a cause of inflation overlooked in the literature. Finally, conflict inflation affects the real exchange rate, which becomes an important distributive variable.
The third essay assesses the connection between global and domestic inflation in Brazil during the period from 1999 to 2020. Input-output linkages have been shown to be an important transmission channel of inflation synchronization of inflation for advanced and emerging economies. International cost shocks have been less studied in the case of Brazil.
We therefore estimate a Structural VAR model with an index for producer prices (PPI) of Brazilian trade partners, in addition to the other relevant determinants of inflation.
The Foreign PPI index is weighted by the yearly average share of each country in Brazilian imports of intermediate and capital goods. Estimates shows a positive effect of Foreign PPI index on Brazilian Consumer Price Index, constituting a relevant explanation for domestic inflation in Brazil during the period 1999-2020.
Impulse Response functions show that the Exchange Rate is the main determinant of domestic CPI in Brazil.
International prices and exchange rate have fundamental implications for the operation of the inflation targeting regime. The results are in line with the literature’s empirical findings showing the overall relevance of international variables in the explanation of inflation. Further research may discuss the transmission channels of cross border inflation as well as evaluate the implications of these results to inflation theory.
The fourth essay extends the baseline Sraffian supermultiplier model for an open economy with government, introducing two autonomous expenditures.
The two sources of autonomous demand correspond to public expenditures and exports. We also analyze the stability conditions for public debt and foreign debt ratios.
Public debt stability requires that the interest rate on public debt is smaller than the output growth rate, as in Domar (1944). Foreign debt is evaluated in proportion to exports, accounting for the availability of foreign currency required to service external liabilities. The foreign debt-to-exports ratio converges to a stable value when the international interest rate is smaller than the growth rate of exports. However, this value may not be compatible 4 INTRODUCTION with the availability of international capital flows. We examine the consequences of a constraint to foreign debt ratio, in line with Bhering et al. (2019), reiterating the importance of a long-term external constraint to economic growth (Thirlwall, 1979). A fiscal policy rule is proposed to keep the foreign debt ratio below an upper limit for this ratio. We simulate five experiments showing the conditions for stability of debt ratios, the execution of the fiscal policy rule, and the alternative of a structural change policy.
Altogether, the chapter provides stability conditions for growth in an open economy paying its international liabilities in foreign currency. Simulations show the fiscal policy successfully reduces the equilibrium foreign debt-to-exports ratio by decreasing the share of public expenditures in autonomous demand.
Experiments also show that industrial policies that cause structural change and increase exports’ growth keep the foreign debt ratio below the threshold with a better growth performance than the fiscal policy rule.
5 Chapter 1 Inflation and Distributive Conflict 1.1 Conflict inflation The regular coexistence of price level increases with labor unemployment and spare capacity is a stylized fact supporting cost-push interpretations of inflation. This view contrasts with the dominant theory of inflation, according to which inflation accelerates (decelerate) when aggregate demand exceeds (falls behind) potential output. The fact that aggregate demand is often below potential output suggests that inflation causes are found in cost shocks.
To analyze the effect of costs on inflation, we must remember that costs consist basically of wages, profit rate, rent, and inputs costs. Inputs are priced according to their own production costs, which, in turn, also depend on the compensations of labor, capital, and land, and input prices. Repeating this process, we conclude that production costs can be decomposed into the basic distributive variables that reward the social groups involved in producing the economy’s output.
The income generated in production is primarily distributed as wages, rewarding the provision of labor, profits for the anticipation of capital, and rent, compensating the hire of scarce and privately-owned resources. After that, these compensations are partially redistributed in the form of taxes, interest, and transfers, whose value depends on tax rates, interest rate, exchange rate, and international prices.
In a monetary economy, distributive variables are set in nominal terms. Altogether, they determine relative prices, the price level, and distributive outcomes (Bastos, 2010).
At this point, the conflict between social groups comes into play in the determination of inflation.
Given productive techniques, a rise in prices can be traced back to the 6 CHAPTER 1. INFLATION AND DISTRIBUTIVE CONFLICT increase in at least one of the distributive variables.1 Indeed, inflation arises when social groups successfully ask for nominal increases, making the price level no longer compatible with the set of distributive variables. By means of nominal increases, classes seek an increase in real terms (which implies a larger income share). When workers obtain wage increases through bargaining, capitalists try to protect their profits, passing higher unit labor costs on to prices.
A higher cost of living leads to further wage demands in a process that may be repeated many times. Along these lines, we can argue that inflation is a consequence of distributive conflict (as Okishio 1977, Rowthorn 1977, Stirati 2001, Lavoie, 2014, p. 541-573).2 In this case, the real outcome for distributive variables is only known at the end of each period, as changes in their nominal value affect the price level and relative prices (Pivetti, 1991; Serrano, 1993).
Structuralist authors underlined that production bottlenecks could cause inflation (Noyola Vázquez, 1956; Pérez Caldentey, 2019; Sunkel, 1958). We can frame a similar argument as follows. In a growing economy, production in certain sectors can approach full capacity within the cost-minimizing technique before other sectors.
Additional production may be carried out with a less productive technique to attend demand in these sectors.
For example, production may employ a technique requiring a higher input of labor per unit of output. When this happens, relative prices will change. The point here is that such adjustments of relative prices can lead to inflation, even when nominal distributive variables are not the original cause of price rise. However, after the price increase, social groups tend to react by asking for increases in their compensations to preserve their purchasing power.
Thus, the inflationary propagation of original shocks through income claims is another manifestation of conflict inflation.3 Conflicting claims models have stressed the race between prices and money wages, in the struggle among capitalists and workers as the main inflationary pressure. “Therefore, roughly speaking, inflation comes about when the capitalist classes are not strong enough to depress the real wage rate without proceeding to raise prices and the labouring classes are not strong enough to win a higher real wage rate at the sacrifice of profits, preventing 1 In the case of an open economy, the rise in prices can be caused by the rise in the international price of inputs or tradable goods. However, an increase in these prices is caused by an increase in foreign distributive variables in the first place. In the next section, we discuss the open economy case in more detail.
2 See also Dutt (1987) Dalziel (1990), Setterfield (2007), Serrano (2010).
3 Historically, a common argument against cost-push explanations of inflation is that relative price adjustments do not necessarily cause inflation. However, these adjustments tend to be inflationary because otherwise, they would require that while some prices and incomes nominally increase, others decrease.
Social groups reject nominal decreases to avoid real losses.
Money wages are rigid downwards, and capitalists will not give up on profits if not compelled by competition.
7 CHAPTER 1. INFLATION AND DISTRIBUTIVE CONFLICT the capitalist classes from raising prices” (Okishio, 1977).
Although the relation between inflation and distributive conflict can be found in previous economic literature,4 Rowthorn’s (1977) contribution is acknowledged as the foundation of conflict claims modeling. Inspired in the original Phillips curve and in the Marxian notion of reserve army of labor, Rowthorn explicitly introduced the unemployment rate among the determinants of workers’ wage claims. The stylized fact that lower unemployment correlates with a higher rate of change in money wages (Phillips, 1958) is consistent with the fact that when unemployment is low, workers experience higher bargaining power in wage negotiations.5 Meanwhile, when unemployment is high, workers fear losing their jobs and face lower bargaining power.
Hence, the rate of change of money wages may depend more on the bargaining conditions than on workers’ aspirations. Indeed, “[w]orkers may feel that the real wage is much too low compared to what they consider to be the just rate, but they may have few means to implement their beliefs” (Lavoie, 2014, p. 550). Apart from labor market conditions, institutional and political factors affect workers’ bargaining position, such as labor laws, strength of unions and employer’s associations, social rights,6 and political representation of class interests.7 While on the one side, fear of unemployment discourages wage claims, on the other, competition imposes boundaries to price increases. If capitalists were able to immediately and completely pass through changes in their costs into prices, workers’ effort to rise real wages would be vain. In this case, an increase in money wages would immediately imply a proportional rise in prices. Real wages would remain unchanged in this process (Tarling and Wilkinson, 1985).
Before a deeper discussion on conflicting claims models, let us briefly review two relevant matters for inflation theory. In the next section, we discuss endogenous money theory, a theoretical development in macroeconomics which is fundamental to the understanding of inflation.
Endogenous money theory rejects the traditional 4 As in Robinson (1938), Aujac (1954), Okishio (1977), Furtado (1954, 1963) and Noyola Vázquez (1956). See also Morlin (2021) for a review on the debate in the US in the post-war period.
5 Indeed, despite later neoclassical reinterpretations of the Phillips curve, the original work of Phillips proposed an explanation of the rate of change of wages based on an interplay between market and institutional variables, compatible with the role played by bargaining power in determining wages (Forder, 2014).
6 According to Esping-Andersen (1990, p.11), “the balance of class power is fundamentally altered when workers enjoy social rights, for the social wage lessens the worker’s dependence on the market and employers, and thus turns into a potential power resource”.
7 Korpi (2002) and Kristal (2010) provide evidence on the impact of workers’ political representations on the wage share.
8 CHAPTER 1. INFLATION AND DISTRIBUTIVE CONFLICT explanation of inflation according to exogenous changes in the quantity of money.
However, it is still compatible with a demand-pull theory of inflation, as in the New-Keynesian Phillips curve.
We therefore dedicate another section to discuss the New-Keynesian approach to inflation theory, which considers excess demand as the most relevant determinant of inflation rather than supply shocks or the distributive struggle. The section also revisits the main critiques against this view, contrasting it with the perspective of conflict inflation.
9 CHAPTER 1. INFLATION AND DISTRIBUTIVE CONFLICT 1.2 Endogenous money Conflict inflation theories contend that prices change autonomously from movements in money supply, relying on some passive adjustment of the money supply to the price rise. Endogenous money theory, as defended by modern post-Keynesian approaches, points that the quantity of money supply is determined by the demand for money, which, in turn, depends on output and the price level (Lavoie, 2014; Wray et al., 1998).
Accordingly, the growth of money aggregates is determined by inflation and output growth.
This framework builds on early contributions of Kaldor (1985) and Moore (1988a,b). Opposing the monetarist view of the exogenous supply of money, Kaldor emphasizes a broad definition of the monetary base in terms of the high liquidity of many financial assets, discussing the money creation by banks through credit.
The author argues that the money supply is determined by the demand for money, given the level of income and the interest rate set by the monetary authority (Kaldor, 1985). This criticism also stressed that central banks are unable to control the money supply. What monetary policy actually determines, according to this view, is the basic interest rate of the economy. Indeed, Moore (1988a, p. 381) states, “[m]onetary endogeneity implies that central banks do not exogenously determine the quantity of credit money in existence, but rather the price at which it is supplied, that is, the short-term interest rate.
The money supply is endogenously determined by market forces.” The notion that money supply is endogenous is widespread in modern macroeconomic theory, currently accepted by post-Keynesian and New-Keynesian approaches. Indeed, this view is also compatible with the operation of Central Banks.
According to McLeay et al. (2014, p. 21), “[t]he supply of both reserves and currency (which together make up base money) is determined by banks’ demand for reserves both for the settlement of payments and to meet demand for currency from their customers — demand that the central bank typically accommodates”.
In turn, commercial banks’ regular operations create and destroy money. In particular, banks create new money supply, in the form of deposits, when lending to borrowers. The decision to make a new loan usually depends on the existence of profitable lending opportunities. The availability of reserves does not constrain this process. However, the requirement of profitability of banks facing a competitive market, the regulatory policy constraint to preserve financial stability, and banks’ restriction to lend due to mitigation of risks may constraint the supply of new loans (McLeay et al., 2014). In other words, the demand for credit drives bank loans. Banks are willing to lend money whenever 10 CHAPTER 1. INFLATION AND DISTRIBUTIVE CONFLICT borrowers are believed to be able to honor their commitments.
The direction of causality between money supply and the price level has long been the object of controversy in economic theory. Arguments supporting the endogeneity of money supply appeared together with cost-push inflation theory in James Steuart and Thomas Tooke (Smith, 2011; Yang, 1999). We can discuss this point in the light of the equation of exchange (see 1.1), according to Green (1992, p. 14) “ a common point of reference for all approaches to the problem of inflation since the relationships it expresses simply constitute a truism and do not in themselves imply causality”.
MSV = PQ [1.1] Suppose the evolution of costs determines both the price level (P) and inflation. In that case, for a given level of real output (Q), movements in prices must be compensated by changes in money supply (the left-hand side of equation 1.1). Note that, here, money supply consists of the product of the nominal amount of money in circulation and the velocity of circulation (i.e., the frequency in which a unit of currency is traded). If the output level (Q) is determined elsewhere and does not directly relate to prices, positive inflation implies growth in MSV. In this regard, we emphasize that classical political economy did not establish a direct relation between output and prices, rather keeping an analytical distinction of the theory of value — determining prices and distribution — on the one hand, and the theory of output and accumulation on the other (Garegnani, 1984). Thus, a separate determination of output and prices support this argument. In terms of post-Keynesian approaches, output level and growth are determined by effective demand level and growth, while costs determine the price level and inflation.
Contrasting with the previous exposition is the Quantitative Theory of Money, which was formulated in different terms over the history of economics. In a simple appraisal, quantitative theory contents that changes in the money supply determine the inflation rate. Thus, this means that in the equation 1.2, the left-hand side is exogenous and determines the right-hand side. Its original formulation reflected an extension of the analysis of the determination of commodity prices through the interaction between supply and demand, as in mercantilist analysis. Thus, money itself — in the form of gold or silver — was also a commodity. Hence, an inflow of metal in a specific country depreciated the metal’s value, increasing the general price of all other commodities with respect to metal (Green, 1992, p. 27-28). Proponents of the quantity theory assumed that 11 CHAPTER 1. INFLATION AND DISTRIBUTIVE CONFLICT both output and velocity of circulation are stable. For this reason, an exogenous increase in m could only cause inflation.8 The stability of V may be due to the fact that institutional factors shaping the time periodicity of transactions tend to change slowly (Snowdon and Vane, 2005, p. 50-53). As a result, nominal variables are affected by changes in money supply, while real variables, as income, remain stable. In this case, money is neutral, and inflation consists strictly of a monetary phenomenon lacking any influence from real causes.
ˆ ms + ˆ v = π + g [1.2] Finally, as pointed previously, the New-Keynesian approach accepts the notion of endogenous money. In this view, inflation arises due to a short-term real disequilibrium caused by an excess of demand with respect to natural output (Pérez Caldentey, 2019, p. 128). In turn, excess demand results from a divergence between the interest rate and the natural interest rate — i.e., the one that brings aggregate supply and demand into equilibrium. Thus, this explanation of inflation contrasts with the distributive conflict perspective.
The last paragraphs show that the widely acknowledged endogeneity of money dismisses explanations of inflation based on changes in the money supply, as in the Quantitative Theory of Money. However, adopting an endogenous money perspective does not necessarily mean endorsing a conflict inflation view. Indeed, a demand-pull inflation theory is consistent with endogenous money supply in the New Keynesian approach. This theory is further explored in the following paragraphs.
8 Still, Green (1992) caveats that some authors allowed for a short-run nonprice effect of a metal inflow.
12 CHAPTER 1. INFLATION AND DISTRIBUTIVE CONFLICT 1.3 Demand-pull inflation in New Keynesian economics The New-Keynesian Phillips Curve describes how past inflation, expected future inflation and aggregate demand drive the current inflation rate. According to Woodford (2003), modern mainstream approach to inflation recalls a “Wicksellian flavor”. Wicksell (1898b) states that inflation arises from the discrepancy between the money interest rate (prevailing in the loans market) and the natural interest rate, which is defined as the rate at which the supply of savings is equal to the demand for loans. Wicksell stresses that banks accommodate additional demand for loans without raising the interest rate in a pure credit economy. The author, thus, introduces a notion of endogenous money. When the money interest rate remains below the natural rate, “entrepreneurs will in the first instance obtain a surplus profit (at the cost of the capitalists) over and above their real entrepreneur profit or wage” (Wicksell, 1898b, p.
127).
Hence, the return on real investment net of interest remains higher than when the interest rate equals the natural rate. In turn, this pushes the expansion of production through investment, raising the demand for raw materials, and capital goods, causing the rise of prices of commodities.
An inflationary process emerges, persisting until the banking system corrects the discrepancy between money and natural rates. Symmetric reasoning describes the case in which the money rate is above the natural rate, causing a cumulative deflation (Wicksell, 1898a,b).9 Likewise, the New Keynesian view posits the existence of a natural interest rate under which the rate of inflation is stable. Nevertheless, we must point out a difference between the two perspectives. While the natural interest rate ensured the stability of the price level in Wicksell, it ensures the stability of the inflation rate in the New Keynesian view.
Inflation expectations explain the stability of inflation around a trend in New Keynesian Phillips Curve. This conclusion is obtained when the coefficient of expected inflation in the Phillips curve is equal to one. In some models, this result requires a complete effect of the past inflation on current inflation through the change in inflation expectations. In this case, increases in inflation rate in the past are fully included in agents’ current expectations. In a hybrid New Keynesian Phillips Curve — as proposed in Galı and Gertler (1999) — both past inflation and forward-looking expectations affect current inflation. Then, the combined effect of past and expected inflation should be 9 A relevant feature of this approach is money neutrality, found in the inability of the money interest rate in influencing the natural interest rate. As pointed before, the money interest rate converges towards the natural interest rate in the long term. Hence, the latter is ultimately determined by real rather than monetary factors. See also Garegnani (1979) and Pivetti (2001).
13 CHAPTER 1. INFLATION AND DISTRIBUTIVE CONFLICT complete, i.e., the sum of the coefficients of past inflation and expected inflation must be equal to one. In both cases, this hypothesis allows new-Keynesians to reject the trade-off between inflation and unemployment in the long term, advocating in favor of the existence of a natural rate of unemployment (see, for instance, Gordon 2011).
The unemployment rate consistent with stable inflation is labeled as the Non-Accelerating Inflation Rate of Unemployment (NAIRU). In the long run, the determination of real values of equilibrium output and unemployment is unaffected by the inflation rate.
According to the New Keynesian approach, when the interest rate set by the monetary authority differs from the natural interest rate, actual output diverges from the potential output determined by supply-side conditions. Thus, an interest rate higher than the natural rate generates an excess of aggregate demand.
Alternatively, if the interest rate is lower than the natural, output remains below potential output, causing the decrease in inflation rate (Woodford, 2003). In sum, in this approach, “the natural real rate of interest is the rate associated with the absence of the nominal rigidities that account for short-run monetary nonneutralities, and the gap between the actual interest rate and the natural real rate represents the key channel through which central bank actions affect the economy” (Walsh, 2005, p. 464).
Considering this, New-Keynesian economists focused on formulating monetary policy rules to stabilize inflation, as, for instance, the inflation targeting policy (Taylor, 1999). Monetary policy is executed by Central Banks and centers on pegging the basic interest rate to stabilize inflation.
According to this view, stabilizing inflation is equivalent to stabilizing the output gap.
Thus, monetary policy should concentrate efforts on the first goal. This property has been labeled as the ‘divine coincidence’ and implies no trade-off between stabilizing inflation and the output gap (Blanchard and Galí, 2007).
Even though an excess of demand is generally considered the cause of inflation acceleration, cost shocks also cause inflation in New-Keynesian models. Asymmetric cost shocks under nominal rigidity of prices and wages can accelerate inflation while relative prices adjust to the change in relative costs (Ball and Mankiw, 1994). An obvious example is the oil-shocks of the 1970s (Gordon, 1984). Moreover, Ravenna and Walsh (2006) argue that monetary policy shocks can temporarily raise inflation due to the effect of the nominal rate of interest on the marginal cost of firms — effect known as the cost-channel. Nevertheless, cost increases consist only of a temporary cause of inflation.
In the long term, inflation is solely caused by the excess of demand, since, in this time 14 CHAPTER 1. INFLATION AND DISTRIBUTIVE CONFLICT horizon, the nominal rigidity of prices and wage is absent, and cost shocks are assumed to have zero mean (Carlin and Soskice, 2014). As shown by Serrano (2019), dropping the zero mean assumption is enough to ensure the existence of cost-push inflation in the New Keynesian Phillips Curve, even in the long term.
An alternative explanation for the absence of cost-push inflation in the long term lies in the effort of reconciling the New Keynesian Phillips Curve with the general equilibrium analysis. In this case, relative prices and income distribution are assumed to converge to general equilibrium positions (Woodford, 2003). Moreover, the economy converges towards the natural output and the natural interest rate, which result from the real equilibrium obtained without the nominal rigidity. Relative prices, wages, and profit rates converge to real equilibrium levels compatible with supply and demand schedules at the microeconomic level. In equilibrium, relative prices match supply and demand, and income distribution depends on factors’ marginal productivity. In this case, the underlying logic of conflict inflation theory cannot operate since it requires that social groups can persistently shift income distribution through changes in their nominal compensations.
Therefore, even though some New-Keynesian scholars aim to introduce conflict inflation within this theoretical framework (as, for instance, Carlin and Soskice, 2014), conflict can only be assigned a transient role and cannot be consistently taken as the main cause of inflation. In contrast, conflict inflation theory rejects that relative prices and income distribution are determined according to supply and demand schedules, as in general equilibrium theory. Rather, conflict inflation admits that nominal shocks can persistently affect distribution.
Distributive variables do not converge to a predetermined equilibrium value in this view. In fact, they may remain persistently different after a nominal shock. If this is the case, cost-shocks tend to be positive on average, so that costs can be considered a recurrent cause of inflation (Serrano, 2019).
Critics have raised many arguments against the New-Keynesian theory of inflation.
First, it relies on the notion of a natural interest rate. In other words, New Keynesians assume that there is a single value for the real interest rate that equalizes investment and full employment savings.
The negative interest-elasticity of investment must be founded on the labor-capital substitution mechanism along the lines of the neoclassical approach (Petri, 2015, p. 316). Thus, the investment curve needs to be derived from a persistent demand for capital, conceived as a single factor (value-capital, rather than a set of capital goods).
Nevertheless, neoclassical labor-capital substitution must be 15 CHAPTER 1. INFLATION AND DISTRIBUTIVE CONFLICT rejected in the light of the Cambridge controversy in capital theory (Garegnani, 1978; Lazzarini, 2011; Petri, 2019).
The idea that “the proportions in which factors are employed vary with the prices of their services” requires that the quantity of factors are defined independently from the relative prices (Garegnani, 1978, p. 349). But this is not the case for the ‘single factor’ capital, which is measured according to the value of the capital goods. In addition, the possibility of reverse capital deepening shows that the value of capital per worker may increase when the interest rate increases. In conclusion, the Cambridge controversy makes it very difficult to rely on the negative interest-elasticity of investment.
Finally, Petri (2015) argues that even if we concede negative interest-elasticity, neoclassical theory cannot determine the flow of investment in each period without assuming full-employment of labor to hold continuously, which is unreasonable.
Critics have also contested assumptions and results of the New-Keynesian Phillips Curve from an empirical perspective.10 Smithin (2004) criticized the arbitrariness of estimates of the value of the natural interest rate. Furthermore, evidence of hysteresis in the unemployment rate challenges the notion of NAIRU (Stanley, 2004, 2013).
Hysteresis means that persistent deviations from equilibrium position affect the equilibrium itself. Hysteresis appears in unemployment in the form of a non-stationary unemployment rate.
Hence, there is not a long-term value towards which the unemployment rate converges.
Critics also find evidence supporting the view that demand shocks have a persistent impact on unemployment (Storm and Naastepad, 2009).
Further empirical findings support the rejection of the accelerationist Phillips curve present in the New-Keynesian models, suggesting the existence of a long-term trade-off between inflation and unemployment (Stirati and Paternesi Meloni, 2018; Summa and Braga).11 These results also challenge important conclusions of the New-Keynesian theory as the long-term neutrality of monetary policy. Finally, recent criticism has focused on the inability of the New-Keynesian Phillips Curve in explaining the contemporary economic events of the so-called Secular Stagnation (Serrano et al., 2020).
10 Serrano (2019) provides an assessment of the New-Keynesian Phillips Curve under the relaxation of its assumptions. By these means, he shows that relaxing the assumption of full inflationary inertia (or, in the hybrid models, a complete combined effect of past and expected inflation over current inflation) is enough to dismiss important conclusions of this approach as the no permanent trade-off between inflation and unemployment.
11 Such findings have also been acknowledged by supporters of the New-Keynesian approach, as Blanchard (2016); Blanchard et al. (2015); Gordon (2018).
16 CHAPTER 1. INFLATION AND DISTRIBUTIVE CONFLICT 1.4 Kaleckian conflicting claims models A deeper understanding of conflict inflation requires introducing an analytical exposition of the relation between prices and distribution. For this reason, we present how this approach describes conflict inflation, mainly based on the exposition of Lavoie (2014), adapting the notation.
Lavoie (2014) departs from the decomposition of aggregate income into wages and average mark-up, as proposed by Weintraub (1978).
By definition, aggregate income (Y) is equal to aggregate output in a closed economy.
From this identity, follows the equation below, where output is measured as the product between real net output (Q) and the price level (Pt).
PtQt = Yt [1.3] Furthermore, aggregate income can be decomposed into the wage bill (W) plus aggregate profit. Weintraub (1978, p.45) also takes the total profits as equal to an average mark-up on labor costs (µt), as presented below.
PtQt = Wt(1 + µt) [1.4] In addition, the wage bill can be written as the product between the number of workers and the average wage rate (w), which gives us the following equation: PtQt = wtLt(1 + µt) [1.5] Finally, dividing both sides of equation 1.5 per Qt, we obtain the equation 1.6.
Pt = wtlt(1 + µt) [1.6] Here, l stands for the labor-output ratio of this economy and is inversely related to labor productivity. Moreover, the term (1 + µt) expresses the ratio of prices to money wages. Rearranging equation 1.6, we find a correspondence between the mark-up and the real wage rate, shown in equation 1.7. Given productivity, a higher mark-up implies 17 CHAPTER 1. INFLATION AND DISTRIBUTIVE CONFLICT a lower real wage.
1 (1 + µt) = wtlt Pt = ωtlt [1.7] The product between the real wage (ωt) and the labor-output ratio (lt) also expresses the wage-share in this economy. A simple operation in equation 1.5 shows that this term corresponds to the ratio between the wage bill and the nominal output ( wtLt PtQt ), which represents the wage-share in this economy. Moreover, it follows that the wage-share can be written as 1 (1+µt). Conversely, the profit share is given by: PtQt −wtLt PtQt = ωtltµt [1.8] Since the total income is given by wtLt(1 + µt), it follows that the profit share also corresponds to wtLtµt wtLt(1+µt), which can be expressed as µt (1+µt).
Therefore, whatever determines the mark-up regulates the functional income distribution under stable prices (Cassetti, 2002; Lavoie, 2014).
In the Kaleckian view, firms set prices by adding a mark-up on unit production costs.
The mark-up depends solely on the degree of monopoly found in the respective market (Kalecki, 1954).
Hence, the aggregation of the mark-ups found in the economy determines income distribution between workers and capitalists. In other words, the average mark-up of the economy is determined by the market power of firms observed in different markets.
As a consequence, workers would not be able to change distribution through wage negotiations. However, Kalecki (1971) eventually admitted that strong labor activism could impose wage increases that would not be passed through prices, reducing firms’ mark-up.
Taking growth rates from equation 1.6, we can derive an interpretation of the causes of inflation (Lavoie, 2014; Weintraub, 1978). Thus, inflation (π) is written as the sum of the rate of change of money wages, rate of change labor-output ratio (so that productivity growth reduces inflation), and the rate of change of the ratio of prices to money wages (caused by changes in the average mark-up)12, all for the same period.
12 Note that the rate of change of (1 + µt) can be rewritten in terms of the change in the average mark-up.
ˆ (1 + µt) = ˙ µt (1 + µt) 18 CHAPTER 1. INFLATION AND DISTRIBUTIVE CONFLICT ˆ πt = ˆ wt + ˆ lt + ˆ (1 + µt) [1.9] Although the market structure determines the baseline mark-up, the mark-up’s real value may change in conflicting claims models. By introducing bargaining power of workers and capitalists, authors13 aim to address the possibility mentioned by Kalecki (1971) that strong unions may compress firm’s mark-ups. Following Lavoie (2014), we can set the general features of these conflicting claims models. First, we assume, for the sake of simplification, that productivity is constant (ˆ l = 0). Hence, the dynamics of inflation is dominated by movements in wages and mark-ups (see equation 1.9).
Moreover, wages negotiated between workers and firms are regulated by the bargaining power of workers (Ω1), the degree of formal and informal wage indexation (Ω2), past inflation rate (πt−1), and the aspiration gap — as shown in equation 1.10. Indexation and past inflation affect wages because workers seek to recover wages‘ purchasing power after price increases. In general, models assume indexation is incomplete — i.e.
Ω2 is less than one. In turn, the aspiration gap corresponds to the difference between worker’s targeted real wage (ωw) and the actual real wage (ωt).
ˆ wt = Ω1(ωw −ωt−1) + Ω2πt−1 [1.10] Firms increase prices to keep the mark-up close to the desired mark-up. As shown above, there is an inverse relationship between the mark-up and the real wage for a given technology.
Thus, the firm’s desired mark-up can be expressed in terms of a targeted real wage (ω f ). The difference between the actual real wage and the targeted real wage defines firm’s aspiration gap (Lavoie, 2014). Whenever the gap is positive, firms raise prices to an extent that also depends on the bargaining power of firms (ψ1).
Furthermore, firms pass through unit labor cost increase — which, in the absence of productivity changes, corresponds to wage increases — into prices, in a degree determined by the price indexation factor (ψ2).
Equation 1.11 summarizes these relations.
where ˙ µt stands for the derivative of µt with respect to time (or the change in time, in the discrete case).
13 See, for instance, Cassetti (2002), Lavoie (2014) 19 CHAPTER 1. INFLATION AND DISTRIBUTIVE CONFLICT πt = ψ1(ωt −ω f ) + ψ2 ˆ wt [1.11] In this model, equilibrium real wage and inflation rate are achieved when prices and wages rise at the same rate, stabilizing income distribution. The following expressions describe the equilibrium.
ω∗= Ω1(1 −ψ2)ωw + ψ1(1 −Ω2)ω f Ω1(1 −ψ2) + ψ1(1 −Ω2) [1.12] π∗= ψ1Ω1(ωw −ω f ) Ω1(1 −ψ2) + ψ1(1 −Ω2) [1.13] In general, firms can only partially index prices to costs, and workers can only partially index wages to the cost of living. Besides, both sides present limited bargaining power, which implies that equilibrium real wage lies within the real wage targets of workers and firms.
In this general case, neither side achieves its desired outcome (Dalziel, 1990; Lavoie, 2014). Thus, the equilibrium real wage is a weighted average of workers’ and firms’ real wage target. It increases with the bargaining power of workers and with wage indexation. It decreases with the bargaining power of firms and the speed of price adjustment after wage increases. Whatever increases the equilibrium real wage decreases the equilibrium level of the mark-up. Therefore, the baseline mark-up determined by the firm’s market power is allowed to change due to inflation. Mark-up varies according to the relative bargaining power of workers and capitalists, the speed of adjustment of prices to wage increases, and the speed at which wage claims react to price increases.
The equilibrium inflation rate depends on the relative bargaining power of both workers and firms. The higher the bargaining power of workers, the higher the inflation rate and the real wage rate.
Conversely, a higher bargaining power in firm’s side generates a combination of higher inflation – since firms raise prices more quickly – and lower equilibrium real wage.
Moreover, a greater discrepancy between real wage targeted by workers and firms, that is, a greater degree of conflict between workers’ and firms’ income claims, leads to a higher rate of inflation (Lavoie, 2014) 20 CHAPTER 1. INFLATION AND DISTRIBUTIVE CONFLICT A particular case of the model is found in Rowthorn’s (1977) inaugural contribution.
The author assumes that firms fully index prices when inflation exceeds a threshold. We introduce this assumption by taking ψ2 = 1 in equation 1.12, when the inflation rate is above the threshold.
In this case, it follows that the equilibrium wage-share exactly corresponds to the wage-share targeted by firms. Thus, the equilibrium wage share is not affected by the workers’ target, and an increase in ωw only affects inflation and money wage growth.
Hence, workers’ aspiration gap is not partially closed by real wage increases.
Rowthorn (1977) further assumes that above the same threshold, workers include expected inflation in their wage claims. Consequently, even though workers cannot affect real wages through money wage increases (because firms can immediately raise prices), a persistent aspiration gap and the impact of expectations lead to increasingly higher wage demands. In these conditions, small cost shocks may drive the economy towards an accelerationist process (Rowthorn, 1977, p.227-229).14 However, since workers are no longer able to affect income distribution, it is hard to justify the rationality behind higher wage demands (in case its real outcome is supposed to be null).
In spite of its contribution, Kaleckian conflicting claim models contain unsolved theoretical shortcomings. Next section discusses critiques on the Kaleckian explanation of distribution embodied in these inflation models.
As an alternative, we explore a Classical-Keynesian contribution to inflation theory.
14 The same outcome of the assumption of complete price indexation emerges if firms have a complete bargaining power, i.e. if the parameter ψ1 tends to infinity. In this case, “[f]irms do not let the margin of profit fall below its target level and they can respond immediately to any increase in their wage costs” (Lavoie, 2014, p. 552). Hence, workers are not able to affect the real wages by means of wage bargaining.
21 CHAPTER 1. INFLATION AND DISTRIBUTIVE CONFLICT 1.5 The Classical-Keynesian approach to inflation 1.5.1 Critique on the Kaleckian distributive closure In the Kaleckian view, firms set prices by adding a mark-up on unit costs.
The magnitude of this mark-up depends on the average of the market prices for the same commodity, weighted by the quantity produced by each firm (Kalecki, 1954, p.12-18).
However, Pivetti (1991, p.105-119) argues that basing the determination of mark-up on the degree of monopoly does not provide a proper explanation of income distribution.
Pivetti (1991) argues that as the competition among firms can change market-shares, the weights would oscillate, leaving the mark-up unstable or indeterminate. Moreover, this measure does not consider the role of potential entrants. A high mark-up (associated with a high profit rate) can attract new entrants to the market. Naturally, we expect that incumbent firms anticipate this possibility and set prices within certain limits to avoid losing market-share15. More importantly, this explanation of the mark-up does not set a lower limit for the profit rate. This framework seems to allow for the possibility of a profit rate equal to zero in a fully competitive industry. It does not explicitly consider the requirement of a minimum positive profit rate for capitalists to carry on production (Pivetti, 1991).
Finally, as shown before, the process of wage bargaining and price inflation sets an equilibrium mark-up different from the one consistent with the degree of monopoly. In this case, the distributive conflict changes the mark-up, even when the market structure across industries remains the same. Hence, the link between the degree of monopoly and income distribution is loosened, weakening the explanatory power of the former.16 Furthermore, Steedman (1992) argues that Kaleckian mark-up pricing overlooks implications of input-output relations. In other words, it does not explicitly consider that firm‘s costs depend on other firm‘s prices — and, hence, on the mark-ups found in 15 See, for instance, Sylos Labini (1957).
16 In his earlier writings Kalecki had shown skepticism regarding the ability of workers to affect functional income distribution through wage bargaining (see, for instance, Kalecki, 1943). In 1954, in Theory of Economics Dynamics, Kalecki broadened the notion of degree of monopoly to include the effect of workers‘ bargaining power. Kalecki (1954) expressed the wage share as an inverse function of (an aggregated measure of) mark-up, determined by the degree of monopoly across different industries. However, he also pondered that strong trade unions may reduce the degree of monopoly (Kalecki, 1954, p. 17-18). According to Kalecki (1954, p. 18), a “high ratio of profits to wages strengthens the bargaining position of trade unions in their demands for wage increases since higher wages are then compatible with ’reasonable profits’ at existing price levels”. This process pressures profits, encouraging firms to adopt a “policy of lower profit margins”.
The argument was further explored in Kalecki (1971). See also López (2010, p. 67-90), Rugitsky (2013) and Pivetti (1991, p. 108-113).
22 CHAPTER 1. INFLATION AND DISTRIBUTIVE CONFLICT other industries. Steedman (1992) adds that the average mark-up of the economy cannot be properly defined as an ‘average‘. The problem is that the weights of this average (the output value of different industries) depend on relative prices and, thus, on the mark-ups.
Finally, competition implies that mark-up in one industry cannot remain persistently independent of the mark-up in other industries. However, if this is the case, the concentration within each industry might no longer be the main determinant of the respective mark-up.
1.5.2 Inflation in the Classical-Keynesian approach The Classical surplus approach, recovered by Sraffa (1960), relies on political and institutional factors to explain income distribution. Relative prices and distribution are determined together in the same step of the analysis (within the same system of equations).
Therefore, the interaction between inflation and distributive conflict emerges straightforwardly.
The Classical approach distinguishes normal and market values for prices and distributive variables in order to introduce a systematic analysis of the long-term regularities observed in capitalist economies (Eatwell, 1982). In this view, competition within markets establishes a normal price for commodities around which market prices tend to gravitate. Normal prices (or production prices) are set according to persistent long-term forces affecting prices and distribution.
