url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://www.exceltrivia.com/2018/07/bin2hex-function.html | 1,716,387,685,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058557.6/warc/CC-MAIN-20240522132433-20240522162433-00277.warc.gz | 661,669,062 | 35,731 | You can convert binary numbers into hexadecimal numbers
The binary numbers contain only 0 and 1 values
Oct numbers contain 0 to 15 values
BIN2HEX Function has a limit of 10 characters
If you write more than 10 characters it will return #NUM error
Syntax
=BIN2HEX(number,[places])
Argument
Number: it is the number you want to convert into octal
Places:[Optional] how many characters you want to show in your result. It will place "0" before the result.
Behavior
Here we have taken an example of the binary number and convert it into HexaDecimal numbers. We have written a formula
=BIN2hex(D4,4) which will convert the binary number into hexadecimal and show result till 4 characters. No character place is filled by 0
On the right side, if you write 10 times "1" it will return 10 times "F"
If you write more than 10 characters it will return #NUM error
Usage
To convert a Binary number into a hexadecimal number
Returns | 236 | 933 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-22 | latest | en | 0.724942 |
https://avatest.org/2022/11/03/complex-network-daixie-math3002-annealed-networks-and-graphons/ | 1,725,909,517,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651133.92/warc/CC-MAIN-20240909170505-20240909200505-00674.warc.gz | 97,548,368 | 19,831 | Posted on Categories:Complex Network, 复杂网络, 数据科学代写, 物理代写
物理代写|复杂网络代写Complex Network代考|MATH3002 Annealed networks and graphons
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物理代写|复杂网络代写Complex Network代考|Annealed networks and graphons
At this point, we mention a useful deterministic construction known as the annealed network approximation based on relations visually similar to formulas in the previous section. This approximation is actually applicable not only to equilibrium or evolving random networks but even to nonrandom graphs. Let us consider a large graph whose vertices have degrees $q_i$, $i=1,2, \ldots, N$. (For the sake of brevity, we assume here that this graph is simple, which is actually not necessary.) This graph is approximated by the ‘annealed network’ that is the fully connected weighted graph of $N$ vertices, whose edges have the weights ${ }^{14}$
$$w_{i j}\left(q_i, q_j\right)=\frac{q_i q_j}{N\langle q\rangle} .$$
Then the sum of the weights of the edges of a vertex coincides with the degree of the original graph,
$$\sum_j w_{i j}=q_i$$
Compare Eqs. (4.18) and (4.20), (4.17) and (4.21). Even if the original graph is random, the annealed network will be deterministic in the limit $N \rightarrow \infty$, which makes it one of the widely used simple instruments in the theory of complex networks. ${ }^{15}$
物理代写|复杂网络代写Complex Network代考|Clustering, cycles, and cliques
Let us look closer at the local tree-likeness of large sparse uncorrelated networks. To do this, we inspect the average number $\mathcal{N}{\text {cycles }}(\ell)$ of cycles of a given length $\ell$ in undirected networks of this kind. Here we consider the configuration model and adapt the derivation from Bianconi and Marsili (2005). The number $\mathcal{N}{\text {cycles }}(\ell)$ can be written as the following product:
$$\mathcal{N}_{\text {cycles }}(\ell)=R(N, \ell,{P(q)}) G(\ell) W(N, \ell) .$$
Here the first term $R(N, \ell,{P(q)})$, is the number of ways to select pairs of edges connected to $\ell$ vertices from the edges of these vertices. This number depends on the form of the degree distribution $P(q)$. We easily get
$$R(N, \ell,{P(q)})=\frac{[N\langle q(q-1)\rangle]^{\ell}}{\ell !}$$
The second term in the product on the right-hand side of Eq. (4.23), $G(\ell)$, is the number of ways to connect $\ell$ given vertices in a cycle. Equally easily, we have
$$G(\ell)=\frac{\ell !}{2 \ell}$$
物理代写|复杂网络代写Complex Network代考|Annealed networks and graphons
$$w_{i j}\left(q_i, q_j\right)=\frac{q_i q_j}{N\langle q\rangle} .$$
$$\sum_j w_{i j}=q_i$$
物理代写|复杂网络代写Complex Network代考|Clustering, cycles, and cliques 中。
$$\mathcal{N}_{\text {cycles }}(\ell)=R(N, \ell, P(q)) G(\ell) W(N, \ell) .$$
$$R(N, \ell, P(q))=\frac{[N\langle q(q-1)\rangle]^{\ell}}{\ell !}$$
$$G(\ell)=\frac{\ell !}{2 \ell}$$
MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 | 1,377 | 3,604 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-38 | latest | en | 0.731567 |
https://lists.cam.ac.uk/pipermail/cl-isabelle-users/2013-October/msg00171.html | 1,638,054,270,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358323.91/warc/CC-MAIN-20211127223710-20211128013710-00065.warc.gz | 447,716,242 | 3,059 | # Re: [isabelle] Recursion through Types
```Hi Tjark,
I also think the general answer is no.
But maybe there is a trick (also called ugly hack) using type classes:
For a function "f" you want to define with primitive recursion on a type
"'a" you introduce a new type class:
class f_def =
fixes f :: "'a ⇒ nat"
Now you instantiate the option type (as the successor type):
instantiation option :: (f_def) f_def
begin
definition
"f_option (x::'a option) = f (undefined :: 'a) + 1"
instance ..
end
and one for the unit type (as termination)
instantiation unit :: f_def
begin
definition f_unit :: "unit ⇒ nat" where
"f_unit (x::unit) = 0"
instance ..
end
But a problem is that numeral types used by the word library and by
Multivariate Analysis use a binary representation instead of a unary
representation.
- Johannes
Am Sonntag, den 20.10.2013, 11:44 +0200 schrieb Jesus Aransay:
> Dear Tjark,
>
> I don't know how far the following solution is applicable to your
> setting, but a somehow similar problem (and two different solutions)
> were proposed in the mailing list by J. Harrison a while ago, for the
> definition of determinants of matrices of dimension n, in terms of
> determinants of matrices of size "n - 1":
>
> https://lists.cam.ac.uk/mailman/htdig/cl-isabelle-users/2012-January/msg00015.html
>
> Hope it helps,
>
> best,
>
> Jesus
>
>
> On Fri, Oct 18, 2013 at 7:29 PM, Tjark Weber <webertj at in.tum.de> wrote:
> > Hi,
> >
> > There is a trick (due to John Harrison) to encode the dimension N of,
> > e.g., N-bit words with a type argument. The word libraries in
> > Isabelle/HOL and HOL4 are based on this approach.
> >
> > In this setting, what is the recommended way to define a function that
> > performs recursion over N, i.e., whose result for an (N+1)-bit word is
> > naturally expressed in terms of its result for an N-bit word?
> >
> > Best,
> > Tjark
> >
> >
> >
>
>
>
```
This archive was generated by a fusion of Pipermail (Mailman edition) and MHonArc. | 558 | 1,997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-49 | latest | en | 0.850095 |
https://submityourpapers.com/a-helicopter-is-moving-horizontally-to-the-right-at-a-constant-velocity-the-weight-of-the-helicopter-is-w-44600-n-the-lift-force-l-generated-by/ | 1,643,047,776,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304572.73/warc/CC-MAIN-20220124155118-20220124185118-00657.warc.gz | 593,831,997 | 10,846 | # A helicopter is moving horizontally to the right at a constant velocity. The weight of the helicopter is W =44600 N. The lift force L generated by…
A helicopter is moving horizontally to the right at a constant velocity. The weight of the helicopter is W=44600 N. The lift force L generated by the rotating blade makes an angle of 21.0° with respect to the vertical. (a) What is the magnitude of the lift force? (b)Determine the magnitude of the air resistance R that opposes the motion. | 116 | 490 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-05 | latest | en | 0.851994 |
https://metanumbers.com/1412366654812131 | 1,585,784,698,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506477.26/warc/CC-MAIN-20200401223807-20200402013807-00115.warc.gz | 551,421,306 | 8,327 | ## 1412366654812131
1,412,366,654,812,131 (one quadrillion four hundred twelve trillion three hundred sixty-six billion six hundred fifty-four million eight hundred twelve thousand one hundred thirty-one) is an odd sixteen-digits composite number following 1412366654812130 and preceding 1412366654812132. In scientific notation, it is written as 1.412366654812131 × 1015. The sum of its digits is 54. It has a total of 8 prime factors and 48 positive divisors. There are 894,065,593,760,064 positive integers (up to 1412366654812131) that are relatively prime to 1412366654812131.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 16
• Sum of Digits 54
• Digital Root 9
## Name
Short name 1 quadrillion 412 trillion 366 billion 654 million 812 thousand 131 one quadrillion four hundred twelve trillion three hundred sixty-six billion six hundred fifty-four million eight hundred twelve thousand one hundred thirty-one
## Notation
Scientific notation 1.412366654812131 × 1015 1.412366654812131 × 1015
## Prime Factorization of 1412366654812131
Prime Factorization 35 × 23 × 137 × 1844559967
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 8 Total number of prime factors rad(n) 17436625368051 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,412,366,654,812,131 is 35 × 23 × 137 × 1844559967. Since it has a total of 8 prime factors, 1,412,366,654,812,131 is a composite number.
## Divisors of 1412366654812131
48 divisors
Even divisors 0 48 24 24
Total Divisors Sum of Divisors Aliquot Sum τ(n) 48 Total number of the positive divisors of n σ(n) 2.22374e+15 Sum of all the positive divisors of n s(n) 8.11376e+14 Sum of the proper positive divisors of n A(n) 4.6328e+13 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 3.75815e+07 Returns the nth root of the product of n divisors H(n) 30.4863 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,412,366,654,812,131 can be divided by 48 positive divisors (out of which 0 are even, and 48 are odd). The sum of these divisors (counting 1,412,366,654,812,131) is 2,223,742,471,501,824, the average is 46,327,968,156,288.
## Other Arithmetic Functions (n = 1412366654812131)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 894065593760064 Total number of positive integers not greater than n that are coprime to n λ(n) 37252733073336 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 41713831716619 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 894,065,593,760,064 positive integers (less than 1,412,366,654,812,131) that are coprime with 1,412,366,654,812,131. And there are approximately 41,713,831,716,619 prime numbers less than or equal to 1,412,366,654,812,131.
## Divisibility of 1412366654812131
m n mod m 2 3 4 5 6 7 8 9 1 0 3 1 3 4 3 0
The number 1,412,366,654,812,131 is divisible by 3 and 9.
• Arithmetic
• Deficient
• Polite
## Base conversion (1412366654812131)
Base System Value
2 Binary 101000001001000101000111100110000001010111111100011
3 Ternary 20212012202221012120012211100000
4 Quaternary 11001020220330300022333203
5 Quinary 2440110203402112442011
6 Senary 21523504141201431043
8 Octal 50110507460127743
10 Decimal 1412366654812131
12 Duodecimal 1124a615303b483
20 Vigesimal 6hiab8jcba6b
36 Base36 dwn4a81r6r
## Basic calculations (n = 1412366654812131)
### Multiplication
n×i
n×2 2824733309624262 4237099964436393 5649466619248524 7061833274060655
### Division
ni
n⁄2 7.06183e+14 4.70789e+14 3.53092e+14 2.82473e+14
### Exponentiation
ni
n2 1994779567625209202818904761161 2817360145014405772870372660560150667880444091 3979145523415016575157906585055406383723543814896154454067921 5620012451916333046686872871378836551617493880687062385694535084442968749651
### Nth Root
i√n
2√n 3.75815e+07 112197 6130.37 1071.49
## 1412366654812131 as geometric shapes
### Circle
Diameter 2.82473e+15 8.87416e+15 6.26678e+30
### Sphere
Volume 1.18013e+46 2.50671e+31 8.87416e+15
### Square
Length = n
Perimeter 5.64947e+15 1.99478e+30 1.99739e+15
### Cube
Length = n
Surface area 1.19687e+31 2.81736e+45 2.44629e+15
### Equilateral Triangle
Length = n
Perimeter 4.2371e+15 8.63765e+29 1.22315e+15
### Triangular Pyramid
Length = n
Surface area 3.45506e+30 3.32029e+44 1.15319e+15
## Cryptographic Hash Functions
md5 10be0816b19c42f0dc021d2ddee80f13 37af012064e574ab840015c25f06b6c7792be0cb 24a5288daccb50a98cb97adedbbab846e47d81ed0096e561648eabf719319819 82cf4138cfcfc9b78a1bd98904e5378029828de97477368d9c2edda761f9dfcacd8c475573c246dfc4f80dd07d453d6b0292f7b80715b8292b1b34d2fcbf7cbd 109b9932cfd1d9b23995f71d658f0ffe3dd43b45 | 1,849 | 5,191 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2020-16 | latest | en | 0.757169 |
https://www.tes.com/news/tes-archive/tes-publication/colour-codedsecondaryreviewsgeneralbooks | 1,512,968,024,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948512121.15/warc/CC-MAIN-20171211033436-20171211053436-00608.warc.gz | 795,413,503 | 15,147 | # Colour coded;Secondary;Reviews;General;Books
15th January 1999 at 00:00
MATHS NOW! BOOK 1 AND TEACHER'S FILE. (Red and Blue Orbit). John Murray pound;10.99 each. BOOK 2. (Green Orbit). John Murray pound;8.99. SUMMIT MATHS BOOK 1. By Ray Allen and Martin Williams. Nelson pound;8.95.
Anyone seeking a change from the books that dominate secondary maths teaching will welcome these publications. Maths Now! has three strands, with Red and Blue books or orbits corresponding to the three tiers of the GCSE, and the Green Orbit aimed at pupils with special needs.
The books are attractive, with excellent colour photographs. The RedBlue book has an interesting start that challenges students to think about the ways maths is used. It includes historical and contemporary references and the style is accessible without being patronising.
The book's main use will be for whole-class teaching. Differentiation is achieved through outcome, using tests of increasing difficulty. Supplementary material is provided in the teacher's resource file.
The teacher's file for the Green Orbit contains good suggestions for teaching pupils with special needs, as well as a wealth of supplementary worksheets. The book is intended for pupils at level 2 of the national curriculum. A "placement test" helps teachers decide what a pupil needs to cover.
Icons tell pupils when to work with others, when to ask for help and when to collect apparatus. There are also many activities such as number bingo and races against time that help pupils choose the correct operations. A system of points for mastering vocabulary, doing well in the tests, presentation of work and mastery of skills is included, and this could be tied into the school's reward system.
Summit Maths is also aimed at pupils below level 3, and the writers have taken care with the language and approach to topics. Multiplication is introduced via calculating the numbers of arms and eyes of friendly aliens, and there are good sections on shape and space. The work on decimals is less satisfactory, and the introduction to algebra progresses too quickly. On the other hand, two excellent "Work-outs" provide extended activities that allow pupils to practise the material in previous chapters.
Ian Wilson is head of Woodcote High School, Croydon, Surrey
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Enter subscription number | 494 | 2,432 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-51 | latest | en | 0.951976 |
https://www.scribd.com/document/92077914/Revised-States-of-Matter | 1,566,195,807,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314667.60/warc/CC-MAIN-20190819052133-20190819074133-00223.warc.gz | 956,146,645 | 110,656 | You are on page 1of 82
# PHYSICAL CHEMISTRY
## States of Matter: Gaseous, Liquid and Solids
Dr. S.P. Singh Ex-Head, Dept. of Chemistry Janta Vedic P.G. College Baraut, Distt Meerut (12.04.2007)
CONTENTS Gaseous State Kinetic Theory of Gases Deviation from Ideal Behaviour Van der Waals Equation of State Critical Phenomenon p - V isotherms of van der Waals equation Van der Waals Reduced Equation of State Law of Corresponding State Molecular Velocities Liquefaction of Gases Liquid State Intermolecular Forces Structure of Liquids Structural Differences between Solids, Liquids and Gases Liquid Crystals Classification of Liquid Crystals Thermography Seven segment cell Solid State Symmetry elements of crystals Braggs Law and Derivation of Braggs Equation Laues Method Power method Crystal Structures of sodium chloride, potassium chloride and cesium chloride
Chemistry deals with different kinds of matter. These can exist in different forms as solids, liquids or gases depending on the conditions of temperature and pressure. These states are characterized by their shape and volume. Solids have definite shape and volume; liquids have definite volume but no definite shape; gases have the volume and shape of the vessel in which it is contained. Apart from these three common states, there is a fourth state of matter called plasma which refers to an ionized gas. To define the state of a system the state variables generally chosen are pressure (p), volume (V), temperature (T) and amount ( number of moles, n ). An equation connecting these variables ( p, V, T, n ) is called an equation of state. The general form of an equation of state is p= f (V, T, n ) The basic features of different states beginning with the gaseous state are discussed below. Gaseous State The gaseous state is characterized by a marked sensitivity of volume to changes of temperature and pressure and has normally no bounding surface and so it tends to fill completely any available space. In gases, cohesive forces are very weak, so the molecules are in chaotic motion and no regular pattern in the arrangement of molecules is observed . The gaseous state has been studied extensively for a very long time. There are some well defined laws from which an equation of state may be derived easily. (a) Boyles Law : Robert Boyle studied the variation of volume of a gas with pressure and in 1662 proposed the following law : At a fixed temperature ,the volume of a fixed amount of a gas varies inversely with the pressure. Mathematically the law is represented as V
or or
1 p 1 p V
## (n and T constant ) ( n and T constant ) ( n and T constant ) (1)
pV = constant
( b) Charles or Gay-Lussacs Law : The effect of temperature on the volume of a gas was investigated in detail by Jacques Charles and Gay-Lussac (1802). The law states that : For a given amount and fixed pressure, the volume of a gas varies directly with its absolute temperature; i.e.,
V T
or
V = constant T
( n and p constant )
(2)
( c )Avogadros Law : This law (1811) states that : At a fixed temperature and pressure, the volume of a gas is directly proportional to the amount of gas, i.e.,
V n
( p and T constant )
(3)
From the principle of joint variation, combining equations (1), (2) and (3) we get a single equation , called combined gas law nT or pV nT or (4) pV = nRT p where R is proportionality constant, known as molar gas constant ( 8.314 J K-1 mol-1 =8.314 x107 erg K-1 mol-1 =0.0821 litre atm. K-1 mol-1 =1.987 cal K -1 mol-1 ). Equation (4) is also known as Ideal Gas Equation or Equation of State for an Ideal Gas. V Kinetic Theory of Gases A simple theoretical model ( Kinetic Theory of Gases ) was developed mainly to explain the gas laws and is based on the following assumptions or postulates: 1. Gases consist of a large number of minute particles called molecules,which are so small and so far apart that the actual volume of molecules is negligible compared to the total volume of the gas. Molecules of a particular gas are identical in all respects. 2. The molecules in a gas are in a state of rapid random motion colliding each other and with the walls of the container. 3. The pressure imparted by the gas originates from the collision of molecules with the walls of the container and the consequent change of momentum. 4. The collisions of the molecules are perfectly elastic, i.e., no energy is lost in collisions, although there may be transfer of energy between the partners in the collision. 5. There is no intermolecular force between the gas molecules and hence the molecules move independent of each other. 6. The effect of gravitational force on the motion of molecules is also negligible. 7. At a particular instant, different molecules of a gas move with different velocities and hence have different kinetic energies. However, the average kinetic energy of the molecules is directly proportional to the absolute temperature of the gas. Computation of Gas Pressure from Kinetic Theory Let us consider N gas molecules, each of mass m, moving with velocities C1, C2, C3,---------, CN respectively and placed in a cubic box of side length l. Resolving the velocity, C1 of a gas molecule into three Cartesian components x1, y1, z1 parallel to three axis X, Y and Z respectively along the sides of the cube, we get , C12 = x12 + y12 + z12 (5) For the motion along X-axis, consider a molecule of momentum mx1 striking with the stationary wall at right angle. Assuming the impact to be elastic, after collision the molecule rebounds with the same velocity in opposite direction. Thus, the momentum after collision is mx1 and the total change of momentum per collision will be mx1 (-mx1) = 2mx1 . After collision the molecule will hit the opposite wall after traveling distance l , in the time l x1 seconds. Thus, the number of collisions per second is x1 l . Therefore, the total change of momentum per second or rate of change of momentum is 2m x12 l . Considering other two Cartesian components, the total rate of change of momentum is 2m ( x12 + y12 + z12 ) l =
2m C12 /l . For all N molecules of the gas , the rate of 2 2 2m (C12 + C 2 + C 32 + + C N ) l (6)
change of momentum
Thus, the total force ( force = rate of change of momentum ) acting on the walls of container = 2 2 2m (C12 + C 2 + C 32 + + C N ) l . Hence, the pressure ( p = Force/ Area ) exerted by the molecules on the walls ( area of six walls = 6 l 2 ), 2 2 2 2 p = 2m (C12 + C 2 + C 32 + +C N ) 6l 2 l = m (C12 + C 2 + C 32 + +C N ) 3V
1 2 m N C rms (7) ( since the volume of the container, V = l3 ) 3 2 2 2 where the mean square velocity, C rms = (C12 + C 2 + +C N ) .
or
pV =
Example 1. Calculate the pressure exerted by 1023 gas molecules each of mass 10 -22 gm in a container of volume 1 litre . The rms velocity is 105 cm / sec.
2 Sol. Given N = 1023 ; m = 10 -22 gm ; rms velocity = 105 cm / sec or C rms = 1010 cm /sec. ; V =1 litre =103 ml.
In S.I. Units m = 10-25 kg rms velocity= 10-3 m/s v= 10-3 m3 1 2 Using the equation, p V = mNC rms 3 23 22 1x10 10 1010 p = = 3.33 x 107 dynes cm-2 3 310
## = 3.33 x 106 N m-2
Average Kinetic Energy of Molecules The pressure of a gas as derived by kinetic theory is,
2 p = m N C rms / 3 V
## 1 NA 2 ( )mC rms 3 Vm constant and Vm = molar volume )
For one mole of the gas, p= We also know that : Hence, or p Vm = R T or
## ( for one mole gas )
1 3 2 N A mC rms = RT 2 2
1 2 N A mC rms = R T 3 1 3 RT 3 2 mC rms = = kT 2 2 NA 2
(8)
## ( k = R N A = Boltzmann constant = 1.38065 x 10-23 J K-1 )
Thus, the average kinetic energy of a molecule is directly proportional to the temperature of a 3 gas. Also the average kinetic energy per mole of the gas = RT 2 Example2. Calculate the translational energy of gas molecules in one mole of nitrogen at 30o C. Sol. Given n = 1 ; R = 8.314 J K-1 mol-1 and T = 30 + 273 = 303 K Hence Etrans =
3 nRT 2 3 = x1x8.314 x303 = 3.7787x 103 2
Example 3. Calculate the total kinetic energy of 33 gm of CO2 at 300o C. 33 Hint : Here n = = 0. 75 mol Ans. ( 5.3594 x 103 ). 44 Example 4. Calculate the kinetic energy of two molecules of CO2 at 300 K, present in an ideal gas. 2 Hint : Here n = Ans. ( 1.24 x 10-20 J ) 23 6.022 x10
Kinetic energy per molecules 3 T 2 = 3 T 2
= 2 x
=3x1.38x10-23x300J = 1.24x10-20J Limitations of ideal gas equation , p V = n R T Gas equation , p V = n R T is valid for ideal gases only. Real gases obey this equation only approximately at low pressures and high temperatures and show considerable deviations otherwise. Ideal and Real Gases An ideal or a perfect gas is one which obeys gas laws or gas equation over entire range of temperature and pressure. Some characteristics of ideal gases are: (i) (ii) (iii) (iv) (v) (vi) The product of pressure ( p ) and volume ( V ) at fixed temperature is constant. The plot of p V against p at constant temperature is a straight line parallel to p-axis. The molar volume ( Vm ) of an ideal gas at STP is 22.414 litres or dm3. The coefficient of expansion of volume ( ) of an ideal gas is 1 ( Gay 273 Lussacs Law ) and at -273o C, the volume should become zero. The compressibility factor, Z = pV is unity. nRT No change in temperature should occur when ideal gas expands in vacuum, because there are no intermolecular forces between the molecules, i.e., dE =0.
No real gas ( H2 , O2 or Cl2 etc. ) is ideal in this sense. Almost all real gases show considerable departure or deviation from ideality at high pressures and low temperatures. Example 5. What will be the volume of 2 moles of CO at 400 atm. and compressibility factor Z is 1.5 at this temperature and pressure . ZnRT pV Sol. Z= or V = = 1.5 x 2 x 0.0821 x 273 / 400 = 0.168 L p nRT 0o C if
Deviation from Ideal Behaviour The extent of deviation from ideal behaviour may be illustrated by the plot of p vs V ( or p V vs p ) at constant temperature called isotherm or in terms of compressibility factor, Z. ( A) p V - p isotherms : Amagats curves and Boyle Temperature. According to the Boyles law, the plot of p V vs p should be a straight line parallel to paxis. In other words, for an ideal gas obeying Boyles law ( p.V ) T =0 (9) p Ragnault and later Amagat studied in detail the p V vs p curves. They found that most gases obey Boyles law only at low pressure ( < 10 atm. ). For most gases, with increase in pressure, the value of p V first decreases, passes through a minimum and then increases ( Fig. 1 ). With rise in temperature, the dip in the curve ( i.e. minimum ) becomes shallower. As the temperature is raised, p V vs p curve becomes horizontal, i.e. parallel to p- axis from p = 0 to a reasonably high pressure at a certain temperature known as Boyle temperature ( TB = a ). Rb Thus, Boyle temperature is defined as the temperature at which or compressibility factor, p.Vm / R T = 1
T3 > T2 > T1 T3 T2 T1
=0
PV
## P Fig. 1. PV-P isotherm for a real gas
( B ) Compression Factor
The deviation of compressibility factor, Z ( Z = p Vm / R T ) from unity is the measure of nonideality of a gas . For all gases Z approaches unity as p 0. However, at high pressures Z deviates from unity and the magnitude of variation depends on the nature of the gas ( Fig. 2 ) and temperature ( Fig. 3 ). Fig. 4 shows that Z for H2 increases continuously with pressure. 6
Helium and neon resemble H2 . Z for N2, CO2, etc. at first decreases ( Z < 1 ) and then increases. The dip in the curve is greatest for the gas which get very easily liquefied.
## Fig. 3. Variation of Z with p for N2 at different temperatures.
The shape of the curves ( Fig. 3. ) reveal that N2 resembles hydrogen at high temperatures while CO2 at low temperatures. It is observed that in general, the deviations are greatest at high pressures and low temperatures and for easily liquefied gases. ( C ) p V isotherms : Andrews Curves Thomas Andrews ( 1869 ) studied the behaviour of gases. The results showing p V isotherms for CO2 ( a real gas ) at different temperatures are summarized in Fig. 4. These isotherms exhibit the following features : ( 1 ) At low temperature,there are three distinct regions : ( i) At very low pressures, the curve AB resembles the Boyles curve. At point B the volume decreases rapidly due to liquefaction. (ii) The portion BC corresponding to one value of pressure is parallel to V- axis and represents the equilibrium of gas with liquid. Different points represent different proportions of gas and liquid. This region is called the liquefaction region and at point C the gas is completely converted into liquid. (iii) The portion CD corresponds to the liquid state and shows that the increase in pressure produces a very small decrease in volume. This curve rises almost vertically.
( 2 ) With increase in temperature the length of the liquefaction region decreases until it reduces
to a point P at critical temperature Tc ( 31.1oC for CO2 ). The isotherm at this temperature is called critical isotherm and the point P is called critical point.
( 3 ) The isotherms above Tc are rectangular hyperbola and approximates to the isotherms fpr ideal gas ( Fig. 5. ) In this region, the matter always exists in gaseous state and the gas cannot be liquefied.
20 18 16 14 12 10
P
8 6 4 2 0
T3 T2 T1
V Fig. 5. P-V. isotherm for an ideal gas Fig. 4. p-V isotherm for a real gas Van der Waals Equation of State van der Waals equation represent one of the early attempts to explain the behaviour of real gases. van der Waals (1873) attributed the deviations of real gases from ideal behaviour to two erroneous postulates of kinetic theory. These are : ( i ) The actual volume of molecules is negligible compared to the total volume of the gas. ( ii ) There are no intermolecular force between the gas molecules. He suggested that since the volume of gas molecules is not negligible, the volume occupied by the gas needs to be corrected. Secondly , there are intermolecular forces of attraction as a result of which the pressure felt does not represent the correct pressure of the gas and so the pressure factor needs correction. If pcorr and Vcorr denote the corrected pressure and volume, the gas equation takes the form pcorr Vcorr = n R T -------------------( 10 ) ( a ) Pressure correction : In the presence of intermolecular forces of attraction the net force on a gas molecule in the interior of the container is zero because it is attracted in all directions by equal forces. However, a molecule near the wall experiences a net inward pull due to the molecules in the bulk of the gas and consequently it will strike the wall with less momentum than it would have in the absence of such forces. The measured pressure ( p ) is thus less than the ideal pressure ( pcorr ) by an amount p. Thus, pcorr = p + p The magnitude of p is proportional to both the number of molecules about to strike the wall and number of molecules engaged in pulling the striking molecules. But the number of molecules is proportional to the density of the gas, i. e., n / V ( n = no. of moles of gas ). Hence, n 2 an 2 p 2 = where a is a constant. V V2 Thus, pcorr = p + an 2 V2
------------------( 11 )
( b ) Volume correction :
Because of the finite volume of gas molecules a portion of the container becomes inaccessible to the gas molecules. If V is the volume of the container and b denotes inaccessible volume or excluded volume for one mole of gas, then Vcorr = V - b . Vcorr = V - n b --------------------( 12 )
## For n moles of the gas,
The value of b is found to be four times the actual volume of gas molecules. Substituting the corrected pressure and volume from equations (11) and (12) in equation (10) we get the van der Waals equation of state for n moles of the gas, an 2 an 2 nRT )( Vnb) = nRT or p = -----------( 13 ) ( p + V nb V2 V2 For one mole ( n = 1 ), equation (13) becomes ( p +
a ) ( Vm - b ) = R T 2 Vm
or
p = R T / ( Vm b ) -
a 2 Vm
------------( 14 )
Parameters a and b which are characteristics of each gas i.e., depend on the nature of gas , are called van der Waals constants or parameters. Calculation of Effective or Excluded Volume Consider the collision of two molecules each of radius r ( Fig. 6 ), then centres of these are separated by a minimum distance 2r . As a result, the centre of one molecule cannot enter into a sphere of radius 2r around the other molecule. This space will not be available to all other molecules of the gas. In other words, this is the excluded volume per pair of molecules. Thus, the excluded volume for the pair of molecules = 4 (2r )3/3 = 8 where = 4 r 3 / 3 i.e., the volume of each molecule. Hence, for single molecule the excluded volume = 8 / 2 = 4 . For one mole of the gas, the excluded volume, b = 4 NA . Therefore, excluded volume is four times the actual volume of molecules of the gas.
Fig. 6.
Units of van der Waals parameters a and b an 2 is the pressure correction and thus has the units of ( i ) In van der Waals equation, V2 an 2 = p ( atm. ) or a = p ( atm. ) x V2 ( dm6 ) / n2 ( mol2 ) = pressure. So, 2 V 2 p.V atm. dm6 mol-2 2 n Thus a is expressed in atm. dm6 mol-2 or atm. litre2 mol-2 units. The S I unit is N m4 mol-2 or Pa m6 mol-2 ( Taking units of pressure, N m-2 and volume, m3 where N = Newton, m = meter and Pa = pascal ). V ( ii ) n b is the excluded volume for n moles of gas. Thus, b = V ( dm3 ) / n ( mol ) = n dm3 mol-1 or litre mol-1 . So, b is expressed in dm3 mol-1 or litre mol-1 units or m3 mol-1 ( S I unit ). Significance of van der Waals Parameters It is observed that the value of a is greater for easily liquefiable gases than those for so called permanent gases ( H2 , He etc. ) and increases with the ease of liquefaction which increases with intermolecular forces of attraction in the gas. So, the term a is a parameter characterizing the nature and extent of intermolecular interactions per mole of the gas. The parameter b is related to the molecular volume by the relation, b = 4 Vm NA = 4 NA { 4 ( /2 )3 / 3} = 2 NA 3 / 3 . Thus knowing the value of b from the compressibility results of a gas, ( diameter of molecule ) can be calculated. Example 6. The van der Waals parameter b for a certain gas is 0.0329 litre mol-1. Calculate the diameter of the molecule, assuming that molecules are spherical ( NA = 6.02 x 1023 ) . Sol. Given b = 0.0329 litre mol-1 = 0.0329 dm3 mol-1 = 32.9 cm3 mol-1 ; NA = 6.02 x 1023 . We know that b = 4 NA x 4 r3 / 3. Thus, r3 = 3 b / 16 NA x = 3 x 32.9 / 16 x 6.02 x 1023 x 3.14 =3.263 x 10-24 cm3 or r = 1.484 x 10-8 cm. Hence, diameter of the molecule = 2r = 2.968 x 10-8 cm = 0. 2968 nm Example 7. The van der Waals constant b for helium is 24 ml / mol. Calculate the molecular diameter of helium NA = 6.02 x 1023. Ans. 2.67 x 10-8 cm . Example 8. Two moles of ammonia are confined to a 5.0 litre flask at 27oC. Calculate its pressure using van der Waals equation ( a = 4.17 atm. litre2 mol-2 and b = 0.0371 litre mol-1 ). Sol. Given n = 2 moles ; R = 0.0821 atm. L mol-1 K-1 ; T = 27 + 273 = 300 K ; a = 4.17 atm. litre2 mol-2 ; b = 0.0371 litre mol-1 and V = 5.0 L
10
Substituting these values in van der Waals equation, 2 x0.0821x300 2 2 x 4.17 5.0 2 x0.0371 5.0 2
10.0004 0.6672
nRT n2a p= V nb V 2
p=
## 2 x 0.0821x300 4 x 4.17 4.9258 25
or p =
= 9.332 atm.
Example 9. Calculate pressure developed in 1 litre vessel containing 10 gms of ammonia at 0oC using ( i ) van der Waals equation ( ii ) ideal gas equation (a = 4.17 atm. litre2 mol-2 and b = 0.0371 litre mol-1 ). Sol. Given V = 1 litre ; mass of ammonia = 10 gm ; a = 4.17 atm. litre2 mol-2 and b = 0.0371 litre mol-1; T = 273 K ; R = 0.0821 atm. L mol-1 K-1 ; Molar mass of NH3 = 17 , Hence n = 10 / 17 ( i ) Using van der Waals equation,
p= 10 x0.0821x 273 10 2 x 4.17 2 10 17 x1.0 2 17(1 x0.0371) 17
=
or
p=
= 13.4784 1.4429
= 12.0355 atm.
## 10 x0.0821x 273 17 x1,0 2
= 13.184 atm.
Example 10. 0.5 mole of carbon dioxide was filled in a vessel of volume 0.6 litre at 47oC. What pressure would be expected on the basis of van der Waals equation ? ( a = 3.36 atm. lit.2 mol-2 ; b = 4.27 x 10-2 lit. mol-1 ) R = 0.0821 atm. L mol-1 K-1 Ans. p = 20.37 atm.
Explanation of the behaviour of real gases by van der Waals equation Almost all observed deviations from ideality by real gases can be explained by van der Waals equation as follows: (1) At very low pressure : Under these conditions, V ( volume ) will become very large and so, b will be negligible and a / V2 will also be very small or negligible. Thus, van der Waals equation for one mole reduces to pVm = R T and gas behaves ideally. (2) At low pressure : When p is small, V will be large and b will be negligible compared to V. So, the van der Waals equation can be written as, a RT a p Vm a p= or p Vm = R T or 2 = 1 Vm Vm Vm RT Vm RT 11
or
Compressibility factor, Z = 1
a Vm RT
or
Z < 1.
The compressibility factor is less than 1. This explains the initial portions of Z p curves for N2 , CO2 etc., which lie below the ideal curve. ( 3 ) At high pressures : At p large, V will be small. In this case b cannot be neglected while in comparison to large p, small added pressure ( a / V2 ) is negligible. Hence, van der Waals equation reduces to, p Vm p b RT = 1+ or p Vm = R T + p b or p= RT RT Vm b
or Z = 1 +
p b RT
or
Z > 1.
This accounts for the rising parts of Z p curves above the ideal gas curve. ( 4 ) At high temperatures : Under these conditions, both p and Vm will be large and the terms a / Vm and b are negligible. So, van der Waals equation reduces to, p Vm = R T and real gases tend to show ideal behaviour. ( 5 ) At low temperature : In this case, both p and Vm will be small so, the term a / Vm would a be large while term p b would be small. Hence, the difference ( p b ) would be a Vm positive quantity and Z < 1. This accounts for the dips at low temperatures. ( 6 ) Exceptional behaviour of hydrogen and helium : The masses of H2 and He molecules are very small and the intermolecular attractions are negligible and so, the term a / V2 would be negligible. Hence, the van der Waals equation becomes, p b p ( Vm b ) = R T or Z = 1+ or Z > 1. RT Since, Z is always greater than 1 and increases as p increases, this explain why Z p curve for these gases lies above ideal gas curve. van der Waals equation thus provides qualitative explanation for observed deviations from ideality by real gases and so it is an improvement over the ideal gas equation. Limitations of van der Waals equation : ( i ) This equation has been found to be very satisfactory over a wide range of temperature and pressure. However, it shows appreciable deviations at too high pressures and too low temperatures. The reason is that the values of a and b vary with temperature. At too low temperature the value of a is found to be about
12
## ( ii ) For many gases, the value of
R Tc 8 as derived by van p c Vc 3
Derivation of Boyle temperature from van der Waals equation : van der Waals equation for one mole may be written as pVm =
RTVm a Vm b Vm
_________________ ( 15 )
or
## pVm Vm Vm a a = = RT Vm b RTVm Vm (1 b Vm ) RTVm
pVm = (! b ) -1 - a RT Vm RTV m
-------------------------- ( 16 )
or
At Boyle temperature p Vm / R T =1 and Vm is much larger than b ( V >> b ) and therefore, the first term on R. H. S. can be expanded in a series with the retention of the first two terms, i.e., pVm b a b a a = 1 = 1+ or or TB = --------(17) = RT Vm Vm RTB Vm Vm RTB Rb Other Equations of State Out of several other equations of state proposed from time to time, some are mentioned below: Clausius Equation : R. Clausius (1880) proposed the following equation accounting for the variation of van der Waals parameter a with temperature, a p+ (Vm b) = R T ------------------------- ( 18 ) 2 T (Vm c) where c is a new constant . This equation gives a better agreement with the experimental data, but due to mathematical complexity did not attract much attention. Berthelot Equation : D. Berthelot eliminated the fourth constant c from Clausius equation , giving a p+ (Vm b) = R T ------------------------ ( !9 ) 2 TVm This equation holds best at low pressures and is applicable over a limited range only. Dieterici Equation : C. Dieterici (1899) suggested the following relationship to account for the effect of molecular attraction on the pressure,
13
p (Vm b) = RT e
a RTVm
----------------------
(20)
This equation is identical to the van der Waals equation at low pressures but differs appreciably at high pressures and gives more satisfactory agreement with the experimental data. Virial Equation of State : H.K. Onnes (1901) proposed the most general equation of state for real gases, B(T ) C (T ) D(T ) p Vm = RT 1 + + 2 + 3 + ----------------- (21) Vm Vm Vm where B(T), C(T), D(T), etc., which are functions of temperature, are called second, third, fourth, etc., Virial Coefficients respectively. The virial equation can also be expressed as a power series in pressure as p Vm = R T { 1 + B(T) p + C(T) p2 + D(T) p3 + ----------- } ------------(22) At low pressures, only the first virial coefficient ( = R T ) is significant and is always positive. It increases with increase in temperature . Apart from the term R T, the second virial coefficient B(T) is the most important term. It is negative at low temperatures, zero at a particular intermediate temperature and becomes increasingly positive as the temperature is raised continuously. The temperature at which B(T) = 0 is called Boyle temperature. Critical Phenomenon The gases can be liquefied by increasing pressure and lowering the temperature. The increase in pressure brings the molecules closer and decrease in temperature decreases their kinetic energies and hence speeds. The slow moving molecules then aggregate due to intermolecular attraction forming liquid state. Although increase in pressure helps in the conversion of gas into liquid, the decrease in temperature is more important. For every gas there is a temperature above which it cannot be liquefied what-so-ever high the pressure is applied. This temperature is known as critical temperature of the gas. Thus, Critical temperature, Tc is the temperature below or at which gas can be liquefied by applying the pressure. For example, Tc for CO2 is 31.1oC or 304.1 K. is the minimum pressure required to liquefy the gas at its critical Critical pressure, pc temperature . For example, at 31.1oC, the pressure required to liquefy CO2 is 72.9 atm. Hence Pc for CO2 is 72.9 atm. Critical volume, Vc is the molar volume of the gas at its critical temperature and critical pressure. For example, volume occupied by one mole of CO2 at 31.1oC and under 72.9 atm. pressure is found to be 94.2 ml. Hence Vc for CO2 is 72.9 ml. per mole. Tc, pc, and Vc are collectively called the critical constants or critical parameters of a gas. All real gases have characteristic critical parameters. At critical temperature and under critical pressure, a gas becomes identical with its liquid and is said to be in critical state and the phenomenon is called critical phenomenon.
14
Continuity of State When a gas placed in a closed vessel with its own liquid is heated, the boundary ( i.e. meniscus ) between these two phases disappears suddenly as the critical temperature is reached. At critical temperature the phases intermix so well that it is impossible to distinguish between them. As explained earlier in case of Andrews curves for CO2 , the horizontal portion of the isotherm at 31.1oC reduces to a point called critical point and isotherm becomes continuous. Here the gas passes into liquid imperceptibly. At critical point, the gas and liquid have the same density and are indistinguishable. The indistinguishable intermixing of the gaseous and liquid phases is known as continuity of state.This phenomenon is utilized in the experimental determination of critical constants of a gas. Experimental determination of critical temperature, Tc and critical pressure, pc Cagniard de la Tours apparatus ( Fig. 7 ) consisting a stout glass U-tube blown into a bulb at lower end, is used to determine Tc and pc. A small quantity of liquid under examination in contact with its vapours is contained in the bulb and the rest of apparatus is filled with mercury leaving a small portion at the upper end which is sealed. The air portion acts as a manometer.
Fig. 7 Method: (i) The bulb is cooled first by circulating a suitable liquid from the thermostate into the jacket so that meniscus between liquid and vapours becomes sharp. (ii) The bulb is then heated by heating the jacket. As meniscus disappers, the corresponding temperature T1 and the pressure in manometer P1 are noted. (iii) Again bulb is cooled till cloudiness due to condensation of vapours appears and meniscus reappears. The corresponding temperature T2 and pressure P2 are also noted. The mean of temperatures T1 and T2 gives the critical temperature, Tc. Knowing the initial and final volumes of air in manometer, the pressure exerted by air ( = p ) can be calculated. Knowing also the difference in the mercury columns ( = h ) in the two limbs, the critical pressure, pc = p + h can be determined. The critical volume cannot be determined by this method, because even a small change in temperature or pressure at critical point produces a large change in volume. Determination of critical volume, Vc
15
Principle : According to Cailletet and Mathias, when mean values of densities of liquid and saturated vapour of a substance are plotted against corresponding temperatures, a straight line is obtained. Amagats method of mean densities is used for most accurate determination of critical volume. Densities of the liquid and its vapours are measured at a number of temperatures near the critical temperature and a graph ( Fig.8 ) is plotted between densities and temperatures.
Fig.8 This gives two curves merging into each other. These curves are extrapolated to get a flat maximum. The line of mean densities is extended to touch the flat maximum at the point, H. The coordinates of H give the critical temperature and critical density ( d ). The critical volume, Vc can be calculated dividing molecular mass by critical density, i.e., Vc = Molecular mass, M / Critical density, d p - V isotherms of van der Waals equation The van der Waals equation for one mole of gas, written as,
p Vm p b + a a b 2 RT = 0 Vm Vm
a p + 2 (Vm b ) = R T Vm
may be
## 2 Multiplying this equation by Vm and dividing by p and rearranging gives
RT 2 aVm ab 3 Vm p b Vm + p p = 0 ---------------
(23)
This is a cubic equation in Vm and so , for given values of p and T, there should be three real solutions of Vm or one real and two imaginary solutions. This behaviour predicted by van der Waals equation is not born out by Andrews experimental isotherms of CO2. Thomson (1871) plotted theoretical isotherms for CO2 at different temperatures ( Fig. 9 ) . These curves indicate that below critical temperature ( 31.1oC for CO2 ) the isotherms consist of three regions a gas region AB, a liquid region CD and an intermediate region represented by BLMNC. First two regions are similar to those of Andrews experimental isotherms. The horizontal portion BC of Andrews experimental isotherms is replaced by a continuous wave like curve BLMNC , so that there are three values of volume represented by B, M and C corresponding to a particular value of pressure. The curve BL represents supper saturated vapours while curve NC represents supper heated liquid. With rise in temperature, the wave
16
like portion decreases until it reduces to a point P at critical temperature Tc and the three values of volume becomes identical to Vc. Thus at T = Tc , p ------------- (24) ( ) T =Tc = 0 V
Fig 9 Calculation of critical constants or parameters in terms of van der Waals parameters The van der Waals equation for one mole of gas is RT a ---------------- (25) p= 2 (Vm b ) Vm At critical temperature, we know that
p =0 V T =Tc
and
2 p =0 V 2 T =Tc
These two derivatives of pressure with respect to volume for a van der Waals gas are given by:
2a RT p = + 3 2 (Vm b ) Vm V T =Tc
## = 0 ----------- (26) ( since dxn / dx = n xn-1 )
and
2 p 2 RT 6a = 4 3 V 2 T =Tc (Vm b ) Vm
=0
(Vc b )
RTc
2a Vc3
(Vc b )
or
2 RTc
6a 4 Vm
Vc b Vc = 2 3
Vc = 3 b
## RTc 2a = 2 (3b b) (3b) 3
or
RTc 2a = 2 4b 27b 3
Substituting the values of Vc and Tc in van der Waals equation (25) at critical point,
17
pc =
a 27b 2
## 8a 27 Rb a = 8a a = 4a a -------------(32) {Hint: p c = (3b b) (3b) 2 27b 2b 9b 2 27b 2 9b 2 R
4a 3a a } = 2 27b 27b 2 It is apparent that values of critical parameters depend on the nature of gas (i. e.values of a and b)
Calculation of van der Waals constants in terms of pc , Vc and Tc Since pc ,Vc and Tc can be determined experimentally, their values may be used to calculate a , b and R. From equation (30) , b = Vc / 3 -------------------- (33) From equations (30) and (32), a = 3 pc Vc2 ------------------- (34)
Substituting the values of a and b from equations (33) and (34) in equation (31), we get Tc = 8 pc Vc / 3 R or R = 8 pc Vc / 3 Tc or R Tc = 8 pc Vc / 3 or R Tc / pc Vc = 8 / 3 = 2.67 or pc Vc / R Tc = 3 / 8 = Zc = 0.375 ---- (35)
where Zc is the critical compressibility factor, a constant independent of the nature of gas. It is observed that for many gases, the experimentally determined value of R Tc / pc Vc lies in the range 3 4 , indicating the deviations from the van der Waals equation. Since the determination of Vc is some what difficult, the values of a and b can also be calculated in terms of pc and Tc only. Substituting the value of Vc from equation (30) in equation (35), b = R Tc / 8 pc ------------- (36) Substituting the value of Vc from equation (35) in equation (34) gives
3RTc a = 3 pc 8p c 3 p 9 R 2T 2 27 R 2T 2 = c = 64 p c 64 p c2
2
------------- (37)
Example 11. For Cl2 , the value of a = 4.37 L2 atm. mol-2 and b = 0.0515 L mol-1 . Calculate pc and Vc. Sol. We know that Vc = 3 b and pc = a / 27 b2 -1 Hence, Vc = 3 x 0.0515 = 0.1545 L mol and pc = 4.37 / 27 x ( 0.0515)2 = 61.024 atm. Example 12. Estimate the critical constants ( pc ,Vc and Tc ) of a gas with van der Waals parameters, a = 0.751 atm. L2 mol-2 and b = 0.0226 L mol-1. Sol. Vc = 3 x 0.0226 = 0.0678 L mol-1 pc = 0.751 / 27 x ( 0.0226 )2 = 54.458 atm. Tc = 8 x 0.751 / 27 x 0.082 x 0.0226 = 120.07 K = - 152.93oC ( R = 0.082 L. atm. K-1 mol-1 )
18
Example 13. The critical temperature and critical pressure of oxygen are 118oC atmospheres. Calculate its van der Waals constants. ( R = 0.082 lit. atm. K-1 mol-1 ).
and 49.7
Sol. Given pc = 49.7 atm. ; Tc = 118oC = - 118 + 373 = 155 K Hence, a = 27 R2 Tc2 / 64 pc = 27 x (0.082)2 x 1552 / 64 x 49.7 = 1.37 atm. L2 mol-2 and b = R Tc / 8 pc = 0.082 x 155 / 8 x 49.7 = 0.03197 L mol-1 . Example 14. Calculate the constant a and b for a gas when critical pressure and critical temperature are 72.8 atm. and 31oC respectively. Ans. ( a = 3.6098 dm6 atm. mol-2 and b = 0.04285 dm3 mol-1 ). Example 15. If Tc = 33.3 K , pc = 12.97 bar and Vc = 6.5 x 10-5 m3 mol-1 for hydrogen, calculate the van der Waals parameters. ( R = 8.314 J K-1 mol-1 ) . Sol. Given Tc = 33.3 K ; pc = 12.97 bar = 12.97 x 105 Pa ( 1 atm. = 1.013 x 105 N m-2 = 1.013 x 105 Pa = 1.013 bar ) ; Vc = 6.5 x 10-5 m3 mol-1 ; R = 8.314 J K-1 mol-1 Hence, a = 3 pc Vc2 = 3 x 12.97 x 105 x ( 6.5 x 10-5)2 = 1.6439 x 10-2 N m4 mol-2 and b = Vc / 3 = 6.5 x 10-5 / 3 = 2.1667 x 10-5 m3 mol-1 Example 16. Critical density of a substance having molecular weight 111 is 0.555 gm / c.c. and pc = 48 atm. Calculate van der Waals parameters a and b . Sol. Critical volume, Vc = Mol. wt. / Critical density = 111 / 0.555 = 200 cm3 = 0.2 L ; Hence, b = 0.2 / 3 = 0.0667 litre mol-1 and pc = 48 atm. a = 3 pc Vc2 = 3 x 48 x 0.22 = 5.76 atm. L2 mol-2 Example 17. Critical temperature of carbon dioxide gas is 31.1oC and its critical density is 0.455 g cm -3 . Calculate the values of a , b , and pc for the gas . ( R = 0.082 L atm. K-1 mol-1 ). Sol. Critical volume, Vc = 44 / 0.455 = 96.7 cm3 = 0.0967 L ; Tc = 31.1oC = 31.1 + 273 =304.1 K ; Hence, pc = 3 R Tc / 8 Vc = 3 x 0.082 x 304.1 / 8 x 0.0967 = 96.7 atm. b = Vc / 3 = 0.0967 / 3 =0.03223 L mol-1 and a = 3 pc Vc2 = 3 x 96.7 x 0.09672 = 2.7127 atm.L2 mol-2 Van der Waals Reduced Equation of State If pressure, volume and temperature of a gas are expressed in terms of critical parameters, then the ratios p / pc = pr , V / Vc = Vr and T / Tc = Tr are termed as reduced pressure, reduced volume and reduced temperature respectively. In terms of reduced variables the van der Waals equation for one mole of gas may be written as, ( pr pc + a / Vr2 Vc2 ) ( Vr Vc - b ) = R Tr Tc --------------- (38) Substituting the values of pc ,Vc and Tc from equations (32) , (30) and (31) we get ( a pr / 27 b2 + a / 9 b2 Vr2 ) ( 3 b Vr b ) = 8 a R Tr / 27 R b Multiplying by 27 b2 on both sides , we have
19
( a pr + 3 a / Vr2 ) ( 3 b Vr b ) = 8 a b Tr = a b 8 Tr or ( pr + 3 / Vr2 ) ( 3 Vr - 1 ) = 8 Tr
or
## a b ( pr + 3 / Vr2 ) ( 3 Vr - 1 ) ------------------------- (39)
This is known as van der Waals reduced equation of state. Since this equation does not involve any of the van der Waals and critical parameters which are characteristic of gases, it is applicable to all substances in liquid or gaseous state irrespective of their specific nature . The same can be shown by ploting Z against pr = p / pc for a variety of gases. The data for different gases fall on the same curve ( Fig. 10 ) at the same temperature showing the independence of the nature of gas . T3 > T2 > T1
T3 T2
T1
Gas 1 Pr
Gas 2
Gas 3
Fig 10 Law of Corresponding State It is clear from the reduced equation of state that different substances having the same reduced pressure and reduced temperature have same reduced volume.This is known as the Law of corresponding state.The substances are said to be in the corresponding state. Importance : In the studies of the relations between physical properties of liquids and the chemical constitution, the properties are determined at the same reduced temperature because pressure has very slight effect on liquids. The boiling points are found to be approximately twothird ( 2 / 3rd ) of critical temperatures and so at boiling points the liquids are in their corresponding states. Thus, the physical properties of substances are determined at their boiling points. Example 18. Calculate the pressure for 1 mole of a gas at 313 K occupying a volume of 0.107 dm3/mol using law of corresponding states. Vc = 0.0957 dm3 mol-1 ; Tc = 31oC ; pc = 73.0 atm. ; R = 0.0821 atm. dm3 K-1 mol-1. Sol. pc = 73.0 atm. ; Vc = 0.0957 dm3 mol-1 ; Tc = 31oC = 31 + 273 = 304 K ; T =313 K ; V = 0.107dm3 mol-1 From the given data pr = p / 73.0 ; Vr = 0.107 / 0.0957 = 1.118 ; Tr = 313 / 304 = 1.0296 Substituting these values in Pr =
8Tr 3 2 3Vr 1 Vr
we get
20
p 8 x1.0296 3 8 x1.0296 3 = = = 3.499 2.400 = 1.099 2 73.0 3 x1.118 1 1.118 2.354 1.118 2 or p = 1.099 x 73.0 = 80.227 atm.
Example 19. Using equation ( pr + 3 / Vr2 ) ( 3 Vr - 1 ) = 8 Tr , calculate the temperature at which 1 mole of gas would exert a pressure of 111.0 bar when confined in a volume of 3 x 10-4 m3. ( R = 8.314 x 10-5 bar m3 K-1 mol-1 ; pc = 74.13 bar ; Tc = 31.2oC ; Vc = 1 x 10-4 m3 mol-1 ) Sol. Given pc = 74.13 bar = 74.13 x 105 Pa ; Tc = 31.2 + 273 = 304.2 K ; Vc = 104 3 -1 -5 3 -1 -1 -1 -1 m mol ; R = 8.314 x 10 bar m K mol = 8.314 Pa K mol ; p = 111.0 bar = 111.0 x 105 Pa ; V = 3 x 10-4 m3. From the given data , pr = p / pc = 111.0 x 105 / 74.13 x 105 =1.4974 ; Vr = V / Vc = 3 x 10-4/ 10-4 = 3; Tr = T / 304.2 ; Substituting these values in ( pr + 3 / Vr2 ) ( 3 Vr - 1 ) = 8 Tr , 2 8 T / 304.2 = ( 1.4974 + 3 / 3 ) ( 3 x 3 1 ) or 8 T / 304.2 = ( 1.4974 + 0.3333) ( 9-1) = 1.8307 x 8 or T = 1.8307 x 304.2 = 556.9 K Molecular Velocities Or Speeds For a quantitative description of the behaviour of gases, three different kinds of molecular speeds are used, viz. average speed ( C ), most probable speed ( C m ) and root mean square speed ( C rms ). Average speed ( C ) is the arithmetic mean of the speeds possessed by the different molecules of a gas at a given temperature. Let C1 , C2 , C3 , ------------- , Cn be the individual speeds of n molecules of a gas, then
C = [ C1 + C 2 + C 3 + + C n ] / n
Most probable speed ( C m ) is the velocity possessed by the maximum number of molecules of a gas at a given temperature. Root mean square speed or RMS speed ( C rms ) is the square root of the mean of squares of speeds of the gas molecules.
2 2 C12 + C 2 + C 32 + + C n n Maxwell Boltzmann Distribution of Molecular Speeds All the molecules in a sample of a gas do not travel with the same speed and their speeds are changing billions of times every second due to intermolecular collisions. It is ,therefore, futile to know the speeds of individual molecules. Maxwell suggested that at a particular temperature, the fractions of molecules possessing particular speeds remain almost constant.
C rms =
The distribution of molecules between different possible speeds was independently derived from statistical considerations by Maxwell and Boltzmann in 1860. According to this distribution, the fraction of molecules 21
## having speed between c and c + dc is given by,
M c dnc M 2 2 RT 2 = 4 c dc e n 2RT 1 dn the probability ( Pc = x c ) n dc 3
2
------------------ (40)
or
## of molecules having speed
is given by
1 dnc M 2 2 RT 2 = 4 x c e n dc 2RT
M c 2
------------------ (41 )
where dnc = number of molecules having speeds between c and c + dc ; n = total number of molecules; M = molecular mass and T = absolute temperature of the gas . This relation is called Maxwells law of distribution of speeds. The plot of probability factor ( Pc ) against speed of molecules (Fig. 11) illustrate the salient features of Maxwell distribution of speeds, which are : (i) The area of the shaded portion gives dnc / n i.e. the fraction of molecules having speed between c and c + dc. The total area of the curve is obviously unity. (ii) Around very low ( c 0 ) or very high speed ( c ) the area is very small. This implies that very few molecules have either very low or very high speeds. (iii) Most of the molecules have speeds near the average velocity. (iv) The curves passes through a maximum. The speed corresponding to the maxima is called most probable speed (Cm) because it is possessed by the maximum number of molecules. The value of Cm increases with rise in temperature. (v) At higher temperatures the distribution function has appreciably higher magnitude, i.e., more molecules have higher speeds. (vi) The fraction of molecules possessing the most probable speed decreases with rise in temperature ( T2 < T1 ).
Fig. 11 Experimental Determination of Molecular Speeds Lammerts Speed Filter Method : The apparatus is shown in Fig. 12. A beam of metal atoms is created by heating a metal like silver or bismuth in an oven. Slits S1 and S2 are used to define the beam. D1 and D2 are the slotted discs with slots exactly in line, mounted on a common axis A. R is a 22
sensitive radiometer for measuring intensity ( number ) of molecular beam. The whole apparatus is kept in an evacuated chamber.
Fig. 12 When the discs are stationary, the beam will pass through the slots of D1 and D2 and finally fall on the vanes of the radiometer. However, when the slotted discs are made to rotate at high speed ( 500600 rpm), a molecule passing through a gap in D1 will pass through a gap in D2 only if the time required by the molecule to cover the distance between the discs is equal to an integral multiple of time required for the discs to rotate from one slot to the next i.e., if ad d 2 r 1 or ------------------- (42) nc = =n x r 2 c a where, c = velocity of the molecule, r = radius of the discs, d = distance between the discs, a = number of slots in each disc, = angular velocity of the discs and n = a small integer. Thus, knowing these parameters, the speed of the molecules passing through can be determined. The number of molecules passing through would be given by deflection of the radiometer vanes . Thus, the speed distribution curve can be obtained, if measurements are made at different disc speeds and the radiometer deflection is plotted against the angular velocity of the discs. Calculation of Most Probable , Average (Mean ) and Root Mean Square Speeds ( a ) Most Probable Speed ( Cm ) : The Maxwells law of distribution of molecular speeds gives,
1 dnc M 2 2 RT 2 x = 4 c e n dc 2RT
Differentiation w.r.t. c gives, 1 dn 3 d x c M c 2 M c 2 n dc = 4 M 2 c 2 Mc e 2 RT + 2c e 2 RT dc 2RT RT
M c 2
## At the maximum point, i.e., where c =Cm ,
MC m 3 MC m 2 RT 2C m e =0 RT
2
1 dn d x c = 0 n dc or
2 MC m Cm 2 RT
Hence,
MCm e 2 RT = 0
23
When left hand side is equal to zero, there are three possibilities, i.e., (i) Cm =
2 MC m =0 RT The true maximum corresponds to (iii), so the most probable speed is given by, 2 RT 2kT = Cm = ----(43) M m where m = mass of individual molecule, k = Boltzmann constant.