Temporary changes in effectual demand, for instance, are assumed to affect market prices without affecting normal prices. Competition among capitals also generates a tendency of convergence towards a normal profit rate.
Thus, production price is defined as the price required for production to be regularly brought to market, standing for a necessary condition for supply.
Hence, production prices must be enough to repay inputs costs, as well as wages, profits, and rents in its normal rates, which hold under competition (Salvadori and Signorino, 2013; Vianello, 1989).
Nevertheless, the system of price equations proposed by Sraffa (1960) analyzes real variables: relative prices, real wages, and (real) profit rate.
This system allows for different distributive closures. Classical economists analyzed distribution taking the real wage as determined according to the socially established level of subsistence (Stirati, 1994). In this case, the profit rate is determined as a residual in the system of price equations, together with relative prices. History has shown, however, that workers may be able to decouple real wages from the subsistence level, obtaining also a share of the 23 CHAPTER 1. INFLATION AND DISTRIBUTIVE CONFLICT surplus (Pivetti, 1999; Sraffa, 1960). For this reason, Sraffa (1960) takes the profit rate as exogenous with respect to the system of price equations, thereby determining real wages endogenously and together with relative prices. The author also suggested that the profit rate is determined according to the level of the money interest rate.17 As we see below, Pivetti (1991) builds on this suggestion, which also allowed for an analysis of inflation.18 Studying inflation requires taking an additional step. We must define not only the set of relative prices but also the price level. Addressing this issue, Pivetti (1991, 2007) introduces money wages, and the interest rate (as a monetary determinant of the profit rate) into the price equations system, obtaining the price level and commodity prices expressed in money proper.19 Stirati (2001) follows the same procedure, setting a framework to discuss inflation along the lines of the Classical surplus approach. Starting with nominal production prices, equation 1.14 presents the price equations in matrix form.
P = PA(1 + i) + PAρ + lw [1.14] In this equation, P represents a row vector of prices. The technical coefficients are given by A, a squared matrix of input coefficients per unit of output, and l, the row vector of labor inputs per unit of output. We will assume these technical coefficients remain unchanged. In turn, w stands for the money wage (uniform across industries, by assumption), and i stands for the nominal interest rate. Finally, ρ represents a diagonal matrix of industry-specific premiums compensating risk and market-power.20 Since ρ 17 As stated by Sraffa (1960, p.33): “The rate of profits, as a ratio, has a significance which is independent of any prices, and can well be ‘given’ before the prices are fixed. It is accordingly susceptible of being determined from outside the system of production, in particular by the level of the money rates of interest”.
18 In the last decades, advanced economies presented low interest rates along with fall in the wage share. This event may have raised interpretations that real wage rate decrease independently from movements in the interest rate (Stirati, 2013). If this is the case, the real wage rate would regain its analytical position as the external variable to the production system. In contrast, the profit rate would be determined as a residual.
Stirati (2013) also explores alternative interpretations to this phenomenon in the light of the Classical surplus approach to distribution.
19 In the author’s words: “Given the rate of interest to be earned on long-term riskless financial assets, and given the money wage, which is the direct outcome of wage bargaining, the price level can be determined in a system of price equations à la Sraffa (in which, however, both the wage rate and commodity prices are expressed in money proper), together with the distribution of income between profits and wages” (Pivetti, 2007, p. 243).
20 The compensation of market-power depends on the existence of persistent barriers to entry restraining the equalization of profit rates by competition. This component is associated with features of the industry, as minimum operational scale and initially required investment. In the Classical surplus approach, competition 24 CHAPTER 1. INFLATION AND DISTRIBUTIVE CONFLICT tends to be stable, we assume that it is equal to zero to focus on the impact of changes on wage rate and interest rate.
Furthermore, equation 1.14 shows that the interest rate is assumed to determine the normal profit rate, following the monetary theory of distribution (Pivetti, 1991). In other words, the money spent in the purchase of inputs in the previous period yields an earning equivalent to the interest rate on the anticipated investment. According to Pivetti (1991), if competition equalizes earnings of different applications of capital, then the normal profit rate cannot persistently differ from the interest rate. Since the nominal interest rate is considered a monetary phenomenon, autonomously set by the monetary authority — accounting for political and economic constraints —, it must determine the normal profit rate rather than the opposite.21 A nominal determination of the (thus, nominal) normal profit rate implies that the real outcome for the profit rate is affected by inflation, as we discuss later.
In equilibrium, the price level and relative prices are stable and are determined as in the equation 1.15 — where I represents the identity matrix.
P = l[I −(1 + i)A]−1w [1.15] By post-multiplying the row vector of prices by a column vector representing quantities (which can be, for instance, proportional to the output or the consumer‘s bundle), we obtain a measure of the price level. Let us consider the column vector Q, as the vector of output. Then, a measure for inflation (π) can be obtained as in equation 1.16. Note that this measure corresponds to the output deflator.
1 + πt = PtQ Pt−1Q [1.16] In this framework, inflation arises during the transition between two long-period positions of relative prices and income distribution. From the previous expressions, it is is related to free mobility of capital across industries, while market-power arises when capital movements face some restraint (Eatwell, 1982; Salvadori and Signorino, 2013).
21 More precisely, the author argues that the normal profit rate of each productive sector is determined by two components (Pivetti, 1991). The first is the interest rate of long-term fixed and risk-free security, best represented by public securities.
The second component corresponds to the remuneration of ’risk and trouble’ of productive investment, changing according to the economic sector. This second component is captured in the price equations (1.14) by the matrix ρ.
25 CHAPTER 1. INFLATION AND DISTRIBUTIVE CONFLICT evident that, in the absence of changes in the technical coefficients, changes in w and i (which correspond to the basic distributive variables) regulate the changes in the price level.
Competition constraints firms to set prices following the price of capital goods at the beginning of the production process: the historical cost of capital. This comes from the following reasoning. The interest rate on the free-risk asset sets the minimum return on advanced capital to allow production to happen. Hence, competition among capitalists will bring the return on advanced capital in production into equality with the risk-free interest rate (abstracting from the industry-specific premium compensating illiquidity).
Different investment opportunities must yield the same return in proportion to the amount of capital advanced in the presence of competition. Hence, capital advanced in production taking place from period t to t + 1 must provide the same return as the interest rate of the risk-free asset. This return must be proportional to the monetary value of the capital advanced in period t, either in the form of capital goods or bonds. In the case of productive investment, this value corresponds to the historical costs of capital (Pivetti, 1991).
As prices may change in time, the cost of inputs required for production can rise. In this case, a discrepancy between the historical cost of capital and reproduction costs of capital arises, i.e., the value of capital at the end of the production period. Reproduction cost of capital “is actually the relevant magnitude as far as the rate of profit is concerned, since profits are what remains of the value of the product after deducting wages and the reproduction cost of capital” (Pivetti, 1991, p. 53). Under these conditions, changes in the production costs (as wage increases) can change the profit rate as measured in terms of the reproduction cost of capital.
As Stirati (2001) shows, assuming the interest rate remains stable at its initial level (i0), the profit rate will be given by equation 1.17. Using the result in equation 1.16, we can obtain equation 1.18, which expresses the profit rate as the ratio between the nominal interest rate and the rate of change in the price level.
(1 + rt) = (1 + i0)Pt−1Q PtQ [1.17] (1 + rt) = (1 + i0) (1 + πt) [1.18] 26 CHAPTER 1. INFLATION AND DISTRIBUTIVE CONFLICT This result sets a relation between inflation and distributive conflict because wage increases affect distribution at least temporarily. After a while, real wages return to their previous value as higher costs are passed through into prices if there are no further wage increases. Nevertheless, a persistent change in distribution can occur if money wages keep increasing at a stable rate (Pivetti, 1991; Stirati, 2001). Suppose the rate of interest determines the normal profit rate. In that case, the real rate of interest is the one consistent with the distributive result after changes in money wages and prices (Pivetti, 1991). Then, the profit rate would not only depend on the nominal interest rate but also on the growth rate of money wages (Pivetti, 2007; Serrano, 1993). Therefore, together with monetary policy, the bargaining power of workers can determine income distribution.
This approach provides a useful framework for discussing the sources of inflation, its impact on distribution, and its relationship with unemployment,22 consistently with both the conflicting claims view and classical theory of distribution (Stirati, 2001). One of its distinctive features corresponds to taking the profit rate as determined according to the monetary theory of distribution. Distribution is also affected by the monopoly power generated by restrictions to competition (which appear in the variable ρ in equation 1.14).
Nevertheless, according to the monetary theory of distribution, monopoly power may affect distribution, but it cannot be considered the ultimate cause of profits (Pivetti, 1991, p. 108-113). This view contrasts with Kaleckian models of inflation in which distribution depends on the degree of monopoly and the relative bargaining power of workers in wage negotiations and on bargaining power of firms in price setting.
22 As in the Phillips curve. See Stirati (2001) and Serrano (2019).
27 CHAPTER 1. INFLATION AND DISTRIBUTIVE CONFLICT 1.6 Final Remarks Inflation emerges from the dispute among social groups over the distribution of the surplus. As incomes are defined in nominal terms, price changes affect their purchasing power.
Prices rise when the claims of social groups are incompatible with the total income generated in production.
The consistency of the view that conflict is the main cause of inflation depends on two important premises. First, earnings are set in nominal terms. That is the case for wage agreements (Keynes, 1937, p. 8-9), but also for profits (Pivetti, 1991). When workers resist the fall in nominal wages, they prevent real losses. Therefore, nominal changes affect income distribution. Nevertheless, models assuming full indexation of prices to money wages do not allow for changes in distribution (as in Rowthorn, 1977). These models have thus been considered incompatible with the conflict inflation view (Serrano, 2010).
Second, income distribution does not converge towards a predetermined real equilibrium.
This premise implies that the conflict over wage agreements and other nominal distributive variables has a persistent effect on distribution.
If otherwise, conflict had a merely temporary effect on distribution, it could only be a transient and eventual cause of inflation (as in Carlin and Soskice, 2014). In that case, the main causes of inflation must be elsewhere (as excess demand and expectations in the New Keynesian Phillips curve). In the Kaleckian perspective, accepting the persistent effect of conflict over income distribution violates the determination of distribution according to the forces of competition. That perspective can be consistent only if one is willing to concede that the wage bargaining process is inside the Kaleckian definition of the degree of monopoly of firms (as done in Reynolds, 1983). Recent contributions rely on firms’ bargaining power to explain the mark-up, combining the role of the forces of competition and firms’ relative power with respect to workers (Lavoie, 2014).
Under these two premises — namely, if income distribution is affected by nominal variables, and is not tied to a predetermined real equilibrium — conflict can be a persistent cause of inflation.
These conditions are unambiguously attended in the Classical-Keynesian approach.
This approach analytically separates the theory of value and distribution from the theory of output and accumulation (Garegnani, 1983). Differently from the marginalist approach, income distribution is not determined by factors’ productivity.
Rather, income distribution does not maintain a necessary functional relation with output or other production variables in the Classical-Keynesian approach (Garegnani, 1983).
28 CHAPTER 1. INFLATION AND DISTRIBUTIVE CONFLICT Distribution is affected by political and institutional factors, which express the dispute of social classes over the surplus. That reveals the contingent and historical nature of income distribution in economic systems, particularly capitalism.
Inflation, wage bargaining, and changes in nominal variables are one sphere of the distributive conflict (Pivetti, 1991; Serrano, 1993; Stirati, 2001).
Expressing the price equations system in money proper connects the Classical approach to distribution with conflict inflation, thereby setting a monetary theory of distribution (Pivetti, 1991).
Therefore, nominal variables can permanently affect distribution.
Inflation is dominantly a cost-push phenomenon in conflict inflation (Morlin, 2021).
Theories of inflation relying mainly on excess demand are not compatible with this view.
We discussed the New Keynesian Phillips curve, exposing its main weaknesses as the notion of a natural interest rate (Petri, 2019), the assumption of a stable NAIRU (Storm and Naastepad, 2009), and the incompatibility with recent economic events (Stirati and Paternesi Meloni, 2018; Summa and Braga).
Conversely, the empirical regularity captured in the Phillips curve can be understood as an outcome of wage bargaining and cost-push inflation. The lower the unemployment rate, the greater workers’ bargaining power is, which speeds up wage growth and thus inflation (Rowthorn, 1977; Serrano, 2019; Stirati, 2001). In contrast with the New-Keynesian view, the interaction between inflation and income distribution in this conflict-augmented Phillips curve is a source of non-neutrality of money both in short and in the long run (Pivetti, 2007; Serrano, 2019).
This chapter reviewed competing explanations for inflation. We focused on conflict inflation theories, comparing the underlying distributive closures. We argued that the Classical-Keynesian approach provides a straightforward foundation to conflict inflation theory, thereby being advantageous with respect to the Kaleckian foundations of conflicting claims models.
After this review on conflict inflation theory, we can move forward to the next chapter. We will discuss open economy models of inflation. New sources of distributive conflict emerge, and additional variables must be considered as the exchange rate and international prices. Finally, we develop a conflicting claims model for a price-taker open economy, building on the conclusions of the present chapter.
29 Chapter 2 Inflation and conflicting claims in the open economy 2.1 Introduction International causes are found in the explanation of inflation since the creation of the word “inflation” during the European price revolution in the sixteenth century.1 As described later in Hume’s specie flow mechanism, price increases were believed to arise because of the vast gold inflow from colonies (Green, 1982). In contrast to these early formulations of the Quantitative Theory of Money, alternative explanations back then linked rises in the price level with rising costs, exchange rate devaluations and income claims of landowners (Arestis and Howells, 2001, p. 188). Later on, during the Napoleon wars, external cost shocks following the English trade blockage appeared among explanations of inflation (Smith, 2011, p. 88-94). The interaction between inflation and the exchange rate was also explicit in the German hyperinflation debate. In contrast with the dominant monetarist explanation, an alternative interpretation claimed that the external debt owing to war reparations caused exchange rate devaluations, leading to hyperinflation (Câmara and Vernengo, 2001; Robinson, 1938). In the post-war period, Latin American structuralist economists emphasized the role of the balance of payments and exchange rate devaluations in inflation (Furtado, 1954, 1963; Noyola Vázquez, 1956; Sunkel, 1958).
The focus on international causes was not exclusive of peripheral 1 During the sixteenth and the first half of the seventeenth century, Western European economies faced a period of slow and continuous rise in price levels. This process stimulated a seminal debate on inflation and money, with some of the first formulations of the quantitative theory of money.
30 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY economies, though, as they were also present in European debates in the context of small-open economies (Aukrust, 1977; Edgren et al., 1969; Frisch, 1977). Afterward, the overall importance of international prices was unavoidable during the oil-shocks in the 1970s (Gordon, 1984). Arising challenges due to external shocks turned the attention of some economists to money wages-exchange rate inflationary spiral.2 In Latin American countries, chronic Balance of Payments deficit and repeated exchange rate devaluations sustained high inflation, provoking episodes of hyperinflation. Once more, international prices and exchange rate devaluations gained centrality among explanations of high inflation. The stability of the exchange rate was among the main policy instruments employed to correct it (Bastos, 2002).
Growing international integration in recent times inspired criticism against the dominance of country-centric approaches to inflation for focusing mainly on domestic variables (Borio and Filardo, 2007). On the contrary, international factors are shown to account for much of national inflation dynamics across countries (Bobeica and Jarocinski, 2019).
Furthermore, global inflation rates act as an attractor for national inflation rates (Ciccarelli and Mojon, 2010).
Therefore, analysis of inflation should account for the high substitutability of tradable goods, pressures of international competition, mobility of labor, and the great importance of cross-border input-output linkages. Acknowledging these effects implies focusing on international prices of goods and inputs, foreign wage rates, and exchange rates (Bugamelli et al., 2015).
In this regard, Auer et al. (2017) found that trade in intermediate goods and services is the main transmission mechanism of global slack to domestic prices. Moreover, Auer et al.
(2019) show the relevance of international input-output linkages in explaining the high synchronization of inflation among different countries.
These results stand out for reclaiming the role of the cost channel as an important transmission mechanism of international shocks to domestic inflation.
Inflation models have typically dealt with open economy concerns by including non-competing imported inputs among production costs — as in Rowthorn (1977), Stirati (2001), Lavoie (2014).
Features of contemporary open economies, however, suggest further properties must be introduced in inflation models. Inter alia, inflation models should consider that many domestic prices follow prices of the international market. More generally, domestically produced tradable goods are exposed to foreign 2 This was the case of Ezio Tarantelli, who proposed different stabilization policies in Italy (Michelagnoli, 2011). In the United Kingdom, search for Balance of Payments equilibrium justified the adoption of incomes policy and wage controls (Tarling and Wilkinson, 1977).
31 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY competition so that their prices cannot be taken as independent from foreign prices. If so, foreign prices must also affect domestic income distribution. Hence, inflation in open economies interacts with income distribution, which is the domain of conflict between social classes.
32 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY 2.2 Inflation in the open economy The introduction of (non-competing) imported inputs in production permitted the extension of closed economy models to open economy (Rowthorn 1977, Stirati 2001, p.
436-9; Hein and Vogel 2008). Alternatively, conflict may also emerge from the presence of imported consumption goods as part of workers’ bundle, as shown by Cassetti (2002).
However, we discuss the case more frequent in the literature in which imported inputs are required for domestic production. Following Hein and Vogel (2008), equation 2.1 introduces imported inputs by means of a small change in equation 1.6. In this case, a unit of output is produced with a combination of labor and an imported input. The input’s price corresponds to a given international price (p∗ µ) times the nominal exchange rate (e). The input is employed in the quantity am.
Pt = (1 + µt)(wtlt + ametp∗ m,t) [2.1] The profit share is described in equation 2.2. For a given wage rate, the profit share increases with the mark-up, the international price, the exchange rate, and the coefficient am.
Πt = 1 1 + 1 µt(1+zt) [2.2] where zt = ametp∗ m,t ltwt Additional sources of inflation and distributive conflict emerge here. Concerning the impact on prices and distribution, both a uniform rise in international prices and an exchange rate devaluation present the same effect in these models.
Either of these shocks proportionally increases the value of imported inputs as measured in domestic currency, thus raising production costs.
For the sake of exposition, let us assume, at this moment, that prices do not adjust instantaneously after an external shock. The immediate consequence is a fall in the profit share due to the higher replacement costs of capital. As higher costs are passed through into prices, the ratio of prices to current reproduction costs is restored to its initial value, 33 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY while the external shock burden is transferred to real wages.
In the equations above, however, the adjustment of prices to costs is assumed to happen instantaneously. Because of this assumption, the real value of the mark-up – i.e., the ratio between prices and current reproduction costs — is constant. In either case, as final prices adjust to larger input costs, profit share becomes larger than before. This effect can be seen in equation 2.2, in which a larger cost of imported inputs (larger value of p∗) implies larger zt and, thus, a larger profit share (Πt). As a counterpart, both the wage share and real wages fall. Then, the burden is increasingly transferred to workers as real wages fall due to the increase in the price level. Putting it differently, the increase in the price of imported inputs (as measured in the domestic currency), ceteris paribus reduces the National Income available to be distributed among workers and capitalists.
If the mark-up is assumed constant, the higher costs are completely pass through into prices leaving all the burden to workers in the form of a lower real wage and a lower wage share. Nevertheless, a lower real wage may trigger reactions in the form of wage claims, leading to an inflationary process.
In sum, higher costs of imported inputs imply either lower mark-up or lower real wages – or an intermediary situation in which losses are shared between capitalists and workers. In Rowthorn (1977), complete and instantaneous pass-through ensures that the burden falls on workers – whose reaction only leads to the acceleration of inflation.
However, the intermediary situation tends to prevail in other models (Bastian and Setterfield, 2020; Blecker, 2011; Lavoie, 2014).
The discussion on inflation in the open economy also considers the role of the exchange rate. In some models (as Vera, 2010; Blecker, 2011; Vernengo and Perry, 2018), a target for the real exchange rate is associated with a policy orientation to avoid — or reduce — deficit in the trade account of the Balance of Payments. However, this policy choice depends on the Marshall-Lerner condition, which establishes the requirement for an exchange rate devaluation to improve the balance of trade account.
A nominal devaluation increases the trade surplus whenever the response of quantities of exports and imports more than compensate for the change in prices associated with the devaluation. Thus, starting from equilibrium in the trade account, an exchange rate devaluation rises the trade surplus if the sum of the absolute value of the demand elasticities of export and imports is larger than one. Let us assume an external shock in the form of a nominal exchange rate devaluation, pursued by a monetary authority targeting a certain value for the real exchange rate. Even if the targeted real exchange 34 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY rate is achieved at the moment of the devaluation, the propagation of inflation through wage and price increases partially offsets the previous real devaluation.
This propagation, in turn, can lead to additional nominal devaluations, imposing a pattern of chronic devaluations and a race between exchange rate and money wages (Bastian and Setterfield, 2020; Blecker, 2011).
Departing from Dornbusch (1980), Vera (2010) explicitly introduces the Balance of Payments. In a first analytical exercise, the author assumes imported goods are part of the workers’ bundle. In a second step, he also includes imported inputs to domestic production. Finally, Vera considers mark-ups to be flexible and positively affected by the real exchange rate. This framework is set up to approach a small economy running Balance of Payments deficits and subsequent exchange rate devaluations, causing persistent inflation.
The argument builds on the interpretation developed (though not explicitly formalized) by the Latin American Structuralist School since the 1950s (Pérez Caldentey, 2019). In this view, underdeveloped economies faced structural deficits in Balance of Payments due to the need for industrial imported goods and the secular decline of terms of trade.3 In turn, this leads to repeated devaluation, which explained the chronic inflation of Latin American economies.
Propagation mechanisms would explain the persistence of inflationary pressures after an external shock. Notably, some structuralists as Furtado (1954, 1963) and Noyola Vázquez (1956) placed conflicting claims among the propagation mechanisms (Pérez Caldentey, 2019). Thus, “[i]f after an inflationary shock a group is dissatisfied with its income share it will try to pass its losses to another group” (Vernengo and Perry, 2018, p. 127).
Along these lines, Vernengo and Perry (2018) discuss the impact of a deterioration in Balance of Payments on inflation in a simple model.
The analysis assumes that monetary authorities target a value for the real exchange rate, pursued by adopting a nominal depreciation rule – namely, crawling pegs. By adopting this policy, authorities aim to stimulate exports, seeking for trade account surplus to pay for external debt service. Hence, an increase in the international interest rate (as the Volcker shock in the 1970s) rises the targeted real exchange rate, leading to exchange rate devaluation. After that, the exchange rate shock is propagated through wage indexation. Nevertheless, the model assumes a given rule of indexation, not addressing the distributive conflict explicitly.
3 Following the Prebisch-Singer hypothesis.
35 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY Charles and Marie (2016) propose a similar effect with a different argument. In this model, firms are indebted to the international capital market and must pay debt service in international currency. Firms’ pricing decision includes the cost of debt service in final prices.
Therefore, if debt-service cost increases either because of exchange rate devaluation or raises in international interest rate, then firms raise prices. Monitoring trade account balance allows firms to form expectations about the future exchange rate.
In this way, firms anticipate the impact of devaluation on costs of debt service and imported inputs, raising final prices.
An external shock can therefore provoke an inflationary spiral. Under extreme conditions, it can lead to hyperinflation (Charles and Marie, 2016). As Vernengo and Perry (2018), Charles and Marie do not explicitly deal with distributive conflict but assume a given indexation of wages to inflation. However, indexation is assumed incomplete in the former and complete in the latter.
In different ways, the models discussed advance in the treatment of inflation in the open economy, dealing with Balance of Payments, and interacting with distribution.
These models try to formally introduce Balance of Payments disequilibria as a source of chronic inflation (as well as high inflation and hyperinflation), which is compatible with the typical situation observed in peripheral economies. Notably, these studies assume the Marshall-Lerner condition to hold, establishing a positive relationship between real exchange rate and trade account balance.
Introducing Balance of Payments concerns in the analysis of inflation reveals the distributive role of the exchange rate and international prices. However, a domestic mechanism of propagation of the external shocks must be present to generate persistent inflation.
When reviewing the work of Bresciani-Turroni on the German hyperinflation in the inter-war period, Joan Robinson (1938) explicitly assigns a central role to money wage increases in the explanation of hyperinflation. Wage increases were a response to the rise in the cost of living due to sharp exchange rate devaluations. Money wages increase further rose production costs, worsening the trade account balance and leading to additional devaluations. This process generated an inflationary spiral of money wages and nominal exchange rate.4 The trigger of the crisis was the external debt created by 4 “With the collapse of the mark in 1921, import prices rose abruptly, dragging home prices after them. The sudden rise in the cost of living led to urgent demands for higher wages. [...] Wage rises had to be granted.
Rising wages, increasing both home costs and home money incomes, counteracted the effect of exchange depreciation in stimulating exports and restricting imports. Each rise in wages, therefore, precipitated a further fall in the exchange rate, and each fall in the exchange rate called forth a further rise in wages”.
(Robinson, 1938, p. 510) 36 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY the reparation payments imposed in the Treaty of Versailles (Câmara and Vernengo, 2001).
Later on, Latin American structuralist scholars argued that distributive conflict is a mechanism of propagation of inflation.
In this view, conflict contributes to the propagation of inflationary shocks coming from structural conditions — as a tendency towards deficit positions in Balance of Payments, productivity bottlenecks, and the inelasticity of supply (especially of food-related products, because of the land concentration and low productivity of agriculture for the domestic market) (Furtado, 1954; Noyola Vázquez, 1956; Sunkel, 1958). International prices obtain relevance in this analysis since they influence the trade account balance. Therefore, a declining trend of terms of trade is placed among the structural conditions that worsened the trade deficit, causing further nominal devaluations and, thus, inflation.
Nevertheless, an improvement of terms of trade can also be inflationary.
Celso Furtado (1963) explains this effect in his analysis of the impact of coffee prices on Brazilian inflation.5 Furtado (1963, 246-258) describes two different transmission mechanisms through which the increase in the international price of an exported commodity (with a relevant share in the output) causes inflation.
First, a direct immediate mechanism in the form of an increase of the domestic price of the same commodity. The second effect concerns a distributive channel. A higher price of the exported commodity increases the compensation of productive resources allocated in the exporting sector — as land rent. As a consequence, a larger portion of land would be dedicated to the exported commodity, reducing the domestic supply of foodstuff. Given the effective demand for food, a smaller supply tends to raise prices. Competition across different productive uses of land leads the economy towards a position in which these prices are enough to cover the higher land rent observed in the exporting sector. In sum, a persistent increase in the coffee price would raise the land rent, increasing prices of other agricultural commodities sold in the domestic market. As Furtado (1963, p. 257) states: "If the exporting sector, as in Brazil, comprises an extremely important sector of agriculture, it is perfectly natural that the factors connected with the domestic market will try to bring their rewards into line with the level established in the exporting sector, at least in a regional basis". Hence, this direct interaction between international prices and income distribution introduces an additional element to the analysis of inflation in open economies.
5 In the period, coffee was the main commodity exported by Brazil.
37 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY The effect of international prices on inflation in small open economies was also discussed in the Scandinavian model of inflation6 (Aukrust, 1977; Edgren et al., 1973).7 This model departs from a distinction between two economic sectors: one tradable and the other non-tradable. The tradable sector has its prices determined in the international market, thereby causing its prices to be insensitive to changes in domestic costs.
In contrast, non-tradable sector prices depend directly on their costs once they are not constraint by foreign competition.
The two sectors also differ with respect to productivity growth since the tradable sector contains industries leading productivity growth (as manufacturing) in contrast with the non-tradable sector (composed of services, among other activities).
The nominal exchange rate is assumed to remain constant in the model, according to the rules of the Bretton Woods agreement, then in force. Moreover, price increases and productivity growth in the tradable commodity are completely absorbed by money wage increases in the same sector. Authors assume that wage growth spreads to the other sector, either because of competition in the labor market or unions activity.
This assumption leads to a higher price increase in the non-tradable industry with respect to tradable prices. Thas occurs because of the slower productivity growth in the non-tradable industry.
Thus, as international inflation directly determines the rate of increase in prices in the tradable sector, it also affects the money wage increase which — discounted of the productivity growth of the non-tradable sector — determines the price increase in the non-tradable sector.
Altogether, this defines the inflation rate of the economy as completely determined by international inflation and domestic differential of productivity (Aukrust, 1977; Edgren et al., 1973).
If wages diverge from the rule imposed by the model, then income distribution changes. However, Edgren et al. (1969, 1973) aimed to prevent changes in distribution (avoiding “excessive” wage claims), claiming this condition would preserve investment, growth, and Balance of Payments equilibrium (refer to Morlin and Bastos (2019b))8.
As emphasized in the Scandinavian model, in the context of price-taker open economies, firms cannot pass through increases in domestic costs into prices of tradable 6 See Canavese (1982) for comparison between European and Latin American structuralist explanations of inflation. See also Frisch (1977) and Morlin and Bastos (2019b).
7 As pointed by its original proponent, the Scandinavian model builds on the “recognition that the developments of prices and incomes in small and medium-sized economies are strongly affected by events in the outside world and that, for this reason, price theory, more than hitherto, should address itself explicitly to the problems of open economies.” (Aukrust, 1977, p. 109) 8 Morlin and Bastos (2019a) discuss the historical context in which this model was developed.
38 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY commodities. Therefore, when wages grow faster than the rise in international prices, profits are compressed in these industries. In a similar reasoning, Blecker (1989a, 2011) introduces flexible mark-ups in price equations for open economies, in contrast with the usual fix mark-ups of other Kaleckian models of inflation.
Introducing flexible mark-ups “ creates the possibility that profit margins may be ’squeezed’ between high domestic costs and low foreign prices” (Blecker, 1989a, p.396).
Hence, mark-ups are assumed to be a positive function of the real exchange rate. A real depreciation makes foreign competing commodities relatively more expensive, allowing domestic firms to raise mark-ups (Blecker, 2011).
Nevertheless, if mark-up pricing presents weaknesses, as discussed in the previous section, the notion of flexible mark-up brings an additional problem. Under flexible mark-ups, it is not clear to which extent the degree of monopoly still determines the mark-up while, at the same time, it is supposed to change with the real exchange rate. An additional problem could arise when considering sectorial mark-ups. A sector with a high degree of monopoly could have a lower mark-up than another sector in which domestic competition is more intense, although being more protected from international competition (by its nature, as non-tradable goods, or institutional features, as trade barriers). In any case, the flexible mark-up brings within an inflationary pressure, generated by a profit-claim of capitalists associated with a real depreciation. In this regard, Blecker (1989a, p. 406) stresses the distributive effect of a real depreciation: “a depreciation is equivalent to a money wage cut, since it redistributes income to profits.” Finally, flexible mark-ups allow for modeling the interaction between international prices and distribution without setting tradable prices as directly determined by international prices — in other words, withdrawing the price-taker hypothesis and disregarding the law of one price.
From an empirical viewpoint, evidence shows that international variables as foreign prices, cross-border input-output linkages, and measures of trade integration play a relevant role in explaining inflation in advanced and developing economies.
These results are discussed in the Chapter 3 of this thesis.
39 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY 2.3 A formal model of inflation in an open economy 2.3.1 Distributive closure in the open economy Developments in trade theory within Classical-Keynesian approaches build on the extension of production prices to the open economy case, focusing on the determination of trade and specialization as a problem of choice of technique (Steedman, 1999).
Although an ultimate closure for the determination of distribution within open economies remains unresolved, this approach can bring useful insights into a conflicting claims modeling. Notably, open economies require accounting for the exchange rate and international prices as relevant determinants of domestic prices and income distribution (Metcalfe and Steedman, 1981). For the purposes of this paper, we depart from a simple setup, which allows us to focus on inflation and its interaction with distributive variables.
Let us consider the case of a price-taker open economy producing two commodities: one tradable machine (commodity 1) and one non-tradable consumption good (commodity 2) — as in Steedman (1999).
Production in both industries requires non-competing imports as inputs (denoted by the subscripts m and µ), and an amount of the tradable machine and labor. Labor is assumed homogeneous, thereby equalizing the wage rate in the two sectors. Moreover, capitalist competition implies that the profit rate is uniform across sectors in the long period position. We assume production takes one period of time. Wages are set at the beginning of this period and paid at the end (post-factum).
p1 = (a1p1 + ampm)(1 + r) + wl1 [2.3] p2 = (a2p1 + aµpµ)(1 + r) + wl2 [2.4] p1 = ep∗ 1 [2.5] The price of commodity 2 follows a customary production price equation as in 2.4, summing unit labor costs and unit costs of circulating capital added of a profit rate. In contrast, the price of the tradable machine is constraint by international competition. An economy is a price-taker in a particular commodity in the international market when it is not able to satisfy the world demand for this commodity with lower costs than the 40 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY other countries.9 Thus, the domestic price for 1 follows that the international price (p∗ 1) as measured in domestic currency (given the nominal exchange rate e) – as in equation 2.5.
In addition, we define the real wage rate as the ratio of money wages to the price of the consumption good.
wR = w p2 [2.6] The distributive closure for this price-taker open economy consists of an endogenous determination of the profit rate in the production of the tradable good.10 This is equivalent to taking the real wage as an exogenous variable in the price equations for the long period position (Metcalfe and Steedman, 1981).
Real wage (and not the profit rate) defines the distributive closure because it is completely determinate once we know money wages and the nominal exchange rate.
We build on two realistic assumptions. First, money wages are determined in the wage bargaining process. Second, the nominal exchange rate is a monetary variable.11 In this regard, Lavoie (2000a, 2002), Smithin (2002) and Vernengo (1999), argue that open economies with a sovereign currency can carry on monetary policy with substantial degrees of freedom. These authors extend the Post-Keynesian view on the autonomy of monetary authority in pegging interest rates to the open economy case. In this case, Central Banks set domestic interest rates, taking into account the international interest rate, the Balance of Payments, the stock of foreign currency reserves, and domestic variables as inflation and employment. As in uncovered interest parity, movements in the nominal exchange rate depend on the monetary policy, which regulates the difference between domestic and foreign interest rates12. For this reason, Smithin (2002) 9 When talking about small economies, we are referring to price-taker open economies. This implicit linkage between these two different ideas derives from a marginalist comprehension of the competition process, in which the atomism of participants of the market is the main characteristic. However, we can alternatively identify (as does Machado (2017)) that the difference between a price-taker or price-maker country (for a specific commodity) depends on a country’s capacity to respond to world demand for the commodity with lower costs than the other countries. This second definition is the one that interests to purposes of the present paper.
10 We avoid here the problem of overdetermination in the price-taker economy by introducing only one tradable commodity. In contrast, if this economy operated with more than one tradable commodity, the price of these commodities, given internationally, would allow for uniformization of the profit rate only by a fluke (Baldone; Steedman, 1999).
11 The case against the conventional determination of the exchange rate by (absolute) Purchasing Power Parity is a well-established empirical result in economics (see, for instance, Engel, 2014).
12 See Lavoie (2000b), and McCallum et al. (1996) 41 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY states that the real exchange rate is determined by monetary factors, subject to manipulation of public policy.
This has important implications in the framework proposed in this chapter because the real exchange rate turns out to be a key determinant of income distribution.
Hence, the price equation of commodity 1 determines the equilibrium profit rate given the money wage rate, the nominal exchange rate, international prices, and technical coefficients.
From equation 2.3 we can derive the expression for the equilibrium profit rate (r∗), as seen in the equation below.
(1 + r∗) = p1 −wl1 a1p1 + ampm [2.7] By substituting p1 per ep∗ 1, and rearranging, we obtain equation 2.8. Note that p∗ m stands for the international price of the input m.
(1 + r∗) = p∗ 1 −w e l1 a1p∗ 1 + amp∗ m [2.8] Capitalist competition leads to the convergence of profit rates between the two sectors. Hence, the profit rate in the non-tradable sector follows the rate established in the tradable sector.13 14 In other words, the classical process of profit rate convergence through competition takes place.15 As an illustration, consider that a nominal exchange rate devaluation rises prices in the tradable sector, leading to a persistently higher profit rate. In this case, there would be a tendency of migration of capitals from the non-tradable sector towards the tradable one. As a consequence, production in the non-tradable sector does not follow the pace of demand increases, which makes the market price for this good larger than its former normal price (that is, the normal price observed before the change in the equilibrium profit rate due to the devaluation). This discrepancy persists until the profit rate obtained with 13 A similar process was described in the previous section in Furtado (1963)‘s discussion on the effect of export prices of agricultural commodities over the land rent in non-exporting crops.
14 Likewise, the authors of the Scandinavian model state that profit rates in tradable and non-tradable sectors either equalize or maintain a normal relationship in the long term. In other words, even if the two profit rates differ, the difference is must be stable. The adjustment mechanism proposed in this model is the change in wages. Thus, money wages absorb any price increase or productivity growth in the tradable sector, keeping its profit rate constant. In the non-tradable sector, a constant profit rate implies that wage increases that exceed productivity growth are passed through into prices. Aukrust (1977); Edgren et al. (1969, 1973) 15 See, for instance, Garegnani (1990), Eatwell (1982) and Shaikh (2016, p. 259-326).
42 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY commodity 2 approaches the new equilibrium profit rate determined in the production of commodity 1.16 Equation 2.8 reveals an inverse relationship between the profit rate and the ratio between money wages and the nominal exchange rate.17 Moreover, as the international price of commodity 1 rises, the equilibrium profit rate increases. In the absence of other changes, the higher profit rate also implies an increase in the price of commodity 2, as convergence takes place.
2.3.2 Inflation To study inflation in this economy, we must check how prices change in time, outside of long-period positions. Let us assume that technical coefficients remain constant, but prices and distributive variables can change. So, we can rewrite the expressions for prices with a time index, as follows below.
pt 1 = (a1pt−1 1 + ampt−1 m )(1 + rt) + wt−1l1 [2.9] pt 2 = (a2pt−1 1 + aµpt−1 µ )(1 + rt−1) + wt−1l2 [2.10] pt 1 = etp∗,t 1 [2.11] Inputs cost follow its prices as at the beginning of the period of production. Likewise, the wage rate is fixed at the beginning of the production period, despite being paid in the end. In contrast, the price of output is set at the moment of selling, which corresponds to the end of the production period. This time gap between the emergence of cost shocks and price adjustments allows inflation to interact with income distribution.
The profit rate in the tradable sector adjusts to the current price (given from the combination of international prices and exchange rate) and the production costs defined in the previous period. It is important to distinguish the movement in the normal profit rate (which concerns equilibrium positions) from the transitory movements in the profit 16 Price adjustments do not necessarily require that migration of capitals occurs. The competition of potential entrants (and thus exits) across sectors can be conceived as sufficient for the transition of prices towards the new equilibrium (along the lines described by Sylos Labini, 1957). Besides, firm‘s willing to preserve their market-share would avoid not satisfying the effective demand. This firms tend to adjust production before any mismatch between supply and effective demand take place.
17 Similar results are shown in Dvoskin and Feldman (2020) and Machado (2017).
43 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY rate, described by what can be labeled as the nominal profit rate.18 Let us discuss this point in light of a short example. Consider a rise in the international price of commodity 1, which, given prices of other inputs and the money wage rate, leads to a higher normal profit rate. However, in the convergence process, the actual profit rate rises above the new normal profit rate before converging towards it. This overshooting happens because of the time lag between the immediate increase in the final price and the delayed increase in costs. Higher international prices for the tradable capital good instantaneously lead to a higher actual profit rate since selling price rose while costs remained fixed. This immediate effect is described in the equation 2.12, where πt 1 represents the rate of change in the price of 1 in the period t. Naturally, in line with the reasoning proposed, πt 1 is positive. The initial condition of equilibrium with stable prices held before the price increase coincides with the one represented by equation 2.7, and could be obtained by setting πt 1 = 0 in equation 2.12.
(1 + rt) = (1 + πt 1)pt−1 1 −wt−1l1 a1pt−1 1 + ampt−1 m [2.12] The same machine, however, consists in part of the production costs.
As a consequence, the total production cost relevant for commodities sold in the following period (t + 1) also increases after the rise in p1. This process reduces the actual profit rate for the following period, as shown in equation 2.13. This profit rate represents the new equilibrium, that is, the normal profit rate consistent with the change in relative prices.
(1 + rt+1) = pt−1 1 (1 + πt 1) −wt−1l1 a1pt−1 1 (1 + πt 1) + ampt−1 m [2.13] It is easy to check that rt+1 is smaller than rt, since the numerator remains the same while the denominator is larger than in period t. Even though, the level of the profit rate is still above the previous equilibrium.19 Thus, the convergence process to a new equilibrium profit rate includes an initial jump in the profit rate observed in this sector since the value of costs (which includes commodity 1) changes only one period after.
As argued above, competition leads to the convergence of the profit rate in the non-18 Here, the normal profit rate corresponds to the relevant value for what concerns income distribution since it takes the replacement costs of capital into account. What we called the actual profit rate could also be labeled as the nominal profit rate and is measured according to the historical costs of capital.
19 Indeed, if we define the initial profit rate as rt−1, we can calculate the difference rt+1 −rt−1, obtaining: 44 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY tradable sector towards the profit rate in the tradable sector. However, this process is not instantaneous. As a simplification, we assume that the convergence across sectors takes only one period. Hence, the profit rate in the price formation of the non-tradable sector is lagged in one period, differently from what happens in the tradable sector. In other words, the profit rate that affects prices of commodity 2 in t is the one obtained in the tradable sector in period t −1. It is reasonable to further assume that the profit rate in the non-tradable sector follows the normal profit rate determined in the tradable sector in the previous period. Introducing this assumption avoids that the transitory overshooting observed in the nominal profit rate of the tradable sector affects the profit rate in the other sector.
These equations allow us to describe different channels through which shocks in international prices and the exchange rate affect domestic prices. First, the domestic prices of tradable commodities are directly increased price abroad rise or the exchange rate devaluates. This direct channel is associated with competition and contestability since domestic goods present some degree of substitutability with goods produced abroad. Second, the increase in production costs due to higher prices of the tradable machine and imported inputs leads to higher prices in the non-tradable sector. Finally, a persistent increase in the profit rate (caused by a change in external conditions) leads to an increase in the prices of the non-tradable sector. This distributive channel builds on capitalist competition and the role of the tradable sector in determining distribution. An exchange rate devaluation triggers all these channels. Domestic factors are also taken into account since a rise in money wages leads to higher relative price for commodity 2.
2.3.3 Exchange rate devaluation As shown elsewhere,20 for given international prices, income distribution changes when the relation between nominal exchange rate and money wages change.
Thus, a devaluation of the exchange rate with respect to money wages leads to a higher rt+1 −rt−1 =pt−1 1 " (1 + πt 1) a1pt−1 1 (1 + πt 1) + ampt−1 m − 1 a1pt−1 1 + ampt−1 m # + −wt−1l1 " 1 a1pt−1 1 (1 + πt 1) + ampt−1 m − 1 a1pt−1 1 + ampt−1 m # Note that the term within the first pair of brackets is positive, and the term within the second pair of brackets is negative, which implies the expression must be positive. We are considering the economically relevant case in which both profit rates are larger than zero and smaller than the maximum profit rate.
20 See, for instance, Metcalfe and Steedman (1981); Steedman (1999) 45 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY equilibrium profit rate. This section discusses how a nominal devaluation affects costs and final prices and distribution before equilibrium. From now on, by exchange rate devaluation, we intend a nominal devaluation, while money wages and foreign prices remain constant unless stated differently.
In this context, a permanent devaluation of the exchange rate has an instantaneous impact on inflation through a proportional increase in the domestic price of commodity 1. Let us consider we depart from an initial situation of equilibrium corresponding to period t −1 and that the exchange rate shock occurs in period t, so that ˆ et > 0. The price of commodity 1 is immediately affected: pt 1 = (1 + ˆ et)pt−1 1 . As discussed above, the higher price of commodity 1 leads to a rise in the profit rate in tradable sector. However, in the case of an exchange rate devaluation, this effect is different from the case of a pure increase in the international price of 1. Now, the domestic price of imported inputs (m, µ) also raises proportionally so that the impact on the profit rate is smaller than in the previous case.
A devaluation also increases production costs in the non-tradable sector due to the rise in the price of the tradable machine and the price – measured in domestic currency – of the non-competing imports. Hence, the price of commodity 2 will include the higher production costs in the following period , as in the expression below. Finally, from the period t + 1 on, the price of commodity 2 rises also because the economy approaches a higher equilibrium profit rate associated with the exchange rate devaluation. Note that unchanged variables keep the initial time index (t −1), shedding light on the exchange rate devaluation effect. Note also that the term (1 + ˆ et) that would multiply the value of the circulating capital cancels with the same term that multiplies the circulating capital of commodity 1 in the denominator of the profit rate.
pt+1 2 = (a2pt−1 1 + aµpt−1 µ ) " pt−1 1 (1 + ˆ et) −wt−1l1 a1pt−1 1 + ampt−1 m # + wt−1l2 In the absence of further shocks, the economy stabilizes with a higher level of prices, higher profit rate, and lower real wages. The relative price of commodity 2 with respect to commodity 1 is lower after the adjustment process. We can derive this result in the following expressions. The first equation expresses the price of commodity one in period t + 1. Since we are focusing on a one-period shock, the price is the same as in period t.
pt+1 1 = pt−1 1 (1 + ˆ et) 46 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY The relative price of commodity 2 with respect to 1 before (that is, period t −1) and after adjusting to the devaluation (completed in t + 1) is expressed in the equations below.
As expected, a lower relative price of 2 is associated with a lower ratio between the money wage rate and the price of commodity 1 — which could be interpreted as the real wage in terms of commodity 1.
pt−1 2 pt−1 1 = (a2pt−1 1 + aµpt−1 µ ) 1 −wt−1 pt−1 1 l1 a1pt−1 1 + ampt−1 m + wt−1 pt−1 1 l2 [2.14] pt+1 2 pt+1 1 = [a2pt−1 1 + aµpt−1 µ ] 1 − wt−1 pt−1 1 (1+ˆ et)l1 a1pt−1 1 + ampt−1 m + wt−1 pt−1 1 (1 + ˆ et) l2 [2.15] If workers react to the rise in prices by asking for higher money wages, there is a further step to be discussed.
Workers react to the fall in real wage in terms of the consumption good, by bargaining for an increase in money wages.
The capacity of workers to obtain the desired wage rate depends on their bargaining power, determined by political and institutional features and by the conditions in the labor market (as the unemployment rate).
If, for instance, the rate of increase in money wages corresponds to a constant (and incomplete) proportion of the rate of the price increase in consumption good’s price, the purchasing power of workers will partially recover from the initial shock. In the following period, however, the money wage increase is passed through into the price of commodity 2. This process can repeat for several rounds of adjustment of prices and money wages. However, as long as the reaction of money wages to the price increase is incomplete, the money wage rate and the price level will converge to a stable equilibrium in the absence of further changes in the nominal exchange rate. In this case, workers still obtain a lower real wage than before the shock, but higher than if they had not reacted.21 Let us consider an extreme case in which the ratio w/p1 recovers completely from the initial shock. This situation arises when the accumulated rate of increase in money wages is equal to the rate of exchange rate devaluation after all adjustments. In this case, 21 Even though it is conceivable that money wages keep rising till the real wage rate becomes larger than the previous equilibrium, it is not expected to obtain this outcome in an inflationary process triggered by an external shock. If workers were dissatisfied with real wages even before the shocks, they would have raised wage demands in the first place, provoking a standard conflicting claims inflation.
47 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY we return to the original relative prices with a higher level of prices and a higher nominal exchange rate. It is easy to see in equation 2.15 that if the money wage rate increases at the same rate as the exchange rate devaluation, we will obtain the same relative prices seen in equation 2.14.
The analysis advanced in the last paragraphs describes the interaction between distributive variables and relative prices after a permanent shock in the exchange rate.
The shock can activate a temporary inflationary process as workers pressure for increases in money wages.
Subsequent devaluations can generate a persistent inflationary process in a wage-exchange rate spiral.
The relations described in detail here are the basis of the model presented in section 2.4.
2.3.4 Price of non-tradable commodity The non-tradable commodity (commodity 2) is the only consumption good. Workers consider its price in wage bargaining. The price of 2 varies with labor and input costs.
While the former depends on the wage rate and the labor coefficient, the latter is determined by technical coefficients, foreign prices, and the nominal exchange rate. As seen in equation 2.10, the price of 2 also varies with the profit rate determined in the tradable sector. Although temporary shocks in prices and costs may deviate profit rates across sectors, competition leads to convergence. It has been argued that this process of gravitation through capital mobility can happen under quite general conditions (Bellino and Serrano, 2018; Garegnani, 1990). Henceforth, we assume the profit rate obtained in the production of commodity 2 follows the equilibrium profit rate of 1 (that is, the one consistent with replacement costs of capital in the production of 1). Then, we can derive the dynamics of prices of commodity 2 by substituting the equilibrium profit rate (assumed to hold in period t −1), given in equation 2.7, in the price equation of 2.
Following this procedure, we obtain the expression that follows below.
pt 2 = (a2pt−1 1 + aµpt−1 µ ) pt−1 1 −wt−1l1 a1pt−1 1 + ampt−1 m !
+ wt−1l2 [2.16] Since prices of commodity 1 and inputs m and µ are determined in the international market, we can substitute each of these terms by its correspondent international price multiplied by the exchange rate. Thus, we obtain: 48 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY pt 2 = (a2et−1p∗,t−1 1 + aµet−1p∗,t−1 µ ) et−1p∗,t−1 1 −wt−1l1 a1et−1p∗,t−1 1 + amet−1p∗,t−1 m !
+ wt−1l2 [2.17] Rearranging and canceling e when possible gives us equation 2.18.
pt 2 = a2p∗,t−1 1 + aµp∗,t−1 µ a1p∗,t−1 1 + amp∗,t−1 m !
et−1p∗,t−1 1 + l2 −l1 a2p∗,t−1 1 + aµp∗,t−1 µ a1p∗,t−1 1 + amp∗,t−1 m !
wt−1 [2.18] For simplicity, we can rewrite 2.18 as in equation 2.19, collapsing the two terms within parenthesis in Φ1 and Φ2. Introducing the assumption that international prices are constant, we get that these terms must also be constant in time. The same holds for the international price of commodity 1. Thus, we can withdraw the time index of this variable.
Consequently, changes in the price of 2 are explained by changes in the exchange rate and in money wages.
pt 2 = Φ1et−1p∗ 1 + Φ2wt−1 [2.19] Note that Φ1 corresponds to the ratio between circulating capital inputs to the production of commodity 2 to the same variable for commodity 1.
In the case of a price-taker open economy with one tradable capital good, this ratio can be expressed in terms of international prices.
It is thus exogenous to the domestic economy.
If the production of 2 requires a larger value of capital input, then Φ1 > 1, which implies that the price of 2 responds more than proportionally to an increase in the exchange rate.
The sign of the Φ2 depends on the relative proportions of labor to circulating capital of each industry. This term is positive if production of commodity two has a higher ratio between labor input and circulating capital, given the prices p∗ 1, p∗ m, p∗ µ.22 We assume this term is positive. This assumption is justified because, in general, non-tradable industries tend to have a smaller capital-labor ratio than tradable industries.23 While tradable 22 It is easy to check this, since Φ2 > 0 if and only if l2 a2p∗ 1+aµp∗ µ > l1 a1p∗ 1+amp∗ m , which corresponds to the interpretation stated in the text.
23 These capital-labor ratios depend on the prices of 1, m and µ. Mathematically, Φ2 may be negative depending on these prices.
However, if this is the case, an increase in money wages would reduce the price of commodity 2. The positive impact in the form of additional labor costs would be more than offset by the decrease in the profit rate. We abstract from this possibility, assuming that Φ2 is positive. Although a negative Φ2 is theoretically possible, it seems less plausible once we acknowledge the stylized facts 49 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY industries include manufacturing and dynamic industries, non-tradable industries include services, among other sectors.24 In addition, it is expected that an increase in wages tends to increase prices in non-tradable industries (what can only happen if this term is positive).
Note that this term contains two effects of wages in the price of commodity 2.
The first one is direct, due to the labor input in the production of commodity 2. For this reason, a wage increase affects production costs in sector 2, which causes a price increase in the subsequent period.
The second effect relates to the negative impact of a wage increase on the profit rate in the production of commodity 1.
By means of this channel, a wage increase reduces the price of 2.
Altogether, the outcome depends on which channel is the strongest. Finally, we can discuss the case in which the combined effect is null, what happens when Φ2 = 0.
In this case, both industries present the same capital-labor ratio. For this reason, changes in distributive variables (either wage rate or profit rate) do not affect relative prices.25 Note that if the term discussed here is null, the price of 2 moves proportionally with the price of 1, maintaining a constant relation between them.
Let us further assume that international relative prices remain stable as well as the international price level. Consequently, the price of commodity 2 varies with changes in the exchange rate and money wages. In this case, we can describe in equation 2.20 the rate of change in the price of commodity two. This expression comes from the rate of change in the price of commodity 2 (as in equation 2.19). Since the international price level is assumed stable, π∗ 1 = 0, international inflation is not included in equation 2.20.
πt 2 = φt 1 ˆ et−1 + φt 2 ˆ wt−1 [2.20] mentioned in the text. Finally, since foreign prices are exogenous, we can also abstract from changes in the value or sign of Φ2.
24 Baumol (1967) stressed the tendency for increasing relative prices of services with respect to manufacturing goods due to the difficulty to reduce labor content in services. Besides, Aukrust (1977), Edgren et al. (1969), and Eatwell et al. (1974) used the fact that tradable industries tend to present higher productivity growth when analyzing inflation. However, the authors do not clearly distinguish if higher productivity results from reductions either in labor or capital input.
25 This is a known property arising from the analysis of price equations with uniform profit rate. As pointed by Sraffa (1960, p. 12), “[t]he key to the movement of relative prices consequent upon a change in the wage lies in the inequality of the proportion in which labour and means of production are employed in the various industries. It is clear that if the proportion were the same in all industries no price-changes could ensue, however great was the diversity of the commodity-composition of the means of production in different industries”. This property is preserved in a small open economy with a tradable capital good (Steedman, 1999).
50 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY where φt 1 = Φ1et−2p∗ 1 Φ1et−2p∗ 1 + Φ2wt−2 and φt 2 = 1 −φt 1.
Note that the weights denoted by φ1 and φ2 vary in time. Clearly, their value is kept positive and smaller than one.26 Since international prices are assumed constant, Φ1 and Φ2 remain constant. Thus, changes in φ1 and φ2 follow changes in the exchange rate and money wages.
Finally, equation 2.20 reveals that the impact of cost shocks on the price of commodity 2 is delayed in one period. This property emerges from the definition of production prices according to historical costs. Hence, there is a delay in the pass-through of costs into prices, which is a desirable property in a conflict inflation model, as discussed in the course of this chapter.
2.3.5 Real exchange rate Analyzing the interactions between inflation and the real exchange rate is important because monetary authorities possibly target a value for the real exchange rate. Such a policy choice can be associated with different objectives.
In some cases, it aims to stimulate exports in order to keep an equilibrium in the balance of payments (as in Vera 2010, Blecker 2011, Vernengo and Perry 2018).
However, this requires that the Marshall-Lerner condition holds, which may not occur. Alternatively, a targeted real exchange rate may be an instrument of structural change in support of industrialization (as in new-structuralist views such as Bresser-Pereira 2008).
Finally, this policy can derive from distributive objectives, once a devaluation of the real exchange rate implies a higher profit rate and a lower real wage rate, as in the model discussed here.27 The real exchange rate is usually defined as the nominal exchange rate adjusted by domestic and foreign price levels (Chinn 2018), as described in the equation 2.21.
et R = etp∗ t pt [2.21] 26 Note that if, in contrast with our assumptions, we had that l2 −l1 a2p∗ 1+aµp∗ µ a1p∗ 1+amp∗ m = 0, then we would obtain φ1 = 1 and φ2 = 0, which is compatible with the description of this particular case (see footnote 25). Hence, by assuming that l2 −l1 a2p∗ 1+aµp∗ µ a1p∗ 1+amp∗ m > 0 we also set that the weights φ1, φ2 must remain between zero and one.
27 See also (Dvoskin et al., 2018).
51 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY In this case, nominal devaluations only succeed in devaluating the real exchange rate if an increase in domestic prices does not completely offset them.
Both foreign and domestic price levels include prices of tradable and non-tradable commodities. Thus, the rate of change in real exchange rate depends on the change in the nominal exchange rate (ˆ e) and the rate of change of tradable and non-tradable prices in both domestic and foreign economies. Equation 2.22 describes a linear approximation of the change in the real exchange rate.28 Note that π stands for domestic inflation while π∗stands for foreign prices inflation.
ˆ et R = ˆ et + π∗ t −πt [2.22] Under the assumption of a price-taker open economy, tradable prices are determined abroad. Its value in domestic currency varies proportionally with the nominal exchange rate. The increase of tradable prices partially offsets the impact of a nominal devaluation on the real exchange rate. Therefore, the performance of non-tradable prices is crucial for the real exchange rate.
The price of the non-tradable commodity depends on the costs of the tradable commodity and the wage rate. As the first is already given, the second variable obtains more relevance in determining the real exchange rate. Thus, the process according to which workers react to nominal exchange rate devaluations by asking wage increase to protect real wages can offset the impact of the nominal devaluation on the real exchange rate.
If monetary authorities target a value for the real exchange rate, this leads to further nominal devaluations.
The economy can experience repeated nominal devaluations followed by money wage increases, generating chronic inflation in a money wages-exchange rate spiral.
Nevertheless, as both sides are unsuccessful in achieving their goals, they tend to moderate their claims (i.e., workers moderate money wage demands, while monetary authorities accept a partially smaller real exchange rate devaluation to avoid inflation).
If this happens, the economy converges to a stable position.
Given technical coefficients, the domestic level of prices is determined by international prices, the nominal exchange rate, and the wage rate. As a consequence, the real exchange rate will be proportional to the ratio of international prices as 28 This approximation is required because we are reasoning in terms of discrete-time. The actual rate of change of the real exchange rate is given by: ˆ et R = ˆ et+π∗ t −πt+ˆ etπ∗ t 1+πt .
52 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY measured in the domestic currency to money wages. That is the relevant real exchange rate for the analysis of distributive conflict. Indeed, other works using similar price equations in the open economy discuss the changes in specialization, relative prices, distribution provoked by changes in the ratio between exchange rate and money wages (Dvoskin and Feldman, 2020; Dvoskin et al., 2018; Machado, 2017). For this reason, let us introduce the variable q, whose definition is presented in equation 2.23 as an equivalent to the real exchange rate, useful for the analysis in the remaining of this chapter. Note that q does not correspond exactly to the real exchange rate as usually defined, but both are certainly affected by the same variables and move in the same direction. We show that in the appendix B. The value of q and its relation with income distribution are defined for given international price level and international relative prices.
qt = et wt [2.23] The rate of change of q is approximately expressed in the equation below.29 ˆ qt = ˆ e −ˆ wt [2.24] The value of q is inversely related to the real wage in terms of commodity 2. By using equation 2.19 as a description of the price of commodity 2, we get that the real wage in period t can be written as follows: wt R = wt Φ1et−1p∗ 1 + Φ2wt−1 [2.25] Thus, by dividing both the numerator and denominator of the fraction in the right-hand side by wt−1, we obtain equation 2.26.
29 As before, this is a linear approximation of the expression: ˆ qt = ˆ et −ˆ wt 1 + ˆ wt 53 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY wt R = 1 + ˆ wt Φ1qt−1p∗ 1 + Φ2 [2.26] The last equation finally gives us a relation between the real wage rate and q, given technical coefficients and international prices. Note also that a higher rate of growth of money wages implies a larger value for real wages. This outcome is consistent with historical cost pricing and the notion of conflict inflation itself, as emphasized by Tarling and Wilkinson (1985), Pivetti (1991), Stirati (2001), Serrano (2010).
54 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY 2.4 Quantities, Unemployment and Conflict In the last section, I have discussed the dynamics of the price level and relative prices following supply shocks. Although the section presented different possible results, the distributive conflict still needs proper treatment. We cannot account for conflict without introducing variables such as output and employment in the analysis. As discussed above, besides political and institutional factors, the unemployment rate affects affects the bargaining power of workers.
Hence, employment affects the rate of change of money wages and, thus, inflation. That makes the employment rate relevant to explain inflation and the distributive outcomes.
Therefore, in this section, we examine the interaction between inflation and conflict given the effect employment on wage bargaining.
We consider output, employment and growth to be determined by the level and growth of effective demand in line with the demand-led growth tradition. The view according to which output is determined by demand both in the short and long run can be combined with conflict inflation.
Indeed, the integration of these two analytical objects is frequent in Post-Keynesian economics.30 Since the potential output is endogenous in the long run, the evolution of productive capacity adjusts to the growth of demand. The perspective that output level and growth are determined by effective demand acknowledges the existence of spare capacity and labor force unemployment as recurrent conditions of capitalist economies. In this case, inflation, as caused by excess of aggregate demand, would be considered a less frequent phenomenon.
That explains why the coexistence of positive inflation rates with unemployment and spare capacity was often reclaimed as an evidence in favor of cost-inflation interpretations (Morlin, 2021). In addition, it is considered that companies operate with planned idle capacity and are therefore able to meet (to some extent) unexpected demand expansions (Ciccone, 2000, 2011; Steindl, 1952). Thus, the excess of aggregate demand can provide an explanation to temporary raises in prices only when it is greater than full capacity output.
Inflation due to excess of aggregate demand occurs when aggregate demand exceeds the potential output, so that a rise in prices adjusts the real value of aggregate demand to the short-term restriction imposed by potential output.
However, demand causes are generally not an explanation for inflation, which is thus mainly a cost-push phenomenon. According to the demand-led 30 Many of these analyses are summarized in section 2.2. However, as I argued, the explicit connection between distributive conflict and inflation is limited in open economy models.
55 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY growth perspective, aggregate supply follows the pace of demand growth. Thus, even though excess of demand may be inflationary in the short term, in general, this effect is not sustained in a longer period, when production and the capacity to produce goods and services grows in line with demand, dissipating the inflationary pressure (Serrano, 2019).
If supply can meet demand increases, the inflationary effect of a situation of excessive demand cannot last.
2.4.1 Output and employment After this introduction, we can set the base of the model for introducing quantities into our analysis of inflation in the open economy.
The model represents a small open economy whose output is composed of two commodities, one tradable another non-tradable, produced by means of domestic and non-competing imported inputs. As it is already clear in the paper, commodity 1 is an input and a tradable good.
Commodity 2 is non-tradable, and it is a consumption good (and it is not an input to any production). Moreover, we have two non-competing imported commodities, which are inputs to commodities one and two. The analysis solely accounts for circulating capital.
We also abstract from public expenditure among the determinants of aggregate demand.
Let us start by introducing the magnitudes Q1 and Q2, which correspond to the quantity produced of commodities 1 and 2. Then, the gross value of output is expressed in equation 2.27. On the left-hand side, we have the nominal output. On the right-hand side, we show output composition according to the sum of the value of production of commodities 1 and 2. In turn, the left-hand side by means of the price equations of both commodities.
Therefore, each circulating capital input and labor input appear in proportion to its technical coefficient in the production of the commodity that requires it, and to the quantity produced of the same commodity.
Q1p1 + Q2p2 = {Q1[(a1p1 + ampm)] + Q2[(a2p1 + aµpµ)]}(1 + r) + {[Q1l1 + Q2l2]}w [2.27] Rearranging equation 2.27, we can obtain equation 2.28. The first term within braces in equation 2.28 corresponds to the Intermediate Consumption in this economy, being composed of domestic produced commodity 1 and non-competing imports m and µ. The second term corresponds to the mass of profits. The third term gives us the wage bill.
56 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY Q1p1 + Q2p2 ={Q1[(a1p1 + ampm)] + Q2[(a2p1 + aµpµ)]}+ {Q1[(a1p1 + ampm)] + Q2[(a2p1 + aµpµ)]}r + {[Q1l1 + Q2l2]}w [2.28] By subtracting the Intermediate Consumption from the value of the gross product, we obtain the economy‘s output value, that is, the value-added product. It is straightforward that the output is equal to the sum of total profits and the wage bill. Therefore, the nominal output is equal to aggregate income.
Q1p1 + Q2p2−{Q1[(a1p1 + ampm)] + Q2[(a2p1 + aµpµ)]} = {Q1[(a1p1 + ampm)] + Q2[(a2p1 + aµpµ)]}r + {[Q1l1 + Q2l2]}w [2.29] In this case, the profit share (Π) is given by the ratio of total profits to aggregate income, as shown in equation 2.30.
Π = {Q1[(a1p1 + ampm)] + Q2[(a2p1 + aµpµ)]}r {Q1[(a1p1 + ampm)] + Q2[(a2p1 + aµpµ)]}r + {[Q1l1 + Q2l2]}w [2.30] In turn, the wage share (1 −Π) is described by the ratio of the wage bill to aggregate income, as in equation 2.31.
1 −Π = {[Q1l1 + Q2l2]}w {Q1[(a1p1 + ampm)] + Q2[(a2p1 + aµpµ)]}r + {[Q1l1 + Q2l2]}w [2.31] Even though these results may not seem particularly relevant, they clarify the connection between income distribution and the basic distributive variables introduced previously. As expected, equation 2.30 shows that the profit share increases with the profit rate.
In contrast, equation 2.31 reveals that the wage share increases with the money wage rate and decreases with the profit rate. Income distribution may also vary due to changes in the technical coefficients of production.
For instance, a higher requirement of capital in the production of either commodity 1 or 2 increases the profit share according to equation 2.30.
However, we abstract from changes in technical coefficients to focus on the conflict inflation.
Thus, we assume that changes in distribution are not large enough to affect the production techniques.
57 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY The aggregate supply in this economy is given by the total output (Y) plus imports (M). Note that imports are composed strictly of non-competing inputs (m, µ) required to produce commodities 1 and 2. Hence, imports of each input will be proportional to the production of each commodity. For this reason, imports are part of the intermediate consumption of this economy. Total uses of goods and services are given by imported Intermediate Consumption (ICM — which corresponds to M), domestically produced Intermediate Consumption (ICD), Consumption (C) and Exports (X).
Thus, the equilibrium condition for the goods market is described by the equation below. Note that all variables in this equation are considered in nominal terms.
Y + M = X + ICM + ICD + C [2.32] Each commodity‘s production has a counterpart in one or more components of demand.
As commodity 1 is exported and used as input to production, the total quantity of 1 must be equal to the sum of exports and domestic intermediate consumption, as in equation 2.33.
Since commodity 2 is produced exclusively for consumption, its quantity is totally determined by aggregate consumption — as in 2.34.
Finally, as stated above, imports correspond to the quantity of inputs imported, which are proportional to the production of commodities one and two. Thus, total imports, or imported intermediate consumption, is described by equation 2.35 We can formalize these lines as follows below: Q1p1 = X + ICD [2.33] Q2p2 = C [2.34] amQ1pm + aµQ2pµ = ICM [2.35] A few further details can be introduced.
Domestic intermediate consumption is proportional to the quantity produced of each commodity, as in the equation below.
58 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY a1Q1p1 + a2Q2p1 = ICD [2.36] Let us further assume that workers consume all their income. As a consequence, consumption is determined by the value of the wage bill, which corresponds to the wage rate times the total employment in the economy (denoted as L), as in 2.37.
C = wL [2.37] The total employment is described by equation 2.38, depending on the quantity of each commodity and on each labor coefficient.
L = l1Q1 + l2Q2 [2.38] Therefore, by substituting equation 2.37 in 2.38, we obtain an expression for the aggregate consumption, as in 2.39. Consumption is increasing in the wage rate and with the employment generated in each industry.
C = w(l1Q1 + l2Q2) [2.39] In the equilibrium of goods market, C = p2Q2. Thus, Q2 = C p2 = wRl1 1 −wRl2 Q1 [2.40] Here, wR denotes the real wage rate, that is the wage rate divided by the price of commodity 2. Therefore, equation 2.40 determines Q2 as a function of Q1. Substituting this result in equation 2.36, we obtain equation 2.41. This expression determines domestic intermediate consumption as a function of Q1. Note that both the quantity produced of commodity 2 and the ICD depend on the quantity of commodity 1 and the real wage.
59 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY ICD p1 = a1Q1 + a2 wRl1 1 −wRl2 Q1 = a1 + a2 wRl1 1 −wRl2 Q1 [2.41] Plugging this last result in equation 2.33, we obtain 2.42.
Q1 = X p1 + a1 + a2 wRl1 1 −wRl2 Q1 [2.42] Denoting the real value of exports by χ and rearranging the former expression, we obtain: Q1 = χ h 1 −a1 −a2 wRl1 1−wRl2 i [2.43] Equation 2.43 shows the quantity produced of commodity 1 is directly determined by the level of exports added of the multiplier effects generated by workers‘ consumption and intermediate consumption of production of both commodities. Given the technical coefficients, the value of Q1 varies increasingly with changes in the level of exports and with changes in the real wage.
By assumption, exports are the only autonomous expenditure in this small open economy. Thus, from the level of exports and their repercussions in consumption, we can get the aggregate production level.
Using these results, we can obtain the level of employment associated with this equilibrium. If we substitute 2.43 and 2.40 in the expression for the level of employment (equation 2.38) and rearranging, we obtain equation 2.44.31 This equation expresses total employment as determined by the demand for labor in production, which, in turn, is determined by the level of aggregate demand.
L = l1 1 −a1 + wR((−l2)(1 −a1) −a2l1) χ [2.44] The denominator of the term within braces is positive.
Otherwise either (i) the economy is not viable, that is, technical conditions are not sufficient for the self-reproduction of the economic system or; (ii) real wage is larger than the maximum 31 The complete mathematical steps are in appendix A.
60 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY real wage admissible for this economy.32 Note that L varies positively with the real wage rate, and this variation increases with the real wage. As shown in equation 2.45, the derivative of L with respect to wR is given in equation 2.45, which is positive and increasing in wR.
dL dwR = l1[(1 −a1)l2 + a2l1]χ {1 −a1 −wR[(1 −a1)l2 + a2l1]}2 [2.45] We can express the term within braces in equation 2.44 as a non-linear function of the real wage (denoted by Λ(wR)), with given technical coefficients. Hence, we can rewrite the expression for the employment level as the product of this function and the level of exports, as follows in equation 2.46.
L = Λ(wR)χ [2.46] Note that Λ varies positively and increasingly with wR.
Once we set the employment determination, we can define the unemployment rate.
Unemployment rate (u) is given by u = 1 −L N, where N stands for to the total labor force.
Note that N entails an upper constrain to employment and, thus, to output. Hence, we must assume that L < N. According to this assumption, effective demand is not large enough to lead the economy towards full employment, which corresponds to the general condition of capitalist economies.
By substituting equation 2.46 in the definition of the unemployment rate, we get equation 2.47. Thereby, we can express the unemployment rate as a function of real wages and exports.
u = 1 − Λ(wR)χ N [2.47] The demand for labor is expressed in equation 2.46. From this equation, we can obtain the following expression for the rate of growth of labor demand (denoted by gL).
32 These results are shown in appendix A.
61 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY gL = \ Λ w p2 + gX + \ Λ w p2 gX [2.48] Which can be approximated linearly by: gL = \ Λ w p2 + gX [2.49] Naturally, for a constant real wage, the growth of labor demand is equal to the growth of exports (gL = gX).
2.4.2 Inflation, wages and exchange rate Now, we are able to establish a relation between employment and money wage growth.
This is approached in a similar way as in Rowthorn’s (1977), and later employed by many authors (Dalziel, 1990, Cassetti, 2002, Hein and Stockhammer, 2010, Lavoie et al., 2002).
In most of these models, workers target an income share expressed either in terms of the wage share or the profit share. For a constant labor productivity and composition of output, there is a direct correspondence between the wage share and real wage rate.33 Hence, we can address this issue by introducing a target for the real wage (wT). Real wage is directly connected with workers interest and is not subject to the effects of changes in the output composition.34 Real wage is inversely related to the normal profit rate.
In a Classical-Keynesian perspective, real wage and normal profit rate are the meaningful variables for discussing distribution. Therefore, we introduce a target for the real wage (wT) so that workers will push for wage increases according to the difference between the actual real wage and this target.
The rate of growth of money wages ( ˆ w) depends on the evolution of the price of commodity 2 (π2). Changes in the cost of living affect wage negotiations, and often 33 However, in a model in which workers target an income share, productivity growth would not affect the distributive conflict (i.e., the degree of incompatibility between income claims of social classes, which, in Kaleckian models, can be measured by the difference between workers and capitalists targeted income shares). We expect, however, that a an increase in the growth rate of productivity relaxes the distributive conflict. Marglin and Schor (1991) and Glyn et al. (1991) exemplify this point by discussing the end of the Golden Age of capitalism. The authors argue that the slowdown in productivity growth exacerbated the distributive conflict in the end of the Golden Age, accelerating the rates of inflation in advanced capitalist economies.
34 An increase in the output of the commodity with a larger labor input coefficient increases the wage-share, for a constant real wage rate. See equation 2.31.
62 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY indexation rules are included in labor agreements. However, indexation is assumed to be incomplete. α1 is thus positive and smaller than 1. Real wages also vary according to workers’ aspiration gap, i.e., the difference between their targeted real wage and the actual real wage. α2 expresses the sensibility of wage increases to the aspiration gap. α2 therefore depends on institutional and political factors, as trade union membership, the structure of unions, labor legislation, and social rights. Considering these effects, we can write equation 2.50.
ˆ wt = α1πt−1 2 + α2(wT R −wt−1 R ) [2.50] Thus, the rate of change of money wages is a function of the rate of change in the price of commodity two, workers’ aspiration gap, and on α1 and α2 (assumed fixed). Equation 2.50 takes workers’ target as given. Later on, we will allow for changes in workers’ target, according to changes in the employment rate.