(ii) e
2 MCm 2 RT Cm
= 0 and (iii)
( b) Average ( Mean ) Speed ( C ) : The average speed of a gas molecule is given by,
C=
C1 + C 2 + C3 + + C n n
c =
or
n C + n 2 C 2 + n3 C 3 + C= 1 1 = n1 + n2 + n3 +
C dn
c =0 i
Where n is the total number of molecules. Since Ci is a continuously varying function, summation can be replaced
c =
C=
c =0
C dn
i
c =
c =0
dn n
3 c =
## dn from Maxwells law n
MC 2 3 2
M 2 M 2 RT 3 C = 4 e 2 RT c dc = 4 2 2RT c =0 2RT M
(since e ax x 3 dx =
2
1 ) 2a 2
or
8RT 8kT 8 RT 2 C = = = M m M
(44)
( c ) Root Mean Square Speed ( C rms ) :The mean square speed of a gas molecule is given by ,
c =
2 rms
n C + n 2 C + n3 C + = 1 = n1 + n2 + n3 +
2 1 2 2 2 3
C
c =0
2 i
dn or
c =
2 rms
c =0
dn n
Substituting for dn
## from Maxwells law
24
2 rms
M = 4 2RT
M = 4 2RT
3 / 2 c =
c =0
5/ 2 1/ 2
MC 2 2 RT
c 4 dc
3 x 4 dx = 5 8a
3/ 2
3 2 RT 8 M
{ since e
ax 2
or
2 C rms =
3RT M
2 C rms = C rms =
3RT 3kT = ---------------- (45) M M Relation between Average, RMS and Most Probable Speeds 8 Average speed (C ) = = 0.9213 From relations (44) and (45), RMS speed (C rms ) 3
Therefore, root mean square speed, or Average speed , C
= 0.9213 x Crms
## 2 x Crms = 0.8165 x Crms 3
Thus, C rms : C : C M =
2 3RT 8 RT 2 RT 8 8 = 3: : : : : 2 = 1: M M M 3 3
------------------------- (46)
## = 1: 0.9213 : 0.8165 = !.2247 : 1.1283 : 1
Example 20. The density of O2 at 0oC and 1 atmosphere pressure is 1.429 gm / cm3. Calculate the root mean square speed of the gas at 0oC. Sol. Given p = 1 atm. = 76 cm of Hg = 76 x 13.6 x 981 g cm-1s-2 = 1.01396 x106 g cm-1 s-1
C rms =
and
3p
## density, = 1.429 g / cm3 = 1.429 x 103 kg m-3 (
M =) V
3RT = M
3 pV = M
Hence,
C rms
3 x1.01396 x10 5 (kg.m 1 .s 1 ) = = 14.59cm.s 1 3 3 1.429 x10 (kg.m ) 3 x1.01396 x10 6 ( g .cm 1 .s 1 ) = 1458.9987cm.s 1 3 1.429( g.cm )
Or
C rms =
Example 21. Calculate the RMS velocity of a carbon dioxide molecule at 1000oC.
25
Sol.
Given T= 1000+273 =1273 K; M = 44 g mol-1; R = 8.314 x 107 ergs K-1 mol-1 3RT 3x8.314 x10 7 x1273 = = 8.495 44 M x 104 cm sec-1.
C rms =
Example 22. Calculate the root mean square velocity of hydrogen at NTP. Sol. Given M = 2 g mol-1 = 0.002 kg mol-1; p = 1 atm. = 1.01396 x 105 kg m-1 s-1; V = 22.4 dm3 mol-1 = 0.0224 m3 mol-1.
C rms =
## 3 pV 3 x1.01396 x10 5 x0.0224 = = 1845.78m.s 1 0.002 M
Example 23. What is the temperature at which an oxygen molecule have the same RMS speed as a hydrogen molecule at 27oC ? Sol. For H2 , M1 = 2 gm mol-1; T1 = 27+273 = 300 K ; for O2, M2 = 32 gm mol-1; T2 =?
Q C rms =
3RT2 3Rx300 = 2 32
or
300 T2 = 2 32
or
T2 =
## 300 x32 = 4800 K = 4527oC 2
Example 24. Calculate the temperature at which the root-mean square speed of a nitrogen molecule is the same as that of a helium molecule at 27oC. Hint. For N2, M1 = 28 g mol-1 and for helium, M2 = 4 g mol-1 Ans. ( 2100 K or o 1827 C ) Example 25. Calculate the root-mean square speed and average speed in cm / sec of hydrogen at 0oC and 760 mm if the density is 9 x 10-5 g / cm3. Sol. Given p = 760 mm = 76 cm = 76 x 13.6 x 981 g cm-1 sec-2 = 1.01396 x 106 g cm-1 sec-2 ; Density, = 9 x 10-5 g / ml.
3p
## 3x1.01396 x10 6 = 1.838 x10 5 cm. / sec 9 x10 5
Average speed, C = 0.9213 x RMS speed = 0.9213x 1.838 x 105 cm / sec= 1.693 x 105 cm / sec
Collision Diameter When two molecules of a gas approach each other, They cannot come closer beyond a certain distance due to mutual repulsion. The closest distance between the centres of two colliding molecules is termed collision diameter.Although, the molecules are infinitesimally small, they have an effective ( appreciable ) collision diameter. For example, the molecular diameters of hydrogen and oxygen are 2.74 and 3.61 respectively.
26
Fig. 13
Mean Free - Path The distance over which any single molecule moves before colliding with other molecule is known as free-path. The free-paths for all the molecules at the same instant and for a particular molecule at different instants cannot be equal . So average of free-paths is considered. The average ( mean ) distance traveled by a molecule between two successive collisions is called mean free-path ( l).
In a gas, molecules during the course of their random motions, undergo very large number of collisions. Suppose that the collision diameter of the molecules in a gas is and all the molecules except one are at rest. Thus, any other molecule whose centre lies within a distance from the moving molecule can be considered touching it. Let l is the mean free-path of the moving molecule, then the effective volume swept out by it = 2 l . For a gas having n molecules per 1 and the values can be equated. Thus, unit volume, the volume per molecule is n 1 1 or l = (47) 2 l = n n 2 When all the molecules are in motion, the mean free path (l) is different from this value. Considering the relation between the actual speed of a molecule and the speed relative to the stationary molecules, it can be shown that l =l 2 , so that l= 1 2 .n 2 (48)
Thus mean free-path is related to n , number of molecules per unit volume (c.c.) and diameter of the gas molecule . The smaller the collision diameter, the larger is the mean free- path. As the diameters of gas molecules do not vary very greatly under the same conditions for all gases, the mean free-path is approximately constant. Since mean free-path is inversely proportional to n, it will also be inversely proportional to pressure and thus it increases as the pressure decreases.
Collision Number The number of collisions of a single molecule with other molecules per unit time ( second ) per unit volume ( ml.) of the gas is called collision number ( Z ). Evidently, Collision number, Z
## =Average speed of the molecule/mean free-path =
2 C n 2
27
Collision frequency: It is the number of molecular collisions per unit time per unit volume of the gas. The total number of molecular collisions per second, ZI = collision number x number
density= 2 C n 2 x n = 2 C 2 n 2
1 is necessary to avoid double counting 2 Z and thus, the number of collisions per second per unit volume will be I . Hence collision 2 frequency is given by,
## Since each collision involves two molecules, a factor
ZII =
2 C 2 n 2 2 ZII = 2 2 n 2 (
C 2 n 2
2
1
2 n 2
2
8RT M
( Since
C =
8 RT M
or
---------------------(49) M Evidently, the collision frequency increases with rise in temperature, molecular size and the number of molecules per unit volume.
)2
RT
sec-1 cm-3
Specific Heats and Molar Heat Capacities Of Gases Specific heat of a substance is the quantity of heat required to raise the temperature of one gram of the substance through 1oC. In case of gases, this definition is not complete, because the value of specific heat may be anything between zero and infinite depending on the heating processes of the gas. For gases only two specific heat values are used, viz., (i) Specific heat at constant volume, cv the quantity of heat required to raise the temperature of 1 gram of the gas through 1o C, when it is heated at constant volume. (ii) Specific heat at constant pressure, cp the quantity of heat required to raise the temperature of 1gram of the gas through 1oC, when it is heated at constant pressure. Molar heat capacieties Cvm and Cpm : Chemical calculations are most frequently made on molar basis i.e., for one mole of the gas and the heat required is called heat capacity per mole at constant volume, Cvm or at constant pressure, Cpm when the gas is heated at constant volume or at constant pressure respectively. Molar heat capacity at constant volume, Cvm = cv ( specific heat ) x M ( molar mass ) and molar heat at constant pressure, Cpm = cp x M.
Similarly, heat capacity ( for n moles ) at constant volume, Cv = cv x n x M = n x Cvm and heat capacity at constant pressure, Cp = cp x n x M = n x Cpm.
Cp > Cv : When the gas is heated at constant volume, the total heat supplied is used up in increasing the internal energy of the gas. On the other hand, when gas is heated at constant pressure, its volume increases and some heat energy is used up in doing the work against external pressure in addition to the heat required to increase the internal energy to the same level. Thus more quantity of heat is required at constant pressure than at constant volume, by an amount equal to mechanical work done. Thus, Cp = Cv + work done in expension or Cp > Cv or Cpm > Cvm or cp > cv
28
## Calculations of Molar Heats of Gases
3 nRT 2 Let n moles of a gas are heated from T K to ( T + 1 ) K at constant volume, then 3 ------------- (iii) at T K, E trans = nRT 2 3 ------------- ( iv) and at ( T + 1) K, E trans = nR (T + 1) 2 3 3 3 Thus, increase in kinetic energy = nR (T + 1) nRT = nR 2 2 2 For mono atomic gases and vapours, excluding rotation and vibration, translational energy is the only type of energy which is to be considered. Hence, increase in kinetic energy = Total heat absorbed. 3 3 Thus, at constant volume heat capacity, Cv = n R ; molar heat capacity, Cvm= R and 2 2 3 R specific heat, cv = 2 M (ii) Molar heat at constant pressure : When gas is heated at constant pressure, heat supplied is utilized in two ways, (1) in increasing the kinetic energy of molecules and is given 3 by Cv = n R . (2) in performing external work in expansion, which can be calculated as 2 follows: Let, when the temperature of n moles of gas is raised from T K to ( T + 1 ) K at constant pressure p ,it expand from volume V1 to V2. Then, from ideal gas law p V1= n R T --------- (v) and p V2 = n R ( T + 1 ) ------------ (vi)
## Subtracting (v) from (vi) p V2 p V1 = n R ( T + 1 ) n R T = n R Hence, work done in expansion = n R
or p ( V2 V1 ) = n R
or p V = n R
( since, work done = pressure x change in volume ) 3 5 Thus, heat capacity at constant pressure, Cp = Cv + n R = n R + n R = n R 2 2
______________
or
C P CV = nR
3 5 R +R = R 2 2
For one mole of gas, work done = R Thus, molar heat at constant pressure, Cpm = Cvm + R = ______________ or
C pm C vm = R
and
cp - cv =
R M
or
M =
R c p cv
Thus knowing the specific heats, cp and cv, the molar mass M can be calculated.
29
Ratio of specific heats : The ratio of specific heats or molar heats or heat capacities is same and is represented by . Thus C pm c p C = = P = CV C vm cv For mono atomic gases, such as He , Ar , Kr etc. and vapours 5 nR 5 = 2 = = 1.66 3 nR 3 2 This value is not valid for diatomic and polyatomic gases, because in these cases, heat supplied is also used to increase rotational and vibrational energies of atoms within the molecule ( = X say ) besides the kinetic energy of translation and consequently more heat is needed to raise the temperature of the gas through 1oC, i.e., 5 3 and Cvm = R + X Cpm = R + X 2 2
where X varies with the complexity or atomicity of gas molecules. 5 R+R 7 R 7 = 2 = = 1 .4 For diatomic gases, like H2, O2, N2 , etc., X = R. Hence, = 2 3 R+R 5 R 5 2 2 5 R+3 R 3 2 = 4 = 1.33 For triatomic gases , like O3, H2S, CO2 etc., X = R . = 2 3 R+3 R 3 2 2 2 The value of decreases as the atomicity of molecules of this gas increases. Specific heat ratio help us (i) to determine the atomicity of gas molecules as above and (ii) in calculating the atomic weights , knowing the atomicity and molecular mass . Atomic weight = Molecular weight or mass M / Atomicity.
Example 26. The specific heat at constant volume of argon is 0.075 and its molecular weight is 40. How many atoms are there in its molecule ? Sol. Molar heat = Specific Heat x molecular mass
Cvm ( molar heat at constant volume ) = 0.075 x 40 = 3.0 and Cpm ( molar heat at constant pressure ) = 3.0 + 2.0 = 5.0 5. 0 Hence, argon is monoatomic. Thus, = = 1.66 3. 0
Example 27. For a gas, the specific heats at constant volume and at constant pressure are 0.152 and 0.215 respectively. Calculate the atomicity and molar mass of the gas. Name the gas . R 2 2 0.215 M = Sol. = and = = = 31.75 = 1.41 c p cv 0.215 0.152 0.063 0.152
## Hence, atomicity of gas = 2 . The gas is O2.
30
Example 28. Calculate molecular weight of a gas if its cp and cv are 0.125 and 0.075 cal. respectively. Ans. ( M = 40 ). Liquefaction of Gases The gases can be liquefied by cooling them well below their critical temperatures along with compression till liquefaction occurs. Freezing mixtures are generally used for cooling. For example, using a mixture of dry ice and ether, temperature of -80oC can be obtained. Gases having sufficiently low critical temperatures are cooled by rapidly evaporating a liquefied gas. Liquid SO2 was used to liquefy carbon dioxide which was next used to liquefy oxygen. This method is called cascade method for the liquefaction of gases. Method based on Joule Thomson Effect : The gases can be liquefied by cooling based on Joule Thomson effect. Principle: When a compressed gas at a certain temperature is allowed to expand adiabatically through a porous plug or a small orifice ( hole ) into a region of low pressure , cooling occurs. An obvious reason is that during this expansion, work is done in overcoming the intermolecular forces of attraction at the cost of the kinetic energy of the molecules. Thus, kinetic energy of molecules decreases and gas temperature falls. This cooling effect is known as Joule Thomson effect.
It is important to note here that for each gas there is a certain temperature known as inversion 2a temperature (Ti = ) below which cooling is caused by expansion but above which expansion Rb results in rise of temperature. At inversion temperature, no temperature change occurs due to expansion. Inversion temperature of H2 is -79oC and that of helium is -240oC. That is why , Joule Thomson expansion of these gases at room temperature leads to heating and can be a fire hazard in case of hydrogen.
Lindes Method: Linde (1825) used an apparatus working on the principle of Joule Thomson for the liquefaction of air ( Fig. 14 ). Pure and dry air is compressed to about 200 atmospheres pressure and cooled by passing through a pipe cooled by a refrigerating liquid. This cooled and compressed air passes through a spiral with a jet at lower end, and escapes through the jet resulting a considerable drop of temperature by Joule Thomson effect. This cooled air passes up cooling the down coming air in the spiral, which is further cooled (by expansion. The up-coming air is again compressed and recirculated till it get liquefied.
(Fig 14)
31
Claudes Method: This process for the liquefaction of air is based on the principle of cooling caused by mechanical work of an engine drived by the gas in addition to cooling by Joule Thomson adiabatic expansion. Pure dry air is compressed to about 200 atmospheres. It is cooled by passing through the heat exchanger and then led into the chamber Y where it expands and does some external work .It cools to a low temperature as a result of this work. A part of air escapes through the jet after passing through the spiral and cools by Joule - Thomson expansion. This cooled air passes up cooling the down-coming air in the spiral and heat exchanger. The gas escaping liquefaction goes back to the compressor and the whole process is repeated over and over again until air is cooled to its liquefaction.
( Fig. 15 )
__________________________
OBJECTIVE QUESTIONS
1. When He is subjected to Joule-Thomson expansion at ordinary temperature, heating effect is observed because (a) He is an inert gas inversion temperature (b) He is an ideal gas (c) He has the lowest boil point (d) He has a low
atm. litre
(b)
(d)
## energy K-1 mol-1
3. At relatively high pressures, the van der Waals equation reduces to (a) p V = R T -
a V
8a 27 Rb
(b) p V = R T + p b
(c) p V = R T
(d) p = R T -
a 2 V
(e) p V =
aRT V2
4. In terms of van der Waals constants the critical volume is given by (a) 3b (b) (c)
a 27b 2
(d) 2b
(e) 3a
5. According to van der Waals two assumptions of kinetic theory of gases were wrong. One of them is that the gas molecules are (a) very large (b) collapsible (c) particles without any significant volume (d) spherical 6. The pressure of real gases is less than that of an ideal gas because of (a) increase in the number of collisions (b) increase in the kinetic energy of molecules (d) finite size of the molecules (e) intermolecular forces (c) viscosity
32
7. The temperature at which a real gas obeys the ideal gas law over a wide range of pressure is called (a) critical temperature (b) inversion temperature (c) transition temperature (d) Boyle temperature (e) reduced temperature 8. The compressibility factor for an ideal gas is (a) unity at all temperatures (b) unity at the Boyle temperature (d) increases with pressure (e) decreases with pressure 9. van der Waals constant a has the dimensions of (a) mole L-1 (b) atm. L2 mol-2 (c) L atm. degree-1 mol-1 (c) unity at critical temperature
(d) L mol-1
## (e) atm. L mol-1
10. If n is the number of molecules per unit volume then the attractive force pulling one molecule which is about to strike the wall inwards, is proportional to (d) (n 1)2 (e) n2 (a) n (b) n - 1 (c) n2- 1 11. At very low pressure real gases obey ideal laws. The reason is that (a) the effective volume of molecules is negligible (b) the attractive force become stronger (c ) the co-volume is appreciable (d) the attraction between the molecules is negligible. 12. Andrews experiments on liquefaction failed to bring about the expected change in case of (a) real gases (b) noble gases (c) permanent gases (d) perfect gases 13. Real gas will approach the behaviour of an ideal gas at (a) low temperatures and low pressures (b) high temperatures and low pressures (c) high temperatures and high pressures (d) the critical point 14. The critical temperature of a gas in terms of van der Waals parameters is equal to (a)
8a 27 Rb
(b)
a 27 Rb
(c)
a 27b 2
(d) 3 b
15. The translational kinetic energy of an ideal gas depends upon (a) the nature of gas (b) the temperature (c) the pressure 16. Molar heat capacity of an inert gas at constant volume (a) increases with temperature (b) is independent of temperature (d) varies with temperature according to the nature of gas
(d) the pressure and nature of the gas. (c) decreases with temperature
17. At the same temperature and pressure the volume of a gas obeying van der Waals equation is (a) greater than that of an ideal gas (b) less than that of an ideal gas (d) greater or less than that of an ideal gas depending on the temperature. 18. The deviations from ideal behaviour are greatest in case of (a) noble gases (b) permanent gases (c) perfect gases 19. The excluded volume of one molecule of a gas is (a) (c) equal to that of an ideal gas
## (d) easily liquefied gases
4 r3 3
(b)
4 8 r3 3
(c)
4 4 r3 3
(d) 4 r
20. Continuity of state is shown by the isotherms of gases (a)above critical temperature (b) at critical temperature (c) below critical temperature (d) at inversion temperature. 21. Critical volume, Vc is equal to (a) Molecular weight x critical density (b) Atomic weight x critical density
33
(c) Molecular weight / critical density 22. The substances are in their corresponding state (a) at their freezing points (c) when their pressure and temperature are same
(d) Critical density / molecular weight (b) at their boiling points (d) when their critical temperature and pressure are same (d) its critical temperature
23. For liquefaction of a gas, it is essential to cool the gas below (a) room temperature (b) its transition temperature (c)its inversion temperature 24. The inversion temperature of a gas is (a)
a Rb a Rb
(b)
2a Rb 2a Rb
(c)
8a 27 Rb 8a 27 Rb
(d)
2a 27 Rb 2a 27 Rb
True / False True / False (d) atm. litre mol-1 (d) dm mol-1
## 25. The Boyle temperature of a gas is (a) (b) (c) (d)
26. The value of Cp / Cv for polyatomic gases is 1.33 . 27. The gas having higher value of a can be liquefied easily. 28. Which one of the following is not the unit of van der Waals parameter a ? (b) dynes cm4 mol-2 (c) atm. litre2mol-2 (a) N m4 mol-2 29. The van der Waals parameter b has not the dimensions of (b) cm3 mol-1 (c) m3 mol-1 (a) litre mol-1
30. The plot of p V against p for an ideal gas is straight line parallel to pressure axis at constant temperature.at constantline parallel( True/Flase
31. The plot of Z/(( = p V / R T ) against pressure/ p is a straight line parallel to pressure axis at constant temperature for an ideal gas. T True/False 32. Heat capacity of mono atomic gas is higher than diatomic gas. 33. Give the value of for a polyatomic gas . 34. The reduced pressure of a gas is not connected to its critical pressure. 35. The ratio of most probable , average and root mean square speed is (a) 1.1283 : 1.2247 : 1 (b) 1 : 1.1283 : 1.2247 (c) 1.2247 : 1.1283 : 1 36. By matching correct the following sentences (a) Molecules of the same gas are perfectly elastic. (b) The molecules are identical in all respects. (c) The molecular collisions are in a state of constant rapid motion. 37. State true or false in the following : (a) The distance traveled by a molecule between two successive collisions is called mean free-path. (b) The distance between the centres of two molecules is termed collision diameter. (c) The number of molecular collisions occurring per unit time per unit volume of the gas is known as collision frequency. (d) The maxima of the curves for distribution of velocities indicate the average speeds. True/False (d) 1 : 1.2247 : 1.1283 True/False
34
38. Fill in the blanks in the following : (a) The gases consist of a large number of ------------ particles called molecules. (b) The pressure of a gas is due to the ------------- of the molecules on the ----------- of the container. (c) Most of the molecules have speeds near the ----------------- speed. (d) The fraction of molecules possessing the most probable speed decreases with ----------- of temperature. 39. The root mean square speed of a gas is 50 m s-1, its most probable speed will be (a) 46.065 m s-1 (b) 40.82 m s-1 (c) 54.27 m s-1 (d) 61.237 m s-1 40. Gases approach ideal behaviour at low pressure and high temperature. 41. The observed pressure of real gas is more than the ideal pressure. 42. The ideal volume of a real gas is lesser than the observed volume. True/False True/False True/False
43.The temperature above and at which a gas can not be liquefied by applying the Joule Thomson effect is called (a) inversion temperature (b) critical temperature (c) Boyle temperature. 44. Inversion temperature is double of the Boyle temperature of a gas. 45. Boyle temperature is higher than critical temperature. True/False True/False
46. Two gases having same reduced temperature and reduced pressure will have the same reduced -------------- . 47. Compressibility factor of an ideal gas is one. 48. The value of critical temperature for carbon dioxide is ---------------- . 49. The specific heat at constant volume, cv is equal to (a) True/False
3 R 2
(b)
3 nR 2
(c)
3R 2M
(c) 1.33
(d)
3R 2n
50. Specific heat ratio for a triatomic gas is (a) 1.40 (b) 1.66 51. Molecular mass of a gas is equal to (a)
(d) 1.20
c p cv R
(b) ( c p c v ) R
(c)
R c p cv
(d)
R cv c p
52. If cp and cv are the specific heat at constant pressure and constant volume respectively, the atomic weight of O3 is equal to (a)
R 2(c p cv )
cp cv
(b)
R cv c p
(c)
3R c p cv
(d)
R 3(c p cv )
## 53. The (a) 2 X
for a gas is 1.4 and atomic weight is X . The molecular weight is (b) 3 X (c) X / 2 (d) 1.4 X
## 54. Molar heat at constant pressure is
3 (a ) R 2
(b)
5 R 2
(c)
5 nR 2
(d)
5R 2n
35
55. Mention the name of two intermolecular energies of a molecule. 56. The liquefaction of gases by Lindes method is based on ----------------------- effect. __________
1. (d) 8. (a) 15. (b) 22. (b) 29. (d) 2. (c) 9. (b) 16. (b) 23. (d) 30. True 3. (b) 10. (a) 17 (a) 24. (b) 31. True 4. (a) 11. (a) 18. (d) 25. (a) 32. Flase 5. (c) 12. (d) 19. (c) 26. Flase 6. (e) 13. (b) 20. (b) 27. True 7. (d) 14. (a) 21. (c) 28. (d) 35. (b)
36. (a) Molecules of the same gas are identical in all respects. 37. (a) Flase (b) The molecules are in a state of constant rapid motion. (c) True (c) The molecular collisions are perfectly elastic. 38. (a) tiny solid (b) bombardment, walls (c) average 39. (b) 40. True 41. Flase 42. True 43. (a) 49. (c) 50. (c) 46. volume 47. True 48. 31.1oC 53. (a) 54. (b) 55. Rotational and vibrational energies
## (b) Flase (d) Flase
(d) increase 44. True 45. True 51. (c) 52. (d) 56. Joule - Thomson
36
Liquid State A liquid like a gas, has no definite shape and always takes up the shape of the vessel wherein it is placed, however, it has a surface limiting the space occupied and hence a definite volume. The molecules in a liquid have strong cohesive forces like solids but the molecules are not as rigidly fixed as in solids. Liquid is much less compressible and far denser than a gas. The densities of liquid and crystalline phases of the same substance are usually only slightly different. Thus, the liquid state is an intermediate between gaseous and solid states. This, however does not mean that properties of a liquid lie mid-way between that of the solid and gas. In most of the cases, it is found that liquid has some solid like and some gas like properties. Internal Pressure As a result of the Joule Thomson experiment, it has been assumed that molecules attract one another. The attractive forces between the molecules are found to vary inversely as the seventh power of their distance apart and so these forces must be of considerable magnitude when the distances between the molecules are of the order of molecular diameter. This is the case in liquids where powerful forces of cohesion operate which are responsible for the main properties of this form of matter. At very short distances of approach repulsive forces become appreciable between the molecules. A balance between the attractive and repulsive forces may be assumed in the interior of the liquid which gives rise to what is called the internal pressure. The quantity ( E / V )T dV is the change of internal energy resulting from an increase of volume at constant temperature and this may be regarded as a measure of the work done in overcoming the internal pressure ( pi ) during the change in volume of the liquid. Thus,
E pi dV = dV V T
or
-----------------(1)
E pi = ------------------(2) V T So that the internal pressure may be defined by ( 2 ) , in terms of the influence of volume on the internal energy of the liquid. It has been deduced thermodynamically that for any substance E p = T p V T T V
------------------( 3 )
This is called the thermodynamic equation of state. Since the external pressure p is generally small relative to the internal pressure , the latter is some times defined by
p pi = T T V
--------------------( 4 )
The internal pressure can thus be determined by studying the increase of pressure with temperature. The values depend to some extent on the external pressure and on the temperature. The internal pressure of liquids are found to be of the order of 103 atmospheres which indicate that the forces of cohesion in liquids are very large.
37
Intermolecular Forces The liquids are characterized by strong cohesion. The forces or interactions, which bind the molecules together are called intermolecular or van der Waals forces. These are of the following types : ( a ) Dipole dipole interaction ( Orientation effect ): When a liquid contains molecules with permanent dipole moment e.g. H Cl , H2O etc., the molecules attract each other by electrostatic attraction. The negative end of one molecule attracts the positive end of another molecule and vice versa ( Fig. 1 ).
Fig. 1 The orientation of molecular dipoles is such that the ends of opposite charges are close to each other while the ends of similar charges are far apart. These interactions are significant even at reasonably large distance and produce orientation effect in the substances. The melting and boiling points of substances are influenced by this type of interaction. The substance with no dipole moment have lower melting or boiling points than polar substances. The dipole dipole interaction leads to a potential energy given by U0 = 2 4 3r 6 kT
where U0 = potential energy; = dipole moment; r = distance of separation of dipoles; k = Boltzmann gas constant; T = absolute temperature. This relation implies that this force is effective at somewhat shorter distance than simple coulombic attraction. This force is mainly responsible for crystal lattice energy of polar substances and must be overcome in the vaporization process.
( b ) Dipole - induced dipole interaction ( Induction effect ) : When a polar molecule approaches a neutral molecule, the dipole of the polar molecule disturbs the electron cloud of the neutral molecule causing a distortion. Thus, polarity is induced in the neutral molecule and the opposite poles attract each other ( Fig. 2 ).
The potential energy of this interaction is given by 2 2 UI = 6 r where = permanent dipole moment; = polarizability of the other molecule and r = distance between the dipoles. Although the dependence of this interaction on distance between the dipoles is same as in the case of dipole dipole interaction, it is usually of much
38
smaller magnitude, since is less than 2 / 3kT . For this reason polar and non polar liquids are often immiscible.
Fig.2
( c ) Instantaneous dipole induced dipole interaction ( Dispersion effect ): The non polar molecules also attract each other, although the attractive force is clearly a weak one. The nature of this force was first explained by F. London. In a non polar molecule or atom, the electrons are constantly moving and it may be regarded as an oscillating system of electrical charges. If the electron cloud, at any instant, is even slightly displaced relative to the nucleous or nuclei a small instantaneous dipole developes which causes instantaneous induced dipole in the nearby molecule resulting a small attractive force, called dispersion force or London force.
## The potential energy due to dispersion force is given by UD = 3h 2 4r 6 =
where h = Plank constant ; = frequency of oscillations of electron cloud and polarizability of the molecule or atom .
Fig.3 The London forces are present in polar molecules also in addition to other van der Waals forces. The strengths of these forces depend as to how easily the electron cloud in a particular molecule is deformed.
(d) Ion dipole interaction: When an ion interact with a dipole, the interaction is called ion-dipole interaction. The cations attract the negative end of the dipole while anions attract the positive end ( Fig. 4 ) . This arrangement is not ordinarily achieved because it is opposed by thermal motions. Ion-dipole interactions are often important in the solvation processes.
39
Fig.4 The appreciable solubility of many ionic solids ( e.g., Na+ Cl- , K+ Br- etc. ) in water is due the fact that the negative ends of water molecules surround the cations ( Na+ or K+ ) and the positive ends surround the anions ( Cl- or Br- ) ( Fig.5 ) . In this way, the lines of electrostatic forces between cations and anions are broken and the ions get solvated and go into the solvent. The potential energy in these interactions depend upon the charge of the ion and the relative orientation of the dipole and for a stable arrangement is given by
U = e r2
## where e = charge on the ion ( anion or cation ).
Fig.5
( e ) Ion induced dipole interaction: When an ion approaches a neutral molecule, disturbs the electron cloud of the molecule causing distortion and producing weaker and shorter range attractions. Here the energy of interaction is given by,
U =
e2
2r 4
This energy can not ordinarily supply the energy required for dissolving an ionic substance and consequently such substances are not readily soluble in non polar solvents.
( f ) Hydrogen bonding: The electrostatic attraction that occurs between molecules containing hydrogen covalently bonded to a very electronegative atom ( O, N or F etc. ) containing nonbonding electron pair, is called hydrogen bonding. H2O is the most common molecule that shows hydrogen bonding. It is the strongest of all intermolecular forces including the dipole dipole attractions and London forces.
40
Fig.6
(g) Overlap repulsion forces: When the distance between any two atoms, ions or molecules is very short, a repulsive force sets in due to the overlap of electron clouds. This overlap repulsion force is responsible for the steeply rising portion of the potential energy curve at small distance of separation ( Fig. 7 ). The net potential energy of interaction of two non polar atoms or molecules is given by a function of the form, U = - A r-6 + B r-n
where n = 9 to 12. The first term describes an attraction while the second a repulsion. Since the exponent of the second term is greater than that of first term, there will be a distance of minimum potential energy, corresponding to the intermolecular distance in the condensed state.
Caloric Equation Of State An equation connecting the total molar energy of a fluid ( gas or liquid ) with its volume and temperature is called Caloric Equation of State. If E is the total molar energy ( energy per mole ), then
E = Ekinetic + Epotential In an ideal gas, the molar energy is entirely kinetic in nature and depends only on the 3 3 temperature of the gas. Hence, E = Ekinetic . But Ekinetic. = RT . Thus, E = RT 2 2 This is the Caloric equation of state for ideal gas . In a real gas or liquid , intermolecular forces exist, which give rise to potential energy. This depends on intermolecular distance. Thus, the potential energy of a real gas or liquid a depends on volume and is given by , where a is a constant, which depends upon the V a 3 intermolecular attraction. Hence, the total molar energy, E = Ekinetic + Epotential = RT . 2 V This equation is known as Caloric equation of State for a real gas or liquid.
Structure of Liquids Observed compactness and cohesion in liquids indicate their structure to be similar to solids while the comparative ease with which the liquids flow and diffuse indicates a disordered random structure like in gases. X- ray examination of many liquids reveals a certain degree of the order in the arrangement of molecules, though to a lesser degree as compared to solids. The structure of a liquid may be described in terms of radial distribution function ( RDF ) by means of which the X- ray diffraction data on liquids are usually interpreted graphically.
41
Radial distribution function ( RDF ) is the probability of finding the number of atoms or molecules i.e., second scattering centers between a distance r and ( r + dr ) irrespective of the angle from an arbitrarily chosen molecule called the reference molecule. For calculating RDF, a series of concentric spheres are drawn around the reference molecule in such a way that the volume interval between two neighbouring spheres is constant. RDF, then is the average number of molecules contained between each spherical shell and thus, is the measure of average particle density as a function of distance from an arbitrary origin. It is calculated to be 4 r2d ( where d is the average density of liquid in atoms or molecules per cc.. If all the intermolecular distances were equally probable i.e., if the liquid were completely devoid of regularity of structure, there would be a continuous scattering of X rays and hence, no diffraction maxima or minima. But a few maxima and minima observed in actual experiments in liquids indicate the existence of regular pattern or order in them. The radial distribution curves for an ideal gas, a liquid and a crystalline solid are shown in Fig. 8. The smooth dashed parabolic line indicates purely random distribution of an ideal gas. The RDF curves for solid are vertical lines, showing regular pattern in the arrangement and distribution of molecules. The maxima and minima in the distribution curve of liquid indicates the intermediate position of liquid state. Very small number of these maxima and minima suggests the existence of a short range order only. It is also clear from this figure that the regularity of ordered structure in liquid is lost beyond 10 , while with solids it persists even at larger distances. Thus, a liquid may be considered to have crystal like structure, except that the well ordered pattern extends over a short range instead of over the entire mass.
Fig.8
Structural Differences between Solids, Liquids and Gases In solids molecules, atoms or ions are tied and arranged in a regular pattern due to very strong cohesive forces and are not free to move about i.e., have no translational or rotational degrees of freedom. In gases the molecules are in chaotic motion and there is no regular pattern in the arrangement of molecules. The liquids are intermediate between solids and gases. Although a regular pattern or order exist in liquids, but the ordered pattern is limited to a short range, instead of over the entire mass. The regularity of ordered structure in liquid is lost beyond 10 . Due to thermal motion, this ordered structure is changing continuously in liquids.
42
Models of Liquids There are following four approaches to the liquid state: (1) Cybotatic group model. (2) Kinetic molecular model. (3) Liquid as a modified solid (4) Random packing model. (1) Cybotatic group model : According to this model of liquids suggested by Stewart, a large number of well- ordered groups called cybotatic groups, each containing a few hundred molecules, are embedded in a mass of disordered molecules. These groups are not, however, permanent and there is a dynamic equilibrium between cybotatic groups and disordered molecules.
Cybotatic groups
Disordered molecules
(2) Kinetic molecular model - Liquid as a modified gas : According to this model the liquids resemble gases in all aspects except closely packed molecules with appreciable attraction among them. Above the critical point, there is no distinction between liquid and gaseous states of a substance and the so called fluid state exists. So, the liquid may be treated as a modified gas. (3) Liquid as a modified solid : In solids, there is well-ordered arrangement of molecules throughout the whole mass and each molecule is surrounded by a set of other molecules in a particular symmetrical pattern. On melting the solid expands only about 10 percent in volume or 3 percent in intermolecular spacing. Therefore, it is reasonable to conclude that in the liquid state a molecule must still remain in the vicinity of some other molecules surrounding it. The ordered arrangement is not , therefore, completely destroyed when the solid changes into a liquid. Thus, a liquid may be regarded as a modified solid. (4) Random packing model : This model was developed by Bernal and Scott and according to this :
(i) The liquid state is a random pack of molecules i.e., the liquid structure is disordered but relatively close-packed. (ii) A pentagonal structure of liquids is assumed to introduce a small disorder factor instead of hexagonal structure as in the solid i.e., only five surrounding atoms to an atom are permitted. (iii) The packing in liquid state is close but still loose. (iv) The packing is greatly affected by repulsive forces whereas there is no influence of attractive forces. (v) This model explains the low and continuously varying coordination number ( ranging between 4 and 11 ) in liquids.
43
Liquid Crystals The ordered arrangement of molecules in some substances is so great that crystalline form does not melt directly to liquid phase at all, but first passes through a turbid intermediate stage ( mesomorphic or paracrystalline state ), which at a higher temperature changes to a clear isotropic liquid ( i.e., a liquid whose optical, mechanical, electrical and magnetic properties have the same value in every direction ). The turbid liquid is found to be doubly refracting and gives interference patterns in polarized light like anisotropic crystals ( crystalline solids which have different values of physical properties in different directions). Cholesteryl benzoate, C6H5COOC27H45 was discovered in 1888 with this peculiar property. O. Lehmann ( 1889 ) proposed the name liquid crystals for the turbid liquids exhibiting these unusual optical effects. This term , however, is not satisfactory, as the substances have no crystalline properties and they are true liquids possessing surface tension and ability to flow. Various other terms, e.g., crystalline liquids, anisotropic liquids and paracrystals have been proposed, but the best is probably mesomorphic state ( Greek: intermediate form ).
The temperature at which the solid melts to form turbid liquid is usually called the transition temperature and the higher temperature at which the true isotropic liquid is obtained is called melting point. On cooling the changes take place in the reverse direction at the same temperatures. The substances showing liquid crystal character are highly stable and do not decompose on heating even on high temperatures. Thus, besides solid, liquid and gaseous states, the fourth state of matter to be considered is mesomorphic state.
Examples :
## (i) Solid form of p azoxyanisole, CH3O- - N = N-
- OCH3
changes at 116oC to liquid crystal, which is stable to 135oC at which it changes to a clear isotropic liquid. (ii) p- anisalaminocinnamate, CH3O - CH=N -CH= CHCOOC2H5
passes through three distinct paracrystalline phases between 83oC and 139oC. (iii) Cholesteryl bromide behaves rather differently. The solid melts at 94oC to an isotropic liquid which can be supercooled to 67oC where it passes over into a metastable liquid crystalline form. Generally all organic substances having long chains terminating in an active group ,e.g., OR, -COOR and often with a mildly active group ,e.g., - C = C - , - C = N - , - N = NO - , in the middle and para-derivatives only of compounds containing benzene ring display mesomorphism.
44
## Differences between solids, liquid crystals and liquids
Solids 1 In solids molecules, atoms or ions are arranged in a regular pattern and are not free to move i.e., translatory and rotatory motions are usually absent but particles can vibrate about a fixed position in the orderly crystalline arrangement.
Liquid Crystals Since mesomorphism is exhibited by liquid form of the substances having long chains, the free rotation of molecules is restricted and they tend to arrange themselves parallel to each other forming a number of groups or swarms with definite orientation. Molecules now have freedom of movement in the direction of the length of molecules remaining parallel to each other.
Liquids In liquids, the molecules are not as rigidly fixed as in solids. They have some freedom of motion and move in a random fashion.
Liquid crystals are turbid liquids Solids have a definite shape obtained by melting solids at transition A liquid has no 2 shape and definite volume. point and have no definite shape but a definite but a definite definite volume. volume. Turbid liquids are found to be doubly 3 Crystalline solids, except refracting and gives interference patterns Most liquids are i.e., cubic crystals are anisotropic in polarized light like anisotropic isotropic their physical i.e., their physical properties crystals. Except this property of crystals, properties have like optical, magnetic they are true liquids in all respects. the same value in properties etc., vary in every direction in magnitude according to the liquid. direction of the crystal. These show double refraction and related phenomena.
Classification of Liquid Crystals Liquid crystals can be classified as smectic ( Greek : soap like ), nematic ( Greek : thread like ) ( G. Friedel, 1922) and cholesteric liquid crystals. (i) Smectic Liquid Crystals : Smectic liquid crystals do not flow as normal liquids. They have limited mobility and flow in layers as if different planes or sheets are gliding over one another. The distribution of velocity in different layers is different from that found in true liquid. When spread over a clean glass surface, they form a series of terraces or strata, as shown in Fig. 10. The edges of these terraces often show a series of fine lines even in ordinary light but more clearly in polarized light . The results suggest a structure consisting of a series of planes, one on the top of other. This is supported by the X-ray diffraction patterns by smectic liquids in one direction only. The spacing in this direction, perpendicular to the parallel planes, is little more than the length of a molecule. When examined in polarized light smectic liquids sometimes have a fan-like appearance (focal conics). They are always uniaxial and are not affected by a magnetic field.
45
Fig. 10 There are several different categories of smectic phase. The two best known of these are smectic A, in which the molecules align perpendicular to the layer planes, and smectic C, where the molecules are tilted at some arbitrary angle to the normal ( Fig. 11 ) . Ethyl pazoxybenzoate ( t. p. 114oC and m. p. 121oC ), ethyl p-azoxycinnamate ( t. p. 140oC and m. p. 249oC ) and n- octyl p-azoxycinnamate ( t. p. 94oC and m. p. 175oC ) etc. form this type of crystals.
Fig. 11
(ii) Nematic Liquid Crystals : Nematic liquid crystals show near normal flow behaviour. They flow more readily than smectic liquids. Mobile thread like structures can often be seen in thick specimens in polarized light.
The nematic phases are also uniaxial but unlike smectic phases, the axis get oriented in strong magnetic field. Like true liquids, nematic liquids give only diffuse haloes with X- ray, showing the absence of layer structure. p-azoxyanisole (t. p. 116oC and m. p. 135oC ), Anisaldazine ( t. p. 137oC and m. p. 167oC ) , p-methoxycinnamic acid ( t. p. 170oC and m. p. 186oC ) etc. show nematic behaviour.
(iii) Cholesteric Liquid Crystals : Certain optically active compounds show many characteristics of nematic behaviour but at the same time display strong colour effects in polarized light indicating layer structure like smectic liquid crystals. The structure differs from smectic liquid crystals in being very much thicker, viz., 500 to 5000 molecules. Such liquid crystals having some nematic and some smectic characters are called cholesteric liquid crystals. A cholestric liquid crystal consists of a stack of sheets or discs on the top of one another which are tilted slightly so that they form a helix. Cholesteryl benzoate ( t. p. 146oC and m. p. 178.5oC) show cholesteric type of mesomorphism. The lcholesteryl esters from formate to myristate give cholesteric phases but the compounds with higher acid chain length are smectic.
46
Compounds with several mesophases. Cholesteryl esters from caprylate to the myristate give smectic liquid on melting which changes sharply to a cholesteric phase at higher temperature and finally a clear liquid is obtained. For example, the three temperatures for myristate are 72o, 78o and 83 oC. A few substances exist in smectic and nematic forms, with a definite transition temperatures and some have more than one smectic phases. Ethyl anisal-p-aminocinnamate, for example, undergoes the following changes, starting with the metastable solid.
Crystal
83 o C
Smectic phase-I
91o C
## Smectic phase-II Nematic Phase
118 o C
139 o C
Liquid
Theory of Liquid Crystals The best explanation for the structure of liquid crystals is based on the swarm theory of E. Bose (1909). Since mesomorphism is shown by substances having long chains, the molecules are supposed to be arranged parallel to each other in a large number of groups or swarms. The molecules in one swarm are parallel to one another but the arrangement in one swarm will not necessarily be parallel to that in another (Fig.12 ).
In a sense, therefore, the mesomorphic liquid may be regarded as consisting of a number of small crystals distributed at random. Although, the molecules in the swarms are not arranged as rigidly as in solids and the size of the swarm is not to be regarded as constant. The turbidity of liquid crystals is attributed to the scattering of light by the swarms.
Fig.12 As the temperature is raised the size of the swarms diminishes due to the increased thermal movements of molecules and ultimately at a certain temperature, the swarms become too small to scatter light. The liquid becomes clear. The double refraction due to orientation within the swarms disappear and the liquid becomes isotropic.
Weakness of the swarm theory (i) According to this theory, the clearance of the turbidity should be gradual as the swarms become smaller and smaller with rise in temperature. But in the actual experiments, it is observed that the turbidity of liquids clears off all of a sudden at a particular temperature ( melting point ).
(ii)
It is not clearly understood why certain long chain molecules which might be expected to form liquid crystals do not do so.
47
Applications of Liquid Crystals: 1. Most of the industrial lubricants exist in mesomorphic state. In fact, it is believed that their functioning as lubricants is due to their liquid crystalline structure.
2. The food constituents proteins and fats also change into mesomorphic state before digestion and are, therefore, easily assimilated in the body. 3. The nematic mesomorphic state are widely used in liquid crystal displays ( LCD ) e. g. in digital watches, electronic calculators and other electronic instruments. In a twisted nematic LCD, the liquid crystal is placed between two flat conducting glass plates and the entire assembly is placed between two polarizers. In the absence of the electric field the light passes through the polarizers. When an electric field is applied the molecules rotate and the light is blocked displaying dark lines. 4. The cholesteric liquid crystal undergoes a series of colour changes with temperature. These crystals are used as indicators to monitor body temperature or to spot areas of over heating in mechanical systems.
Thermography The method of detection and recording as visual image of variations in the amount of heat being emitted as infra-red radiation from different structures or parts of the body is known as thermography.
Heat produced by cellular metabolism in body, e.g., by malignant cells, is distributed by blood and lymps ( colourless fluid in tissues ) throughout the body is lost by radiation ( 45% ), evaporation ( 25% ) and convection ( 30% ). Heat lost by radiation in the infra-red part of the spectrum can be measured accurately, reducing the heat lost due to evaporation and convection by lowering the temperature of the part of the body being examined before thermography is carried out. A system of mirrors focuses the infra-red emission from the subject on a sensitive detector which converts the emission into an electronic impulse which is displayed on a cathode-ray tube as a thermal image of the area being scanned. Many new thermal imaging systems are also video compatible. On most equipment, area of increased infra-red emission are displayed as white and area of lower infra-red emission as grey or black.
Colour thermography : Colour thermography is achieved by digitization through a micro or mini computer. The display is recorded by photography, using either process or instant film. Inkject computer printers are also used to produce hard copy . Contact thermography : Contact thermography is achieved with nematic or cholesteric liquid crystal detectors. Of particular interest are tilted phases of chiral molecules, which possess permanent polarizations and are thus ferroelectric. The ferroelectric liquid crystals ( FLCs ) respond much more quickly to applied fields than nematics do and can be used to make fast bistable electro-optic devices. They are encapsulated in a flexible film forming one side of an air-inflated pillow . The colour changes are photographed through the transparent window. The detectors are re-usable, losing colour when removed from the skin. Each detector panel is limited to 3 5 oC , so a set of detectors is used to cover the entire
48
required temperature range. A portable frame with a camera and flash unit is used to mount them on the skin site. Quantization is poor of this technique but it is inexpensive and portable. It can be used to identify vascular patterns or localized lesions in a qualitative manner.
Uses of thermography : Thermography is used in the detection of various diseases, as mentioned below. 1. Assessment of superficial lymph node or bone involvement in metastatic disease. 2. Detection of malignant disease in the breast. 3. Malignant melanomas are sometimes monitored by infra-red or liquid crystal thermography. A thermal flare around the melanoma is usually indicative of an active lesion with poor prognosis. 4. Assessment of the degree of inflammation in peripheral joint disease and connective tissue disease. 5. Assessment of anti-rheumatic drug therapy and surgery. 6. Monitoring the effects of treatment on soft tissue injuries, algodystrophy etc. 7. Demonstration and monitoring of nerve root lesions and root compression. 8. Assessment of the extent of pressure sores and ulcers. 9. Assessment of burns, e.g., to distinguish between full thickness burns and partial thickness burns. It makes earlier grafting possible, thus saving time and the risk of infection. 10. Demonstration of thermal patterns in the hands in Raynauds disease. 11. Selection of optimum level for amputation in ischaemia. 12. Demonstration of deep vein thrombosis. 13. Investigation of incompetent valves in perforating veins. 14. Non-invasive investigation and evaluation of arterial disease. Seven segment cell One common requirement of digital devices is a visual numeric display for which lightemitting diodes ( LEDs ) and liquid crystal displays ( LCDs ) are frequently used. Because LCD require much less power than LED, LEDs are replaced by LCDs in many digital displays. To display any number, a combination of seven segment cell of LCDs is used ( Fig. 13 ) . The seven segments are arranged in a rectangle of two vertical segments on each side and one horizontal segment on the top, one at bottom and one in the middle bisecting the rectangle horizontally. Individually on or off, these can be combined to produce representations of any digit 0 - 9 .
## Fig.13: A seven segment cell displays the digit 4.
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OBJECTIVE QUESTIONS
1. The properties of a liquid lie mid-way between that of the solid and gas. True / False 2. The internal pressure of the order of 103 atmospheres in the liquids indicates small forces of cohesion in the liquids. True / False 3. The forces or interactions, which bind the molecules together are called (a) intramolecular forces (b) interatomic forces (c) intermolecular forces (d) repulsive forces 4. Liquid crystals are (a) a form of crystalline solid (c) isotropic
## (b) a form of clear liquid (d) anisotropic.
5. Mesomorphism is displayed by substances having (a) long chains terminating in an inactive group and with a mildly active group in the middle. (b) long chains terminating in an active group and with an inactive group in the middle. (c) ortho- derivatives of compounds containing benzene ring (d) para- derivatives of compounds containing benzene ring 6. Smectic liquid crystals (a) show normal flow behaviour (c) have thread like structures
(b) do not flow as normal liquids (d) are affected by magnetic field (c) cholesteric liquid crystals (d)
7. p- Azoxyanisole form (a) smectic liquid crystals (b) nematic liquid crystals none of the above
8. Nematic liquid crystals have limited mobility and flow in layers. Ture / False 9. Cholesteric liquid crystals display strong colour effects in polarized light. True / False 10. When examined in polarized light nematic substances sometimes have a fan-like appearance. True / False 11. The temperature at which the solid melts to form turbid liquid is called (a) melting point (b) transition point (c) critical point (d) triple point 12. Contact thermography is achieved by digitization through a micro or mini computer. True / False 13. Cholesteric liquid crystal detectors are re-usable. True / False
14. Malignant cells produce a rise in temperature in the overlying skin, because of their
50
## (b) increased cellular metabolism (d) none of the above (d)
15. When an ion interact with a dipole, the interaction is called (a) ion induced dipole interaction (b) induction effect (c) orientation effect ion- dipole interaction
16. Malignant disease in the breast can be detected using the technique of (a) thrombosis (b) thermography (c) thermal change (d) thermometry
1. False 2.
False
3. ( c )
4. ( d )
(d)
6. ( b )
7.
(b)
8. False 16. ( b )
## 13. True 14. ( b ) 15. ( d )
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Solid State Characteristics of Solids: Solids are characterised by definite shape, definite volume, regular pattern of arrangement of structural units (atoms, ions or molecules), incompressibility, high density, rigidity and mechanical strength due to their closely packed structural units. Types of solids : The solids are of the following types: (a) Crystalline solids: The solids in which structural units are arranged in a definite pattern, constantly repeated, giving a definite geometrical configuration, characteristic of the nature of the substance are called crystalline solids. In a crystal, the arrangement of atoms is in a periodically repeating pattern. These are anisotropic i.e., their physical properties have different values in different directions. Examples: NaCl, KCl etc. (b) Amorphous solids: The solids which have no definite geometrical form are called amorphous solids. These are isotropic i.e., their physical properties have same value in all directions. In amorphous solids, the arrangement of particles are random and disordered. Examples: glass, rubber, plastic etc. (c) Polycrystalline (microcrystalline) solids: The solids in which many microsized crystals are tightly packed together without any specific order are termed polycrystalline solids. Since the micro-crystals are randomly oriented in polycrystalline substances it will appear to be isotropic although each single crystal is anisotropic. Examples: Metals and alloys. Difference between crystalline and amorphous solids: (i) Crystalline solids have a perfect and well ordered arrangement of structural units throughout the entire mass while in amorphous solids, the arrangement of particles is random and disordered. (ii) Crystalline solids are anisotropic while amorphous solids are isotropic. (iii) Crystalline solids possess sharp melting points while amorphous solids do not. (iv) On being cut with a sharp edged knife, crystalline solid gives a clean cut while amosphous solid break irregularly.
It is important to note that the distinction between crystalline and amorphous solids is not absolute. When examined by x-ray methods, the amorphous solids are also found to possess definite crystal lattices. Therefore, from the viewpoint of internal structure, the solids are essentially crystalline i.e., they are aggregates of very small crystals packed together in a more or less random fashion. Thus, solid state refers to a crystalline state. Strictly speaking, only a crystalline material can be considered as a true solid.
Crystal: A crystal may be defined as a homogeneous portion of a solid having a very regular structure bounded by plane faces which make definite angles with each other and having a typical and distinctive symmetry and geometrical form characteristics of the solid substance. Types of Crystalline Solids: Based on the nature of bonding in crystalline solids, these are classified as follows:
(a) Ionic Crystals : In ionic crystals, the positively and negatively charged repeating units are arranged so that the potential energy of the ions in the lattice is minimum. The ion of a given
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sign is bonded through coulombic forces to all the ions of the opposite sign. Ionic crystals possess the following characteristics: These are hard and brittle. (i) (ii) The vaporization energy of ionic substances is relatively higher. (iii) They possess very small vapour pressure. (iv) Their melting and boiling points are high. (v) They are insulators but in molten form conduct electricity. (b) Covalent Crystals: In these crystals, the repeating units are atoms of the same or different kind bonded through electron pair bonds. The most common example is diamond in which each C-atom is surrounded by four carbon atoms tetrahedrally. General characteristics of covalent crystals are: (i) They are extremely nonvolatile and possess very high melting points. (ii) They are hard and incompressible. (iii) They are electrically insulators even above their melting points. (iv) They require high energies to separate them into constituent atoms. (c) Molecular Crystals: In these crystals the repeating units are atoms or molecules bonded by weak van der Waals type forces. General characteristics of such crystals are: (i) These are soft, compressible and can be distorted easily. (ii) They are volatile. (iii) Their melting and boiling points are low. (iv) They are electrical insulators. (d) Metallic Crystals: These crystals are considered to be an arrangement of nuclei with free electrons moving in interstices. These possess the characteristics: (i) Silvery lusture and high reflectivity. (ii) High thermal and electrical conductivity. (iii) They can be easily drawn into wires, hammered into leaves and bent without fracture. Crystallography The branch of science which deals with the study of geomtry, properties and structure of crystals and crystalline substances is known as crystallography.
External Features of Crystals From the external features of the crystals, we can get first hand hint about the nature of the internal structure of crystals. The external features of crystals are as follows: (i) Face: The crystals are bounded by plane surfaces which are arranged in definite pattern. These surfaces are called faces of the crystal (Fig.1). (ii) Edge : The line along the intersection of the two adjacent faces is called the edge (Fig.1). (ii) Form: All the faces of a crystal are said to constitute a form. (iii)Zone and Zone-axis: Faces are found to occur in sets which meet in parallel edges or would do so if the planes of faces were extended. Such a set of faces is known as Zone.A line passing through the centre of a crystal in a direction parallel to the edges of a zone is called zone axis. (iv) Crystal habit: The external shape of the crystal is called the crystal habit.
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(v) Interfacial angle: The angle between the lines perpendiculars to any two adjacent faces of a crystal is called interfacial angle.
Relation between faces, edges and interfacial angles: These elements are inter-related by the following expression: f + c = e + 2 ----(1)
## where f = Number of faces c = Number of corners e = Number of edges
Symmetry elements of crystals A crystal may be symmetrical about a plane, an axis or a centre, i.e., on rotating the crystal about that axis or reflection across that plane, the appearance of the crystal remains unchanged. The total number of planes, axes and centre of symmetries possessed by a crystal is termed its elements of symmetry: (i) Plane of symmetry: An imaginary plane passing through the centre of the crystal and dividing the crystal into two parts in such a way that one part is the mirror image of the other, is called plane of symmetry. A crystal can have any number of planes of symmetry. (ii) Centre of symmetry: It is a point in crystal such that any imaginary line passing through this point meet the opposite faces of the crystal at equal distances on either side. A crystal can not have more than one centre of symmetry.
The centre of symmetry is also called the centre of inversion because the crystal structure remains unchanged when the crystal is inverted through the centre of symmetry.
(iii)Axis of symmetry: An axis of symmetry is the imaginary line passing through the centre of the crystal, such that if the crystal is rotated about this line, it presents similar appearance more than once in one complete revolution, i.e., through an angle of 360. If similar appearance occurs twice in one revolution (i.e., after every 180), the axis is called two fold axis of symmetry or a diads. In the same way three fold (triad), four fold (tetrad) and six fold (hexad) axis of symmetry exist if in one complete revolution similar appearance of the crystal occurs three, four and six times respectively. In general, if similar appearance of a crystal is repeated on rotation through an angle of 360/n, around an imaginary axis, the axis is called an n-fold axis. Five fold symmetry does not exist in crystals.
Symmetry Elements of a Cubical Crystal: A cubic crystal has the greatest number (23) symmetry elements as follows:
of
(i)Centre of symmetry (Fig. 2f) (located at the centre of gravity of the cube) = 01
54
Fig 2
(ii)Axes of symmery Four-fold axes of symmetry (Fig.2c) Three-fold axes of symmetry (Fig.2d) Two-fold axes of symmetry (Fig.2e) (iii) Planes of symmetry Rectangular planes of symmetry (Fig.2a) Diagonal planes of symmetry (Fig.2b)
3 4 6 3 6
= 13
= 09
Point Groups or Crystal systems: Any crystal will have some combinations of elements of symmetry. These combinations of symmetry elements are called point groups or classes or crystal systems. Theoretically, there can be 32 different combinations of elements of symmetry. These 32 point groups can be conveniently grouped into seven different categories, known as seven basic crystal systems which are given below:
System
Cubic or regular 2 3 Tetragonal Orthorhombic or Rhombic
Axial characteristics
= = = 90 a=b=c = = = 90 a=bc = = = 90 abc
Maximum symmetry
Planes = 9 Axes = 13 Planes = 5 Axes = 5 Planes = 3 Axes = 3
Examples
NaCl, KCL, ZnS, Diamond, Pb, Ag, Au Sn, SnO2 , TiO2, KH2PO4, ZrSiO4 KNO3, K2SO4, BaSO4, PbCO3, Mg2SiO4, Rhombic sulphur Na2SO4.10H2O, CaSO4.2H2O,
Monoclinic
= = = 90 abc
Planes = 1 Axes = 1
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Monoclinic sulphur 5 6 7 Triclinic Rhombohedral or Trigonal Hexagonal 90 abc = = 90 a=b=c = = 90 = 120 a = b c Planes = nil Axes = nil Planes = 7 Axes = 7 Planes = 7 Axes = 7 CuSO4.5H2O, K2Cr2O7, H3BO3 NaNO3, As, Sb, Bi, Quartz, Calcite ZnO, CdS, HgS Graphite, Mg, Zn, Cd
Six of these systems are represented by three crystallographic axes, but one the hexagonal system, is conveniently refered to a set of four axes.
Crystallographic terms: For describing the geometry of a crystal, the knowledge of the following terms called elements of crystal, is essential. (a) Crystallographic axes: In order to describe complete structure of a crystal including planes present in the crystal, relative direction and orientation of the faces, three non-coplanar coordinate axes called crystallographic axes are chosen arbitrarily which meet at a point. Choice of these axes can be done in a number of ways but the best choice is usually the lines either coinciding with or parallel to the edges between the principal faces. These axes are represented by X, Y and Z and the angle between the axes opposite to intercepts are represented by , and respectively (Fig.3).