Let us introduce a rule of exchange rate devaluation. We assume that the monetary authorities follow a rule for changing the nominal exchange to pursue a target real exchange rate (qT). Since there is a direct association between the real exchange rate and income distribution, this target can be associated with distributive purposes. Therefore, the rate of change in the exchange rate can be given by equation 2.51. This rule was introduced by Blecker (1989a, 2011).
ˆ et = λ(qT −qt−1) [2.51] As we have shown, q and real wage rate are inversely related. Hence, we can express equation 2.51 in terms of a target for the real wage rate. In this case, the equation can be written as follows below. The target for the real wage corresponding to qT is denoted as wq R.
ˆ et = λ(wt−1 R −wq R) [2.52] Collecting the equations for inflation, money wages, and exchange rate, we obtain the following system: 63 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY ˆ wt = α1πt−1 2 + α2(wT R −wt−1 R ) [2.53] πt 2 = φt 1 ˆ et−1 + φt 2 ˆ wt−1 [2.54] ˆ et = λ(wt−1 R −wq R) [2.55] We can rewrite the system in terms of real variables, restricting it to two equations describing real wages and real exchange rate.
The rate of change in real wages is defined as the difference between change in money wages and prices of commodity two: ˆ wRt = ˆ wt −ˆ p2t. Combining this definition with equation 2.50 gives us equation 2.56.
ˆ wRt = −πt 2 + α1πt−1 2 + α2(wT R −wt−1 R ) [2.56] The rate of change in the variable q is described in equation 2.24, rewritten below.
ˆ qt = ˆ et −ˆ wt [2.57] By substituting the expression for the rate of change in the exchange rate given in equation 2.55, we obtain equation 2.58. Together, equations 2.56 and 2.58 form the basis of the dynamic system.
ˆ qt = λ(wt−1 R −wq R) −ˆ wt [2.58] 2.4.3 Equilibrium The economy achieves equilibrium when both real wages and the real exchange rate achieve their equilibrium points. In other words, when both equations 2.56 and 2.58 are 64 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY equal to zero. Equilibrium positions are denoted by the superscript .
From equation 2.57, we obtain that in equilibrium ˆ w = ˆ e, since the level of q must remain stable. Besides, a constant real wage requires that ˆ w = π2. By setting ˆ wR = 0 in equation 2.56, we can obtain the equilibrium value for the rate of change in prices of commodity two, expressed below. That expression also gives us the results for the rate of change of the nominal exchange rate and the rate of increase in money wages.
In the absence of changes in international prices, the rate of change of prices of sectors 1 and 2 is the same, since the nominal exchange rate and money wages rise at the same rate (in the equilibrium). As a consequence, the inflation rate is given by the following expression — that is, the same expression as the one for the rate of growth of money wages, rate of change in the exchange rate, and rate of change in prices in each industry (1,2). In other words, in the equilibrium, π2 = π = ˆ w. The equilibrium inflation rate is described by the equation below.
π∗∗= λα2(wT R −wq R) α2 + λ(1 −α1) [2.59] We can rewrite it as follows.
π∗∗= wT R −wq R 1 λ + (1−α1) α2 [2.60] In turn, equilibrium real wage rate is described by the equation below.
w∗∗ R = α2wT R + λ(1 −α1)wq R α2 + λ(1 −α1) [2.61] Similarly to other conflicting claim models, equilibrium real wage is an weighted average of the two different targeted real wage. That equilibrium value is sensitive to changes in the parameters α1, α2, and λ. For instance, if wage indexation parameter (α1) increases, then equilibrium real wage gets closer to workers’ target. The same result is obtained if the parameter related to workers’ bargaining power increases. In turn, if monetary authorities accelerate the speed of adjustment of the nominal exchange rate (i.e., λ increases), then the equilibrium real exchange rate will come closer to its targeted value wq R.
65 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY If inflation emerges from wage claims (a greater wT R) or the strengthening of workers position (as a greater α2), a higher inflation rate is associated with a larger value for the equilibrium real wage rate. This outcome is usually obtained in conflict inflation models.
Finally, a larger real wage rate implies a loss in capitalist’s share in distribution. It follows, thus, that a higher equilibrium real wage and a a lower equilibrium q are associated with a lower profit rate and lower profit-share.
Equilibrium with endogenous worker’s claim Let us now consider the case in which workers’ income claim is affected by the labor market. Previous conflicting claim models have set workers’ wage target as dependent of changes in the rate of unemployment (Cassetti, 2002; Lavoie, 2014). As shown in section 2.4.1, persistent changes in unemployment rate depend on the difference between the growth rate of exports gX and the growth rate of labor force gL. Hence, we can define that workers’ target for the real wage depends on an autonomous component, reflecting institutional and political factors, and a second term that expresses the effect of changes in the unemployment rate.
wT R = θ0 + θ1(gX −gL) [2.62] Now we can plug equation 2.62 in the equilibrium results obtained above.
Equilibrium real wage rate is now given by: w∗∗ R = α2[θ0 + θ1(gX −gL)] + λ(1 −α1)wq R α2 + λ(1 −α1) [2.63] The result for the inflation rate is given by: π∗∗= [θ0 + θ1(gX −gL)] −wq R 1 λ + (1−α1) α2 [2.64] Therefore, a situation of increasing unemployment will reduce both the equilibrium real wage and inflation. In contrast, falling unemployment implies an increase in both the equilibrium real wage and inflation.
66 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY Closure in the aspiration gap Rowthorn (1977, p. 217) defines the aspiration gap as the difference between the target share in income and the actual income shares resulting from the processes of wage bargaining and inflation.
In the model presented here, workers’ aspiration gap corresponds to the difference between workers’ targeted real wage (wT R) and the equilibrium real wage (w∗∗ R ).
Capitalists’ aspiration gap can be described by the difference between the targeted real wage coming from monetary authorities’ target for the ratio between nominal exchange rate and money wages (wq R) and the equilibrium real wage (w∗∗ R ). A positive aspiration gap is the fundamental cause of conflicting claims inflation (Lavoie, 2014; Rowthorn, 1977). If both sides give up on claiming for a real wage different than the one resulting from the process of bargaining and inflation, the aspiration gap goes to zero. In that case, inflation rate is also zero in the absence of cost shocks. This theoretical scenario allow us to describe the long period position associated with the distributive equilibrium.
We do not expect the aspiration gap to persist over time. Hence, conflict inflation is treated as a short period phenomenon, tending to dissipate in the absence of shocks in costs or employment (which affect the workers’ target).35 The equilibrium associated with the closure of the aspiration gaps is described by the real wage expressed in equation 2.63 and stability in the price level. Thus, we can obtain the equilibrium ratio between nominal exchange rate and money wages (q).
As argued before, the result for q is inversely related to the real wage rate (see section 2.3.5). Hence, we obtain the equilibrium for the ratio between the exchange rate and money wages from the relation between the real exchange rate and the real wage, as defined previously in equation 2.26.36 35 Naturally this is a theoretical possibility not accomplished in the real world as different sources of disequilibrium continuously emerge.
Still, it is worth exploring this scenario to fully understand the outcomes of the model.
36 Note that q must be positive. We can show that in a few steps. The left hand side of equation 2.65 is larger than zero as long as 1 Φ1p∗ 1 1 w∗∗ R −Φ2 > 0.
Since Φ1p∗ 1 > 0, we can rewrite this condition as follows: 1 w∗∗ R −Φ2 > 0 That, in turn, implies that p2 −Φ2w w > 0 which is trivially truth since production of 2 requires that its price is larger than the direct and indirect labor 67 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY q∗∗= 1 Φ1p∗ 1w∗∗ R − Φ2 Φ1p∗ 1 [2.65] Substituting the equilibrium real wage in the former equation, we obtain the equation below.
q∗∗= α2 + λ(1 −α1) Φ1p∗ 1{α2[θ0 + θ1(gX −gL)] + λ(1 −α1)wq R} − Φ2 Φ1p∗ 1 [2.66] As expected, whatever increases the real wage rate reduces the equilibrium ratio between the exchange rate and money wages.
An increase in workers’ bargaining power (α2) and in wage indexation (α1) reduce equilibrium q by bringing the equilibrium real wage closer to workers’ targeted value [θ0 + θ1(gX −gL)]. In turn, an increase in λ brings the equilibrium real wage closer to wq R, thereby increasing the equilibrium ratio between nominal exchange rate to money wages.
Finally, an increasing rate of unemployment increases workers’ target, reducing the equilibrium q.
costs (p2 > Φ2w). The reader can refer to section 2.3.4, where we define Φ2.
68 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY 2.5 Conflict inflation and distribution in the open economy This chapter contributes to the literature of conflict inflation in the open economy by setting up a model of inflation based on the classical approach to distribution, as in its applications to the case of a small open economy. By these means, we explicitly discuss the distributive impact of the exchange rate and international prices. Once international prices directly determine part of domestic prices, they also affect distribution and conflict. Although this fact is generally acknowledged, its consequences have not been fully explored in inflation models. This section compares the different approaches to conflict inflation in the open economy with the one proposed in this chapter.
In general, open economy conflict inflation models either assume a given level of the mark-up or introduce an effect of the real exchange rate over mark-ups. In the first case, mark-up (or firm’s targeted mark-up) is exogenous with respect to prices, money wages, exchange rate, and interest rate, being solely determined by the degree of monopoly. To make this mark-up compatible with the presence of international competition, these models assume the law of one price does not hold. In other words, the domestic price of a certain commodity is independent of the foreign price for the same commodity.
Hence, domestic prices depend on production costs, wages, and mark-up, all determined according to the conditions of the domestic economy. International prices matter only as they explain the cost of inputs — or the price of imported consumption goods, as in Cassetti (2002) and Vera (2010). This framework is sufficient to generate a conflict between real wages and either the real exchange rate or international prices of imported commodities.
Such a conflict can cause inflation in open economies, as discussed by Rowthorn (1977) and Cassetti (2002).
Stirati (2001) also introduces imported inputs as a source of inflation and conflict. Although the author rejects the mark-up pricing framework existing in the other works, she assumes the profit rate is unaffected by foreign prices and the exchange rate.
Another strand of the literature considers the mark-up cannot be independent of the exchange rate and foreign prices. Two different perspectives are developed along these lines. The first considers the mark-up to be directly affected by the real exchange rate.
This approach, labeled as flexible mark-up, is proposed in Blecker (1989b) and later used by Vera (2010).
Thereby, Blecker (1989b) introduces a capitalists’ distributive claim associated with exchange rate devaluation, because it loosens the pressure of foreign competition.
Nevertheless, it raises additional problems because, in this case, the mark-up is contemporaneously determined by the degree of monopoly and by the real 69 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY exchange rate. Thus, the role of each variable in the explanation of distribution is not clearly defined.
The second view departs from the Kaleckian model of inflation for the open economy. It introduces the assumption that the real exchange rate affects firms’ and workers’ income claims by impacting their targeted distributive variables.
In other words, the real exchange rate increases firms’ targeted mark-up (or reduces the firm’s targeted wage share), and in some cases, it also reduces worker’s targeted mark-up (or increases its targeted wage share).
This approach is discussed by Lavoie (2014) and further developed by Bastian and Setterfield (2020).
Once again, the ultimate determinant of the mark-up becomes blurry.
These difficulties add up to other shortcomings of mark-up pricing, especially concerning the role of the degree of monopoly as a determinant of the profit rate (criticized along this thesis).
Altogether, these shortcomings suggest that conflict inflation theory should build on an alternative explanation of distribution.
Such an alternative can be found in the classical surplus approach, extended to the open economy context.
In addition to avoiding the mentioned shortcomings, the distributive closure presented in this chapter allows us to consider that international prices directly determines tradable prices in open economies. Moreover, the profit rate is positively affected by the exchange rate and the international price of the tradable commodity.
This property is analogous to outcomes of other models, in the form of the increase in the mark-up in the flexible mark-up approach and in the case in which the real exchange rate affects income claims.
In section 2.2, we discussed different models of inflation and conflicting claims in the open economy. Neither of these models explicitly includes the effect of international prices on the domestic price of tradable commodities.
Indeed, most of them acknowledge the law of one price is disregarded.
Finally, they reason in terms of mark-up pricing, and firm’s targeted mark-up, whose value would be affected by the real exchange rate.
In contrast, the Scandinavian model of inflation (Aukrust, 1977; Edgren et al., 1969) explicitly introduces the role of tradable prices. However, this model does not formally include the effect of distributive conflict on wage bargaining and inflation. In fact, the Scandinavian model assumed that wages change according to a rule that offsets the positive impact of international inflation (and of increases in productivity) on the profitability of tradable industries.
Other authors focused on the impact of chronic deficits in the Balance of Payments as 70 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY a cause of inflation. Vernengo and Perry (2018) consider a simple structure but include a target for the real exchange rate, pursued by monetary authorities according to Balance of Payments concerns. Thus, the authors can model a money wage-exchange rate spiral, describing a typical condition of Latin American economies. However, Vernengo and Perry (2018), as well as Charles and Marie (2016), assume wages are indexed by a fixed parameter so that they do not explicitly account for wage claims in the model. For this reason, the relationship between worker’s bargaining power and the conditions in the labor market is left aside. In the model presented by Charles and Marie (2016), one unstable equilibrium among multiple equilibria may lead to hyperinflation. Vera (2010) also discusses the Balance of Payments and its relation with exchange rate devaluations and conflict inflation.
This chapter introduces a target for the ratio of the nominal exchange rate to money wages owing to distributive concerns of the monetary authority. In this way, we do not explicitly cope with the Balance of Payments. However, future research can account for the Balance of Payments as an additional source of inflation.
In order to avoid the shortcomings found in the literature, this chapter seeks an alternative approach to model conflict inflation in an open economy. For this reason, we introduced a framework based on classical production prices for the open economy case.
In section 2.3, we showed how the wages, profit rate, exchange rate interact according to these production prices. This setup also laid the foundation for deriving the expressions for the rate of inflation and obtaining the model’s equilibrium — according to the relation between prices and exchange rate, on the one hand, with profit rate and real wage rate, on the other.
In contrast with the Kaleckian approaches, here, capitalists cannot target a desired income share. In fact, they set prices constrained by competition, and thereby the profit rate is regulated by the interaction between money wages, nominal exchange rate, and international prices.
The real outcome of distribution is known only after the inflationary process. Moreover, we explicitly account for tradable prices in a conflict inflation model, contrasting with the literature reviewed. A consequence emerging from this feature is the raise of profit rate after an exchange rate devaluation. Competition generates a tendency for a rise in prices of non-tradable commodities, allowing for the adjustment towards a higher profit rate in the tradable commodity production. That consists of an additional cause of inflation related to an income claim of capitalists of non-tradable sectors driven by competition.
Thereby, we account for an inflationary 71 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY pressure neglected in the literature. The model also account for other causes of inflation related to open economy concerns. First, the direct impact of tradable prices. As argued before, this channel is often disregarded. Second, the channel of imported inputs, whose costs transmitted to final prices and interact with distribution.
Several results of our analysis are in line with expected outcomes of conflict inflation models. Stronger bargaining power of workers, due to the fall in unemployment or larger institutional power (i.e., a higher θ0), is associated with higher real wages and a higher rate of inflation. It is also associated with a lower value of q, and a lower real exchange rate.
Indeed, Cherkasky and Abeles (2019) show that the weakening of worker’s bargaining power, in the form of weaker trade unions and the retreat of labor market institutions, reduce the pass-through after exchange rate devaluations. Hence, the authors argue that the lower inflationary impact of an exchange rate devaluation is caused by the lower wage resistance given the weaker institutional support of workers.
In our model, this can be shown by imposing a lower value of θ0, which leads to a lower rate of inflation, lower real wage, and higher real exchange rate.
Finally, we contend this chapter brings new insights into the distributive closures within the classical surplus approach. Developments in the surplus approach rejected any mechanistic approach to the theory of distribution and emphasized the role of institutions and political factors (Pivetti, 1991; Sraffa, 1960; Stirati, 1994). Insights from this approach were applied in the discussion of inflation by Pivetti (1991); Serrano (2010); Stirati (2001). However, the price-taker open economy analytical framework had not been employed in an analysis of inflation till now.
Within the Classical surplus approach, the monetary theory of distribution stressed the role of nominal variables in determining income distribution. Pivetti, following a hint of Sraffa, takes the nominal interest rate as the exogenous variable in the system of prices equations determining relative prices and distribution. However, the actual outcome (i.e., real profit rate and real wages) of distribution can only be known after wage bargaining and the inflationary process take place. As argued by Pivetti (1991, p. 52), “ given a policy-determined nominal interest rate, competition among firms within each industry should tend to cause the rate of profit to move in sympathy with the real rate of interest, rather than with the nominal one, because it is the former which constitutes the actual price for the use of capital in production or its opportunity cost”. In other words, since the price of labor and capital goods has changed at the end of each period, the relevant value of the profit rate must be evaluated with respect to these new prices (and new 72 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY production costs) rather than the initial ones.
Based on this point, Serrano (1993) emphasized that even though the monetary authority can peg the nominal interest rate of the economy, it cannot fully determine distribution since workers can react to increases in the nominal interest rate preserving their income share. Thus, the real interest rate remains to be determined by the nominal interest rate and wage bargaining (in a closed economy).37 The results presented in this chapter reveal an analogy with the monetary theory of distribution.
Nevertheless, rather than the nominal interest rate, distribution is regulated by the nominal exchange rate, international prices, and money wage rate. In line with Post-Keynesian developments, we consider the nominal exchange rate to be strongly influenced by monetary policy within certain constraints. Our price equations explicitly show the exchange rate is a relevant distributive variable in this small open economy context. Hence, there is a relation between the nominal exchange rate and the ratio between prices and money wages. We try to show, however, that as wages can react to an exchange rate depreciation, the outcome depends on the reaction of wage bargaining and the inflationary process.
As a consequence, the real exchange rate, known at the end of the period, is relevant to the outcome in distribution, rather than the nominal exchange rate.
From this perspective, the chapter contributes to this analysis of distribution in the open economy context, considering the distributive variables determining income distribution, wage bargaining, and the distributive conflict.
37 When discussing the monetary theory of distribution, Serrano (1993, p.
123) argues that “[Pivetti] has convincingly shown that (again, in a fiat money context) the nominal rate of interest is determined independently from the ’real side’ of the economy. This complete ’autonomy’ of the nominal rate of interest means that this variable might not always be compatible with the bargaining position of the workers or the possibilities of the technology of the economy”. Garegnani had anticipated the same point: “Indeed, it seems reasonable to suppose that, as a result of competition in product markets, the average rate of profit and the average rate of interest on long-term loans will tend, over a sufficiently long period of time, to move in step with one another. If, then, the rate of interest depends on the policy of the monetary authorities, both the long-term movement of the average rate of profit and, through the relation just mentioned, that of real wages are explained by that policy. This does not entail maintaining afresh that the wage bargain has no power to change real wages: the policy of the monetary authorities is not conducted in a vacuum and the movement of prices and of the money wages determined in the wage bargain will be amongst the most important considerations in the formulation of that policy” (Garegnani, 1979, p. 81).
73 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY 2.6 Final Remarks We build on open economy price equations to develop a conflict inflation model following a Classical-Keynesian perspective. The extension of price equations to the open economy defined the relation among money wages, exchange rate, international prices, and the profit rate. Based on this setup, we further derived expressions for the inflation rate, the real wage rate, and its relation with the real exchange rate. An increase in international prices or an exchange rate devaluation raises production costs through more expensive inputs. That pushes prices up, possibly leading to a persistent inflation as workers respond to protect their purchasing power.
Since the model includes tradable prices, these shocks also have a direct impact on prices and distribution. All else being equal, the tradable sector’s profit rate increases after an exchange rate devaluation.
Competition pushes prices of non-tradable commodities due to the tendency of convergence of profit rates. That describes an additional cause of inflation included in the model.
Finally, the chapter brings new insights into the distributive closures within the Classical-Keynesian approach. Contributions in this approach highlighted the interest rate as an independent monetary variable that regulates income distribution. However, the real interest rate is the meaningful variable, being known after inflation occurs.
Hence, the social conflict causing inflation also shapes distribution (Pivetti, 1991; Serrano, 1993; Stirati, 2001).
Our model provides analogous results for the open economy setup. We conclude that the nominal exchange rate becomes a central variable for distribution in small open economies. However, its ultimate impact on distribution is conditional on the reactions of workers through bargaining and inflation.
74 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY Appendix A Mathematical Appendix Determination of the level of employment This appendix presents omitted mathematical passages to obtain the level of employment, as in equation 2.4.
We depart from the expression from the level of employment, presented in equation 2.38 and rewritten below.
L = l1Q1 + l2Q2 By substituting 2.43 and 2.40 in the expression for the level of employment and rearranging, we obtain the following equation: L = l1 X h 1 −a1 −a2 wRl1 1−wRl2 i + l2 wRl1 1 −wRl2 X h 1 −a1 −a2 wRl1 1−wRl2 i [2.67] Rearranging gives us the following expression L = l1 + l2 wRl1 1 −wRl2 X h 1 −a1 −a2 wRl1 1−wRl2 i [2.68] We can rewrite as follows in equation 2.69. In this case, we are able to distinguish the impact of exports and of real wages on employment.
L = h l1 + l2 wRl1 1−wRl2 i h 1 −a1 −a2 wRl1 1−wRl2 i X [2.69] Thus, L = l1(1−wRl2)+l1l2wR (1−wRl2) 1−wRl2−a1+a1wRl2−a2wRl1 (1−wRl2) X = l1 1 −a1 + wR(a1l2 −a2l1 −l2) X [2.70] Finally, L can be written as 75 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY L = l1 1 −a1 + wR((−l2)(1 −a1) −a2l1) X [2.71] This is the expression presented in the chapter as equation 2.44. The term within braces can be understood as the employment multiplier compatible with the input-output structure of the model and the assumptions — in particular the assumption that workers spend all their income in commodity 2.
As expected, the multiplier is positive. We can show that this term is positive by showing that the denominator of this fraction is positive. Note that all the technical coefficients (ai, li) are positive.
Let us assume, ab absurdo, that the denominator of the last equation is equal to zero.
Viability of this economy requires that a1 < 1.38 Then, using the fact that wR = w/p2, we would have: 1 −a1 + w p2 ((−l2)(1 −a1) −a2l1) = 0 In this case, w = p2 1 −a1 l2(1 −a1) + a2l1 If we divide both sides per p1, we obtain: w p1 = p2 p1 1 −a1 l2(1 −a1) + a2l1 By setting p1 = 1, w p2 = 1 −a1 l2(1 −a1) + a2l1 [2.72] However, this value for the real wage is larger than the maximum value for the real wage admissible in this economy. Indeed, this expression coincides with the maximum real wage in terms of 2 that would be observed if there were no imported inputs. As this value for the real wage is not possible in the open economy case, we know that wR is smaller than the one expressed above, what implies that the denominator of the fraction studied is larger than zero.
The maximum real wage compatible with the system of price equations can be obtained by setting r = 0 in the system constituted by equations 2.3 and 2.4. Setting a 38 See, for instance, (Kurz and Salvadori, 1997, p. 64-65).
76 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY zero profit rate in equation 2.3 and rearranging gives us: w = p1 −a1p1 −ampm l1 [2.73] Now, we can find an expression for the price of commodity 2 compatible with this value for the maximum wage rate obtain in production of commodity 1. By substituting the last result in the price equation for 2 after setting also r = 0, we obtain: p2 = a2p1 + aµpµ + p1 −a1p1 −ampm l1 [2.74] Rearranging, we get: p2 = l1(a2p1 + aµpµ) + l2[p1(1 −a1) −ampm] l1 [2.75] Dividing the right-hand side of equation 2.73 per the right-hand side of equation 2.75 gives us an expression for the maximum real wage rate in terms of commodity 2.
w p2 = p1 −a1p1 −ampm l1(a2p1 + aµpµ) + l2[p1(1 −a1) −ampm] [2.76] As before, we set p1 = 1, obtaining an expression comparable to the real wage obtain previously in equation 2.72.
After this procedure, the price of each commodity correspond to the relative price with respect to commodity one.
w p2 = 1 −a1 −ampm l1(a2 + aµpµ) + l2[1 −a1 −ampm] [2.77] It is easy to check that this value for the maximum real wage is smaller than the one established in equation 2.72, whenever (1 −a1)aµpµ + a2ampm > 0. This condition is verified when at least in one sector production requires imported inputs. Indeed, this result can be immediately interpreted from an economic viewpoint, since the maximum real wage must be smaller in the second case, because production requires a larger quantity of inputs.
77 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY B Real exchange rate The purpose of this note is to show the relation between the real exchange rate and the distributive variable q defined in equation 2.23. The discussion related to this appendix is presented in the subsection 2.3.5. The most common definition of real exchange rate corresponds to the nominal exchange rate adjusted by the price levels. Let us consider here our price-taker open economy and another economy labeled as the Rest of the World, whose variables will be signed with an ∗.
For simplicity, we consider commodities 1 and 2 to be the only final goods in those economies. However, the same argument could be developed in a more complex set up. Since commodity 1 is tradable, its price must be equal across countries.
Therefore, the price index in each country consists in a weighed average of the prices of these commodities. This means we can write the expression for the real exchange rate as follows.
eR = e γp∗ 1 + (1 −γ)p∗ 2 δep∗ 1 + (1 −δ)p2 [2.78] As demonstrated in subsection 2.3.4, in particular in equation 2.19, the price of commodity 2 can be decomposed into the domestic price of 1 and labor. The share of each part, however, depends on the technical coefficients and on relative prices of the tradable commodity and imported inputs. Even though these relative prices are equal in both economies, by assumption, we do not impose that technical coefficients must be the same. Consequently, these shares differ across these two economies. In the case of domestic economy, we have the value expressed in equation 2.19 rewritten below.
p2 = Φ1ep∗ 1 + Φ2w [2.79] In the case of the Rest of the World, assume these weights are given by ζ1 and ζ2.
Thus, we can write: p∗ 2 = ζ1p∗ 1 + ζ2w∗ [2.80] Therefore, the real exchange rate can be expressed as: eR = e [γ + (1 −γ)ζ1]p∗ 1 + (1 −γ)ζ2w∗ [δ + (1 −δ)Φ1]ep∗ 1 + (1 −δ)Φ2w [2.81] 78 CHAPTER 2. INFLATION AND CONFLICTING CLAIMS IN THE OPEN ECONOMY We can rewrite the last equation as follows below, by collapsing the coefficients into a simpler notation.
eR = e γ ′ 1p∗ 1 + γ ′ 2w∗ δ ′ 1ep∗ 1 + δ ′ 2w !
[2.82] Finally, the last equation can be rewritten as follows below. This shows that the real exchange rate varies inversely with the ratio w e , so that it varies proportionally with the ratio e w, which corresponds to our definition for q.
eR = γ ′ 1p∗ 1 + γ ′ 2w∗ δ ′ 1p∗ 1 + δ ′ 2 w e !
[2.83] 79 Chapter 3 International inflation and trade linkages in Brazil under inflation targeting 3.1 Introduction Inflation theory has focused mainly on the role of domestic variables, addressing the role of international variables only to a lesser extent.
Although international prices imposed strong pressures on inflationary processes historically, they are often overlooked during periods of price stability. Recently, a growing literature shows that inflation is a global phenomenon, suggestion inflation models should add focus for international prices, trade, and foreign competition.
International price shocks are often assumed to be transmitted to domestic prices exclusively through imports. Borio and Filardo (2007) argue that this belief results from the dominance of a “country-centric” perspective.
The authors argue that country-centric inflation models implicitly assume that domestic produced goods and foreign goods are very imperfect substitutes – otherwise one would interfere directly in the price formation of the other.
In addition, those models often assume (whether explicitly or not) very limited substitutability between domestic and foreign labor inputs. Hence, international measures of economic slack or international costs of labor are usually neglected in empirical analysis for a specific country.
Borio and Filardo (2007) argue that the larger stability of prices and inflation in the 1990s and 2000s was spread across countries. In this view, the global nature of the decline of inflation requires 80 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING a global explanation.
The larger integration of international markets and the constitution of global value chains makes domestic inflation more sensitive to international developments (Bobeica and Jarocinski, 2019; Ciccarelli and Mojon, 2010; Mumtaz and Surico, 2012).
International competition and foreign prices also explain a great share of the inflation across countries (Auer et al., 2017, 2019; Bugamelli et al., 2015). Still, Auer et al. (2019) shows that cross-country trade input-output linkages is the most important factor in explaining producer inflation synchronization across countries. Although many of these results refer to advanced economies, the role of international variables tend to be even more prominent in emerging economies. In fact, exchange rate shocks and commodity prices have usually been considered the main causes of inflation in developing countries.
Studies focusing on the Brazilian economy have pointed the exchange rate and commodity prices as the main determinants of domestic inflation (Bastos et al., 2015; Braga and Summa, 2016; Modenesi and Araújo, 2013; Nassif et al., 2020).
The dominance of international factors constrains the operation of the inflation targeting regime.
Theoretically, inflation targeting seeks to control inflation through the manipulation of aggregate demand by changing the interest rate. The exchange rate is expected to play an auxiliary role in open economy inflation targeting regimes by directly affecting import prices and reducing the demand for exports (Svensson, 2000).
Nevertheless, the exchange rate has been the dominant transmission mechanism of monetary policy shocks to domestic prices in Brazil (Nassif et al., 2020). Exchange rate appreciation following monetary policy shocks directly impact the prices of tradable goods and imports.
However, a cost-channel related to international input-output linkages has not been analyzed in the literature. Although the importance of exchange rate and commodity prices is established for the Brazilian case, the role of imported costs has not been properly addressed.
For this reason, this paper introduces an index of foreign cost shocks, following Auer et al. (2017).
We compute an average of other countries’ producer prices weighted by the yearly share of each country in Brazilian imports of capital and intermediate goods.
The chapter investigates the domestic and international causes of Brazilian inflation, focusing on the impact of other countries’ inflation on Brazilian inflation. We estimate a Structural Vector Autoregression model to test the impact of international cost-shocks 81 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING on Brazilian inflation. We therefore included a variable that combines the Producer Price Index (PPI) of Brazil’s trade partners, in addition to the usual variables that identify real and monetary shocks. Foreign PPI explains a relevant share of inflation in Brazil during the period 1999-2020. We conclude that this variable provides an additional channel through which external shocks affect domestic inflation in Brazil, contributing to the literature on the matter. Such result raises additional concerns on the effectiveness of monetary policy as the single instrument of the inflation targeting regime.
82 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING 3.2 Inflation and International Shocks 3.2.1 The globalization of inflation Empirical evidence shows the growing importance of the international component in the explanation of inflation rates across countries (Bobeica and Jarocinski, 2019; Borio and Filardo, 2007; Ciccarelli and Mojon, 2010) .
Meanwhile, conventional empirical specifications of the Phillips curve centered on measures of the domestic output slack have been loosing explanatory power in the last decades (Borio and Filardo, 2007).
These "country-centric" approaches have focused mainly on domestic slack, overlooking international variables (Auer et al., 2017; Borio and Filardo, 2007). A country-centric perspective implicitly assumes that: i) production supply chains are fully domestic; ii) foreign and domestic goods are highly imperfect substitutes and priced in domestic currency; iii) there is limited mobility of labor (Auer et al., 2017).
Growing international integration in recent times has inspired criticism against the dominance of country-centric approaches to inflation for focusing mainly on domestic variables (Borio and Filardo, 2007). On the contrary, international factors are shown to account for much of national inflation dynamics across countries (Bobeica and Jarocinski, 2019).
Empirical evidence shows that international prices, cross-border input-output linkages, and measures of trade integration play a relevant role in explaining inflation in advanced and developing economies.
Indeed, the decline in inflation in advanced economies seems to have benefited from the competition of imports from low-wage countries (Bugamelli et al., 2015). Studies also report that the increasing economic integration intensified the importance of foreign inflation in explaining domestic inflation and the increased synchronicity of inflation rates across countries (Bobeica and Jarocinski, 2019; Borio and Filardo, 2007; Ciccarelli and Mojon, 2010). A detailed analysis according to product category confirms these results, showing that international factors have gained importance for most of the product categories (Karagedikli et al., 2010).
Furthermore, global inflation rates act as an attractor for national inflation rates (Ciccarelli and Mojon, 2010).
These results suggest empirical studies should attribute greater importance to international variables in order to explain domestic inflation in both advanced and developing economies. Therefore, analysis of inflation should account for the high substitutability of tradable goods, pressures of international competition, mobility of labor, and the great importance of cross-border input-output linkages. Acknowledging these effects also implies focusing on the impact 83 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING of international prices of goods and inputs, foreign wage rates, and exchange rates (Bugamelli et al., 2015).
Supply chains are global and connect a wide range of countries. Greater intergration in Global Value Chains cause greater contestability at each stage of the production process, meaning that firms can avoid the price effect of supply and demand shocks from one country as they alter their sourcing decisions towards another country. This process dampens the impact of inflationary shocks at an international scale. Auer et al.
(2017) show that the more intense integration in Global Value Chains, the more reduced the sensitiveness of final prices to domestic slack across different countries.
International competition is thus manifested not only through higher interrelations in supply chains but also by the greater contestability at each stage of production. The importance of supply chains is reinforced by the finding that trade in intermediate goods and services is the main transmission mechanism of global shocks to domestic prices (Auer et al., 2017).
Moreover, Auer et al. (2019) show the relevance of international input-output linkages in explaining the high synchronization of inflation across countries. These results stand out for reclaiming the role of the cost channel as an important transmission mechanism of international shocks to domestic inflation.
The exchange rate also plays a crucial role in the relation between international shocks and domestic prices.
Exchange rate pass through into import prices is high across different countries (except for the US). Gopinath et al. (2020) show that the majority of trade is invoiced in a few currencies, with an outstanding role of the US dollar. Firms set trade prices in these currencies, and pass-through exchange rate shocks to final prices.
Devaluations with respect to these currencies therefore imply a proportional increase in the domestic price of imports. This regularity supports the view that the dollar is the dominant currency in the international payments system, profiting from an asymmetric position. According to this view, (mostly dollar) (Gopinath et al., 2020).
In contrast, from the domestic viewpoint, exports tend to be less sensitive to exchange rate movements. In this case, an exchange rate depreciation does not lower export prices, as they are invoiced in foreign currency, but increases the mark-up of exporting firms (Gopinath, 2015). These conclusions confirm the asymmetric position of the US due to dollar’s invoicing currency status in world trade. While exchange rate devaluations in the US strongly affect inflation in other countries, inflation in the US is unaffected by others’ exchange rate movements (Gopinath, 2015).
Furthermore, evidence shows that these features hold for both advanced and developing economies 84 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING (Giuliano and Luttini, 2020; Gopinath et al., 2020). This conclusion suggests the case of the price-taker open economy (advanced in the second chapter of this thesis) may not be restricted to peripheral or small economies, but holds more generally, at least concerning some commodities.
3.2.2 Inflation targeting and international shocks in Brazil Brazil adopted the inflation targeting policy regime in 1999 after five years of stabilization policies based on exchange rate targeting regime, which ended on a major currency crisis (Barbosa-Filho, 2008). The introduction of the inflation targeting regime shifted the policy instrument in the classical "trilemma" of open economies.1 In other words, Brazil chose to have an independent monetary policy, free capital flows and a floating exchange rate (Barbosa-Filho, 2008). By pegging the short-term interest rate, monetary policy became the main macroeconomic policy. The interest rate, in turn, became the main instrument of macroeconomic policy (Nassif et al., 2020).2 The Central Bank of Brazil sets the basic interest rate aiming to make inflation converge towards the inflation target established by the National Monetary Council.
Brazilian inflation targeting regime works according to an annual target for full (rather than core) inflation to be achieved within each calendar year. The target has a tolerance range, which varied in the last two decades. In this period, inflation rate was outside the tolerance range in six years: 2001, 2002, 2003, 2015, 2017, and 2021. With the exception of 2017, in which inflation rate was below the target, in all the other cases the inflation rate exceeded the inflation target, owing to substantial shocks in the exchange rate or in commodity prices.3 The success of the inflation target regime has frequently relied on currency appreciation, especially in years of international price shocks (Barbosa-Filho, 2008; Nassif et al., 2020).
Despite keeping inflation within a foreseeable range, inflation targeting has been associated with a poor performance in terms of growth, and with the persistence of high real interest rates (Modenesi and Araújo, 2013; Nassif et al., 2020).
1 See Mundell (1960).
2 This policy regime is thus in line with the guidelines of the New Macroeconomic Consensus. In this view, by properly anchoring inflation expectations, leading the inflation to its target would imply also closing the output gap in the so-called divine coincidence (Blanchard and Galí, 2007). A brief discussion on the New Keynesian Phillips Curve can be found in the first chapter of this thesis. In this framework, a flexible exchange rate would play an auxiliary role.
3 Although the interest rate is the relevant policy instrument in the inflation targeting regime, in open economies, a complementary role is played by the nominal exchange rate through its direct impact on prices and foreign demand (Svensson, 2000).
85 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING Estimates highlighted the role of the exchange rate and commodity prices as the main explanations of inflation in Brazil.
Braga and Bastos (2010) estimate an Autoregressive Distributed Lags Model with data from 1999 to 2008 to analyze the causes of inflation in Brazil. While the international price of basic commodities and the exchange rate were the main causes of inflation, excess demand and money wages were not significant in this estimate. Other studies disaggregated the components of inflation, searching for industry-specific explanations of price changes (Bastos et al., 2015; Braga, 2013; Braga and Summa, 2016; Martinez and Cerqueira, 2013). These studies verify if aggregated results are also observed in the industry level. Besides, they capture the relation between inflation and the change in relative prices. The relative importance of each sector on Brazilian inflation has varied since 1999. In times of international price shocks, foodstuff and other tradable dominate the price index. In a recent period of labor market tightening, inflation was concentrated in services rather than in industrial goods (Santos et al., 2018). Disaggregating inflation among durable goods, semi-durable goods non-durable goods and services, Braga (2013) shows that the main cause of inflation was imported inflation – a measure that combines the index for international price of basic commodities and the Brazilian exchange rate. The author also found a structural break in the series of prices monitored by the government.
This series achieved greater stability in 2006, confirming the success of regulatory changes introduced in this period in order to reduce the impact of public services in total inflation (Braga, 2013; Martinez and Cerqueira, 2013).
Braga and Summa (2016) disaggregated inflation in four components: prices monitored by government, foodstuff, industrialized goods and services.
In their analysis, based on data for the period 1999-2012, the exchange rate and international prices affected all four components of inflation. Another result obtained is that inflation of the services sector presented a great degree of inertia.4 Finally, Bastos et al. (2015) adopted a model of auto regressive distributed lags to investigate the behavior of inflation over seventeen distinct industries (from extractive industry and manufacturing), using data for the period 1996-2011. Results reiterated the importance of the exchange rate and of international prices of basic commodities among the determinants of the inflation in industrial goods.
The industries in which imported inflation presented a higher impact were those with greater interaction with 4 The role of past inflation in explaining current inflation has been highlighted in different studies. In general, the autoregressive component of inflation has been found to be significant and strong. Nevertheless, inertia is not large enough to justify an accelerationist interpretation of the Phillips curve (Summa, 2011).
86 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING international markets and with larger diffusion of imported inputs and final goods (Bastos et al., 2015).
3.3 Data Our dataset comprises the variables usually included in time series analysis of inflation and monetary policy (Miranda-Agrippino and Ricco, 2021) Consumer Price Index, Interest rate, GDP, Exchange Rate, and Commodity Prices. The novelty of this paper is to introduce an index of imported cost-shocks at the intermediate goods level. That is done with the inclusion of a Foreign PPI, following Auer and Mehrotra (2014), and Auer et al. (2019). These authors employed this index to analyze price co-movements across countries, showing that input-output linkages are fundamental to explain the synchronization of inflation rates across advanced economies. We introduce the Foreign PPI in a time series analysis of Brazilian inflation. Imported cost-shocks may generate an additional source of inflation in open economies, in addition to commodity prices and exchange rates. Next section provides further details about the construction of the measure of the Foreign PPI for Brazil.
The measure of CPI is the National Price Index to the Broad Consumer (IPCA), released monthly by the Brazilian National Bureau of Geography and Statistics (IBGE).
The index measures the inflation of a bundle of goods and services traded in retail. The index’s bundle corresponds to the household consumption of 90% of the population in the areas covered by the National System of Consumer Price Indexes. It is noteworthy that the inflation target pursued by the Central Bank is set in terms of the IPCA, which is therefore the most relevant price index for monetary policy.
Recent studies of inflation and monetary policy include an index of industrial production as a measure of output slack, identifying demand shocks (Miranda-Agrippino and Ricco, 2021; Ramey, 2016). However, the growing importance of services suggest that industrial production may not completely capture demand shocks.
Services have dominated inflation in Brazil in the early 2010s (Santos et al., 2018). Therefore, we include real GDP to account for domestic shocks of supply and demand. The monthly series of the real GDP is generated by the Central Bank of Brazil, through an interpolation of the Quarterly GDP series released by the Brazilian National Bureau of Geography and Statistics (IBGE). The interpolation into a monthly basis is based on additional data on industrial production, electric energy consumption, primary goods exports, and price indexes. As a robustness check, we replace real GDP 87 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING with an index of industrial production in the VAR estimate. We find similar results, as reported in section 3.6.1.
As for the interest rate, we choose the one year base nominal rate as our policy variable as in Gertler and Karadi (2015). The variable for the interest rate is the nominal policy rate set by the Central Bank of Brazil, on a yearly basis. Figure 3.1 shows that the interest rate series was at a peak in the first years of the inflation targeting regime, as a response to adverse external shocks. In the following years, the interest rate decreased progressively, in spite of the occasional policy shocks in response to inflation.
The measure of the exchange rate is the monthly average of the nominal dollar-real exchange rate. The exchange rate is expressed in dollars per real (US$/R$). An increase in the exchange rate variable thus implies a depreciation of the Brazilian currency with respect to the U.S. dollar. The choice of the exchange rate variable is consistent with the central role of the U.S. currency in price setting globally (Casas et al., 2017; Gopinath, 2015). Gopinath et al. (2020, p. 678) argue that the “vast majority of trade is invoiced in a small number of currencies, with the US dollar playing a dominant role”. Therefore, the dollar exchange rate dominates the bilateral exchange rate in price pass-through estimates. The US dollar is also a key predictor of trade volume and inflation in the rest-of-world (Casas et al., 2017; Gopinath et al., 2020).
The preference for the nominal dollar-real exchange rate over measures of the effective exchange rate is also due to the correlation of the effective exchange rate with the Foreign PPI. Both variables consist of an average weighted according to the share of countries in Brazilian trade. Hence we include the nominal dollar-real exchange rate, to capture the pure exchange rate effect in the baseline model.
Consumer Price Index, GDP, Commodity Price index and PPI are included in logs.
Interest rate is expressed in the nominal units for 12 month period. All the series are introduced in a monthly basis from January of 1999 until December of 2020. The sample, thus, includes 264 observations for six variables. Data sources are summarized in table 3.1.
88 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING TABLE 3.1: Summary of Data Variable Description Source CPI Consumer Price Index (IPCA) IBGE Interest Rate Base Interest Rate in Brazilian Central Bank accumulated in 12 months BCB Real GDP Real GDP 12 months IBGE Exchange Rate Nominal Effective Exchange Rate BCB Foreign PPI Producer Price Index of Brazilian Imports Own Elaboration.
Data from MDIC, EuroStat, OECDStat and National Sources.∗∗ Commodity Prices Commodity Price Index IMF IBGE stands for the Brazilian National Bureau of Geography and Statistics. BCB stands for the Central Bank of Brazil. MDIC is the Ministry for Development, Industry and International Trade.
IMF, International Monetary Fund.
National Sources are detailed in the Appendix.
3.3.1 Stationarity Analysis We analyzed the stationarity of the series by running Augmented Dickey Fuller and Phillips Perron tests. We conclude that CPI, GDP, Exchange rate, Commodity prices and Foreign PPI have a unit root. The first-difference of all those series is stationary. The level of Interest Rate is stationary.
Test statistics and p-values for level and first-difference unit root tests are reported in the appendix A.
Johansen trace test indicates the existence of two cointegration relationships among the model variables at the 5% level.5 The presence of cointegration relations suggest the preference for including cointegration vectors in a Vector Error Correction models with variables in first difference. Nevertheless, an alternative approach lies in the estimate of VARs in levels without imposing restrictions related to the cointegration relations.
Levels VARs are often chosen in the literature (Miranda-Agrippino and Ricco, 2021) since they are robust to cointegration and provide consistent impulse-response functions in the short run.
5 Johansen trace test is reported in Appendix B.
89 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING (A) GDP (B) CPI (C) Echange Rate (D) Interest Rate (E) Foreign PPI (F) Commodity Prices FIGURE 3.1: Summary of Data, 1999-2020.
Note: GDP, CPI and Foreign PPI in percentage change over 12 months. Exchange Rate, Interest Rate and Commodity Prices in levels.
90 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING 3.4 Methodology 3.4.1 Foreign PPI We build a Foreign Producers Price Index for Brazil, inspired by Auer and Mehrotra (2014) and Auer et al. (2019).
The authors derive country-level indexes of imported producer prices to capture cost-shocks at sectorial and aggregate levels.
The ultimate object of interest is the country-level rather than sector-level inflation.
Thus, we aggregate sectoral PPI series and cost shocks using sectoral output weights.
Previous studies have examined the role of input linkages in inflation synchronization (Auer et al., 2017; Auer and Saure, 2013). Other studies have used data on global input linkages, domestic prices and nominal exchange rates to obtain measures of real exchange rates, as Bems and Johnson (2017) and Patel et al. (2019).
PPIt = ∑ c∈C ωc,t PPIc,t εc,t [3.1] Equation 3.1 shows how we computed the Foreign PPI. The variable is a weighted average of the producer price indexes of Brazilian trade partners. The ratio between each countries’ producer price index (PPIc,t) and its exchange rate with respect to the dollar (εc,t) gives us each countries’ producer inflation in dollar terms. εc,t is the average for period t of the rate defined as the unit of the currency of country c required to buy one dollar. The inclusion of the exchange rate is an refinement on the measure proposed by Auer and Mehrotra (2014) and Auer et al. (2019). These authors studied a sample mainly formed by countries with great stability of prices and exchange rates. While omitting the exchange rate is unlikely to be a problem in studies focusing on advanced economies, that is not the case for emerging market economies. A few Brazilian trade partners presented very high inflation rates in the period, so that they dominate the index if we omit the exchange rate. These high inflation rates often emerge from high exchange rate devaluations. In that case, the change in prices as measured in foreign currency is the net outcome of the two contradictory effects. This net effect, captured in our index, is the measure coherent with the economic transmission mechanism since the final impact of a country’s inflation on other economies depends on the combined effect of domestic inflation and the exchange rate.6 6 In other words, the impact of inflation in country A on inflation in country B also depends on the changes in 91 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING The weights (ωc,t) are given by the annual share of each country (c) in Brazilian imports of Intermediate and Capital goods. Brazilian trade data is processed by the Special Secretary of Trade and International Affairs (Secint), being classified into four groups: capital goods, intermediate goods, consumption goods, fuels and lubricants. A small portion of traded goods fall in the non-specified residual classification.
The countries included in the sample for computing Brazilian Foreign PPI were responsible for more than 80% of Brazilian imports of Intermediate and Capital goods during the period 1999-2020.
Table 3.2 presents the average share of each country included in the index in Brazilian imports of intermediate and Capital goods in the whole period. The table shows that these imports are distributed among advanced and developing economies.
TABLE 3.2: Average share in Brazialian Imports of Intermediate and Capital goods, per country, 1999-2020.
United States 19.33% Russia 1.58% Turkey 0.30% China 13.00% India 1.37% Norway 0.30% Germany 8.51% Sweden 1.28% Czech Republic 0.29% Argentina 7.32% Belgium 1.09% Ireland 0.27% Japan 4.70% Thailand 1.04% Portugal 0.24% Korea 3.76% Netherlands 0.89% Hungary 0.19% Italy 3.55% Paraguay 0.62% Slovakia 0.07% France 3.42% Finland 0.60% Luxembourg 0.06% Spain 1.95% South Africa 0.44% Greece 0.05% Mexico 1.88% Denmark 0.37% Slovenia 0.04% United Kingdom 1.75% Poland 0.32% Estonia 0.02% The countries add up to 80.6% of Brazilian imports of Intermediate and Capital goods in the period 1999-2020.
The evolution of the share of the main countries since 1999 can be seen in Figure 3.2.
The figure reveals the trends of Brazilian trade found in the last two decades as the rise of Chinese share and the relative loss of importance of the United States. For our purposes, this implies that inflation in China is increasingly more influential to explain inflation in Brazil (as may be the case for many countries).
The implicit hypothesis that foreign producer price shocks are passed through to domestic prices is compatible with marginal cost pricing and constant markups over average costs (Auer et al., 2019). This hypothesis is also compatible with the theoretical framework presented in the second chapter of this thesis.
the exchange rate between the currencies of A and B.
92 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING FIGURE 3.2: Share in Brazilian imports of Intermediate and Capital goods Recent results showed that cost shock pass-through at the border is high and close to one. Ahn et al. (2017) analyze the pass through of imported inputs into producer prices at a sector-level. The authors construct an effective imported input price index, based on the sector-level import price index and data from the input-output table, finding a long run pass through of 0.7 for Korea, and a nearly complete pass through for France, Germany and Netherlands.
In turn, Berman et al. (2012) find a nearly complete pass-through of exchange rate shocks into import prices, exceeding the pass-through into consumer goods prices.
Cost indexes are rarely employed in empirical analysis of Brazilian inflation.
Recently, Pimentel et al. (2020) constructed cost indexes for 21 industrial activities, based on intermediate consumption from the National Accounts.
The authors showed the pass-through of positive cost shocks is higher than for the negative cost-shocks in all industries. The result evidences the existence of asymmetry in the pass-through of cost into prices.
Prompted by the cost shocks following the outbreak of the COVID-19 pandemic crisis, the Central Bank of Brazil analyzed the impact of costs on inflation in 24 industrial activities BCB (2021).
Cost indexes were built by combining the intermediate consumption structure from the Input-Output Matrix released in 2015 with price series from domestic price indexes and imports data. The prices of these industrial 93 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING activities follow closely the cost indexes. Evidence shows a rapid pass-through of costs into final prices in the manufacturing industry in the period 2005-2020.
Still, the pass-through was high and fast in 2020, surpassing what would be expected considering the historical pattern (BCB, 2021).
3.4.2 VARS and SVARS The study of the propagation of macroeconomic shocks still generates controversy among economists.
Since the seminal work ok Sims (1980), Structural Vector Autoregressive (SVAR) have become one of the most used methodologies in empirical research with time series data, in particular in empirical macroeconomic analysis (Kilian and Lütkepohl, 2017).
Vector Autoregressive (VAR) models consist of multivariate autoregressions in which each model variable is regressed on its own lags and lags of the other model variables. The innovation vector in a reduced-form VAR, ut in equation 3.2, represents the innovations of the model, given the information set formed by the lags of the variables in yt.
ut can also be interpreted as a linear combination of the underlying structural shocks (Gambetti, 2021).
yt = ∑Dpyt−p + ut [3.2] The reduced-form VAR is therefore the basis for the identification of the economic structural shocks.
The vector of structural shocks, wt in equation 3.3, contains uncorrelated shocks with a well-defined economic interpretation.
Many different methods can be used to identify these shocks. Structural Vector Autoregressive models express endogenous variables as a combination of current and past economic shocks.
The study of VAR in the structural form allows researchers to quantify causal relations in the data. SVARs therefore provide a useful framework for macroeconomic analysis in which the dynamic response of each variable to economic shocks can be represented by impulse response functions (Gambetti, 2021; Kilian and Lütkepohl, 2017).
B0yt = ∑Bpyt−p + wt [3.3] A common methodology to identify structural shocks is found in the imposition of short-run restrictions through the Cholesky decomposition (Blanchard and Quah, 1988).
94 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING As for the domestic variables, we follow a usual ordering found in the empirical literature on of inflation and monetary policy (Sims, 1992), testing alternative orders as robustness check.7 Considering that Brazil is a price-taker in international markets allows us to assume that Foreign PPI and Commodity Prices are not affected by contemporaneous shocks in the domestic economy. However, we consider that Foreign PPI may be affected by changes in Commodity Prices.
The order of variables adopted in our analysis is expressed in equation 3.4. The model includes 12 lags for each variable since it relies on monthly data. The model’s estimate also includes a constant term.
B0yt = − 0 0 0 0 0 − − 0 0 0 0 − − − 0 0 0 − − − − 0 0 − − − − − 0 − − − − − − Commodity ForeignPPI InterestRate ExchangeRate CPI GDP [3.4] 7 Alternatively, we may consider that the information set underlying monetary policy decisions is updated with the contemporaneous variables. We thus consider an alternative ordering of the Cholesky matrix, setting the interest rate as the most endogenous variable to contemporaneous shocks. This robustness test, presented in section 3.6.3, confirms the results obtained here.
95 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING 3.5 Foreign PPI Shocks and Inflation in Brazil We tested the hypothesis that Foreign PPI explains domestic CPI for Brazil in the period 1999-2020. Foreign PPI may be an additional transmission channel of external shocks to domestic inflation. Figure 3.3 reports the response of CPI to structural shocks identified in the model.8 Shocks in Foreign PPI increase the CPI in the first twelve months. The impact of the shocks is weakly reverted in the following periods. Cumulative impulse response functions (figure 3.4) show that the positive effect is persistent over the time horizon. The positive effect is significant within the 90% level until period 13.
FIGURE 3.3: Non Cummulative Response of CPI to Structural Shocks Note: Non Cummulative Impulse Response Functions of CPI to Shocks in GDP, CPI, Exchange Rate, Interest Rate, Foreign PPI, and Commodity Prices.
Impulse response functions confirm the importance of the exchange rate and 8 The complete impulse response functions are presented in the appendix.
96 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING commodity prices to explain the CPI. Real domestic shocks as measured by the real GDP cause increases in the price level.
In the baseline specification, interest rate shocks increase the price level in the whole period, presenting a persistent price puzzle. The significant positive effect of interest rate shocks on CPI means that monetary policy does not show the expected effect in the impulse response functions. However, the expected negative effect reappear (sometimes after a short lived price puzzle) in the impulse response functions when we replace the real GDP with an Industrial Production index or when we drop the extreme observations associated with the COVID-19 pandemic crisis.
FIGURE 3.4: Cummulative Response of CPI to Structural Shocks Note: Cummulative Impulse Response Functions of CPI to shocks in GDP, CPI, Exchange Rate, Interest Rate, Foreign PPI, and Commodity Prices.
In addition to the impulse response functions, the Structural VAR methodology 97 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING allows us to describe how much of the forecast error variance (or prediction mean squared error) of the dependent variable is accounted for by each structural shock at different time horizons (Kilian and Lütkepohl, 2017; Kilian and Park, 2009).
Table 3.3 reports the forecast error variance decomposition for CPI. Foreign PPI shocks account for 11.03% of the forecast error variance of Brazilian inflation at the 12-month horizon. Its explanatory power decreases to 7.47% at the 36-month horizon. It is noteworthy that until the 36-month horizon, Foreign PPI shocks are approximately as important as Commodity Prices shocks to explain the forecast error variance of CPI. The results confirm that the Exchange Rate shocks have the largest explanatory power over the variance of CPI. Exchange Rate shocks account for 31.29% of the forecast error variance at the 12-month horizon, and for 54.40% of the forecast error variance at the 36-month horizon. In the long run, almost half of the variance of Brazilian inflation is explained by the exchange rate.
TABLE 3.3: Forecast Error Variance Decomposition for CPI.
Time Horizon GDP CPI Exchange Rate Interest Rate Foreign PPI Commodity Prices 12 1.82 45.02 31.29 3.71 11.03 7.13 24 4.91 34.11 45.87 2.14 6.76 6.23 36 5.14 19.08 54.40 5.54 7.47 8.37 120 13.14 11.79 47.75 15.07 10.18 22.08 Percentage of Ahead Forecast Error Variance Explained by structural shocks on GDP, CPI, Exchange Rate, Interest Rate, Foreign PPI, Commodity Prices at the respective time horizon.
Figure 3.5 plots the historical decomposition of the stochastic component of CPI according to the structural shocks. Shocks related to the external sector (captured by the variables Exchange Rate, Foreign PPI, and Commodity Prices) dominated the explanation of the stochastic component of the CPI series. The exchange rate presented an outstanding role, confirming the results of previous research.
This variable responded for 27.17% of the total shocks affecting CPI. The CPI itself responded for 29.0% of the shocks. Foreign PPI shocks explain 12.05% of the stochastic component of CPI in the period.
Finally, GDP, Interest Rate, and Commodity Prices represented, respectively, 10.17%, 7.87%, and 13.75%. The monetary policy instrument presents the weakest effect on the CPI series.
However, monetary policy becomes much more effective once we consider that it also operates through the exchange rate effect.
98 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING FIGURE 3.5: Historical Decomposition of CPI according to shocks on GDP, CPI, Exchange Rate, Interest Rate, Foreign PPI, and Commodity Prices.
We conclude that international shocks have been fundamental to explain inflation in Brazil in the last decades.
In particular, shocks on producer prices of Brazil’s trade partners are shown to be a relevant transmission mechanism of international shocks to domestic inflation. Impulse response functions show that Foreign PPI has a positive and significant effect on domestic CPI. Foreign PPI shocks are also an important explanation of the variance of CPI’s forecast error, especially in the short-term.
Historical decomposition confirmed the relative importance of the effect of Foreign PPI shocks on CPI. We will show in the next section that this result is robust to alternative specifications and change in variables.
The prominent role of shocks related to the external sector undermines the power of monetary policy to control inflation.
The only effective instrument to avoid the transmission of international shocks is the appreciation of the exchange rate.
This variable, however, must be defined according to a broader policy perspective, since it affects financial stability, besides of interacting with distribution and trade (Nassif et al., 2020). Therefore, our results seem to confirm that the inflation targeting regime relied on the exchange rate effect of contractionary monetary policy. Finally, the pass through of foreign costs shocks into final prices may be avoided through the stabilization of domestic basic input prices. In Brazil, a great deal of basic input prices are monitored by the government or are directly set by state owned companies. Such prices may be key to mitigate the volatility coming form foreign shocks.
99 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING 3.6 Robustness Analysis We performed different analyses to test the robustness of the results of the baseline model.
We tested if results are robust to (a) replacing real GDP with an Industrial Production index; (b) dropping the observations of 2020 (due to the pandemic crisis) from the sample; (c) making monetary policy the most endogenous variable in the identification of the contemporaneous shocks; (d) estimating the impulse response functions with local projections.
The sign of the expected effect of shocks in Foreign PPI on CPI did not change in these analyses. The magnitude of the effect became smaller in specification (a), and greater in specification (c). Foreign PPI shocks are shown to have a persistent positive effect on CPI in most cumulative impulse response functions. The main conclusions of the model are therefore robust to changes in the variable measuring real domestic shocks, dropping extreme observations related to the pandemic crisis and changing the recursive ordering of structural shocks. Local projections estimate showed a positive and significant effect only in the initial periods, but this effect soon becomes non significant afterward.9 3.6.1 Industrial Production We tested if the results are robust to the use of an index of industrial production as a measure of domestic real shocks rather than the GDP. Industrial production index is often preferred in the literature (Gertler and Karadi, 2015; Ramey, 2016), since it is a volume index measured in a monthly basis while a monthly GDP series relies on the interpolation of quarterly series. Nevertheless, in the particular case of Brazil, the real GDP series provides a longer time horizon, which allowed us to estimate the model from 1999-2020.
In contrast, the index of industrial production dates back to 2002, reducing the sample size to 228 observations.
Figure 3.7 plots the industrial production index in the period 2002-2020. The series is released by the Brazilian Bureau of Geography and Statistics (IBGE).
9 Future research may explore Bayesian Local Projections estimate to improve the efficiency of the impulse response functions.
100 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING FIGURE 3.6: Industrial Production Impulse response functions show that the results obtained in the baseline model are robust to the substitution of real GDP with a series of Industrial Production. Foreign PPI shocks have a positive effect on CPI. The non cumulative impulse response converges to zero. The cumulative effect of Foreign PPI shocks on CPI is positive and persistent, as shown in figure 3.7. The cumulative effect is significant at the 90% level until the 12th month following the shock. Appendix D reports the complete impulse response functions.
FIGURE 3.7: Non cumulative (left) and cumulative (right) response of CPI to shocks on the Foreign PPI.
3.6.2 COVID Shocks The COVID-19 pandemic engendered the deepest recession of the last decades. The outbreak of the pandemic crisis thus generated a sequence of extreme observations after March 2020.
The pandemic shocks caused an immediate fall in international trade, 101 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING disruption in Global Value Chains, sharp changes in commodity prices, and a huge depreciation of the Brazilian currency (Hevia et al., 2020; Kohlscheen et al., 2020). The huge data variation in the months following the pandemic outbreak constitutes a challenge for the estimation of parameters of VAR models (Lenza and Primiceri, 2020).
Dropping the extreme observations related to the pandemic can be an alternative for parameter estimation in time series analysis (Lenza and Primiceri, 2020). Nevertheless, dropping such observations can harm the predictive power of the model, since it underestimates the variance of the Data Generating Process.
We therefore test if the observations following the pandemic crisis affect the conclusions of the baseline model.
Hence, we dropped the observations of 2020, remaining with a dataset of 252 observations. We then reestimated the model with the reduced sample by following the same procedure as in the baseline model. Figure 3.8 shows the response of CPI to Foreign PPI shocks in the new estimate. The complete impulse response functions are reported in the appendix.
FIGURE 3.8: Non cumulative (left) and cumulative (right) response of CPI to shocks on the Foreign PPI.
We conclude that our main result is robust to the drop of the extreme observations following the pandemic crisis. Foreign PPI still presents a positive and significant effect on CPI. The response of CPI to Foreign PPI shocks becomes negative, but not significant from the 13th period onward. Notwithstanding this change in sign, the cumulative effect remains positive in the 24 months horizon, and significant at the 90% level until the 12th month after the shock.
102 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING 3.6.3 Monetary Policy The baseline model relied on the recursive ordering of the orthogonalized shocks as described in section 4.3. The ordering imposed that Foreign PPI and Commodity Prices are not affected by domestic contemporaneous shocks. That assumption ensured that real international shocks are independent from contemporaneous domestic shocks.
However, this strategy has the downside of assuming that monetary policy has a delayed response to domestic shocks.
The recursive ordering of structural shocks in VARs models usually considers that monetary policy respond contemporaneously to both domestic and external shocks, but has a lagged effect on these variables. Therefore, the standard identification strategy imposes zero contemporaneous restrictions to the monetary policy variable (Christiano et al., 1998). In fact, Central Banks are likely to consider a broader informational set, anticipating the impact of economic events before they affect the meaningful variables (Romer and Romer, 2000; Sims, 1992).
We therefore test if the results from the baseline model are robust to the change in the identification of structural shocks. Namely, we reordered the Cholesky matrix, setting the Interest Rate as the most endogenous variable to contemporaneous shocks. Figure 3.9 report the response of CPI to Foreign PPI shocks in the alternative specification of the model.
Impulse response functions show that the effect obtained in the baseline model is robust to the alternative specification. The sign of the effect of Foreign PPI shocks on CPI follows the expected result. The impulse response follows a similar patter as in the baseline case, with an initial increase in CPI, stabilizing in a longer time horizon. The effect is has a greater magnitude with respect to the baseline model. The positive effect is significant (at the 90% level) until eight periods following the shocks in the non cumulative impulse response function.
The cumulative impulse response function reveals a positive and significant effect until the 12th period after the shock.
The complete impulse response functions can be found in Appendix D.
103 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING FIGURE 3.9: Non cumulative (left) and cumulative (right) response of CPI to shocks on the Foreign PPI.
3.6.4 Local Projections Local projection estimations can be employed to test the robustness of impulse response functions estimated by VARs. Jordà (2005) introduced local projections as an alternative method to Structural VARs. Local projections are robust to misspecification of the Data Generating Process. If the VAR adequately captures the data generating process, it is the optimal model to recover impulse responses at all horizons. However, if the VAR is misspecified, then the specification errors will be compounded at the increasingly distant horizons of the impulse response functions (Jordà, 2005). Alternatively, Jordà (2005) suggests collecting projections local to each forecast horizon. Local projections are analogous to performing a series of direct forecasts, while the VAR method is analogous to the iterated forecasting method (Ramey, 2016). Once local projections impose fewer restrictions than VAR models, “the estimates are often less precisely estimated and are sometimes erratic. Nevertheless, this procedure is more robust than standard methods, so it can be very useful as a heuristic check on the standard methods.” (Ramey, 2016).
We thus tested the robustness of our results by estimating the local projections impulse response function.
We estimated the impulse response of CPI to Foreign PPI shocks at the horizon h according to the following model: CPIt+h = θi,hϵ1t + ControlVariablest + ζt+h [3.5] ϵ stands for the Foreign PPI shocks. Shocks were recovered by means of a Structural 104 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING VAR model that regressed Foreign PPI against Commodity Prices. Foreign PPI shocks are therefore the structural residuals of the Foreign PPI equation. ControlVariables include a deterministic term (constant), the lagged dependent variable (CPI) and current and lagged variables included in the baseline model (that is, GDP, Interest Rate, Exchange Rate, Commodity Prices). We used a lag length of 12 periods for the shock, the dependent variable and the controls. Local projections may generate serial correlation in the error terms owing to the successive leading of the dependent variable (Jordà, 2005). Therefore, we use the Newey-West correction for the standard errors (Newey and West, 1987).
Figure 3.10 plots the impulse response function of Foreign PPI shocks on CPI. Impulse response function shows a initial positive and significant effect smoothed in the following periods. The cumulative impulse response shows a positive, although it is significant only in the initial periods following the shock. We conclude that estimating the impact of Foreign PPI shocks on CPI with Local Projections method provides the expected sign, as in the baseline model. However, widening confidence intervals make the effect non significant after the initial periods following the shock. In fact, local projections are likely to be less efficient than iterated methods, thereby being subject to volatile and imprecise estimates (Ramey, 2012). Future research may improve the identification of Foreign PPI shocks and estimate with a more efficient methodology as Bayesian Local Projections (Miranda-Agrippino and Ricco, 2021).
FIGURE 3.10: Non cumulative (left) and cumulative (right) response of CPI to shocks on the Foreign PPI.
105 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING 3.7 Final Remarks We analyzed if cost shocks in the rest of the world affect inflation in Brazil. The effect of foreign producer price shocks had not been addressed by the literature on Brazilian inflation yet, although the effect was tested for other countries (Auer et al., 2019; Auer and Mehrotra, 2014). To measure foreign producer price shocks, we built a Foreign PPI according to the producer price indexes of Brazilian trade partners and their share on Brazilian imports of intermediate and capital goods.
Impulse response functions confirm the expected effect of the Foreign PPI on the domestic CPI. Impulse response functions also confirmed the overall importance of exchange rate and commodity prices as explanations of CPI. We conclude that the index measuring international cost shocks was a significant transmission channel of external shocks to domestic CPI in Brazil in the last two decades. The result is robust to the replacement of GDP with an Industrial Production index, dropping the observations related to the pandemic crisis, and making monetary policy the most endogenous variable in the identification of the contemporaneous shocks.
Local Projections estimate of impulse response functions shows the expected sign, although the result becomes non-significant a few periods after the shock.
Overall, our results reveal the dominance of shocks related to the external sector (Exchange Rate, Foreign PPI, and Commodity Prices) over domestic shocks (GDP and Interest Rate) to explain inflation in Brazil. The importance of international shocks and of Foreign PPI in particular has important implications for monetary policy.
International shocks are not affected by the policy rate pegged by the Central Bank of Brazil.
However, the impact of these shocks on Brazilian prices also depend on the exchange rate. Therefore, our results seem to confirm that the inflation targeting regime relied mainly on the exchange rate effect of interest rate increases, confirming previous results.
Finally, this chapter provides an additional variable explaining the effect of external shocks on domestic inflation in Brazil.
106 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING Appendix A Unit Root tests The table below presents test statistics and p-values for levels unit root tests.
TABLE A.1: Unit Root tests in levels Variable Test Statistic ADF P-value ADF Test Statistic PP P-value PP GDP −0.6769 0.9717 1.4063 0.9900 CPI −1.8219 0.6509 −4.7095 0.8461 ExchangeRate −1.4474 0.8086 −3.3310 0.9189 InterestRate −3.7174 0.0236 −10.4646 0.5232 ForeignPPI −1.5936 0.7471 −3.9482 0.8889 CommodityPrices −1.5003 0.7864 −4.6251 0.8509 TABLE A.2: Unit Root tests in first difference Variable Test Statistic ADF P-value ADF Test Statistic PP P-value PP GDP −3.5300 0.0403 −51.4616 0.01 CPI −5.8647 0.01 −102.6247 0.01 ExchangeRate −6.1267 0.01 −191.1040 0.01 InterestRate −7.1256 0.01 −41.1915 0.01 ForeignPPI −6.0357 0.01 −126.9793 0.01 CommodityPrices −6.2574 0.01 −191.5341 0.01 B Cointegration Analysis TABLE B.3: Johansen trace test H0 Test Statistic 10% 5% 1% r ≤5 0.32 6.50 8.18 11.65 r ≤4 8.26 15.66 17.95 23.52 r ≤3 19.23 28.71 31.52 37.22 r ≤2 41.49 45.23 48.28 55.43 r ≤1 81.82 66.49 70.60 78.87 r = 0 136.40 85.18 90.39 104.20 107 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING C Baseline Model Impulse Response Functions FIGURE 3.11: Non Cummulative Impulse Response Functions of the Baseline Model 108 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING FIGURE 3.12: Cummulative Impulse Response Functions of the Baseline Model 109 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING Forecast Error Variance Decomposition FIGURE 3.13: Forecast Error Variance Decomposition of the Baseline Model 110 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING D Impulse Response Functions of the Robustness Tests Industrial Production FIGURE 3.14: Non Cumulative Impulse Response Functions 111 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING FIGURE 3.15: Cumulative Impulse Response Functions 112 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING COVID FIGURE 3.16: Non Cumulative Impulse Response Functions 113 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING FIGURE 3.17: Cumulative Impulse Response Functions 114 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING Monetary Policy FIGURE 3.18: Non Cumulative Impulse Response Functions 115 CHAPTER 3. INTERNATIONAL INFLATION AND TRADE LINKAGES IN BRAZIL UNDER INFLATION TARGETING FIGURE 3.19: Cumulative Impulse Response Functions 116 Chapter 4 Growth and debt stability in a supermultiplier model with public expenditures and foreign trade 4.1 Introduction Post-Keynesian Economics assigns a prominent role to government and foreign trade as drivers of demand. Government spending and exports are typical examples of "external markets" (Kalecki, 2016), fundamental for the absorption of growing output in capitalist economies. External to the capitalist income circuit, these expenditures can be defined as autonomous demand (Cesaratto, 2015). In the Sraffian supermultiplier class of models, autonomous expenditures are the ultimate drivers of demand and growth. Because of this key role, they must be carefully appreciated. Thus, we investigate two dimensions of the sustainability of foreign trade and government spending. The first one concerns the balance of payments and the external constraint, and the second, the long-run stability of public finance.
The Sraffian supermultiplier provides a useful framework for analyzing the consequences of autonomous demand to economic growth (Freitas and Serrano, 2015; Serrano, 1995a).
The supermultiplier literature has discussed growth models led by different sources of autonomous demand (Allain, 2015; Hein, 2018; Lavoie, 2016; Nah and Lavoie, 2017; Nomaler et al., 2021).
However, growth in the supermultiplier depends on the long-run sustainability of the financial stocks behind the autonomous components of aggregate demand (Hein and Woodgate, 2020; Pariboni, 2016b). Steady growth in an open economy with government thus requires the stability of the financial imbalances of the public and the external sectors.
This chapter aims to analyze the sustainability of growth led by exports and public 117 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE expenditure. Hence, we examine the stability of external and public debt ratios in a supermultiplier model with exports and public expenditure. Debt ratios are stable when the growth rates (of output or exports) are larger than the interest rates on the (public or external) debt service. Nevertheless, the growth of public expenditures may threaten the stability of the external debt in the presence of international financial constraints.
Therefore, we propose a fiscal policy rule that contains foreign indebtedness.
Five simulated experiments support the analytical conclusions of the model and compare policy alternatives.
We contribute to the Sraffian supermultiplier literature by modeling an open economy with government and assessing the sustainability of its growth path. Differently from most of the supermultiplier literature, we explicitly introduce more than one autonomous expenditure and allow for changes in the composition of autonomous demand. We also contribute to the literature on growth and the external constraint (Barbosa-Filho, 2001; Moreno-Brid, 1998; Thirlwall, 1979) by discussing how this constraint emerges in an open economy supermultiplier model. Thirlwall’s Law reappears as an upper limit to growth in the long run. Nonetheless, introducing a domestic autonomous expenditure implies that the external constraint is not always binding given that demand is not driven only by exports. Finally, we show that treating the external constraint as a long run debt-stability condition also may relax this constraint in the short run, which confirms the importance of an out of the steady-state analysis.
The remainder of this chapter is structured as follows. Section 2 presents the Sraffian supermultiplier growth model, describing its contribution to demand-led growth theory. Section 3 extends the model for an open economy with government, in which exports and public expenditure compose autonomous demand.
Section 4 discusses public and external debt stability in demand-led approaches.
The same section introduces debt dynamics in the model, defining the equilibrium values for debt ratios and the conditions for their stability. The section also proposes a fiscal policy rule, and compare the results for external debt with Thirlwall’s tradition. Section 5 presents the simulations of the model, exploring five different experiments.
A final section summarizes the main findings of the chapter.