Fig 3 (b) Standard or unit plane: Any face of the crystal will cut one or more of the three crystallographic axes. The face which cuts all the three axes is called standard or unit plane. The distance of points of intersects of three axes from the origion are called intercepts. The intercepts a, b and c on X, Y and Z axes are generally called crystal parameters. (c) Axial ratio: The ratio of intercepts made by the unit plane on the crystallographic axes is called axial ratio and is represented as a : b : c or a/b: 1 : c/b. In fig.3 OX, OY and OZ represent three crystallographic axes inclined at angles , and . The ABC is the standard plane, which cuts the axes at A, B and C, giving intercepts of length a, b and c on X, Y and Z axes respectively.
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Laws of Crystallography There are three basic laws of crystallography, regarding the forms of the crystals. (1) Stenos law of constancy of interfacial angles. (2) Law of constancy of symmetry. (3) Hays law of rational indices or intercepts. 1. Stenos Law of Constancy of Interfacial Angles: The size and shape of the crystals of one and same substance may vary with the conditions under which crystallization occurs. In spite of the different sizes and shapes of the crystals of the same substance, the interfacial angles between the corresponding faces remain constant. This is called the law of constancy of interfacial angles and was first realized by N. Steno (1666) for quartz crystals. The measurement of interfacial angles in crystals is, therefore, important in the study of crystals. The interfacial angles may change with change in temperature of the crystals.
The instrument which is used for the measurement of interfacial angles is called goniometer.
2. Law of Constancy of Symmetry: This law states that all the crystals of a particular substance always possess same elements of symmetry. 3. Law of rational indices or intercepts (R.J. Hay, 1784): This law states that the intercepts made by any face of the crystal on the crystallographic axes are either (i) same as those of a unit plane (denoted by a, b, c) or (ii) some simple whole number multiples of them, e.g., la, mb and nc etc., where l, m and n are simple whole numbers, or (iii) one or two intercepts may be infinity if the face is parallel to one or two axes.
For example, in Fig.3 the intercepts made by the face LMN are 2a, 2b and 2c which are simple whole number multiples of those of unit plane.
Weiss indices : The intercepts of unit plane on X, Y and Z axes, i.e., crystal parameters are a, b and c respectively. If a face parallel to unit plane cuts the three axes with intercepts x, y and z respectively, then according to law of rational indices x = l.a y = m.b and z = n.c
The coefficients (or multiples) l, m and n are called Weiss indices of the plane. The corresponding plane is described as (l m n) plane. These are generally small whole number. But this is not always true. These may be fractions of whole numbers as well as infinity.
Millers indices (W.H. Miller, 1839): The Miller indices of a face are the reciprocals of Weiss indices (i.e. coefficients of unit intercepts of a, b and c) expressed as integers. Example: Suppose the intercepts made by a plane of the crystallographic axes are 3a, 3b, 2c then (i) Weiss indices are 3, 3, 2 (ii) Reciprocals of Weiss indices 1/3, 1/3, (iii) Multiply reciprocals by the least common factor i.e., 6 to get small whole numbers 2, 2, 3
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(iv) (v)
Hence, Miller indices are 2, 2, 3 In general, Miller indices are denoted by h, k and l. Then the plane is represented by (h k l). Miller indices of crystal faces are, in general, small integers, as well as zero. Miller indices of the standard plane are always 1, 1, 1. These do not define merely, a particular plane, but a set of parallel planes. It is only the ratio of the Miller indices which is of importance, e.g., the (422) plane is the same as (211) plane; (200) plane is the same as (100) plane and so on.
The faces of a cubic crystal represented by Miller indices (100), (110) and (111) are displayed in Fig.4.
Z Z Z
X Y (100)
Fig. 4.
X (110 Y (111 )
## Schematic description of (100), (110) and (111) planes.
The face BCEG or any plane parallel to it, has an intercept OG on X-axis but it is parallel to the Y- and Z- axes. If OG=1 (unity) then the Miller indices are reciprocals of (1, , ) i.e., (1, 0, 0). In the same manner ABCD is a (0, 1, 0) plane and CDFE is (0, 0, 1) plane. The diagonal plane AGED will have equal intercepts on OX and OY of unity, with an intercept of on the Z-axis. Hence, this is (1, 1, 0) plane. The plane AGF makes equal intercepts with the three axes, so its Miller indices are (1, 1, 1).
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Example 1. Calculate the Miller indices of crystal planes which cut through the crystal axes at (i) (2a, 3b, c) (ii) (a,b,c) (iii) (6a, 3b, 3c) (iv) (2a, -3b, -3c). Sol. (i) The unit intercepts are a, b, and c a b c
2 1/2 3
3 1/3 2
1 1 6
## Hence Miller indices are (3 2 6) (ii) a 1 1 1 b 1 1 1 c 1 1 1 intercepts reciprocals Miller indices
Hence Miller indices are (1 1 1) (iii) a 6 1/6 1 b 3 1/3 2 c 3 1/3 2 intercepts reciprocals Miller indices
Hence Miller indices are (1 22) (iv) a 2 1/2 3 b -3 -1/3 -2 c -3 -1/3 -2 intercepts reciprocals Miller indices
## Hence Miller indices are (3 2 2 )
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Example 2. Index the corresponding planes, whose intercepts on the crystallographic axes are (i) a, -b, (ii) a/2, b/4, (iii) 2a, 3b, 4c Sol. (i)
-a -1 -1
1
-b -1 -1
1
0
0
(ii) Intercepts
a 2
1 2
2 2
b 4
1 4 4 4
Weiss indices
## Reciprocals Miller indices
0 0
Hence, the plane is indexed as (2 4 0) plane. (iii) Intercepts Weiss indices Reciprocals 2a 2 3b 3 4c 4
1 2
6
1 3
4
1 4 3
Miller indices
## Hence, the plane is indexed as (6 4 3) plane.
Isomorphism E. Mitscherlich (1819) observed that a pair of salts of similar constitution, e.g., KH2PO4.H2O and KH2AsO4.H2O, in which phosphorus atom of one was replaced by an atom of arsenic in the
60
other, had the same crystalline form. To describe this replacement, he used the term isomorphism (Greek: same shape) and the elements were said to be isomorphous. These expressions are now applied more specifically, however, to the crystalline substances rather than to the elements contained in them. As a results of his work on the phosphates and arsenates, Mitscherlish proposed a generalization known as Mitscherlichs law of isomorphism as an equal number of atoms combined in the same manner produce the same crystalline form which is independent of the chemical nature of the combined atoms and determined only by their number and mode of combination. There are, however, many exceptions to this rule and a modified form of this law can be expressed as the substances which are similar in crystalline form and in chemical properties can usually be represented by similar formulae. A better understanding of isomorphism has resulted from the investigation of crystals structure by x-ray methods. In order that two salts may be isomorphous it is not necessary that they should be chemically similar. The conditions of isomorphism are (i) the compounds must have the same formula type (ii) the respective structural units (atoms or ions) need not necessarily be of the same size but their relative sizes should be little different and should have the same polarization properties. The crystals will then have the same type of space lattices with almost identical axial ratios.
Polymorphism Many substances exist in more than onecrystalline form. This phenomenon is known as polymorphism. The resulting shapes of the crystals depend upon the conditions of crystallization, e.g. temperature, pressure, concentration of solution, rate of cooling, etc. In case of elements, the polymorphism is called allotropy, e.g. carbon, phosphorus, sulfur, etc., exist in different allotropic form. Allotropic forms differ from one another in the number and arrangement of the structural units in the crystal lattice. Internal arrangement of the crystals The crystalline substances are characterized by the regular well-ordered arrangement of their constituents (ions, atoms or molecules). The arrangement can be best described in terms of space lattice, unit cell and lattice plane. (a) Space Lattice or Crystal Lattice: A space lattice is defined as an infinite regular three dimensional array of points in which every point has surrounding identical to that of every other point. For example, a two dimensional square array of points is shown in Fig. 5. By repeated translation of two vectors a and b on the plane of paper, we can generate the square array. For this array magnitudes of a and b are equal and can be taken to be unity. These are called the fundamental translation vectors. The angle between them is 90. Select any point in the array and look due north from this point, we will find another point at a distance of 1 unit, along north-east the nearest point at a distance of 2 units and along north north-east, the next nearest point is at a distance of 5 units. As this is true of every point in the array, the array can be called a two dimensional square lattice.
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Fig 5: Two dimensional lattice where the distance between any two points is equal to na ( where n=1,2,3..)
A typical three dimensional arrangement of points i.e. space lattice is shown in Fig. 6. It is characterized by equal distance between the successive points along each of three axes. It must be noted here that a point is an imaginary and infinitesimal spot in space and hence the space lattice is an imaginary concept.
Fig 6
(b) Basis : A crystal structure is formed by associating an identical unit of assembly of particles (ions, atoms or molecules) with every space lattice point. This unit assembly is called the basis. Hence, Crystal structure = Lattice + Basis
The planes and the inter-planar distances, dhkl which are calculated by substituting the values of h, k, and l in equations(4) for SC(simple cube), BCC(body centred cube) and FCC(face centred cube) lattices are shown in Fig. 7.
## Fig. 7(i) d100 = a
Fig.7(ii) d110 = a2
Fig.7(iii) d111 = a3
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## Fig.7(v) d220 = a/22
Fig.7(vi) d111 = a3
## Fig. 7(vii) d200 = a/2
Fig.7(viii) d110 = a2
Fig.7(ix) d222 = a3
Fig. 7 (i) The shaded planes are (100) in the simple cubic lattice. d100 = a.. Fig. 7 (ii) The shaded planes are (110) in the simple cubic lattice. d110 = a2 Fig. 7 (iii) The shaded planes are (111) in the simple cubic lattice. d111 = a3 Fig. 7 (iv) The shaded planes are (200) in the face centred cubic lattice. d200 = a/2 Fig. 7 (v) The shaded planes are (220) in the face-centred cubic lattice. d220 = a/22 Fig. 7 (vi) The shaded planes are (111) in the face-centred cubic lattice. d111 = a3 Fig. (vii) The shaded planes are (200) in the body centred cubic lattice d200 = a/2 Fig. 7 (viii) The shaded planes are (110) in the body-centred cubic lattice. d110 = a2 Fig. 7 (ix) The shaded planes are (222) in the body-centred cubic lattice. d222 = a3
(c) Unit cell: A unit cell is the smallest unit or geometrical portion which when repeated in space indefinitely will generate the space lattice or complete crystal. Thus, a unit cell is the fundamental building block of the lattice. For example, a unit cell is the square obtained by joining four neighbouring lattice points (Fig5). Alternatively, unit cell can be visualized with one lattice point at the centre of the square and with none at the corners (Fig. 5). In three dimensional space lattice A box represents unit cell (Fig. 6).
The number of points per unit cell (N) is given by the expression N= Nc N f + + Ni 8 2 (2)
where Nc = points on the corners, Nf = points on the faces and Ni = point in the interior of the unit cell. The unit cells are of the following types: 1. Simple or Primitive unit cell (P) which have the lattice points at the corners only. 2. Non-Primitive or Multiple Unit cell which have more than one lattice point. These are of the following types: (i) Face-centred unit cell (F) which contains one point in the centre of each face besides the points at the corners of the cell.
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(ii) (iii)
Body-centred unit cell (I) which contains one point at the centre within its body in addition to the points at the corners of the cell. Side-centred or End-centred unit cell (C) which contain points at the centre of any two parallel faces end faces in addition to the points at the corners of the cell.
(d) Bravais Lattices: Bravais (1848) showed that there can be only fourteen (14) different ways in which lattice points can be arranged in a three-dimensional space. These 14 arrangements are known as Bravais lattices. In a cubic system, for example, maximum three kinds of Bravais lattices are possible. These are: (i) Simple or primate cubic (SC) (ii) Face-centred cubic (FCC) and (iii) Body-centred cubic (BCC) lattices. Calculation of number of particles in the unit cells of cubic lattices The number of particles (ions, atoms or molecules) in the unit cell of simple, face-centred and body-centred cubic lattice are calculated as follows: (i) In the simple unit of cell of cubic lattice: In this cell, there is one particle at each corner of the cube and each particle is equally shared between eight unit cubes. Hence the corner particle contributes only 1/8 to each cube. Hence, number of particle in one cell = 8 x 1/8 = 1 (ii) In the face-centred unit cell of cubic lattice: In this cell, there are six face-particles, one on each face and eight corner-particles, one on each corner. The fcc-centred particle is shared equally between two unit cubes and hence contributes only to each cube. Hence, the total number of particles in one face-centred unit cell = (8 x 1/8) + (6 x ) = 1 + 3 = 4 (iii) In the body-centred unit cell of cubic lattice: The body-centred particle belongs to only the single unit cell and contributes 1 to each cell. In addition to this, there are eight corner particles, each contributing 1/8 to each cell. Hence, the total number of particles in a body-centred cubic cell = (8 x 1/8) + 1 = 1 + 1 = 2 Example 3. A metallic element exists as a cubic lattice with edge length of 2.88 . The density of metal is 7.20 g cm-3. Calculate the number of unit cells in 100 g of the metal. Sol.
Volume of unit cell = (2.88 )3 = 23.888 10-24 cm3 100 Volume of 100 g of metal = = 13.889 cm3 7.20 13.889 cm3 3 = 5.814 10 23 Number of unit cells in 13.889 cm = 24 3 23.88810 cm
Example 4. At room temperature, sodium crystallizes in a body-centred cubic cell with edge length, a = 4.24 . Calculate the theoretical density of sodium, if molar mass, Mm of sodium is equal to 23.08 mol-1. n Mm Sol. Theoretical density, e= NA V where n = number of molecules or atoms or ions in a unit cell, Mm = molar mass, NA = Avogadros number and V = volume of the cell.
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For b.c.c. unit cell, n = 2 Volume of unit cell, V = (4.24 )3 = 76.225 10-24 cm3 2 23.08 = 1.0056 gcm-3 Hence, p = 23 24 6.02210 76.22510
Example 5. A certain solid X (atomic mass 27) crystallizes in a fcc structure. The density of X is 2.70 g/cm3. Calculate the edge length a of the unit cell of X. Sol.
Molar volume of x = 27/2.70 = 10 cm3 10 cm3 Volume occupied by one molecule = 23 6.02210 As it is a face-centred cube (fcc), each unit cell contains 4 molecules. Hence, volume of each unit 410 cell = cm3 23 6.02210
If a is the edge length, then volume of each cell = a3 410 = 66.423 10-24 cm3 Hence, a3 = 6.0221023 or a = 4.05 10-8 cm = 4.05
Example 6. Gold has a face-centred cubic lattice with a unit length 4.07 . Its density is 19.3 g/cm3. Calculate the Avogadros number from the data (at. wt. of Au = 197) Sol. Number of atoms in gold fcc unit cell = 4 Let Avogadros number = NA Thus, weight of one Au atom = 197/NA g Weight of one unit cell = 4 (197/NA) g The volume of one unit cell = (4.07 x 10-8)3 cm3 4 (197/NA ) g = 19.3 g/cm3 Hence, density of gold cell = 8 3 3 (4.0710 ) cm 4197 = 6.056 x 1023 NA = 19.3 (4.07108 )3 Example 7. Nickel crystallizes in a face centred cubic crystal, the edge of the unit cell is 3.52 ; the atomic weight of nickel is 58.7 and its density is 8.948/cm3. Calculate Avogadros number. Ans. NA= 6.02 x 1023 Example 8. An organic compound crystallizes in an orthorhombic system with two molecules per unit cell. The unit cell dimensions are 12.05, 15.05 and 2.69 . If the density of the crystal is 1.419 g cm-3, calculate the molar mass of the organic compound. NA V Sol. From equation Mm= n
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Mm =
## (1.419 g cm3 ) (6.0221023 mol1 ) (12.515.05 2.6910-24 cm3 ) 2 -1 = 216.2 gmol
Example 9. An organic compound forms crystals of orthorhombic type with unit cell dimensions of 13.0 x 7.48 x 3.09 nm. If the density of the crystal is 1.315 x 103 kgm-3 and there are six molecules per unit cell, what is the molar mass of the compound?
## Ans. 39.7 kg mol-1
Example 10 : Ag crystallizes in a cubic lattice. The density is 10.7 103 kg m-3. If the edge length of the unit cell is 406 pm, determine the type of the lattice.
Sol: Volume of the cell = (406 10-12m)3 Density of Ag = 10.7 103 kg m-3 Mass of the unit cell = 4063 10-36 10.7 103 = 7.1608 10-25 kg = 7.1608 10-22 g 107 g mol1 =1.77681022 g Mass of one atom of Ag = 23 1 6.02210 mol 7.16081022 = 4.03 4 Hence no. of atoms of Ag per unit cell = 1.77681022 Therefore, the lattice is f.c.c. type.
Example 11. The density of TlCl which crystallizes out in a cesium chloride lattice is 7.00 g cm3 . What is the length of the edge of the unit cell? What is the shortest distance between Tl+ and Cl-? (Mm of TlCl = 239.82 g mol-1). Sol. Since TlCl crystallizes out in a cesium chloride lattice which is a body-centred cubic lattice, it contains one molecule of TlCl per unit cell. Hence, volume of the unit, V = Mm n 239.821 = = 56.88191024 cm3 23 NA 7.006.02210 Length of the edge of unit cell = 3.8458 10-8 cm 3 Hence, shortest distance between Tl+ and Cl- = 3.8458108 = 3.33 10-8cm 2 Example 12. The edge of the unit cell of KCl is 6.29 x 10-10 m and the density of KCl is 1.99 x 10-3kg m-3. If the KCl unit cell is face centred cubic and contains 4K+ and 4Cl- ions per unit cell, then calculate the number of K+ ions in one kg of KCl. Sol. Given edge of the unit cell of KCl 6.29 x 10-10m; density of KCl = 1.99 x 103kg m-3 Volume of the unit cell = (6.29 x 10-10)3m3 Hence weight of one unit cell = (6.29 x 10-10)3 x 1.99 x 103kg 1 1 kg of KCl contains = unit cells. 10 3 (6.29 10 ) 1.99 10 3
Since each unit cell contains 4 K+ ions, therefore, the number of K+ ions per kg of KCl
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## 4 1 = 8.077 10 24 3 (6.29 10 ) 1.99 10
10 3
Lattice Planes or Net-Planes: All the points in a space lattice can be included in a set of (e) parallel and equally spaced planes, known as lattice planes. Such a set of planes may be chosen in a large number of ways. The distance between two successive planes is represented by hkl or interplanar spacing in a crystal system is represented by the symbol, dhkl and in a cubic tetragonal and orthorhombic crystal systems, it is given by the expression
1 h k l = + + .3 2 (dhkl) a b c where h, k, l = Millers indices. For cubic system, a = b = c, so this equation becomes a dh,k,l = 2 2 2 1 ..4 (h + k + l ) 2 where a = unit cell edge length The ratio of inter-planar distances of different faces in the three cubic lattices are: Simple cube d100:d110:d111 = 1:1/2:1/3 = 1:0.707:0.577 Face centred cube d200 : d220 : d111 = 1:0.707:1.1547 Body-centred cube d200:d110:d222 = 1:414:0.577
Example 13. Calculate the interplanar spacing (dkl) for a cubic system between the following sets of planes: (a) 110 (b) 111 (c) 222. Assume that a is the edge length of the unit cell. a Ans: Using equation dkl = 2 2 2 1 (h + k + l ) 2 a a a d110 = 2 2 2 1 = 1 = 2 2 2 (1 +1 + 0 ) 2 a a a d111 = 2 2 2 1 = 1 = 3 (1 +1 +1 ) 2 32 a a a and d222 = 2 2 2 1 = 1 = 2 3 (2 + 2 + 2 ) 2 (12) 2 Example 14: The parameters of an orthorhombic unit cell are a=50 pm, b=100 pm, c=150 pm. Determine the spacing between the (123) planes.
h k l = + + Sol. 2 (dhkl ) a b c 1
2 2 2
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or
## 1 3 50 pm = or d123 = = 28.87 pm d123 50 pm 3
Example 15. The density of Li metal is 0.53 g cm-3 and the separation of the (100) planes of the metal is 350 pm. Determine whether the lattice is f.c.c. or b.c.c. Mm (Li) = 6.941 g mol-1 Sol. Density of Li = 0.53 g cm-3 = 530 kg m-3 a a For the cubic system, dhkl = = = 350 pm 2 2 2 2 h +k +l 1 + 02 + 02
n=
## NA a3 (530 kg m3 ) (6.0221023 mol1 ) (3501012 m)3 = =1.97 2 Mm 6.941103 kg mol1
Since, we know that for a f.c.c. lattice, n = 4 and for a b.c.c. lattice, n = 2. Hence, lithium has a b.c.c. lattice.
Example 16. The only metal that crystallizes in a primitive cubic lattice is polonium which has a unit cell side of 334.5 pm. What are the perpendicular distances between planes with Miller Indices (110), (111), (210) and (211)? a Sol. For a primitive cubic lattice dhkl = 2 2 2 1 (h + k + l ) 2
Given a = 334.5 pm
d110 = 334.5 (1 +1 + 0 )
2 2 2
1 2
d111 =
2 2 2
1 2 1 2
d210 =
1 2
d211 =
## 334.5 334.5 = =136.56 pm 6 2.4495
Example 17. Pd crystallizes in f.c.c. form. Its density is 11.9 gm/cm3. Calculate the distance between two consecutive 111 planes. Mass number of Pd = 106.4 Sol. Since Pd crystallizes in a f.c.c. form, the number of atoms of Pd in a unit cell = 4; density = 11.9 gm/cm3; Mass number of Pd = 106.4
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## a = 59.39 10 24 = 3.9015 10 8 cm a 3.905 10 8 d 111 = = = 2.2526 10 8 cm 1.732 3
(f) Atomic Radius (r):In simple or primitive cubic cell, the two adjacent corner atoms are supposed to touch each other. In b.c.c. cell, the atom at the centre of the cube is supposed to touch corner atoms. In f.c.c. cell, the atom at the face-centred is supposed to touch its adjacent corner atoms. The atomic radius may thus be calculated by applying simple geometry shown below: a (i) Simple or primitive cubic cell r=a/2 or r = 4 . 4 3a a (ii) Body-centred cubic cell r= or r = 3 . 4 4 2a a (iii) Face-centred cubic cell r= or r = 2 . 4 4
## Face centred cube d200 : d220 : d111 =
1 1 1 1 2 =1: = 1:0.707:1.1547 :2 : : 2 2 3 2 3
## Body-centred cube d200:d110:d222 =
1 1 1 1 = 1:414:0.577 : : 2 =1: 2 : 2 2 3 3
Example 18. Copper has f.c.c. structure. Its interatomic spacing is 2.54. Calculate (a) the atomic radius and (b) lattice constant for Cu. Sol. Interatomic spacing, 2r=2.54 2.54 Hence atomic radium, r = =1.27 2 9 In a f.c.c. structure, r= 2 . 4 4r 41.27 Or a = = = 3.59 2 1.414 Example 19. Aluminium has a f.c.c. structure having density as 2700 Kg m-3 and atomic weight as 27. Calculate its radius. Sol. Given P=2700 kg m-3 = 2.7 g cm-3 ; Atomic wt. Am=27; Na =6.022x10-23 mol 1 ; n=4
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## 1 n A.Wt. 4 27 3 24 8 = a = = (66.423 10 ) 3 = 4.05 10 cm = 4.05 23 P N 6.022 10 2.7 A
Example 20. Copper has f.c.c. structure with atomic radius as 1.278. Calculate its density (At. Wt. of Cu = 63.5) a 4r 4 1.278 Sol. For f.c.c. structure, r = 2 . or a = = = 3.615 = 3.615 10 -8 cm 4 1.414 2 n At. Wt. 4 63.5 = = 8.928 cm -3 = 8.928 10 3 kgm 3 Hence density = 3 23 8 3 NA a 6.022 10 (3.615 10 ) Example 21. Ni has a f.c.c. structure. Its density is 8.8 gm/cm3. Find out the closest distance between two Ni atoms. Mass no. of Ni = 58.71. Sol. Since Ni has a f.c.c structure, number of Ni atoms in unit cell, n = 4; Density,l=8.8 gm/cm3; Mass number of Ni=58.71; M n 58.71 4 = V = a3 = m Hence, volume of a unit cell, N A 8.8 6.022 10 23 = 4.4315 10 23 cm 3 = 44.315 10 24 cm 3
## or a = 44.315 10 24 = 3.5387 10 8 cm The closest distance between two Ni atoms
= a 3.5387 10 8 cm = = 2.5026 10 8 cm = 2.5026 10 -10 m 1.414 2
Packing Fraction and Empty Space in the Closest Packing: In the closest packing, (g) spherical balls must have some vacant space in the crystal. The fraction of the total volume of the unit cell occupied by the atom(s) is known as packing fraction.
(i)
In simple or primitive cubic cell, there is only one atom per unit cell.
3
4 3 4 a a3 r = = 3 3 2 6 3 Volume of one unit cell = a a3 /6 = = 0.5236 = 52.36% Hence, packing fraction = a3 6 and vacant space in the cell = 1-0.5236 = 0.4764 = 47.64% Volume of an atom = (ii) In body-centred cubic cell, there are two atoms per unit cell.
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4 8 3a 3 3 Volume of two atoms= 2 r3 = 4 = 8 a 3 3 3 3 a 3 =0.6802=68.02% Hence, paking fraction= 8 3 = a 8 and vacant space in the cell=10.6802=0.3198=31.98% (iii) In face-centred cubic cell, there are four atoms per unit cell.
3
4 16 2a 2 3 a Volume of four atoms = 4 r3 = = 3 3 4 8 2 3 a 2 =0.7405=74.05% Hence, packing fraction = 6 3 = a 6 and vacant space in the cell = 10.7405=0.2595=25.95%
Coordination Number (CN) : The number of nearest neighbour particles that a particle (h) has in a crystal is called its coordination number.
(i) In a simple cubic unit cell, each particle has 6 equally spaced nearest neighbour particles. Thus, the CN = 6 (ii) In a body-centred cubic unit cell, the particle at the centre of the cell is surrounded by 8 nearest particles situated at the corners of the cube. Thus, CN = 8 (iii) The two arrangements of closest packed layers are hexagonal close-packed structure (hcp) and cubic close-packed (ccp) or face-centred cubic structure (fcc). In each of these structures, every sphere is in contact with twelve others; six in its own layer, three in the layer above and three in the layer below. Thus, the CN = 12
X-ray Crystallography The determination of internal crystal structure with the help of x-rays is called x-ray crystallography. Max Von Laue (1912) suggested that crystal can act as grating to x-ray i.e., xray might be diffracted by the crystals, because wavelength of x-rays is of the same order (10-8 cm) as the interplanar distances in crystals. This idea was confirmed by W.L. Bragg and others. The observation has proved to be highly useful in determining the structures and dimensions of crystals as well as in the study of a number of properties of x-rays themselves. Braggs Law and Derivation of Braggs Equation for the Diffraction of X-rays by crystal Lattice In 1913 W.L. Bragg and W.H. Bragg were the first to develop a practical and simple way of analysis of crystal structure using x-ray diffraction. They pointed out that: The x-ray diffracted from atoms in the crystal planes obey the laws of reflection. (i)
(ii)
The reflection of x-rays can take place only at certain angles which depend on the wavelength of x-rays and the interplanar distances in the crystals.
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(iii)
The two rays reflected by successive planes will be in the phase if the extra distance travelled by the second ray is an integral multiple of wavelength. Bragg developed a mathematical equation between the wavelength of X-rays, the interplanar distance in the crystal and the angle of reflection. This equation is called Braggs equation.
Derivation of Braggs equation A crystal may be considered to be made up of a number of parallel equidistant atomic planes aa, bb, cc, etc. as shown in Fig. 8. Two successive atomic planes are separated by a distance d. Let the X-rays of wavelength falls on the surface of the crystal, at an angle . Some of the rays will be reflected at the same angle. Some of the rays will penetrate and suffer reflections from successive planes. If the waves reflected from the inner lattice planes of the crystal will be in phase, the intensity of the reflected beam will be maximum. If a photographic plate is placed to receive the reflected rays, diffraction pattern will be obtained.
Fig 8 In the Fig. 8 for a wave front suffering reflection from the planes aa and bb, the path difference is given by: Path-difference = CB + BD But for diffraction maxima, path difference of two ways = n where n is an integer. Hence, CB + BD = n .......................(1) Geometry shows that CB = BD = ABsin..(2) From (1) and (2), it follows that 2ABsin=n.......................................(3) But AB = d, Hence, 2dsin = n.............................(4) This is known as Bragg equation. Thus, if , and n are known, the inter planar distance (d) can be calculated and hence crystal structure can be derived. The angle 2 is called the diffraction angle.
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Example 22. Calculate the angles at which first, second and third order reflections are obtained from planes 500 pm apart, using X-rays of wavelength 100 pm. Sol. (i)
For first order reflection n = 1 and the Braggs equation is 2dSin = 100 pm 1 o =Sin1 = Sin 1 = Sin (0.1)=5 44 2d 2500 pm (ii) For second order reflection n = 2 and the Braggs equation is 2dSin = 2 2100 pm 2 1 o or =Sin1 = Sin1 = Sin (0.2)=11 32 2d 2500 pm (iii) For third order reflection n = 3, and the Braggs equation is 2dSin = 3 3100 pm 3 1 o or =Sin1 = Sin1 = Sin (0.3)=17 28 2d 2500 pm
Example 23. The density of NaCl at 25C is 2.163 x 103 kg m3. When X-rays from a palladium target having a wave length of 58.1 pm are used, the 200 reflection of NaCl occurs at an angle of 5.91. Calculate the number of Na+ and Cl- ions in the unit cell. Sol. 2dhklSinhkl = n
For first reflection n = 1 d200 = 58.1 pm 58.1 pm = = = 281.77 pm 2 Sin 200 2 Sin 5.910 2 0.1031
## Also for the cubic system, d200 = a 2 +0 +0
2 2 2
a a = 4 2
So that a = 2 d200 = 2 x 281.77 = 563.53 pm Mm(NaCl) = 58.5 g mol-1 = 58.5 x 10-3 kg mol-1 d = 2.163 x 103 kg m-3; Na = 6.023 x 1023 mol-1 n= NA a3 2.163103 kg m3 6.0221023 (563.531012 )3 m3 = = 3.985 4 Mm 58.5103 kg mol+-1
Since n = 4, hence no. of molecules of NaCl in a unit cell = 4 or there are 4 Na+ and 4 Cl- ions in the unit cell. 73
Example 24. HgCl2 crystallizes in orthorhombic lattice. Using X-rays of wavelength 154 pm, the first order reflections from (100), (010) and (001) planes of HgCl2 occur at 725, 328 and 1013 respectively. If the density of the crystal is 5.42 g cm-3, calculate the dimensions of the unit cell and the number of HgCl2 molecules per unit cell. [Mm(HgCl2] = 271.5 g mol-1 Sol. For first order reflection, n = 1,
## 2 dhkl Sin hkl = For orthorhombic system,
2 2
h k l = + + 2 (dhkl ) a b c 1
2 2
1 0 0 1 = + + = 2 (d100 ) a b c a 1
or d100=a=
2 154 pm = = 593 pm 2 Sin 100 2 Sin 70251 154 pm 154 pm = =1275 pm 2 Sin 010 2 Sin 3o 28
Similarly d010=b=
And d011=c=
## 154 pm 154 pm = = 434 pm 2 Sin 001 2 Sin 10o13
Number of molecules in unit cell, n= NA V (5.42 g cm3 ) (6.0221023 mol1 ) (5931275 4341030 cm3 ) = = 3.945 4 Mm 271.5 g mol1
## Since n=4, therefore 4 molecules of HgCl2 per unit cell.
Example 25. The first order reflection of x-ray for a crystal plane occurs at a glancing angle 7.5 with the wavelength 0.75 x 10-10m. Find out the interplaner distance. Sol. The interplanar distance, d=
## 0.751010 m 0.751010 m = = = 2.8741010 m o 2 Sin 2Sin 7.5 2 0.1305
Example 26. The distance between two consecutive 110 planes of a crystal is 1.678 x 1010m. What will be the glancing angle for an x-ray of wave length 0.65 x 10-10 m incident on the planes for first order reflection? Sol. Given d110=1.6781010 m, =0.651010 m, n=1
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## Hence, 110=Sin1 0.1937=1110=11.17
Example 27. the glancing angle for the first order X-ray reflections is 11.2 and the interplanar distance is 2.01 x 10-10m. What is the wavelength of the x-ray used? = 2dsin 2 2.01 10 10 msin11.2 = = = 2 2.01 10 10 m 0.1942 Sol. n 1 = 0.78 10 10 m Example 28. KNO3 crystallizes in orthorhombic system with the unit cell dimensions a = 542 pm, b = 917 pm and c = 645 pm. Calculate the diffraction angles for first-order X-ray reflections from (100), (010), and (111) planes using radiation with wave length = 154.1 pm. Sol. We know that 2dnklSin = n and for an orthorhombic system
h k l = + + 2 (dhkl ) a b c 1
2 2 2
1 d1002
## 1 0 0 1 = + + = 542 917 645 542
d100
= a = 542 pm
Similarly d010 = b = 917 pm and d 111 = 378 pm For first order reflection, n = 1 and = 154.1 pm
Sin 100 = 1154.1 pm 1154.1 pm = 0.142; Sin 010 = = 0.084 2542 pm 2917 pm 1154.1 pm = 0.2038 2378 pm
## Hence, 100 = 810 010 = 449 and 111 =1146
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Laues Method A simple arrangement for the apparatus used in Laues method is shown in Fig. 9. A beam of Xrays of known wavelength, after passing through a narrow hole of a lead shield is allowed to fall upon the crystal placed on an adjustable mount. Every atom in the path of X-rays reflect some of the rays at certain angles. The X-rays strike the photographic plate placed at some distance from the crystal. The photographic plate shows on development spots of different size and intensity characteristics of the crystal. From the analysis of these photographs the crystal structure can be determined.
Fig 9
Power method In this method the crystalline material contained in a capillary tube is placed in the camera containing a film strip (Fig. 10). The sample is rotated by means of a motor. The X-rays of known wavelength fall on the sample after passing through the gap between the ends of the of the film. Since the powdered sample contains small crystals arranged in all orientations, the Xrays will be reflected from each lattice plane at the same time. The reflected X-rays will make an angle 2 with the original direction. From the geometry of the camera, can be calculated for different crystal planes.
Fig 10
Structures of sodium chloride, potassium chloride and cesium chloride crystals According to Braggs equation, 2dsin = n
For a given wavelength X-rays and a given order of reflection, n and are constant and hence the above equation becomes d 1/sin
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In order to find out the type of the cubic lattice in the given crystal, the angles at which intensities of reflection are maximum, for different orders for different planes are studied and the values of d100, d110 and d111 are calculated.
(a) Structure of NaCl: In case of sodium chloride (NaCl, rock salt), the first order reflections from (100), (110) and (111) faces using K line from palladium anti-cathode are 5.9, 8.4 and 5.2 respectively. Hence the ratio 1 1 1 1 1 1 d100 : d110 and d111 = : : = : : o o o sin 5.9 sin 8.4 sin 5.2 0.103 0.146 0.091
## d100 : d110 and d111 = 1:0.704:1.14=1:
1 2 : 2 3
This experimental ratio of interplanar distances is the same as that theoretically predicted from face-centred cubic lattice. Hence sodium chloride has face-centred cubic lattice (d = 2.819 , a = 5.636 )
Fig. 11 The structure of NaCl crystal is really composed of two interpenetrating FCC lattices, one composed of Na+ ions and other of Cl- ions. The positions of Na+ ions and Cl- ions in cubic lattice of sodium chloride are emphasized in Fig. 11. The black dots and circles in Fig. 11 represent the centres of Na+ and Cl- ions respectively. Notice in Fig. 11 that: 1. In the face-centred cubic lattice of NaCl, Na+ and Cl- ions are arranged alternatively in all directions. 2. 3. Each Na+ ion is surrounded by six octahedrally arranged Cl- ions and each Cl- ion is surrounded by six octahedrally arranged Na+ ions, in three dimensions. Each unit cell of sodium chloride consists of 14 Na+ one at each of the eight corners and one in the middle of each of the six faces, and 13 Cl- ions one at the center and twelve at the middle of each edges, or 14 Cl- ions and 13 Na+ ions. The eight Na+ particles at the corners are shared equally by eight cubes meeting at each corner. 77
Similarly, each face centred particle is shared equally between two unit cells. So, the number of Na+ particles in one cell = 8 1/8 + 6 = 1 + 3 = 4. Similarly, one Clparticle at the centre belongs exclusively to the unit cell and the other twelve are each equally shared between four cubes meeting at the edges. So, the number of Clparticles in one unit cell = 1 + 12 = 1 + 3 = 4. Thus, four Na+ ions and four Clions i.e. four NaCl molecules are associated with each unit cell of length of a = (2d) and the volume a3 = (8d3 ). 4. The structure of sodium chloride is said to have 6:6 coordination because Na+ ions have a coordination number of 6 and the Cl- ions also have a coordination number of f6.
(b) Structure of KCl (Sylvine): Potassium chloride is isomorphous with sodium chloride. The first order reflection from (100) plane occurs at 5.30, so that d100 (KCl) sin 5.9o = =1.11 d100 (NaCl) sin 5.30o
Since the volume of a small lattice cube is (d100)3, it follows that if KCl and NaCl have the same crystal structure, the quantity (1.11)3 = 1.37 should give the ratio of the molecular volumes of the two salts. The experimental value of 1.39 suggests that the space lattices of the chlorides are in fact same, i.e. face centred cubic lattice. X-ray diffraction studies, however, show that KCl has simple cubic lattice. The ratio, 1 1 1 1 1 d(100) :d(110) :d(111) = : : =1: : o o o sin 5.38 sin 7.61 sin 9.38 2 3 is in full agreement with that calculated for a simple cubic structure. The anomaly in the structures can be explained when it is realized that scattering power of an atom or an ion is governed by the number of extra-nuclear electrons. In KCl, K+and Cl- ion possess the same number of electrons (18 in each), hence, their scattering power will be equal. Consequently, X-rays can not distinguish them. The first order of reflection from (111) planes is extinguished due to interference and hence show a simple cubic unit instead of face centred cubic lattice. The edge length of the unit cell is 6.2930 .
(c) Structure of cesium chloride (CsCl): X-ray diffraction results show that cesium chloride has a body-centred cubic structure. In its crystal lattice each Cs+ ion is located at the centre of a cube at whose corners are ions of the other kind, i.e. Cl- ions and its coordination number is 8. In the same way, the coordination number of Cl- ions is also eight. In fact, the lattice could be redrawn with the Cl- ions exchanged for Cs+ ions. The structure of cesium chloride is, therefore, said to have 8:8 coordination. The unit cell of cesium chloride contains 8 1/8 = 1 Cl- and 1 Cs+ ion, i.e., one molecule of CsCl. Thus, each ion has eight nearest neighbours of the other kind.
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Fig 12 OBJECTIVE QUESTIONS (1) The solids which have no definite geometrical form are called (a) Crystalline solids (b) Polycrystalline solids (c) Amorphous solids (d) None of the above (2) Crystalline solids possess melting points. (3) Since the micro-crystals are randomly oriented in a polycrystalline substance, it will be (a) anisotropic (b) isotropic (c) None of the above (4) A space lattice is defined as a finite array of points in three dimensions in which every point has surroundings identical to that of every other point in the array. True or False (5) Bravis lattices are of types. (6) Laws of crystallography are (i).. (ii). (iii).. (7) The instrument used to measure the interfacial angies is called (8) The total number of elements of symmetry possessed by a cubic crystal are: (a) 9 (b) 13 (c) 23 (d) 7 (9) The rectangular planes of symmetry in a cubic crystal are: (a) 9 (b) 6 (c) 4 (d) 3 (10) The basic crystal systems are of the type: (a)32 (b)7 (c)13 (d)9 (11) The intercepts made by any of the crystal on the crystallographic axes are called crystal.. (12) The Miller indices are: (a) coefficients of crystal parameters
79
(b) equal to the crystal parameters (c ) reciprocals of Weiss indices expressed as integers (d) none of the above (13) The Miller indices for the face having intercepts a/2, b/4, are: 1 1 (a) , , (b) 2,4,0 2 4 2 4 (d) None of the above (c) , , 0 a b (14)The plane, whose intercepts on crystallographic axes are: a, -b, c is: (a) 1 T 1 (b) 1 1 1 (c ) 1 _1 1 (d) T T T (15)The ratio of the interplanar spacing of the planes (100), (110) and (111) for simple cubic lattice is: 1 2 1 1 (a) 1 : : (b) 1 : : 2 3 2 3 1 (c) 1 : 2 : (d) 1 : 2 : 3 3
(16)The wavelength of x-rays are the order of: (a) 10-9 cm (b) 10-8 cm -10 8 (c ) 10 cm (d) 10 cm (17) According to Braggs equation, the interplanar distance(d) is proportional to: 1 (a) sin (b) sin 1 (c) (d) sin 2 sin 2 (18) Sodium chloride has a: (a) simple cubic lattice (b) body-centred cubic lattice (c) face-centred cubic lattice (19) The interplanar distance dhkl ratio is 1:0.707:0.577, identify the cubic lattice. (20) For a crystal system, abc, = = =900. Identify the system. (21) CsCl has a body-centred cubic structure. What is the coordination number of Cs+? (22) NaCl has a face-centred cubic structure. How many Na+ and Cl- ions are there in the unit cell?
80
8 6 Hint : + + 0 = 4 8 2 (23) A compound alloy of gold and copper crystallizes in a cubic lattice in which gold atoms occupy the lattice points at the corners of a cube and the copper atoms occupy the centers of each of the cube faces. What is the formula of the compound. (Hint: 8/8 = 1 Au atom and 6/2 = 3 Cu atoms per unit cell. Hence the empirical formula is AuCu3). (24) CsCl has a b.c.c.structure. How many Cs+ and Cl- ions are there in the unit cell? (25) The 8:8 type of packing is present in (a) CsCl (b) NaCl (c) CuCl2 (d) KCl (26) A unit cell has 4 atoms. The type of cell is (a) Simple or Primitive (b) Face-centred (c) Body-centred (d) End-centred (27) Braggs Law is given by the equation (a) n = dSin (b) n = 2Sin (c) = 2dSin (d) n = 2dSin (28) In a face-centred cubic cell an atom at the face centre is shared by (a) 4 unit cells (b) One unit cell (c) 2 unit cells (d) 8 unit cells (29) The ratio of radius of the atoms to the lattice parameter for fcc unit cell is (a) 1:2 (b) 2:1 (c) 1:2 (d) 1:4 (30) Which of the following is an amorphous solid? (a) Diamond (b) Graphite (c) Glass (d) Common salt (31) The number of Cl- ions surrounding each Na+ ion in NaCl crystal is (a) 5 (b) 7 (c) 8 (d) 6 (32) The number of atoms in a fcc unit cell is (a) 2 (b) 1 (c) 4 (d) 3 (33) Each unit cell of NaCl consists of 13 Na+ ions and (a) 13 Cl-ions (b) 14 Cl- ions (c) 6 Cl- ions (d) 8 Cl- ions (34) The ability of a given substance to assume two or more crystalline structures is called (a) isomorphism (b) polymorphism (c) isomerism (d) amorphism (35) An example of covalent crystal is (a) LiF (b) KCl (c) CsCl (d) Diamond (36) The unit cell of a KCl lattice is 81
(a) fcc (b) bcc (c) simple cubic (d) End-centred cubic (37) Each Cs+ ion in CsCl lattice is surrounded by (a) 2 Cl- ions (b) 6 Cl- ions (c) 8 Cl- ions (d) 4 Cl- ions (38) In a body centred cubic cell, the central atom touches (a) 4 atoms (b) 5 atoms (c) 8 atoms (d) 2 atoms (39) KH2PO4.H2O and KH2AsO4.H2O are: (a) polymorphous (b) amorphous (c) isomorphous (d) isotropic (40) The relation between faces, edges and corners of a crystal is (a) f + e = c + 2 (b) f + c = e + 1 (c) f + c = e + 2 (d) f + 2 = c + e (41) The crystalline solids are (a) anisotropic (b) amorphous (c) isotropic (d) polytropic (42) The fraction of the total volume occupied by atoms in a bcc cell is 2 3 (a ) ( b) (c) (d ) 4 6 6 8 (43) The atomic radius in a simple cubic cell of edge a is 3 2 3 a (a ) a ( b) a (c) a (d ) 4 4 8 2 (44) In a body-centred cubic cell, an atom at the corner is shared by (a) 2 unit cells (b) 8 unit cells (c) 1 unit cell (d) 4 unit cells (45) The volume occupied by an atom in a face-centred cubic cell is a 3 2 3 3a 3 2a 3 a (d ) (c ) (a) (b) 6 8 6 6 (46) In a body-centred cubic lattice, the ion A+ occupies the centre while the ion B- occupies the corners of the cube. The formula of crystal is (a) AB8 (b) AB2 (c) AB (d) AB4
Answers 1(c) 2 Sharp 3 (b) 4 False 5(14) 6 (i) Law of constancy of interfacial angles (ii) Law of symmetry (iii) Law of rationality of indices 7 Goniometer 8 (c ) 9 (d) 10 (b) 11 parameters 12 13 (b) 14 (a) 15 (b) 16 (b) 17 (b) 18 (c ) 19 simple cubic lattice 20 Rhombic 21 (8) 22 Na+ = 4, Cl- = 4 23 AuCu3 24 Cs+ = 1, Cl- =1 25 (a) 26 (b) 27 (d) 28 (c ) 29 (c ) 30 (c ) 31 (d) 32 (c ) 33 (b) 34 (b) 35 (d) 36 (a) 37 (c ) 38 (c ) 39 (c ) 40 (c ) 41 (a) 42 (d) 43 (d) 44 (b) 45 (b) 46 (c )
82 | 43,055 | 155,945 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2019-35 | latest | en | 0.891602 |
https://discuss.codechef.com/t/givcandy-editorial/12628 | 1,686,347,923,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656833.99/warc/CC-MAIN-20230609201549-20230609231549-00029.warc.gz | 257,753,370 | 8,295 | # GIVCANDY - Editorial
### PROBLEM LINK:
Author: Kevin Atienza
Testers: Sergey Kulik and Vasya Antoniuk
Translators: Vasya Antoniuk (Russian), Team VNOI (Vietnamese) and Hu Zecong (Mandarin)
Editorialist: Kevin Atienza
Simple
### PREREQUISITES:
Greatest common divisors, Euclid’s algorithm, Linear combinations, Bézout’s identity
### PROBLEM:
Alvin has A candies and Berto has B candies. You can give C candies to Alvin at a time, and D candies to Berto at a time. What is the minimum possible absolute difference between their candies that can be achieved?
### QUICK EXPLANATION:
Let G = \gcd(C,D). Then every possible absolute difference is of the form |A - B + kG| for some integer k, and every difference of that form can be achieved. The lowest difference that can be achieved is \min((A - B) \bmod G, (B - A) \bmod G).
### EXPLANATION:
First, any absolute value is nonnegative, so the lowest possible answer in every test case is 0. Sometimes, this can be achieved, like in the first sample input. But in some cases this cannot be achieved. For example, in the second sample input, A = 1 and B = C = D = 2. In this case, you can only increase A and B by an even amount, so you can’t make them have the same parity.
Here’s another example. Suppose A = 11, B = 35, C = 50 and D = 130. Notice that \gcd(C,D) = 10. Thus, no matter what you do, you cannot change the last digit of A and B. The difference between any pair of numbers whose last digits are 1 and 5 respectively ends in either 4 or 6. Thus, the lowest possible difference we can get is 4, and it turns out we can achieve this difference by adding C to A three times and D to B one time: |(11 + 3\cdot 50) - (35 + 1\cdot 130)| = |165 - 161| = 4.
We can generalize the insight we got from the previous example. Suppose we add C to A m times and D to B n times. Then the absolute difference that we get is |(A + mC) - (B + nD)| = |(A - B) + (mC - nD)|. Every number of the form mC - nD (a linear combination) is divisible by \gcd(C,D)! Thus, we can state the following:
If G = \gcd(C,D), then every possible absolute difference is of the form |A - B + kG| for some integer k.
The smallest possible value of the form |A - B + kG| depends on the sign of the number inside the absolute value. If it’s possible, then the smallest you can get is
(A - B + kG) \bmod G = (A - B)\bmod G,
and if it’s negative, then it is
-(A - B + kG)\bmod G = (B - A)\bmod G.
(Remember that x \bmod G is always nonnegative even if x is negative.) Thus, the answer is at least
\min((A - B)\bmod G, (B - A)\bmod G).
Unfortunately, we haven’t yet proved that every integer of the form |A - B + kG| can be achieved as a difference, so “\min((A - B)\bmod G, (B - A)\bmod G)” is just a lower bound at this point. Thus, we want to know whether the following is true:
Let G = \gcd(C,D). Can every possible value of the form |A - B + kG| for any integer k be expressed in the form |A - B + (mC - nD)| for some nonnegative integers m and n?
If we can show this, then we establish that the answer is \min((A - B)\bmod G, (B - A)\bmod G). If not, then we’ll have to find other ways.
Note that the word “nonnegative” is pretty important; Without it, then we can easily answer the problem by using Bézout’s identity:
If G = \gcd(C,D), then there exist integers x and y such that Cx + Dy = G.
Using this theorem, we can now achieve any difference |A - B + kG| by setting m = kx and n = -ky. Unfortunately, the theorem doesn’t guarantee anything about the signs of x and y.
But actually, there’s a way to convert any choice (m,n) to a new choice (m',n') such that both are nonnegative, by noticing that if (m,n) is a valid choice, then (m+D,n+C) is also valid! It’s simply because
(m+D)C - (n+C)D = mC + DC - nD - CD = mC - nD.
But m+D and n+C are strictly greater than m and n, respectively, so we can apply this rule over and over again until both m and n become positive!
This last piece proves that the answer is \min((A - B) \bmod G, (B - A) \bmod G).
Another way to look at it is that we can (partially) strengthen Bézout’s identity to the following two statements:
1. If G = \gcd(C,D) and C and D are positive, then there exist integers x > 0 and y < 0 such that Cx + Dy = G.
2. If G = \gcd(C,D) and C and D are positive, then there exist integers x < 0 and y > 0 such that Cx + Dy = G.
The proof goes similarly: For any (x,y) satisfying Cx + Dy = G, the pairs (x + D, y - C) and (x - D, y + C) also satisfies it, so we can keep on applying these until we get the signs that we want. And since we choose (m,n) = (kx,-ky), then:
• If k \ge 0, then we can choose x > 0 and y < 0 to make (m,n) nonnegative.
• If k < 0, then we can choose x < 0 and y > 0 to make (m,n) nonnegative.
# Appendix: Bézout’s identity
Here we’ll prove Bézout’s identity:
If G = \gcd(C,D), then there exist integers x and y such that Cx + Dy = G.
We will actually prove this constructively, that is, by giving an algorithm that finds such x and y.
Our solution here will involve Euclid’s algorithm to compute GCD’s. The basic Euclid recursion for GCD goes:
\gcd(a,b) = \gcd(b, a\bmod b)
Euclid’s algorithm simply applies this over and over again until b becomes 0, in which case \gcd(a,0) = a.
Another way of looking at it is that we’re generating a sequence of numbers r_0, r_1, r_2, \ldots such that:
• r_0 = a
• r_1 = b
• r_i = r_{i-2} \bmod r_{i-1}
And we halt this sequence when r_i becomes 0, in which case r_{i-1} is the GCD!
For example, suppose we want to find the GCD of C = 602 and D = 427. Then:
\begin{array}{r|rrrrr} i & & & & & r_i \\ \hline 0 & & & & & 602 \\ 1 & & & & & 427 \\ 2 & 602 &\bmod& 427 &=& 175 \\ 3 & 427 &\bmod& 175 &=& 77 \\ 4 & 175 &\bmod& 77 &=& 21 \\ 5 & 77 &\bmod& 21 &=& 14 \\ 6 & 21 &\bmod& 14 &=& 7 \\ 7 & 14 &\bmod& 7 &=& 0 \end{array}
Since r_7 = 0, we find that the GCD is r_6, or 7.
But notice that “a\bmod b” is equivalent to “a - qb”, where q = \lfloor a/b \rfloor. Thus, every number is a linear combination of the numbers above it: r_i = r_{i-2} - qr_{i-1}, where q = \left\lfloor \frac{r_{i-2}}{r_{i-1}} \right\rfloor. By also noticing that C and D can be written as linear combinations of themselves: C = C\cdot 1 + D\cdot 0 and D = C\cdot 0 + D\cdot 1, we can express each subsequent value as a linear combination of C and D too, and by doing so we’re able to extend Euclid’s algorithm to actually find the coefficients x and y!
For example, using C = 602 and D = 427 again:
\begin{array}{r|r|rrrrr|rrrrr} i & \left\lfloor \frac{r_{i-2}}{r_{i-1}} \right\rfloor & & & & & r_i & & & & & \text{linear combination} \\ \hline 0 & & & & & & 602 & & & & & C\cdot 1 + D\cdot 0 \\ 1 & & & & & & 427 & & & & & C\cdot 0 + D\cdot 1 \\ 2 & 1 & 602 &{}-1\,\cdot& 427 &=& 175 & (C\cdot 1 + D\cdot 0) &{}-1\,\cdot& (C\cdot 0 + D\cdot 1) &=& C\cdot 1 + D\cdot -1 \\ 3 & 2 & 427 &{}-2\,\cdot& 175 &=& 77 & (C\cdot 0 + D\cdot 1) &{}-2\,\cdot& (C\cdot 1 + D\cdot -1) &=& C\cdot -2 + D\cdot 3 \\ 4 & 2 & 175 &{}-2\,\cdot& 77 &=& 21 & (C\cdot 1 + D\cdot -1) &{}-2\,\cdot& (C\cdot -2 + D\cdot 3) &=& C\cdot 5 + D\cdot -7 \\ 5 & 3 & 77 &{}-3\,\cdot& 21 &=& 14 & (C\cdot -2 + D\cdot 3) &{}-3\,\cdot& (C\cdot 5 + D\cdot -7) &=& C\cdot -17 + D\cdot 24 \\ 6 & 1 & 21 &{}-1\,\cdot& 14 &=& 7 & (C\cdot 5 + D\cdot -7) &{}-1\,\cdot& (C\cdot -17 + D\cdot 24) &=& C\cdot 22 + D\cdot -31 \\ 7 & 2 & 14 &{}-2\,\cdot& 7 &=& 0 & (C\cdot -17 + D\cdot 24) &{}-2\,\cdot& (C\cdot 22 + D\cdot -31) &=& C\cdot -61 + D\cdot 86 \end{array}
Thus, we establish that \gcd(C,D) = 7, and C\cdot 22 + D\cdot -31 = 7! It’s easy to see that this method can also be applied for any pair (C,D). This is one version of the extended Euclid’s algorithm.
### Time Complexity:
O(\log \min(C, D))
### AUTHOR’S AND TESTER’S SOLUTIONS:
4 Likes
from fractions import gcd
ls = []
for dfj in range(input()):
a,b,c,d = [int(i) for i in raw_input().split()]
g = gcd(c,d)
ls.append(min((a-b)%g,(b-a)%g))
for i in ls:
print i
Hi,I didn’t get this line .“If it’s possible, then the smallest you can get is : (A-B+kG) mod G”. Please explain it.
What if we also want to know the sequence in which the candies were given? How to print the sequence?
@bhagirathi08 ,suppose that the initial apples and bananas be 0, then the difference that can be created is kg,
but since initially they were a and b the difference which can be created is (|A-B+kg|) k can also be negative
how smallest value of |(A−B+kG)|=(A−B)modG? I didn’t get it. Please explain | 2,864 | 8,483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2023-23 | latest | en | 0.878793 |
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## C. Duties
Time limit: 1.0 second
Memory limit: 64 MB
School principal introduced a new system of duties to maintain order in all classrooms. Each student is assigned one classroom he is responsible for. Each day two students responsible for different classrooms should be on duty. These students should water the flowers in their classrooms and ensure that the highlighters in computer classrooms are not dry. The principal wants you to make a plan of the duties for the first m days of study. Each of the students should be on duty at least once during these days. You are to determine for each student the classroom he will be responsible for and the days this student will be on duty. Of course, each classroom should be assigned to at least one student. In addition to that, the principal requires that no pair of students should be on duty twice. To make your task easier, the principal allows you to distribute the duties unevenly — the students who misbehave will have more duties than those who are diligent.
### Input
The first line contains three space-separated integers: n — the number of students in the school, k — the number of classrooms, and m — the number of days in the required plan of duties. These numbers satisfy the constraint 2 ≤ kn ≤ 100. You may assume that there is at least one correct plan of the duties for given n, k and m.
### Output
Let the students be numbered from 1 to n, and the classrooms numbered from 1 to k. Output the assignments of the students to the classrooms on the first n lines: i-th line should contain the number of classroom i-th student is responsible for. The next m lines should contain the pairs of numbers of students who will be on duty on each of m days. Remember that these pairs should be unique, and the students who are responsible for the same classroom can't be paired.
### Samples
inputoutput
`5 3 4`
```1
2
1
2
3
1 2
3 4
3 5
4 5
```
`4 2 4`
```1
2
1
2
1 2
3 4
1 4
2 3```
Problem Author: Pavel Atnashev
Problem Source: USU Junior Contest, October 2008
To submit the solution for this problem go to the Problem set: 1641. Duties | 565 | 2,343 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-33 | latest | en | 0.92036 |
https://studysoup.com/tsg/math/166/introduction-to-linear-algebra/chapter/20267/7-2 | 1,586,210,781,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371660550.75/warc/CC-MAIN-20200406200320-20200406230820-00146.warc.gz | 691,942,905 | 10,111 | ×
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# Solutions for Chapter 7.2: The Matrix of a Linear Transformation
## Full solutions for Introduction to Linear Algebra | 4th Edition
ISBN: 9780980232714
Solutions for Chapter 7.2: The Matrix of a Linear Transformation
Solutions for Chapter 7.2
4 5 0 428 Reviews
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##### ISBN: 9780980232714
Since 38 problems in chapter 7.2: The Matrix of a Linear Transformation have been answered, more than 8145 students have viewed full step-by-step solutions from this chapter. This textbook survival guide was created for the textbook: Introduction to Linear Algebra, edition: 4. Chapter 7.2: The Matrix of a Linear Transformation includes 38 full step-by-step solutions. This expansive textbook survival guide covers the following chapters and their solutions. Introduction to Linear Algebra was written by and is associated to the ISBN: 9780980232714.
Key Math Terms and definitions covered in this textbook
• Cayley-Hamilton Theorem.
peA) = det(A - AI) has peA) = zero matrix.
• Cross product u xv in R3:
Vector perpendicular to u and v, length Ilullllvlll sin el = area of parallelogram, u x v = "determinant" of [i j k; UI U2 U3; VI V2 V3].
• Free variable Xi.
Column i has no pivot in elimination. We can give the n - r free variables any values, then Ax = b determines the r pivot variables (if solvable!).
• Hermitian matrix A H = AT = A.
Complex analog a j i = aU of a symmetric matrix.
• Hypercube matrix pl.
Row n + 1 counts corners, edges, faces, ... of a cube in Rn.
• Incidence matrix of a directed graph.
The m by n edge-node incidence matrix has a row for each edge (node i to node j), with entries -1 and 1 in columns i and j .
• Inverse matrix A-I.
Square matrix with A-I A = I and AA-l = I. No inverse if det A = 0 and rank(A) < n and Ax = 0 for a nonzero vector x. The inverses of AB and AT are B-1 A-I and (A-I)T. Cofactor formula (A-l)ij = Cji! detA.
• Least squares solution X.
The vector x that minimizes the error lie 112 solves AT Ax = ATb. Then e = b - Ax is orthogonal to all columns of A.
• Left nullspace N (AT).
Nullspace of AT = "left nullspace" of A because y T A = OT.
• Lucas numbers
Ln = 2,J, 3, 4, ... satisfy Ln = L n- l +Ln- 2 = A1 +A~, with AI, A2 = (1 ± -/5)/2 from the Fibonacci matrix U~]' Compare Lo = 2 with Fo = O.
• Multiplicities AM and G M.
The algebraic multiplicity A M of A is the number of times A appears as a root of det(A - AI) = O. The geometric multiplicity GM is the number of independent eigenvectors for A (= dimension of the eigenspace).
• Pivot.
The diagonal entry (first nonzero) at the time when a row is used in elimination.