118 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE 4.2 The Sraffian supermultiplier The Sraffian Supermultiplier is a contribution to demand-led growth theory originally proposed by Serrano (1995a, 1996b), extended lately by Freitas and Serrano (2015) Serrano and Freitas (2017).
By highlighting the role of autonomous components of aggregate demand that do not generate productive capacity, the Sraffian Supermultiplier coped with long-lasting shortcomings of demand-led models of growth.
In particular, this framework generates an endogenous convergence of the degree of capacity utilization towards normal capacity without necessarily triggering Harrodian instability neither relying on endogenous adjustments in income distribution. For this reason, the supermultiplier has achieved greater acceptance within the demand-led growth tradition, being endorsed by some Kaleckian scholars (Allain, 2015; Dutt, 2019; Hein, 2018; Lavoie, 2016).
The Sraffian supermultiplier combines an exogenous determination of distribution and an investment function following the capital stock adjustment principle in a stable path of growth which ensures convergence towards normal capacity. The exogenous determination of distribution means that the model does not require any functional relation between growth and distribution.
Assuming distribution is exogenous to growth means that the supermultiplier model can be integrated with the Classical surplus approach to value and distribution. This approach analytically distinguishes the theory of value and distribution from the theory of output and accumulation (Garegnani, 1984).
In this view, political and institutional factors, together with economic factors, explain income distribution (Pivetti, 1991; Sraffa, 1960; Stirati, 1994).
This feature of the supermultiplier contrasts with the Cambridge growth model, in which movements in functional income distribution allow for the equilibrium between aggregate savings and investment.
In the Sraffian supermultiplier, investment adjusts to effective demand, in a way similar to the accelerator investment function originally proposed by Harrod (1939). In simple terms, the accelerator suggests that increases in productive capacity are a consequence of higher use of existing capacity.
Firms regularly operate with spare capacity, thereby being able to attend to unexpected peaks of demand.
Technical conditions of productive efficiency define a degree of capacity utilization labeled as the normal degree (Ciccone, 1986, 2011).
If sales increase, capacity utilization will also increase.
If firms expect demand to remain persistently higher, they will expand production capacity by investing in capital goods. It is noteworthy that the introduction of the adjustment of capacity to demand does not necessarily trigger an unstable pattern in the supermultiplier, in which it differs from Harrod‘s growth model.1 Harrodian 1 See Freitas and Serrano (2015) for an analysis of the stability condition of the Sraffian supermultiplier.
Estimating the model’s parameters for the US economy, Haluska et al. (2020) shows that Harrodian 119 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE instability is solved with the presence of autonomous expenditures, which adjust the economy‘s average propensity to save to the investment share in output (Freitas and Serrano, 2015; Serrano et al., 2019).
Harrodian instability is also avoided due to the assumption of a slow adjustment of capacity to demand (Freitas and Serrano, 2015).
Contrasting with Neo-Kaleckian growth models, in the Supermultiplier, there is no direct relation between distributive variables (as profit rate) and investment. Hence, a shift in distribution in favor of profits that is not followed by an increase in demand (and, thus, use of capacity) will not push investment in this model.2 The main conclusion of the Sraffian supermultiplier posits that the growth rate of non-capacity generating autonomous expenditures determines the growth rate of aggregate demand, productive capacity, and output in the long run (Serrano, 1995b).
Demand drives output growth and capital accumulation, fitting in with desirable features of a demand-led growth model.3 Putting it differently, the supermultiplier is consistent with the Keynesian Hypothesis – i.e., the notion that “in the long period, in which productive capacity changes, no less than in the short period analyzed by Keynes, it is an independently determined level of investment that generates the corresponding amount of savings, rather than an autonomous propensity to save that generates the level of investment as in the traditional answer” (Garegnani, 1992, p. 47).
Finally, capacity utilization converges towards an exogenous normal degree of capacity utilization in a stable process.
The Sraffian Supermultiplier has stimulated the debate among heterodox macroeconomists in the past few years (Allain, 2015; Cesaratto, 2015; Lavoie, 2016). As the model achieved greater acceptance as a relevant contribution to demand-led growth theory, further developments were introduced into the original framework.
Neo-Kaleckian scholars advanced an alternative version of the model, preserving an autonomous parameter as a determinant of investment — together with deviations in capacity utilization — in the short and the medium-run (Lavoie, 2016). This parameter is often associated with capitalists ‘animal spirits’ (Amadeo, 1986). The long-run of this Neo-Kaleckain supermultiplier coincides with the fully adjusted position of Serrano’s (1995a) model.
Furthermore, different sources of autonomous demand, as public expenditures (Allain, 2015; Hein, 2018) and exports (Nah and Lavoie, 2017) were introduced in the supermultiplier model. Fiebiger and Lavoie (2019); Pariboni (2016b) discussed connections between autonomous (or semi-autonomous) private instability is unlikely to happen in the supermultiplier.
2 See Pariboni (2016a) for a critique to the Neo-Kaleckian growth model in light of the Sraffian supermultiplier.
Another divergence from the Neo-Kaleckian growth model lies in the (univocal) convergence of capacity utilization towards the normal degree of capacity utilization (Girardi and Pariboni, 2019).
3 In the supermultiplier, the autonomous components of demand are the proximate cause of growth.
Nevertheless, the ultimate causes of growth in the supermultiplier are not found in the intricacies of economic modeling but in the political and social determinants of autonomous demand (Morlin et al., 2021; Passos and Morlin, 2020).
120 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE consumption and financial variables. Brochier and Silva (2019); Cassetti et al. (2017); Mandarino et al. (2020) approached the supermultiplier in a stock-flow consistent framework. Fazzari et al. (2020); Palley (2019) discuss unemployment and labor market concerns in light of the Sraffian supermultiplier growth model.
Few works have explicitly dealt with models with two sources of autonomous demand.
With different objectives, Freitas and Christianes (2020) and Hein and Woodgate (2020) account for private consumption and public expenditure.
Pariboni (2016b) studies how changes in one (non-specified) autonomous expenditure affects the stability of private debt associated with autonomous consumption.
In the current chapter, we develop a supermultiplier model with two sources of autonomous demand: public expenditure and exports.
We show how these expenditures determine the growth path in a fully adjusted position and discuss the long-run conditions of public and foreign debt sustainability.
4.2.1 Sraffian superrmultiplier model After this summary, we can present the basic analytical framework proposed by Serrano (1995b) of the Sraffian supermultiplier as a demand-led growth model.
I follow the exposition of Freitas and Serrano (2015) closely.
Let us start by introducing some useful relations. In this framework, the level of output corresponding to full capacity utilization (YK) depends on the capital stock (K) and the capital-output ratio (v). Note that the capital-output ratio is exogenous and assumed to be fixed in time since there is no substitution between labor and capital in production and we are not coping with technical progress. Thus, we can describe the potential output with equation 4.1.
YKt = 1 v Kt [4.1] Since v is constant, the growth rate of full capacity output coincides with the rate of capital accumulation (i.e., the growth rate of the capital stock), which we denote as gK. In each period, capital stock changes increasingly with investment, while part of it depreciates. For this reason, we can express the rate of accumulation as in equation 4.2.
Note that δ stands for the rate of depreciation of the capital stock, which is assumed fixed for simplicity.
gKt := ˙ Kt Kt = It Kt −δ [4.2] 121 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE With few operations we can obtain equation 4.3.4 Note that h stands for the investment share in aggregate output, corresponding to the ratio between gross investment and the level of output (Freitas and Serrano, 2015). Hereafter, we may also refer to this variable as the propensity to invest.
gKt = ht v ut −δ [4.3] In turn, ut stands for the degree of productive capacity utilization. It is given by the ratio between current output and full capacity output. The change in capacity utilization ( ˆ u) follows the difference between the growth rate of output (gt) and the growth rate of the capital stock, as in equation 4.4.5 ˙ ut = ut(gt −gKt) [4.4] We can now determine the level of output for a closed economy without government. In this case, aggregate demand is solely composed of private consumption and private investment. Serrano (1995b) considers that a part of consumption is induced by current income as in the Keynesian multiplier, with a marginal propensity to consume out of income (c) smaller than one.
Another part of consumption is autonomous, which means it is neither financed nor directly caused by current income.
Within the supermultiplier framework, autonomous demand includes “all those expenditures that are neither financed by the contractual (wage and salary) income generated by production decisions, nor are capable of [directly] affecting the productive 4 This is done with few manipulations departing from equation 4.2. We start by dividing both elements of the fraction by the variable YKt, obtaining: gKt = It/YKt Kt/YKt −δ Then, we can multiply the numerator by the term Yt Yt without changing its value. Rearranging, we obtain: gKt = It/Yt.Yt/YKt Kt/YKt −δ Finally, substituting the ratios by the variables defined, we can rewrite the previous expression as in equation 4.3.
5 The rate of capacity utilization is defined as the ratio between current output and full capacity output: ut = Yt/YKt. By simply taking natural logarithms and derivatives with respect to time in both sides of the previous equation, we obtain the following expression for the rate of change in ut.
ˆ ut = gt −gtK By multiplying both sides by ut, we can finally obtain the equation 4.4 presented in the text.
122 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE capacity of the capitalist sector of the economy" (Serrano, 1995, p. 71). In this simple version, we assume that autonomous consumption can be financed out of accumulated wealth. Thus, we can write equation 4.5 as the equilibrium condition in the market of goods.
Yt = CAt + cYt + It [4.5] Investment is assumed to be completely induced in the supermultiplier model once firms adjust the capital stock to demand. We abstract from residential investment and investment in innovation, which are not directly caused by current demand.
As described by Serrano and Freitas (2017), the Sraffian supermultiplier assumes that aggregate investment is determined according to the capital-stock adjustment principle.
In this view, “inter-capitalist competition influences the process of investment leading to the tendency towards the adjustment of productive capacity to meet demand at a price that covers the production expenses and allows, at least, the obtainment of a minimum required profitability. Thus, the capital stock adjustment principle conceives the demand for capital goods as a derived demand with the objective of creating capacity to meet profitable (or effective) demand”. The supermultiplier accounts for this principle by setting investment as induced by current income according to a propensity to invest (h).
Thus, by substituting It per htYt in equation 4.5 and rearranging we obtain equation 4.6.
Yt = 1 1 −ct −ht CAt [4.6] The term within parenthesis is the supermultiplier, which corresponds to the magnitude of the indirect impact of autonomous consumption on aggregate income due the recursive effect of income increase on induced consumption and private investment.
Freitas and Serrano (2015) emphasize that the level of output is determined by aggregate demand as long as the marginal propensity to spend is smaller than one (c + ht < 1). That has often been labeled as the Keynesian stability condition. We expect this condition to hold in capitalist economies generally.
Otherwise, i.e., if the propensity to spend equals one, then the level of output is not demand constrained, but capacity constrained. To explain this point, let us suppose that the marginal propensity to spend is equal to one (c + ht = 1). In that case, any income increase would immediately cause an increase in demand, which, in turn, would further increase income.
This process would be repeated until the economy achieved full capacity. We can see in equation 4.6 that as the marginal propensity to spend approaches one, Yt, as determined according to autonomous demand and the supermultiplier effect, approaches infinity. If this is the case, there is no lack of aggregate demand. The upper 123 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE limit to output, given by the availability of productive capacity, is easily reached. In other words, if the marginal propensity to spend is equal to one, the economy follows Say’s Law.
Freitas and Serrano (2015) and Serrano and Freitas (2017) show that the investment share must be flexible, allowing capacity utilization to converge towards the normal degree of capacity utilization. On the contrary, if ht were fixed, changes in the growth rate of capital stock would always lag behind changes in aggregate demand. In this case, the equilibrium rate of capacity utilization would persistently differ from normal capacity. We can better explain this point with a brief example. Assume, for instance, that the growth of autonomous demand accelerates. Then, aggregate demand will grow at a higher rate than before because gt = gCAt in the long run.
In equilibrium, investment grows at the same rate as demand. However, at least for a short period, demand grows faster than the capital stock. The assumption of a fixed ht implies that the rate of accumulation follows the growth rate of demand without ever exceeding it.
Consequently, the rate of capacity utilization remains permanently above the normal capacity utilization after an acceleration of the growth of demand.
The opposite outcome (i.e., u∗< un) is obtained in the case of a slow down in the pace of growth of demand (Freitas and Serrano, 2015).
We conclude that convergence towards normal capacity utilization requires that movements in the accumulation rate overreact to changes in the rate of growth of demand. That is possible by means of a flexible ht. In other words, investment follows a flexible accelerator investment function in which propensity to invest slowly adjusts whenever current capacity utilization differs from normal capacity.
It follows that propensity to invest increases in periods of higher growth of demand, and decreases when aggregate demand slows down. In equilibrium, a larger propensity to invest is associated with a higher growth rate. These results are consistent with expected features arising from competition.6 Firms tend to keep a certain degree of spare capacity to attend unexpected peaks of demand. However, spare capacity must lie within a limit.
Otherwise, it generates unnecessary costs for firms.
Hence, if capacity utilization is below the normal degree, firms slow down capital accumulation to avoid the costs of spare capacity (Freitas and Serrano, 2015).
This discussion allows us to write down the expression for the adjustment in the investment share as shown in equation 4.7. Here, γ stands for the sensitivity of the investment share to deviations of capacity utilization with respect to normal capacity, being larger than zero and lower than one. In turn, un represents the normal degree of capacity utilization.
6 Moreover, Girardi and Pariboni (2020) provide empirical evidence supporting the conclusion that the investment share correlates with the rate of growth of autonomous expenditures.
124 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE ˙ ht = htγ(ut −un) [4.7] From equations 4.6 and 4.7, we can obtain the expression for the growth rate of output presented in equation 4.8 7. Note that gt stands for the growth rate of output, and gCAt stands for the growth rate of the autonomous consumption.
gt = htγ(ut −un) 1 −ct −ht + gCAt [4.8] Equation 4.8 shows that output growth is determined by the growth of autonomous consumption plus the change in the propensity to invest when utilized capacity differs from normal capacity. As the rate of capacity utilization converges towards the normal rate, the propensity to invest stabilizes. When this happens, output growth is solely determined by the growth of autonomous consumption.
Now, taking equation 4.4, and substituting in equations 4.3 and 4.8, we obtain: ˙ ut = ut gCAt + htγ(ut −un) 1 −ct −ht − ht v ut + δ [4.9] The system formed by equations 4.7 and 4.9 is sufficient for the dynamic analysis of the basic model presented in this section. In equilibrium, ˙ ut = ˙ ht = 0. As a consequence, capacity utilization converges to normal capacity. Both the output growth rate and the accumulation rate converge to the growth rate of autonomous demand (Freitas and Serrano, 2015; Serrano, 1995b). Putting it differently, in equilibrium, g∗ t = g∗ Kt = g∗ CAt. As stressed by Freitas and Serrano (2015), the conclusion that the growth of demand determines the growth of output and capital stock is in line with demand-led growth theory.
Plugging the equilibrium values in equation 4.3 gives us the equilibrium investment share, shown in equation 4.10. The equilibrium propensity to invest is a positive function of the rate of growth of autonomous demand.
7 The equation is obtained from equations 4.6 and 4.7. By taking natural logarithm of equation 4.6, we obtain: lnYt = ln(1) −ln(1 −ct −ht) + ln(CAt) Then, by canceling the term ln(1) and after taking derivatives with respect to time on both sides, we obtain the expression below.
gt = ˙ ht 1 −ct −ht + gCAt Finally, we substitute the value for ˙ ht, as given by equation 4.7, and then we obtain equation 4.8.
125 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE h∗ t = v un (gCAt + δ) [4.10] Freitas and Serrano (2015) show the model to be stable, as long as the “expanded” marginal propensity to spend is smaller than one. Stated differently, stability requires that c + v + v un (gCAt + δ) + γv < 1. This expression represents the marginal propensity to spend added of the term related to changes in the propensity to investment out of equilibrium. If propensity to invest’s reaction to deviations of capacity utilization from normal capacity (given by γ) is sufficiently small, then the model is stable.
126 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE 4.3 The Sraffian supermultiplier in an open economy with government In this section, we extend the Sraffian supermultiplier to analyze an open economy with government.
In this case, we introduce two autonomous expenditures: exports and public expenditure. Both expenditures are autonomous once they are neither financed out of current income nor caused by production decisions.
Indeed, exports do not depend on domestic demand but rather on foreign demand. In turn, government counts with several degrees of freedom to run deficits and expand (or compress) public expenditures independently of current income and taxation. That is particularly true for countries whose public debt is invoiced in their sovereign currency (Cesaratto, 2016; Wray, 2015). Introducing foreign trade and government expenditures implies that we must also deal with taxation and propensity to import. For simplicity, we restraint the analysis to the case of two autonomous expenditures, abstracting from autonomous consumption.
Hence, hereafter, consumption is assumed to be fully induced by disposable income.
Let us start deriving the equilibrium condition for the goods market.
We keep notation whenever possible, identifying only the new variables.
Yt = Ct + It + Gt + Xt −Mt [4.11] Here, G stands for public expenditure; X, exports and M, imports. We carry on setting consumption as induced by the current level of disposable income, as in equation 4.12. τ denotes the income tax rate. As before, c stands for the marginal propensity to consume from disposable income. Moreover, both c and τ are assumed to be constant in time.
Ct = c(1 −τ)Yt [4.12] Along the lines of the capital stock adjustment principle, investment depends on both propensity to invest and current income.
It = htYt [4.13] Imported goods are either inputs to domestic production or part of the induced consumption. For this reason, we assume that imports are proportional to the level of income, with a constant propensity to import (m).
127 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE Mt = mYt [4.14] Both autonomous expenditures (G and X) can be grouped in a single term Z, so that Zt = Gt + Xt. In this case, output determination can be written as in equation 4.15. The level of output is determined by total autonomous demand and by the supermultiplier.
The value of the supermultiplier is smaller than the one in previous section, given the inclusion of taxation and imports.
Increases in taxation reduce disposable income, decreasing the impact of the consumption multiplier on aggregate income. In turn, with the inclusion of foreign trade, part of demand leaks to imported goods, mitigating the impact of autonomous demand on domestic income.
Yt = 1 1 −c(1 −τ) −ht + m Zt [4.15] The propensity to invest behaves according to the same process described in the previous section. We repeat it in equation 4.16.
˙ ht = htγ(u −un) [4.16] Finally, we can determine output growth (g) according to the growth of aggregate demand. We obtain equation 4.17 by taking log and derivatives from equation 4.15, and then substituting equation 4.16.
gt = htγ(ut −un) 1 −c(1 −τ) −ht + m + gZt [4.17] As before, output growth depends on the growth of autonomous demand and changes in the propensity to invest. In the present case, however, autonomous demand is composed of exports and public expenditure. Another difference is that the impact of h is weighted by a supermultiplier that includes taxation and the propensity to import.
If the growth rate of autonomous expenditures (gZ) remains stable for sufficient time, the degree of capacity utilization converges towards the normal capacity (u = un). Hence, the investment share stabilizes (˙ h = 0), and the growth rate of output converges to the growth rate of autonomous expenditures (Freitas and Serrano, 2015; Serrano, 1995a,b) (as in equation 4.18).
128 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE g = gZ [4.18] We can now proceed to the discussion of the growth of autonomous demand.
Hereafter, we depart from the baseline supermultiplier model, exploring the results from accounting for autonomous public expenditures and exports. That is part of the novelty proposed in this chapter.
gZ depends on the growth rate of government expenditures and the growth rate of exports. From the definition of Z, we obtain that gZ is given by the average of the growth rates of exports and public expenditure, weighted by the each expenditure’s share on Z. This relation is expressed in equation 4.19.
gZ = σtgGt + (1 −σt)gXt [4.19] Where, σt = Gt Zt [4.20] Note that the ratio between government expenditure and total autonomous expenditure (σt) varies in time. Departing from equation 4.20, we can take logs and derivative with respect to time, obtaining equation 4.21.
Finally, by substituting the definition for gZt given in equation 4.19, we can define the rate of change of σ as follows in equation 4.22 below. The equation shows that σ changes whenever the rate of growth of exports and public expenditure differ.
˙ σt = σt(gGt −gZt) [4.21] Thus, ˙ σt = σt(1 −σt)(gGt −gXt) [4.22] If public expenditure grows faster (slower) than exports, then σt continuously rises (falls). We can conceive that one of the expenditures keeps growing faster than the other until σ converges to one of the extreme positions — that is, either σ = 0 or σ = 1. In this case, one of the autonomous expenditures dominates the explanation of the growth of autonomous demand. However, as modern economies usually count with positive demand from both exports and public expenditures, we argue that this condition does 129 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE not usually hold.
The fully adjusted position of the supermultiplier model is only obtained under a persistently stable growth rate of autonomous expenditures (Freitas and Serrano, 2015).
In the model presented in last section, steady-state was achieved when the autonomous consumption kept persistently growing at a constant rate. However, when we model two autonomous expenditures, the fully adjusted position requires that both expenditures grow persistently at the same rate. Only in this case we have a persistent growth rate for the total autonomous expenditures, in the model with two sources of autonomous demand. That also implies that each autonomous expenditures keeps a constant share in total autonomous demand. Thus, σ is constant in the fully adjusted position.
Appendix A shows that a constant growth rate for the total autonomous expenditures can only be obtained either if both expenditures constantly grow at the same rate or under a very specific growth pattern. In the appendix, we show the second alternative relies on specific assumptions concerning the growth rate of each autonomous component.8 We abstract from it in the remaining of this chapter.
When modeling growth led by autonomous consumption and autonomous public expenditure, Freitas and Christianes (2020) focus on the steady-state analysis. Therefore, the authors assume each expenditure keeps a stable share in total autonomous expenditures. The level of public expenditure is set as Gt = σZt, with a constant σ.
Likewise, the level of autonomous consumption is defined as Cat = (1 −σ)Zt. As a consequence, public expenditure growth rate and autonomous consumption growth rate are always equal in this analysis.
In this chapter, we will assume both public expenditures and exports grow at the same rate when the objective is to analyze the steady-state of this growth model. That is done in section 4.3.1, and in section 4.4.3.
Nevertheless, we also allow for different growth rates and changes in the share of each expenditure in autonomous demand.
Section 4.7 presents the simulation of different experiments, exploring the dynamics out of the fully adjusted position. We show in sections 4.5 and in section 4.7 that the share of each expenditure on autonomous demand matters for the equilibrium value of both public and external debt ratios.
Managing the share of public expenditure in autonomous demand is a way fiscal policy can bring the equilibrium foreign debt-to-exports ratio towards the limit acceptable by the international financial market.
The first step, however, is to analyze the equilibrium of the model. We do that in the next section.
8 As shown in Appendix A, autonomous expenditures must grow at time-varying rates, and one of the expenditures must exactly compensate from movements in the other and from changes in σ to keep a constant value for gZ. This possibility is disregarded in this chapter since it requires an arbitrary growth pattern for each autonomous expenditure.
130 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE 4.3.1 Fully adjusted position We can describe the dynamics of the model with the equations below.
˙ ut = ut σtgGt + (1 −σt)gXt + htγ(ut −un) 1 −c(1 −τ) −ht + m − ht v ut + δ [4.23] ˙ ht = htγ(ut −un) [4.24] ˙ σt = σt(1 −σt)(gGt −gXt) [4.25] The model achieves the fully adjusted position when ˙ ht = ˙ ut = ˙ σt = 0. Hence, we are adding one condition to the basic model presented in section 4.2.1.
According to the previous discussion, it follows that the condition ˙ σt = 0 gives us g∗∗ Z = gGt = gXt (as in equation 4.25).
Note that the equilibrium level of σt is not relevant.
In this case, gZ is sufficiently persistent to lead the propensity to invest towards equilibrium. Thus, as shown in 4.24, u∗∗= un. Finally, equation 4.23 gives us an expression for the equilibrium propensity to invest similar to the one presented in section 4.2.1: h∗∗= v un (gZ + δ).
4.3.2 Stability analysis The equilibrium of the system described by equations 4.23, 4.24, and 4.25 is locally stable under the same conditions as the supermultiplier model. The model is therefore locally stable when the expanded marginal propensity to spend (that is, the propensity to spend added of the term γv associated to the accelerator investment function) is smaller than one (Freitas and Serrano, 2015). The condition is expressed as follows: γv + c(1 −τ) + v un (gZ + δ) −m < 1 [4.26] We conclude that the simple inclusion of another autonomous expenditure does not affect the local stability of the supermultiplier growth model. The complete stability analysis can be found in the appendix.
131 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE 4.4 Public Expenditure, Exports and Debt stability The original supermultiplier model introduced autonomous demand in the form of autonomous consumption, which corresponds to the consumption of capitalists.
Another strand of the literature on demand-led growth highlighted exports as a source of autonomous demand.
This view acknowledges the twofold role of exports in economic growth.
On the one hand, exports constitute part of aggregate demand, affecting output and employment, and growth.
On the other hand, exports provide foreign currency, loosening the external constraint and allowing the country to pay for the import of the inputs and capital goods required to economic growth (Kaldor, 1966).
Based on this premise, the theory balance of payments constraint growth was developed by Thirlwall (1979).9 Thirlwall (1979) argues that the balance of payments represents the dominant constraint to demand expansion in open economies. In this view, demand and income changes bring the balance of payments into equilibrium rather than changes in exchange rates or relative prices (Thirlwall, 2012). Thirlwall expressed his preference for the parsimonious model that builds on long-run balanced trade assumption.
That implies, though, assuming that trade flows dominate the explanation of the Balance of Payments in the long run.10 Along these lines, Thirlwall (1979) introduces a balanced trade account condition, meaning that nominal exports must be equal to nominal imports, as a long-run constraint to demand-led growth. From this condition, the author derives the result known as Thirlwall’s Law. Abstracting from changes in the terms of trade, the growth rate compatible with balanced trade is given by the growth rate of exports divided by import’s income-elasticity. This result is obtained in few steps. First, balanced trade implies a long-run equality between imports and exports.
Assuming a two-good economy, being one imported and the other domestically produced and exported, we can express a balanced trade account as in the following equation: ep∗ MMt = pXXt.
p∗ M stands for the international price of the imported commodity, e is the exchange rate expressed as units of domestic currency per unit of foreign currency, and pX corresponds to the price of the exported commodity in domestic currency. Then, we can define both imports and exports as demand functions, specified as multiplicative with constant elasticities. Therefore, quantities exported or imported vary according to the 9 See Thirlwall (2012) for a historical overview of this approach.
10 However, this is often not the case. Ocampo (2016) argues that the Balance of Payments of emerging markets are dominated by liquidity cycles of international financial markets. In this regard, de Medeiros (2020) posits that: “[t]he history of BoP crises in developing countries since the nineteenth century shows that overborrowing and overlending by bankers have played an autonomous role in the rapid deterioration of external solvency”.
132 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE income (domestic income affects the demand for imports, and foreign income affects exports), prices, and both income and price-elasticity of demand (Thirlwall, 1979).
X = α pX epM η (Y∗)ϵ M = β epM pX ψ (Y)π Taking natural logarithms of the last two equations and then differentiating with respect to time, and, finally, substituting in the balanced trade condition gives us the growth rate compatible with the external constraint (denoted as gb).11 Hats denote the rate of change in variables.
gb = (1 + η + ψ)( ˆ pX −ˆ pM −ˆ e) + ϵg∗ π Finally, abstracting from changes in relative prices gives us the following expression, known as Thirlwall’s Law: gb = ϵg∗ π = gX π [4.27] Therefore, under constant terms of trade, the growth rate compatible with balanced trade is given by the growth rate of exports divided by the income elasticity of demand for imports. Thirlwall (1979) labels this as the balance-of-payments constrained growth rate. The original model does not account for the role of income and capital flows in determining the balance of payments result.
For this reason, Thirlwall and Hussain (1982) introduce capital flows in the balance of payments constraint growth model.
They argue that countries with positive net capital inflows can finance trade deficits in the initial period.
However, only increasing capital flows would allow for the persistence of trade deficit in a growing economy. If capital flows remain constant, the growth rate consistent with balance-of-payments equilibrium will be lower than the original Thirlwall’s Law (Thirlwall and Hussain, 1982, p. 502-4). However, rather than considering the absolute level of capital flows, Moreno-Brid (1998) suggests considering the ratio between capital flows and domestic income. Hence, Moreno-Brid (1998) finds the growth rate compatible with a constant ratio between the trade deficit and income.
11 Taking natural logarithm and derivatives with respect to time of the three equations gives us: ˆ e + ˆ p∗ M + gM = ˆ pX + gX gX = η( ˆ pX −ˆ p∗ M −ˆ e) + ϵg∗ gM = −ψ( ˆ pX −ˆ p∗ M −ˆ e) + πg By substituting the last two equations in the first one, we can obtain the expression in the text.
133 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE This model shows that capital flows can finance trade account deficits in non-explosive dynamics, keeping a stable ratio between capital inflows and income. Growth is still constrained by the balance of payments but under different conditions (Moreno-Brid, 1998). Building on this framework, Moreno-Brid (2003) further develops this model, including net interest payments. Finally, Caldentey and Moreno-Brid (2019) reexamine both Thirlwall (1979) and Moreno-Brid (1998) contributions, emphasizing the effect of long-run trends in terms of trade.
Indeed, the previous models abstracted from the impact of changes in the terms of trade in either tightening or loosening the external constraint.
Nevertheless, Caldentey and Moreno-Brid (2019, p.
464) argue that the “inclusion of the terms of trade (if they are significant) implies that the rates of growth compatible with balance-of-payments equilibrium should be calculated on the basis of income rather than gross domestic product (GDP). The inclusion of capital flows implies that the basic balance rather the current account is the correct measure of a country’s external position.” Departing from the previous contributions, Barbosa-Filho (2001) suggests framing the balance of payments constraint in terms of long-run sustainability of external debt rather than in a constant deficit or balanced trade condition.
The accumulation of foreign liabilities leads to a convergent and stable external debt-to-income ratio whenever the income growth rate is larger than the interest rate on debt service.12 Building on this approach, Bhering et al. (2019) point out that the foreign debt should be compared to the level of exports rather than domestic income. The authors emphasize the currency mismatch between domestic income and foreign debt, stressing that the value of exports consists of a proper measure of solvency of a country for liabilities taken in foreign currency. Thus, the authors follow Barbosa-Filho (2001), assessing the long-run convergence of external debt. However, they normalize debt and balance of payments variables by exports rather than domestic output.
Models following Thirlwall’s tradition usually assume that exports are the only source of autonomous demand in the long run.13 As pointed by Bhering et al. (2019), the 12 Thus, given the initial values of exports and imports, “it is straightforward that d [that is, the foreign debt-to-income ratio,] is stable as long as the home growth rate exceeds the real cost of foreign debt in home currency” (Barbosa-Filho, 2001, p. 396).
13 Thirlwall (1997, p.
378) claims Balance-of-Payments constrained growth model build on the same assumptions as Harrod’s foreign trade multiplier, “namely, that exports are the only component of autonomous demand, that trade is balanced, and the terms of trade remain unchanged“. McCombie (1985) introduces other source of autonomous expenditures, though he posits that it must adjust to the growth of exports in the long run. The assumption that exports are the only autonomous expenditure is often associated with Kaldor‘s remarks on growth. According to the author, “[f]rom the point of view of any particular region, the ’autonomous component of demand’ is the demand emanating from outside the region; and the Hicks’ notion of the ’super-multiplier’ can be applied so as to express the doctrine of the foreign trade multiplier in a dynamic setting. So expressed, the doctrine asserts that the rate of economic development of a region is fundamentally governed by the rate of growth of its exports. For the exports, via the ’accelerator’; will govern the rate of growth of industrial capacity, as well as the growth of consumption; it will also serve to adjust (again under rather severe simplifying assumptions) both the level, and the rate 134 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE conclusion that growth is always constrained by the Balance of Payments depends on this assumption. Introducing another source of autonomous demand means that the external constraint may not be binding. Growth is also affected by the growth rate of the other autonomous expenditure. Hence, if the other autonomous expenditure grows at a pace slower than exports, the economy (and, thus, imports) grows less than exports, obtaining trade surpluses.
Within the supermultiplier literature, exports have been less explored in comparison with other autonomous expenditures.
Dejuán (2017) includes exports as the only autonomous expenditure in an open economy version of the supermultiplier model.
However, the author focuses on the stability of growth in models with an accelerator investment function. Nah and Lavoie (2017) include exports in a supermultiplier model, further assuming that exports growth rate is given by an exogenous trend plus a negative effect of the wage share. The authors focus on the relation between growth and distribution in the ‘traverse‘ to the fully adjusted position. Finally, exploring different sources of autonomous demand, Dutt (2019) briefly discusses the case of a supermultiplier model with exports as the only autonomous expenditure.
Unlike the mentioned contributions, in this chapter, we account for exports and public expenditure as sources of autonomous demand. We discuss how the model presented in this chapter relates with the Balance of Payments constraint to economic growth. We also define conditions for the long run stability of both public and external debt ratios.
Convergence of debt ratios to stable values supports the feasibility of growth being led by public expenditure and exports in the long run.
The Sraffian supermultiplier as proposed by Serrano (1995a) relied on autonomous consumption as the source of autonomous demand in a closed economy without government. Serrano (1995a,b) assumed autonomous consumption to be financed out of accumulated wealth. We may also conceived it as financed out of credit.14 In any case, the long-term sustainability of the autonomous expenditure must be evaluated, ensuring a stable stock of wealth or a sustainable level of debt are consistent with the growth of the autonomous expenditure in the long term.15 Along these lines, Pariboni (2016b) assessed the sustainability of private debt in a supermultiplier model led by consumer credit. The author also showed that problems to the long-run sustainability of private debt arise when another autonomous expenditure is introduced in the analysis.
Thus, if debt-financed consumption grows faster than the other component of autonomous demand, “the accumulation of debt is faster than the growth of the whole of growth, of imports to that of exports” (Kaldor, 1978, p. 146). See Palumbo (2009) for an interpretation of Kaldor’s ideas on foreign trade and growth.
14 See Cesaratto (2017).
15 Recently, Hein and Woodgate (2020) raised this point. They claim that financial stocks (as wealth and debt) should be deeply analyzed in growth models following the supermultiplier framework since the long-run stability of growth depends on these stocks.
135 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE autonomous part of demand, which determines the rate of growth of output” (Pariboni, 2016b, p. 227). Therefore, when more than one autonomous expenditure is considered, the long run stability of debt also depends on the other autonomous components’ growth rate. We will show in the next pages that this is also the case in an integrated analysis of public and foreign debt sustainability.
The stability of public debt has long been a concern in theoretical debates, particularly in approaches compatible with demand-led growth perspective (Domar, 1944). Public debt dynamics has been analyzed in terms of the ratio between public debt and aggregate income, what we label as the public debt-to-income ratio. Domar (1944) concluded inter alia that the public debt-to-income ratio converges to an equilibrium level in the long run as long as the growth rate is larger than the interest rate. Hereafter, we refer to this result as the Domar stability condition. For this reason, Domar (1944) suggests conceiving the issue of public debt as a matter of economic growth.16 In light of the supermultiplier growth model, we can account for public expenditure as a relevant component of autonomous demand.
Indeed, countries with monetary sovereignty can finance deficits with taxes, debt issue, and monetary emission.17 Being rather independent of taxation, public expenditure growth can be considered exogenous to current income and firm’s production decisions, thereby being an autonomous expenditure. Allain (2015) modeled public expenditure as an autonomous expenditure, showing that public expenditure may act as a stabilizer of economic growth in the supermultiplier model.18 However, the author does not account for the public deficit and public debt, since he assumes the public budget is kept balanced by tax rate adjustments.
Cassetti et al. (2017) simulates a supermultiplier model with public expenditure as the only source of autonomous demand.
The author discusses the interaction of growth with inflation, distribution, and public debt. In turn, Hein (2018) introduces public expenditure, debt dynamics, and the related distributive concerns in a Kaleckain version of the supermultiplier model.
The analysis considers only one autonomous expenditure in a closed economy setting.
Hein (2018) concludes that primary public deficit (or surplus) and public debt converge to stable values as long as the growth rate of public expenditure is larger than the interest rate on debt service and that the supermultiplier’s stability condition holds.
16 Indeed, as stated by the author, “the problem of the debt burden [that is, the problem of the magnitude of primary budget surplus required to pay for debt service,] is essentially a problem of achieving a growing national income” (Domar, 1944, p. 822).
17 See, for instance, Wray (2015) and Cesaratto (2016) for descriptions of the circuit of expenditures decisions and government finance in economies with monetary sovereignty.
18 Allain (2015) points out that autonomous demand had not been properly accounted for in Kaleckian growth literature yet, even in the case of expenditures that may be typically considered autonomous as public expenditure and exports. In this literature, “government expenditure or public deficits are assumed to be proportional to capital stock and then to grow at the same rate. When exports are introduced, they are partly autonomous, but the results of the models do not fully take into account the consequences of this exogeneity” (Allain, 2015, p. 5-6).