• Plane (or hyperplane) in Rn.
Vectors x with aT x = O. Plane is perpendicular to a =1= O.
• Rank one matrix A = uvT f=. O.
Column and row spaces = lines cu and cv.
• Rank r (A)
= number of pivots = dimension of column space = dimension of row space.
• Rayleigh quotient q (x) = X T Ax I x T x for symmetric A: Amin < q (x) < Amax.
Those extremes are reached at the eigenvectors x for Amin(A) and Amax(A).
• Singular Value Decomposition
(SVD) A = U:E VT = (orthogonal) ( diag)( orthogonal) First r columns of U and V are orthonormal bases of C (A) and C (AT), AVi = O'iUi with singular value O'i > O. Last columns are orthonormal bases of nullspaces.
• Solvable system Ax = b.
The right side b is in the column space of A.
• Subspace S of V.
Any vector space inside V, including V and Z = {zero vector only}.
• Sum V + W of subs paces.
Space of all (v in V) + (w in W). Direct sum: V n W = to}.
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Reset your password | 1,080 | 3,860 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2020-16 | latest | en | 0.82777 |
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## Further Details: Error Bounds for the Singular Value Decomposition
The usual error analysis of the SVD algorithms xGESVD and xGESDD in LAPACK (see subsection 2.3.4) or the routines in LINPACK and EISPACK is as follows [25,55]:
The SVD algorithm is backward stable. This means that the computed SVD, , is nearly the exact SVD of A+E where , and p(m,n) is a modestly growing function of m and n. This means is the true SVD, so that and are both orthogonal, where , and . Each computed singular value differs from true by at most
(we take p(m,n)=1 in the code fragment). Thus large singular values (those near ) are computed to high relative accuracy and small ones may not be.
There are two questions to ask about the computed singular vectors: Are they orthogonal?'' and How much do they differ from the true eigenvectors?'' The answer to the first question is yes, the computed singular vectors are always nearly orthogonal to working precision, independent of how much they differ from the true singular vectors. In other words
for .
Here is the answer to the second question about singular vectors. The angular difference between the computed left singular vector and a true ui satisfies the approximate bound
where is the absolute gap between and the nearest other singular value. We take p(m,n)=1 in the code fragment. Thus, if is close to other singular values, its corresponding singular vector ui may be inaccurate. When n < m, then must be redefined as . The gaps may be easily computed from the array of computed singular values using function SDISNA. The gaps computed by SDISNA are ensured not to be so small as to cause overflow when used as divisors. The same bound applies to the computed right singular vector and a true vector vi.
Let be the space spanned by a collection of computed left singular vectors , where is a subset of the integers from 1 to n. Let be the corresponding true space. Then
where
is the absolute gap between the singular values in and the nearest other singular value. Thus, a cluster of close singular values which is far away from any other singular value may have a well determined space even if its individual singular vectors are ill-conditioned. The same bound applies to a set of right singular vectors 4.1.
In the special case of bidiagonal matrices, the singular values and singular vectors may be computed much more accurately. A bidiagonal matrix B has nonzero entries only on the main diagonal and the diagonal immediately above it (or immediately below it). xGESVD computes the SVD of a general matrix by first reducing it to bidiagonal form B, and then calling xBDSQR (subsection 2.4.6) to compute the SVD of B. xGESDD is similar, but calls xBDSDC to compute the SVD of B. Reduction of a dense matrix to bidiagonal form B can introduce additional errors, so the following bounds for the bidiagonal case do not apply to the dense case.
Each computed singular value of a bidiagonal matrix is accurate to nearly full relative accuracy, no matter how tiny it is:
The following bounds apply only to xBDSQR. The computed left singular vector has an angular error at most about
where is the relative gap between and the nearest other singular value. The same bound applies to the right singular vector and vi. Since the relative gap may be much larger than the absolute gap, this error bound may be much smaller than the previous one. The relative gaps may be easily computed from the array of computed singular values.
In the very special case of 2-by-2 bidiagonal matrices, xBDSQR and xBDSDC call auxiliary routine xLASV2 to compute the SVD. xLASV2 will actually compute nearly correctly rounded singular vectors independent of the relative gap, but this requires accurate computer arithmetic: if leading digits cancel during floating-point subtraction, the resulting difference must be exact. On machines without guard digits one has the slightly weaker result that the algorithm is componentwise relatively backward stable, and therefore the accuracy of the singular vectors depends on the relative gap as described above.
Jacobi's method [34,99,91] is another algorithm for finding singular values and singular vectors of matrices. It is slower than the algorithms based on first bidiagonalizing the matrix, but is capable of computing more accurate answers in several important cases.
Next: Error Bounds for the Up: Error Bounds for the Previous: Error Bounds for the Contents Index
Susan Blackford
1999-10-01 | 970 | 4,588 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2019-22 | latest | en | 0.910126 |
https://www.mrexcel.com/archive/formulas/another-brainteaser/ | 1,542,402,493,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743184.39/warc/CC-MAIN-20181116194306-20181116220306-00103.warc.gz | 935,203,416 | 9,875 | # Another Brainteaser
Posted by Mark W. on January 10, 2001 2:48 PM
Write a formula using only built-in Excel functions
that returns a boolean value that indicates if a word
in cell A1 is an anagram of another word in cell B1.
For example, if A1 contained "slime" and B1 contained
"smile" the formula would return TRUE (other examples
include: "looter","retool"; "looks","spook"). Your
solution should work with words of any length.
Posted by Aladin Akyurek on January 10, 2001 4:26 PM
Array-enter:
=AND(LEN(A1)=LEN(B1)*(SUM(CODE(LOWER(MID(A1,ROW(INDIRECT("1:"&LEN(A1))),1))))=SUM(CODE(LOWER(MID(B1,ROW(INDIRECT("1:"&LEN(B1))),1))))))
Hope it works.
Posted by dd on January 11, 2001 12:36 AM
No, doesn't work. For example :-
"ad" is equivalent to 197 and "bc" is also equivalent to 197.
Posted by Mark W. on January 11, 2001 3:24 AM
dd is right! Also, your outer AND() function is
superfluous. Care to try again?
Posted by Mark W. on January 11, 2001 3:26 AM
dd, would you like to take a stab at it?
Posted by dd on January 11, 2001 3:32 AM
dd, would you like to take a stab at it?
Still grinding the knife.
Posted by Tim Francis-Wright on January 11, 2001 7:34 AM
Write a formula using only built-in Excel functions
=IF(LEN(A1)<>LEN(B1),FALSE,SMALL(CODE(LOWER(MID(A1,ROW(INDIRECT("1:"&LEN(A1))),1))),ROW(INDIRECT("1:"&LEN(A1))))=SMALL(CODE(LOWER(MID(B1,ROW(INDIRECT("1:"&LEN(B1))),1))),ROW(INDIRECT("1:"&LEN(B1)))))
?
Posted by dd on January 11, 2001 8:45 AM
Yes! That's great!
I was trying a different route. Couldn't get it down to one formula. Purely for interest, is there any way of getting one formula out of the following?
If one word is in A1 and the other in A2, enter in B1 and fill to B26 :
=IF(LEN(UPPER(\$A\$1))-LEN(SUBSTITUTE(UPPER(\$A\$1),CHAR(ROW()+62),""))=LEN(UPPER(\$A\$2))-LEN(SUBSTITUTE(UPPER(\$A\$2),CHAR(ROW()+62),"")),TRUE,FALSE)
In C1 enter =IF(COUNTIF(B1:B26,TRUE)=26,TRUE,FALSE)
Posted by Mark W. on January 11, 2001 11:40 AM
Yep! It works!!!
It works!!!
Posted by Mark W. on January 11, 2001 11:42 AM
Here's what I came up with...
{=AND(LEN(A1)=LEN(B1),ISNUMBER(SEARCH(MID(A1,ROW(INDIRECT("1:"&LEN(A1))),1),B1)))}
Posted by Tim Francis-Wright on January 11, 2001 12:34 PM
Back to the drawing board... anyone have a solution?
{=AND(LEN(A1)=LEN(B1),ISNUMBER(SEARCH(MID(A1,ROW(INDIRECT("1:"&LEN(A1))),1),B1)))}
This gives some false positives:
A1 = state and
B1 = sates
returns TRUE (because each letter in A1 is in B1.
Unfortunately, the same problem plagues my answer
(for reasons I don't quite fathom). Any ideas?
Posted by Mark W. on January 11, 2001 12:50 PM
Good grief!!! Processing...
...
Posted by Tim Francis-Wright on January 11, 2001 12:55 PM
From the drawing board:
=(LINEST(SMALL(CODE(LOWER(MID(A1,ROW(INDIRECT("1:"&LEN(A1))),1))),ROW(INDIRECT("1:"&LEN(A1)))),SMALL(CODE(LOWER(MID(B1,ROW(INDIRECT("1:"&LEN(A1))),1))),ROW(INDIRECT("1:"&LEN(A1)))),FALSE)=1)*(LEN(A1)=LEN(B1))=1
LINEST(yarr, xarr, FALSE) fits the data to
y = mx + b. The FALSE argument fixes b = 0.
In order for m to equal 1, I believe that the
sorted elements must equal each other.
Can anyone prove or disprove this? Or come
up with a better way of comparing the two arrays?
Posted by Mark W. on January 11, 2001 1:00 PM
Here's My Fix...
{=NOT(ISERROR(AND(SEARCH(MID(A7,ROW(INDIRECT("1:"&LEN(A7))),1),B7)=SEARCH(MID(B7,ROW(INDIRECT("1:"&LEN(B7))),1),A7))))}
Posted by Mark W. on January 11, 2001 1:07 PM
Not Quite!!
...still doesn't catch {"state","sates"}!
Posted by dd on January 11, 2001 4:52 PM
Re: Not Quite!!
...still doesn't catch {"state","sates"}!
Ian's seems to be the one. However, Aladin's formula could be incorporated with yours but I'm still not sure if it would then cover all situations:
=AND(LEN(A1)=LEN(B1),ISNUMBER(SEARCH(MID(A1,ROW(INDIRECT("1:"&LEN(A1))),1),B1)),SUM(CODE(LOWER(MID(A1,ROW(INDIRECT("1:"&LEN(A1))),1))))=SUM(CODE(LOWER(MID(B1,ROW(INDIRECT("1:"&LEN(B1))),1))))) | 1,350 | 3,961 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2018-47 | latest | en | 0.899138 |
https://howto.org/how-many-grams-is-2-cups-of-milk-67212/ | 1,726,891,575,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701427996.97/warc/CC-MAIN-20240921015054-20240921045054-00819.warc.gz | 275,607,140 | 12,491 | ## What are 2 cups of milk?
US cooking measurements vs UK cooking measurements
US cups US fl oz UK ml
2 cups 16 fl oz = 1 US pint 450 ml
500 ml
3 cups 750 ml
4 cups 32 fl oz = 2 US pints 1 litre
## How many grams Makes 2 cups?
Packed Brown Sugar
Cups Grams Ounces
1/4 cup 55 g 1.9 oz
1/3 cup 73 g 2.58 oz
1/2 cup 110 g 3.88 oz
1 cup 220 g 7.75 oz
Nov 19, 2020
## How much is 250 grams of milk in cups?
Water
WATER – GRAMS TO CUPS
Grams Cups
200g ¾ cup + 1 tbsp
250g 1 cup + 1 tbsp
300g 1¼ cups
Sep 20, 2018
## How much does 1 cup of 2% milk weigh in grams?
Rather they refer to what percentage of the total weight is milk fat. For example, one cup of milk weighs about 225 grams. Of that weight, 2% milk holds 5 grams of fat and whole milk contains 8 grams.
## How much is a cup of milk in grams?
240 grams
Common Measurement Conversions for Baking
1 cup flour 120 grams 4.2 oz
1 cup whole milk 240 grams 8.6 oz
1 cup sour cream 240 grams 8.6 oz
1 cup sugar 200 grams 7.1 oz
1 cup brown sugar 220 grams 7.8 oz
## How many grams is a cup of liquid?
240 grams
1 US cup (c) of water = 240 milliliters (mL) or 240 grams (g).
## How much milk is in a cup?
Volume Equivalents (liquid)*
16 tablespoons 1 cup 8 fluid ounces
2 cups 1 pint 16 fluid ounces
2 pints 1 quart 32 fluid ounces
4 quarts 1 gallon 128 fluid ounces
## What is 100ml milk?
So, in this case 100 ml milk = 100×1.03 = 103 gram milk.
## How many grams is 3/4 cup milk?
Cups Grams Ounces
1/2 c 170 g 6 oz
2/3 c 227 g 8 oz
3/4 c 255 g 9 oz
1 c 340 g 12 oz
## How much is milk?
January 2022 Highlights: U.S. simple average prices are: \$3.82 per gallon for conventional whole milk, \$3.78 per gallon for conventional reduced fat 2% milk, \$4.26 per half gallon organic whole milk, and \$4.26 per half gallon organic reduced fat 2% milk.
## How many grams is a serving of milk?
One cup (249 grams) of whole cow’s milk with 3.25% fat provides ( 1 ): Calories: 152. Water: 88% Protein: 8.14 grams.
## When a recipe says 1 cup How much is that?
Dry/Weight Measure
Ounces
10 tablespoons plus 2 teaspoons 2/3 cup 5.2 ounces
12 tablespoons 3/4 cup 6 ounces
16 tablespoons 1 cup 8 ounces
32 tablespoons 2 cups 16 ounces
## How much is a jug of milk?
According to the USDA, the average gallon of whole milk is \$3.59 a gallon.
## Why is milk 2%?
2% means that the entire weight of the milk contains 2% milkfat. The dairy processor skims the fat off the top of the vat, and adds it back in, after calculating the weight of fat needed. Excess fat is turned into butter or cream.
## How much is milk in New York?
Cost of Living in New York
Restaurants Edit
Milk (regular), (1 gallon) 4.55\$
Loaf of Fresh White Bread (1 lb) 3.65\$
Rice (white), (1 lb) 2.42\$
Eggs (regular) (12) 3.38\$
## What size are milk jugs?
Most milk jugs are one of two sizes, 12 oz and 20 oz. However, it’s possible to find even smaller or larger pitchers, should your coffee bar need them. Generally speaking, the 12 oz and 20 oz jugs should have similar base sizes, so width shouldn’t come into that choice.
## How much is a gallon of milk weigh?
8.6 lb
A gallon of milk weighs 8.6 lb, and a quart of milk weighs 2.15 lb.
## Is one jug of milk a gallon?
A jug of milk isn’t exactly one gallon, nor is a bottle of wine exactly 750 milliliters. [1] A US gallon is 231 cubic inches, and an inch is 2.54 centimeters. | 1,078 | 3,368 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-38 | latest | en | 0.850714 |
https://the-equivalent.com/equivalent-impedance-calculator/ | 1,679,491,162,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943809.76/warc/CC-MAIN-20230322114226-20230322144226-00221.warc.gz | 655,053,379 | 15,963 | # Equivalent impedance calculator
## How to find equivalent impedance?
Find the equivalent impedance between points A and B in the circuit given below and write it in exponential and polar form. . Solution to Example 1 Let ( Z_1 ) be the impedance of resistor R and hence ( Z_1 = R) Let ( Z_2 ) be the impedance of the capacitor ( C ) and the inductor ( L ) that are in parallel.
## How do you calculate circuit impedance?
• Since the value of frequency and inductor are known, so firstly calculate the value of inductive reactance X L: X L = 2πfL ohms.
• From the value of X L and R, calculate the total impedance of the circuit which is given by.
• Calculate the total phase angle for the circuit θ = tan – 1 (X L / R).
## How to calculate total impedance of two parallel conductors?
• Z = R + jX, where j is the imaginary component: √ (-1). Use j instead of i to avoid confusion with I for current.
• You cannot combine the two numbers. For example, an impedance might be expressed as 60Ω + j120Ω.
• If you have two circuits like this one in series, you can add the real and imaginary components together separately. …
## How to calculate exponent using calculator?
Using the Exponent Key. Suppose you want the value y x. On most calculators, you enter the base, press the exponent key and enter the exponent. Here’s an example: Enter 10, press the exponent key, then press 5 and enter. (10^5=) The calculator should display the number 100,000, because that’s equal to 10 5.
## How do you calculate equivalent impedance?
Hence the equivalent impedance, Zeq, of two parallel elements equals the product of the two impedances divided by their sum.
## What is the impedance of RC parallel circuit?
The impedance (Z) of a parallel RC circuit is similar to that of a parallel RL circuit and is summarized as follows: Impedance can be calculated directly from the resistance and capacitive reactance values using the equation.
## How do you combine parallel impedances?
2:0114:57Parallel Complex Impedances (Part 1 of 2) – YouTubeYouTubeStart of suggested clipEnd of suggested clipAnd may necessitate the use of parentheses. If we perform this calculation in one entry on theMoreAnd may necessitate the use of parentheses. If we perform this calculation in one entry on the scientific calculator.
## How do you find the impedance of a resistor and inductor in parallel?
Ohm’s Law for AC circuits: E = IZ ; I = E/Z ; Z = E/I. When resistors and inductors are mixed together in parallel circuits (just as in series circuits), the total impedance will have a phase angle somewhere between 0° and +90°. The circuit current will have a phase angle somewhere between 0° and -90°.
## What is the formula of impedance in a RC circuit?
A purely capacitive impedance will always have a phase angle of exactly -90° (ZC = XC Ω ∠ -90°). Ohm’s Law for AC circuits: E = IZ ; I = E/Z ; Z = E/I. When resistors and capacitors are mixed together in circuits, the total impedance will have a phase angle somewhere between 0°- and -90°.
## What is RC formula?
This transient response time, T, is expressed in seconds as τ= R.C, where R is the resistor value in ohms and C is the capacitor value in Farads. This then serves as the foundation for an RC charging circuit, with 5T standing for “5 x RC.”
## How do you add impedance in series and parallel?
The first step is to combine L and C2 as a series combination of impedances, by adding their impedances together. Then, that impedance will be combined in parallel with the impedance of the resistor, to arrive at another combination of impedances.
## Is impedance the same as resistance?
Impedance extends the concept of resistance to alternating current (AC) circuits, and possesses both magnitude and phase, unlike resistance, which has only magnitude. Impedance is a complex number, with the same units as resistance, for which the SI unit is the ohm (Ω).
## How do you calculate impedance of an RLC circuit?
The impedance of the circuit is the total opposition to the flow of current. For a series RLC circuit, and impedance triangle can be drawn by dividing each side of the voltage triangle by its current, I.
## What is impedance in RLC circuit?
Impedance is the term that describes the characteristics of an electronic component in resisting current flow. It sounds similar to resistance, but impedance is proportional to frequency change. An RLC circuit consists of a resistor, inductor, and capacitor.
## How do you calculate impedance of an inductor?
The inductor impedance calculator calculates the impedance of an inductor based on the value of the inductance, L, of the inductor and the frequency, f, of the signal passing through the inductor, according to the formula, XL= 2πfL.
## How do you calculate impedance in RL parallel circuit?
11:1412:15Parallel RL Circuit Impedance and Current – YouTubeYouTubeStart of suggested clipEnd of suggested clipUnderstand that those two currents are ninety degrees out of phase with each other. So we add themMoreUnderstand that those two currents are ninety degrees out of phase with each other. So we add them vectorially pythagoras lee and with our total current use Ohm’s law to derive our impedance.
## What is the impedance of an AC RC parallel circuit if the resistance is 12 ohms and the capacitive reactance equals 5 ohms?
What is the impedance of an ac RC parallel circuit if the resistance is 12 ohms and the capacitive reactance equals 5 ohms?…Exercise :: Capacitors – General Questions.A.0.2 ohmsB.3.5 ohmsC.4.6 ohmsD.13 ohms
## What is the admittance of parallel RC circuit?
Admittance ( Y ) : The admittance of a parallel circuit is the ratio of phasor current to phasor voltage with the angle of the admittance being the negative to that of impedance.
## How do you find the impedance of a series parallel circuit?
There are two strategies for calculating the total current and total impedance. First, we could calculate total impedance from all the individual impedances in parallel (ZTotal = 1/(1/ZR + 1/ZL + 1/ZC), and then calculate total current by dividing source voltage by total impedance (I=E/Z).
## What is the impedance of a capacitor?
The impedance of an ideal capacitor is equal in magnitude to its reactance, but these two quantities are not identical. Reactance is expressed as an ordinary number with the unit ohms, whereas the impedance of a capacitor is the reactance multiplied by -j, i.e., Z = -jX.
## How many ohms is a low impedance speaker?
3. Enter the impedance of the “low impedance” loudspeaker, typically either 4 ohms or 8 ohms. (LZ)
## What are the trade offs of a low Z speaker?
There are often performance trade offs associated with transformer insertion loss, bandwidth limitations and line losses. Of course the low Z speaker can sometimes suffer from extreme line losses, particularly when the resistance of the line becomes an appreciable percentage of the nominal impedance of the low Z speaker. In order to make a good comparison between low impedance loudspeaker systems and “high impedance”, or transformer based loudspeakers several user inputs are required.
## Can a high Z speaker be a transformer?
One thing to remember is that the high Z speaker really limits you to the power rating of the transformer, where the low Z speaker power can be increase to near the power rating of the speaker enclosure itself.
## Example of Parallel Impedances
There are 4 parallel impedances in the circuit below,
In complex form, the impedances Z 1 , Z 1 2 , Z 3 and Z 4 are written as
Z 1 = R 1 + 0 j
Z 2 = 0 + ω L 1 j
Z 3 = R 2 − 1 ω C 1 j
Z 4 = R 3 + ( ω L 2 − 1 ω C 1) j
where j is the imaginary unit.
The equivalent impedance Z seen between points A and B is given by
Z = 1 1 Z 1 + 1 Z 2 + 1 Z 3 + 1 Z 3
The numerical values of the resistances, capacitances and inductances included in the circuit and the frequency of the input voltage are needed to express the impedances numerically and use them in the calculator below.
More AC circuits calculators and solvers are included..
## How to Use the Calculator
Enter the number n of impedances in parallel as a whole number and press “Enter”.
Then enter the values of the impedances as complex numbers of the form a + b j , where the real part a is on the left column of the table and the imaginary part b is on the right column of the table and press “Update/Calculate”.
The outputs include all the impedances entered which may be checked and modified if necessary as well as the equivalent impedance Z in complex standard and polar forms.
Note that each impedance must have a real part and an imaginary part and if one of the two parts is equal to 0 you must enter the number 0 for that part.
Number of Impedances: n =.
## How does the power system equivalent impedance affect hybrid filter compensation?
The influence of the power system equivalent impedance on the hybrid filter compensation performance is related with its effects on the passive filter, since if the system equivalent impedance is lower compared to the passive filter equivalent impedance at the resonant frequency , most of the load current harmonics will flow mainly to the power distribution system. In order to compensate this negative effect on the hybrid filter compensation performance, K must be increased, as shown in Eq. (39.57), increasing the active power filter rated power.
## What is the equivalent reactance of a circuit?
The equivalent reactance of the whole circuit is positive, indicating an effective inductive reactance so that the overall power factor is lagging and its value is given by cos ϕ = cos 17.36 = 0.9544.
## How does the system equivalent impedance affect the relation between the system current THD and the active filter gain?
41.57 shows how the system equivalent impedance affects the relation between the system current THD with the active filter gain, K, in a power distribution system with passive filters tuned at the fifth and seventh harmonics. If Zs decreases , the current system THD increases, so in order to keep the same compensation performance of the hybrid scheme , the active power filter gain, K, must be increased. On the other hand, if Zs is high, it is not necessary to increase K in order to ensure a low THD value in the system current.
## What is the maximum allowable fault current with the two auxiliaries transformers operating under the minimum impedance condition?
The maximum allowable fault current with the two auxiliaries transformers operating under the minimum impedance condition in parallel=12.5−2.274=10.226 kA. (Note: The transformers in this example were specified at the outset with an internal design to cater for later unforeseen load growth by the addition of oil pumps and forced oil cooling to give a possible future 45 MVA OFAF rating.)
## What is Figure 14.19?
Figure 14.19 shows the reduced network equivalent impedance for breaking and making duties. Chapter 1 describes simplifications that may be used for hand network reduction calculations. Neglecting the system resistance the reactance of the auxiliaries transformers is calculated as follows:
## What is the fault level of a 132 kV transformer?
The maximum source fault level on the 132 kV primary side of the transformers is 2,015 MVA. If XT is the transformer impedance on a 45 MVA base, then:
## What principle is used to find the required circuit variable?
In Problems 24–29, use the superposition principle to find the required circuit variable.
## How to tell the impedance of a capacitor?
Reactance is expressed as an ordinary number with the unit ohms, whereas the impedance of a capacitor is the reactance multiplied by -j, i.e., Z = -jX.
## What is reactance in AC?
Reactance is a more straightforward value; it tells you how much resistance a capac itor will have at a certain frequency. Impedance, however, is needed for comprehensive AC circuit analysis. As you can see from the above equation, a capacitor’s reactance is inversely proportional to both frequency and capacitance: higher frequency …
## Formulae for series RLC Circuit Used in the Calculator and their Units
We first give the formulas used in the series RLC calculator and the proof of these formulas is presented in the bottom part of the page.
## Use of the calculator
Enter the resistance, the capacitance, the inductance and the frequency as positive real numbers with the given units then press “calculate”.
## Formulae for Parallel R C Circuit Impedance Used in the Calculator and their Units
We first give the formulas used in the parallel RC calculator and the proof of these formulas is presented in the bottom part of the page.
## Use of the calculator
Enter the resistance, the capacitance and the frequency as positive real numbers with the given units then press “calculate”.
## Formulae for Parallel LC Circuit Impedance Used in Calculator and their Units
Let f be the frequency, in Hertz, of the source voltage supplying the circuit.
## Use of the calculator
Enter the resistance, the capacitance and the frequency as positive real numbers with the given units then press “calculate”. | 2,936 | 13,151 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2023-14 | latest | en | 0.835191 |
https://calculate.co.nz/logistic-regression-calculator.php | 1,720,836,624,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514484.89/warc/CC-MAIN-20240713020211-20240713050211-00344.warc.gz | 124,055,244 | 10,819 | # Logistic Regression Calculator
In certain situations, the linear regression model isn't appropriate. You may find yourself in a situation where the predicted values are probabilities and thus must always be between zero and one. A linear regression assumes no such restriction. A logistic regression does.
The best way to explain logistic regression analysis may be with an example. Say we are interested in modelling the probability of a newborn having a birth defect given a mother's age. One could survey a number of mothers at various ages (X in our table), counting both those with normal births (coded as 0 in our table) and those with abnormal ones (coded as 1 in our table).
After collecting some data, it is time to run the logistic regression. To do this you need the total of all observations (both normal and abnormal births), the probability of observing some phenomenon (a birth defect in our case), the odds of occurence (infants with birth defects over those born normally), and the natural logarithm of the calculated odds.
A weighted linear regression is then produced using the total number of observations in each category (again, both normal births and those with defects per age) as weights. This linear regression is created with X (age) as the independent variable and the observed logarithm of the calculated odds. Again, using the total number of observations in each X (age) as weights.
This weighted linear regression produces an equation of the form ln(odds) = intercept + slope * X. (i.e., we regressed the Xs with the logarithm of the calculated odds). If we take both sides to be the exponent of e, we get odds = eintercept + slope * X. The probability of observing our phenomenon (again, a birth with a defect in this case) is then equal to odds / (1 + odds).
Notice from the graph that our logistic model appears to do a very good job of modelling our actual data. In fact, you can't even see the actual (blue dot) data behind its predicted counterparts (orange dots). But they are there. Also notice that the data is somewhat sparce. We have no probability of birth defects corresponding to women aged 39, for instance. But now we have a logistic equation we can use to help us predict what that probability would be. We just put an age (39) in for X and the model equation will generate the corresponding probability: e-13.595 + 0.237 * (39) / (1 + e-13.595 + 0.237 * (39) ) = 0.013. So, we can say that a woman of age 39 can expect to have a 1.3% chance of having a child with a birth defect.
Enter the number of data points you have below, then the table will show. Then enter your observed Xs, Ys coded as 0 and Ys coded as 1. Your results will automatically be generated.
Enter the number of data points:
I II III IV V VI VII VIII
X
Instances of Y Coded as 0 1
Total II + III Y as Observed Probability Y as Odds Y as Log Odds Predicted Probability | 651 | 2,899 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-30 | latest | en | 0.947511 |
https://ncatlab.org/nlab/show/differential+graded+Hopf+algebra | 1,721,289,400,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2024-30/segments/1720763514826.9/warc/CC-MAIN-20240718064851-20240718094851-00526.warc.gz | 381,368,208 | 11,173 | # nLab differential graded Hopf algebra
Contents
under construction
### Context
#### Higher algebra
higher algebra
universal algebra
# Contents
## Definitions
A $\mathbb{Z}$-graded Hopf algebra (pre-gha) is a $\mathbb{Z}$-graded vector space, which, for that grading, is both a $\mathbb{Z}$-graded algebra, $(A,\mu)$, with unity, $\eta : K \to A$, and a $\mathbb{Z}$-graded coalgebra $(A, \Delta, \varepsilon)$ such that:
• $\eta : K \to A$ is a morphism of $\mathbb{Z}$-graded coalgebras;
• $\varepsilon : A \to K$ is a morphism of $\mathbb{Z}$-graded algebras;
• $\mu : A \otimes A \to A$ is a morphism of $\mathbb{Z}$-graded coalgebras
Remark
We can replace the third condition by:
• $\Delta : A \to A \otimes A$ is a morphism of $\mathbb{Z}$-graded algebras.
Of course, wherever possible, we will abbreviate $(A,\Delta,\mu,\epsilon,\eta)$ to $A$.
A homomorphism of pre-ghas is a linear map of degree zero compatible with both the algebra and coalgebra structures. We may write $pre GHA$ for the resulting category.
If $A$ and $A'$ are two pre-ghas, $A\otimes A'$ is a pre-gha for the algebra and coalgebra structures already defined.
### Derivations on Hopf algebras
Let $A$ be a pre-gha. A Hopf algebra derivation of $A$ of degree $p\in \mathbb{Z}$ is a linear mapping $\theta \in Hom_p(A,A)$, defining both an algebra and a coalgebra derivation.
A differential $\partial$ of pre-ghas is a Hopf algebra derivation of degree -1 such that $\partial\circ \partial = 0$. The pair $(A,\partial)$ is called a differential $\mathbb{Z}$-graded Hopf algebra (pre-dgha). Its homology $H(A,\partial)$ is also a pre-gha. A morphism of pre-dghas is a morphism, at the same time, of pre-ghas and pre-dgvs. This gives a category $pre DGHA$.
A pre-gha $(A,\Delta,\mu,\epsilon,\eta)$ is commutative if $(A,\mu)$ is commutative and is cocommutative if $(A,\Delta,\varepsilon)$ is cocommutative.
This gives categories $pre CDGHA$ and $pre CoDGHA$ respectively.
A cocommutative (resp. commutative) dgha is an object of $pre CoDGHA$ (resp. $pre CDGHA$, which has a lower (resp. upper) grading.
A cocommutative (resp. commutative) dgha $A$ is $n$-connected if $\bar{A}_p = 0$ (resp $\bar{A}^p = 0$) for $p\leq n$.
### Shuffle product on $T(V)$
Let $V$ be a pre-gvs. The gvs $T(V)$ is a pre-cga for the shuffle product defined by
$(v_1\otimes \ldots \otimes v_p)\star (v_{p+1}\otimes\ldots \otimes v_n) = \sum_\sigma \varepsilon(\sigma)v_{\sigma^{-1}(1)}\otimes\ldots \otimes v_{\sigma^{-1}(n)},$
where the sum is over all $(p,n-p)$ shuffles, $\varepsilon(\sigma)$ is the Koszul sign of $\sigma$ and the elements $v_i$ of $V$ are all homogeneous.
### Graded-commutative Hopf algebra structure on $T(V)$
The underlying algebra structure is $T(V)$ with the shuffle product. The reduced diagonal is given by
$\bar{\Delta}(v_1\otimes \ldots \otimes v_n) = \sum_{p=1}^{n-1} (v_1\otimes \ldots \otimes v_p)\otimes(v_{p+1}\otimes \ldots \otimes v_n).$
### Graded-cocommutative Hopf algebra structure on $T(V)$
The underlying algebra structure this time is $T(V)$ with the usual product
$(v_1\otimes \ldots \otimes v_p)\cdot(v_{p+1}\otimes \ldots \otimes v_n) = v_1\otimes \ldots \otimes v_p\otimes v_{p+1}\otimes \ldots \otimes v_n,$
but with the reduced diagonal given by
$\bar{\Delta}(v_1\otimes \ldots \otimes v_n) = \sum_{p=1}^{n-1}\sum_\sigma \varepsilon(\sigma) (v_{\sigma(1)}\otimes \ldots \otimes v_{\sigma(p)})\otimes(v_{{\sigma(p+1)}}\otimes \ldots \otimes v_{\sigma(n)}),$
where the sum is over all $p$ and all $(p,n-p)$-shuffles and, as usual, $\varepsilon(\sigma)$ is the Koszul sign.
The diagonal $\Delta$ is thus defined by the conditions
• $\Delta v = v\otimes 1 + 1\otimes v$ if $v \in V$;
• $\Delta$ is a morphism of $\mathbb{Z}$-graded algebras.
A commutative and cocommutative $\mathbb{Z}$-graded Hopf algebra structure on $\bigwedge V$ is obtained by using the algebra and coalgebra structures defined in differential graded algebra and differential graded coalgebra. respectively.
### The enveloping algebra of a Lie algebra, $U(L)$.
Let $L$ be a pre-gla, $U(L)$, is the quotient algebra of the tensor algebra $T(L)$ by the two sided ideal generated by the elements
$x\otimes y - (-1)^{|y||x|}y\otimes x - [x,y], \quad x,y,\in L.$
The diagonal $\Delta : L \to L\times L$, with $\Delta(x) = (x,x)$ defines a homomorphism of pre-gas,
$U(\Delta) : U(L)\to U(L\times L) \cong U(L)\otimes U(L),$
which makes $U(L)$ a pre-gha which is cocommutative and conilpotent.
If $L$ is a free Lie algebra on $V$, then the enveloping algebra is the tensor algebra: $U\mathbb{L}(V) \cong T(V)$.
Let $(L,\partial)$ be a pre-dgla, the differential $\partial$ extends to an algebra differential on $T(L)$. With the quotient differential, $U(L)$ becomes a cocommutative pre-dgha, which will be denoted $U(L,\partial)$.
The differential $\partial$ determines a differential, also denoted $\partial$, on the cocommutative pre-gca $\bigwedge' L$, (for which gca see differential graded coalgebra). It satisfies:
$\bigwedge' H(L,\partial) \cong H(\bigwedge' L,\partial).$
Let $i : L \to U(L)$ be the linear mapping $L\to T(L) \to U(L)$, then define $e: \bigwedge' L \to U(L)$ by
$e(x_1\wedge \ldots x_n) = \frac{1}{n!}\sum_\sigma \varepsilon(\sigma)i(x_{\sigma(1)})\ldots i(x_{\sigma(n)}),$
where the sum is over all permutations and $\varepsilon(\sigma)$ is the Koszul sign.
###### Theorem
(Poincaré-Birkhoff-Witt)(cf. Quillen)
The mapping $e$ is an isomorphism of pre-dgcas.
###### Corollary
$i : L \to U(L)$ defines an isomorphism between $L$ and the space of primitives of $U(L)$.
###### Corollary
The natural map $UH(L,\partial)\to H(U(L,\partial)$ is an isomorphism of cocommutative pre-ghas.
### The Lie algebra of primitives, $P$
Let $(A,\partial)$ be a cocommutative pre-dgha. The vector space $P(A)$ of primitive elements (for the coalgebra structure, cf. differential graded coalgebra), is not stable under the multiplication, however the commutator $[\alpha,\beta]$ of two elements of $P(A)$ is again in $P(A)$. This defines a pre-gla structure on $P(A)$ and we can put the induced differential on it to obtain $P(A,\partial)$.
The inclusion $P(A)\to A$ extends to a morphism of cocommutative pre-dghas $\sigma: UP(A)\to A.$
###### Theorem (Quillen, Quillen)
If $A$ is conilpotent, $\sigma$ is an isomorphism.
The above theorem and earlier corollary show that $U$ and $P$ are inverse equivalences between the category, $pre DGLA$ and that of cocommutative, conilpotent pre-dghas.
Remark
The enveloping algebra of a free Lie algebra $\mathbb{L}(V)$ coincides with the tensor algebra, $T(V)$. It is conilpotent from which one gets $PT(V) = \mathbb{L}(V)$.
## References
• D. Tanré, Homotopie rationnelle: Modèles de Chen, Quillen, Sullivan, Lecture Notes in Maths No. 1025, Springer, 1983.
• Daniel Quillen, Rational Homotopy Theory, Ann. of Math., (2) 90 (1969), 205-295.
Discussion of bar-cobar construction for dg-Hopf algebras:
Last revised on June 11, 2022 at 10:57:18. See the history of this page for a list of all contributions to it. | 2,298 | 7,096 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 123, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-30 | latest | en | 0.839763 |
https://www.vogelsxxl.nl/07V1uF | 1,597,389,842,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739182.35/warc/CC-MAIN-20200814070558-20200814100558-00221.warc.gz | 851,233,486 | 9,067 | # Calculation Of Steam Bagasse Based Boiler, diagram of horizontal boiler
## IEEE Latin America Transactions: Mathematical Modeling Based on Exergy Analysis for a Bagasse Boiler
Modelación Matemática basada en el Análisis de Exergía para una Caldera de Bagazo (Mathematical Modeling Based on Exergy Analysis for a Bagasse Boiler) Ducardo Leon Molina ([email protected]) 1, Juan Ricardo Vidal ([email protected]) 1, Felix González ([email protected]) 2 ...
## A Method for Exergy Analysis of Sugarcane Bagasse Boilers
Abstract - This work presents a method to conduct a thermodynamic analysis of sugarcane bagasse boilers. The method is based on the standard and actual reactions which allows the calculation of the enthalpies of each process subequation and the exergies of each of the main flowrates participating in the combustion.
## Calculation Of Steam Bagasse Based Boiler In Uae
Calculation Of Steam Bagasse Based Boiler In Uae. If you run into any problems with the boiler, please contact us directly by email [email protected] or fill out the online service request. We will do our best to resolve any problems you might have within 24 hours.
## TWO PROPOSALS TO DETERMINE THE EFFICIENCY OF BAGASSE BOILER
based on HHV, highlights the effect of bagasse moisture content upon boiler efficiency. This effect, in the Beatón and Lora proposal, is hidden, because the energy required to evaporate the bagasse moisture content and the water vapour from hydrogen contained in the fuel are discounted in the LHV calculation. Three types of boilers, with ...
## Efficient cogeneration scheme for sugar industry
Steam turbine cogeneration system can offer a large range of ... condenser, 75 TPH; and steam / bagasse ratio, 2.43. Old boilers are to be replaced by a new high-pressure boiler . ... steam to process). Process Heat Calculation Steam values are as follows: steam entering into turbine, 64TPH; steam extraction at 8 ata, 5 TPH; steam ...
## Performance of Wet and Dry Bagasse Combustion in Assalaya
Abstract This Paper aims to study performances of dry and wet bagasse combustion to generate 51.58 kg/s of superheated steam in Assalaya Sugar Factory thermal power plant, wherefore, 26 kg/s of bagasse flows with constant rate (proposed design). Boilers
## Sample calculations of Boiler Pumps and ID / FD Fans
Aug 13, 2008 · Sample sizing calculations for BFW pumps and Fans for a typical Coal fired Boiler generating steam of 50,000 Kg/hr at 67 kg/cm2 and 485 degC. (110,000 lb/hr at 950 PSI & 905 F). Feed Water inle…
## cost calculation for high pressure bagasse fired boilers – Industrial Boiler
calculation of steam bagasse based flow while corrected for sudden Bagasse based highpressure co bagasse fired CFB boilers,Bagasse Analysis of New Boiler Technologies Thermal Dr Mike Inkson … File Format: PDF/Adobe Acrobat Analysis of New Boiler Technologies the additional capital cost outstrips the additional A bagasse fired boiler is really a plant in its own
## bagasse fired boiler best boiler
bagasse fired boilers best residential hot water. vertical bagasse fired hot water boiler prices for laundry stoker 1 ton steam boiler price for slaughter plant The best fuel for steam boiler is rice husk, which can save cost. manufacturer of vertical (floor standing
## Bagasse Cogeneration Plant Manufacturer
Recognizing this, Shrijee group, took to the cause of producing energy from bagasse based cogeneration power plant. Until a few years back, bagasse,( the sugar cane waste after extracting juice to make sugar ), was burnt as fuel in the boilers to generate steam and power just sufficient for captive consumption for the plant.
## List of Steam Boiler Calculator | Boiler Calculators & Formulas
Steam Boiler Calculator Based on Formulas Our Steam Boiler Calculators making boiler system related calculations easy, the thermodyne team created several online calculators and tools. Use following Essential industrial Steam boiler calculator to solve boiler industry related calculations for free. ...
## Cogeneration of Bagasse
The prime technology for sugar mill cogeneration is the conventional steam-Rankine cycle design for conversion of fuel into electricity. A combination of stored and fresh bagasse is usually fed to a specially designed furnace to generate steam in a boiler at typical
## PDF Properties and Operating Experience with Bagasse as a Boiler Fuel
PROPERTIES AND OPERATING EXPERIENCE WITH BAGASSE AS A BOILER FUEL T. N. ADAMS The University of British Columbia Vancouver, B.C., Canada ABSTRACT Data obtained in two investigations of bagasse as a boiler fuel are interpreted to obtain informa tion on the properties and combustion of bagasse. Data from these investigations show that over a
## Efficiency Calculations of Bagasse Fired Boiler on the Pages 1
Efficiency Calculations of Bagasse Fired Boiler on the Basis of Flue Gases ... calorific value of bagasse. A considerable loss (c), which remains to be (Pak. j. life soc. sci. (2004
## Encontro de Energia no Meio Rural
This work analyzes and compares two proposals for determination of the bagasse boiler efficiency, one of it based on bagasse higher heating value (HHV), the other one based on bagasse lower heating value (LHV). The methodology of calculation, for both proposals, uses the heat loss method.
## wet bagasse boiler
1/29/2019 · Bagasse cogeneration (which is the term most countries use for combined heat and power) was initiated in Mauritius and Hawaii.Steam Superheater in Boiler: Role, Benefits, Types …2019-1-24 · Steam superheater is a coil type heat exchanger which is used to produce superheated steam or to convert the wet steam to dry steam, generated by a boiler.
## Performance evaluation Of Boiler In 46mw Bagasse Based
compared three different boiler configurations viz. i) boiler with no air preheater or bagasse dryer ii) boiler with air preheater and iii) boiler with bagasse dryer. He showed that the overall increase in steam production using a bagasse dryer is not
## Calculation Of Steam Bagasse Based Boiler
Steam Calculators: Boiler Calculator - US Department of Boiler Calculator watch tutorial view guide Determines the amount of fuel energy required to produce steam with specified properties at a given flow rate using general boiler operational characteristics.
## 2001 wienese BOILERS, BOILER FUEL AND
boiler fuel and discusses the calculation of boiler efficiency. The figures that are used are generic and are not to be taken definitively. Boilers Typical modern boilers in the South African sugar industry pro-duce superheated steam at a pressure of 3100 kPa (abs) and a temperature of 400°C. They are designed to burn bagasse, coal
## Biomass Power: High Efficiency Boiler Technology for Sugar
Advantages of High Pressure Boiler (Calculations shown above are based on a 100 TPH Travelling Grate boiler with 69% efficiency (on GCV 2270 Kcal/Kg Basis) and Turbine exhaust at 0.1 Kg/Cm2 ( a) 90% of sugar mill boilers in Thailand are below this category 9
## Efficiency Calculations of Bagasse Fired Boiler on the
Pak. j. life soc. sci. (2004), 2(1): 36-39 Pakistan Journal of Life and Social SciencesEfficiency Calculations of Bagasse Fired Boiler on the Basis of Flue GasesTemperature and Total Heat Values of SteamAnjum Munir, A.R.Tahir, M.Shafi Sabir, Khuram EjazDepartment of Farm Machinery and Power, University of Agriculture, Faisalabad-PakistanAbstract Moreover, to meet World Trade Organization (WTO ...
## avant garde bagasse fired power plants
Bio Mass Bagasse Boiler - lodgings-hamburg.de Bio-mass and Bagasse based energy issued during the year 2012 has depreciated the value of plant and machinery to 90% of the Get a Quote led to sub-optimal functioning of the bio-mass based power plants
## how to calculate oil fired boiler
How to Calculate 10 Ton Boiler Gas Consumption 0 5 … 2019-2-11 · 4 ton gas steam boiler fuel consumption. Taking 4 ton gas fired steam boiler fuel consumption for example, and introduce the calculation method of gas consumption, how much gas is consumed by 4 ton gas steam boiler, and how much natural gas is consumed by a 4 ton gas hot water boiler.. calculate cost of gas boiler 1.5 ton .
## lcv calculation example for boiler – Industrial Boiler Supplier
THERMAL POWER: Boiler Efficiency calculations For example, efficiency of a Bagasse fired boiler is about 70% where as that of oil fired boilers is about 85 %. Higher moisture content in Bagasse reduces it’s efficiency . So better criterion is efficiency based on ...
## Improve the Efficiency of Boiler by Reduce the Moisture in Bagasse
Improve the Efficiency of Boiler by Reduce the Moisture in Bagasse - written by Periyasamy K, R Karthikeyan, Gopinath R N published on 2018/04/24 download full article with reference data and citations Use Equipment Steam conditions Remarks Elect rical Pow
## How to Calculate Boiler Steam Flows
Use boiler horsepower to calculate steam flow. Boiler horsepower is not related to mechanical horsepower. It is a boiler industry rating that predicates the amount of saturated steam a boiler will generate starting with water at 212 degrees Fahrenheit and 0 pounds per square inch gauge (psig--meaning atmospheric pressure) and ending with steam at 212 F and 0 psig.
## CO27: DEVELOPMENT OF AN EFFICIENT SLOP FIRED BOILER
S ISGEC Heavy Engineering Ltd, Noida, India At : 29th ISSCT, Chiang Mai , Thailand, 7 Dec 2016 DEVELOPMENT OF AN EFFICIENT SLOP FIRED BOILER Speaker
## Energy, Environment & Efficiency Measures of Bagasse fired
an extra 6 TPH steam. Bagasse consumption has also reduced due to increase in boiler efficiency. 3. ID Fan installed with improved material of Hardox – 500 which have significantly wear resistance against erosion. 4. Bagasse to Steam ratio improved from 1.81 1.99 5. Bagasse Consumption reduce 1.69 TPH for unit TPH of Steam generation. 6.
## (PDF) Efficiency calculations of bagasse fired boiler on the
Ultimate analysis of bagasse was performed and the actual air supplied to the boiler was calculated to be 4.05 kg per kg of bagasse under the available resources of the plant.
## PERFORMANCE EVALUATION OF BAGASSE UNIT IN A SUGAR INDUSTRY; A
biofuel and in the manufacture of pulp but also in paper goods and building materials. The bagasse produced in a sugar factory is however used for generation of steam which in turn is used as a fuel source and the surplus generation is exported to the power grids of state governments. Bagasse burned in quantity produces adequate
## A METHOD FOR EXERGY ANALYSIS OF SUGARCANE BAGASSE BOILERS
Abstract - This work presents a method to conduct a thermodynamic analysis of sugarcane bagasse boilers. The method is based on the standard and actual reactions which allows the calculation of the enthalpies of each process subequation and the exergies of each of the main flowrates participating in the combustion.
## calculation of steam production for boiler from bagasse – CFBC
3.How about the quality of calculation of steam production for boiler from bagasse ? Our industrial steam boiler manufactured strictly according to national and international standard. our products have approved ASME, IBR, ISO9001-2008, etc. calculation of
## bagasse based 110 ton boiler suppliers – Industrial Boiler
7/12/2018 · 8t bagasse boiler – Fire Tube Boiler Supplier Page 8 Bagasse Boiler, Bagasse Boiler Suppliers and Manufacturers at Bagasse Boiler, … 1 ton gas-fired steam boiler 2 4 6 8 bagasse fired to 10 capacity. 38 ton per hour sugarcane bagasse fired boiler price – Water Tube … 4 нояб. 2017 г.
## calculation and design of sugarcane bagasse boiler in sugar
Background Document: Bagasse Combustion in Sugar Mills – epa nepis In order for the sugar cane mill to avoid a large solid waste disposal problem and to … in the reviewed emission tests generally use this type of furnace design. …. bagasse were calculated by dividing the steam load by 60% boiler efficiency, …
## A METHOD FOR EXERGY ANALYSIS OF SUGARCANE BAGASSE
The method is based on the standard and actual reactions which allows the calculation of the enthalpies of each process subequation and the exergies of each of the main flowrates participating in the combustion. The method is presented using an example with real data from a sugarcane bagasse boiler.
## Pollution Control Guidelines for Conversion of Boilers
make a policy for the conversion of Natural Gas based Boiler/Utility to Solid fuel based Boiler/Utility, which would at the same time address the issues related to Air pollution . in both, the areas covered underthe CEPI and rest of the areas without CEPI. This has lead to formation of a committee that would prepare guideline for this purpose.
## Boiler Efficiency: Introduction and Methods of Calculation
Shifting to real time efficiency monitoring can improve the boiler efficiency significantly depending upon the boiler type and actual conditions at site. In nutshell, monitoring and maintaining the boiler efficiency for the overall operational life of the boiler is a must to cut fuel bills and reduce carbon footprint.
## Calculator: Boiler Efficiency | TLV
Online calculator to quickly determine Boiler Efficiency. Includes 53 different calculations. Equations displayed for easy reference.
## How much fuel required for 1TPH boiler?
I will give you an example, so you can take this as reference and calculate fuel required for any capacity of Boiler for any fuel. Boiler Capacity - 1000 kg/hr Steam Pressure - 10.5 kg/cm2(g) Steam Temperature - 185.5 Deg c (Saturated Steam)- You ... | 3,019 | 13,597 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-34 | latest | en | 0.757654 |
https://www.onlinemath4all.com/Lagrange_theorem.html | 1,603,232,069,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107874340.10/warc/CC-MAIN-20201020221156-20201021011156-00516.warc.gz | 838,542,635 | 7,974 | ## Lagrange Theorem
In this page Lagrange theorem we are going to see how to check la grange's theorem in a function.
If f (x) be a real valued function that satisfies the following conditions.
1 f(x) is defined and continuous on the closed interval [a,b] 2 f(x) is differentiable on the open interval (a,b).
Then there exists at least one point c ∊ (a,b) such that
f ' (c) = f (b) - f (a) / (b - a)
Example 1:
Using Lagrange-theorem find the values of c.
f (x) = x² + 4 x + 3 , 1 ≤ x ≤ 3
Solution:
(i) f (x) is continuous on [1 ,3].
(ii) f(x) is differentiable (1,3).
f ' (x) = 2 x + 4 (1)
= 2 x + 4
f ' (c) = 2 x + 4
f(1) = 1² + 4 (1) + 3
= 1 + 4 + 3
f (1) = 8
f(3) = 3² + 4 (3) + 3
= 9 + 12 + 3
= 12 + 12
f (3) = 24
Here a = 1 and b = 3
f ' (c) = [f (b) - f (a)] / (b - a)
2x + 4 = (24 - 8)/(3 -1)
2x + 4 = 16/2
2x + 4 = 8
2x = 8 - 4
2x = 4
x = 4/2
x = 2 Lagrange theorem
Example 2:
Using Lagrange theorem find the values of c.
f (x) = 2x³ + x² - x - 1, 0 ≤ x ≤ 2
Solution:
(i) f (x) is continuous on [0 ,2].
(ii) f(x) is differentiable (0,2).
f ' (x) = 2(3x²) + 2 x - 1 - 0
= 6 x² + 2 x - 1
f (x) = 2x³ + x² - x - 1
f (0) = 2 (0)³ + (0)² - 0 - 1
f (0) = -1
f (2) = 2 (2)³ + (2)² - 2 - 1
= 2 (8) + 4 - 2 - 1
= 16 + 4 - 3
= 20 - 3
f (0) = 17
Here a = 0 and b = 2
f ' (c) = [f (b) - f (a)] / (b - a)
6 x² + 2 x - 1 = [17 - (-1)]/(0 + 2)
6 x² + 2 x - 1 = (17 +1)/2
6 x² + 2 x - 1 = 18/2
6 x² + 2 x - 1 = 9
6 x² + 2 x - 1 - 9 = 0
6 x² + 2 x - 10 = 0
3 x² + x - 5 = 0
This equation cannot be solved using a = 3, b = 1, c = -5
x = - 1 ± √ [(1)² - 4 (3) ( -5 )]/2 (3)
x = - 1 ± √ [1 + 64]/6
x = [- 1 ± √(65)]/6
x = (- 1 ± 8.06)/6
x = (- 1 + 8.06)/6 , x = (- 1 - 8.06)/6
x = 7.06/6 , x = - 9.06/6
x = 1.17 , x = - 1.51
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Aptitude Test | 1,010 | 2,120 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2020-45 | latest | en | 0.629292 |
https://www.storyofmathematics.com/fractions-to-decimals/2-64-as-a-decimal/ | 1,721,011,888,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514655.27/warc/CC-MAIN-20240715010519-20240715040519-00715.warc.gz | 872,199,800 | 39,081 | # What Is 2/64 as a Decimal + Solution With Free Steps
The fraction 2/64 as a decimal is equal to 0.031.
The division of two numbers p and q is one of the four primary arithmetic operations, the others being addition, subtraction, and multiplication. It is the inverse of multiplication, and therefore answers the question “how much are p parts of q?” The result of a division can be either an integer or decimal value.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 2/64.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 2
Divisor = 64
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 2 $\div$ 64
This is when we go through the Long Division solution to our problem.
Figure 1
## 2/64 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 2 and 64, we can see how 2 is Smaller than 64, and to solve this division, we require that 2 be Bigger than 64.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
However, in our case, 2 x 10 = 20, which is still smaller than 64. Therefore, we again multiply by 10 to get 20 x 10 = 200, which is bigger than 64. To indicate the double multiplication, we add a decimal point “.” followed by a 0 to our quotient.
Now, we begin solving for our dividend 2, which after getting multiplied by 10 becomes 200.
We take this 200 and divide it by 64; this can be done as follows:
200 $\div$ 64 $\approx$ 3
Where:
64 x 3 = 192
This will lead to the generation of a Remainder equal to 200 – 192 = 8. Now this means we have to repeat the process by Converting the 8 into 80 and solving for that:
80 $\div$ 64 $\approx$ 1
Where:
64 x 1 = 64
This, therefore, produces another Remainder which is equal to 80 – 64 = 16. We now have the three decimal places for our quotient, so we stop the division process. Our final Quotient is 0.031 with a final remainder of 16. | 726 | 2,973 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.9375 | 5 | CC-MAIN-2024-30 | latest | en | 0.922984 |
https://www.aimsciences.org/article/doi/10.3934/jimo.2020070?viewType=html | 1,631,809,112,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780053657.29/warc/CC-MAIN-20210916145123-20210916175123-00201.warc.gz | 665,519,179 | 20,208 | # American Institute of Mathematical Sciences
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September 2021, 17(5): 2307-2329. doi: 10.3934/jimo.2020070
## Computing shadow prices with multiple Lagrange multipliers
516 Jungong Road, Business School, University of Shanghai for Science and Technology, Shanghai 200093, China
* Corresponding author: Gao Yan
Received August 2019 Revised November 2019 Published September 2021 Early access March 2020
Fund Project: Tao Jie is supported by National Natural Science Foundation of China grant No. 71601117 and Soft Science Foundation of Shanghai grant No. 19692104600
There is a wide consensus that the shadow prices of certain resources in an economic system are equal to Lagrange multipliers. However, this is misleading with respect to multiple Lagrange multipliers. In this paper, we propose a new type of Lagrange multiplier, the weighted minimum norm Lagrange multiplier, which is a type of shadow price. An attractive aspect of this type of Lagrange multiplier is that it conveys the sensitivity information when resources are required to be proportionally input. To compute the weighted minimum norm Lagrange multiplier, we propose two algorithms. One is the penalty function method with numeric stability, and the other is the accelerated gradient method with fewer arithmetic operations and a convergence rate of $O(\frac{1}{k^2})$. Furthermore, we propose a two-phase procedure to compute a particular subset of shadow prices that belongs to the set of bounded Lagrange multipliers. This subset is particularly attractive since all its elements are computable shadow prices. We report the numerical results for randomly generated problems.
Citation: Tao Jie, Gao Yan. Computing shadow prices with multiple Lagrange multipliers. Journal of Industrial & Management Optimization, 2021, 17 (5) : 2307-2329. doi: 10.3934/jimo.2020070
##### References:
show all references
##### References:
Numerical Example in Schttfkowski (1987)
Relationship of Lagrange Multiplier and Shadow Price
Numerical Example
Convergence of the $\mathcal{PFA}$ algorithm with different penalty parameters
Example of Multiple Lagrange Multipliers
Computational Times of the $\mathcal{AGM}$ algorithm on Large - scale Data Sets
$m$ $n$ Computational Time 500 5000 10.7504 500 10000 11.4222 500 20000 12.5574 500 50000 15.5700 1000 5000 21.4259 1000 10000 21.8054 1000 20000 22.2733 1000 50000 25.2937 5000 5000 101.1437 5000 10000 102.1762 5000 20000 106.8786 5000 50000 107.3515
$m$ $n$ Computational Time 500 5000 10.7504 500 10000 11.4222 500 20000 12.5574 500 50000 15.5700 1000 5000 21.4259 1000 10000 21.8054 1000 20000 22.2733 1000 50000 25.2937 5000 5000 101.1437 5000 10000 102.1762 5000 20000 106.8786 5000 50000 107.3515
Result of the 2-phase Procedure with 23 Vertices of the Bounded Lagrange Multiplier Set
Vertices Elements $v_1$ 0.0000 1.6118 0.0000 0.0000 0.0000 0.0000 0.4939 0.0000 0.0000 $v_2$ 0.0000 2.1057 0.4939 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 $v_3$ 0.0000 3.0834 0.0000 0.0000 0.1352 0.0000 0.0000 0.6957 0.0000 $v_4$ 0.0000 3.4207 0.0000 0.0000 0.0026 0.0000 0.0064 0.7603 0.1801 $v_5$ 0.0000 5.7145 0.0000 0.0000 3.1147 2.5431 3.6277 0.0000 0.0000 $v_6$ 0.0000 7.1430 0.0000 4.9601 0.9637 0.0000 1.9331 0.0000 0.0000 $v_7$ 0.0000 7.2983 0.0000 0.0000 0.0000 1.7188 3.3700 0.0000 2.6128 $v_8$ 0.0000 7.5898 0.0000 4.3192 0.0000 0.0000 2.0493 0.0000 1.0416 $v_9$ 0.0000 7.7043 2.9184 0.0000 2.4098 1.9675 0.0000 0.0000 0.0000 $v_{10}$ 0.0000 8.1361 1.7390 4.2912 0.8338 0.0000 0.0000 0.0000 0.0000 $v_{11}$ 0.0000 8.5707 1.8287 3.7067 0.0000 0.0000 0.0000 0.0000 0.8939 $v_{12}$ 0.0000 8.8386 2.7554 0.0000 0.0000 1.3515 0.0000 0.0000 2.0545 $v_{13}$ 0.0000 9.0761 0.0000 3.0271 0.9637 0.0000 0.0000 1.9331 0.0000 $v_{14}$ 0.0000 9.1971 0.0000 0.0000 1.9422 1.1568 0.0000 2.7039 0.0000 $v_{15}$ 0.0000 9.6392 0.0000 2.2699 0.0000 0.0000 0.0000 2.0493 1.0416 $v_{16}$ 0.0000 10.0534 0.0000 0.0000 0.0000 0.6918 0.0000 2.5809 1.6740 $v_{17}$ 0.0000 3.4315 0.0050 0.0000 0.0027 0.0000 0.0000 0.7627 0.1807 $v_{18}$ 0.6494 10.2980 0.0000 0.0000 0.0000 0.0000 0.0000 2.4700 1.5827 $v_{19}$ 1.0475 9.6669 0.0000 0.0000 1.7714 0.0000 0.0000 2.5142 0.0000 $v_{20}$ 1.2187 9.3956 2.5332 0.0000 0.0000 0.0000 0.0000 0.0000 1.8526 $v_{21}$ 1.5166 8.1461 0.0000 0.0000 0.0000 0.0000 3.0317 0.0000 2.3055 $v_{22}$ 1.6981 8.6358 2.5864 0.0000 2.0798 0.0000 0.0000 0.0000 0.0000 $v_{23}$ 2.1242 7.1628 0.0000 0.0000 2.6016 0.0000 3.1114 0.0000 0.0000
Vertices Elements $v_1$ 0.0000 1.6118 0.0000 0.0000 0.0000 0.0000 0.4939 0.0000 0.0000 $v_2$ 0.0000 2.1057 0.4939 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 $v_3$ 0.0000 3.0834 0.0000 0.0000 0.1352 0.0000 0.0000 0.6957 0.0000 $v_4$ 0.0000 3.4207 0.0000 0.0000 0.0026 0.0000 0.0064 0.7603 0.1801 $v_5$ 0.0000 5.7145 0.0000 0.0000 3.1147 2.5431 3.6277 0.0000 0.0000 $v_6$ 0.0000 7.1430 0.0000 4.9601 0.9637 0.0000 1.9331 0.0000 0.0000 $v_7$ 0.0000 7.2983 0.0000 0.0000 0.0000 1.7188 3.3700 0.0000 2.6128 $v_8$ 0.0000 7.5898 0.0000 4.3192 0.0000 0.0000 2.0493 0.0000 1.0416 $v_9$ 0.0000 7.7043 2.9184 0.0000 2.4098 1.9675 0.0000 0.0000 0.0000 $v_{10}$ 0.0000 8.1361 1.7390 4.2912 0.8338 0.0000 0.0000 0.0000 0.0000 $v_{11}$ 0.0000 8.5707 1.8287 3.7067 0.0000 0.0000 0.0000 0.0000 0.8939 $v_{12}$ 0.0000 8.8386 2.7554 0.0000 0.0000 1.3515 0.0000 0.0000 2.0545 $v_{13}$ 0.0000 9.0761 0.0000 3.0271 0.9637 0.0000 0.0000 1.9331 0.0000 $v_{14}$ 0.0000 9.1971 0.0000 0.0000 1.9422 1.1568 0.0000 2.7039 0.0000 $v_{15}$ 0.0000 9.6392 0.0000 2.2699 0.0000 0.0000 0.0000 2.0493 1.0416 $v_{16}$ 0.0000 10.0534 0.0000 0.0000 0.0000 0.6918 0.0000 2.5809 1.6740 $v_{17}$ 0.0000 3.4315 0.0050 0.0000 0.0027 0.0000 0.0000 0.7627 0.1807 $v_{18}$ 0.6494 10.2980 0.0000 0.0000 0.0000 0.0000 0.0000 2.4700 1.5827 $v_{19}$ 1.0475 9.6669 0.0000 0.0000 1.7714 0.0000 0.0000 2.5142 0.0000 $v_{20}$ 1.2187 9.3956 2.5332 0.0000 0.0000 0.0000 0.0000 0.0000 1.8526 $v_{21}$ 1.5166 8.1461 0.0000 0.0000 0.0000 0.0000 3.0317 0.0000 2.3055 $v_{22}$ 1.6981 8.6358 2.5864 0.0000 2.0798 0.0000 0.0000 0.0000 0.0000 $v_{23}$ 2.1242 7.1628 0.0000 0.0000 2.6016 0.0000 3.1114 0.0000 0.0000
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2020 Impact Factor: 1.801
## Tools
Article outline
Figures and Tables | 4,565 | 10,823 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-39 | latest | en | 0.824959 |
https://findthefactors.com/tag/unit/ | 1,701,374,157,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100232.63/warc/CC-MAIN-20231130193829-20231130223829-00154.warc.gz | 302,564,709 | 19,498 | # 1 Perfect Square
1 has 1 factor. 2 has 2 factors…that is the end of that pattern because no number greater than 2 equals its number of factors.
• 1 is not a prime number, and 1 is not a composite number. 1 is in a category all by itself. It is classified as a unit.
• 1 has no Prime factorization.
• p⁰ = 1, where p is any prime number, so 1 is a factor of every prime number and every composite number.
• 1 is also the only number to have exactly 1 factor.
• Factors of 1: 1
• Factor pairs: 1 = 1 x 1
• √1 = 1. Since its square root is a whole number, 1 is a perfect square.
1(n) = n and n ÷ 1 = n for every number n.
Also 1⁰ = 1, 1¹ = 1, 1² = 1, 1³ = 1, 1⁴ = 1, 1⁵ = 1, 1⁶ = 1, 1⁷ = 1, 1⁸ = 1, 1⁹ = 1. In fact, 1 raised to any power equals 1. Even 1⁻⁹⁸⁷⁶⁵⁴³²¹⁰ = 1.
Not only that, but any number (EXCEPT 0) raised to the zeroth power is equal to 1.
One of my college professors wrote something like the following on the board to show why 0º is NOT defined:
When 1 is a clue in the FIND THE FACTORS puzzle, write 1 in both the corresponding factor row and the corresponding factor column.
————————————————————————
Like most people, you probably know how to fill in a multiplication table even if it looks like this:
The numbers that are given on a table can be called clues. The table above has 20 clues. What is the least number of clues that a table could have and still only have one way to fill it out?
Although the table above has just nine clues, there is still only one way to complete it. Nine is the fewest number of clues that will still yield a unique solution. All of those clues would have to be perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100. One of the clues will be missing, but it isn’t difficult to figure out where the missing clue should go. Always find the factors on the outside of the puzzle BEFORE writing down the products on the inside of the puzzle.
The puzzle above is rated difficulty level ONE because you only need to know 10 multiplication facts to find all the factors. If this puzzle is too easy for you, you can try a more difficult puzzle. Levels FOUR, FIVE, or SIX will be much more challenging, even for adults.
This link, 10 Factors 2013-10-28, will bring up an excel file with the puzzles that are on this post. After you enable editing, you can print the puzzles or type the factor answers directly onto the excel file.
An answer key will be posted one week after a puzzle is published.
If you don’t want to open the excel file, the rest of the puzzles will be printed below. If you cut and paste them on a document, you can make them any size you want.
If you want to check your work, the answers are given in a tab of the excel file that was published a week later: 10 factors 2013-11-04. | 785 | 2,764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2023-50 | latest | en | 0.940478 |
https://mail.haskell.org/pipermail/haskell-cafe/2007-June/026796.html | 1,709,310,880,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475422.71/warc/CC-MAIN-20240301161412-20240301191412-00251.warc.gz | 362,686,372 | 2,599 | # [Haskell-cafe] Construct all possible trees
Lennart Augustsson lennart at augustsson.net
Wed Jun 13 18:50:48 EDT 2007
```I now realize that my solution is needlessly complicated. Here's a simpler
one.
module Trees where
data Tree = Leaf Int | Branch Tree Tree
deriving (Show)
insert x t@(Leaf y) = [Branch s t, Branch t s] where s = Leaf x
insert x (Branch l r) = [Branch l' r | l' <- insert x l] ++
[Branch l r' | r' <- insert x r]
allTrees [] = []
allTrees (x:xs) = Leaf x : ts ++ [ s | t <- ts, s <- insert x t ]
where ts = allTrees xs
-- Lennart
On 6/13/07, Lennart Augustsson <lennart at augustsson.net> wrote:
>
> This doesn't enumerate them in the order you want, but maybe it doesn't
> matter.
>
> module Trees where
>
> combinations :: [a] -> [[a]]
> combinations [] = [[]]
> combinations (x:xs)
> = combinations xs ++ [ x:xs' | xs' <- combinations xs ]
>
> data Tree = Leaf Int | Branch Tree Tree
> deriving (Show)
>
> trees [x] = [Leaf x]
> trees (x:xs) = [ s | t <- trees xs, s <- insert x t ]
>
> insert x t@(Leaf y) = [Branch s t, Branch t s] where s = Leaf x
> insert x (Branch l r) = [Branch l' r | l' <- insert x l] ++
> [Branch l r' | r' <- insert x r]
>
> allTrees xs = [ t | ys <- combinations xs, not (null ys), t <- trees ys ]
>
> -- Lennart
>
>
> On 6/12/07, Andrew Coppin <andrewcoppin at btinternet.com> wrote:
> >
> > I'm trying to construct a function
> >
> > all_trees :: [Int] -> [Tree]
> >
> > such that all_trees [1,2,3] will yield
> >
> > [
> > Leaf 1,
> > Leaf 2,
> > Leaf 3,
> > Branch (Leaf 1) (Leaf 2),
> > Branch (Leaf 1) (Leaf 3),
> > Branch (Leaf 2) (Leaf 1),
> > Branch (Leaf 2) (Leaf 3),
> > Branch (Leaf 3) (Leaf 1),
> > Branch (Leaf 3) (Leaf 2),
> > Branch (Branch (Leaf 1) (Leaf 2)) (Leaf 3),
> > Branch (Branch (Leaf 1) (Leaf 3)) (Leaf 1),
> > Branch (Branch (Leaf 2) (Leaf 1)) (Leaf 3),
> > Branch (Branch (Leaf 2) (Leaf 3)) (Leaf 1),
> > Branch (Branch (Leaf 3) (Leaf 1)) (Leaf 2),
> > Branch (Branch (Leaf 3) (Leaf 2)) (Leaf 1),
> > Branch (Leaf 1) (Branch (Leaf 2) (Leaf 3)),
> > Branch (Leaf 1) (Branch (Leaf 3) (Leaf 2)),
> > Branch (Leaf 2) (Branch (Leaf 1) (Leaf 3)),
> > Branch (Leaf 2) (Branch (Leaf 3) (Leaf 2)),
> > Branch (Leaf 3) (Branch (Leaf 1) (Leaf 2)),
> > Branch (Leaf 3) (Branch (Leaf 2) (Leaf 1))
> > ]
> >
> >
> >
> > So far I'm not doing too well. Here's what I've got:
> >
> > data Tree = Leaf Int | Branch Tree Tree
> >
> > pick :: [x] -> [(x,[x])]
> > pick = pick_from []
> >
> > pick_from :: [x] -> [x] -> [(x,[x])]
> > pick_from ks [] = []
> > pick_from ks [x] = []
> > pick_from ks xs = (head xs, ks ++ tail xs) : pick_from (ks ++ [head xs])
> >
> > (tail xs)
> >
> > setup :: [Int] -> [Tree]
> > setup = map Leaf
> >
> > tree2 :: [Tree] -> [Tree]
> > tree2 xs = do
> > (x0,xs0) <- pick xs
> > (x1,xs1) <- pick xs0
> > return (Branch x0 x1)
> >
> > all_trees ns = (setup ns) ++ (tree2 \$ setup ns)
> >
> > Clearly I need another layer of recursion here. (The input list is of
> > arbitrary length.) However, I need to somehow avoid creating duplicate
> > subtrees...
> >
> > (BTW, I'm really impressed with how useful the list monad is for
> > constructing tree2...)
> >
> > _______________________________________________ | 1,148 | 3,252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-10 | latest | en | 0.828999 |
http://math.stackexchange.com/questions/101498/explain-why-a-system-of-more-than-2-equations-has-only-one-solution | 1,469,372,004,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824109.37/warc/CC-MAIN-20160723071024-00152-ip-10-185-27-174.ec2.internal.warc.gz | 167,477,288 | 19,374 | # Explain why a system of more than 2 equations has only one solution
This is a homework problem that I would love some direction on!
I'm given $$A = \begin{bmatrix} 3 & 7 & -4\\ 5 & -2 & 6\\ 2 & 1 & -1\\ 4 & 1 & 2 \end{bmatrix}$$
The question: Let $\vec{b}$ be a vector in $R^4$ such that the system $A\vec{x} = \vec{b}$ has a solution. Explain why it has only one solution.
Now, I've started off attempting to actually solve the system using the vector $b$:
$$\begin{bmatrix} 3 & 7 & -4\\ 5 & -2 & 6\\ 2 & 1 & -1\\ 4 & 1 & 2 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} b_{1} \\ b_{2} \\ b_{3} \\ b_{4} \end{bmatrix}$$
This proved to be a huge mess so I'm going to guess that this was the wrong way to go about it. Then I thought about relating it to pivots/pivot positions but I don't fully understand all of that yet. Can anyone offer me some suggestions?
EDIT:
$$A =\begin{bmatrix} 3 & 7 & -4\\ 5 & -2 & 6\\ 2 & 1 & -1\\ 4 & 1 & 2 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$$
So based on the reduced form above, can I assume this matrix only has one solution because there are no free variables?
-
What do you know about rank or invertability? You can ditch one of the equations and see that a system based on the remaining three always has a unique solution. – ncmathsadist Jan 23 '12 at 2:16
Something is amiss. You have a $4\times 3$ matrix multiplied by a $4 \times 1$ matrix, which is undefined. – Austin Mohr Jan 23 '12 at 2:17
Try writing $A$ in its reduced-row-echelon form. – JavaMan Jan 23 '12 at 2:19
AustinMohr I think I fixed that issue, sorry. ncmathsadist, I know about rank, but have not learned about invertability yet. JavaMan, I've posted what I think is the correct idea based on your suggestion. – intervade Jan 23 '12 at 3:41
{\bf Hint}: If there are two different solutions, then their difference would be a non-trivial solution for the corresponding homogeneous system. Can you solve $Ax=0$? – N. S. Jan 23 '12 at 3:42
If $c \in \mathbf R^3$ is a vector such that $Ac = b$, then the solutions of $Ax = b$ are precisely $$c + \operatorname{null} A = \{c + d : d \in \mathbf R^3 \text{ and } Ad = 0\}.$$ If you can use or justify this, then all you need to do is show that the homogeneous system $Ax = 0$ has only the trivial solution $x = (0, 0, 0)^T$. This is true if and only if after performing elementary row operations to $A$ to get a matrix in row echelon form there are exactly three (the maximum possible number) pivots. If you had fewer pivots, then there would be free variables.
@Dalton Interesting. Perhaps there is another way, then? To your question: if $c_1$ and $c_2$ are such that $Ac_1 = Ac_2 = b$, then $A(c_1 - c_2) = b - b = 0$, so $c_1 - c_2 \in \operatorname{null}A$. If the kernel is trivial, then this means that $c_1 = c_2$, so the solution is unique. – Dylan Moreland Jan 23 '12 at 4:11 | 974 | 2,947 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2016-30 | latest | en | 0.897788 |
https://wiki.ubc.ca/Science:Math_Exam_Resources/Courses/MATH110/April_2011/Question_07_(c) | 1,718,924,475,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862006.96/warc/CC-MAIN-20240620204310-20240620234310-00797.warc.gz | 547,805,800 | 12,241 | # Science:Math Exam Resources/Courses/MATH110/April 2011/Question 07 (c)
MATH110 April 2011
Other MATH110 Exams
### Question 07 (c)
As long as ${\displaystyle m<0}$, the line you gave as an answer in part (a) cuts out a triangle in the upper righthand quadrant, whose vertices are at the origin, the x-intercept, and the y-intercept. Find m so that the area of this triangle is as small as possible.
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!
Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result. | 377 | 1,608 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 12, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-26 | latest | en | 0.968831 |
https://or.stackexchange.com/questions/7169/estimate-of-mean-lead-time-demand | 1,721,248,928,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514801.32/warc/CC-MAIN-20240717182340-20240717212340-00853.warc.gz | 408,360,192 | 29,635 | # Estimate of mean lead time demand
I am trying to estimate the order-up-to level of inventory, $$y_t$$, according to ref [1]
$$y_t = \hat{D}_t^L + z \hat{\sigma}^L_{et}$$
where $$\hat{D}_t^L$$ is an estimate of the mean lead-time demand, $$\hat{\sigma}^L_{et}$$ is n estimate of the standard deviation of the $$L$$ period forecast error, and $$z$$ is a constant chosen to meet a desired service level. Assume that the retailer uses moving average to estimate $$\hat{D}_t^L$$ based on the demand observations from the previous $$p$$ periods, then
$$\hat{D}_{t}^{L}=L\left(\frac{\sum_{i=1}^{p} D_{t-i}}{p}\right)$$.
According to this equation, the estimated mean lead-time demand seems to only cover the demand during the lead time.
However, I checked out the definition from Lokad Quantitative Supply Chain[2], "The lead demand (also called lead time demand) is the total demand between now and the anticipated time for the delivery after the next one if a reorder is made now to replenish the inventory." This seems to mean that the mean lead-time demand should cover the demand during the lead time in this period plus the demand during the next period. This definition makes sense because assuming the safety stock factor is zero, the estimated order-up-to level is equal to the estimated lead-time demand, which should cover the demand from the current reorder time to the delivery time after the next one.
Can anyone help to clarify the definition of lead-time demand?
[1] Chen, F., Drezner, Z., Ryan, J. K., & Simchi-Levi, D. (2000). Quantifying the bullwhip effect in a simple supply chain: The impact of forecasting, lead times, and information. Management science, 46(3), 436-443.
The lead-time demand is the cumulative demand in $$L+R$$ consecutive periods, where $$L$$ is the lead time and $$R$$ is the reorder interval (the number of periods between orders). (At least, that's how we define it in our book, and I think that definition is consistent with other usages elsewhere.) Often $$R=1$$ and the lead-time demand is the demand over $$L+1$$ periods.
In the Chen et al. paper, they use a somewhat non-standard sequence of events, in which the demand is observed before the order is placed. A lead time of $$L$$ in the Chen sequence of events is effectively like a lead time of $$L-1$$ in the "usual" sequence of events (in which you observe the demand after the order is placed).
So, for Chen et al., we have a reorder interval of $$R$$ and an "effective" lead time of $$L-1$$, so the lead-time demand is the demand over $$L$$ periods.
I don't know why the website you linked to says "the delivery after the next one". Maybe they mean "the delivery after the next order", i.e., "the next delivery"?
I ran this past a colleague in supply chain management who confirmed my understanding of "demand during lead time" to be the demand that occurs between when a replenishment order is placed and when it arrives. I don't recall seeing it defined to include the demand occurring between orders, and in particular I think that in the (unusual? purely hypothetical?) case of immediate replenishment, the lead time demand is considered to be zero.
Regarding "the delivery after the next one", there are cases where orders overlap (you place a replenishment order and, before it shows up, place another one). As my colleague reminded me, this is usually handled by basing your reorder point on "inventory position", which includes both on-hand inventory and replenishment orders already made, rather than just on on-hand inventory.
• Thanks! I also think lead-time demand is the cumulative demand during the lead time. Per LarrySnyder610's answer, the estimated mean lead-time demand is used to determine the order-up-to level, so it should cover the cumulative demand during the lead time and the next review period/reorder interval. As such, the estimated mean lead-time demand is different from the demand during lead time. Commented Oct 23, 2021 at 14:21 | 933 | 3,979 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 19, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-30 | latest | en | 0.91979 |
https://eurekamathanswerkeys.com/eureka-math-grade-8-module-7-lesson-9/ | 1,718,307,239,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861488.82/warc/CC-MAIN-20240613190234-20240613220234-00753.warc.gz | 214,694,645 | 11,663 | ## Engage NY Eureka Math 8th Grade Module 7 Lesson 9 Answer Key
### Eureka Math Grade 8 Module 7 Lesson 9 Exercise Answer Key
Opening Exercise
a. Compute the decimal expansions of $$\frac{5}{6}$$ and $$\frac{7}{9}$$.
$$\frac{5}{6}$$ = 0.8333… and $$\frac{7}{9}$$ = 0.7777…
b. What is $$\frac{5}{6}$$ + $$\frac{7}{9}$$ as a fraction? What is the decimal expansion of this fraction?
$$\frac{5}{6}$$ + $$\frac{7}{9}$$ = $$\frac{15 + 14}{18}$$ = $$\frac{29}{18}$$
$$\frac{29}{18}$$ = 1.61111″…”
c. What is $$\frac{5}{6}$$ × $$\frac{7}{9}$$ as a fraction? According to a calculator, what is the decimal expansion of the answer?
$$\frac{5}{6}$$ × $$\frac{7}{9}$$ = $$\frac{35}{54}$$ = 0.648 148 148 1″…”
d. If you were given just the decimal expansions of $$\frac{5}{6}$$ and $$\frac{7}{9}$$, without knowing which fractions produced them, do you think you could easily add the two decimals to find the decimal expansion of their sum? Could you easily multiply the two decimals to find the decimal expansion of their product?
No. To add 0.8333… and 0.777…, we need to start by adding together their rightmost digits. But these decimals are infinitely long, and there are no rightmost digits. It is not clear how we can start the addition.
Thinking about how to multiply the two decimals, 0.8333… × 0.77777…, is even more confusing!
Exercise 1.
Two irrational numbers x and y have infinite decimal expansions that begin 0.67035267… for x and 0.84991341… for y.
a. Explain why 0.670 is an approximation for x with an error of less than one thousandth. Explain why 0.849 is an approximation for y with an error of less than one thousandth.
The difference between 0.670 and 0.67035267… is 0.00035267…, which is less than 0.001, a thousandth.
The difference between 0.849 and 0.84991341… is 0.00091341…, which is less than 0.001, a thousandth.
b. Using the approximations given in part (a), what is an approximate value for x + y, for x × y, and for x2 + 7y2?
x + y is approximately 1.519 because 0.670 + 0.849 = 1.519.
x × y is approximately 0.56883 because 0.670 × 0.849 = 0.56883.
x2 + 7y2 is approximately 5.494507 because (0.670)2 + 7(0.849)2 = 5.494 507.
c. Repeat part (b), but use approximations for x and y that have errors less than $$\frac{1}{10^{5}}$$.
We want the error in the approximation to be less than 0.00001.
If we approximate x by truncating to five decimal places, that is, as 0.67035, then the error is 0.00000267…, which is indeed less than 0.00001.
Truncating y to five decimal places, that is, as 0.84991, gives an error of 0.00000341…, which is indeed less than 0.00001.
Now:
x + y is approximately 1.52026 because 0.67035 + 0.84991 = 1.52026.
x × y is approximately 0.5697371685 because 0.67035 × 0.84991 = 0.5697371685.
x2 + 7y2 is approximately 5.505798179 because (0.67035)2 + 7(0.84991)2 = 5.505798179.
Exercise 2.
Two real numbers have decimal expansions that begin with the following:
x = 0.1538461…
y = 0.3076923…
a. Using approximations for x and y that are accurate within a measure of $$\frac{1}{10^{3}}$$, find approximate values for x + y and y-2x.
Using x ≈ 0.153 and y ≈ 0.307, we obtain x + y ≈ 0.460 and y-2x ≈ 0.001.
b. Using approximations for x and y that are accurate within a measure of $$\frac{1}{10^{7}}$$, find approximate values for x + y and y-2x.
Using x ≈ 0.1538461 and y ≈ 0.3076923, we obtain x + y ≈ 0.4615384 and y-2x ≈ 0.0000001.
c. We now reveal that x = $$\frac{2}{13}$$ and y = $$\frac{4}{13}$$. How accurate is your approximate value to y-2x from part (a)? From part (b)?
The error in part (a) is 0.001. The error in part (b) is 0.0000001.
d. Compute the first seven decimal places of $$\frac{6}{13}$$. How accurate is your approximate value to x + y from part (a)? From part (b)?
$$\frac{6}{13}$$ = 0.4615384…
The error in part (a) is 0.4615384…-0.460 = 0.0015384…, which is less than 0.01.
Our approximate answer in part (b) and the exact answer match in the first seven decimal places. There is likely a mismatch from the eighth decimal place onward. This means that the error is no larger than 0.0000001, or $$\frac{1}{10^{7}}$$.
### Eureka Math Grade 8 Module 7 Lesson 9 Problem Set Answer Key
Question 1.
Two irrational numbers x and y have infinite decimal expansions that begin 0.3338117… for x and 0.9769112… for y.
a. Explain why 0.33 is an approximation for x with an error of less than one hundredth. Explain why 0.97 is an approximation for y with an error of less than one hundredth.
The difference between 0.33 and 0.3338117… is 0.0038117…, which is less than 0.01, a hundredth.
The difference between 0.97 and 0.9769112… is 0.0069112…, which is less than 0.01, a hundredth.
b. Using the approximations given in part (a), what is an approximate value for 2x(y + 1)?
2x(y + 1) is approximately 1.3002 because 2 × 0.33 × 1.97 = 1.3002.
c. Repeat part (b), but use approximations for x and y that have errors less than $$\frac{1}{10^{6}}$$.
We want the error in the approximation to be less than 0.000001.
If we approximate x by truncating to six decimal places, that is, as 0.333811, then the error is 0.0000007…, which is indeed less than 0.000001.
Truncating y to six decimal places, that is, as 0.976911, gives an error of 0.0000002…, which is indeed less than 0.000001.
Now:
2x(y + 1) is approximately 1.319829276, which is a rounding of 2 × 0.333811 × 1.976911.
Question 2.
Two real numbers have decimal expansions that begin with the following:
x = 0.70588…
y = 0.23529…
a. Using approximations for x and y that are accurate within a measure of $$\frac{1}{10^{2}}$$, find approximate values for x + 1.25y and x/y.
Using x ≈ 0.70 and y ≈ 0.23, we obtain x + 1.25y ≈ 0.9875 and x/y ≈ 3.0434….
b. Using approximations for x and y that are accurate within a measure of $$\frac{1}{10^{4}}$$, find approximate values for x + 1.25y and $$\frac{x}{y}$$.
Using x ≈ 0.7058 and y ≈ 0.2352, we obtain x + 1.25y ≈ 0.9998 and $$\frac{x}{y}$$ ≈ 3.000850….
c. We now reveal that x and y are rational numbers with the property that each of the values x + 1.25y and $$\frac{x}{y}$$ is a whole number. Make a guess as to what whole numbers these values are, and use your guesses to find what fractions x and y might be.
It looks like x + 1.25y = 1 and $$\frac{x}{y}$$ = 3. Thus, we guess x = 3y and so 3y + 1.25y = 1, that is, 4.25y = 1, so y = $$\frac{1}{4.25}$$ = $$\frac{100}{425}$$ = $$\frac{4}{17}$$ and x = 3y = $$\frac{12}{17}$$.
### Eureka Math Grade 8 Module 7 Lesson 9 Exit Ticket Answer Key
Question 1.
Suppose x = $$\frac{2}{3}$$ = 0.6666… and y = $$\frac{5}{9}$$ = 0.5555….
a. Using 0.666 as an approximation for x and 0.555 as an approximation for y, find an approximate value for x + y.
x + y = $$\frac{2}{3}$$ + $$\frac{5}{9}$$ = $$\frac{11}{9}$$ = 1 + $$\frac{2}{9}$$ = 1.22222…
c. Use approximations for x and y, each accurate to within an error of $$\frac{1}{10^{5}}$$, to estimate a value of the product x × y. | 2,371 | 6,905 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2024-26 | latest | en | 0.879475 |
https://unlearningmath.com/172.72-cm-to-feet/ | 1,721,314,847,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514831.13/warc/CC-MAIN-20240718130417-20240718160417-00659.warc.gz | 524,816,511 | 11,050 | # Understanding Conversion: 172.72 cm to feet
By / Under Centimeter To Feet / Published on
Explore a detailed breakdown of the conversion from 172.72 cm to feet, including the importance and applications of accurate measurements.
## A Comprehensive Guide to Converting 172.72 cm to Feet
172.72 centimeters is equal to approximately 5.67 feet. This conversion is essential in a multitude of realms including architecture, fashion, and even in the fitness industry.
Understanding conversion units between different measurement systems, especially the basics like transforming centimeters to feet, is a crucial skill in today's internationally connected world. Whether you're checking body height, measuring room space, or even shopping online for furniture or clothing, getting your numbers right means making accurate decisions.
Did you know? About 95% of the world's population use the Metric system (i.e., centimeters, meters), while the UK and the US still predominantly use the Imperial system (i.e., inches, feet). This starkly different usage illustrates the importance of understanding both.
Now, you might wonder why the conversion number seems so random? Well, it's because one foot equals 30.48 centimeters exactly, thus deriving the proportion of 5.67 feet when converted from 172.72 centimeters.
It's like a language - while you might speak English, understanding German can help you communicate with a broader population. Similarly, comprehending both the Imperial and Metric systems of measurement can bridge potential gaps in understanding and calculating dimensions.
Let's dig a bit deeper. Converting 172.72 centimeters to feet can be visualized as stacking twelve 1-foot rulers (since 1 foot equals 12 inches or approximately 30.48 cm). This gives us a sense of how height or lengths compare in real-world objects.
### FAQ
Why is it important to convert 172.72 cm to feet? Converting units makes it easier for people who are more familiar with a particular system to understand the measurement. In this case, converting 172.72 cm to feet aids people accustomed to the imperial system to envision the actual height or length.
How can I convert cm to feet easily? You can use a standard conversion factor. Since one foot is 30.48 cm, you can multiply the number of feet by 30.48 to get the measurement in centimeters. To convert the other way, divide the number of centimeters by 30.48. So for 172.72 cm, you would divide by 30.48 to get approximately 5.67 feet.
Where is this conversion mostly used? Industries like architecture, interior design, construction, healthcare, and even aviation often utilize the conversion from cm to feet. Internationally produced products like furniture, clothing, and sporting goods often provide measurements in both units for global understanding.
What is the formula to convert cm to feet? The formula to convert cm to feet is `Number of cm ÷ 30.48 = length in feet`. In this case, `172.72 ÷ 30.48 equals approximately 5.67 feet`.
Feet: 0 | 646 | 3,008 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2024-30 | latest | en | 0.902759 |
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- FLUENT (https://www.cfd-online.com/Forums/fluent/)
Leo February 20, 2001 02:35
Cool system ,Convection + Radiation problem.
Hi guys,
I am trying to model a cooling system in a cavity. That is, a 2D model with a solid region containing different materials and a fluid region inside. In the solid region there exists a cool surface , say at about 260 K while the outer the environment is at 290 K. I am trying to find out the flow inside due to convection, radiation and bouyant forces.Also the heat transfer rates due to conduction and finally the temperature distribution in the model..
I used a mapped and fine 2D mesh of quadrilateral elements., segregated solver,steady state analyis, k-e modelling. When I choose bousinesq for the density of the fluid in the region, the code hangs up. .?? Firs order discretization for momentum and turbulence params. Standart handling for presusre and SIMPLE for velocity-pressure coupling.
Now, for the operating conditions, I defined a gravitaional accel. But for the operating temp, pressure and density, which values should I choose.
I have applied the case in the tutorials but it seems to be very brief..
Anybody has any experience with such a case.. Any suggestions are welcome!!!
Regards
Leo
Ashutosh February 21, 2001 04:39
Re: Cool system ,Convection + Radiation problem.
Try to calculate density of air using ideal gas most of the time. If the temperature difference is very small, Ra no. is small then Boussinesq method is advisable. Pressure can be atmospheric pressure. Operating temp can be kept at average temperature( use defaults initially, if ambient is not different from default 300K). For radiation you have to choose one method out of four depending on the optical thickness. Most complex method is DO method which even captures radiation through semi-transparent material( like glass). You need to specify the surface emissivities of all boundaries, emissivity etc. Material Properties have to be changed for DO Radiation-model. Start solution with default solution parameters. You will see it works. Ashutosh
All times are GMT -4. The time now is 06:46. | 491 | 2,204 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-30 | latest | en | 0.894497 |
https://b-ok.org/book/3427596/5818fc | 1,553,480,956,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203547.62/warc/CC-MAIN-20190325010547-20190325032547-00446.warc.gz | 425,114,953 | 37,629 | Main Fundamentals of Engineering Thermodynamics - Solution Manual 8th
# Fundamentals of Engineering Thermodynamics - Solution Manual 8th
,
Edition: 8th
Language: english
Pages: 2475
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### Contexts for Amos: Prophetic Poetics in Latin American Perspective
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### Controversial Issues in Energy Policy
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```1.4
Perform the following unit conversions:
(a) 1 L
0.0353 ft 3 12 in.
1L
1 ft
(b) 650 J
61 in.3 ←
1 kJ
1 Btu
0.616 Btu ←
103 J 1.0551 kJ
(c) 0.135 kW
(d) 378
3
3413 Btu/h 1 h 778.17 ft lbf
ft lbf
99.596
1 kW
3600 s
1 Btu
s
g 1 kg
1 lb
60 s
lb
50
s 10 3 g 0.4536 kg 1 min
min
←
1 lbf/in. 2 10 3 Pa
lbf
44.09
6894.8 Pa 1 kPa
in. 2
←
(e) 304 kPa
3
(f) 55
m 3 3.2808 ft
h
1m
(g) 50
km 10 3 m 3.2808 ft 1 h
ft
45.57 ←
h 1 km
1m
3600 s
s
(h) 8896 N
1h
ft 3
0.54
3600 s
s
←
1 lbf
1 ton
1 ton ←
4.4482 N 2000 lbf
1
←
1.5
Perform the following unit conversions:
(a) 122 in.
3
1 cm 3
(d) 1000
(g) 75
1L
2L←
1 kJ
1.0551 kJ ←
737.56 ft lbf
1 kW
74.57 kW ←
1.341 hp
lb 1 h
1 kg
kg
←
0.126
h 3600 s 2.2046 lb
s
(e) 29.392
(f) 2500
3
0.061024 in.3 10 2 cm 10 - 3 m 3
(b) 778.17 ft lbf
(c) 100 hp
1m
lbf 6894.8 Pa 1 N/m2 1 bar
2.027 bar ←
in.2 1 lbf/in. 2 1 Pa 105 N/m2
ft 3 0.028317 m3 1 min
m3
←
1
.
18
min
60 s
s
1 ft 3
mile 1.6093 km/h
km
←
120.7
h
1 mile/h
h
(h) 1 ton
2000 lbf 4.4482 N
8896 N ←
1 ton
1 lbf
1
1.6
Which of the following food items weighs approximately one newton?
a. a grain of rice
b. a small strawberry
c. a medium-sized apple
d. a large watermelon
1
1.7 A person whose mass is 150 lb weights 144.4 lbf. Determine (a) the local acceleration of
gravity, in ft/s2, and (b) the person’s mass, in lb, and weight, in lbf, if g = 32.174 ft/s2.
(a) Fgrav = mg →
g
Fgrav
m
144.4 lbf 32.174 lb ft/s2
= 30.97 ft/s2
150 lb
1 lbf
(b) Mass value remains the same. So
ft
1 lbf
Fgrav = mg = (150 lb) 32.174 2
= 150 lbf
s 32.174 lb ft/s2
1
1.8 The Phoenix with a mass of 350 kg was a spacecraft used for exploration of Mars.
Determine the weight of the Phoenix, in N, (a) on the surface of Mars where the acceleration of
gravity is 3.73 m/s2 and (b) on Earth where the acceleration of gravity is 9.81 m/s2.
KNOWN: Phoenix spacecraft has mass of 350 kg.
FIND: (a) Weight of Phoenix on Mars, in N, and (b) weight of Phoenix on Earth, in N.
SCHEMATIC AND GIVEN DATA:
m = 350 kg
gMars = 3.73 m/s2
gEarth = 9.81 m/s2
ENGINEERING MODEL:
1. Acceleration of gravity is constant at the surface of both Mars and Earth.
ANALYSIS: Weight is the force of gravity. Applying Newton’s second law using the mass of
the Phoenix and the local acceleration of gravity
F = mg
(a) On Mars,
m 1N
= 1305.5 N
F (350 kg) 3.73 2
s 1 kg m/s 2
(b) On Earth,
m
1N
= 3433.5 N
F (350 kg) 9.81 2
s 1 kg m/s 2
Although the mass of the Phoenix is constant, the weight of the Phoenix is less on Mars than on
Earth since the acceleration due to gravity is less on Mars than on Earth.
1
1.11 At the grocery store you place a pumpkin with a mass of 12.5 lb on the produce spring
scale. The spring in the scale operates such that for each 4.7 lbf applied, the spring elongates one
inch. If local acceleration of gravity is 32.2 ft/s2, what distance, in inches, did the spring
elongate?
KNOWN: Pumpkin placed on a spring scale causes the spring to elongate.
FIND: Distance spring elongated, in inches.
SCHEMATIC AND GIVEN DATA:
m = 12.5 m
x
ENGINEERING MODEL:
1. Spring constant is 4.7 lbf/in.
2. Local acceleration of gravity is 32.2 ft/s2.
ANALYSIS:
The force applied to the spring to cause it to elongate can be expressed as the spring constant, k,
times the elongation, x.
F = kx
The applied force is due to the weight of the pumpkin, which can be expressed as the mass (m) of
the pumpkin times acceleration of gravity, (g).
F = Weight = mg = kx
Solving for elongation, x, substituting values for pumpkin mass, acceleration of gravity, and
spring constant, and applying the appropriate conversion factor yield
x
mg
k
12.5 lb 32.2 ft2
lbf
4.7
in.
s
1 lbf
= 2.66 in.
lb ft
32.174 2
s
1.17 A communications satellite weighs 4400 N on Earth where g = 9.81 m/s2. What is the
weight of the satellite, in N, as it orbits Earth where the acceleration of gravity is 0.224 m/s2?
Express each weight in lbf.
KNOWN: Weight of communications satellite on Earth.
FIND: Determine weight of the satellite, in N, as it orbits Earth where the acceleration of
gravity is 0.224 m/s2. Express the satellite weight, in lbf, on Earth and in orbit.
SCHEMATIC AND GIVEN DATA:
WSat(Earth) = 4400 N
gEarth = 9.81 m/s2
gorbit = 0.224 m/s2
ENGINEERING MODEL:
1. Gravitational acceleration on Earth is constant at 9.81 m/s2.
2. Gravitational acceleration at orbital altitude is constant at 0.224 m/s2.
ANALYSIS: Weight of the satellite is the force of gravity and varies with altitude. Mass of the
satellite remains constant. Applying Newton’s second law to solve for the mass of the satellite
yields
W = mg → m = W/g
On Earth,
m = WSat(Earth)/gEarth
m
(4400 N) 1 kg m/s 2
= 448.5 kg
1N
m
9.81
s2
Solving for the satellite weight in orbit,
WSat(orbit) = mgorbit
m 1N
= 100.5 N
WSat(orbit) (448.5 kg) 0.224 2
s 1 kg m/s 2
Although the mass of the communications satellite is constant, the weight of the satellite is less at
orbital altitude than on Earth since the acceleration due to gravity is less at orbital altitude than
on Earth.
1
To determine the corresponding weights in lbf, apply the conversion factor, 1 lbf = 4.4482 N.
WSat(Earth) (4400 N)
1 lbf
= 989.2 lbf
4.4482 N
WSat(orbit) (100.5 N)
2
1 lbf
= 22.6 lbf
4.4482 N
1.21 A 2-lb sample of an unknown liquid occupies a volume of 62.6 in.3 For the liquid
determine (a) the specific volume, in ft3/lb, and (b) the density, in lb/ft3.
KNOWN: Volume and mass of an unknown liquid sample.
FIND: Determine (a) the specific volume, in ft3/lb, and (b) the density, in lb/ft3.
SCHEMATIC AND GIVEN DATA:
m = 2 lb
V = 62.6 in.3
ENGINEERING MODEL:
1. The liquid can be treated as continuous.
ANALYSIS:
(a) The specific volume is volume per unit mass and can be determined from the total volume
and the mass of the liquid
v
V 62.6 in.3 1 ft 3
= 0.0181 ft3/lb
m
2 lb 1728 in.3
(b) Density is the reciprocal of specific volume. Thus,
1
v
1
3
0.0181
ft
lb
= 55.2 lb/ft3
1.23 The specific volume of 5 kg of water vapor at 1.5 MPa, 440oC is 0.2160 m3/kg. Determine
(a) the volume, in m3, occupied by the water vapor, (b) the amount of water vapor present, in
gram moles, and (c) the number of molecules.
KNOWN: Mass, pressure, temperature, and specific volume of water vapor.
FIND: Determine (a) the volume, in m3, occupied by the water vapor, (b) the amount of water
vapor present, in gram moles, and (c) the number of molecules.
SCHEMATIC AND GIVEN DATA:
m = 5 kg
p = 1.5 MPa
T = 440oC
v = 0.2160 m3/kg
ENGINEERING MODEL:
1. The water vapor is a closed system.
ANALYSIS:
(a) The specific volume is volume per unit mass. Thus, the volume occupied by the water vapor
can be determined by multiplying its mass by its specific volume.
m 3
= 1.08 m3
V mv (5 kg) 0.2160
kg
(b) Using molecular weight of water from Table A-1 and applying the appropriate relation to
convert the water vapor mass to gram moles gives
1000 moles
m
5 kg
= 277.5 moles
n
kg 1 kmol
M
18.02
kmol
(c) Using Avogadro’s number to determine the number of molecules yields
molecules
# Molecules Avogadro' s Number # moles 6.022 10 23
(277.5 moles)
mole
# Molecules = 1.671×1026 molecules
1.27 Three kg of gas in a piston-cylinder assembly undergo a process during which the
relationship between pressure and specific volume is pv0.5 = constant. The process begins with
p1 = 250 kPa and V1 = 1.5 m3 and ends with p2 = 100 kPa. Determine the final specific volume,
in m3/kg. Plot the process on a graph of pressure versus specific volume.
KNOWN: A gas of known mass undergoes a process from a known initial state to a specified
final pressure. The pressure-specific volume relationship for the process is given.
FIND: Determine the final specific volume and plot the process on a pressure versus specific
volume graph.
SCHEMATIC AND GIVEN DATA:
State 1
State 2
Process 1→ 2
Gas
Gas
pv0.5 = constant
m2 = 3 kg
p2 = 100 kPa
m1 = 3 kg
p1 = 250 kPa
V1 = 1.5 m3
ENGINEERING MODEL:
1. The gas is a closed system.
2. The system undergoes a polytropic process in which pv0.5 = constant.
ANALYSIS:
The final specific volume, v2, can be determined from the polytropic process equation
p1v10.5= p2v20.5
Solving for v2 yields
1
p 0.5
v2 = v1 1
p2
Specific volume at the initial state, v1, can be determined by dividing the volume at the initial
state, V1, by the mass, m, of the system
V1 1.5 m 3
v1 =
= 0.5 m3/kg
m
3 kg
1
Substituting values for pressures and specific volume yields
1
m 3 250 kPa 0.5
3
v2 = 0.5
= 3.125 m /kg
kg 100 kPa
The volume of the system increased while pressure decreased during the process.
A plot of the process on a pressure versus specific volume graph is as follows:
Pressure versus Specific Volume
260
Pressure (kPa)
240
220
200
180
160
140
120
100
0.50
1.00
1.50
2.00
2.50
Specific Volume (m^3/kg)
2
3.00
3.50
1.28 A closed system consisting of 4 lb of a gas undergoes a process during which the relation
between pressure and volume is pVn = constant. The process begins with p1 = 15 lbf/in.2, v1 =
1.25 ft3/lb and ends with p2 = 53 lbf/in.2, v2 = 0.5 ft3/lb. Determine (a) the volume, in ft3,
occupied by the gas at states 1 and 2 and (b) the value of n. (c) Sketch Process 1-2 on pressurevolume coordinates.
KNOWN: Gas undergoes a process from a known initial pressure and specific volume to a
known final pressure and specific volume.
FIND: Determine (a) the volume, in ft3, occupied by the gas at states 1 and 2 and (b) the value
of n. (c) Sketch Process 1-2 on pressure-volume coordinates.
SCHEMATIC AND GIVEN DATA:
State 1
State 2
Gas
Gas
p2 = 53 lbf/in.2
v2 = 0.5 ft3/lb
m = 4 lb
p1 = 15 lbf/in.2
v1 = 1.25 ft3/lb
ENGINEERING MODEL:
1. The gas is a closed system.
2. The relation between pressure and volume is pVn = constant during process 1-2.
ANALYSIS:
(a) The specific volume is volume per unit mass. Thus, the volume occupied by the gas can be
determined by multiplying its mass by its specific volume.
V = mv
For state 1
ft 3
= 5 ft3
V1 mv1 (4 lb)1.25
lb
For state 2
ft 3
V2 mv2 (4 lb) 0.5 = 2 ft3
lb
1
(b) The value of n can be determined by substituting values into the relationship:
p1(V1)n = constant = p2(V2)n
Solving for n
p1 V2
p2 V1
ln
n
V
p1
n ln 2
p2
V1
2
p1 ln 15 lbf/in.
ln
53 lbf/in. 2
p
= 1.38
n 2
3
V
2 ft
ln 2
ln 3
5 ft
V1
(c) Process 1-2 is shown on pressure-volume coordinates below:
Process 1-2
60
State 2
Pressure (lbf/in.2)
50
40
30
20
State 1
10
0
0
1
2
3
Volume (ft3)
2
4
5
6
1.31 A gas contained within a piston-cylinder assembly undergoes four processes in series:
Process 1-2: Constant-pressure expansion at 1 bar from V1 = 0.5 m3 to V2 = 2 m3
Process 2-3: Constant volume to 2 bar
Process 3-4: Constant-pressure compression to 1 m3
Process 4-1: Compression with pV−1 = constant
Sketch the process in series on a p-V diagram labeled with pressure and volume values at each
numbered state.
p (bar)
4•
2
1
•3
•
•
1
0.5
1
2
2
1
V (m3)
1.36 Liquid kerosene flows through a Venturi meter, as shown in Fig. P1.36. The pressure of
the kerosene in the pipe supports columns of kerosene that differ in height by 12 cm. Determine
the difference in pressure between points a and b, in kPa. Does the pressure increase or decrease
as the kerosene flows from point a to point b as the pipe diameter decreases? The atmospheric
pressure is 101 kPa, the specific volume of kerosene is 0.00122 m3/kg, and the acceleration of
gravity is g = 9.81 m/s2.
KNOWN: Kerosene flows through a Venturi meter.
FIND: The pressure difference between points a and b, in kPa and whether pressure increases or
decreases as the kerosene flows from point a to point b as the pipe diameter decreases.
SCHEMATIC AND GIVEN DATA:
patm = 101 kPa
g = 9.81 m/s2
L = 12 cm
Kerosene
v = 0.00122 kg/m3
•
b
•
a
ENGINEERING MODEL:
1. The kerosene is incompressible.
2. Atmospheric pressure is exerted at the open end of the fluid columns.
ANALYSIS:
Equation 1.11 applies to both columns of fluid (a and b). Let hb be the height of the fluid above
point b. Then hb + L is the height of the fluid above point a. Applying Eq. 1.11 to each column
yields
pa = patm + g(hb + L) = patm + ghb + gL
and
pb = patm + ghb
Thus, the difference in pressure between point a and point b is
p = pb – pa = (patm + ghb) – (patm + ghb + gL)
p = –gL
1
Density of kerosene is the reciprocal of its specific volume
= 1/v = 1/0.00122 m3/kg = 820 kg/m3
Solving for the difference in pressure yields
kg
m
1N
1m
1 kPa
= −0.965 kPa
p 820 3 9.81 2 (12 cm)
kg
m
N
m
s
1 2 100 cm 1000 2
s
m
Since points a and b are at the same elevation in the flow, the difference in pressure is indicated
by the difference in height between the two columns. The negative sign indicates pressure
decreases as the kerosene flows from point a to point b as the pipe diameter decreases.
2
1.37 Figure P1.37 shows a tank within a tank, each containing air. Pressure gage A, which
indicates pressure inside tank A, is located inside tank B and reads 5 psig (vacuum). The U-tube
manometer connected to tank B contains water with a column length of 10 in. Using data on the
diagram, determine the absolute pressure of the air inside tank B and inside tank A, both in psia.
The atmospheric pressure surrounding tank B is 14.7 psia. The acceleration of gravity is g =
32.2 ft/s2.
KNOWN: A tank within a tank, each containing air.
FIND: Absolute pressure of air in tank B and in tank A, both in psia.
SCHEMATIC AND GIVEN DATA:
patm = 14.7 psia
Tank B
L = 10 in.
Tank A
Gage A
Water ( = 62.4 lb/ft3)
g = 32.2 ft/s2
pgage, A = 5 psig (vacuum)
ENGINEERING MODEL:
1. The gas is a closed system.
2. Atmospheric pressure is exerted at the open end of the manometer.
3. The manometer fluid is water with a density of 62.4 lb/ft3.
ANALYSIS:
(a) Applying Eq. 1.11
pgas,B = patm + gL
where patm is the local atmospheric pressure to tank B, is the density of the manometer fluid
(water), g is the acceleration due to gravity, and L is the column length of the manometer fluid.
Substituting values
pgas, B 14.7
lb
ft
1 lbf
1 ft 3
32.2 (10 in.)
62.4
= 15.1 lbf/in.2
2
3
2
3
lbm
ft
in.
ft
s
1728 in.
32.2
2
s
lbf
1
Since the gage pressure of the air in tank A is a vacuum, Eq. 1.15 applies.
p(vacuum) = patm(absolute) – p(absolute)
The pressure of the gas in tank B is the local atmospheric pressure to tank A. Solving for p
(absolute) and substituting values yield
p(absolute) = patm(absolute) – p(vacuum) = 15.1 psia – 5 psig = 10.1 psia
2
1.39 Show that a standard atmospheric pressure of 760 mmHg is equivalent to 101.3 kPa. The
density of mercury is 13,590 kg/m3 and g = 9.81 m/s2.
KNOWN: Standard atmospheric pressure of 760 mmHg.
FIND: Show that 760 mmHg is equivalent to 101.3 kPa.
SCHEMATIC AND GIVEN DATA:
Mercury vapor
Hg = 13,590 kg/m3
L = 760 mm
Mercury (Hg)
ENGINEERING MODEL:
1. Local gravitational acceleration is 9.81 m/s2.
2. Pressure of mercury vapor is much less than that of the atmosphere and can be neglected.
ANALYSIS:
Equation 1.12 applies.
patm = pvapor + HggL = HggL
Neglecting the pressure of mercury vapor and applying appropriate conversion factors yield
kg
m
1N
1m
1 kPa
patm 13,590 3 9.81 2 (760 mm)
= 101.3 kPa
kg m 1000 mm
N
m
s
1 2
1000 2
s
m
1.41 As shown in Figure P1.41, air is contained in a vertical piston-cylinder assembly such
that the piston is in static equilibrium. The atmosphere exerts a pressure of 14.7 lbf/in.2 on top of
the 6-in.-diameter piston. The absolute pressure of the air inside the cylinder is 16 lbf/in.2 The
local acceleration of gravity is g = 32.2 ft/s2. Determine (a) the mass of the piston, in lb, and (b)
the gage pressure of the air in the cylinder, in psig.
KNOWN: A piston-cylinder assembly contains air such that the piston is in static equilibrium.
FIND: (a) The mass of the piston, in lb, and (b) the gage pressure of the air in the cylinder, in
psig.
SCHEMATIC AND GIVEN DATA:
patm = 14.7 lbf/in.2
g = 32.2 ft/s2
Dpiston = 6 in.
Air
pAir = 16 lbf/in.2
ENGINEERING MODEL:
1. The air is a closed system.
2. The piston is in static equilibrium.
3. Atmospheric pressure is exerted on the top of the piston.
4. Local gravitational acceleration is 32.2 ft/s2.
ANALYSIS:
(a) Draw a free body diagram indicating all forces acting on the piston. Taking upward as the
positive y-direction, the sum of the forces acting on the piston in the y-direction must equal zero
for static equilibrium of the piston.
Free Body Diagram
Fy 0
mpistong
patmApiston
pAirApiston – patmApiston – mpistong = 0
y
Solving for the mass of the piston,
mpiston
pAir A piston patm A piston
pAirApiston
g
1
mpiston
pAir patm Apiston
g
The area of the piston is determined from the piston diameter
A piston
4
D2
4
(6 in.) 2 = 28.3 in.2
Substituting values and solving for the mass of the piston,
lbf
lbf
lb ft
16 2 14.7 2 28.3 in.2 32.2
in.
s 2 = 36.8 lb
mpiston in.
ft
1 lbf
32.2 2
s
(b) Gage pressure of the air is given by Eq. 1.14
p(gage) = p(absolute) – patm(absolute) = 16.0 psia – 14.7 psia = 1.3 psig
2
1.42 Air is contained in a vertical piston-cylinder assembly such that the piston is in static
equilibrium. The atmosphere exerts a pressure of 101 kPa on top of the 0.5-meter-diameter
piston. The gage pressure of the air inside the cylinder is 1.2 kPa. The local acceleration of
gravity is g = 9.81 m/s2. Subsequently, a weight is placed on top of the piston causing the piston
to fall until reaching a new static equilibrium position. At this position, the gage pressure of the
air inside the cylinder is 2.8 kPa. Determine (a) the mass of the piston, in kg, and (b) the mass of
KNOWN: A piston-cylinder assembly contains air such that the piston is in static equilibrium.
Upon addition of a weight, the piston falls until reaching a new position of static equilibrium.
FIND: (a) The mass of the piston, in kg, and (b) the mass of the added weight, in kg.
SCHEMATIC AND GIVEN DATA:
Part (a)
Part (b)
patm = 101 kPa
g = 9.81 m/s2
patm = 101 kPa
g = 9.81 m/s2
Apiston = 0.5 m2
Apiston = 0.5 m2
Weight
Air
Air
pAir = 1.2 kPa (gage)
pAir = 2.8 kPa (gage)
ENGINEERING MODEL:
1. The air is a closed system.
2. The piston is in static equilibrium for both part (a) and part (b).
3. Atmospheric pressure is exerted on the top of the piston.
4. Local gravitational acceleration is 9.81 m/s2.
ANALYSIS:
(a) Draw a free body diagram indicating all forces acting on the piston. Taking upward as the
positive y-direction, the sum of the forces acting on the piston in the y-direction must equal zero
for static equilibrium of the piston.
1
Free Body Diagram
Fy 0
mpistong
patmApiston
pAirApiston – patmApiston – mpistong = 0
y
Solving for the mass of the piston,
mpiston
mpiston
pAir A piston patm A piston
pAirApiston
g
pAir patm Apiston
g
From Eq. 1.14, the quantity in parenthesis is the gage pressure of the air in the cylinder.
Rewriting the equation above
mpiston
pAir(gage) A piston
g
Substituting values and solving for the mass of the piston,
mpiston
1.2 kPa 0.5 m
m
9.81 2
s
2
N kg m
1000 2 1 2
s
m
= 61.2 kg
1 kPa
1N
2
(b) Draw a second free body diagram indicating all forces acting on the piston including the
newly added weight expressed as the product of its mass and gravitational acceleration. Taking
upward as the positive y-direction, the sum of the forces acting on the piston in the y-direction
must equal zero for static equilibrium of the piston.
Free Body Diagram
Fy 0
mpistong
patmApiston
pAirApiston – patmApiston – mpistong – mweightg = 0
mweightg
Solving for the mass of the weight,
mweight
mweight
pAir A piston patm A piston
g
pAir patm Apiston
g
pAirApiston
mpiston
mpiston
From Eq. 1.14, the quantity in parenthesis is the gage pressure of the air in the cylinder.
Rewriting the equation above
mweight
pAir(gage) A piston
g
mpiston
Substituting values and solving for the mass of the weight,
mweight
2.8 kPa 0.5 m
m
9.81 2
s
2
N kg m
1000 2 1 2
s
m
61.2 kg = 81.5 kg
1 kPa
1N
3
y
1.46 As shown in Figure P1.46, an inclined manometer is used to measure the pressure of the
gas within the reservoir. (a) Using data on the figure, determine the gas pressure, in lbf/in.2 (b)
Express the pressure as a gage or a vacuum pressure, as appropriate, in lbf/in.2 (c) What
advantage does an inclined manometer have over the U-tube manometer shown in Figure 1.7?
KNOWN: A gas contained in a reservoir with inclined manometer attached.
FIND: (a) Pressure of gas within the reservoir, in lbf/in.2 (b) Pressure expressed as gage or
vacuum pressure, as appropriate, in lbf/in.2 (c) Advantage of inclined manometer over the Utube manometer.
SCHEMATIC AND GIVEN DATA:
patm = 14.7 lbf/in.2
g = 32.2 ft/s2
•
Gas
b•
a•
15 in.
40o
Oil ( = 54.2 lb/ft3)
ENGINEERING MODEL:
1. The gas is a closed system.
2. Atmospheric pressure is exerted at the open end of the manometer.
3. The manometer fluid is oil with a density of 54.2 lb/ft3.
ANALYSIS:
(a) Applying Eq. 1.11
pgas = patm + gL
where patm is the local atmospheric pressure, is the density of the manometer fluid (oil), g is the
acceleration due to gravity, and L is the vertical difference in liquid levels. Since level a is the
same as level b, applying trigonometry to determine the vertical difference in liquid levels
between level b and the liquid level at the free surface with the atmosphere yields
pgas = patm + gL(sin 40o)
Substituting values
1
p gas 14.7
lb
ft
1 lbf
1 ft 3
= 15.0 lbf/in.2
54.2 32.2 (15 in.)(sin 40)
2
2
3
lbm ft 1728 in.
ft
in.
s
32.2
s2
lbf
(b) Since the pressure of the gas is greater than atmospheric pressure, gage pressure is given by
Eq. 1.14
p(gage) = p(absolute) – patm(absolute) = 15.0 psia – 14.7 psia = 0.3 psig
(c) The advantage of the inclined manometer is its easier readability since the surface of the
liquid is wider than with a same diameter U-tube manometer. The scale on the inclined
manometer is much more precise since more graduations are possible compared with the U-tube
manometer.
2
Substituting values for pressures and specific volume yields
1
m 3 250 kPa 0.5
3
v2 = 0.5
= 3.125 m /kg
kg
100
kPa
The volume of the system increased while pressure decreased during the process.
A plot of the process on a pressure versus specific volume graph is as follows:
Pressure versus Specific Volume
260
Pressure (kPa)
240
220
200
180
160
140
120
100
0.50
1.00
1.50
2.00
2.50
Specific Volume (m^3/kg)
3
3.00
3.50
1.52 Water in a swimming pool has a temperature of 24oC. Express this temperature in K, oF,
and oR.
KNOWN: Water is at a specified temperature in oC.
FIND: Equivalent temperature in K, oF, and oR.
SCHEMATIC AND GIVEN DATA:
T = 24oC
ANALYSIS:
First convert temperature from oC to K by rearranging Eq. 1.17 to solve for temperature in K
T(oC) = T(K) – 273.15
→
T(K) = T(oC) + 273.15
Twater (K) = 24oC + 273.15 = 297.15 K
Next apply Eq. 1.16 to solve for temperature in oR
T(oR) = 1.8T(K)
Twater (oR) = (1.8)(297.15 K) = 534.87 oR
Finally, apply Eq. 1.18 to solve for temperatures in oF
T(oF) = T(oR) – 459.67
Twater (oF) = 534.87 oR – 459.67 = 75.2oF
1.53 A cake recipe specifies an oven temperature of 350oF. Express this temperature in oR, K,
and oC.
KNOWN: Oven temperature is specified in oF.
FIND: Equivalent temperature in oR, K, and oC.
SCHEMATIC AND GIVEN DATA:
T = 350oF
ANALYSIS:
First convert temperature from oF to oR using Eq. 1.18 to solve for temperature in oR
T(oF) = T(oR) – 459.67 → T(oR) = T(oF) + 459.67
Toven (oR) = 350oF + 459.67 = 809.67oR
Next apply Eq. 1.16 to solve for temperature in K
T(oR) = 1.8T(K) → T(K) = T(oR)/1.8
Toven (K) = 809.67oR/1.8 = 449.82 K
Finally, apply Eq. 1.17 to solve for temperature in oC
T(oC) = T(K) – 273.15
Toven (oC) = 449.82 K – 273.15 = 176.67oC
1.56 Left for independent study using the Internet.
1.57 Air temperature rises from a morning low of 42oF to an afternoon high of 70oF.
(a) Express these temperatures in oR, K, and oC.
(b) Determine the temperature change in oF, oR, K, and oC from morning low to afternoon high.
(c) What conclusion do you draw about temperature change for oF and oR scales?
(d) What conclusion do you draw about temperature change for oC and K scales?
KNOWN: Morning low temperature and afternoon high temperature, both in oF.
FIND: (a) Express these temperatures in oR, K, and oC, (b) temperature change in oF, oR, K, and
o
C from morning low to afternoon high, (c) conclusion about temperature change for oF and oR
scales, (d) conclusion about temperature change for oC and K scales.