136 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE These three contributions consider government expenditure as the only source of autonomous demand, avoiding the complications arising from the inclusion of more than one autonomous expenditure. However, such a shortcoming was addressed by Freitas and Christianes (2020), and Hein and Woodgate (2020), who include autonomous private consumption in the analysis. Freitas and Christianes (2020) restrain their analysis to the steady-state. They show that, assuming Domar stability condition holds, the growth rate of autonomous demand and the composition of autonomous demand (that is, the share of each expenditure) affect the equilibrium level of public debt-to-income ratio and the equilibrium primary deficit-to-income ratio. In this model, an increase in the equilibrium growth rate of autonomous demand reduces the equilibrium public debt-to-income ratio.
In contrast, an increase in the government share in autonomous demand increases the equilibrium public debt-to-income ratio.
Freitas and Christianes (2020) also discuss the relation between the level of debt and the impact of interest rate and taxation on distribution, showing that changes in distribution have a feedback effect on the level of output (but not in its equilibrium growth rate).19 In turn, Hein and Woodgate (2020) also account for both autonomous consumption and government expenditure in a supermultiplier model framed within the Kaleckian perspective.
The authors focus on the relation between the consumption out of the financial income of public debt creditors and Harrodian instability.
What happens when we combine exports and public expenditures as sources of autonomous demand?
In the next sections, we discuss how introducing public expenditure affects the external constraint given by the stability of foreign debt. Thus, we carry on in the analysis of a supermultiplier model with exports and public expenditure as components of autonomous demand, as presented in section 4.3. In the next section, we approach public and foreign debt dynamics in this approach.
4.4.1 Public debt Let us start considering the case of public debt. The main reference consistent with the model presented here is Freitas and Christianes (2020). As discussed above, Freitas and Christianes (2020) develop a supermultiplier model with autonomous private consumption and autonomous public expenditure, presenting conditions for public debt stability. Differently from these authors, we initially allow for the share of each autonomous expenditure (in total autonomous demand) to change in time. Changes in these shares also occur in the simulations presented in section 4.7.
The change in total public debt, denoted by ˙ B, is described in equation 4.28. The change in the public debt depends on the government expenditure (Gt), the total amount 19 This result is consistent with previous conclusions of the supermultiplier growth model (Freitas and Serrano, 2015).
137 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE of taxes — given by a proportional tax rate (τ) times current income level (Yt) — and the total debt service — depending on the interest rate (i) and the total debt (Bt).
˙ Bt = Gt −τYt + iBt [4.28] The relevant economic variable for the analysis of public debt is the ratio between debt and income. We can denote it as b, as defined below.
bt = Bt Yt [4.29] From this definition, we can deduce the rate of change in the debt-income ratio as follows: ˙ bt = ˙ BY −B ˙ Y Y2 = ˙ Bt Y − ˙ Yt Y b [4.30] Then, we can substitute 4.28 in 4.30, obtaining equation 4.31.
˙ bt = Gt Yt −τYt Yt + iBt Yt −gtbt [4.31] By further simplifying, we get the expression below.
˙ bt = σtzt −τ + ib −gtbt [4.32] Remember that σt stands for the share of government expenditure in total autonomous expenditures. In turn, zt stands for the ratio of autonomous expenditures to total output.
As shown by Freitas and Christianes (2020), the term σtzt −τ corresponds to the primary government deficit (or surplus) to output ratio.
4.4.2 Foreign debt We introduce the analysis of foreign debt’s long-run stability, following Barbosa-Filho (2001). We accept the suggestion of Bhering et al. (2019) to focus on the debt-to-exports ratio, acknowledging the existence of a currency mismatch between foreign liabilities and domestic income. We assume the Balance of Payments is composed of the trade balance, factor income balance and capital account. We also assume the factor income balance is solely composed of external debt service payments. Thus, trade balance plus 138 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE external debt service constitute the current account. Note that while Bhering et al. (2019) develop their model in discrete time, we express the same relations in continuous time.
Still following Bhering et al. (2019), we assume that capital flows (denoted by F) are just enough to cover the deficit in the current account. Hence, the result of the Balance of Payments is equal to zero so that the country neither accumulates nor looses reserves.
This assumption is related to the analytical purposes of the model and does not posit any automatic adjustment of the Balance of Payments as a feature of real economies.
Then, if capital flows are equal to the current account deficit, we can write equation 4.33.
Ft = Mt −Xt + Rt [4.33] As usual, M and X stand for imports and exports. In its turn, R represents the external debt service payments. Hence, it is proportional to the international interest rate (r) and the total debt (D). For simplicity, we assume r to be fixed in time.
Rt = rDt [4.34] The change in external debt will be given by the net entrance of capitals to finance the current account deficit. Thus, ˙ Dt = Ft [4.35] Finally, gathering the previous relations in equations 4.33, 4.34, 4.35, we get: ˙ Dt = Mt −Xt + rDt [4.36] Following Bhering et al. (2019), we study the dynamics of the ratio between external debt and exports. The underlying reasoning is related to the solvency conditions of a country in terms of foreign currency. For this reason, Bhering et al. (2019) reject analyses centered in the external debt-to-output ratio (as done by Moreno-Brid, 1998; and Barbosa-Filho, 2001), defining exports as the relevant variable to explain the long-term solvency of the external debt. Thus, we can define: dt = Dt Xt [4.37] 139 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE Its derivative with respect to time is given by.
˙ dt = ˙ DtXt −Dt ˙ Xt X2 t [4.38] We can rewrite as: ˙ dt = ˙ Dt Xt − ˙ Xt Xt Dt Xt [4.39] Let us set that ˙ Xt Xt = gXt, where gXt is the growth rate of exports, assumed to be constant in time.
By substituting the result for ˙ Dt of equation 4.36, we obtain the expression below.
˙ dt = Mt −Xt + rDt Xt −gXtdt [4.40] Thus, with simple manipulations we can write: ˙ dt = Mt Xt −1 + rdt −gXtdt [4.41] Finally, we obtain the expression 4.42.
˙ dt = Mt Xt −1 + (r −gXt)dt [4.42] As pointed by Bhering et al. (2019), the equation shows that r < gX is a necessary condition for the stability of the ratio of the external debt-to-exports. That implies that exports must raise at a higher rate than the debt service — and it is analogous to the condition pointed in the case of the public debt. In other words, for debt to be sustainable, the growth of debt due to debt service must be slower than exports growth. Otherwise, the rise in debt caused by interests is enough to generate a path of increasing external indebtedness.
Still, we can rewrite expression 4.42 as follows in equation 4.43. In this result, it becomes evident that then the path of foreign debt-to-exports ratio diverges if σt approaches 1.
140 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE ˙ dt = m (1 −σt)zt −1 + (r −gXt)dt [4.43] 4.4.3 Fully adjusted position and debt stability We can summarize the system in five equations, which are presented below. Note that the first three come from the supermultiplier growth model, considering the presence of the two autonomous expenditures. The last two are related to public and external debt.
˙ ut = ut σtgGt + (1 −σt)gXt + htγ(ut −un) 1 −c(1 −τ) −ht + m − ht v ut + δ [4.44] ˙ ht = htγ(ut −un) [4.45] ˙ σt = σt(1 −σt)(gGt −gXt) [4.46] ˙ bt = σtzt −τ + ib −gtbt [4.47] ˙ dt = m (1 −σt)zt −1 + (r −gXt)dt [4.48] Equation 4.46 shows that stability of σ imposes that gX = gG. As pointed in section 4.3.1, the fully adjusted position requires that both autonomous expenditures grow at the same rate (gX = gG). Otherwise, the total autonomous expenditures (Z) cannot grow at a stable rate, which would prevent a formal description of the equilibrium. When both autonomous expenditures grow at the same rate, σ achieves a constant value. In this case, ˙ σt = 0, as can be seen in equation 4.46. Note also that this can happen at any level of σ.
We denote this constant value for σ as σ∗, even though it does not correspond to a specific value. Given this condition, a constant σ allows for a persistently stable growth rate of autonomous demand (gZ). In that case, the economy converges to the fully adjusted position, as in the baseline supermultiplier model. In this equilibrium, the growth rate converges to the growth rate of autonomous demand, propensity to invest stabilizes, and capacity utilization converges to normal capacity. The equilibrium positions for output 141 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE growth, propensity to invest and capacity utilization were presented in section 4.3.1. In this section we discuss the equilibrium levels for public debt-to-income ratio and foreign debt-to-exports ratio.
By setting equation 4.47 equal to zero, we obtain the equilibrium level of the public debt-to-income ratio. The equilibrium for this ratio is expressed in the equation below.
b∗= σ∗z∗−τ gZ −i [4.49] Note that the value for z∗emerges from the system of equations describing the supermultiplier growth model. z∗is given by the inverse of the equilibrium value of the supermultiplier, as shown in the equation below.
z∗= 1 −c(1 −τ) −(gZ + δ) v un + m [4.50] Public debt-to-income ratio converges to a positive value if the public budget is in deficit in the fully adjusted position, that is, if σ∗z∗> τ. The equilibrium level of public debt-to-income ratio is larger when the share of government expenditure in autonomous demand is larger. The condition that the equilibrium growth rate, given by the growth of autonomous demand, is larger than the interest rate is fundamental for the long-term stability of the public debt-to-income ratio. This result was recalled by Freitas and Christianes (2020), being considered a version of Domar‘s stability condition (Domar, 1944). Moreover, note that public debt stability does not depend on the value of σ∗. That is true even in the extreme case in which government expenditure dominates autonomous demand, and σ∗is equal to 1.
In contrast, if exports fully explain autonomous demand with σ∗equal to 0, the public debt-to-income ratio converges to a negative stable value.
Let us now consider the case of foreign debt. By setting equation 4.48 equal to zero, we can obtain the equilibrium level for the foreign debt-to-exports ratio, expressed in equation 4.51. Note that for simplicity, gX was substituted per gZ, since the equilibrium requires that both variables have the same value.
d∗= m (1 −σ∗)z∗−1 1 gZ −r [4.51] Equation 4.51 shows that the country will accumulate debt if, in equilibrium, the term within brackets is positive. This occurs when the country faces persisting trade deficits.
given the propensity to import and the share of exports in output. Otherwise, if the country faces a trade surplus in the equilibrium, it will accumulate foreign credit rather 142 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE than external debt.
The condition that the growth rate of exports is larger than the interest rate on foreign debt is fundamental for the stability of d. If this condition does not hold, then the growth of exports is not enough to offset the increase in the foreign debt caused by the debt service, leading to an increasing foreign debt-to-exports ratio. Moreover, equation 4.51 shows that if the ratio between government expenditure and income is equal to one (i.e., σ∗= 1), debt-to-exports ratio diverges. In the steady-state derived from this condition, government expenditure dominates the determination of autonomous demand growth.
As a consequence of domestic income growth, the economy accumulate trade deficits, since it imports goods and services but it does not export. Hence, an increasing amount of foreign debt is accumulated in each period.
4.4.4 Sensitivity analysis We performed a sensitivity analysis to test the robustness of the complete model, inclusing the supermultiplier model and the dynamics of the public and external debt ratios — i.e., the dynamic system formed by equations 4.23, 4.24, 4.25, 4.47, and 4.48. The sensitivity analysis shows that the model is robust to shocks in its main parameters, provided that the growth rates of public expenditure and exports do not remain persistently different.
We performed a sensitivity analysis through Monte Carlo experiments of random shocks in the growth rate of government and exports, domestic and international interest rates, propensity to import, tax rate, and the sensitivity of propensity to invest to deviations of capacity utilization. Detailed results are reported in the appendix.
Convergence to steady-state of capacity utilization, propensity to invest, government’s share on autonomous demand (σ), public debt-to-income ratio (b), and external debt-to-exports ratio (d) are robust to shocks in parameters as the interest rate (i), and the international rate (r), propensity to import (m), tax rate (τ), and the sensitivity parameter in the investment function (γ). Nevertheless, the convergence of σ, b, and d depends crucially on the growth rate of each of the autonomous expenditures.
Permanent shocks in the growth rate of government expenditure (gG) and exports (gX) that keep the two growth rates persistently different from each other lead to divergence in the three variables.
However, when we assume that the shocks affect gG and gX in the same way, thereby preserving the equality between the two growth rates, σ, b, and d converge to the equilibrium. Noteworthy, capacity utilization and the propensity to invest converge to the steady-state equilibrium regardless of the equality between the growth rates.
143 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE 4.5 The external constraint and Thirlwall’s Law Until now, we have assumed that creditors will finance any level of foreign debt.
However, let us introduce an upper limit to the debt-to-exports ratio — similarly to what has been done in Bhering et al. (2019). This limit may be due to the imposition of an international institution or a threshold effect considered in international capital markets, meaning that above such limit, capital inflows cease. We denote the bound to foreign debt as ¯ d. Naturally, such a constraint does not raise concerns if it is larger than the equilibrium level of foreign debt. Let us focus, thus, on the case in which ¯ d < d∗.
An exogenous bound to external debt generates policy concerns for the domestic economy.
Given this ceiling, the variables that determine the equilibrium external debt-to-exports ratio (that is, the variables in the right-hand side of equation 4.51) must be changed, pushing the equilibrium level of d towards ¯ d. Suppose the current foreign debt-to-exports ratio is sufficiently below ¯ d. In that case, there is room for structural change policies that may affect the composition of output, the propensity to import, and even stimulate export growth as technical progress improves the income elasticity of exports.20 In terms of the model’s parameters, these types of industrial and innovation policies inducing structural change may lead to a lower m, increase the share of exports in output (reducing σ) or even increase the growth of exports (gX). These changes affect the equilibrium foreign-debt-to exports ratio, contributing to lead it towards the upper limit level. Since these policies take time to affect the parameters, their implementation requires that the economy is still able to maintain increasing indebtedness before their results appear.
In contrast, it may be the case that current indebtedness is too close to ¯ d or that policymakers reject structural change policies in favor of slowing the pace of economic growth. Section 4.6 develops a fiscal policy rule compatible with the external constraint.
Previous discussions on the balance of payments constraint to growth have emphasized that corrections in the balance of payments happen through changes in aggregate income rather than exchange rates or relative prices (Thirlwall, 1979, 2012).
Furthermore, Barbosa-Filho (2001) argued that the equilibrium consistent with the Balance of Payments constraint can be achieved by means of demand management, that is, changes in fiscal policy.
Our result also shows how a long-term limit to external finance constrains the growth of public expenditure. This result is in line with Hein and Woodgate (2020), who 20 Araujo and Lima (2007) shows that structural change affecting the composition of output and exports towards a larger share of industries with a larger demand income-elasticity can loosen the external constraint, increasing the growth rate compatible with the Balance of Payments constraint. Cimoli et al.
(2009) argue that some developing economies performed better than others in terms of growth because of structural change in favor of more technological and efficient sectors which allow for a larger output growth within the external constraint.
144 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE suggested the assessment of how financial cycles or constraints affect the sustainability of autonomous expenditures. The external constraint consists of a financial constraint to the domestic economy once it relies on imported inputs or final goods that can only be purchased with an internationally accepted currency. In our model, this currency can be obtained either through exports or by accepting foreign liabilities.
Suppose the availability of capital flows depends on the level of debt, or, in particular, on the foreign debt-to-exports ratio. In that case, the possibilities of economic growth are also shaped by this condition. When the foreign debt-to-exports ratio surpasses the acceptable upper level, capital flows cease, interrupting the availability of the imported inputs and final goods required for production and growth to take place.
The next section analytically shows that introducing a limit to the external debt brings back Thirlwall’s Law as an upper limit to growth in the long run.
4.5.1 Thirlwall’s Law in the supermultiplier In section 4.4.3, foreign debt-to-exports ratio converges to a stable value under two conditions. First, the growth rate of exports must be larger than the interest rate on foreign debt. Second, the share of public expenditures in autonomous demand must be smaller than one (σ < 1). However, introducing a limit to foreign debt-to-exports ratio brings back the external constraint to economic growth. Bhering et al. (2019) argue that if there is a ceiling to foreign debt ratio, the growth rate compatible with the external constraint is given by Thirlwall’s Law.21 We explicitly show that this result holds in a supermultiplier model with autonomous public expenditure and exports.
Therefore, Thirlwall’s Law imposes an upper limit to growth in the long term, although it does not determine output growth rate, which depends on the growth of both components of autonomous demand. We also show that a positive difference between exports growth rate and the international interest rate relaxes the external constraint in the short run. In this case, the economy can grow faster than the rate defined by Thirlwall’s Law even if it must keep a constant foreign debt-to-exports ratio.
Let us carry on now on this analysis. We start from the expression for the change in foreign debt-to-exports ratio (equation 4.42). Since we assumed a constant propensity to import m, imports grow at the same rate as output. Then, we can rewrite the current value of imports as the initial value multiplied by the accumulated output growth. The same procedure can be done with exports’ initial value and exports growth rate. By these 21 After exploring the effects of the introduction of a limit in the foreign debt-to-exports ratio, the authors conclude that “in the longer run, the BoP-constrained rate of growth remains the same: Thirlwall’s law. A result easily understood once we consider that while the level of imports may be (if b > 0, [where b stands for the ratio between the capacity to import and exports]) permanently higher than exports, those imports cannot grow permanently at a faster rate than exports without making the current account and external debt relative to exports grow without limit. (Bhering et al., 2019, p. 491)” 145 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE means, we obtain equation 4.52. The initial values of imports and exports are denoted by the subscript 0.
˙ dt = M0e R t 0 g.dt X0e R t 0 gX.dt !
−1 + (r −gXt)dt [4.52] Suppose we must keep a constant ratio of foreign debt-to-exports. In that case, we set ˙ d = 0.
We keep the assumption that r < gX, otherwise d would be inherently unstable. With simple operations in the last equation, after making ˙ d = 0, we can obtain the expression below.
e R t 0 g.dt = e R t 0 gX.dt X0 M0 [1 −(r −gX)dt] [4.53] And then, by taking natural logarithms of both sides, we obtain equation 4.54.
Z t 0 g.dt = Z t 0 gX.dt + ln X0 M0 + ln[1 −(r −gX)dt] [4.54] The integral of the growth rates over the period [0, t] must be equal to the average growth rates in the same period times t. Hence, we can substitute both integrals by its averages as in equation 4.55. The superscript µ denotes the average over the period [0, t].
gµt = gµ X.t + ln X0 M0 + ln[1 −(r −gXt)dt] [4.55] In this case, the average output growth rate that is compatible with a constant d is given by the equation 4.56.
gµ = gµ X + 1 t {ln X0 M0 + ln[1 −(r −gXt)dt]} [4.56] Therefore, the average growth rate compatible with a constant d is equal to the average growth rate of exports plus a term depending on the initial trade balance and the difference between the international interest rate and the exports growth rate.
Suppose the economy presented a trade deficit in period 0 (with X0 < M0). In that case, the term within braces is positive whenever the condition in inequality 4.57 holds.
According to this inequality, average output growth can exceed average growth of exports when the initial trade account deficit normalized by exports is smaller than the 146 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE effect of exports growth and interest rate on the foreign debt-to-exports ratio.22 M0 −X0 X0 < (gX −r)dt [4.57] The difference between exports growth and the international interest rate has a negative impact on the change in the foreign debt-to-exports ratio. Inequality 4.57 holds if this effect more than compensates the positive impact of the initial trade deficit, relaxing the external constraint.
Therefore, according to this condition, domestic expenditures can grow faster than exports without affecting the foreign debt-to-exports ratio. This result provides further degrees of freedom to gG since the growth of output (g) depends on the growth rate of autonomous demand, which is an weighted average of the growth of the two autonomous expenditures. In this case, if output can grow more than exports, public expenditures are allowed to grow even faster. The smaller the σ, the larger the possible difference between the growth rate of the two autonomous expenditures.
Nevertheless, we can also see in equation 4.56 that the average output growth rate that maintains d constant approaches the average exports growth rate in the long run. As t approaches +∞, gµ converges to gµ X.
That result is not different from the Thirlwall’s Law. According to Thirlwall’s Law, balance of payments constrained growth rate is given by the growth rate of exports divided by the income-elasticity of imports (see equation 4.58, discussed in section 4.4).
In the present case, the income-elasticity of imports (π) is constant and equal to one since we assume the propensity to import to be constant.23 In the equation above, that implies that the growth rate of output must be equal to the growth rate of exports.
gb = gX π [4.58] 22 Inequality 4.57 is obtained as follows.
In equation 4.56, we can see that if the average output growth compatible with a constant d can be larger than exports growth, then the term within braces must be positive.
That is, ln X0 M0 + ln[1 −(r −gXt)dt] > 0 Since we are assuming that X0 < M0, the term ln(X0/M0) is negative. In this case, the previous inequality holds if: ln[1 −(r −gXt)dt] > −ln X0 M0 By operating the exponential function of e in both sides, we obtain the inequality 4.57 shown in the text.
23 Total imports are given by M = mY. Thus, we can calculate the income elasticity of imports as follows: π = dM dY Y M = m m = 1 147 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE Therefore, the introduction of a ceiling for the foreign debt-to-exports ratio brings back the external constraint to economic growth. The growth rate compatible with the external constraint (gb) represents an upper limit to growth. We showed that this growth rate corresponds to the one given by Thirlwall’s Law. As discussed above, for large values of t, gµ gets closer to gµ X, which, for a constant propensity to import, corresponds to the growth rate known as Thirlwall’s Law (Thirlwall, 1979).
Even though it imposes an upper limit to economic growth, the external constraint is not necessarily binding. If public expenditures continuously grow less than exports, the economy does not reach the growth rate given gb. In that case, the foreign debt-to-exports ratio tends to fall. Therefore, the model presented in this chapter allows for two different growth regimes, exploring two alternative regimes included in demand-led growth. As stated by Freitas and Dweck (2013, p. 168), the supermultiplier “theoretical framework conceives two possible growth regimes: a balance of payments constrained demand-led growth process, and a policy constrained (or pure) demand-led growth process”.
148 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE 4.6 Fiscal Policy and the External Debt The sustainability of growth in the long term requires the economy to remain within the external debt limit.
Assuming a given growth rate of exports means that public expenditure growth must be moderated to avoid that the external debt exceeds the threshold.
The external constraint sets a boundary for the growth of domestic autonomous expenditures (in this model, public expenditure). Therefore, fiscal policy management can prevent the convergence of the debt-to-exports ratio towards an excessively high ratio.
Fiscal policy intervention can change the value of σ to keep the economic within the external debt limit. This outcome can be achieved by keeping the growth rate of public expenditure below the growth rate of exports until σ reaches the value of ¯ σ expressed in equation 4.59. ¯ σ corresponds to the ratio between public expenditure and income that is compatible with the upper limit to foreign debt-to-exports ratio, given the growth rate of exports, the propensities to import and to consume, the tax rate, and the international interest rate. Putting it differently, fiscal policy can keep gG < gX, which leads to a decreasing σ, as shown previously in equation 4.46.
This regime must be kept long enough, allowing σ to reach ¯ σ.
When this happens, the growth rate of public expenditure can return to the steady-state value (equal to gX).
Then, the economy achieves a new steady-state compatible with the constraint to external indebtedness.
¯ σ = 1 − m z∗(1 + ¯ d)(gX −r) [4.59] We define a rule for fiscal policy according to which the growth rate of public expenditure is compatible with the limit to foreign indebtedness.
This rule requires evaluating the necessary change in σ in each period and adjusting the growth rate of public expenditures accordingly. Knowing ¯ d and gX allow us to obtain ¯ σ as in equation 4.59. Note that the value of z∗is also known since we are assuming a constant gX. The export growth rate, in turn, gives us the steady-state value for h, the only time-varying parameter in the determination of z∗. However, defining both σt and ¯ σ is not enough to establish the rule for the growth rate of public expenditures. We still need to find the speed of adjustment of σt that is compatible with the limit in foreign-debt-to exports ratio.
This speed is proportional to the distance between the current foreign-debt-to export ratio and the ratio’s upper limit. In other words, the sooner the current path of debt tends to achieve ¯ d, the faster must be the convergence of σt towards ¯ σ.
Let us further explain this mechanism with a brief example, represented in figure 4.1.
Assume that according to current trends, the economy will converge to a steady-state compatible with the limit to the foreign debt-to-export ratio ( ¯ d0). Further, assume that, in 149 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE FIGURE 4.1: Foreign Debt-to-Exports Ratio and Shock in ¯ d period t0, an international shock in global financial liquidity reduces the upper limit to external indebtedness from ¯ d0 to ¯ d1. The new upper limit for d is no longer consistent with the equilibrium ratio. Hence, fiscal policy management must bring public expenditure share in autonomous demand towards a value compatible with the new limit to foreign debt-to-income ratio.
The speed of this adjustment depends on the period in which debt would surpass the threshold ¯ d1. An approximation for this period can be obtained through a Taylor expansion of the foreign debt-to-exports ratio. We denote this linear approximation as ¯ t.24 Evaluating ˙ d in the period of the shock gives us the expression below.
¯ t = t0 + ¯ d1 −d0 ˙ d0 [4.60] Where ˙ d0 corresponds to the change in d evaluated in period t0.
˙ d0 = m (1 −σ0)z0 −1 + (r −gX)d0 [4.61] Equation 4.60 defines the time horizon in which fiscal policy can act before the debt-to-exports ratio achieves its upper limit. Thus, the difference ¯ t −t0 defines the time interval 24 A higher-order Taylor expansion can provide a more precise estimate. We use a linear approximation in order to keep the analysis simple and emphasize the theoretical argument.
150 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE within which such changes must take place. We can, thus, define that the ratio between public expenditures and total autonomous demand changes continuously at a rate given by equation 4.62. This equation sets the proportional change in σ, happening in each instant of time, required to avoid that d exceeds ¯ d. Hence, this fiscal policy provides a gradual adjustment in σ since the change in this variable is distributed through several periods.
˙ σ0 = ¯ σ −σ0 ¯ t −t0 [4.62] We can now substitute ˙ σ0 in equation 4.46 to obtain a rule for gGt. Equation 4.46 determines ˙ σ according to the current level of the parameter and the difference between gGt and gXt. By isolating gGt in this equation, we obtain the following rule for the fiscal policy.
gGt = gX + ˙ σ0 σ0(1 −σ0) [4.63] The second term of the right-hand-side is negative, according to the definition of the problem. Note that the concern with external debt convergence appears when ¯ σ < σ0, which, in turn, means that ˙ σ0 must be negative. Therefore, equation 4.63 defines the growth rate of public expenditure consistent with the objective of reducing the share of public expenditure in autonomous demand. It is given by the growth rate of exports minus the required change in the share for each period (that is, ˙ σ0) divided by the product σ0(1 −σ0).
Nevertheless, we can conceive an opposite case in which the economy converges towards a d smaller than ¯ d. If this is the case, fiscal policy can lead σ to a larger value of ¯ σ by setting the growth rate of public expenditures temporarily above the growth rate of exports. The rule can follow the same principle presented above, with the opposite sign.
If such policy is executed, it leads the economy to a permanently higher output than if it had not been executed.
Returning to our example, the rule in equation 4.63 defines a growth rate for public expenditure smaller than the growth rate of exports. In turn, that leads to the desired outcome of this policy, i.e., the decrease in the share of government in autonomous expenditures (σ).
Nevertheless, each instant σ falls, the whole curve d is shifted downwards, as shown in figure 4.2.
This shift, in turn, changes the estimate of the period in which the d achieves the limit.
Now, this period corresponds to ¯ t1.
Considering this change, the process described above can be repeated periodically, making the adjustment of σ even smoother. In this case, the fiscal policy rule is updated 151 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE FIGURE 4.2: Foreign Debt-to-Exports Ratio and Fiscal Policy periodically, changing gGt according to the shifts in the d curve.
This policy alternative has the advantage of being very gradual, avoiding a sudden interruption on the growth path, with its economic and social consequences. However, such policy requires framing the issue of external debt as a matter of long term equilibrium, which is possible when the short-term liquidity in foreign currency is guaranteed.
152 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE 4.7 Simulations We rely on numerical simulations to analyze five different experiments. All of them start from a steady-state position, in which both components of autonomous demand grow at a 4 percent rate. Initially, the degree of capacity utilization is equal to the normal degree, and the propensity to invest is equal to its steady-state value. We discuss the evolution of foreign debt and public debt ratios, and growth of autonomous demand and output in each of the five cases.
First, we present a baseline case. In this case, both foreign debt and public debt converge to a stable positive value.
Foreign-debt-to exports ratios converges to a positive value because [m/(1 −σ∗z∗)] −1 > 0.
In other words, the economy accumulates external debt because it maintains trade account deficits in the equilibrium.
Foreign debt-to-exports ratio converges to a stable value because gX > r in the baseline case.
In turn, the public debt-to-income ratio converges to a positive value because σ∗z∗−τ > 0. That means that the public budget presents deficits in the equilibrium.
Convergence of public debt-to-income ratio results from the assumption that g > i, that is, output growth rate is larger than the interest rate on public debt.
The other four cases explore different features within the same framework. They keep most of the parameters of the baseline case, varying with respect to one particular characteristic to examine the effect of such change. The particularities of each case are summarized in table 4.1, with the results of the simulation for key variables.
The parameters of the experiments are presented in a table in appendix E.25 TABLE 4.1: Summary of the Results of the Simulations Case 1 Case 2 Case 3 Case 4 Case 5 Name Baseline Explosive d Explosive b Fiscal Policy Structural Change Change -r = 0.05 i = 0.045 Policy rule (p.30) gX = 0.045 (p.40) g∗ 0.04 0.04 0.04 0.04 0.045 b∗ 1.0126 1.0126 +∞ 0.8022 0.3777 d∗ 4.2944 +∞ 4.2944 2.5000 2.3927 h∗ 0.2029 0.2029 0.2029 0.2029 0.2118 z∗ 0.4096 0.4096 0.4096 0.4096 0.4007 σ∗ 0.4138 0.4138 0.4138 0.4039 0.3969 Figure 4.3 compares the baseline case with cases 2 and 3. Case 2 reveals the explosive dynamics for foreign debt-to-exports ratio, obtained when the interest rate on foreign debt service (r) is higher than the growth rate of exports. Case 3 shows the divergent pattern for the public debt-to-income ratio. Similarly to the previous case, the interest rate on public debt (i) is higher than the output growth rate.
25 Simulation exercises employed the software MATLAB 2019b. The code is available upon request.
153 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE FIGURE 4.3: Evolution of foreign debt-to-exports ratio (right) and public debt-to-income ratio (left) for Cases 1,2, and 3.
Note that according to table 4.1, the experiments show that other than the particular debt ratio that follows an explosive path, the remaining variables converge towards equilibrium values. In fact, they converge towards the same equilibrium as the baseline case. Nevertheless, this conclusion depends on the assumption that the fast increase in debt ratios does not affect the parameters. That may not be the case in reality. A fast increase in public debt, for instance, may cause a fall in the growth rate of public expenditures, changing the pattern observed in the simulation. Instead, it may motivate monetary authorities to reduce i, pushing the public debt-to-income ratio to a stable path.
If foreign debt-to-exports is the one following a divergent path, then above a threshold value for d capital inflows may cease. That makes the economy unable to pay for its external liabilities.
Consequently, it may run for the support of international organizations. Sometimes that means a debt renegotiation with a lower interest rate.
Sometimes financial support also requires the imposition of large real exchange rate devaluations and cuts in public expenditures. If the real devaluation is successful, it reduces the propensity to import, and reduce the propensity to consume due to its regressive impact on distribution.
In any case, these different scenarios show that parameters do not tend to remain still while debt ratios follow and explosive dynamics.
However, we abstract from those complex feedback effects since they do not arise from any mechanical change in the parameters but are a contingent outcome from policy decisions. We take that parameters as given, not assuming any type of policy response in these cases.26 We can, thus, move forward to the policy choices presented in cases 4 and 5.
26 This is not equivalent to stating that the parameters must remain constant. However, it would not be accurate to introduce mechanical adjustment in the parameters due to an explosive debt dynamics. Future research can explore alternative scenarios of economic policy in the face of increasing debt ratios.
154 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE Case 4 describes the case in which there is a ceiling to foreign debt-to-exports ratio, and fiscal policy follows the rule presented in section 4.6. This experiment builds on the same parameters as the baseline case. The only difference is that the fiscal policy rule is executed from period 30 onward. Figure 4.4 compares the performance of both debt ratios in the baseline case with case 4. It shows that fiscal policy intervention is successful in keeping the foreign debt-to-exports ratio below the threshold of ¯ d. The objective of fiscal policy is to reduce the share of public expenditures on autonomous demand by temporarily slowing the pace of public expenditures growth. The fall in the steady-state value for σ implies also a fall in the equilibrium public debt-to-income ratio.
In the baseline case, b converges to 1.01, while after the intervention of fiscal policy it goes to 0.80.
FIGURE 4.4: Debt ratios in the baseline case (left) and the fiscal policy case (right) Additional details on case 4 are presented in figure 4.5.
The figure shows the continuous fall in the share of public expenditures in autonomous demand (σ) after the intervention of fiscal policy. This share slowly converges to ¯ σ, which corresponds to the share of public expenditure in autonomous demand that is compatible with the ceiling to foreign debt-to-exports ratio.
The graph at the bottom-right shows the evolution of the growth rates of output and each component of autonomous demand.
Public expenditures’ growth rate falls immediately after the intervention of fiscal policy. Output growth also falls in the same period, but less than public expenditures, since exports keep growing at the same rate.
Both public expenditures and output growth rates converge to the growth rate of exports. This convergence happens as fiscal policy brings σ closer to ¯ σ, implying the stabilization of the level of d. The picture also shows this adjustment is very gradual. In spite of the fall in public expenditures growth rate, it remains very close to the growth 155 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE FIGURE 4.5: Evolution of debt ratios, share of public expenditure in autonomous demand, and growth rates for case 4.
rate of exports. The adjustment in σ is also shown to be progressive and slow. That is a desirable property of this type of policy.
In contrast with case 4, authorities may choose to execute industrial policies that cause a structural change.
We address this possibility with the experiment 5, summarized in figure 4.6. Suppose that the government execute this type of policy in period 30, after rejecting the fiscal policy of case 4. If industrial policy succeeds, then the growth rate of exports accelerates a few periods after their implementation. Assume that the industrial policies do work, and take ten periods to affect the economy. In that case, from period 40 onward, the growth rate of exports accelerates. In the experiment that means increasing the growth rate of exports from 4 percent to 4.5 percent. In line with the Balance of Payments constrained growth tradition, a higher growth of exports may be associated with an increase of the income-elasticity of exports. That may happen as the economy develops production in more technologically intensive industries (Araujo and Lima, 2007; Cimoli et al., 2020, 2009).27 27 Naturally, the relation among industrial policies, structural change, and growth is complex. We cannot 156 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE FIGURE 4.6: Evolution of debt ratios, share of public expenditure in autonomous demand, and growth rates for case 5.
In this case there is a new ¯ σ (value for the share of public expenditure compatible with the ceiling to foreign debt-to-exports ratio), given the higher growth rate of exports for the same ¯ d. Note that the ¯ σ of case 5 is smaller than the ¯ σ of case 4. A new steady-state can only be achieved if public expenditures growth rate accelerates, reaching the new growth rate of exports. However, that can only happen after σ becomes smaller or equal to ¯ σ. In the simulation, that happens in period 51. Then, after this period the economy converge to a new steady-state.
Initially, output growth rate follows the original steady-state. It accelerates after the increase in the growth rate of exports. In the following periods, it keeps accelerating, as σ falls, increasing the impact of exports on output growth. Finally, when the public ensure that the required growth of exports included in the experiment is necessarily obtained.
The effectiveness of these policies varies with the type of measures executed and with the intensity of such measures. In fact, this simulation can also be interpreted as providing an indication of what is the extent of the boost given to exports required to avoid fiscal austerity and still achieve stability of the debt ratios.
In practice, policymakers can choose among many different combinations of the two pure cases (4 and 5) compared here.
157 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE expenditures growth rate adjusts to exports growth, output growth converges to a new steady-state.
It becomes temporarily higher than the growth rate of autonomous demand due to the increase in the propensity to invest, which is necessary to restore normal utilization after an acceleration of growth.
After a while, output growth approaches the growth rate of autonomous expenditures in a new fully adjusted position.
The foreign debt-to-exports ratio also converges and remains below ¯ d in case 5. In turn, the public debt-to-income ratio converges to a lower value than in case 4. Lower b∗ in case 5 is due to the higher steady-state growth rate of output, the lower share of public expenditure in autonomous demand (σ∗), and the lower ratio of autonomous demand to total output (z∗). The lower value for z∗is also associated with a higher h∗.
As shown in table 4.1, equilibrium propensity to invest is higher in case 5 than in the other cases because the economy presents a higher output growth rate. Therefore, if the chosen industrial policies successfully affect exports growth, then case 5 provides a better performance in terms of output growth than the fiscal policy executed in case 4.
The growth rates of these two cases are compared in figure 4.7.
FIGURE 4.7: Output growth rate in Case 4 (Fiscal Policy) and Case 5 (Structural Change).
The simulated experiments confirm theoretical results presented in the previous section. Interest rate smaller than growth rate is crucial for long run stability of public debt. International interest rate smaller than exports growth. The fiscal policy proposed in section 4.5 successfully keeps the foreign debt-to-exports ratio below ¯ d by adjusting the share of public expenditures in autonomous demand. After the adjustment in σ is finished, public expenditure can follow the same pace as exports, keeping a constant σ and allowing the economy to converge to a steady-state compatible with the limit to external debt.