SCHEMATIC AND GIVEN DATA:
Tlow = 42oF
Thigh = 70oF
ANALYSIS:
(a) First convert temperatures from oF to oR using Eq. 1.18 to solve for temperatures in oR
T(oF) = T(oR) – 459.67 → T(oR) = T(oF) + 459.67
Tlow (oR) = 42 oF + 459.67 = 501.67oR
Thigh (oR) = 70 oF + 459.67 = 529.67oR
Next apply Eq. 1.16 to solve for temperature in K
T(oR) = 1.8T(K) → T(K) = T(oR)/1.8
Tlow (K) = 501.67oR/1.8 = 278.71 K
Thigh (K) = 529.67oR/1.8 = 294.26 K
Finally, apply Eq. 1.17 to solve for temperature in oC
T(oC) = T(K) – 273.15
Tlow (oC) = 278.71 K – 273.15 = 5.56oC
Thigh (oC) = 294.26 K – 273.15 = 21.11oC
(b) Temperature change, T, is Thigh – Tlow. Calculating the differences yields
T(oF) = 70oF – 42oF = 28oF
1
T(oR) = 529.67oR – 501.67oR = 28oR
T(K) = 294.26 K – 278.71 K = 15.55 K
T(oC) = 21.11 oC – 5.56 oC = 15.55oC
(c) For oF and oR scales, the temperature change is the same since a Rankine degree and a
Fahrenheit degree are the same temperature unit.
(d) For oC and K scales, the temperature change is the same since a Kelvin degree and a Celsius
degree are the same temperature unit.
2
1.58 Left for independent study using the Internet.
Problem 2.2
Determine the gravitational potential energy, in kJ, of 2 m3 of liquid water at an elevation of
30 m above the surface of Earth. The acceleration of gravity is constant at 9.7 m/s2 and the
density of the water is uniform at 1000 kg/m3. Determine the change in gravitational potential
energy as the elevation decreased by 15 m.
KNOWN: The elevation of a known quantity of water is decreased from a given initial value by
a given amount.
FIND: Determine the initial gravitational potential energy and the change in gravitational
potential energy.
V = 2 m3
g = 9.7 m/s2
ρ = 1000 kg/m3
SCHEMATIC AND GIVEN DATA:
ENGINERING MODEL:
(1) The water is a closed system. (2) The acceleration of
gravity is constant. (3) The density of water is uniform.
ANALYSIS: The initial gravitational potential energy is
= 582 kJ
The change in potential energy is
= -291 kJ
Δz = - 15 m
z1 = 30 m
Problem 2.4
A construction crane weighing 12,000 lbf fell from a height of 400 ft to the street below during a
severe storm. For g = 32.05 ft/s2, determine mass, in lb, and the change in gravitational potential
energy of the crane, in ft∙lbf.
KNOWN: A crane of known weight falls from a known elevation to the street below.
FIND: Determine the change in gravitational potential energy of the crane.
SCHMATIC AND GIVEN DATA:
Fgrav = 12,000 lbf
g = 32.05 ft/s2
z2 = 0
z1 = 400 ft
ENGINEERING MODEL: (1) The crane is the closed system. (2) The acceleration of gravity
is constant.
ANALYSIS:
To get the mass, note that Fgrrav = mg. Thus
= 12,046 lb
The change in gravitational potential energy is
Problem 2.5
A automobile weighing 2500-lbf increases its gravitational potential energy by 2.25 x 104 Btu in
going from an elevation of 5,183 ft in Denver to the highest elevation on Trail Ridge road in the
Rocky Mountains. What is the elevation at the high point of the road, in ft?
KNOWN: An automobile of known weight increases its gravitational potential energy by a
given amount. The initial elevation is known.
FIND: Determine the final elevation.
Fgrav = 2500 lbf
ΔPE = 2.25 x 104 Btu
z2 = ?
z1 = 5183 ft
ENGINEERING MODEL: (1) The automobile is the closed system. (2) The acceleration of
gravity is constant.
ANALYSIS: The change in gravitational potential energy is: ΔPE = mg(z2 – z1). With
Fgrav = mg, we get
ΔPE = Fgrav(z2 – z1)
Solving for z2
+ 5183 ft = 12,185 ft
Problem 2.10
An object whose mass is 300 lb experiences changes in kinetic and potential energies owing to
the action of a resultant force R. The work done on the object by the resultant force is 140 lbf.
There are no other interactions between the object and its surroundings. It the object’s elevation
increases by 100 ft and its final velocity is 200 ft/s, what is the initial velocity, in ft/s? Let g =
32.2 ft/s2.
KNOWN: An object of known mass experiences changes in kinetic and potential energy due to
the action of a resultant force. The final velocity, the change in elevation, and the work done by
the force are given.
m = 300 lb
V2 = 200 ft/s
FIND: Determine the final velocity.
z2 – z1 = +100 ft
R
g = 32.2 ft/s2
SCHEMATIC AND GIVEN DATA:
ENGINEEIRNG MODEL: (1) The object is a closed
system. (2) The force of gravity acts on the object, and
g = 32.2 ft/s2. (3) The resultant force accounts for all
interactions between the system and its surroundings.
Work done by resultant force = 140 Btu
ANALYSIS: By modeling assumption (3), the work of the resultant force must equal the sum of
the changes in kinetic and gravitational potential energies. Thus, with Eq. 2.9
Work = ½ m(
Solving for
) + mg(z2 – z1)
and inserting values
+
First
mg(z2 – z1) = (300 lb)(32.2 ft/s2)(100 ft)
= 38.6 Btu
So
+ 2002 ft2/s2 = 23065 ft2/s2
or
1
V1 = 151.9 ft/s
1. The increase in velocity reflects the increase in kinetic energy of the object as a result of
energy transferred to it by the work of the resultant force. Carefully observe that in Eq. 2.9 the
work of the resultant force acting on the body is positive.
Problem 2.13
Two objects having different masses are propelled vertically from the surface of Earth, each with
the same initial velocities. Assuming the objects are acted upon only by the force of gravity,
show that they reach zero velocity at the same height.
KNOWN: Two objects are propelled upward from the surface of Earth with the same initial
velocities and are acted upon only by the force of gravity.
FIND: Show that the reach zero velocity at the same height.
V2 = 0
SCHEMATIC AND GIVEN DATA:
ENGINEERING MODEL: (1) Each object is a closed system.
(2) The acceleration of gravity is constant. (3) The only force
acting is the force of gravity.
z2
V1
ANALYSIS: For an object moving vertically under the
influence of gravity only, Eq. 2.11 applies
z1 = 0
m1
V1
m2
For V2 = 0 and z1 = 0
Thus
Since the final height doesn’t depend on mass, both objects will reach zero velocity at the same
final height.
Problem 2.14
An object whose mass is 100 lb falls freely under the influence of gravity from an initial
elevation of 600 ft above the surface of Earth. The initial velocity is downward with a
magnitude of 50 ft/s. The effect of air resistance is negligible. Determine the velocity, in ft/s, of
the object just before it strikes Earth. Assume g = 31.5 ft/s2.
KNOWN: An object of known mass falls freely from a known elevation and with a given initial
velocity. The only force acting is the force of gravity.
m = 100 lb
g = 31.5 ft/s2
FIND: Determine the velocity of the object just before it strikes Earth.
SCHEMATIC AND GIVEN DATA:
ENGINEERING MODEL: (1) The object is a closed system.
(2) The acceleration of gravity is constant. (3) The only force
acting on the object is the force of gravity.
ANALYSIS: Since the only force acting on the object is the force of
gravity, Eq. 2.11 applies. Thus
0
V1 = 50 ft/s
z1 = 600 ft
z2 = 0
V2
1
Solving for V2
Inserting values
1. Note that the mass cancels out. Any object falling freely under the influence of gravity, with
no effects of air resistance, would reach the same final velocity.
Problem 2.18
An object initially at an elevation of 5 m relative to Earth’s surface and with a velocity of 50 m/s
is acted on by an applied force R and moves along a path. Its final elevation is 20 m and its
velocity is 100 m/s. The acceleration of gravity is 9.81 m/s2. Determine the work done on the
object by the applied force, in kJ.
KNOWN: An object moves along a path due to the action of an applied force. The elevation
and velocities are known initially and finally.
V2 = 100 m/s
FIND: Determine the work of the applied force.
R
SCHEMATIC AND GIVEN DATA:
m = 50 kg
V1 = 50 m/s
ENGINEERING MODEL: (1) The object is a closed
system. (2) R is the only force acting on the object other
z
than the force of gravity. (3) g = 9.81 m/s2 and is constant.
ANALYSIS: To find the work of force R we use
Work =
Inserting values and converting units
= 187.5 + 7.36 = 194.9 kJ
z2 = 20 m
z1 = 5 m
Problem 2.19
An object of mass 10 kg, initially at rest, experiences a constant horizontal acceleration of 4 m/s2
due to the action of a resultant force applied for 20 s. Determine the total amount of energy
transfer by work, in kJ.
KNOWN:
A system of known mass experiences a constant horizontal acceleration due to an applied force
for a specified length of time.
m = 10 kg
ax = 4 m/s2
FIND: Determine the amount of energy transfer by work.
Fx
Δt = 20 s
V1 = 0
SCHEMATIC AND GIVEN DATA:
x
ENGINEERING MODEL: (1) The object is a closed system.
(2) The horizontal acceleration is constant.
ANALYSIS: The work of the resultant force is determined using Eq. 2.6
0
To find V2, use the fact that the acceleration is constant
→
Ax =
or
dV = axdt
→
0
(V2 – V1) = ax(t2 – t1) = axΔt
Thus
V2 = (4 m/s2) (20 s) = 80 m/s
Finally, the work of the resultant force is
= 32 kJ
Problem 2.23
The two major forces opposing the motion of a vehicle are the rolling resistance of the tires, Fr,
and the aerodynamic drag force of the air flowing around the vehicle, Fd, given respectively by
Fd = CdA(1/2)ρV2
Fr = fW,
where f and Cd are constants known as rolling resistance and drag coefficient, respectively, W
and A are the vehicle weight and projected frontal area, respectively, V is the vehicle velocity, ,
and ρ is the air density. For a popular gasoline hybrid car with W = 3040 lbf, A = 6.24 ft2 and Cd
= 0.25, when f = 0.02 and ρ= 0/08 lb/ft3.
(a) determine the power required, in hp, to overcome rolling resistance and aerodynamic drag
when V is 55 mph.
(b) plot versus vehicle velocity ranging from 0 to 90 mi/h (i) the power to overcome rolling
resistance, (ii) the power to overcome aerodynamic drag, and (iii) the total power, all in hp.
What implications for vehicle fuel economy can be deduced from the results of part (b)?
KNOWN: The drag force and the force associated with rolling resistance are known as functions
of variables associated with a vehicle in motion.
FIND: (a) Determine the power required to overcome drag and rolling resistance when the
vehicle is moving at 55 mi/h. (b) Plot the quantities of part (a) and their sum versus vehicle
velocity ranging from 0 to 90 mi/h. Discuss implications for fuel economy.
SCHMATIC AND GIVEN DATA:
W = 3040 lbf
A = 6.24 ft2
Cd = 0.25
f = 0.01
ENGINEERING MODEL: The vehicle is
the closed system.
ρ = 0.08 lb/ft3
ANALYSIS: Applying Eq. 2.13, the power, in hp, required to overcome aerodynamic drag is
(*)
-5
3
= 1.11 x 10 [V ] (where V is in mi/h)
The power, in hp, required to overcome rolling resistance is
Problem 2.23 (Continued) – Page 2
W
(**)
(a) When V = 55 mi/h, we get
= 1.847 hp
= 4.461 hp
= 6.308 hp
(b) The plots are developed by letting V vary from 0 to 90 mi/h:
Power (hp)
Wd
Wr
Wtot
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
0
10
20
30
40
50
60
70
80
90
V (mi/h)
We see from the plots that up to about 87 mi/h, the power required to overcome rolling resistance
exceeds the power to overcome aerodynamic drag. However, the total power required increases
dramatically with velocity. The aerodynamic drag varies as the cube of velocity, so it increases
rapidly and contributes much more significantly as speed increases. The total power required by
the engine increases about 5-fold from 30 mi/h to 90mi/h. Since the power is developed by the
engine from fuel stored on board the vehicle, high-speed driving has a significant negative effect
on fuel consumption.
PROBLEM 2.24
PROBLEM 2.24 (Cont’d)
PROBLEM 2.25
PROBLEM 2.25 (Cont’d)
PROBLEM 2.26
Problem 2.27
Carbon dioxide (CO2) gas within a piston-cylinder assembly undergoes a process from a state
where p1 = 5 lbf/in.2, V1 = 2.5 ft3 to a state where p2 = 20 lbf/in.2, V2 = 0.5 ft3. The relationship
between pressure and volume during the process is given by p = 23.75 – 7.5V, where V is in ft3
and p is in lbf/in.2 Determine the work for the process, in Btu.
KNOWN: CO2 gas within a piston-cylinder assembly undergoes a process where the p-V
relation is given. The initial and final states are specified.
FIND: Determine the work for the process.
SCHEMATIC AND GIVEN DATA:
25
p1 = 5 lbf/in.2
V1 = 2.5 ft3
CO2
p (lbf/in^2)
p2 = 20 lbf/in.
V2 = 0.5 ft3
2
p = 23/75 – 7.5 V
2
.
20
15
10
.
1
W
5
ENGINEERING MODEL: (1) The CO2 is the
closed system. (2) The p-V relation during the
process is linear. (3) Volume change is the only
work mode.
0
0
0.5
1
1.5
2
2.5
3
V (ft^3)
ANALYSIS: The given p-V relation can be used with Eq. 2.17 as follows:
= -3600 ft∙lbf
=
(negative sign denotes energy transfer in.)
Alternative Solution
Since the p-V relation is linear, W can also be evaluated geometrically as the area under the
process line:
= -4.63 Btu
PROBLEM 2.28
Problem 2.29
Nitrogen (N2) gas within a piston-cylinder assembly undergoes a process from p1 = 20 bar, V1 =
0.5 m3 to a state where V2 = 2.75 m3. The relationship between pressure and volume during the
process is pV1.35 = constant. For the N2, determine (a) the pressure at state 2, in bar, and (b) the
work, in kJ.
KNOWN: N2 gas within a piston-cylinder assembly undergoes a process where the p-V relation
is pV1.35 = constant. Data are given at the initial and final states.
FIND: Determine the pressure at the final state and the work.
SCHEMATIC AND GIVEN DATA:
pV 1.35= constant
p1 = 20 bar, V1 = 0.5 m3
V2 = 2.75 m3
N2
ANALYSIS: (a)
→
ENGINEERING MODEL: (1) The
N2 is the closed system. (2) The p-v
relation is specified for the process.
(3) Volume change is the only work
mode.
; n = 1.35. Thus
(b) Since volume change is the only work mode, Eq. 2.17 applies. Following the procedure of
part (a) of Example 2.1, we have
W=
= 1285.7 kJ
PROBLEM 2.30
+B
PROBLEM 2.31
Problem 2.31 (Cont’d)
Problem 2.32
Air contained within a piston-cylinder assembly is slowly compressed. As shown in Fig P2.32,
during this first process the pressure first varies linearly with volume and then remains constant.
Determine the total work, in kJ.
KNOWN: Air within a piston-cylinder assembly undergoes two processes in series.
FIND: Determine the total work.
SCEMATIC AND GIVEN DATA:
ENGINEERING MODEL: (1) The air within the
piston-cylinder assembly is the closed system. (2) The
two-step p-V relation is specified graphically. (3)
Volume change is the only work mode.
3
150
2
+
+
1
p
(kPa)
+
100
1
2
50
0.015
0.055
0.07
V (m3)
ANALYSIS: Since volume change is the work mode, Eq. 2.17 applies. Furthermore, the
integral can be evaluated geometrically in terms of the total area under process lines:
1
2
= (-1.875 kJ) + (-6 kJ) = -7.875 kJ (in)
PROBLEM 2.33
Note: Minus signs
signify energy transfer
by work to the gas.
Problem 2.34
Carbon monoxide gas (CO) contained within a piston-cylinder assembly undergoes three
processes in series:
Process 1-2: Constant pressure expansion at 5 bar from V1 = 0.2 m3 to V2 = 1 m3.
Process 2-3: Constant volume cooling from state 2 to state 3 where p3 = 1 bar.
Process 3-1: Compression from state 3 to the initial state during which the pressure-volume
relationship is pV = constant.
Sketch the processes in series on p-V coordinates and evaluate the work for each process, in kJ.
KNOWN: Carbon monoxide gas within a piston-cylinder assembly undergoes three processes
in series.
FIND: Sketch the processes in series on a p-V diagram and evaluate the work for each process.
SCHEMATIC AND GIVEN DATA:
p
(bar)
CO
5
Process 1-2: Constant pressure expansion at 5
bar from V1 = 0.2 m3 to V2 = 1 m3.
Process 2-3: Constant volume cooling from
state 2 to state 3 where p3 = 1 bar.
Process 3-1: Compression from state 3 to the
initial state during which the pressure-volume
relationship is pV = constant.
1
1
2
3
pV = constant
0.2
1
ENGINEERING MODEL: (1) The gas is the closed system. (2) Volume change is the only
work mode. (3) Each of the three processes is specified.
ANALYSIS: Since volume change is the only work mode, Eq. 2.17 applies.
Process 1-2: Constant pressure processes:
(out)
Process 2-3: Constant volume (piston does not move). Thus W23 = 0
V (m3)
Problem 2.33 (Continued)
Process 3-1: For process 3-1, pV = constant = p1V1 . Noting that V3 = V2, we get
Inserting values and converting units
1
= -160.9 kJ (in)
1. The net work for the three process is
Wnet = W12 + W23 + W31 = (+400) + 0 + (-160.9) = 239.1 kJ (net work is positive - out)
Problem 2.35
Air contained within a piston-cylinder assembly undergoes three processes in series:
Process 1-2: Compression during which the pressure-volume relationship is pV = constant from
p1 = 10 lbf/in.2, V1 = 4 ft3 to p2 = 50 lbf/in.2
Process 2-3: Constant volume from state 2 to state 3 where p = 10 lbf/in.2
Process 3-1: Constant pressure expansion to the initial state.
Sketch the processes in series on a p-V diagram. Evaluate (a) the volume at state 2, in ft3, and (b)
the work for each process, in Btu.
KNOWN: Air within a piston-cylinder assembly undergoes three processes in series.
FIND: Sketch the processes in series on a p-V diagram. Evaluate (a) the volume at state 2, and
(b) the work for each process.
SCHEMATIC AND GIVEN DATA:
p
2
Air
pV = constant
Process 1-2: Compression during which the
pressure-volume relationship is pV = constant
from p1 = 10 lbf/in.2, V1 = 4 ft3 to p2 = 50 lbf/in.2
Process 2-3: Constant volume from state 2 to state
3 where p = 10 lbf/in.2
Process 3-1: Constant pressure expansion to the
initial state.
3
ENGINEERING MODEL: (1) The gas is the closed system. (2) Volume change is the only
work mode. (3) Each of the three processes is specified.
ANALYSIS: (a) For process 1-2; pV = constant. Thus p1V1 = p2V2, and
= 0.8 ft3
(b) Since volume change is the only work mode, Eq. 2.17 applies.
Process 1-2: For process 1-2, pV = constant = p1V1 . Thus
1
V
Problem 2.35 (Continued)
Inserting values and converting units
= -11.92 Btu (in)
Process 2-3: Constant volume (piston does not move). Thus W23 = 0
Process 3-1: Constant pressure processes (p3 = p1):
Noting that V3 = V2
= 5.92 Btu (out)
1. The net work for the three process is
Wnet = W12 + W23 + W31 = (-11.92) + 0 + (5.92)) = - 6 kJ (net work is negative - in)
PROBLEM 2.36
PROBLEM 2.37
PROBLEM 2.37
Problem 2.39
An electric heater draws a constant current of 6 amp, with an applied voltage of 220 V, for 24 h.
Determine the instantaneous electric power provided to the heater, in kW, and the total amount
of energy supplied to the heater by electrical work, in kW∙h. If electric power is valued at
\$0.08/kW∙h, determine the cost of operation for one day.
KNOWN: An electric heater draws a constant current at a specified voltage for a given length
of time. The cost of electricity is specified.
FIND: Determine the instantaneous power provided to the heater and the total amount of energy
supplied by electrical work. Determine the cost of operation for one day.
SHEMATIC AND GIVEN DATA:
i = 6 amp
E = 220 V
Δt = 10 h
ENGINEERING MODEL: The current and voltage are
constant.
ANALYSIS: The constant power input to the heater is
given by Eq. 2.21
EI
Thus, the total energy input is
Using the specified cost of electricity
Cost per day = (31.68 kW∙h) (\$0.08/kW∙h) = \$2.53
PROBLEM 2.40
PROBLEM 2.41
PROBLEM 2.42
Figure P2.42 shows an object whose mass is 5 lb attached to a rope wound around a pulley. The
radius of the pulley is 3 in. If the mass falls at a constant velocity of 5 ft/s, determine the power
transmitted to the pulley, in hp, and the rotational speed of the shaft, in revolutions per minute
(RPM). The acceleration of gravity is 32.2 ft/ s2.
KNOWN: An object attached to a rope wound around a pulley falls at a constant velocity.
FIND: Find the power transmitted to the pulley and the rotational speed.
SCHEMATIC AND GIVEN DATA:
ENGINEERING MODEL: (1) the object falls at a constant
speed. (2) The acceleration of gravity is constant.
ANALYSIS: The power is obtained using Eq. 2.13
= 25 ft∙lb/s
Converting to horsepower
= 0.0455 hp
The rotational speed of the pulley is related to the velocity of the object and the radius by
V = Rω. Thus
= 191 rev/min
PROBLEM 2.43
PROBLEM 2.43 (Continued)
PROBLEM 2.44
PROBLEM 2.45
PROBLEM 2.46
PROBLEM 2.47
PROBLEM 2.48
As shown in Fig. P2.48, an oven wall consists of a 0.635-cm-thick layer of steel (κs = 15.1
W/m∙K) and a layer of brick (κb = 0.72 W/m∙K). At steady state, a temperature decrease of 0.7oC
occurs over the steel layer. The inner temperature of the steel layer is 300 K. If the temperature
of the outer surface of the brick must be no greater than 40oC, determine the thickness of brick,
in cm, that ensures this limit is met. What is the rate of conduction, in kW per m2 of wall surface
area?
KNOWN: Steady-state data are provided for a composite wall formed from a steel layer and a
brick layer.
FIND: Determine the minimum thickness of the brick layer to keep the outer surface
temperature of the brick at or below a specified value.
SCHEMATIC AND GIVEN DATA:
Ti = 300 K
Tm
o
ENGINEERING MODEL: (1) The wall is
ΔT = -0.7 C
the system at steady state. (2) The temperature
varies linearly through each layer.
ANALYSIS: Using Eq. 2.31 together with
model assumption 2
To ≤ 40oC
κs = 15.1 W/m∙K
κb = 0.72 W/m∙K
Ls = 0.635 cm
and
where Tm denotes the temperature at the steel-brick interface.
At steady state, the rate of conduction to the interface through the steel must equal the rate of
conduction from the interface through the brick:
. Thus
And solving for Lb we get
(300 – 0.7) = 299.3 oC
= – 0.7 oC
PROBLEM 2.48 (Continued)
Since To ≤ 40 oC
Lb ≥ 11.22 cm
The rate of conduction is
or
The slight difference is due to round-off.
PROBLEM 2.49
PROBLEM 2.49 (CONTINUED)
Problem 2.50
A composite plane wall consists of a 3-in.-thick layer of insulation (κi = 0.029 Btu/h∙ft∙oR) and a
0.75-in.-thick layer of siding (κs = 0.058 Btu/h∙ft∙oR). The inner temperature of the insulation is
67oF. The outer temperature of the siding is -8oF. Determine at steady state (a) the temperature
at the interface of the two layers, in oF, and (b) the rate of heat transfer through the wall in Btu
per ft2 of surface area.
KNOWN: Energy transfer by conduction occurs through a composite wall consisting of two
layers.
FIND: Determine the temperature at the interface between the two layers and the rate of heat
transfer per unit area through the wall.
SCHEMATIC AND GIVEN DATA:
T1 = 67 oF
ENGINEERING MODEL: (1) The wall is
the system at steady state. (2) The temperature
varies linearly through each layer.
T3 = -8 oF
T2 = ?
κi = 0.029 Btu/h∙ft∙oR
ANALYSIS: With Eq. 2.17, and recognizing that
at steady state the rates of energy conduction must be
equal through each layer
insulation
siding
κs = 0.058 Btu/h∙ft∙oR
Li = 3 in.
(*)
Ls = 0.75 in.
Solving for T2
= 0.116 Btu/h∙oR
= 0.928 Btu/h∙oR
Thus
= 460.3 oR = 0.33 oF
Thus, using Eq. (*)
= 7.73 Btu/ft2
= 7.73 Btu/ft2
PROBLEM 2.51
Problem 2.52
Complete the following exercise using heat transfer relations:
(a) Referring to Fig. 2.12, determine the rate of conduction heat transfer, in W, for κ = 0.07
W/m∙K, A = 0.125 m2, T1 = 298 K, T2 = 273 K.
(b) Referring to Fig. 2.14, determine the rate of convection heat transfer from the surface to the
air, in W, for h = 10 W/m2, A = 0.125 m2, Tb = 305 K, Tf = 298 K.
(a) Referring to Fig. 2.12, determine the rate of conduction heat transfer, in W, for κ = 0.07
W/m∙K, A = 0.125 m2, T1 = 298 K, T2 = 273 K.
Using Eq. 2.31 and noting that the temperature varies linearly
through the wall
= 1.722 W
(b) Referring to Fig. 2.14, determine the rate of convection heat transfer from the surface to the
air, in W, for h = 10 W/m2, A = 0.125 m2, Tb = 305 K, Tf = 298 K.
Using Eq. 2.34
= hA[Tb – Tf]
=
= 8.75 W
PROBLEM 2.53
PROBLEM 2.54
net
PROBLEM 2.55
Problem 2.56
Each line of the following table gives data for a process of a closed system. Each entry has the
same energy units. Determine the missing entries.
Process
a
b
c
d
e
Process
a
b
c
d
e
Q
+50
W
E1
-20
+20
-60
-40
+50
Q
+50
+50
-40
-90
+50
E2
+50
+60
+40
+50
0
+150
W
-20
+20
-60
-90
+150
E1
-20
+20
+40
+50
+20
Process a:
W = Q - ΔE = +50 – (+ 70) = -20
ΔE = E2 – E1
E2 = ΔE + E1 = +70 + (-20) = +50
Process b:
Q = ΔE + W = +30 + (+20) = +50
ΔE = E2 – E1
E1 = E2 – ΔE = +50 – (+30) = +20
Process c:
ΔE = E2 – E1 = +60 – (+40) = +20
Q = ΔE + W = +20 + (-60) = -40
Process d:
W = Q – ΔE = (-90) – 0 = -90
ΔE = E2 – E1
E2 = ΔE + E1 = 0 +50 = +50
Process e:
ΔE = Q – W = +50 – (+150) = -100
E1 = E2 – ΔE = (-80) – (-100) = +20
E2
+50
+50
+60
+50
-80
-80
ΔE
+70
+30
+20
0
-100
ΔE
+70
+30
ΔE = Q - W
PROBLEM 2.57
Each line of the following table gives data, in Btu, for a process of a closed system. Determine
the missing table entries, in Btu.
Process
a
b
c
d
e
Process
a
b
c
d
e
Q
+40
W
+5
+10
-4
-10
+3
Q
+40
+20
-4
-10
+3
E1
+15
+7
E1
+15
+7
+6
-10
+8
E2
+30
+22
-8
+10
+14
Process a:
W = Q - ΔE = +40 – (+ 15) = -25 Btu
ΔE = E2 – E1
E2 = ΔE + E1 = +15 + (+15) = +30 Btu
Process b:
ΔE = E2 – E1 = 22 – 7 = +15 Btu
Q = ΔE + W = +15 + 5 = +20 Btu
Process c:
ΔE = Q – W = (-4) – (10) = -14 Btu
E1 = E2 – ΔE = (-8) – (-14) = 6 Btu
Process d:
W = Q – ΔE = (-10) – (+20) = -30 Btu
ΔE = E2 – E1
E2 = ΔE + E1 = +20 + (-10) = +10 Btu
Process e:
ΔE = Q – W = +3 – (-3) = +6 Btu
E2 = ΔE + E1 = (+6) + (+8) = +14 Btu
ΔE
+15
+22
-8
-10
+8
-3
W
-25
+5
+10
-30
-3
E2
+20
ΔE
+15
+15
-14
+20
+6
ΔE = Q - W
Problem 2.58
A closed system of mass of 10 kg undergoes a process during which there is energy transfer by
work from the system of 0.147 kJ per kg, an elevation decrease of 50 m, and an increase in
velocity from 15 m/s to 30 m/s. The specific internal energy decreases by 5 kJ/kg and the
acceleration of gravity is constant at 9.7 m/s2. Determine the heat transfer for the process, in kJ.
KNOWN: Data are provided for a closed system undergoing a process involving work, heat
transfer, change in elevation, and change in velocity.
FIND: Determine the heat transfer for the process.
V1 = 15 m/s
SCHEMATIC AND GIVEN DATA:
Δu = -5 kJ/kg
W/m = + 0.147 kJ/kg
50 m
V2 = 30 m/s
ENGINEERING MODEL: (1) The system is a
closed system. (2) The acceleration of gravity is
constant.
z
m = 10 kg
g = 9.7 m/s2
ANALYSIS:
ΔU + ΔPE + ΔKE = Q - W
→
Q = ΔU + ΔPE + ΔKE - W
W = m [W/m] = 10 kg [-0.147 kJ/kg] = -1.47 kJ
ΔU = mΔu = 10 kg [- 5 kJ/kg] = -50 kJ
ΔKE =
ΔPE = mg(z2 – z1) = (10 kg) (9.7 m/s2)(-50 m)
Q = (-50) + (-4.85) + (3.38) – (-1.47) = -50 kJ (out)
= +3.38 kJ
= - 4.85 kJ
PROBLEM 2.59
PROBLEM 2.60
PROBLEM 2.61
Problem 2.62
An electric motor draws a current of 10 amp with a voltage of 110 V, as shown in Fig. P2.62.
The output shaft develops a torque of 9.7 N∙m and a rotational speed of 1000 RPM. For
operation at steady state, determine for the motor
(a) the electric power required, in kW.
(b) the power developed by the output shaft, in kW.
(c) the average surface temperature, Ts, in oC, if heat transfer occurs by convection to the
surroundings at Tf = 21oC.
KNOWN: Operating data are provided for an electric motor at steady state.
FIND: Determine (a) the electric power required, (b) the power developed by the output shaft,
and (c) average the surface temperature.
Tf = 21oC
Ts
T = 9.7 N∙m
1000 RPM
10 amp
110 V
hA = 3.9 W/K
ENGINEERING MODEL: (1) The motor is the closed system. (2) The system is at steady
state.
ANALYSIS: (a) Using Eq. 2.21
- (voltage) (current) = - (110 V)(10 amp)
= -1.1 kW (in)
(b) Using Eq. 2.20
= (torque) (angular velocity)
= 1.016 kW (out)
(c) To determine the surface temperature, first find the rate of energy transfer by heat using the
energy balance
0
= (-1.1 kW) + (1.016 kW) = -0.084 kW
The surface temperature of the motor is
Problem 2.62 (Continued)
= (-0.084 kW)/(3.9 W/K)
= 315.5 K = 42.5 oC
+ 294 K
Fig. P2.62 – 8e
Pick-up motor graphic from Fig. E2.6 – 7e
PROBLEM 6.63
PROBLEM 2.64
PROBLEM 2.65
PROBLEM 2.65 (Continued)
Problem 2.66
A gas undergoes a process in a piston-cylinder assembly during which the pressure-specific
volume relation is pv1.2 = constant. The mass of the gas is 0.4 lb and the following data are
known: p1 = 160 lbf/in.2, V1 = 1 ft3, and p2 = 390 lbf/in.2 During the process, heat transfer from
the gas is 2.1 Btu. Kinetic and potential energy effects are negligible. Determine the change in
specific internal energy of the gas, in Btu/lb.
KNOWN: A gas is compressed in a piston-cylinder assembly. The pressure-specific volume
relation is specified.
FIND: Determine the change in specific internal energy.
p
2
.
pv1.2 = constnat
SCHEMATIC AND GIVEN DATA:
Q = -2.1 Btu
gas
m = 0.4 lb
W
.
p1 = 160 lbf/in.2
V1 = 1 ft3
p2 = 390 lbf/in.2
1
v
ENGINEERING MODEL: (1) The gas is a closed system.
(2) The process follows pv1.2 = constant. (3) Kinetic and
potential energy effects are negligible.
ANALYSIS: The change in specific internal energy will be found from an energy balance.
First, determine the work. Since volume change is the only work mode, Eq. 2.17 applies:
W=
=
Evaluating V2
V2 =
= 0.4759 ft3
Thus
W
= -23.69 Btu (in)
Now, writing the energy balance:
With ΔU = mΔu
1
Δu
= 54.0 Btu
1. The amount of energy transfer in by work exceeds the amount of energy transfer out by heat,
resulting in a net increase in internal energy.
PROBLEM 2.67
PROBLEM 2.68
PROBLEM 2.69
Steam in a piston-cylinder assembly undergoes a polytropic process. Data for the initial and
final states are given in the accompanying table. Kinetic and potential energy effects are
negligible. For the process, determine the work and heat transfer, each in Btu per lb of steam.
State
1
2
p (lbf/in.2)
100
40
v (ft3/lb)
4.934
11.04
u (Btu/lb)
1136.2
1124.2
KNOWN: Steam undergoes a polytropic process in a piston-cylinder assembly. Data are
known at the initial and final states.
p
FIND: Determine the work and heat transfer, each per unit mass of steam.
State
1
2
Steam
p (lbf/in.2)
100
40
v (ft3/lb)
4.934
11.04
1
.
pvn = constant
u (Btu/lb)
1136.2
1124.2
.
2
v
ENGINEERING MODEL: (1) The steam is a closed system. (2) The process is polytropic
process, and volume change is the only work mode. (3) Kinetic and potential energy effects are
negligible.
ANALYSIS: Since the process is polytropic, Eq 2.17 applies for the work:
W/m =
The pressures and specific volumes are known at each state, but n is unknown. To find n, pvn =
constant , as follows:
→
→
=
= 1.1377
Thus
W/m =
The heat transfer is obtained using the energy balance.
= 69.63 Btu/lb (out)
PROBLEM 2.69 (CONTINUED)
ΔU + ΔKE + ΔPE = Q – W →
Q = ΔU + W
With ΔU = m Δu = m(u2 – u1)
Q/m = (u2 – u1) + (W/m) = (1124.2 – 1136.2) Btu/lb + (69.63 Btu/lb)
= 57.63 Btu/lb (in)
PROBLEM 2.70
Air expands adiabatically in a piston-cylinder assembly from an initial state where p1 = 100
lbf/in.2, v1 = 3.704 ft3/lb, and T1 = 1000 oR, to a final state where p2 = 50 lbf/in.2 The process is
polytropic with n = 1.4. The change in specific internal energy, in Btu/lb, can be expressed in
terms of temperature change as Δu = (0.171)(T2 – T1). Determine the final temperature, in oR.
Kinetic and potential energy effects can be neglected.
KNOWN: Air undergoes a polytropic process with known n in a piston-cylinder assembly. Data
are known at the initial and final states, and the change in specific internal energy is expressed as
a function of temperature change.
FIND: Determine the final temperature.
p
SCHEMATIC AND GIVEN DATA:
p1 = 100 lbf/in.2
v1 = 3.704 ft3/lb
Air
T1 = 1000 oR
p2 = 50 lbf/in.2
Δu = (0.171Btu/lb∙oR)(T2 – T1)
Q=0
1
.
pv1.4 = constant
.2
ENGINEERING MODEL: (1) The air is a closed system.
(2) The process is polytropic with n = 1.4 and volume change
is the only work mode. (3) The process is adiabatic: Q = 0.
(4) Kinetic and potential energy effects are negligible.
v
ANALYSIS: To find the final temperature, we will use the energy balance with the given
expression for change in specific internal energy as a function of temperature change. First,
determine the work using Eq. 2.17
W/m =
For the polytropic proess,
. Thus
= 6.077 ft3/lb
So, the work is
W/m =
The energy balance is: ΔU + ΔKE + ΔPE = Q – W. With ΔU = m(u2 – u1)
= 30.794 Btu/lb
PROBLEM 2.70 (CONTINUED)
(u2 – u1) =
W/m
Inserting values
(0.171 Btu/lb∙oR)(T2 – 1000 oR) =
(30.794 Btu/lb)
Solving; T2 = (-30.794)/(0.171) + 1000 = 819.9 oR
PROBLEM 2.71
PROBLEM 2.71 (Continued)
PROBLEM 2.72
PROBLEM 2.73
PROBLEM 2.74
Process
1-2
2-3
3-4
4-1
ΔE
-1200
400
Q
0
800
-200
W
-200
400
PROBLEM 2.75
PROBLEM 2.76
.
PROBLEM 2.77
Process 1-2: Compression with pV = constant, W12 = -104 kJ,
U1 = 512 kJ, U2 = 690 kJ
Process 2-3: W23 = 0, Q23 = -150 kJ
Process 3-1: W31 = +50 kJ
PROBLEM 2.78
PROBLEM 2.78 (Continued) – Page 2
Problem 2.78 (Continued) – Page 2
Problem 2.79
A gas undergoes a cycle in a piston-cylinder assembly consisting of the following three
processes:
Process 1-2: Constant pressure, p = 1.4 bar, V1 = 0.028 m3, W12 = 10.5 kJ
Process 2-3: Compression with pV = constant, U3 = U2
Process 3-1: Constant volume, U1 – U3 = -26.4 kJ
There are no significant changes in kinetic or potential energy.
(a) Sketch the cycle on a p-V diagram.
(b) Calculate the net work for the cycle, in kJ.
(c) Calculate the heat transfer for process 1-2, in kJ
KNOWN: A gas undergoes a cycle consisting of three processes.
FIND: Sketch the cycle on a p-V diagram and determine the net work for the cycle and the heat
transfer for process 1-2.
SCHEMATIC AND GIVEN DATA:
Process 1-2: Constant pressure, p = 1.4 bar, V1 = 0.028 m3,
W12 = 10.5 kJ
Process 2-3:
Compression with pV = constant, U3 = U2
Process 3-1:
Constant volume, U1 – U3 = -26.4 kJ
Gas
ENGINEERING MODEL: (1) The gas is a closed system. (2) Kinetic and potential energy
effects are negligible. (3) The compression from state 2 to 3 is a polytropic process.
ANALYSIS: (a) Since W12 > 0, the process is an expansion. Thus
p
3
1
.
.
.
2
V
Problem 2.79 (Continued)
0
(b) The net work for the cycle is Wcycle = W12 +W23 + W31. W12 = 10.5 kJ, so we need W23.
W23
(*)
where V3 = V1 has been incorporated. But, we still need to evaluate V2. For Process 1-2 at
constant pressure
W12 =
or
V2 =
+ 0.028 m3 = 0.103 m3
Thus, with Eq. (*)
W23 =
= -18.78 kJ
Thus
Wcycle = 10.5 kJ + (-18.78 kJ) + 0 = -8.28 kJ
0
0
(c) To get Q12, we apply the energy balance to process 1-2: ΔKE + ΔPE + (U2 – U1) = Q12 – W12
With U2 = U3,
Q12 = (U3 – U1) + W12 = (+26.4 kJ) + (10.5 kJ) = 36.9 kJ
PROBLEM 2.80
Problem 2.81
The net work of a power cycle operating as in Fig. 2.17a is 10,000 kJ, and the thermal efficiency
is 0.4. Determine the heat transfers Qin and Qout, each in kJ.
→
Qin
Qin = (10,000 kJ) / (0.4) = 25,000 kJ
Wcycle = 10,000 kJ
η = 0.4
Wcycle = Qcycle = Qin - Qout
Thus
Qout = Qin – Wcycle = 25,000 – 10,000 = 15,000 kJ
Problem 2.82
For a power cycle operating as shown in Fig. 2.17a, the energy transfer by heat into the cycle,
Qin, is 500 MJ. What is the net work developed, in MJ, if the cycle thermal efficiency is 30%?
What is the value of Qout, in MJ?
η
Wcycle = ηQin = (0.3)(500 MJ) = 150 MJ
Qin = 500 MJ
Wcycle = Qcycle = Qin - Qout
n
Thus
η = 30%
Qout = Qin – wcycle = 500 MJ – 150 MJ = 350 MJ
Problem 2.83
For a power cycle operating as in fig. 2.17a, Qin = 17 x 106 Btu and Qout = 12 x 106 Btu.
Determine Wcycle, in Btu, and η.
Wcycle = Qcycle = Qin – Qout
= (17 x 106) – (12 x 106) = 5 x 106 Btu
Qin = 17 x 106 Btu
η=
=
= 0.294 (29.4%)
Alternatively
Qout = 12 x 106 Btu
η=1
=1
= 0.294
PROBLEM 2.84
)
Problem 2.85
A concentrating solar collector system, as shown in Fig. P2.85, provides energy by heat transfer
to a power cycle at a rate of 2 MW. The cycle thermal efficiency is 36%. Determine the power
developed by the cycle, in MW. What is the work output, in MW∙h, for 4380 hours of steadystate operation? If the work is valued at \$0.08/kW∙h, what is the total dollar value of the work
output?
Power
Cycle
= 2 MW
Atmosphere
The power developed is
=η
= (0.36) (2 MW) = 0.72 MW
For 4380 hours of steady-state operation
Wcycle =
Δt = (0.72 MW)(4380 h) = 3153.6 MW∙h
The total dollar value is
\$ Value = (3153.6 MW∙h)(\$0.08/kW∙h)
= \$252,300
PROBLEM 2.86
PROBLEM 2.87
PROBLEM 2.88
Problem 2.89
PROBLEM 2.90
PROBLEM 2.91
Problem 2.92
A window-mounted room air conditioner removes energy by heat transfer from a room and
rejects energy by heat transfer to the outside air. For steady operation, the air conditioner cycle
requires a power input of 0.434 kW and has a coefficient of performance of 6.22. Determine the
rate that energy is removed from the room air, in kW. If electricity is valued at \$0.1/kW∙h,
determine the cost of operation for 24 hours of operation.
KNOWN: Steady-state operating data are provided for an air conditioner.
FIND: Determine the rate energy is removed from the room and air the cost of 24 hours of
operation.
SCHEMATIC AND GIVEN DATA:
Room air
Outside air
Air
Conditioner
= 0.434 kW
ENGINEERING MODEL:
(1) The system shown in the schematic
undergoes a refrigeration cycle.
(2) Energy transfers are positive in the
directions of the arrows.
(3) The cycle operates steadily for 24
hours.
(4) Electricity is valued at \$0.1/kW∙h.
Refrigeration
Cycle, β =
6.22
Electric cost:
\$0.1/kW∙h
ANALYSIS: Using Eq. 2.45 on a time rate basis
β = 6.22
β=
→
(6.22)(0.434 kW) = 2.70 kW
The total amount of electric energy input by work for 24 h of operation is
Wcycle =
= (0.434 kW)(24 h) = 10.42 kW∙h
Thus, the total cost is
Total cost = (10.42 kW∙h)(\$0.1/kW∙h) = \$1.04 (for 24 hours)
PROBLEM 2.93
PROBLEM 2.94
PROBLEM 2.95
PROBLEM 2.96
PROBLEM 3.1
A system consisting of liquid water and ice undergoes a process. At the end of the process, the
ice has melted and the system contains only liquid water. Can the system be considered a pure
substance during the process? Explain.
liquid water
→
+ ice
→
liquid water
The system is a pure substance. Although
the phases change, the system remains of
fixed chemical composition and is
chemically homogenous.
PROBLEM 3.2
A system consists of liquid nitrogen in equilibrium with nitrogen vapor. How many phases are
present? The system undergoes a process during which all of the liquid is vaporized. Can the
system be viewed as a pure substance during the process? Explain.
vapor
nitrogen
Initially, two phases are present: saturated liquid
and saturated vapor.
→
liquid
nitrogen
The system is a pure substance. Although liquid
is vaporized, the system remains fixed in
chemical composition and is chemically
homogenous.
PROBLEM 3.3
A system consists of liquid water in equilibrium with a gaseous mixture of air and water vapor.
How many phases are present? Does the system consist of a pure substance? Explain. Repeat
for a system consisting ice and liquid water in equilibrium with a gaseous mixture of air and
water vapor.
PROBLEM 3.4
An open container of pure ethanol (ethyl alcohol) liquid is placed on a table in a room.
Evaporation occurs until all of the ethanol is gone. Where did the alcohol go? If the ethanol
liquid and the room air are taken to be a closed system, can the system be regarded as a pure
substance during the process? How many phases are present initially and finally? Explain.
The ethanol vaporizes and diffuses into the room air.
air in room
ethanol
The system is not a pure substance during the process
since the composition of the gas phase changes as
ethanol evaporates into the air. Also, the liquid and
gas phases each have different chemical compositions,
so the system is not chemically homogenous
Once all of the ethanol evaporates, the gas phase
comes to equilibrium and the composition becomes
homogeneous. At this point, the gas phase can be
treated as a pure substance.
PROBLEM 3.5
Determine the phase or phases in a system consisting of H2O at the following conditions and
sketch the p-v and T-v diagrams showing the location of each state.
(a)
(b)
(c)
(d)
(e)
p = 100 lbf/in.2, T = 327.86oF
p = 100 lbf/in.2, T = 240oF
T = 212oF, p = 10 lbf/in.2
T = 70oF, p = 20 lbf/in.2
p = 14.7 lbf/in.2, T = 20oF
(a) p = 100 lbf/in.2, T = 327.86oF
100 lbf/in.2
p
Two-phase
liquid-vapor
mixture
T
327.86oF
100 lbf/in.2
327.86oF (Table A-3E)
v
v
(b) p = 100 lbf/in.2, T = 240oF
p
T<Tsat@p
sub-cooled liquid
100 lbf/in.2
T
327.86oF
.
100 lbf/in.2
327.86oF (Table A-3E)
240oF
v
.
240oF
v
Problem 3.5 (Continued)
(c) T = 212oF, p = 10 lbf/in.2
p<psat@T
superheated
vapor
p
T
14.7 lbf/in.2
10 lbf/in.2
.
14.7 lbf/in.2
.
o
212oF (Table A-3E)
10 lbf/in.2
212 F
v
v
(d) T = 70oF, p = 20 lbf/in.2
p>psat@T
sub-cooled liquid
p
.
T
20 lbf/in.2
0.3632 lbf/in.2 (Table A-2E)
20 lbf/in.2
0.3632 lbf/in.
70oF
.
2
v
2
70oF
v
o
(e) p = 14.7 lbf/in. , T = 20 F
p>psat@T
solid
(T is below the triple
point temperature)
p
.
14.7 lbf/in.
T
14.7 lbf/in.2
0.0505 lbf/in.2 (Table A-5E)
2
0.0505 lbf/in.2
20oF
v
.
20oF
v
PROBLEM 3.6
Determine the phase or phases in a system consisting of H2O at the following conditions and
sketch the p-v and T-v diagrams showing the location of each state.
(a)
(b)
(c)
(d)
(e)
p = 10 bar, T = 179.9oC
p = 10 bar, T = 150oC
T = 100oC, p = 0.5 bar
T = 20oC, p = 50 bar
p = 1 bar, T = - 6oC
(a) p = 10 bar, T = 179.9oC
10 bar
p
Two-phase
liquid-vapor
mixture
T
179.9oC
10 bar
179.9oC
(Table A-3)
v
v
(b) p = 10 bar, T = 150oC
p
T<Tsat@p
sub-cooled liquid
10 bar
T
179.9oC
.
10 bar
179.9oC
150oC
v
.
150oC
v
Problem 3.5 (Continued)
(c) T = 100oC, p = 0.5 bar
p<psat@T
superheated
vapor
p
.
T
1.014 bar (Table A-2)
0.5 bar
.
1.014 bar
100oC
0.5 bar
o
100 C
v
v
(d) T = 20oC, p = 50 bar
p>psat@T
sub-cooled liquid
p
.
T
50 bar
0.02339 bar (Table A-2)
50 bar
.
0.02339 bar
20oC
20oC
v
v
(e) p = 1 bar, T = - 6oC
p>psat@T
solid
(T is below the triple
point temperature)
p
.
T
1 bar
0.003689 bar (Table A-5)
1 bar
0.003689 bar
-6 oC
v
.
-6 oC
v
PROBLEM 3.7
PROBLEM 3.8
PROBLEM 3.9
COMMENT: As the pressure increases, the difference in specific volume between saturated
vapor and saturated liquid decreases. At the critical pressure, the two states
coincide and the difference is zero.
PROBLEM 3.10
For H2O, determine the specified property at the indicated state. Locate the state on a sketch of
the T-v diagram.
(a)
(b)
(c)
(d)
T = 140oC, v = 0.5 m3/kg. Find p, in bar.
p = 30 MPa, T = 100oC. Find v, in m3/kg.
p = 10 MPa, T = 485oC. Find v, in m3/kg.
T = 80oC, x = 0.75. Find p, in bar, and v, in m3/kg.
(a) T = 140oC, v = 0.5 m3/kg. Find p, in bar.
p = psat(140oC) = 3.613 bar
T
.
f
Table A-3: vf = 1.0435 x 10-3 m3/kg , vg = 1.673
m3/kg. Since vf < v < vg, the state is in the twophase liquid-vapor region, as shown.
140oC
g
From Table A-3, the pressure is the saturation
pressure at 140oC: p = 3.613 bar.
v
(b) p = 30 MPa, T = 100oC. Find v, in m3/kg.
30 MPa = 300 bar
pc = 220.9 bar (Table A-3)
T
.
.
Tc = 374.14oC (Table A-3)
The pressure is higher than the critical
pressure, as shown on the diagram. Hence, the
state is in the compressed liquid region.
100oC
From Table A-5: v = 1.0290 m3/kg.
v
Problem 3.10 (Continued)
(c) p = 10 MPa, T = 485oC. Find v, in m3/kg.
10 MPa = 100 bar
485oC
.
T
.
Tc = 374.14oC
Since the temperature is higher than Tsat at 100
bar, the state is superheated vapor.
Interpolating in Table A-4, we get
Tsat (100 bar) = 311.1oC
(Table A-3)
v = 0.03160 +
(485 – 480)/(520 – 480)*(0.02343 – 0.3160)
= 0.03058 m3/kg
v
(d) T = 80oC, x = 0.75. Find p, in bar, and v, in m3/kg.
T
psat (80oC) = 0.4739 bar
(Table A-2)
Eq. 3.2: vx = vf + x(vg – vf)
.
With data from Table A-2 at 80oC
80oC
v = 1.0291 x 10-3 + (0.75)(3.407 - 1.0291 x 10-3)
x = 0.75
= 2.556 m3/kg
v
PROBLEM 3.11
For each case, determine the specific volume at the indicated state. Locate the state on a sketch
of the T-v diagram.
(a) Water at p = 1 bar, T = 20oC. Find v, in m3/kg.
(b) Refrigerant-22 at p = 40 lbf/in.2, x = 0.6. Find v, in ft3/lb.
(c) Ammonia at p = 200 lbf/in.2, T = 195oF. Find v, in ft3/lb.
(a) Water at p = 1 bar, T = 20oC. Find v, in m3/kg.
T
1 bar
..f
The state is in the compressed liquid region.
Since p < 25 bar, Table A-5 cannot provide the
needed data.
Tsat = 99.63oC
(Table A-3)
Using Eq. 3.11 as an approximation with data
from Table A-2:
o
20 C
v ≈ vf (20oC) = 1.0018 x 10-3 m3/kg
v
(b) Refrigerant-22 at p = 40 lbf/in.2, x = 0.6. Find v, in ft3/lb.
Table A-8E at 40 lbf/in.2 gives:
vf = 0.01198 ft3/lb, vg = 1.3277 ft3/lb
T
With Eq. 3.2
vx = vf + x(vg - vf)
40 lbf/in.2
.f
.
.
g
x = 0.6
= 0.01198 + (0.6)(1.3277 – 0.01198)
o
Tsat = 1.54 F
(Table A-8E)
v
= 0.8014 ft3/lb
PROBLEM 3.11 (CONTINUED)
(c) Ammonia at p = 200 lbf/in.2, T = 195oF. Find v, in ft3/lb.
T
200 lbf/in.2
.
Since T > Tsat, the state is in the superheated vapor
region. Interpolating in Table A-9E:
o
195 F
Tsat(200 lbf/in.2)
= 96.24oF
(Table A-8E)
v
v = 0.3497 +
(195 – 180)/(200 – 180)*(0.3661 – 0.3497)
= 0.3620 ft3/lb
PROBLEM 3.12
PROBLEM 3.13
PROBLEM 3.14
PROBLEM 3.15
PROBLEM 3.16
A 1-m3 tank holds a two-phase liquid-vapor mixture of carbon dioxide at – 17oC. The quality of
the mixture is 70%. For saturated carbon dioxide at – 17oC, vf = 0.9827 x 10-3 m3/kg and vg =
1.756 x 10-2 m3/kg. Determine the masses of saturated liquid and saturated vapor, each in kg.
What is the percent of the total volume occupied by saturated liquid?
saturated vapor
carbon
dioxide
T = -17oC
x = 0.7
V = 1 m3
saturated liquid
vf = 0.9827 x 10-3 m3/kg
vg = 1.756 x 10-2 m3/kg
First, find the total mass as follows:
vx = vf +x (vg - vf) = 0.9827 x 10 -3 + (0.7)(1.756 x 10-2 - 0.9827 x 10 -3) = 0.01258 m3/kg
Thus
m = V/vx = (1 m3)/(0.01258 m3/kg) = 79.46 kg
Now, using the definition of quality
mg = x m = (0.7) ( 79.46 kg) = 55.62 kg
mf = (1 - x) m = (1 – 0.7) (79.46 kg) = 23.84 kg
The volume of saturated liquid is
Vf = vf ∙mf = (0.9827 x 10-3 m3/kg) ∙ (23.84 kg) = 0.0234 m3
The total volume is 1 m3, so the percent of the total volume occupied saturated liquid is 2.34%.
Note: Although the liquid is 30% of the total mass, its specific volume is much less than that of
the vapor. Consequently, the liquid occupies a very small fraction of the total volume.
PROBLEM 3.17
Determine the volume, in ft3, of 2 lb of a two-phase liquid-vapor mixture of Refrigerant 134A at
40oF with a quality of 20%. What is the pressure, in lbf/in.2
saturated vapor
R-134A
T = 40oF
x = 0.2
m = 2 lb
saturated liquid
First, find the specific volume using Eq. 3.2 and data from Table A-10E at 40oF.
v = vf + x (vf - vg) = 0.01251 + (0.2) (0.9470 – 0.1251) = 0.28948 ft3/lb
Now
V = v m = (0.1994 ft3/lb) (2 lb) = 0.579 ft3
p
.f x=. 0.2 g.
p = 49.738 lbf/in.2 (Table A-10E)
T = 40oF
v
PROBLEM 3.18
PROBLEM 3.19
A tank contains a two-phase liquid-vapor mixture of Refrigerant 22 at 10 bar. The mass of
saturated liquid in the tank is 25 kg and the quality is 60%. Determine the volume of the tank, in
m3, and the fraction of the total volume occupied by saturated vapor.
R – 22
saturated vapor
p = 10 bar
T
p = 10 bar
x = 0.6
saturated liquid
mf = 25 kg
.
x= 0.6
T = 23.40oC
v
First, determine the specific volume using Eq. 3.2 and data from Table A-8 at 10 bar.
vx = vx + x (vg – vf) = 0.8352 x 10-3 + (0.6) (0.0236 - 0.8352 x 10-3) = 0.01449 m3/kg
The total mass is determined from the mass of saturated liquid and the definition of quality, as
follows.
m = mf / (1 – x) = (25 kg) / (1 – 0.6) = 62.5 kg
Now, the volume is
V = vx m = (0.01449 m3/kg) (62.5 kg) = 0.9056 m3
The volume occupied by saturated vapor is
Vg = vg (m – mf) = (0.0236 m3/kg) (62.5 – 25)kg = 0.885 m3
fraction occupied by vapor = (0.885)/(0.9056) = 0.977 (97.7%)
Note: Even though the vapor is only 60% of the mixture by mass, it occupies nearly the entire
volume because the specific volume of saturated liquid is much smaller than the specific volume
of saturated vapor at this pressure.
PROBLEM 3.20
PROBLEM 3.21
PROBLEM 3.22
Ammonia, initially at 6 bar, 40oC, undergoes a constant volume process in a closed system to a
final pressure of 3 bar. At the final state, determine the temperature, in oC, and the quality.
Locate each state on a sketch of the T-v diagram.
constant
volume
6 bar
T
1
Ammonia
.
40oC
3 bar
.
2
-9.24oC
v
The initial state is in the superheated vapor region. From Table A-25, v1 = 0.24118 m3/kg. The
system is a closed system (constant mass) and the volume is constant. Therefore, v2 = v1. From
Table A-14 at v2 = 0.24118 m3/kg, the state is in the two-phase liquid-vapor region, and
T2 = Tsat(3 bar) = -9.24oC
The quality is
x2 =
= 0.5924 (59.24%)
PROBLEM 3.23
PROBLEM 3.24
PROBLEM 3.25
COMMENT: The values obtained using IT:Interactive Thermodynamics compare favorably
with those obtained using table data.
PROBLEM 3.26
A closed, rigid tank contains a two-phase liquid-vapor mixture of Refrigerant-22 initially at -20oC
with a quality of 50.36%. Energy transfer by heat into the tank occurs until the refrigerant is at a
final pressure of 6 bar. Determine the final temperature, in oC. If the final state is in the superheated
vapor region, at what temperature, in oC, does the tank contain only saturated vapor?
State 1
State 2
R-22
T1 = -20oC
x1 = 0.5036
V constant
m constant
R-22
p2 = 6 bar
Therefore, v2 = v1
T
6 bar
.
2
.
1
x1 = 0.5036
0oC (vg = 0.0470 m3/kg)
-20oC
v
First, using data from Table A-7 and Eq. 3.2, we can determine v1 as follows
v1 = vf1 + x1(vg1 – vf1) = 0.7427 x 10-3 + (0.5036)(0.0926 - 0.7427 x 10-3) = 0.0470 m3/kg
Since v2 = v1, State 2 is in the superheated vapor region (v2 > vg@6bar). Thus, interpolating at 6 bar
with v2 = 0.0470 m3/kg in Table A-9 we get
T2 ≈ 43.75oC
Since State 2 is superheated vapor, the tank contains only saturated vapor at the condition where vg =
0.0470 m3/kg. Referring to Table A-7, this occurs at T = 0oC.
PROBLEM 3.27
PROBLEM 3.28
PROBLEM 3.29
PROBLEM 3.30
PROBLEM 3.31
A piston-cylinder assembly contains a two-phase liquid-vapor mixture of H2O at 200 lbf/in.2
with a quality of 80%. The mixture is heated and expands at constant pressure until a final
temperature of 480oF is reached. Determine the work for the process, in Btu per lb of H2O
present.
SCHEMATIC AND GIVEN DATA:
p
H2O
p = 200 lbf/in.2
.2
.
x = 0.8
1
1
200 lbf/in.2
480o
f
v
ENGINEERING MODEL: 1. The water is a
closed system. 2. The pressure is constant.
ANALYSIS: Since the pressure is constant, Eq. 2.17 can be used to determine the work per unit
mass of H2O. First, fix each state and find the respective specific volumes.
For State 1, the specific volume can be determined using Eq. 3.2 with data from Table A-3E, as
follows.
v1 = vf1 + x1(vg1 – vf1) = 0.01839 + (0.8)(2.289 – 0.01839) = 1.835 ft3/lb
At State 2, T2 > Tsat(200 lbf/in.2), so the state is in the superheated vapor region. Interpolating in
Table A-4e at 200 lbf/in.2 and 480oF we get
v2 ≈ 2.548 + [(480 – 450)/(500 – 450)]*(2.724 – 2.548) = 2.654 ft3/lb
Now, with Eq. 2.17
W=
Thus
W/m =
= 30.32 Btu/lb (out)
Note, the work is positive, indicating that the transfer of energy is out of the system as expected.
PROBLEM 3.32
Seven lb of propane in a piston-cylinder assembly, initially at p1 = 200 lbf/in.2 and T1 = 200oF,
undergoes a constant-pressure process to a final state. The work for the process is – 88.84 Btu.
At the final state, determine the temperature, in oF, if superheated, and the quality if saturated.
KNOWN: Propane undergoes a process at constant pressure in a piston-cylinder assembly for
which data are provided at the initial state and the work is specified.
FIND: Determine specified data at the final state.
SCHEMAITC AND GIVEN DATA:
ENGINEERING MODEL: (1) the
given mass of propane is the closed
system. (2) Volume change is the only
work mode. (3) The process occurs at
constant pressure.
Propane
m = 7 lb
p1 = 200 lbf/in.2
T1 = 200oF
W = -88.84 Btu
ANALYSIS: Two properties are required to fix the State 2. One of these is the pressure, p2 = p1
= 200 lbf/in.2 The other is specific volume found from the given values for work as follows.
Since volume change is the only work mode and the pressure is constant, we can use Eq. 2.17 to
get
W=
= mp(v2 – v1)
Solving for v2
v2 = W/mp + v1
From Table A-18E, v1 = 0.7026 ft3/lb. Thus
v2 =
+ 0.7025 ft3/lb = 0.3597 ft3/lb
From Table A-17E we see that at 200 lbf/in.2, vf < v2 < vg. thus, State 2 is a two-phase liquidvapor mixture, and the quality is
x2 =
=
= 0.662 (66.2%)
p
.
2
.
1
200 lbf/in.2
200oF
v
PROBLEM 3.33
Two kg of Refrigerant 134A undergoes a polytropic process in a piston-cylinder assembly from
an initial state of saturated vapor at 2 bar to a final state of 12 bar, 80oC. Determine the work for
the process, in kJ.
KNOWN: Refrigerant 134A undergoes a polytropic process in a piston-cylinder assembly.
FIND: Determine the work.
SCHEMATIC AND GIVEN DATA:
p
R-134A
m = 2 kg
.2
ENGINEERING MODEL: 1. The refrigerant is a
closed system. 2. The process is polytropic: pvn =
constant.
12 bar
80oC
.
1
pvn = constant
2 bar
v
ANALYSIS: The work for the polytropic process is determined using Eq. 2.17, with pvn =
constant. Following the procedure of part (a) of Ex. 2.1
(*)
W=
In order to evaluate this expression, we need to determine the specific volumes and the
polytropic exponent, n.
State 1: From Table A-11; v1 = vg1 = 0.0993 m3/kg
State 2: From Table A-12, at 12 bar, 80oC; v2 = 0.02051 m3/kg
The polytropic exponent is found from pvn = constant as follows.
→
→
n = ln(p1/p2) / ln(v2/v1)
n = ln(2/12) / ln(0.02051/0.0993) = 1.136
Inserting values in Eq. (*) and converting units, we get
PROBLEM 3.33 (CONTINUED)
W=
= -69.88 kJ (in)
PROBLEM 3.34
PROBLEM 3.35
PROBLEM 3.35 (Continued)
PROBLEM 3.36
PROBLEM 3.37
PROBLEM 3.38
For each of the following cases, determine the specified properties and show the states on a
sketch of the T-v diagram.
(a) For Refrigerant 22 at p = 3 bar and v = 0.05 m3/kg, determine T in oC and u in kJ/kg.
(b) For water at T = 200oC and v = 0.2429 m3/kg, determine p in bar and h in kJ/kg.
(c) For ammonia at p = 5 bar and u = 1400 kJ.kg, determine T in oC and v in m3/kg.
(a) For Refrigerant 22 at p = 3 bar and v = 0.05 m3/kg, determine T in oC and u in kJ/kg.
From Table A-8, vf < v < vg, so the state is in the two-phase
liquid-vapor region and T = -14.66oC.
T
3 bar
.f
The quality is
x = (v - vf) / (vg - vf)
-14.66 oC
= (0.05 – 0.7521 x 10-3)/( 0.0765 – 0.7521 x 10-3) = 0.65
.x g.
v
The specific internal energy is
u = uf + x(ug – uf)
= 27.99 + (0.65)(221.34 – 27.99) = 153.67 kJ/kg
(b) For water at T = 200oC and v = 0.4249 m3/kg, determine p in bar and h in kJ/kg.
T
. .
200 oC
g
From Table A-2, v > vg, so the state is in the
superheated vapor region. Scanning across at a
constant temperature of 200oC in Table A-4, we find
that v = 0.4249 m3/kg at p = 5 bar.
The specific enthalpy is
h = 2855.4 kJ/kg
v
(c) For ammonia at p = 5 bar and u = 1400 kJ.kg, determine T in oC and v in m3/kg.
T
.
g
5 bar
44.46 oC
.
v
From Table A-14, u > ug at 5 bar , so the state is in the superheated
vapor region. Interpolating in Table A-15 with p = 5 bar and u =
1400 kJ/kg, we get
T ≈ 40 oC + [(1400 – 1391.74)/(1428.76 – 1391.74)] (60 – 40oC)
= 44.46oC
The specific volume is
v ≈ 0.29227 +
[(1400 – 1391.74)/(1428.76 – 1391.74)] (0.31410 – 0.29277)
= 0.2971 m3/kg
PROBLEM 3.39
Determine the values of the specified properties at each of the following conditions and show the
states on a sketch of the T-v diagram.
(a) For Refrigerant 22 at p = 60 lbf/in.2 and u = 50 Btu/lb, determine T in oF and v in ft3/lb.
(b) For Refrigerant 134a at T = 120 oF and u = 114 Btu/lb, determine p in lbf/in.2 and v in ft3/lb.
(c) For water vapor at p = 100 lbf/in.2 and h = 1240 Btu/lb, determine T in oF, v in ft3/lb, and u
in Btu/lb.
(a) For Refrigerant 22 at p = 60 lbf/in.2 and u = 50Btu/lb in kJ/kg, determine T in oF and v in
ft3/lb.
From Table A-8E, uf < u < ug, so the state is in the two-phase
liquid-vapor region and T = 21.96 oF.
T
60 lbf/in.2
.f
.x g.
21.96 oF
v
The quality is
x = (u - uf) / (ug - uf)
= (50 – 16.48)/( 96.62 – 16.48) = 0.418
The specific volume is
v = vf + x(vg –vf)
= 0.01232 + (0.418)(0.9014 – 0.01232) = 0.384 ft3/lb
(b) For Refrigerant 134a at T = 120oF and u = 114 Btu/lb, determine p in lbf/in.2 and v in ft3/lb.
T
. .
g
120 oF
From Table A-10E at 120oF, u > ug, so the state is in the
superheated vapor region. Scanning across at a constant
temperature of 120oF in Table A-12E, we find that u = 114 m3/kg
between pressures of 60 and 70 lbf/in.2
Interpolating at T = 120oF and u = 114 Btu/lb, we get
v
p ≈ 60 + [(114 – 113.96)/(114.35 - 113.96)] (70 – 60)
= 61.03 lbf/in.2
The specific volume is
v ≈ 0.9482 + [(114 – 113.96)/(114.35 - 113.96)](0.8023 - 0.9482)
= 0.9332 ft3/lb
PROBLEM 3.39 (CONTINUED) – PAGE 2
(c) For water vapor at p = 100 lbf/in.2 and h = 1240 Btu/lb, determine T in oF, v in ft3/lb, and u
in Btu/lb.
T
p = 100 lbf/in.2
423.9 oF
.
g
.
v
From Table A-3E, h > hg at 100 lbf/in.2, so the state is in the
superheated vapor region. Interpolating in Table A-4E with p =
100 lbf/in.2 and h = 1240 Btu/lb, we get
T ≈ 400oF + [(1240 – 1227.5)/(1253.6 – 1227.5)] (450 – 400)oF
= 423.9oF
The specific volume is
v ≈ 4.934 +
[(1240 – 1227.5)/(1253.6 – 1227.5)] (5.265 – 4.934)
= 5.093 ft3/lb
PROBLEM 3.40
Using IT, determine the specified property data at the indicated states. Compare with the results
from the appropriate table.
(a) Cases (a), (b), and (c) of Problem 3.38.
(b) Cases (a), (b), and (c) of Problem 3.39.
Problem 3.38
Case (a)
p=3 // bar
v = 0.05 // m^3/kg
x = x_vP("R22", v, p)
u = usat_Px("R22", p, x)
IT Results
T = -14.66oC
x = 0.65
u = 153.7 kJ/kg
Table Results
T = -14.66oC
x = 0.65
u = 153.67 kJ/kg
Case (b)
T = 200 // C
v = 0.4229 // m^3/kg
v = v_PT("Water/Steam", p, T)
h = h_PT("Water/Steam", p, T)
IT Results
p = 5.023 bar
h = 2855 kJ/kg
Table Results
p = 5 bar
h = 2855.4 kJ/kg
Case (c)
p = 5 // bar
u = 1400 // kJ/kg
u = u_PT("Ammonia", p, T)
v = v_PT("Ammonia", p, T)
Problem 3.39
Case (a)
p = 60 // lbf/in^2
u = 50 // Btu/lb
T = Tsat_P("R22", p)
u = usat_Px("R22", p, x)
v = vsat_Px("R22", p, x)
IT Results
T = 44.42oC
v = 0.2972 m3/kg
Table Results
T = 44.46oC
v = 0.2971 m3/kg
IT Results
T = 21.96oF
x = 0.4187
v = 0.3846 ft3/lb
Table Results
T = 21.96oF
x = 0.418
v = 0.384 ft3/lb
IT Results
p = 62 lbf/in.2
v = 0.9153 ft3/lb
Table Results
p = 61.03 lbf/in.2
v = 0.9332 ft3/lb
Case (b)
T = 120 // oF
u = 114 // Btu/lb
u = u_PT("R134A", p, T)
v = v_PT("R134A", p, T)
Case (c)
p = 100 // lbf/in.^2
h = 1240 // Btu/lb
T = T_Ph("Water/Steam", p, h)
v = v_Ph("Water/Steam", p, h)
IT Results
T = 424.1oF
v = 5.095 ft3/lb
Table Results
T = 423.9oF
v = 5.093 ft3/lb
The results compare favorably. Note that finding the states using IT requires identifying the
region (e.g. superheated vapor, two-phase liquid-vapor,…) in order to choose the correct
functional forms for the data expressions.
PROBLEM 3.41
PROBLEM 3.42
PROBLEM 3.43
PROBLEM 3.44
(a)
(b)
(c)
(d)
PROBLEM 3.44 (CONTINUED)
(e)
PROBLEM 3.45
PROBLEM 3.45 (CONTINUED)
PROBLEM 3.46
PROBLEM 3.47
4. The volume is constant.
PROBLEM 3.48
PROBLEM 3.49
Revised 11-14
PROBLEM 3.50
Refrigerant 22 undergoes a constant pressure process within a piston-cylinder assembly from
saturated vapor at 4 bar to a final temperature of 30oC. Kinetic and potential energy effects are
negligible. For the refrigerant, show the process on a p-v diagram. Evaluate the work and the
heat transfer, each in kJ per kg of refrigerant.
KNOWN: Data are provided for a process of Refrigerant 22 in a piston-cylinder assembly.
FIND: Show the process on a p-v diagram and evaluate the work and heat transfer, each per unit
mass of refrigerant.
SCEMATIC AND GIVEN DATA:
p
R - 22
constant pressure process
.
1
ENGINEERING MODEL: (1) The refrigerant
is the closed system. (2) The refrigerant expands
at constant pressure. (3) Volume change is the
only work mode. (4) Kinetic and potential
energy effects are negligible.
.
2
30oC
-6.56oC
(Table A-8)
v
ANALYSIS: The process is shown on the p-v diagram. Since volume change is the only work
mode and pressure is constant, Eq. 2.17 is used to evaluate the work, as follows.
→
From Table A-8, v1 = vg(4 bar) = 0.0581 m3/kg, and from Table A-9, v2 = 0.06872 m3/kg.
Inserting values and converting units, we get
W12/m = (4 bar) (0.06872 – 0.0581) m3/kg
0
Now, the energy balance reduces to
= 4.248 kJ/kg (out)
0
, or, with ΔU = m(u2 – u1)
Q12/m = (u2 – u1) + W12/m
From Table A-8, u1 = ug(4 bar) = 224.24 kJ/kg, and from Table A-9, u2 = 245.73 kJ/kg. Thus
Q12/m = (245.73 – 224.24) kJ/kg + (4.248 kJ/kg) = 25.738 kJ/kg (in)
Revised 11-14
PROBLEM 3.51
For the system of Problem 3.26, determine the amount of energy transfer by heat, in kJ per kg of
refrigerant. Kinetic and potential energy effects can be neglected.
KNOWN: Refrigerant 22 undergoes a process involving energy transfer by heat in a closed,
rigid tank.
FIND: Determine the amount of energy transfer by heat per kg of refrigerant.
SCHEMATIC AND GIVEN DATA:
State 1
R-22
T1 = -20oC
x1 = 0.5036
State 2
V constant
m constant
R-22
p2 = 6 bar
Therefore, v2 = v1
T
6 bar
.
2
.
1
x1 = 0.5036
-20oC
ENGINEERING MODEL: (1) The refrigerant is the
closed system. (2) The tank is rigid and there is no
mechanism for work. (3) Kinetic and potential energy
effect can be neglected.
v
ANALYSIS: To find the heat transfer, we start with the closed system energy balance:
0
0
0
→
Q/m = (u2 – u1)
(*)
Using data from Table A-7 with the given initial quality to determine the value of u1
u1 = uf1 + x1(ug1 – uf1) = 21.99 + 0.5036(219.37 – 21.99) = 121.39 kJ/kg
From the solution to Problem 3.26; v2 = 0.0470 m3/kg. Interpolating in Table A-9 at p2 = 6 bar,
we get u2 ≈ 252.0 kJ/kg.
Inserting values into Eq. (*)
Q/m = (252.0 – 121.39) = 130.6 kJ/kg (in)
PROBLEM 3.52
For the system of Problem 3.31, determine the amount of energy transfer by heat, in Btu, if the
mass is 2 lb. Kinetic and potential energy effects can be neglected.
SCHEMATIC AND GIVEN DATA:
p
H2O
p = 200 lbf/in.2
m = 2 lb
.2
.
x = 0.8
1
1
200 lbf/in.2
480o
f
ENGINEERING MODEL: 1. The water is a closed
system. 2. The pressure is constant. 3. Kinetic and
potential energy effects can be neglected.
v
ANALYSIS: To find the heat transfer, we start with the closed system energy balance:
0
0
→
Q = m(u2 – u1) + W
(*)
The initial specific internal energy can be determined using data from Table A-3E, as follows.
u1 = uf1 + x1(ug1 – uf1) = 354.9 + (0.8)(1114.6 – 354.9) = 962.7 Btu/lb
Interpolating in Table A-4e at 200 lbf/in.2 and 480oF, we get u2 ≈ 1167.9 Btu/lb
From the solution to Problem 3.31, W/m = 29.58 Btu/lb. Therefore, W = (2 lb) (29.58 Btu/lb) =
59.16 Btu. Inserting values into Eq. (*)
Q = (2 lb)(1167.9 – 962.7) Btu/lb + (59.16 Btu) = 469.6 Btu (in)
PROBLEM 3.53
For the system of Problem 3.33, determine the amount of energy transfer by heat, in kJ. Kinetic
and potential energy effects can be neglected.
SCHEMATIC AND GIVEN DATA:
p
R-134A
m = 2 kg
.2
ENGINEERING MODEL: 1. The refrigerant is a
closed system. 2. The process is polytropic: pvn =
constant. 3. Kinetic and potential energy effects can
be neglected.
12 bar
80oC
.
1
pvn = constant
2 bar
v
ANALYSIS: To find the heat transfer, we start with the closed system energy balance:
0
0
→
Q = m(u2 – u1) + W
Next, we find the specific internal energy values, as follows.
State 1: u1 = ug(2 bar) = 221.43 kJ/kg (Table A-11)
State 2: p2 = 12 bar, T2 = 80oC: u2 = 285.62 kJ (Table A=12)
With W = -69.88 kJ (See the solution to Problem 3.33.), we get
Q = (2 kg) (285.62 – 221.43) kJ + (-69.88 kJ) = 58.5 kJ (in)
(*)
PROBLEM 3.54
PROBLEM 3.55
PROBLEM 3.56
PROBLEM 3.57
PROBLEM 3.58
PROBLEM 3.59
A well-insulated, rigid tank contains 1.5 kg of Refrigerant 134A initially a two-phase liquidvapor mixture with a quality of 60% and a temperature of 0oC. An electrical resistor transfers
energy to the contents of the tank at a rate of 2 kW until the tank contains only saturated vapor.
For the refrigerant, locate the initial and final states on a T-v diagram and determine the time it
takes, in s, for the process.
KNOWN: An electrical resistor transfers energy to refrigerant in a well-insulated, rigid tank at a
known rate. Data are given for the initial and final state of the refrigerant.
FIND: Determine the time it takes for the process.
Refrigerant
134A
m = 1.5 kg
x1 = 0.6
V is constant
m is constant
T
v1 = v2
→ v1 = v2
.
.
1
+
2
-
15.16oC
0oC
ENGINEERING MODEL: (1)The refrigerant is a
closed system. (2) For the process, Q = 0. (3) The
volume is constant. (4) Kinetic and potential energy
effects can be neglected.
v
ANALYSIS: Since the volume is constant for the closed system, v1 = v2. At state 1, using data
from Table A-10 at 0oC:
v1 = vf1 + x1( vg1 – vf1) = (0.7721 x 10-3) + (0.6)(0.0689 – 0.7721 x 10-3) = 0.04165 m3/kg
and
u1 = uf1+ x1(ug1 – uf1) = 49.79 + (0.6)(227.06 - 49.79) = 156.15 kJ/kg
To fix state 2, v2 = v1 = vg(T2). Interpolating in Table A-10 with vg(T2) = 0.04165 m3/kg; T2 ≈
15.16oC, and u2 = ug(15.16oC) ≈ 235.33 kJ/kg.
0
0
0
Using the energy balance with assumptions 2 and 4, we get ΔKE + ΔPE + ΔU = Q – Welec
Thus
Welec = m(u1 – u2) = (1.5 kg)(156.15 – 235.33)kJ/kg = -118.77 kJ
The time is determined from Welec =
Δt =
= (-118.77 kJ)/(-2 kW)
Δt as follows
= 59.4
PROBLEM 3.60
PROBLEM 3.61
PROBLEM 3.62
PROBLEM 3.63
PROBLEM 3.64
PROBLEM 3.65
Five lb of propane is contained in a closed, rigid tank initially at 80 lbf/in.2, 110oF. Heat transfer
occurs until the final temperature in the tank is 0oF. Kinetic and potential energy effects are
negligible. Show the initial and final states on a T- v diagram and determine the amount of
energy transfer by heat, in Btu.
Propane
m = 5 lb
p1 = 80 lbf/in.2
T1 = 110oF
T
80 lbf/in.2
V is constant
m is constant → v1 = v2
1
.
Q
110o
F
T2 = 0oF
2
ENGINEERING MODEL: 1. The propane is a closed
system. 2. Volume is constant. 3. Volume change is
the only work mode, so W = 0. 4. Kinetic and potential
energy effects can be neglected.
.
0o
F
v
ANALYSIS: Since the volume and mass are constant, v1 = v2. At State 1, from Table A-18E;
v1 = 1.599 ft3/lb and u1 = 210.9 Btu/lb
From Table A-16E at 0oF, v2 = 1.599 ft3/lb is between vf and vg. Thus, the state is a two-phase
liquid-vapor mixture, and
x2 =
= 0.5878
u2 = uf2 + x2(ug2 – uf2) = 21.98 + (0.5878)(174.01 – 21.98) = 111.34 Btu/lb
0
0
0
Applying the energy balance; ΔKE + ΔPE + ΔU = Q – W. Thus
Q = m(u2 – u1) = (5 lb)(111.34 – 210.9) Btu/lb = - 497.8 Btu (out)
PROBLEM 3.66
PROBLEM 3.67
PROBLEM 3.68
PROBLEM 3.69
T2 = 300 oF
PROBLEM 3.70
PROBLEM 3.70 (CONTINUED)
1. An iterative solution can be developed using Eq. (2) together with Steam Table data. However, this
is less direct than Method #1, which centers only on interpolation with h2.
PROBLEM 3.71
PROBLEM 3.72
REVISED 12-14
A piston-cylinder assembly contains 2 lb of water, initially at 100 lbf/in.2 and 400oF. The water
undergoes two processes in series: a constant pressure process followed by a constant volume
process. At the of the constant volume process, the temperature is 300oF and the water is a twophase liquid-vapor mixture with a quality of 60%. Neglect kinetic and potential energy effects.
T
(a) Sketch T-v and p-v diagrams showing the key states and the processes.
(b) Determine the work and heat transfer for each process, all in Btu.
KNOWN: Water contained in a pistoncylinder assembly undergoes two processes
in series.
p
2
FIND: Sketch the T-v and p-v diagrams and
for each process determine Q and W.
SCHEMATIC AND GIVEN DATA:
3
. .1
100 lbf/in.2
400oF
.
x3 = 0.6
300oF
Water
m = 2 lb
v
T
ENGINEERING MODEL: 1. The `water is a
closed system. 2. Volume change is the only work
mode. 3. Process 1-2 occurs at constant pressure and
Process 2-3 occurs at constant volume. 4. Kinetic and
potential energy effects can be neglected.
ANALYSIS: First, we fix each state. State 1 is in
the superheated vapor region. From Table A-4E;
v1 = 4.934 ft3/lb and u1 = 1136.2 Btu/lb.
.
.
3
2
100 lbf/in.2
1
400oF
.
300oF
x3 = 0.6
With T3 = 300oF and x3 = 0.6, we can evaluate v3 and
u3 using data from Table A-2E at 300oF as follows.
v3 = vf3 +x3(vg3 – vf3 ) = 0.01745 + (0.6)(6.472 – 0.01745) = 3.89 ft3/lb
u3 = uf3 + x3(ug3 – uf3) = 269.5 + (0.6)(1100.0 – 269.5) = 767.8 Btu/lb
Note that v2 = v3 = 3.89 ft3/lb, and from Table A-3E we see that v2 < vg(100 lbf/in.2). Thus
x2 =
=
= 0.8768
v
T
PROBLEM 3.72 – (CONTINUED) PAGE 2
and
u2 = uf2 + x2(ug2 – uf2) = 298.3 + (0.8768)(1105.8 – 298.3) = 1006.3 Btu/lb
Now, for Process 1-2 the pressure is constant. Thus
W12 =
= 38.65 Btu (in)
An energy balance reduces to give
Q12 = m(u2 – u1) + W12 = (2 lb)(1006.3 – 1136.2) Btu/lb + (– 38.65 Btu) = – 298.5 Btu (out)
Now, for Process 2-3, the volume is constant, so W23 = 0
And, the energy balance reduces to give
Q23 = m(u3 – u2) = (2 lb)(767.8 – 1006.3)But/lb = – 477 Btu (out)
REVISED 12-14
PROBLEM 3.73
PROBLEM 3.73 (CONTINUED) – PAGE 2
PROBLEM 3.73 (CONTINUED) – PAGE 3
PROBLEM 3.74
PROBLEM 3.74 (CONTINUED)
PROBLEM 3.75
PROBLEM 3.76
PROBLEM 3.77
PROBLEM 3.78
A system consisting of 1 kg of H2O undergoes a heat pump cycle composed of the following
processes.
Process 1-2: Constant-volume heating from p1 = 5 bar, T1 = 160oC to p2 = 10 bar.
Process 2-3: Constant-pressure cooling to saturated vapor.
Process 3-4: Constant-volume cooling to T4 = 160oC.
Process 4-1: Isothermal expansion with Q41 = 815.8 kJ.
Sketch the cycle on T-v and p-v diagrams. Neglecting kinetic and potential energy effects,
determine the coefficient of performance.
KNOWN: One kg of water undergoes a heat pump cycle consisting of four processes.
FIND: Sketch the cycle on T-v and p-v diagrams and determine the coefficient of performance.
SCHEMATIC AND GIVEN DATA:
Process 1-2: Constant-volume heating from
p1 = 5 bar, T1 = 160oC to p2 = 10 bar.
Process 2-3: Constant-pressure cooling to
saturated vapor.
Process 3-4: Constant-volume cooling to T4
= 160oC.
Process 4-1: Isothermal expansion with Q41
= 815.8 kJ.
10 bar
2
.
T
.
4.
p
10 bar
3
3
.
.2
.
.1
5 bar
.1
5 bar
4
160oC
v
ENGINEERING MODEL: 1. The water is the closed system. 2. Volume change is the
only work mode. 3. Kinetic and potential energy effects can be ignored.
v
PROBLEM 3.78 (CONTINUED) – PAGE 2
ANALYSIS: Referring to Sec. 2.6.3, the coefficient of performance of a heat pump cycle is
γ = Qout/Wcycle, where Wcycle = W12 + W23 + W34 + W41.
Process 1-2: Referring to Table A-2 at 160oC, p1 < psat so State 1 is in the superheated vapor
region. From Table A-4; v1 = 0.3835 m3/kg and u1 = 2575.2 kJ/kg.
State is fixed by p2 = 10 bar and v2 = v1 = 0.3835 m3/kg. Interpolating in Table A-4, we get
u2 = 3231.8 kJ/kg.
Since the volume is constant, W12 = 0, and the energy balance reduces to give
Q12 = m(u2 – u1) = (1 kg)(3231.8 – 2575.2) kJ/kg = 656.6 kJ
Process 2-3: State 3 is saturated vapor at 10 bar. From Table A-3; u3 = ug3 = 2583.6 kJ/kg and
v3 = vg3 = 0.1944 m3/kg.
For the constant-pressure cooling
= m p2(v3 – v2) = (1 kg)(
W23 =
= - 189.1 kJ
The energy balance reduces to
Q23 = m(u3 – u2) + W23 = (1 kg)(2583.6 – 3231.8) kJ/kg + (-189.1kJ) = -837.3 kJ
Process 3-4: State 4 is fixed by T4 = 160oC and v4 = v3 = 0.1944 m3/kg. The state is in the twophase liquid-vapor region, so
x4 =
= 0.6317
and
u4 = uf4 + x4(ug4 – uf4) = 674.86 + (0.6317)(2568.4 – 674.86) = 1871 kJ/kg
With W34 = 0, the energy balance reduces to give
Q34 = m(u4 – u3) = (1 kg)(1871 – 2583.6)kJ/kg = -712.6 kJ
Process 4-1: Q41 = 815.8 kJ (given). From the energy balance, m(u1 – u4) = Q41 – W41, so
1
W41 = m(u4 – u1) + Q41 = (1 kg)(1871 - 2575.2)kJ/kg + (815.8 kJ) = 111.6 kJ
Finally, the net work is Wcycle = W12 + W23 + W34 + W41 = (0) + (-189.1) + (0) + (111.6) = -77.5 kJ
PROBLEM 3.78 (CONTINUED) – PAGE 3
Now, Qout = Q23 + Q34 = (-837.3) + (-712.6) = -1549.9 kJ
So, the coefficient of performance is
γ = (-1549.9 kJ)/(-77.5 kJ) = 20.0
1
As a check, note that for every cycle, Qcycle = Wcycle. So
Qcycle = Q12 + Q23 + Q134 + Q41 = (656.6) + (-837.3) + (-712.6) + (815.8) = -77.5 kJ
Which agrees with the value of Wcycle calculated using the work quantities, as expected.
1
PROBLEM 3.79
PROBLEM 3.80
PROBLEM 3.81
PROBLEM 3.82
PROBLEM 3.83
PROBLEM 3.84
PROBLEM 3.85
PROBLEM 8.86
PROBLEM 3.87
PROBLEM 3.88
PROBLEM 3.89
PROBLEM 3.90
PROBLEM 3.91
PROBLEM 3.92
Determine the volume, in m3, occupied by 2 kg of H2O at 100 bar, 400oC, using (a) data from the
compressibility chart, (b) data from the steam tables.
Compare the results of parts (a) and (b) and discuss.
H2O
V = ??
m = 2 kg
p = 100 bar
T = 400oC
(a) Using the compressibility chart, first we need to determine the reduced pressure and
temperature. From Table A-1:
pc = 220.9 bar and Tc = 647.3 K
pR = p/pc = (100)/(220.9) = 0.45
→ (Figure A-1): z ≈ 0.86
TR = T/Tc = (400 + 273.15)/(647.3) = 1.04
Now, we can calculate the specific volume as follows.
v=
= (0.86)
= 0.0267 m3/kg
So, the volume is: V = m v = (2)(0.0267) = 0.0534 m3
(b) From Table A-4 at 100 bar, 400oC; v = 0.02641 m3/kg
Thus, V = m v = (2)(0.02641) = 0.05282 m3
Comments: The compressibility chart gives a fairly accurate value considering the relative
imprecision of reading values from the chart. The percent difference is approximately 1.1%.
Note also that the value of z is 0.86. Hence, the ideal gas model is not particularly applicable at
this state. The ideal gas model would predict a volume of 0.03105 m3, which is about 15% low.
PROBLEM 3.93
Five kmol of oxygen (O2) gas undergoes a process in a closed system from p1 = 50 bar, T1 = 170
K to p2 = 25 bar, T2 = 200 K. Determine the change in volume, in m3.
Oxygen
(O2)
n = 5 kmol
p1 = 50 bar
T1 = 170 K
→
P2 = 25 bar
T2 = 200 K
ΔV = ?
ANALYSIS: Find the volumes at each state using V = Zn T/p, with data for Z from the
compressibility chart.
From Table A-1 for oxygen (O2): pc = 50.5 bar, Tc = 154 K.
State 1: pR1 = p1/pc = (50)/(50.5) ≈ 1 and TR1 = T1/Tc = (170)/(154) ≈ 1.1
From Figure A-1: Z ≈ 0.71
Thus
V1 =
= 1.00 m3
State 2: pR2 = p2/pc = (25)/(50.5) ≈ 0.5 and TR1 = T1/Tc = (200)/(154) ≈ 1.13
From Figure A-1: Z ≈ 0.93
Thus
V2 =
= 3.09 m3
Finally
ΔV = 3.09 – 1.00 = 2.09 m3
Note that z ranges from 0.71 to 0.93. Hence, the ideal gas model would not be particularly
suitable for this calculation.
PROBLEM 3.94
PROBLEM 3.95
PROBLEM 3.96
PROBLEM 3.97
PROBLEM 3.98
Five kg of butane (C4H10) in a piston-cylinder assembly undergo a process from p1 = 5 MPa, T1 =
500 K to p2 = 3 MPa during which the relationship between pressure and specific volume is pv =
constant. Determine the work, in kJ.
KNOWN: Five kg of C4H10 undergo a process for which pv = constant in a piston-cylinder
assembly between two states.
FIND: Determine the work.
SCHEMATIC AND GIVEN DATA:
Butane
(C4H10)
m = 5 kg
p1 = 5 MPa = 50 bar
T1 = 500 K
→
p1 = 3 MPa = 30 bar
pv = constant
ENGINEERING MODEL: 1. The Butane is a closed system. 2. The process is polytropic with
pv = constant. 3. Volume change is the only work mode.
ANALYSIS: The work is given by W = m
. With pv = constant = p1v1
= (p1v1) m ln
W=m
(*)
To evaluate W requires v1 and v2. The compressibility chart can be used to obtain v1: From
Table A-1; pc = 38 bar, Tc = 425 K, M = 58.12 kg/kmol.
pR1 = p1/pc = (50)/(38) = 1.32
Fig. A-2: Z1 ≈ 0.67
TR1 = T/Tc = (500)/(425) = 1.18
Accordingly, with v1 = Z1 (RT1/p1), we get
v1 = (0.67)
= 0.0096 m3/kg
Now, with pv = constant,
= [(5 MPa)(0.0096 m3)]/(3 MPa) = 0.016 m3
Now, inserting values into (*), we get
PROBLEM 3.98 (CONTINUED) – PAGE 2
W = (5 MPa)(0.0096 m3/kg)(5 kg) ln [(0.016)/(0.0096)]
= 122.6 kJ (out)
Alternative Evaluation of
pR1 = p1/pc = (50)/(38) = 1.32
Fig. A-2:
≈ 0.6
TR1 = T/Tc = (500)/(425) = 1.18
v1 =
= (0.6)
= 0.0096 m3/g
PROBLEM 3.99
In a cryogenic application, carbon monoxide (CO) gas undergoes a constant pressure process at
1000 lbf/in.2 in a piston-cylinder assembly from T1 = -100oF to T2 = -30oF. Determine the work
for the process, in Btu per lb of carbon monoxide present.
KNOWN: Carbon monoxide undergoes a constant pressure process in a piston-cylinder
assembly. Temperatures are known at the initial and final states.
FIND: Determine the work per unit mass of carbon monoxide present.
p
SCHMEATIC AND GIVEN DATA:
o
.1
o
T1 = -100 F = 360 R
T2 = -30oF = 430oR
Carbon
Monoxide
(CO)
T1
.2
p = 1000 lbf/in.2 = 68.0 atm
v
ENGINEERING MODEL: 1. The CO is a closed system.
2. The process is at constant pressure and volume change is the
only work mode.
ANALYSIS: To evaluate the work, we note that W/m
determine the specific volumes at the initial and final states.
. We need to
From Table A-1E, for carbon monoxide: Tc = 239oR, pc = 34.5 atm and M = 28.01 lb/lbmol.
Thus, using data from Figure A-1:
State 1
TR1 = (360)/(239) = 1.5
pR1 = (68.0)/(34.5) = 1.97 ≈ 2
State 2
T2
TR2 = (430)/(239) = 1.8
pR2 = (68.0)/(34.5) = 1.97 ≈ 2
Z1 ≈ 0.84
Z1 ≈ 0.94
Now, the specific volumes are
v1 = Z1RT1/p1 =
= 11.58 ft3/lb
v2 = Z2RT2/p2 =
= 13.84 ft3/lb
and
PROBLEM 3.99 (CONTIN``` | 37,269 | 102,806 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2019-13 | longest | en | 0.617006 |
https://jjthetutor.com/physics-1-final-exam-study-guide-review-multiple-choice-practice-problems/ | 1,561,536,545,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000231.40/warc/CC-MAIN-20190626073946-20190626095946-00202.warc.gz | 473,263,938 | 16,978 | # Physics 1 Final Exam Study Guide Review – Multiple Choice Practice Problems
This physics video tutorial is for high school and college students studying for their physics midterm exam or the physics final exam. This study guide review tutorial contains 50 multiple choice practice problems.
Algebra Final Exam Review:
Calculus Final Exam Review:
General Chemistry 1 Review:
Organic Chemistry 1 Final Review:
Physics 2 Final Exam Review:
You can access the full video here:
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New Physics Video Playlist:
Here is a list of topics:
1. One Dimension Kinematic Problems
2. Average Speed and Average Velocity Problems
3. Displacement, Velocity and Acceleration Problems
4. Projectile Motion Physics Problems
5. Magnitude and Direction Angle of Vectors
6. Newton’s Second Law of Motion
7. Inclined Plane Physics Problems & Free Body Diagrams
8. Static Friction and Kinetic Friction
9. Net Force Problems
10. Tension Force and Normal Force Physics Problems
11. Maximum Height and Range of a Projectile
12. Launch Angle of a Projectile
13. Work Energy Principle
14. How To Calculate The Work Required To Lift an Object
15. Kinetic Energy and Potential Energy
16. Newton’s Third Law of Motion
17. Pulley Problems and Acceleration
18. Circular Motion and Tension In a Rope
19. Car Rounding a Curve – Static Friction and Maximum Speed
20. Centripetal Force and Circular Motion
21. Gravitational Force
22. How To Calculate The Speed of a Satellite
23. Conservation of Energy Problems
24. Elastic Potential Energy and Final Speed
25. Average Power, Force, and Velocity
26. Cost of Electricity and Kilowatt hours
27. Velocity Time Graphs, Slope, and Acceleration
28. Inelastic Collisions and Conservation of Momentum
29. Torque, Rotational Inertia, and Angular Acceleration
30. Conservation of Angular Momentum
31. Seesaw Problem With Torques & Mechanical Advantage
32. Rotational Kinematic Problems
33. Angular Speed and Linear Speed Problems
34. Net Force and Momentum
35. How To Calculate The Work Done Given a Force Displacement Graph
36. Work Done By a Force | 475 | 2,098 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2019-26 | longest | en | 0.660945 |
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# Decimal in Wrong Place
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## Decimal in Wrong Place
(OP)
I am trying to summarize multiple bill amounts and I am getting the right numbers but the decimal is in the wrong spot. When I use this statement to get bill amounts for a case, it returns the correct amounts:
BILL_TOTAL= sum(CASE [TRAN_TYPE] WHEN 'BL' THEN([FEES_AMT]+[HARD_AMT]+[SOFT_AMT])*[SIGN] WHEN 'BLX' THEN([FEES_AMT]+[HARD_AMT]+[SOFT_AMT])*[SIGN] ELSE 0 end)
Results: 3475.18000000 and 29740.70000000
However, when I sum those 2 amounts by client, I get 332158.80000000 instead of 33215.88000000. All the fields being added in my statement are EXPCURRENCY(decimal(25,10)),null)
How do I fix this??
### RE: Decimal in Wrong Place
Hard to tell with that field in isolation, but most likely, there's something in your code that's causing you to add these values in multiple times.
Tamar
### RE: Decimal in Wrong Place
#### Quote:
I am trying to summarize multiple bill amounts and I am getting the right numbers but...
Could it be that you are summarizing 10 rows, resulting in your result being 10 times as great as what you expect?
First return the DISTINCT values and then SUM().
Skip,
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for a NUance!
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Register now while it's still free! | 618 | 2,552 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-51 | latest | en | 0.844038 |
https://kr.mathworks.com/matlabcentral/cody/problems/57-summing-digits-within-text/solutions/1421614 | 1,582,236,723,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145282.57/warc/CC-MAIN-20200220193228-20200220223228-00470.warc.gz | 382,781,338 | 15,718 | Cody
# Problem 57. Summing Digits within Text
Solution 1421614
Submitted on 21 Jan 2018 by Jacob Lamb
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
str = '4 and 20 blackbirds baked in a pie'; total = 24; assert(isequal(number_sum(str),total))
2 Pass
str = '2 4 6 8 who do we appreciate?'; total = 20; assert(isequal(number_sum(str),total))
3 Pass
str = 'He worked at the 7-11 for \$10 an hour'; total = 28; assert(isequal(number_sum(str),total))
4 Pass
str = 'that is 6 of one and a half dozen of the other'; total = 6; assert(isequal(number_sum(str),total)) | 208 | 692 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2020-10 | latest | en | 0.751125 |
https://codeforces.com/blog/entry/72142 | 1,600,445,723,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400188049.8/warc/CC-MAIN-20200918155203-20200918185203-00033.warc.gz | 336,944,626 | 19,101 | spirited_away_'s blog
By spirited_away_, history, 9 months ago,
How to handle 0's in this problem. Like when you have input like 230005450569. When there is no 0's the answer will be 2^n-1 when C is infinity. Now for C we will count how many numbers we have to subtract from answer which are >= length of C. But how to handle 0's in this case like sample input above.
• +4
» 9 months ago, # | 0 Why don’t you just count the answer itself with DP? The length of possible number is maximum 10, so for every position you’d only need to check 10 possible numbers for validity.
• » » 9 months ago, # ^ | 0 How can we do that. When we look for C length then what about rest of digits. How we will handle them . And specially zeros.Since 12 can be answer but 00 can't be.Can you describe your approach in more detail.
• » » » 9 months ago, # ^ | ← Rev. 2 → 0 $a_i$ – the amount of splits considering only the first $i$ digits.$a_0 = 1$$a_i = \sum_{j=\max(0, i-10)}^{i-1} a_j$, if number formed by digits $[j..i)$ is valid (has no leading zeroes and is $\leq C$)
• » » » » 9 months ago, # ^ | 0 if the number formed by digits[j..i) is not valid then a[i] = 0. is it correct?
• » » » » » 9 months ago, # ^ | 0 I think you meant $a_j$, but yes, for that it’s enough to check first digit. Don’t forget that you still need to compare the whole number as well though.
• » » » » » » 9 months ago, # ^ | ← Rev. 4 → 0 yes, sorry. if j to i-1 is not valid then a[ j ] = 0. and by max(0,i-10) you mean max(0,i-C). Please correct me if wrong.
• » » » » » » » 9 months ago, # ^ | 0 C is not the boundary of the length of a number. It is a boundary of a number. 10 is coming from the fact that the maximum allowed value of C has 10 digits.
• » » » » » » » » 9 months ago, # ^ | 0 Ok, thanks i understood it.
» 9 months ago, # | ← Rev. 2 → 0 can you provide problem link?
» 9 months ago, # | +28 I don't trust authors who use Latex but still manage to write $<=$ instead of $\le$. | 627 | 1,992 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2020-40 | latest | en | 0.853576 |
https://www.transtutors.com/questions/plum-corporation-began-the-month-of-may-with-1-000-000-of-current-assets-a-current-r-2564484.htm | 1,537,887,006,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267161661.80/warc/CC-MAIN-20180925143000-20180925163400-00453.warc.gz | 896,971,603 | 19,088 | # Plum Corporation began the month of May with $1,000,000 of current assets, a current ratio of 2.7... Plum Corporation began the month of May with$1,000,000 of current assets, a current ratio of 2.70:1, and an acid-test ratio of 1.70:1. During the month, it completed the following transactions (the company uses a perpetual inventory system).
May 2 Purchased $55,000 of merchandise inventory on credit. 8 Sold merchandise inventory that cost$50,000 for $160,000 cash. 10 Collected$26,000 cash on an account receivable. 15 Paid $30,500 cash to settle an account payable. 17 Wrote off a$5,000 bad debt against the Allowance for Doubtful Accounts account. 22 Declared a $1 per share cash dividend on its 56,000 shares of outstanding common stock. 26 Paid the dividend declared on May 22. 27 Borrowed$115,000 cash by giving the bank a 30-day, 10% note. 28 Borrowed $125,000 cash by signing a long-term secured note. 29 Used the$240,000 cash proceeds from the notes to buy new machinery.
Required: Prepare a table showing Plum's (1) current ratio, (2) acid-test ratio, and (3) working capital, after each transaction. (Do not round intermediate calculations. Round your ratios to 2 decimal places and the working capitals to nearest dollar amount.Subtracted amount should be indicated with a minus sign.)
Answer is not complete Acid-Test Ratio Transaction Current Assets Quick Assets Current Liabilities Current Ratio Working Capital 1,000,000 629,630 1.70 Beginning 370,370 2.70 629,630 248V 148V 1,055,000 629,630 629,630 425,370 Balance May 8 (50,000) 160,000M 160,000 1,165,000 789,630 Balance (26,000) 26,000 425.370 274 1.92 x 815,630 1,165,000 Balance 739,630 785,130 1,134,500 394,870 Balance 739,630 Balance 1,134,500 780,130 389,870 2902 200x 744,630X Balance 1,134,500 780,130 744,630X 389,870 2.91X 2.00X (56,000) (56,000) (56,000) Balance 1,078,500 724,130 33 870 3.23) 2.1 x 744,630X | 588 | 1,900 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-39 | longest | en | 0.839737 |
http://mymathforum.com/physics/345391-probability-temperature-system-has-energy.html | 1,566,128,864,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313803.9/warc/CC-MAIN-20190818104019-20190818130019-00239.warc.gz | 129,838,705 | 8,471 | My Math Forum Probability at temperature in system has energy
Physics Physics Forum
December 1st, 2018, 08:27 PM #1 Newbie Joined: Dec 2018 From: Colombia Posts: 6 Thanks: 0 Probability at temperature in system has energy Salutations, I'm starting in statistical mechanics and reviewing some related studying cases I would like to understand what occurs in small systems with normal modes of vibration, for example, a small system that has 2 normal modes of vibration, with natural frequencies $\omega_1$ and $\omega_2=2\omega_1$. So, what would be the probability at a temperature T that the system would get an energy less than $5\omega_1/2$? if it's assumed the zero of energy is taken as its value at $T=0$. I would like any guidance for better understanding of that kind of cases because those problems are very interesting, specially how to approach them. Thanks for your attention.
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https://oeis.org/A041439 | 1,718,567,145,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861670.48/warc/CC-MAIN-20240616172129-20240616202129-00633.warc.gz | 392,357,158 | 4,162 | The OEIS mourns the passing of Jim Simons and is grateful to the Simons Foundation for its support of research in many branches of science, including the OEIS.
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A041439 Denominators of continued fraction convergents to sqrt(235). 2
1, 3, 91, 276, 8371, 25389, 770041, 2335512, 70835401, 214841715, 6516086851, 19763102268, 599409154891, 1817990566941, 55139126163121, 167235369056304, 5072200197852241, 15383835962613027, 466587279076243051, 1415145673191342180, 42920957474816508451 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..200 Index entries for linear recurrences with constant coefficients, signature (0, 92, 0, -1). FORMULA G.f.: -(x^2-3*x-1) / (x^4-92*x^2+1). - Colin Barker, Nov 17 2013 a(n) = 92*a(n-2) - a(n-4). - Vincenzo Librandi, Dec 17 2013 MATHEMATICA Denominator[Convergents[Sqrt[235], 30]] (* Vincenzo Librandi, Dec 17 2013 *) LinearRecurrence[{0, 92, 0, -1}, {1, 3, 91, 276}, 30] (* Harvey P. Dale, May 05 2018 *) PROG (Magma) I:=[1, 3, 91, 276]; [n le 4 select I[n] else 92*Self(n-2)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Dec 17 2013 CROSSREFS Cf. A041438, A176524, A040219. Sequence in context: A115704 A066751 A115886 * A366219 A300419 A156754 Adjacent sequences: A041436 A041437 A041438 * A041440 A041441 A041442 KEYWORD nonn,frac,easy AUTHOR N. J. A. Sloane. EXTENSIONS More terms from Colin Barker, Nov 17 2013 STATUS approved
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Geologists believe that the Bering land bridge, over which
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325. Geologists believe that the Bering land bridge, over which human beings are thought to have first entered the Americans, disappeared about 14,000 years ago when massive glaciers melted and caused the sea level to rise several hundred feet worldwide.
(A) are thought to have first entered
(B) were thought first to enter
(C) were thought at first to enter
(D) are thought of as first entering
(E) were thought to first enter
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16 Nov 2006, 11:34
jyotsnasarabu wrote:
dwivedys wrote:
A OKAY...
wokay
No! I said Aokay not Wokay
OK...Just kidding
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Display posts from previous: Sort by | 631 | 2,281 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2016-36 | longest | en | 0.930162 |
https://sourceforge.net/p/chessx/mailman/message/32411083/ | 1,469,940,823,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469258948913.96/warc/CC-MAIN-20160723072908-00008-ip-10-185-27-174.ec2.internal.warc.gz | 865,470,694 | 8,644 | ## Re: [Chessx-users] MultiPV inverting values
Re: [Chessx-users] MultiPV inverting values From: Fex Xef - 2014-06-03 00:13:55 Attachments: Message as HTML ```Luca, thanks for the reply to the problem of chessx not showing MultiPV lines correctly with Stockfish. Specifically, I am using the Stockfish that comes with ChessX, "Stockfish-3-64". As an example, try the position: 1q3R2/8/8/8/8/K1kBP3/3pP3/8 w - - 0 1 Set the "Lines" field to "2" in the Analysis window. Mine shows two lines. The first has an evaluation of 84.23 and is in blue - White wins with R:b8. The second line shows "Mate in 4" with 1. Rc8# , and is also in blue. But in that line, it's White who is getting mated, not black. And in fact the line shown ends with 4...Qa7#. So why is the second line in blue and not in red? ```
Re: [Chessx-users] MultiPV inverting values From: Fex Xef - 2014-06-03 00:13:55 Attachments: Message as HTML ```Luca, thanks for the reply to the problem of chessx not showing MultiPV lines correctly with Stockfish. Specifically, I am using the Stockfish that comes with ChessX, "Stockfish-3-64". As an example, try the position: 1q3R2/8/8/8/8/K1kBP3/3pP3/8 w - - 0 1 Set the "Lines" field to "2" in the Analysis window. Mine shows two lines. The first has an evaluation of 84.23 and is in blue - White wins with R:b8. The second line shows "Mate in 4" with 1. Rc8# , and is also in blue. But in that line, it's White who is getting mated, not black. And in fact the line shown ends with 4...Qa7#. So why is the second line in blue and not in red? ```
Re: [Chessx-users] MultiPV inverting values From: Jens Nissen - 2014-06-03 06:11:33 ```The + and the * symbol are not part of the evaluation, they are hyperlinks. If you click on one of them, the first move (+) or the entire variation (*) is taken over as line into the game text. So the evaluation is perfectly ok. Jens Am 03.06.2014 um 02:13 schrieb Fex Xef : > Luca, thanks for the reply to the problem of chessx not showing MultiPV lines correctly with Stockfish. > > Specifically, I am using the Stockfish that comes with ChessX, "Stockfish-3-64". > > As an example, try the position: > 1q3R2/8/8/8/8/K1kBP3/3pP3/8 w - - 0 1 > > Set the "Lines" field to "2" in the Analysis window. > > Mine shows two lines. The first has an evaluation of 84.23 and is in blue - White wins with R:b8. > > The second line shows "Mate in 4" with 1. Rc8# , and is also in blue. But in that line, it's White who is getting mated, not black. And in fact the line shown ends with 4...Qa7#. > > So why is the second line in blue and not in red? > > > ------------------------------------------------------------------------------ > Learn Graph Databases - Download FREE O'Reilly Book > "Graph Databases" is the definitive new guide to graph databases and their > applications. Written by three acclaimed leaders in the field, > this first edition is now available. Download your free book today! > http://p.sf.net/sfu/NeoTech_______________________________________________ > Chessx-users mailing list > Chessx-users@... > https://lists.sourceforge.net/lists/listinfo/chessx-users ``` | 905 | 3,120 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2016-30 | latest | en | 0.900729 |
http://www.math.colostate.edu/~lewis/m151/Lab04/Lab04_solutions.html | 1,519,050,170,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812665.41/warc/CC-MAIN-20180219131951-20180219151951-00184.warc.gz | 495,553,477 | 4,749 | # Math 151 Lab 04 - 01/29/2018:
DUE: Wednesday, 31, 2018 (by 4:00pm before class starts)
INSTRUCTIONS: Save this Matlab script with the filename LastName_Lab04.m (example: Lewis_Lab04.m). Complete each question using MATLAB code. When you are ready to have your assignment graded, use the PUBLISH command to generate a pdf file, and submit the resulting file to CANVAS.
Please make sure to leave code uncommented unless I ask for a comment. You do not need to label your steps for each problem, but I should be able to see you do each step I asked for.
## #1
WRITE CODE BELOW:
There is a famous algorithm called the "bubble sort" algorithm, which we're going to implement here. It will sort a vector until the entries are all in ascending order.
Step 1: Set x=[1,8,3,6,2,9,7,5,4]
Step 2: Use nested loops (i.e., two for loops) to sort x. The inner loop isn't bad: loop over elements of x from 1 to 8, and if x(j) > x(j+1), then swap them. You may need an extra variable to store one of the values while you swap them.
Step 3: Since you have suppressed the output within the loops, verify that your code sorted x so you have x=[1,2,3,4,5,6,7,8,9] (this means print x to the screen).
------------------------------------------------------------------------
```x=[1,8,3,6,2,9,7,5,4];
for i = 1:length(x)
for j = 1:(length(x)-1)
if (x(j)>x(j+1))
tmp = x(j);
x(j) = x(j+1);
x(j+1) = tmp;
end
end
end
x
```
```x =
1 2 3 4 5 6 7 8 9
```
## #2
WRITE CODE BELOW:
Write a function called "collatz" below where you take an input n, and if n is odd, return 3*n+1. If n is even, return n/2. (You need to compute mod(n,2) to test for even/odd. If mod(n,2)==1, n is odd, otherwise n is even.)
Note: You cannot use the keyword function when you publish. To avoid errors, paste your function code below as comments.
------------------------------------------------------------------------
```function ret = collatz(n)
if (mod(n,2)==0)
ret = n/2;
else
ret = 3*n+1;
end
```
## #3
WRITE CODE BELOW:
Please make sure your collatz function from the previous problem is correct first.
Step 1: Set up a list of 100 blank counters (i.e., c=zeros(1,100)) and a list of x values x=1:100.
Step 2: Use a while loop inside a for loop to reassign x(i) to the value collatz(x(i)), and every time it reassigns x(i), add 1 to counter c(i). (Hint: Your while loop should run as long as x(i) is not equal to 1, and your for loop should loop over all entries of x). What you are checking is how many times you have to apply the collatz function in order for x to reach 1 given different initial conditions.
------------------------------------------------------------------------
```c=zeros(1,100);
x=1:100;
for i=1:length(x)
while(x(i)~=1)
x(i)=collatz(x(i));
c(i)=c(i)+1;
end
end
c
```
```c =
Columns 1 through 13
0 1 7 2 5 8 16 3 19 6 14 9 9
Columns 14 through 26
17 17 4 12 20 20 7 7 15 15 10 23 10
Columns 27 through 39
111 18 18 18 106 5 26 13 13 21 21 21 34
Columns 40 through 52
8 109 8 29 16 16 16 104 11 24 24 24 11
Columns 53 through 65
11 112 112 19 32 19 32 19 19 107 107 6 27
Columns 66 through 78
27 27 14 14 14 102 22 115 22 14 22 22 35
Columns 79 through 91
35 9 22 110 110 9 9 30 30 17 30 17 92
Columns 92 through 100
17 17 105 105 12 118 25 25 25
```
## #4
WRITE CODE BELOW:
Plot your list of counters from #3 as a scatter plot of x (the integers from 1 to 100) vs. c. Instead of using the markers Matlab gave you, please plot the points as red diamonds when you use the scatter or plot commands. Label your vertical and horizontal axes and give the plot a title (These can be whatever you wish. I just want to see that you know how to set them.)
WRITE A COMMENT BELOW:
Does there seem to be any pattern to the number of tries and the x-value or does it seem to be completely random?
------------------------------------------------------------------------
```scatter(1:100,c,'rd') | 1,299 | 4,225 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-09 | latest | en | 0.800743 |
https://calculife.com/m2-to-in2-online-converter/ | 1,726,597,153,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651829.57/warc/CC-MAIN-20240917172631-20240917202631-00671.warc.gz | 136,323,255 | 119,743 | This online converter allows users to accurately convert square meters (m²) to square inches (in²) and vice versa. It is ideal for various applications, such as construction, real estate, interior design, or any task that requires area conversion between metric and imperial units. The tool automatically handles the conversion in both directions, making it convenient for users who need quick and reliable results.
## Square Meters to Square Inches Converter
This converter automatically converts between square meters (m²) and square inches (in²) both ways. You can also set the number of decimal places to round the results.
## How to Use the Converter
To use the square meters to square inches converter, simply enter the value in square meters (m²) in the provided input field. The corresponding value in square inches (in²) will be calculated automatically and displayed in the output field. Similarly, if you need to convert square inches to square meters, input the value in the square inches field, and the equivalent in square meters will be shown.
## Formula Used
The conversion formula used to convert square meters to square inches is:
Square Inches (in²) = Square Meters (m²) × 1550.0031
For the reverse conversion, you would divide the square inches by 1550.0031 to get the value in square meters.
## Precalculated Values
Here are some commonly converted real-world values:
• 1 m² = 1550.0031 in² (Approximate size of a standard desk)
• 5 m² = 7750.0155 in² (Typical size of a small room)
• 10 m² = 15500.031 in² (Small living room size)
• 20 m² = 31000.062 in² (Medium-sized living room)
• 30 m² = 46500.093 in² (Large living room or small office)
• 50 m² = 77500.155 in² (Studio apartment or large office space)
• 100 m² = 155000.31 in² (Small house or large open space)
Square inch is approximately 1550 times smaller than square meter.
## General Info
Square meters (m²) and square inches (in²) are both units of area used to measure the size of surfaces. Square meters are commonly used in countries that follow the metric system, while square inches are used in countries that adhere to the imperial system, such as the United States. Understanding how to convert between these two units is important in fields like construction, architecture, real estate, and design, where accurate area measurements are required. | 542 | 2,350 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-38 | latest | en | 0.835511 |
http://new.transum.com/Maths/Exam/Question.asp?Q=137 | 1,544,743,500,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825112.63/warc/CC-MAIN-20181213215347-20181214000847-00561.warc.gz | 193,347,617 | 6,028 | # Exam Style Question
## Worked solutions to typical exam type questions that you can reveal gradually
##### List Of QuestionsExam Style QuestionMore Proof QuestionsOther Topics
Question id: 137. This question is similar to one that appeared in a GCSE Higher paper (specimen) for 2017. The use of a calculator is allowed.
### Proof
(a) Prove that the recurring decimal $$0.\dot2 \dot1$$ has the value $$\frac{7}{33}$$
(b) The value of $$x$$ is given as:
$$x=\frac{1}{5^{120}\times2^{123}}$$
Show that, when $$x$$ is written as a terminating decimal, there are 120 zeros after the decimal point.
(c) The reciprocal of any prime number $$p$$ (where $$p$$ is neither 2 nor 5) when written as a decimal, is always a recurring decimal.
A theorem in mathematics states:
The period of a recurring decimal is the least value of $$n$$ for which $$p$$ is a factor of $$10^n – 1$$
Marilou tests this theorem for the reciprocal of eleven.
She uses her calculator to show that 11 is a factor of $$10^2 – 1$$ then makes this statement:
"The period of the recurring decimal is 2 because 11 is a factor of $$10^2-1$$. This shows the theorem to be true in this case."
Explain why Marilou's statement is incomplete.
The worked solutions to these exam-style questions are only available to those who have a Transum Subscription. Subscribers can drag down the panel to reveal the solution line by line. This is a very helpful strategy for the student who does not know how to do the question but given a clue, a peep at the beginnings of a method, they may be able to make progress themselves. This could be a great resource for a teacher using a projector or for a parent helping their child work through the solution to this question. The worked solutions also contain screen shots (where needed) of the step by step calculator procedures. A subscription also opens up the answers to all of the other online exercises, puzzles and lesson starters on Transum Mathematics and provides an ad-free browsing experience.
Drag this panel down to reveal the solution
The exam-style questions appearing on this site are based on those set in previous examinations (or sample assessment papers for future examinations) by the major examination boards. The wording, diagrams and figures used in these questions have been changed from the originals so that students can have fresh, relevant problem solving practice even if they have previously worked through the related exam paper.
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To search the entire Transum website use the search box in the grey area below. | 604 | 2,767 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2018-51 | latest | en | 0.926432 |
https://www.cfd-online.com/Forums/flow-3d/137204-problem-general-moving-object-collision-modeling.html | 1,502,937,849,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102819.58/warc/CC-MAIN-20170817013033-20170817033033-00493.warc.gz | 882,543,714 | 15,760 | # problem in general moving object and collision modeling
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June 11, 2014, 15:40
problem in general moving object and collision modeling
#1
New Member
Join Date: Jun 2014
Posts: 15
Rep Power: 5
Hello everyone,
I have a big problem with my modeling.
I want to model some concrete block and see how it interact with water in the flume.
I put GMO(and collision) & Viscous and turbulence & Gravity in physics of model.
coefficient of friction=0.5
coefficient of restitution=0.53
Coupled motion with mass density=2000 (initial velocities=0)
boundary condition: y min & y max & z min = wall , z max = symmetry , x min= wave , x max = wall.
but blocks move and behave unreasonable as you can see in attached image, the blocks immerge together!!!
Attached Images
1.jpg (29.9 KB, 29 views) 2.jpg (34.2 KB, 29 views)
June 14, 2014, 02:03 #2 New Member Saeid Ahmadvand Join Date: Jun 2014 Posts: 15 Rep Power: 5 Is there anyone who can solve my problem???
June 27, 2014, 03:44 #3 New Member frode gimsøy Join Date: Jun 2014 Posts: 9 Rep Power: 5 Did you try smaller cells in your mesh?