The fall in σ also imply a decrease in the equilibrium level of public 158 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE debt-to-income ratio in the scenario with the fiscal policy. The graphs in figure 4.5 show the adjustment in σ is slow and gradual.
The impact of the fiscal policy on output growth also reveals a gradual adjustment of this variable. The last scenario shows that debt ratios can converge under a better economic performance when structural policies are successfully employed. Authors have emphasized the importance of these policies to relax the external constraint by increasing the income-elasticity of exports (Araujo and Lima, 2007; Cimoli et al., 2020, 2009; Thirlwall, 2019). Such policies allow for the improvement of the economic performance. Indeed, case 5 presented a higher output growth rate than previous cases.
It also ensured convergence of the foreign debt-to-export ratio to a value below ¯ d. The public-debt-to income ratio converged to a smaller value than in the other cases, since output growth was larger and both σ and z∗ were smaller.
Overall, the five experiments show that there is no a priori obstacle to the long run sustainability of growth led by autonomous demand as determined by public expenditures and exports.
159 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE 4.8 Final Remarks This chapter presented a supermultiplier model for an open economy with government in which autonomous demand is composed of public expenditure and exports.
The framework allows for assessing the long run stability of growth, and public and foreign debt ratios. The convergence of public debt-to-income ratio requires that output growth rate is larger than the interest rate on debt service. In turn, the convergence of foreign debt-to-exports ratio depends on an analogous condition, i.e., that exports growth is larger than the interest rate on foreign debt service. The model also showed that, in the presence of two sources of autonomous demand, equilibrium values for debt ratios vary with the share of each expenditure in total autonomous demand. The equilibrium value for the public debt-to-income ratio increases with the share of government in autonomous demand. However, it converges to a stable value even in the extreme case happening when government fully dominates autonomous demand.
Equilibrium foreign debt-to-exports ratio also increases with the share of exports in autonomous demand, but it diverges to infinity in the extreme case in which this share equals zero.
The relevance of those shares to equilibrium debt ratios means that fiscal policy can manage public expenditure to bring debt ratios towards desirable levels. Hence, a fiscal policy rule is proposed in the chapter, according to which government moderates the growth of public expenditure to meet a ceiling in the foreign debt-to-exports ratio, associated with constraints in international liquidity. The external constraint to growth in open economies appears here in the form of a limit to foreign indebtedness.
We further show that in this supermultiplier model, Thirlwall’s Law emerges as a particular result of the long run growth path.
Five simulated experiments support our main conclusions.
Overall, there is no a priori obstacle to the long run sustainability of growth led by autonomous demand as determined by public expenditures and exports. Furthermore, the fiscal policy rule succeeds in keeping foreign debt-to-exports ratio below the given threshold. This policy, however, has a cost in terms of average growth rate. Such a cost may be mitigated with the introduction of industrial policies that increase the income elasticity of exports. Although the effect of industrial policies on growth and structural change is not guaranteed, if this approach is successful it may ensure the convergence of debt ratios with a better performance in terms of growth.
Naturally, a variety of combinations of those (and other) policies are available in reality.
160 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE Appendix A Equilibrium growth and the share of each autonomous expenditure This appendix briefly describe some relations between the proportion of public expenditures in autonomous expenditures (σ) and the rate of growth of public expenditures (gG) and exports (gX). Those relations are useful to justify the analysis developed in the chapter.
I start rewriting here equations 4.19 and 4.20.
gZ = σtgGt + (1 −σt)gXt σt = Gt Zt Operating derivatives with respect to time in both sides of the last equation gives us: ˙ σt = ˙ GtZt −Gt ˙ Zt Z2 t Then, ˙ σt = ˙ Gt Zt −σtgZt By rearranging the relation in equation 4.20, to substitute Zt in the last equation, we obtain: ˙ σt = σtgGt −σtgZt [4.64] Substituting the gZt and rearranging, we obtain: ˙ σt = σt(1 −σt)(gGt −gXt) [4.65] Now, let us take the derivatives with respect to time of the expression for the growth rate of autonomous demand, in equation 4.19.
˙ gZt = ˙ σtgGt + σt ˙ gGt −˙ σtgXt + (1 −σt) ˙ gXt 161 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE Rearranging, we get: ˙ gZt = ˙ σt(gGt −gXt) + σt ˙ gGt + (1 −σt) ˙ gXt Finally, by substituting the value for ˙ σt as in equation 4.65, we obtain: ˙ gZt = σt(1 −σt)(gGt −gXt)2 + σt ˙ gGt + (1 −σt) ˙ gXt [4.66] Let us now discuss under which conditions we obtain that the growth rate of autonomous expenditures is constant, so that the derivative in equation 4.66 is equal to zero. It is easy to see that if public expenditure is null (and thus, σt = 0), stability in total autonomous expenditures growth rate is exactly the same as stability in exports growth rate — putting it differently, ˙ gZt = 0 requires that ˙ gXt = 0. In the contrary, if exports are null (σt = 1), the same reasoning show us that the stability in autonomous demand requires stability in public expenditures.
Apart from these two extreme cases, there are two further conditions that can define stability in the growth rate of autonomous demand. First, this can happen if both exports and public expenditures grow at the same constant rate. In this case, ˙ gGt = ˙ gXt = 0 and gGt = gXt, so that clearly in equation 4.66, the derivative of Zt with respect to time goes to zero.
The last possibility arises if we allow the growth rate of exports and public expenditure to differ from each other. We can see in equation 4.65 that this difference in the two expenditures growth rate implies continuous changes in the value of σ, given that a higher (smaller) growth rate of public expenditures with respect to exports will cause a higher (lower) ratio of public expenditures to autonomous demand. In this case, we can have a constant value for gZt only under a specific condition.
Rearranging equation 4.66, we obtain that this holds as long as ˙ gGt and ˙ gXt move in opposite directions and according with the rule in equation 4.67: ˙ gGt = −(1 −σt)(gGt −gXt)2 − (1 −σt) σt ˙ gXt [4.67] This is clearly an arbitrary condition to be imposed on the dynamics of public expenditure. Thus, we can conclude that the more general situation, the stability in the growth rate of autonomous expenditures requires that public expenditures and exports grow at the same constant rate, implying also that σt remains stable. This is a basic feature to determine the equilibrium within the supermultiplier model, which requires 162 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE the stability of gZt. It is important to point that this relations do not imply causality from gX to gG or vice-versa. The results presented here only discuss the equilibrium and stability conditions, and do not claim any type of constraint to the autonomous expenditures.
B Stability of the supermultiplier in an open economy with public expenditure In this appendix, we analyze the stability of the version of the Sraffian supermultiplier model presented in section 4.3.1.
This analysis follows closely the stability analysis provided by Freitas and Serrano (2015).
Nevertheless, we consider an additional equation describing the evolution of government’s share in autonomous demand (σ).
Moreover, as we deal with an open economy with government, we must also account for a tax rate (τ) and the propensity to import (m) as part of the supermultiplier.
The three equations relevant for the stability analysis were presented in section 2.1 and are rewritten below.
˙ ut = ut σtgGt + (1 −σt)gXt + htγ(ut −un) 1 −c(1 −τ) −ht + m − ht v ut + δ [4.68] ˙ ht = htγ(ut −un) [4.69] ˙ σt = σt(1 −σt)(gGt −gXt) [4.70] We also repeat here the equilibrium values for these three variables.
Capacity utilization converges to the normal rate, u∗= un. In turn, propensity to invest depends on the equilibrium growth rate, depreciation rate, normal rate of capacity utilization and the capital-output ratio: h∗= v un (gZ + δ). σ achieves equilibrium at any level when the two autonomous expenditures grow at the same rate. We denote this equilibrium level as σ∗.
The Jacobian matrix evaluated in the equilibrium is defined by the expression below.
163 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE J∗= h d ˙ h dh i h∗,u∗,σ∗ h d ˙ h du i h∗,u∗,σ∗ h d ˙ h dσ i h∗,u∗,σ∗ h d ˙ u dh i h∗,u∗,σ∗ h d ˙ u du i h∗,u∗,σ∗ h d ˙ u dσ i h∗,u∗,σ∗ h d ˙ σ dh i h∗,u∗,σ∗ h d ˙ σ du i h∗,u∗,σ∗ h d ˙ σ dσ i h∗,u∗,σ∗ [4.71] The elements h d ˙ h dh i h∗,u∗,σ∗ h d ˙ h du i h∗,u∗,σ∗, h d ˙ u dh i h∗,u∗,σ∗, and h d ˙ u du i h∗,u∗,σ∗are trivially obtain and coincide with the results of Freitas and Serrano (2015).
We are thus left to compute the remaining elements of the Jacobian matrix, that is, the elements of the third row and third column.
It is easy to see that: d ˙ h dσ = 0 [4.72] In turn: d ˙ u dσ = utgGt −utgXt = ut(gGt −gXt) [4.73] Evaluating this derivative in the equilibrium means substituting ut = un and making gGt = gXt. In that case, d ˙ u dσ h∗,u∗,σ∗= 0 [4.74] As σ does not depend on h or u (see equation 4.70), then the partial derivatives h d ˙ σ dh i and h d ˙ σ du i are trivially equal to zero.
Finally, we must compute h d ˙ σ dσ i . First, let us rewrite equation 4.70 as follows: ˙ σ = σt(gGt −gXt) −σ2 t (gGt −gXt) [4.75] Then, we have: d ˙ σ dσ = (gGt −gXt) −2σt(gGt −gXt) = (1 −2σt)(gGt −gXt) [4.76] Evaluating this derivative in the equilibrium implies plugging σt = σ∗and making gGt = gXt.
164 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE Hence: d ˙ σ dσ h∗,u∗,σ∗= 0 [4.77] Thus, we obtain: J∗= 0 γ(gZ + δ) v un 0 −u2 n v (gZ + δ) γv 1−c(1−τ)−( v un )(gZ+δ)+m −1 0 0 0 0 [4.78] Local stability requires obtaining the eigenvalues of the matrix 4.78. The characteristic polynomial of 4.78 is given by the equation below.
λ3 −λ2 (gZ + δ) γv 1 −c(1 −τ) − v un (gZ + δ) + m −1 + λunγ(gZ + δ) = 0 [4.79] From the characteristic polynomial, we obtain that λ = 0 is one of the eigenvalues of the matrix 4.78. The other two roots can be obtained from the following equation.
λ2 −λ (gZ + δ) γv 1 −c(1 −τ) − v un (gZ + δ) + m −1 + unγ(gZ + δ) = 0 [4.80] Since the first and the third coefficients of the second degree equation (4.80 are positive, its two roots are negative as long as the second coefficient is positive. In other words, the two remaining eigenvalues are negative if the following inequality holds: − (gZ + δ) γv 1 −c(1 −τ) − v un (gZ + δ) + m −1 > 0 [4.81] The inequality above holds whenever the condition below is attended.
165 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE γv + c(1 −τ) + v un (gZ + δ) −m < 1 [4.82] 4.82 is analogous to the condition found by Freitas and Serrano (2015). In fact, if we set τ = m = 0, we obtain the same condition they found in the model of a closed economy without government.
Condition 4.82 ensures that the other two roots of the characteristic polynomial, and thus the other two eigenvalues, are negative.
Therefore, the matrix 4.78 has one eigenvalue equal to zero and two eigenvalues that are negative.
The eigenvalue equal to zero shifts the level of the equilibrium of the system, but does not affect the conditions of asymptotic convergence. The fact that the other two eigenvalues are smaller than zero, however, ensures the asymptotic convergence of the system.
Form an economic interpretation, inequality 4.82 implies that the model is locally stable when the expanded marginal propensity to spend (that is, the marginal propensity to spend added of the term γv associated to the accelerator investment function) is lower than one (Freitas and Serrano, 2015).
Taxation and imports represent two leakages of induced demand. The presence of taxation and propensity to import reduces the value of the expanded marginal propensity to spend, easing the satisfaction of inequality 4.82. Therefore, an open economy with government is less vulnerable to instability coming from economic growth and the capital stock adjustment principle (i.e., the flexible accelerator).
C Sensitivity Analysis This appendix reports the results of the sensitivity analysis performed in the dynamic system presented in section 3.3.
We tested the robustness of the equilibrium of the model in face of random shocks in the main variables determining the dynamics: the interest rate (i), the international rate (r), the growth rate of exports (gX), the growth rate of public expenditure (gG), propensity to import (m), tax rate (τ), and the sensitivity of the investment share to deviations of capacity utilization concerning normal capacity (γ).
We analyzed three scenarios.
In the first scenario, a permanent random shocks is added to four parameters: i, r, gG, and gX.
Figure 4.8 presents the results of the Monte Carlo simulations of this scenario.
While capacity utilization and the propensity to invest converge, the debt 166 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE ratios and the government’s share in autonomous demand present a widening range of values in the time interval, due to the permanent difference between the growth rates of the two autonomous expenditures (gG, gX).
The second scenario maintains the random shocks in the same parameters, but establishes the condition that the shocks affect gG and gX in the same way. Therefore, in this scenario the growth rate of autonomous demand varies, due to the random shocks.
However, the growth rate of government and exports is always equal. As a consequence of this condition, all the variables present a convergence within a stable interval in the Monte Carlo simulations. The results can be seen in figure 4.9.
Finally, the third scenario includes random shocks in additional parameters, but maintains the equality between gG and gX.
Now we test the sensitivity of the equilibrium to changes in the interest rate, the international interest rate and the growth rate of autonomous demand, as before, in addition to changes in the propensity to import (m), tax rate (τ), and the sensitivity of the investment share to deviations of capacity utilization concerning normal capacity (γ). Once again, the variables converge to equilibrium within a stable interval, as seen in figure 4.10.
167 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE FIGURE 4.8: Sensitivity analysis. Case 1: shocks in i, r, gG, and gX.
168 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE FIGURE 4.9: Sensitivity analysis. Case 2: shocks in i, r, gZ, keeping gG = gX.
169 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE FIGURE 4.10: Sensitivity analysis. Case 3: shocks in i, r, gZ, m, τ, γ, keeping gG = gX.
170 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE D Discrete time This appendix presents a discrete time version of the model developed in this chapter.
The discrete time model follows the same structure and the same assumptions of the original model. This version is required for the simulation exercise.
Output Level As in the original model, equilibrium in the goods market implies aggregate output equals aggregate demand, which is given by consumption, investment, public expenditure, exports and imports.
Yt = Ct + It + Gt + Xt −Mt Consumption depends on propensity to consume (c), and on current disposable income.
Ct = c(1 −τ)Yt Likewise, imports depend on the current income and the propensity to import (m): Mt = mYt Investment, in turn, is given by the propensity to invest times current income.
It = htYt Public expenditure and exports are autonomous with respect to current income, being added in the term Zt. We can, then, determine according to the level of autonomous expenditures and the supermultiplier.
Yt = 1 1 −c(1 −τ) −ht + m Zt 171 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE Capacity Utilization The degree of capacity utilization is defined as the ratio between output and potential output.
u = Yt YKt Since the capital-output ratio is constant, the growth rate of potential output is equal to the growth rate of capital stock, i.e., the accumulation rate. Hence, the rate of change in capacity utilization follows the expression below.
ˆ ut = gt −gKt 1 + gKt And the change in the degree of capacity utilization in each period is given by the expression below.
∆ut = ut−1 gt −gKt 1 + gKt However, we still need to determine gKt.
gKt := Kt −Kt−1 Kt−1 = It−1 Kt−1 −δ Finally, with the same operations of the footnote 4, we can obtain the equation below.
gKt = ht−1 v ut−1 −δ The propensity to invest Propensity to invest changes when the degree of capacity utilization deviates from normal capacity, according to the parameter γ.
∆ht = ht−1γ(ut−1 −un) Thus, the rate of change in the propensity to invest can be described as follows.
172 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE ˆ ht = γ(ut−1 −un) Output Growth Now, we are able to determine output growth (gt), according to the change in the supermultiplier due to the variation in the propensity to invest and to the growth in autonomous expenditures.
gt = ht−1ˆ ht 1 −c(1 −τ) −ht−1(1 + ˆ ht) + m + gZt + " ht−1ˆ ht 1 −c(1 −τ) −ht−1(1 + ˆ ht) + m # gZt Autonomous demand varies proportionally with public expenditures and exports, and to the share of each expenditure in autonomous demand. We define σ as the share of public expenditure in autonomous demand. Thus, σt = Gt/Yt.
gZt = σt−1gGt + (1 −σt−1)gXt Change in σ follows the expression below.
∆σt = σt−1(1 −σt−1)(gGt −gXt) 1 + σt−1gGt + (1 −σt−1)gXt Which can be written as: ∆σt = σt−1(1 −σt−1)(gGt −gXt) 1 + gZt Public Debt The relevant variable here is public debt-to-income ratio, denoted as bt. Changes in public debt in each period depend on the primary surplus/deficit and on the interest payments of the debt service.
∆Bt = Gt −τYt + iBt−1 173 CHAPTER 4. GROWTH AND DEBT STABILITY IN A SUPERMULTIPLIER MODEL WITH PUBLIC EXPENDITURES AND FOREIGN TRADE Hence, bt = σtzt −τ + (1 + i) (1 + gt) bt−1 Finally, b∗= (σ∗z∗−τ) 1 + g∗ g∗−i Foreign debt We focus in the foreign debt-to-exports ratio, denoted as dt.
According to the assumptions of the model, foreign debt change in each period depends on the trade balance and on foreign debt service payments.
∆Dt = Mt −Xt + rDt−1 Hence, the debt-to-exports ratio can be expressed as in the equation below.
dt = Mt Xt −1 + (1 + r) (1 + gXt) dt−1 Finally, by setting dt −dt−1 = 0, we obtain the equilibrium foreign debt-to-exports ratio.
d∗= m (1 −σ∗)z∗−1 1 + gX gX −r E Simulation Parameters and Equilibrium values Table E.1 presents the parameters employed in the simulations.
Table E.2 displays the initial values of the simulations. Finally, table E.3 presents the equilibrium values for the main variables. The different results for equilibrium come from the change in each of the four cases with respect to the baseline case, which is also shown in the table.
174 BIBLIOGRAPHY TABLE E.1: Parameters Parameter Value gG 0.04 gX 0.04 c 0.075 τ 0.15 m 0.25 γ 0.05 un 0.85 v 1.50 δ 0.075 i 0.02 r 0.03 TABLE E.2: Initial Values Variable Value G0 120.000 X0 173.000 σ0 0.4138 h0 0.2029 u0 0.85 b0 0.5 d0 1.6 TABLE E.3: Simulations Result Case 1 Case 2 Case 3 Case 4 Case 5 Name Baseline Explosive d Explosive b Fiscal Policy Structural Change Change -r = 0.05 i = 0.045 Policy rule (p.30) gX = 0.045 (p.40) g∗ 0.04 0.04 0.04 0.04 0.045 b∗ 1.0126 1.0126 +∞ 0.8022 0.3777 d∗ 4.2944 +∞ 4.2944 2.5000 2.3927 h∗ 0.2029 0.2029 0.2029 0.2029 0.2118 z∗ 0.4096 0.4096 0.4096 0.4096 0.4007 σ∗ 0.4138 0.4138 0.4138 0.4039 0.3969 175 Bibliography Ahn, J., Park, C.-G., and Park, C. (2017).
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Home / Knowledge base / Technical Information / Torr (Tr) - Pressure Unit - Definition, Application, Conversions
Torr (Tr) - Pressure Unit - Definition, Application, Conversions
What is a torr?
A torr is a unit of pressure measurement named after the Italian physicist Evangelista Torricelli, a pioneer in the study of atmospheric pressure. One torr is defined as 1/760 of an atmosphere (atm).
In most cases, the unit torr is considered equivalent to a millimetre column of mercury (mmHg), but there is a minimal difference between the two (less than 0.000015%).
The torr is not part of the SI system, but is used in some specialised applications.
Torr is a unit mainly used in the context of vacuum, rather than in industrial hoses. However, the working pressure of vacuum hoses must be precisely determined to prevent collapse under low pressure. Industrial hoses must be designed to operate in specific vacuum conditions.
When is a torr used to measure pressure?
Torr is used to measure pressure mainly in specialised applications such as:
Vacuum testing- Torr is traditionally and widely used in vacuum testing, including physics and engineering. It is crucial in the semiconductor industry, research laboratories and wherever precise vacuum pressure measurements are required.
Medicine- Although mmHg is more commonly used in medicine, the torr can be used to measure blood pressure and other medical applications requiring precise measurements.
Meteorology- In meteorology, the torr can be used to measure atmospheric pressure, especially in older instruments and in some specialised applications.
Physics and chemistry- In laboratory research, torr is used to measure pressure in experiments with gases, vacuum and under controlled pressure conditions.
Aviation and aerospace- In aerospace research, torr is used in simulation and pressure measurements at altitude and space conditions.
How to measure pressure in torr?
To measure pressure in torr, you need the right measuring equipment, such as a manometer calibrated in this unit or another that can be easily converted to torr.
Here are the steps to take to measure pressure in torr:
Choose the right manometer - You can use a mercury manometer for the measurement, since 1 torr is almost equal to 1 mmHg, and these types of devices measure pressure directly in mmHg. An alternative is digital manometers, which often allow you to choose different units, including torr.
Identify the measuring point - Determine where you want to measure the pressure (e.g. atmospheric pressure, pressure in vacuum systems or in industrial systems).
Connect the manometer - Place the mercury manometer vertically and let the pressure act on the mercury column. With a digital manometer, you simply connect the device to the system.
Read the result - In a mercury manometer, the height of the mercury column will give you the result in mmHg, which is equal to the same value in torr. In turn, in a digital manometer, the display will show the pressure directly in Tr, if you have the option to select this unit.
If your device measures pressure in other pressure units, you can convert it to Tr using the appropriate pressure converters.
Torr pressure unit conversions
1 torr = 0,00133322 bar
1 torr= 133,322 Pa
1 torr= 1,33322 hPa
1 torr= 0,133322 kPa
1 torr= 0,000133322 MPa
1 torr= 0,00131579 atm
1 torr= 0,0193368 PSI
1 torr= 13,5951 mmH2O
1 torr= 0,53524 inH2O
1 torr= 1 mmHg
1 torr= 0,0393701 inHg
1 torr= 2,7845 PSF
1 torr= 0,00135951 at
1 torr (Tr) = 0,00133322 bar = 133,322 Pa = 1,33322 hPa = 0,133322 kPa = 0,000133322 MPa = 0,00131579 atm = 0,0193368 PSI = 13,5951 mmH2O = 0,53524 inH2O = 1 mmHg = 0,0393701 inHg = 2,7845 PSF = 0,00135951 at
torr to Bar pressure unit conversion
To convert pressure from torr (Tr) toBarmultiply the pressure in torr (Tr) by 0,00133322.
1 torr to bar? 1 torr is equal to 0,0013332 bar
2 torr to bar? 2 torr is equal to 0,0026664 bar
5 torr to bar? 5 torr is equal to 0,0066661 bar
10 torr to bar? 10 torr is equal to 0,0133322 bar
20 torr to bar? 20 torr is equal to 0,0266644 bar
50 torr to bar? 50 torr is equal to 0,066661 bar
100 torr to bar? 100 torr is equal to 0,133322 bar
200 torr to bar? 200 torr is equal to 0,266644 bar
500 torr to bar? 500 torr is equal to 0,66661 bar
1000 torr to bar? 1000 torr is equal to 1,33322 bar
torr to Pa - torr to Pascal pressure unit conversion
To convert pressure from torr (Tr) toPascal (Pa)multiply the pressure in torr (Tr) by 133,322.
1 torr to Pa? 1 torr is equal to 133,322 Pa
2 torr to Pa? 2 torr is equal to 266,644 Pa
5 torr to Pa? 5 torr is equal to 666,61 Pa
10 torr to Pa? 10 torr is equal to 1333,22 Pa
20 torr to Pa? 20 torr is equal to 2666,44 Pa
40 torr to Pa? 40 torr is equal to 5332,88 Pa
50 torr to Pa? 50 torr is equal to 6666,1 Pa
100 torr to Pa? 100 torr is equal to 13332,2 Pa
200 torr to Pa? 200 torr is equal to 26664,4 Pa
500 torr to Pa? 500 torr is equal to 66661 Pa
760 torr to Pa? 760 torr is equal to 101324,72 Pa
1000 torr to Pa? 1000 torr is equal to 133322 Pa
torr to hPa - torr to Hectopascal pressure unit conversion
To convert pressure from torr (Tr) toHectopascal (hPa)multiply the pressure in torr (Tr) by 1,33322.
1 torr to hPa? 1 torr is equal to 1,33322 hPa
2 torr to hPa? 2 torr is equal to 2,66644 hPa
5 torr to hPa? 5 torr is equal to 6,6661 hPa
10 torr to hPa? 10 torr is equal to 13,3322 hPa
20 torr to hPa? 20 torr is equal to 26,6644 hPa
50 torr to hPa? 50 torr is equal to 66,661 hPa
100 torr to hPa? 100 torr is equal to 133,322 hPa
200 torr to hPa? 200 torr is equal to 266,644 hPa
500 torr to hPa? 500 torr is equal to 666,61 hPa
1000 torr to hPa? 1000 torr is equal to 1333,22 hPa
torr to kPa - torr to Kilopascal pressure unit conversion
To convert pressure from torr (Tr) toKilopascal (kPa)multiply the pressure in torr (Tr) by 0,133322.
1 torr to kPa? 1 torr is equal to 0,133322 kPa
2 torr to kPa? 2 torr is equal to 0,266644 kPa
5 torr to kPa? 5 torr is equal to 0,66661 kPa
10 torr to kPa? 10 torr is equal to 1,33322 kPa
20 torr to kPa? 20 torr is equal to 2,66644 kPa
40 torr to kPa? 40 torr is equal to 5,33288 kPa
50 torr to kPa? 50 torr is equal to 6,6661 kPa
100 torr to kPa? 100 torr is equal to 13,3322 kPa
200 torr to kPa? 200 torr is equal to 26,6644 kPa
500 torr to kPa? 500 torr is equal to 66,661 kPa
1000 torr to kPa? 1000 torr is equal to 133,322 kPa
torr to MPa - torr to Megapascal pressure unit conversion
To convert pressure from torr (Tr) toMegapascal (MPa)multiply the pressure in torr (Tr) by 0,000133322.
1 torr to MPa? 1 torr is equal to 0,0001333 MPa
2 torr to MPa? 2 torr is equal to 0,0002666 MPa
5 torr to MPa? 5 torr is equal to 0,0006666 MPa
10 torr to MPa? 10 torr is equal to 0,0013332 MPa
20 torr to MPa? 20 torr is equal to 0,0026664 MPa
50 torr to MPa? 50 torr is equal to 0,0066661 MPa
100 torr to MPa? 100 torr is equal to 0,0133322 MPa
200 torr to MPa? 200 torr is equal to 0,0266644 MPa
500 torr to MPa? 500 torr is equal to 0,066661 MPa
1000 torr to MPa? 1000 torr is equal to 0,133322 MPa
1000 torr to MPa? 1000 torr is equal to 0,133322 MPa
torr to atm - torr to Atmosphere pressure unit conversion
To convert pressure from torr (Tr) to Atmosphere (atm) multiply the pressure in torr (Tr) by 0,00131579.
1 torr to atm? 1 torr is equal to 0,001316 atm
2 torr to atm? 2 torr is equal to 0,002632 atm
5 torr to atm? 5 torr is equal to 0,006579 atm
10 torr to atm? 10 torr is equal to 0,013158 atm
20 torr to atm? 20 torr is equal to 0,026316 atm
50 torr to atm? 50 torr is equal to 0,06579 atm
100 torr to atm? 100 torr is equal to 0,131579 atm
200 torr to atm? 200 torr is equal to 0,263158 atm
250 torr to atm? 250 torr is equal to 0,328948 atm
300 torr to atm? 300 torr is equal to 0,394737 atm
340 torr to atm? 340 torr is equal to 0,447369 atm
380 torr to atm? 380 torr is equal to 0,5 atm
500 torr to atm? 500 torr is equal to 0,657895 atm
650 torr to atm? 650 torr is equal to 0,855264 atm
710 torr to atm? 710 torr is equal to 0,934211 atm
720 torr to atm? 720 torr is equal to 0,947369 atm
725 torr to atm? 725 torr is equal to 0,953948 atm
730 torr to atm? 730 torr is equal to 0,960527 atm
745 torr to atm? 745 torr is equal to 0,980264 atm
750 torr to atm? 750 torr is equal to 0,986843 atm
754 torr to atm? 754 torr is equal to 0,992106 atm
755 torr to atm? 755 torr is equal to 0,993421 atm
760 torr to atm? 760 torr is equal to 1 atm
765 torr to atm? 765 torr is equal to 1,006579 atm
780 torr to atm? 780 torr is equal to 1,026316 atm
800 torr to atm? 800 torr is equal to 1,052632 atm
850 torr to atm? 850 torr is equal to 1,118422 atm
912 torr to atm? 912 torr is equal to 1,2 atm
1000 torr to atm? 1000 torr is equal to 1,31579 atm
torr to PSI pressure unit conversion
To convert pressure from torr (Tr) toPSImultiply the pressure in torr (Tr) by 0,0193368.
1 torr to PSI? 1 torr is equal to 0,019337 PSI
2 torr to PSI? 2 torr is equal to 0,038674 PSI
5 torr to PSI? 5 torr is equal to 0,096684 PSI
10 torr to PSI? 10 torr is equal to 0,193368 PSI
20 torr to PSI? 20 torr is equal to 0,386736 PSI
50 torr to PSI? 50 torr is equal to 0,96684 PSI
100 torr to PSI? 100 torr is equal to 1,93368 PSI
200 torr to PSI? 200 torr is equal to 3,86736 PSI
500 torr to PSI? 500 torr is equal to 9,6684 PSI
755 torr to PSI? 755 torr is equal to 14,599284 PSI
760 torr to PSI? 760 torr is equal to 14,695968 PSI
1000 torr to PSI? 1000 torr is equal to 19,3368 PSI
torr to mmH2O - torr to Milimetre of water column pressure unit conversion
To convert pressure from torr (Tr) to Milimetre of water column (mmH2O) multiply the pressure in torr (Tr) by 13,5951.
1 torr to mmH2O? 1 torr is equal to 13,5951 mmH2O
2 torr to mmH2O? 2 torr is equal to 27,1902 mmH2O
5 torr to mmH2O? 5 torr is equal to 67,9755 mmH2O
10 torr to mmH2O? 10 torr is equal to 135,951 mmH2O
20 torr to mmH2O? 20 torr is equal to 271,902 mmH2O
50 torr to mmH2O? 50 torr is equal to 679,755 mmH2O
100 torr to mmH2O? 100 torr is equal to 1359,51 mmH2O
200 torr to mmH2O? 200 torr is equal to 2719,02 mmH2O
500 torr to mmH2O? 500 torr is equal to 6797,55 mmH2O
1000 torr to mmH2O? 1000 torr is equal to 13595,1 mmH2O
torr to inH2O - torr to Inch of water column pressure unit conversion
To convert pressure from torr (Tr) toInch of water column (inH2O)multiply the pressure in torr (Tr) by 0,53524.
1 torr to inH2O? 1 torr is equal to 0,53524 inH2O
2 torr to inH2O? 2 torr is equal to 1,07048 inH2O
5 torr to inH2O? 5 torr is equal to 2,6762 inH2O
10 torr to inH2O? 10 torr is equal to 5,3524 inH2O
20 torr to inH2O? 20 torr is equal to 10,7048 inH2O
50 torr to inH2O? 50 torr is equal to 26,762 inH2O
100 torr to inH2O? 100 torr is equal to 53,524 inH2O
200 torr to inH2O? 200 torr is equal to 107,048 inH2O
500 torr to inH2O? 500 torr is equal to 267,62 inH2O
1000 torr to inH2O? 1000 torr is equal to 535,24 inH2O
torr to mmHg - torr to Milimetre of mercury column pressure unit conversion
To convert pressure from torr (Tr) to Milimetre of mercury column (mmHg) multiply the pressure in torr (Tr) by 1.
1 torr to mmHg? 1 torr is equal to 1 mmHg
2 torr to mmHg? 2 torr is equal to 2 mmHg
5 torr to mmHg? 5 torr is equal to 5 mmHg
10 torr to mmHg? 10 torr is equal to 10 mmHg
20 torr to mmHg? 20 torr is equal to 20 mmHg
50 torr to mmHg? 50 torr is equal to 50 mmHg
100 torr to mmHg? 100 torr is equal to 100 mmHg
200 torr to mmHg? 200 torr is equal to 200 mmHg
340 torr to mmHg? 340 torr is equal to 340 mmHg
500 torr to mmHg? 500 torr is equal to 500 mmHg
720 torr to mmHg? 720 torr is equal to 720 mmHg
1000 torr to mmHg? 1000 torr is equal to 1000 mmHg
torr to inHg - torr to Inch of mercury column pressure unit conversion
To convert pressure from torr (Tr) to Inch of mercury column (inHg)multiply the pressure in torr (Tr) by 0,0393701.
1 torr to inHg? 1 torr is equal to 0,03937 inHg
2 torr to inHg? 2 torr is equal to 0,07874 inHg
5 torr to inHg? 5 torr is equal to 0,196851 inHg
10 torr to inHg? 10 torr is equal to 0,393701 inHg
20 torr to inHg? 20 torr is equal to 0,787402 inHg
50 torr to inHg? 50 torr is equal to 1,968505 inHg
100 torr to inHg? 100 torr is equal to 3,93701 inHg
200 torr to inHg? 200 torr is equal to 7,87402 inHg
500 torr to inHg? 500 torr is equal to 19,68505 inHg
1000 torr to inHg? 1000 torr is equal to 39,3701 inHg
torr to PSF pressure unit conversion
To convert pressure from torr (Tr) toPSFmultiply the pressure in torr (Tr) by 2,7845.
1 torr to PSF? 1 torr is equal to 2,7845 PSF
2 torr to PSF? 2 torr is equal to 5,569 PSF
5 torr to PSF? 5 torr is equal to 13,9225 PSF
10 torr to PSF? 10 torr is equal to 27,845 PSF
20 torr to PSF? 20 torr is equal to 55,69 PSF
50 torr to PSF? 50 torr is equal to 139,225 PSF
100 torr to PSF? 100 torr is equal to 278,45 PSF
200 torr to PSF? 200 torr is equal to 556,9 PSF
500 torr to PSF? 500 torr is equal to 1392,25 PSF
1000 torr to PSF? 1000 torr is equal to 2784,5 PSF
torr to at - torr to Technical atmosphere pressure unit conversion
To convert pressure from torr (Tr) to Technical Atmosphere (at) multiply the pressure in torr (Tr) by 0,00135951.
1 torr to at? 1 torr is equal to 0,00136 at
2 torr to at? 2 torr is equal to 0,002719 at
5 torr to at? 5 torr is equal to 0,006798 at
10 torr to at? 10 torr is equal to 0,013595 at
20 torr to at? 20 torr is equal to 0,02719 at
50 torr to at? 50 torr is equal to 0,067976 at
100 torr to at? 100 torr is equal to 0,135951 at
200 torr to at? 200 torr is equal to 0,271902 at
500 torr to at? 500 torr is equal to 0,679755 at
1000 torr to at? 1000 torr is equal to 1,35951 at
Articles author
Bartosz Kułakowski
CEO of Hosetech Sp. z o.o.
Bartosz Kulakowski is an industrial hoses and couplings specialist with over 10 years of experience. Bartosz has been present in the technical industry since 2013. He gained experience as a technical and commercial advisor in the sector of plastic conveyor belts, steel structures, industrial hoses, and connectors. Since 2016, he has specialized exclusively in hoses and connectors. In 2019 he opened his own business under the HOSETECH Bartosz Kulakowski brand and since July 2022 he has been the CEO of the capital company HOSETECH Sp z o. o. (LLC).
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Table of Contents
What is a torr?
When is a torr used to measure pressure?
How to measure pressure in torr?
Torr pressure unit conversions
torr to Bar pressure unit conversion
torr to Pa - torr to Pascal pressure unit conversion
torr to hPa - torr to Hectopascal pressure unit conversion
torr to kPa - torr to Kilopascal pressure unit conversion
torr to MPa - torr to Megapascal pressure unit conversion
torr to atm - torr to Atmosphere pressure unit conversion
torr to PSI pressure unit conversion
torr to mmH2O - torr to Milimetre of water column pressure unit conversion
torr to inH2O - torr to Inch of water column pressure unit conversion
torr to mmHg - torr to Milimetre of mercury column pressure unit conversion
torr to inHg - torr to Inch of mercury column pressure unit conversion
torr to PSF pressure unit conversion
torr to at - torr to Technical atmosphere pressure unit conversion
8 min ##### Bend radius - Definition, How To Calculate, Types Bend radius is the distance to the inside edge of the hose (not the centerline) when making a 90° turn. Measurements of bend radius are made under controlled laboratory conditions. The bend radius is a key parameter when selecting a hose, especially ... Read
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