June 27, 2014, 04:15 #4 New Member Saeid Ahmadvand Join Date: Jun 2014 Posts: 15 Rep Power: 5 Hi, yes, as you seen at those picture, it is very fine mesh. This problem is not related to the mesh.
June 28, 2014, 11:39 #5 New Member frode gimsøy Join Date: Jun 2014 Posts: 9 Rep Power: 5 Are the small blocks separated from the big block? Did you try placing them further away before starting the simulation?
June 28, 2014, 17:18 #6 New Member Saeid Ahmadvand Join Date: Jun 2014 Posts: 15 Rep Power: 5 yes, I do it.
August 30, 2016, 04:10 #7 New Member Join Date: Aug 2016 Posts: 3 Rep Power: 2 have you ever use restitution coefficient 1?
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#### Resources tagged with Working systematically similar to Purr-fection:
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### There are 71 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically
### Purr-fection
##### Age 16 to 18 Challenge Level:
What is the smallest perfect square that ends with the four digits 9009?
### Latin Squares
##### Age 11 to 18
A Latin square of order n is an array of n symbols in which each symbol occurs exactly once in each row and exactly once in each column.
### LCM Sudoku
##### Age 14 to 16 Challenge Level:
Here is a Sudoku with a difference! Use information about lowest common multiples to help you solve it.
### LCM Sudoku II
##### Age 11 to 18 Challenge Level:
You are given the Lowest Common Multiples of sets of digits. Find the digits and then solve the Sudoku.
### Diagonal Product Sudoku
##### Age 11 to 16 Challenge Level:
Given the products of diagonally opposite cells - can you complete this Sudoku?
### Star Product Sudoku
##### Age 11 to 16 Challenge Level:
The puzzle can be solved by finding the values of the unknown digits (all indicated by asterisks) in the squares of the $9\times9$ grid.
##### Age 11 to 16 Challenge Level:
The items in the shopping basket add and multiply to give the same amount. What could their prices be?
### Integrated Sums Sudoku
##### Age 11 to 16 Challenge Level:
The puzzle can be solved with the help of small clue-numbers which are either placed on the border lines between selected pairs of neighbouring squares of the grid or placed after slash marks on. . . .
### Twin Corresponding Sudoku
##### Age 11 to 18 Challenge Level:
This sudoku requires you to have "double vision" - two Sudoku's for the price of one
### Wallpaper Sudoku
##### Age 11 to 16 Challenge Level:
A Sudoku that uses transformations as supporting clues.
### Integrated Product Sudoku
##### Age 11 to 16 Challenge Level:
This Sudoku puzzle can be solved with the help of small clue-numbers on the border lines between pairs of neighbouring squares of the grid.
### Diagonal Sums Sudoku
##### Age 7 to 16 Challenge Level:
Solve this Sudoku puzzle whose clues are in the form of sums of the numbers which should appear in diagonal opposite cells.
### Constellation Sudoku
##### Age 14 to 18 Challenge Level:
Special clue numbers related to the difference between numbers in two adjacent cells and values of the stars in the "constellation" make this a doubly interesting problem.
### Magnetic Personality
##### Age 7 to 16 Challenge Level:
60 pieces and a challenge. What can you make and how many of the pieces can you use creating skeleton polyhedra?
### Rainstorm Sudoku
##### Age 14 to 16 Challenge Level:
Use the clues about the shaded areas to help solve this sudoku
### Difference Sudoku
##### Age 14 to 16 Challenge Level:
Use the differences to find the solution to this Sudoku.
##### Age 11 to 16 Challenge Level:
Four small numbers give the clue to the contents of the four surrounding cells.
### Intersection Sudoku 1
##### Age 11 to 16 Challenge Level:
A Sudoku with a twist.
### Seasonal Twin Sudokus
##### Age 11 to 16 Challenge Level:
This pair of linked Sudokus matches letters with numbers and hides a seasonal greeting. Can you find it?
### Difference Dynamics
##### Age 14 to 18 Challenge Level:
Take three whole numbers. The differences between them give you three new numbers. Find the differences between the new numbers and keep repeating this. What happens?
### Olympic Logic
##### Age 11 to 16 Challenge Level:
Can you use your powers of logic and deduction to work out the missing information in these sporty situations?
### LOGO Challenge - Sequences and Pentagrams
##### Age 11 to 18 Challenge Level:
Explore this how this program produces the sequences it does. What are you controlling when you change the values of the variables?
### The Best Card Trick?
##### Age 11 to 16 Challenge Level:
Time for a little mathemagic! Choose any five cards from a pack and show four of them to your partner. How can they work out the fifth?
### A Long Time at the Till
##### Age 14 to 18 Challenge Level:
Try to solve this very difficult problem and then study our two suggested solutions. How would you use your knowledge to try to solve variants on the original problem?
### Ratio Sudoku 2
##### Age 11 to 16 Challenge Level:
A Sudoku with clues as ratios.
### An Introduction to Magic Squares
##### Age 11 to 16 Challenge Level:
This is a variation of sudoku which contains a set of special clue-numbers. Each set of 4 small digits stands for the numbers in the four cells of the grid adjacent to this set.
### Twin Line-swapping Sudoku
##### Age 14 to 16 Challenge Level:
A pair of Sudoku puzzles that together lead to a complete solution.
### I've Submitted a Solution - What Next?
##### Age 5 to 18
In this article, the NRICH team describe the process of selecting solutions for publication on the site.
### Bochap Sudoku
##### Age 11 to 16 Challenge Level:
This Sudoku combines all four arithmetic operations.
### Function Pyramids
##### Age 16 to 18 Challenge Level:
A function pyramid is a structure where each entry in the pyramid is determined by the two entries below it. Can you figure out how the pyramid is generated?
##### Age 11 to 16 Challenge Level:
Four numbers on an intersection that need to be placed in the surrounding cells. That is all you need to know to solve this sudoku.
### Pole Star Sudoku 2
##### Age 11 to 16 Challenge Level:
This Sudoku, based on differences. Using the one clue number can you find the solution?
### Rectangle Outline Sudoku
##### Age 11 to 16 Challenge Level:
Each of the main diagonals of this sudoku must contain the numbers 1 to 9 and each rectangle width the numbers 1 to 4.
### Twin Corresponding Sudokus II
##### Age 11 to 16 Challenge Level:
Two sudokus in one. Challenge yourself to make the necessary connections.
### Twin Corresponding Sudoku III
##### Age 11 to 16 Challenge Level:
Two sudokus in one. Challenge yourself to make the necessary connections.
### Advent Calendar 2011 - Secondary
##### Age 11 to 18 Challenge Level:
Advent Calendar 2011 - a mathematical activity for each day during the run-up to Christmas.
### Colour Islands Sudoku 2
##### Age 11 to 18 Challenge Level:
In this Sudoku, there are three coloured "islands" in the 9x9 grid. Within each "island" EVERY group of nine cells that form a 3x3 square must contain the numbers 1 through 9.
### Magic Caterpillars
##### Age 14 to 18 Challenge Level:
Label the joints and legs of these graph theory caterpillars so that the vertex sums are all equal.
### All-variables Sudoku
##### Age 11 to 18 Challenge Level:
The challenge is to find the values of the variables if you are to solve this Sudoku.
### Alphabetti Sudoku
##### Age 11 to 16 Challenge Level:
This Sudoku requires you to do some working backwards before working forwards.
### W Mates
##### Age 16 to 18 Challenge Level:
Show there are exactly 12 magic labellings of the Magic W using the numbers 1 to 9. Prove that for every labelling with a magic total T there is a corresponding labelling with a magic total 30-T.
### One Out One Under
##### Age 14 to 16 Challenge Level:
Imagine a stack of numbered cards with one on top. Discard the top, put the next card to the bottom and repeat continuously. Can you predict the last card?
### Pole Star Sudoku
##### Age 14 to 18 Challenge Level:
A Sudoku based on clues that give the differences between adjacent cells.
### Intersection Sudoku 2
##### Age 11 to 16 Challenge Level:
A Sudoku with a twist.
### Simultaneous Equations Sudoku
##### Age 11 to 16 Challenge Level:
Solve the equations to identify the clue numbers in this Sudoku problem.
### Olympic Magic
##### Age 14 to 16 Challenge Level:
in how many ways can you place the numbers 1, 2, 3 … 9 in the nine regions of the Olympic Emblem (5 overlapping circles) so that the amount in each ring is the same?
### Exhaustion
##### Age 16 to 18 Challenge Level:
Find the positive integer solutions of the equation (1+1/a)(1+1/b)(1+1/c) = 2
### Twin Chute-swapping Sudoku
##### Age 14 to 18 Challenge Level:
A pair of Sudokus with lots in common. In fact they are the same problem but rearranged. Can you find how they relate to solve them both?
### The Naked Pair in Sudoku
##### Age 7 to 16
A particular technique for solving Sudoku puzzles, known as "naked pair", is explained in this easy-to-read article. | 1,940 | 8,592 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2018-47 | latest | en | 0.830756 |
https://atmospheric_sciences_en_ch.enacademic.com/1722/Bernoulli_probability | 1,600,752,992,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400203096.42/warc/CC-MAIN-20200922031902-20200922061902-00460.warc.gz | 286,561,444 | 10,436 |
# Bernoulli probability
Atmospheric Sciences (English-Chinese) dictionary. 2014.
### Look at other dictionaries:
• Bernoulli distribution — Probability distribution name =Bernoulli type =mass pdf cdf parameters =1>p>0, pinR support =k={0,1}, pdf = egin{matrix} q=(1 p) mbox{for }k=0 p mbox{for }k=1 end{matrix} cdf = egin{matrix} 0 mbox{for }k … Wikipedia
• Probability theory — is the branch of mathematics concerned with analysis of random phenomena.[1] The central objects of probability theory are random variables, stochastic processes, and events: mathematical abstractions of non deterministic events or measured… … Wikipedia
• Probability — is the likelihood or chance that something is the case or will happen. Probability theory is used extensively in areas such as statistics, mathematics, science and philosophy to draw conclusions about the likelihood of potential events and the… … Wikipedia
• Bernoulli , Jakob I — Bernoulli , Jakob (or Jacques) I (1654–1705) Swiss mathematician Jakob I was the first of the Bernoulli family of scientists to achieve fame as a mathematician. As with the two other particularly outstanding Bernoullis – his brother, Johann I and … Scientists
• Bernoulli process — In probability and statistics, a Bernoulli processis a discrete time stochastic process consisting ofa sequence of independent random variables taking values over two symbols. Prosaically, a Bernoulli process is coin flipping, possibly with an… … Wikipedia
• Probability distribution — This article is about probability distribution. For generalized functions in mathematical analysis, see Distribution (mathematics). For other uses, see Distribution (disambiguation). In probability theory, a probability mass, probability density … Wikipedia
• probability theory — Math., Statistics. the theory of analyzing and making statements concerning the probability of the occurrence of uncertain events. Cf. probability (def. 4). [1830 40] * * * Branch of mathematics that deals with analysis of random events.… … Universalium
• probability and statistics — ▪ mathematics Introduction the branches of mathematics concerned with the laws governing random events, including the collection, analysis, interpretation, and display of numerical data. Probability has its origin in the study of gambling… … Universalium
• Bernoulli trial — In the theory of probability and statistics, a Bernoulli trial is an experiment whose outcome is random and can be either of two possible outcomes, success and failure .In practice it refers to a single experiment which can have one of two… … Wikipedia
• Bernoulli-Versuch — Zufallsgrößen mit einer Null Eins Verteilung bzw. Bernoulli Verteilung benutzt man zur Beschreibung von zufälligen Ereignissen, bei denen nur zwei mögliche Versuchsausgänge interessieren, das zufällige Ereignis (Erfolg) und sein komplementäres… … Deutsch Wikipedia
• Bernoulli-Verteilung — Wahrscheinlichkeitsfunktion der Bernoulli Verteilung für p = 0.2 (blau), p = 0.5 (grün) und p = 0.8 (rot) Zufallsgrößen mit einer Null Eins Verteilung bzw. Bernoulli Verteilung benutzt man zur Beschreibung von zufälligen Ereignissen, bei denen es … Deutsch Wikipedia | 753 | 3,216 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2020-40 | latest | en | 0.833798 |
https://boyslife.org/hobbies-projects/funstuff/1483/optical-illusions/comment-page-86/?replytocom=29605 | 1,596,698,145,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439736883.40/warc/CC-MAIN-20200806061804-20200806091804-00270.warc.gz | 247,349,997 | 19,880 | Your eyes can play tricks on you. Pictures that confuse your eyes and brain, tricking them into seeing something differently, are called optical illusions. See if you can figure out these optical illusions.
## WHICH OBJECT IS TALLER?
None: They’re all the same size.
## WHICH LINE IS LONGER?
Neither: They’re both the same.
## IS THIS GRAY HAZE SHRINKING?
Stare at the black dot. After a while the gray haze will appear to shrink.
## DO THESE COLORED LINES BEND?
No, they’re perfectly straight—but try telling your eyes that!
## ARE THESE WHEELS SPINNING?
Stare at the center. Now move your head back and forth toward and away from the page. The circles will appear to spin.
## HOW MANY BLACK DOTS CAN YOU COUNT?
Look closely and you will see them.
## DO YOU SEE TWO FACES OR A VASE?
If you see one, close your eyes for a moment, then look for the other.
## ARE THESE LINES STRAIGHT OR CROOKED?
Yep, you guessed it. The horizontal lines look crooked, but they are perfectly straight.
## ARE YOU SEEING RED, WHITE & BLUE?
Stare at the center of this flag for one minute. Then look at an empty white sheet and you’ll see a red, white and blue flag.
HOW THIS FLAG WORKS
Your eyes see color as measures of red or green, blue or yellow and bright or dark. When you look at a green object for a long time, your eyes get tired and start seeing red. When you look at yellow, after a while you’ll start to see blue. And darkness turns into brightness. The result: Even this wacky flag can be good, old red, white and blue—after a while!
#### 13 Comments on 9 Optical Illusions to Confuse Your Brain
1. The face/vase one was neat and i cant get enough!!!!!!!!
2. why did none of these work
3. WE WANT MORE! WE WANT MORE!!!!!!!!!
4. The flag did not work. All I saw was a blank sheet of paper.
5. Spaced Out // July 26, 2016 at 12:34 pm // Reply
The black dot one was awesome!
6. I love the Black dots one! So cool!
7. Easter bunny // June 16, 2016 at 7:58 am // Reply
That was awesome =)
8. Right, more
9. this is awesome please do more
10. The tire one is really cool! please do more more! MORE! | 545 | 2,119 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2020-34 | latest | en | 0.87556 |
https://www.reference.com/math/explore/math-calculators?qo=cdpArticles | 1,537,582,403,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158001.44/warc/CC-MAIN-20180922005340-20180922025740-00469.warc.gz | 848,374,634 | 16,785 | # Math Calculators
Q:
## How Do You Convert Square Feet to Square Meters?
A:
Changing square feet to square meters requires use of basic conversion where one square meter equals 10.76 square feet. Hence, one square foot equals approximately 0.093 square meters. A house that has 2,500 square feet is 232.5 square meters in size.
Filed Under:
• #### Q:What Do the Keys on a Calculator Mean?
A: There are many kinds of calculators, but most calculators have the same standard keys that allow users to complete a variety of calculations. The number keys and addition, subtraction, multiplication, division and equal signs are all very straightforward, but the functions of the other keys are not so well-known.
Filed Under:
• #### Q:What Is the Difference Between Qualitative and Quantitative Analysis?
A: Qualitative analysis is the scientific study of data that can be observed, but not measured. It is concerned with cataloguing the qualities of what is studied. Quantitative analysis is the study of data that can be measured, the quantities of a category of data. Each type of inquiry contributes important insights to scientific study that the other cannot.
Filed Under:
• #### Q:What Does the Word "term" Mean in Mathematics?
A: The word "term" is used in mathematical equations to describe either a single number, or numbers and variables multiplied together. Numerical terms are then grouped into expressions to come up with a definitive answer to the equation.
Filed Under:
• #### Q:What Is a Real World Example for a Absolute Function?
A: A real world example of an absolute value function is any calculation in which the total depends upon simple addition of all the quantities in the function, regardless of sign. One example of such a calculation is the number of feet traveled by an athlete running around a race track.
Filed Under:
• #### Q:What Are Some Early Counting Devices?
A: Two early counting devices were the abacus and the Antikythera mechanism. The abacus and similar counting devices were in use across many nations and cultures. The Antikythera mechanism hails from the Greek island of Antikythera.
Filed Under:
• #### Q:How Can You Calculate the Mass of a Known Number of Atoms of an Element?
A: To calculate the mass of a specific number of atoms of an element, divide the number of atoms by 6.02 x 1023 and then multiply by the molar mass of the element. The process for calculating mass from the number of atoms is the same for all elements.
Filed Under:
• #### Q:How Do You Convert Square Feet to Square Meters?
A: Changing square feet to square meters requires use of basic conversion where one square meter equals 10.76 square feet. Hence, one square foot equals approximately 0.093 square meters. A house that has 2,500 square feet is 232.5 square meters in size.
Filed Under:
• #### Q:What Is the Formula to Convert Lbs to Kg?
A: The formula to convert kilograms to pounds is pound = kilogram/0.45359237. Conversely, the formula is kilogram = pound x 0.45359237. One kilogram is equal to roughly 2.2 pounds, and 1 pound is roughly 454 grams.
Filed Under:
• #### Q:Where Is the Abacus Still Used Today?
A: The abacus is a basic mathematical computing device. It is still widely used today in Africa, Japan, Russia, China and other countries as a visual aid to facilitate the learning of mathematical concepts.
Filed Under:
• #### Q:What Is the Formula Used to Find Percent Recovery?
A: When purifying a substance in chemistry, use (collected mass/starting mass)*100 to calculate percent recovery. This formula is also commonly stated as (pure product recovered/crude material used)*100.
Filed Under:
• #### Q:What Is the Formula Used to Find the Area of an Isosceles Triangle?
A: The common formula to find the area of an isosceles triangle is area equals 1/2 * base * height of triangle. The height is the line bisecting the two equal sides of the triangle and intersects the midpoint of the base.
Filed Under:
• #### Q:What Activities Does Think Through Math Offer?
A: Think Through Math offers a number of activities designed to motivate students in grades 3 and above to want to learn math, including games and theme-based contests. Students also have the ability to earn points that they can use to customize avatars and program dashboards or donate to charities.
Filed Under:
• #### Q:How Do You Use a Number Pattern Calculator?
A: Simple-to-use number-pattern calculators for data sets such as arithmetic sequences and geometric sequences can be found on Calculator.net. In these calculators and similar tools, a missing value of a number sequence is calculated when a user specifies the defining pattern behind the set of numbers and selects the appropriate variables. AlteredQualia.com contains a number sequence solver that computes the defining pattern behind a valid set of numbers entered by a user.
Filed Under:
• #### Q:What Is the Value of 68 Centimeters in Inches?
A: The equivalent value of 68 centimeters is approximately 26.77 or 26 49/64 inches. It is also the equivalent of 2.23 feet. One inch is the equivalent of 2.54 centimeters, and one foot equals 30.48 centimeters.
Filed Under:
• #### Q:What Is 48 Inches in Feet?
A: A measurement of 48 inches is equivalent to 4 feet. Converting from inches to feet can be done by dividing the number of inches by 12, as 12 inches is equivalent to 1 foot.
Filed Under:
• #### Q:What Does Water Do at a Temperature of 273 Degrees Kelvin?
A: Water freezes at 273 Kelvin. Kelvin is a temperature scale named after William Thomson, also called Lord Kelvin. The Kelvin scale starts at zero, the lowest possible temperature.
Filed Under:
• #### Q:How Do You Use an Online Calculator?
A: Online calculators are used in the same manner as hand-held calculators. Calculations can be formulated online or the calculator can be downloaded to your computer. The numbers that require calculation can be inputted from the computer keyboard. Enlarged versions are available for individuals with weak eyesight.
Filed Under:
• #### Q:How Do You Convert 161 Lbs to Kg?
A: To convert 161 pounds (lbs) to kilograms (kg) simply divide 161 by 2.2046, which equals 73.028. The conversion rate between pounds and kilograms is 1 kg = 2.2046 lbs.
Filed Under:
• #### Q:What Is 400 Meters Equal To?
A: A length of 400 meters is usually used to approximate 1/4 of a mile, especially in track races. Converting 400 meters to miles results in a value of exactly 0.2486 miles, according to Engineering Toolbox.
Filed Under:
• #### Q:How Do You Calculate the Length of a Stair Stringer?
A: In general terms, the length of a stringer for a stairs is 14 inches for every step. For a more precise calculation, you need the know the height of the riser and the width of the tread for the steps.
Filed Under:
• #### Q:What Are Some Good Calculators for Math?
A: Some good math calculators are the TI-Nspire CX Cas, TI-84 Plus C Silver Edition, Casio Prizm, HP Prime and TI-89 Titanium. These calculators are capable of handling basic arithmetic, complex scientific formulas and graphing.
Filed Under: | 1,607 | 7,086 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2018-39 | longest | en | 0.909136 |
https://electronics.stackexchange.com/questions/283024/in-ac-rl-circuit-why-does-current-increase-when-voltage-decreases | 1,679,829,172,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945472.93/warc/CC-MAIN-20230326111045-20230326141045-00415.warc.gz | 305,835,892 | 39,264 | # In AC RL circuit why does current increase when voltage decreases?
We know that the inductor creates an emf trying to oppose the change in ΔΦ. For an AC LR circuit I found the following diagram.
My question is why doesn't the current reach its peak a little earlier than π/2? When the voltage is at its peak I suppose the emf is smaller than the source voltage so the current keeps increasing, trying to flow through the inductor. The emf is big at this point based on what I know for the DC RL circuit and decreases through time. We can tell from the diagram that the current increases until π/2 when the inductor comes into play to oppose its decrease this time.
However, I can't understand why this is not happening earlier. I get the increase in current at first when voltage is at maximum value but when it gets really small the current has already reached the maximum value for that particular voltage. So won't the emf change polarity at that point?
In case I wasn't clear I give you this example. Say we have a DC LR circuit. Time has passed and the current has reached its maximum value given by Ohm's law. Now I disconnect the source and connect another source with a lower voltage(this happens instantly). The current will slowly decrease because the emf tries to keep it constant. So the decrease starts now and that is because the voltage is lower.
Isn't there a time before π/2 in the AC circuit when voltage is lower than the voltage given by Ohm's law for the current at that moment?
• this is how current energy is stored in an inductor with a 90 deg lag in current. thus the real power V(t)*I(t) over 1 cycle= 0 and this is stored reactive energy Jan 28, 2017 at 23:59
• Hmm. Are you trying for a deep understanding that "sings in your mind" so to speak? Or would it be acceptable to just tell you to look at the curves and notice that the peak magnitude of one takes place when the other one is at its maximum rate of change?
– jonk
Jan 29, 2017 at 0:15
• Deep understanding. I can read the diagram. Jan 29, 2017 at 2:58
• The voltage across the inductor is a result of the rate of change of current through the inductor. Not the other way round. Ohms law is not a good way to analyse it. $V = -L\frac{di}{dt}$ Jan 29, 2017 at 4:49
• You might want to look at my answers to these two questions. The first shows plots of V and I, but starting from zero, so you see the difference between transient/DC behaviour, and long term AC electronics.stackexchange.com/questions/281476/…. The second talks about capacitors as the dual of inductors, to aid understanding, if you grok the first, this might help you grok the second. electronics.stackexchange.com/questions/282053/… Jan 29, 2017 at 6:21
The diagram is for the pure inductance only. In other words, the voltage shown there is the voltage across the inductor only, and the current is only the inductor current. These values are always 90° out of phase.
When considering the series or parallel combination of an inductor with a resistor, it gets a little more complicated. In a series circuit, the current through both components is identical, but the voltages have different magnitudes and phases, and their sum has a phase angle of somewhere between 0° and 90° relative to the current.
Similarly, in a parallel circuit, the voltage across both components is identical, but the currents have different magnitudes and phases, and their sum has a phase angle of somewhere between 0° and 90° relative to the voltage.
As already commented, your image is for pure inductance, no resistors included. But that does not change the explanation.
The inductor has slow response. When a voltage is applied over it, the current starts to rise.
It rises as long as the voltage has the original polarity. The voltage can get lower, but as long as it has the original polarity, the current in the inductor gets higher.
The tide turns when the polarity of the voltage changes. Then inductor's current starts to decrease. It can still have its original direction quite long time.
In mathematics we say that the growth rate (amperes/second) of inductor's current is = the voltage over the inductor divided by the inductance (Henrys)
In theory one volt connected to one Henry inductor should result infinitely growing current that has growth rate = 1 amperes /second. In practice all electric parts have some resistance. Thus the growth rate of the current tails off and the current never goes over the Ohmic limit - that's equal to voltage/total resistance.
probably you have seen this:
How does the inductor ''really'' induce voltage? | 1,052 | 4,605 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2023-14 | latest | en | 0.964186 |
https://cpp.openfoam.org/v6/polynomialFunction_8H_source.html | 1,670,369,106,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711121.31/warc/CC-MAIN-20221206225143-20221207015143-00562.warc.gz | 220,349,651 | 8,041 | The OpenFOAM Foundation
polynomialFunction.H
Go to the documentation of this file.
1 /*---------------------------------------------------------------------------*\
2 ========= |
3 \\ / F ield | OpenFOAM: The Open Source CFD Toolbox
4 \\ / O peration | Website: https://openfoam.org
5 \\ / A nd | Copyright (C) 2011-2018 OpenFOAM Foundation
6 \\/ M anipulation |
7 -------------------------------------------------------------------------------
9 This file is part of OpenFOAM.
10
11 OpenFOAM is free software: you can redistribute it and/or modify it
13 the Free Software Foundation, either version 3 of the License, or
14 (at your option) any later version.
15
16 OpenFOAM is distributed in the hope that it will be useful, but WITHOUT
17 ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
18 FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License
19 for more details.
20
21 You should have received a copy of the GNU General Public License
22 along with OpenFOAM. If not, see <http://www.gnu.org/licenses/>.
23
24 Class
25 Foam::polynomialFunction
26
27 Description
28 Polynomial function representation
29
30 \verbatim
31 poly = logCoeff*log(x) + sum(coeffs[i]*x^i)
32 \endverbatim
33
34 where <tt> 0 <= i <= N </tt>
35
36 - integer powers, starting at zero
37 - \c value(x) to evaluate the poly for a given value
38 - \c integrate(x1, x2) between two scalar values
39 - \c integral() to return a new, integral coeff polynomial
40 - increases the size (order)
41 - \c integralMinus1() to return a new, integral coeff polynomial where
42 the base poly starts at order -1
43
45 Foam::Polynomial for a templated implementation
46
47 SourceFiles
48 polynomialFunction.C
49
50 \*---------------------------------------------------------------------------*/
51
52 #ifndef polynomialFunction_H
53 #define polynomialFunction_H
54
55 #include "scalarList.H"
56 #include "Ostream.H"
57 #include "runTimeSelectionTables.H"
58
59 // * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * //
60
61 namespace Foam
62 {
63
64 // Forward declaration of classes
65 class polynomialFunction;
66
67 // Forward declaration of friend functions
68 Ostream& operator<<(Ostream&, const polynomialFunction&);
69
70
71 /*---------------------------------------------------------------------------*\
72 Class polynomialFunction Declaration
73 \*---------------------------------------------------------------------------*/
76 :
77 private scalarList
78 {
79 // Private data
80
81 //- Include the log term? - only activated using integralMinus1()
82 bool logActive_;
83
84 //- Log coefficient - only activated using integralMinus1()
85 scalar logCoeff_;
86
87
88 // Private Member Functions
89
90 //- Return integral coefficients.
91 // Argument becomes zeroth element (constant of integration)
92 static polynomialFunction cloneIntegral
93 (
94 const polynomialFunction&,
95 const scalar intConstant = 0.0
96 );
97
98 //- Return integral coefficients when lowest order is -1.
99 // Argument becomes zeroth element (constant of integration)
100 static polynomialFunction cloneIntegralMinus1
101 (
102 const polynomialFunction&,
103 const scalar intConstant = 0.0
104 );
105
106
107 //- Disallow default bitwise assignment
108 void operator=(const polynomialFunction&);
109
110
111
112 public:
113
114 //- Runtime type information
115 TypeName("polynomialFunction");
116
117
118 // Constructors
119
120 //- Construct a particular size, with all coefficients = 0.0
121 explicit polynomialFunction(const label);
122
123 //- Copy constructor
125
126 //- Construct from a list of coefficients
127 explicit polynomialFunction(const UList<scalar>& coeffs);
128
129 //- Construct from Istream
131
132
133 //- Destructor
134 virtual ~polynomialFunction();
135
136
137 // Member Functions
138
139 //- Return the number of coefficients
140 using scalarList::size;
141
142 //- Return coefficient
143 using scalarList::operator[];
144
145
146 // Access
147
148
149 //- Return true if the log term is active
150 bool logActive() const;
151
152 //- Return the log coefficient
153 scalar logCoeff() const;
154
155
156 // Evaluation
157
158 //- Return polynomial value
159 scalar value(const scalar x) const;
160
161 //- Integrate between two values
162 scalar integrate(const scalar x1, const scalar x2) const;
163
164
165 //- Return integral coefficients.
166 // Argument becomes zeroth element (constant of integration)
168 (
169 const scalar intConstant = 0.0
170 ) const;
171
172 //- Return integral coefficients when lowest order is -1.
173 // Argument becomes zeroth element (constant of integration)
175 (
176 const scalar intConstant = 0.0
177 ) const;
178
179
180 // Member Operators
181
184
185 polynomialFunction& operator*=(const scalar);
186 polynomialFunction& operator/=(const scalar);
187
188
189 //- Ostream Operator
190 friend Ostream& operator<<(Ostream&, const polynomialFunction&);
191 };
192
193
194 // * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * //
195
196 // * * * * * * * * * * * * * * * Global Operators * * * * * * * * * * * * * //
197
198 polynomialFunction operator+
199 (
200 const polynomialFunction&,
201 const polynomialFunction&
202 );
203
204
205 polynomialFunction operator-
206 (
207 const polynomialFunction&,
208 const polynomialFunction&
209 );
210
211
212 polynomialFunction operator*
213 (
214 const scalar,
215 const polynomialFunction&
216 );
217
218
219 polynomialFunction operator/
220 (
221 const scalar,
222 const polynomialFunction&
223 );
224
225
226 polynomialFunction operator*
227 (
228 const polynomialFunction&,
229 const scalar
230 );
231
232
233 polynomialFunction operator/
234 (
235 const polynomialFunction&,
236 const scalar
237 );
238
239
240 // * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * //
241
242 } // End namespace Foam
243
244 // * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * //
245
246 #endif
247
248 // ************************************************************************* //
polynomialFunction & operator/=(const scalar)
polynomialFunction & operator+=(const polynomialFunction &)
polynomialFunction integral(const scalar intConstant=0.0) const
Return integral coefficients.
intWM_LABEL_SIZE_t label
A label is an int32_t or int64_t as specified by the pre-processor macro WM_LABEL_SIZE.
Definition: label.H:59
friend Ostream & operator(Ostream &, const UList< T > &)
TypeName("polynomialFunction")
Runtime type information.
An Istream is an abstract base class for all input systems (streams, files, token lists etc)...
Definition: Istream.H:57
polynomialFunction(const label)
Construct a particular size, with all coefficients = 0.0.
scalar integrate(const scalar x1, const scalar x2) const
Integrate between two values.
polynomialFunction & operator*=(const scalar)
Polynomial function representation.
scalar value(const scalar x) const
Return polynomial value.
friend Ostream & operator<<(Ostream &, const polynomialFunction &)
Ostream Operator.
A 1D vector of objects of type <T>, where the size of the vector is known and can be used for subscri...
Definition: HashTable.H:61
An Ostream is an abstract base class for all output systems (streams, files, token lists...
Definition: Ostream.H:53
scalar logCoeff() const
Return the log coefficient.
bool logActive() const
Return true if the log term is active.
polynomialFunction & operator-=(const polynomialFunction &)
polynomialFunction integralMinus1(const scalar intConstant=0.0) const
Return integral coefficients when lowest order is -1.
Ostream & operator<<(Ostream &, const ensightPart &)
Macros to ease declaration of run-time selection tables.
virtual ~polynomialFunction()
Destructor.
label size() const
Return the number of elements in the UList.
Definition: ListI.H:170
Namespace for OpenFOAM. | 2,043 | 7,941 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-49 | latest | en | 0.491473 |
https://db0nus869y26v.cloudfront.net/en/Discrete_phase-type_distribution | 1,696,192,087,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510924.74/warc/CC-MAIN-20231001173415-20231001203415-00863.warc.gz | 212,550,011 | 11,234 | The discrete phase-type distribution is a probability distribution that results from a system of one or more inter-related geometric distributions occurring in sequence, or phases. The sequence in which each of the phases occur may itself be a stochastic process. The distribution can be represented by a random variable describing the time until absorption of an absorbing Markov chain with one absorbing state. Each of the states of the Markov chain represents one of the phases.
It has continuous time equivalent in the phase-type distribution.
## Definition
A terminating Markov chain is a Markov chain where all states are transient, except one which is absorbing. Reordering the states, the transition probability matrix of a terminating Markov chain with ${\displaystyle m}$ transient states is
${\displaystyle {P}=\left[{\begin{matrix}{T}&\mathbf {T} ^{0}\\\mathbf {0} ^{\mathsf {T))&1\end{matrix))\right],}$
where ${\displaystyle {T))$ is a ${\displaystyle m\times m}$ matrix, ${\displaystyle \mathbf {T} ^{0))$ and ${\displaystyle \mathbf {0} }$ are column vectors with ${\displaystyle m}$ entries, and ${\displaystyle \mathbf {T} ^{0}+{T}\mathbf {1} =\mathbf {1} }$. The transition matrix is characterized entirely by its upper-left block ${\displaystyle {T))$.
Definition. A distribution on ${\displaystyle \{0,1,2,...\))$ is a discrete phase-type distribution if it is the distribution of the first passage time to the absorbing state of a terminating Markov chain with finitely many states.
## Characterization
Fix a terminating Markov chain. Denote ${\displaystyle {T))$ the upper-left block of its transition matrix and ${\displaystyle \tau }$ the initial distribution. The distribution of the first time to the absorbing state is denoted ${\displaystyle \mathrm {PH} _{d}({\boldsymbol {\tau )),{T})}$ or ${\displaystyle \mathrm {DPH} ({\boldsymbol {\tau )),{T})}$.
Its cumulative distribution function is
${\displaystyle F(k)=1-{\boldsymbol {\tau )){T}^{k}\mathbf {1} ,}$
for ${\displaystyle k=1,2,...}$, and its density function is
${\displaystyle f(k)={\boldsymbol {\tau )){T}^{k-1}\mathbf {T^{0)) ,}$
for ${\displaystyle k=1,2,...}$. It is assumed the probability of process starting in the absorbing state is zero. The factorial moments of the distribution function are given by,
${\displaystyle E[K(K-1)...(K-n+1)]=n!{\boldsymbol {\tau ))(I-{T})^{-n}{T}^{n-1}\mathbf {1} ,}$
where ${\displaystyle I}$ is the appropriate dimension identity matrix.
## Special cases
Just as the continuous time distribution is a generalisation of the exponential distribution, the discrete time distribution is a generalisation of the geometric distribution, for example: | 690 | 2,691 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 20, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2023-40 | latest | en | 0.827838 |
https://benvitalenum3ers.wordpress.com/2016/10/26/triangle-abc-inradius-and-circumradius/ | 1,500,828,672,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424575.75/warc/CC-MAIN-20170723162614-20170723182614-00247.warc.gz | 620,322,887 | 35,174 | ## Triangle (a,b,c) – inradius and circumradius — (Part 1)
Establish the following
For any triangle $(a, b, c)$
$a \, b \; + \; b \, c \; + \; c \, a \; = \; s^2 \; + \; 4 \, R \, r \; + \; r^2$, and
$a \, b \, c \; = \; 4 \, R \, r \, s$
where $R$ is the circumradius, $r$ the inradius and $s$ the semiperimeter.
Then $a, b, c$ are in arithmetic progression iff
$s^2 \; + \; 9 r^2 \; = \; 18 \, R \, r$
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## About benvitalis
math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged , . Bookmark the permalink. | 222 | 583 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 8, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2017-30 | longest | en | 0.570316 |
http://www.econometricsbysimulation.com/2013/08/study-design-test-power-test-rejection.html | 1,726,011,828,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651323.1/warc/CC-MAIN-20240910224659-20240911014659-00854.warc.gz | 41,018,454 | 27,598 | ## Friday, August 16, 2013
### Study design - test power, test rejection rate, permutation test
In an experiment involving social policy, a question has arisen as to what statistical method to use. The outcome is 0/1 -- let's say, whether a criminal defendant shows up for trial or not. Then there are a control group and 4 treatment groups (thus, 5 indicator variables showing which group each defendant is in). So the point is to predict the 0/1 outcome based on which treatment group the defendant is in.
One possibility is to use ANOVA, but I'm not sure of that due to the assumption that the outcome is normally distributed, which an indicator variable is not.
I'd like to simulate a dataset such that the outcome (0/1) is as follows: Control group, 20% chance of not showing up (label that as 1, so 20% of the controls should show 1 in the outcome column and 80% should show zeros); treatment group 1, 18% chance of not showing up), treatment group 2, 16% chance of not showing up; treatment group 3, 14% chance of not showing up; and treatment group 4, 12% chance of not showing up. Then I'd like to be able to compare the results from ANOVA as well as a permutation test's
p-values.
Stata Code:
```
* Define a program called sampler to generate our data and run our tests.
cap program drop sampler
program define sampler , rclass
* The first argument of the sampler command is the sample
* size to be generated.
gl samp_size = `1'
* Start generating data
clear
set obs 5
* A token observation for each group defined by treatment status
gen treat = _n
* Generate the numbers in each group size
gen grp_size = \${samp_size} * \${ntreat1} if treat==1
replace grp_size = \${samp_size} * \${ntreat2} if treat==2
replace grp_size = \${samp_size} * \${ntreat3} if treat==3
replace grp_size = \${samp_size} * \${ntreat4} if treat==4
replace grp_size = \${samp_size} * \${ntreat5} if treat==5
tab treat, gen(treat)
* Generate a variable to store the true probability that a defendent
* will show up for trial.
gen p_show_up = \${treat1_trial} if treat==1
replace p_show_up = \${treat2_trial} if treat==2
replace p_show_up = \${treat3_trial} if treat==3
replace p_show_up = \${treat4_trial} if treat==4
replace p_show_up = \${treat5_trial} if treat==5
* Now let's expand our token observations to a full data set
expand grp_size
* Now we have the individuals make the actual decisions to show up
gen show_up = rbinomial(1,p_show_up)
* We can do an simple anova by using a multiple regression
reg show_up treat2-treat5
test treat2=treat3=treat4=treat5=0
return scalar F05 = `=r(p)<.05'
return scalar F10 = `=r(p)<.1'
test treat2=0
return scalar t2_05 = `=r(p)<.05'
return scalar t2_1 = `=r(p)<.1'
test treat3=0
return scalar t3_05 = `=r(p)<.05'
return scalar t3_1 = `=r(p)<.1'
test treat4=0
return scalar t4_05 = `=r(p)<.05'
return scalar t4_1 = `=r(p)<.1'
test treat5=0
return scalar t5_05 = `=r(p)<.05'
return scalar t5_1 = `=r(p)<.1'
end
* Define a program for defining an exact statistics for false rejection by
* randomly sorting the dependent variable.
cap program drop perm
program define perm, rclass
* Perm takes three arguments
* command: the command to be run after the dependent variables
* have been scrambled
* dep_var: a list of dependent variables to be scrambled
* tests: a list of tests to be run after running the command
preserve
syntax [anything], COmmand(string) DEp_var(varlist) tests(string)
foreach v in `dep_var' {
tempvar sorter orderer
gen `sorter' = rnormal()
gen `orderer' = _n
sort `sorter'
replace `v' = `v'[`orderer']
di "reorder `v'"
}
`command'
local t=1
gl return_list
foreach v in `tests' {
test `v'
return scalar t`t'_05 = `=r(p)<.05'
return scalar t`t'_10 = `=r(p)<.1'
gl return_list \${return_list} t`t'_05=r(t`t'_05) t`t'_10=r(t`t'_10)
local t=1+`t'
}
restore
end
* Begin Estimation
* Treatment one is the control group
* Define the fraction of subjects in each group
gl ntreat1 = 1/5
gl ntreat2 = 1/5
gl ntreat3 = 1/5
gl ntreat4 = 1/5
gl ntreat5 = 1/5
* Probability that a criminal defendent shows up for trial
gl treat1_trial = .20
gl treat2_trial = .18
gl treat3_trial = .16
gl treat4_trial = .14
gl treat5_trial = .12
sampler 5500
* Let's see how the permutation test works
perm, co(reg show_up treat2-treat5) de(show_up) ///
tests("treat2=treat3=treat4=treat5=0" "treat2=0" ///
"treat3=0" "treat4=0" "treat5=0")
simulate \${return_list}, rep(10000): ///
perm, co(reg show_up treat2-treat5) de(show_up) ///
tests("treat2=treat3=treat4=treat5=0" "treat2=0" ///
"treat3=0" "treat4=0" "treat5=0")
sum
/* Variable | Obs Mean Std. Dev. Min Max
-------------+--------------------------------------------------------
t1_05 | 10000 .0507 .2193954 0 1
t1_10 | 10000 .1002 .3002815 0 1
t2_05 | 10000 .0524 .2228435 0 1
t2_10 | 10000 .1015 .3020048 0 1
t3_05 | 10000 .054 .226029 0 1
-------------+--------------------------------------------------------
t3_10 | 10000 .1049 .3064398 0 1
t4_05 | 10000 .0519 .2218362 0 1
t4_10 | 10000 .1008 .3010788 0 1
t5_05 | 10000 .0514 .2208233 0 1
t5_10 | 10000 .1018 .3024002 0 1
These numbers are looking pretty good. I would suspect any bias in rejection
rates is very small from these estimates. */
gl save_values F05=r(F05) F10=r(F10) t2_05=r(t2_05) t2_1=r(t2_1) ///
t3_05=r(t3_05) t3_1=r(t3_1) t4_05=r(t4_05) ///
t4_1=r(t4_1) t5_05=r(t5_05) ///
b2=_b[treat2] b3=_b[treat3] b4=_b[treat4] b5=_b[treat5]
* Now let's use a Monte Carlo resampler to see how frequently we reject the null.
simulate \${save_values} , rep(5000): sampler 5500
sum
/* Variable | Obs Mean Std. Dev. Min Max
-------------+--------------------------------------------------------
F05 | 5000 .999 .0316101 0 1
F10 | 5000 .9994 .02449 0 1
t2_05 | 5000 .2672 .4425419 0 1
t2_1 | 5000 .381 .4856811 0 1
t3_05 | 5000 .7202 .4489457 0 1
-------------+--------------------------------------------------------
t3_1 | 5000 .8114 .3912297 0 1
t4_05 | 5000 .9686 .1744137 0 1
t4_1 | 5000 .9846 .1231498 0 1
t5_05 | 5000 .9994 .02449 0 1
b2 | 5000 -.0200947 .0163725 -.0754545 .0527273
-------------+--------------------------------------------------------
b3 | 5000 -.0399916 .0161132 -.1045455 .0254545
b4 | 5000 -.0601151 .0158372 -.1154545 -.0045455
b5 | 5000 -.0801233 .0155707 -.1354546 -.0254545
Looking at the estiamates of the coeficients b2-b5 the estimates seem
to be unbiased since their divergence from -.02, -.04, -.06, and -.08
is very small. However, we can test this as well.
*/
reg b2
test _cons==-.02
reg b3
test _cons==-.04
reg b4
test _cons==-.06
reg b5
test _cons==-.08
* All of the p values are well above the 10% rejection rate.
Formatted By EconometricsbySimulation.com
``` | 2,369 | 7,510 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-38 | latest | en | 0.877048 |
https://www.statisticshowto.com/probability-and-statistics/f-statistic-value-test/#ANOVA | 1,653,499,512,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662588661.65/warc/CC-MAIN-20220525151311-20220525181311-00516.warc.gz | 1,187,426,413 | 23,239 | # F Statistic / F Value: Simple Definition and Interpretation
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Contents :
## What is an F Statistic?
Image: U of Michigan.
Watch the video for an overview:
Intro To The F Statistic
An F statistic is a value you get when you run an ANOVA test or a regression analysis to find out if the means between two populations are significantly different. It’s similar to a T statistic from a T-Test; A T-test will tell you if a single variable is statistically significant and an F test will tell you if a group of variables are jointly significant.
### What is “Statistically Significant“?
Simply put, if you have significant result, it means that your results likely did not happen by chance. If you don’t have statistically significant results, you throw your test data out (as it doesn’t show anything!); in other words, you can’t reject the null hypothesis.
## Using The F Statistic.
You can use the F statistic when deciding to support or reject the null hypothesis. In your F test results, you’ll have both an F value and an F critical value.
• The value you calculate from your data is called the F Statistic or F value (without the “critical” part).
• The F critical value is a specific value you compare your f-value to.
• In general, if your calculated F value in a test is larger than your F critical value, you can reject the null hypothesis. However, the statistic is only one measure of significance in an F Test. You should also consider the p value. The p value is determined by the F statistic and is the probability your results could have happened by chance.
## The F Statistic and P Value
The F statistic must be used in combination with the p value when you are deciding if your overall results are significant. Why? If you have a significant result, it doesn’t mean that all your variables are significant. The statistic is just comparing the joint effect of all the variables together.
For example, if you are using the F Statistic in regression analysis (perhaps for a change in R Squared, the Coefficient of Determination), you would use the p value to get the “big picture.”
1. If the p value is less than the alpha level, go to Step 2 (otherwise your results are not significant and you cannot reject the null hypothesis). A common alpha level for tests is 0.05.
2. Study the individual p values to find out which of the individual variables are statistically significant.
## The F Value in ANOVA
The F value in one way ANOVA is a tool to help you answer the question “Is the variance between the means of two populations significantly different?” The F value in the ANOVA test also determines the P value; The P value is the probability of getting a result at least as extreme as the one that was actually observed, given that the null hypothesis is true.
The p value is a probability, while the f ratio is a test statistic, calculated as:
F value = variance of the group means (Mean Square Between) / mean of the within group variances (Mean Squared Error)
## When Do I Reject the Null Hypothesis?
Reject the null when your p value is smaller than your alpha level. You should not reject the null if your critical f value is smaller than your F Value, unless you also have a small p-value.
Where this could get confusing is where one of these values seems to indicate that you should reject the null hypothesis and one of the values indicates you should not. For example, let’s say your One Way ANOVA has a p value of 0.68 and an alpha level of 0.05. As the p value is large, you should not reject the null hypothesis. However, your f value is 4.0 with an f critical value of 3.2. Should you now reject the null hypothesis? The answer is NO.
Why?
The F value should always be used along with the p value in deciding whether your results are significant enough to reject the null hypothesis. If you get a large f value (one that is bigger than the F critical value found in a table), it means something is significant, while a small p value means all your results are significant. The F statistic just compares the joint effect of all the variables together. To put it simply, reject the null hypothesis only if your alpha level is larger than your p value.
Caution: If you are running an F Test in Excel, make sure your variance 1 is smaller than variance 2. This “quirk” can give you an incorrect f ratio if you put the variances in the wrong place. See the bottom of this article for an example: F Test Two Sample Variances in Excel.
## F Value in Regression
The F value in regression is the result of a test where the null hypothesis is that all of the regression coefficients are equal to zero. In other words, the model has no predictive capability. Basically, the f-test compares your model with zero predictor variables (the intercept only model), and decides whether your added coefficients improved the model. If you get a significant result, then whatever coefficients you included in your model improved the model’s fit.
Read your p-value first. If the p-value is small (less than your alpha level), you can reject the null hypothesis. Only then should you consider the f-value. If you don’t reject the null, ignore the f-value.
Many authors recommend ignoring the P values for individual regression coefficients if the overall F ratio is not statistically significant. This is because of the multiple testing problem. In other words, your p-value and f-value should both be statistically significant in order to correctly interpret the results.
If you want to know whether your regression F-value is significant, you’ll need to find the critical value in the f-table. For example, let’s say you had 3 regression degrees of freedom (df1) and 120 residual degrees of freedom (df2). An F statistic of at least 3.95 is needed to reject the null hypothesis at an alpha level of 0.1. At this level, you stand a 1% chance of being wrong (Archdeacon, 1994, p.168). For more details on how to do this, see: F Test. F Values will range from 0 to an arbitrarily large number.
## F Distribution
The F Distribution is a probability distribution of the F Statistic. In other words, it’s a distribution of all possible values of the f statistic.
F Distribution.
The distribution is an asymmetric distribution usually used for ANOVA . It has a minimum value of zero; there is no maximum value. The distribution’s peak happens just to the right of zero and the higher the f-value after that point, the lower the curve. The F distribution is actually a collection of distribution curves. The F distribution is related to chi-square, because the f distribution is the ratio of two chi-square distributions with degrees of freedom ν1 and ν2 (note: each chi-square is first been divided by its degrees of freedom). Each curve depends on the degrees of freedom in the numerator (dfn) and the denominator (dfd). These depend upon your sample characteristics.
For example, in a simple one-way ANOVA between-groups,
• Dfn = a – 1
• dfd = N – a
where:
• a = the number of groups
• n = the total number of subjects in the experiment
The degrees of freedom in the denominator (dfd) is also referred to as the degrees of freedom error (dfe).
The F Distribution is also called the Snedecor’s F, Fisher’s F or the Fisher–Snedecor distribution.
## F Distribution on the TI 89: Overview
There are two types of main problem you’ll encounter with the F-Distribution you might be asked to find the area under a F curve given numerator degrees of freedom (ndf), denominator degrees of freedom (ddf), and a certain range (for example, P( 1 ≤ X ≤ 2 ), or you might be asked to find the F value with area to the left, a certain ndf and ddf (useful for finding critical values for hypotheses tests).
## F Distribution on TI 89: Steps
Example problem: find the area under a F curve with numerator degrees of freedom (ndf) 4 and
denominator degrees of freedom (ddf) 10 for For, P( 1≤ X ≤ 2 ):
Step 1: Press APPS.
Step 2:Press ENTER twice to get to the list entry screen.
Step 3: Press F5 for “F5-Distr.”
Step 4: Scroll down to “A:F Cdf” and press ENTER.
Step 5: Enter 1 in the box for “Lower Value,” then press the down arrow key.
Step 6: Enter 2 in the box for “Upper Value,” then press the down arrow key.
Step 7: Enter 4 in the “Num df” box, then press the down arrow key.
Step 8: Enter 5 in the “Den df” box.
Step 9: Press ENTER. The calculator will return .281 as the answer.
Example problem: to find the F value with area to the left, with ndf = 5, ddf = 8, and an area of .99:
Step 1: Press APPS.
Step 2: Press ENTER twice to get to the list entry screen.
Step 3: Press F5 for “F5-Distr.”
Step 4: Press 2 for “Inverse.”
Step 5: Press 4 for “Inverse F…,” then press ENTER.
Step 6: Enter .99 in the “Area” box, then press the down arrow key.
Step 7: Enter 5 in the “Num df,” box, then press the down arrow key.
Step 8: Enter 8 in the “Den df.” box, then press ENTER. This returns the answer (63183).
Tip: For P( X ≥ 1 ), enter 1 in the box for Lower Value and 10 ^ 99 in the box for
Upper Value, and for For P( X ≤ 1 ), enter 0 in the box for Lower Value, then enter 1 in the box for Upper
Value.
## The F Statistic Table
The F Table is a collection of tables that give you the probability for a certain alpha level. The F Table is actually a collection of tables, for four alpha levels: .10. .5, .025 and .01.
The three f tables you can find on this site are for alpha levels of .10, .0 and .01. When using the F dist. table, always put the numerator degrees of freedom first; if you switch the numerator and denominator around, you’ll get a different result. The table gives you the area in the right tail. Instead of a table, you can use a calculator — which will give you more accurate results.
## What is the F Statistic Table Used for?
When you have found the F value, you can compare it with an f critical value in the table. If your observed value of F is larger than the value in the F table, then you can reject the null hypothesis with 95 percent confidence that the variance between your two populations isn’t due to random chance.
## How to use the F Statistic Table
Watch the video for an overview:
How to read an F table in statistics
The F Statistic Table is actually a collection of tables. Which specific table you use will depend on which alpha level you use. For example, if you have an alpha level of .05, then your right tail area is .05 (5 percent), and you’ll look up the f critical value in the alpha level = .05 table. The rows in the F Distribution Table represent denominator degrees of freedom and the columns represent numerator degrees of freedom.
For example, to determine the .10 f critical value for an F distribution with 6 and 6 degrees of freedom, look in the 6 column (numerator) and the 6 row (denominator) of the F Table for alpha=.10. F(.10, 6, 6) = 3.05455.
Why Use the F Statistic Table? Why not just use a calculator?
A calculator will certainly give you a fast answer. But with many scenarios in statistics, you will look at a range of possibilities and a table is much better for visualizing a large number of probabilities at the same time.
## References
Archdeacon, T. (1994). Correlation and Regression Analysis: A Historian’s Guide. Univ of Wisconsin Press.
CITE THIS AS:
Stephanie Glen. "F Statistic / F Value: Simple Definition and Interpretation" From StatisticsHowTo.com: Elementary Statistics for the rest of us! https://www.statisticshowto.com/probability-and-statistics/f-statistic-value-test/
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A275381 Number of prime factors (with multiplicity) of generalized Fermat number 10^(2^n) + 1. 2
1, 1, 2, 2, 5, 4, 3, 4, 5 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 LINKS FORMULA a(n) = A001222(A080176(n)). - Felix Fröhlich, Jul 25 2016 EXAMPLE b(n) = 10^(2^n) + 1. Complete Factorizations b(0) = 11 b(1) = 101 b(2) = 73*137 b(3) = 17*5882353 b(4) = 353*449*641*1409*69857 b(5) = 19841*976193*6187457*834427406578561 b(6) = 1265011073* 15343168188889137818369*515217525265213267447869906815873 b(7) = 257*15361*453377*P116 b(8) = 10753*8253953*9524994049*73171503617*P225 MATHEMATICA Table[PrimeOmega[10^(2^n) + 1], {n, 0, 6}] (* Michael De Vlieger, Jul 26 2016 *) PROG (PARI) a(n) = bigomega(factor(10^(2^n)+1)) CROSSREFS Cf. A072982, A080176. Sequence in context: A217876 A209771 A209751 * A283235 A209763 A209761 Adjacent sequences: A275378 A275379 A275380 * A275382 A275383 A275384 KEYWORD nonn,hard,more AUTHOR Arkadiusz Wesolowski, Jul 25 2016 STATUS approved
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Last modified January 29 10:34 EST 2020. Contains 331337 sequences. (Running on oeis4.) | 547 | 1,523 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-05 | latest | en | 0.523857 |
https://phys.org/news/2013-12-santa-logistics-maths.html | 1,713,326,722,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817128.7/warc/CC-MAIN-20240417013540-20240417043540-00292.warc.gz | 441,057,740 | 18,042 | # Help solve Santa's logistics troubles with a little maths
In just one night, Santa has to visit millions of homes to deliver presents. If he could travel at the speed of light, the task would be simple.
However, Einstein's formula, E=MC², tells us that anything with mass cannot travel faster than the speed of light. And as we all know, Santa has mass. That's before you even count all the presents that have to be transported along with his sleigh. Then add Rudolph and co and what you get is a lot of flying mass that makes Santa's chances of travelling faster than light pretty slim.
Luckily, there are other options available to Santa to help increase his chances of delivering all the presents on time. And they relate to what is known in maths as The Travelling Salesman Problem.
In this problem, a salesman has to plan a through a number of cities. He has to start and end at the same city and visit every other city in between just once, while minimising the distance he travels. If we replace the salesman with Santa and the cities with chimneys, then Santa's problem is a variant of the Travelling Salesman Problem.
Unfortunately, this isn't much consolation to Santa. The Travelling Salesman Problem is known to be NP-Complete. This means that there is no known efficient algorithm that always returns the optimal solution in a reasonable time.
In fact, most mathematicians and computer scientists believe that no such algorithm exists, although this is yet to be proved. Anyone who can prove that it does exist (or indeed that it doesn't) stands to win \$1 million for solving one of the Millennium Problems and proving whether P=NP (or not).
Back to Santa. What can he do to plan his route? Although we do not know of an algorithm that is guaranteed to tell Santa the best route to take, there are algorithms that attempt to do this in reasonable time.
The Route Santa application, from Napier University, is one. The app shows an example of an algorithm solving the Travelling Santa Problem. Those hoping to receive gifts on the 25 December can contribute to the efficient running of Santa's rounds by uploading their address into an interactive map. The Route Santa software will then add that address to its list and work out the best route for Rudolph. The more hopeful recipients that sign up, the more efficient Santa's journey will be.
One type of the used for these type of problems are known as genetic algorithms because they are based on natural phenomena such as evolution, inheritance and selection.
Santa's problem is your problem too
We tend to take it for granted that Santa will make his millions of deliveries on time every year but the Travelling Salesman Problem affects our daily lives, particularly now that so much of what we buy, from our food to our clothes and technology is delivered to our doors. If we can work out the best route for Santa, we can also contribute to thinking about how to make other delivery services more efficient for the other 11 months of the year.
This story is published courtesy of The Conversation (under Creative Commons-Attribution/No derivatives).
Explore further
Woman's secret Santa turns out to be Bill Gates
Feedback to editors | 655 | 3,219 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2024-18 | latest | en | 0.968682 |
https://www.gigacalculator.com/converters/oil-to-butter-converter.php | 1,685,387,865,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644907.31/warc/CC-MAIN-20230529173312-20230529203312-00335.warc.gz | 863,330,837 | 15,334 | # Oil to Butter Conversion
Some recipes list an amount of vegetable oil and you may want to replace it with butter since "Everything's better with butter"! This oil to butter converter allows you to easily substitute vegetable oil with butter. The converter supports conversion in cups, teaspoons (tsp), tablespoons (tbsp), ounces (oz), pounds (lb), sticks of butter, grams (g), and milliliters (ml).
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## Substituting vegetable oil with butter
Vegetable oils are oils extracted from a variety of seeds and other plant parts with popular vegetable oils including sunflower oil, olive oil, palm oil, soybean oil, grape seed oil, and coconut oil. Butter is a dairy product typically produced from the fats and proteins in cow's milk and has a yellowish to white color. Both are used in many cooking recipes.
A recipe may call for a certain amount of vegetable oil in its ingredients list, but if you want to use butter instead, you can definitely substitute oil with butter. The unique flavor added by butter will be well worth it!
## Oil to butter conversion
Vegetable oils of all kinds consist of close to 100% fat while butter consists of only about 80% fat. Given this, an easy rule of thumb is that to replace a unit of oil you need 4/3 the amount of butter. This can be done by multiplying by 4/3 (1.333), or by dividing by 3/4 (0.75). For example, if a recipe calls for one cup of oil you'll need 1 · 1.333 = 1⅓ cups of butter to replace it, or roughly 2½ sticks of butter.
To substitute an amount of oil specified in a volume unit such as tablespoons, cups, teaspoons, or milliliters with butter, the difference in density between oil and butter needs to be accounted for to achieve a precise oil to butter conversion. Since the differences are quite small, the division by 3/4 rule works just fine for most recipes prepared in a home kitchen.
Note that the kind of vegetable oil does not really matter. Olive oil, sunflower oil, soybean oil, and all other popular ones have roughly the same consistency and density therefore converting from any kind of vegetable oil to butter works the same way.
## Oil to butter conversion examples
Here are some example vegetable oil to butter conversion calculations to get you started. As noted, oil to butter conversion which includes switching the measurement unit is more difficult and using our oil to butter conversion calculator above is recommended so you do not need to remember the ratios between cups and grams, cups and ounces, tablespoons and grams, and so on.
### Cups of oil to sticks of butter
Sample task: convert 1 cup of oil to sticks of butter, knowing that a standard sticks of butter is 1/4 lb or 4 oz and one cup of oil weighs 7.704 oz. Solution:
Formula:
cups of oil * 2.5906666666667 = sticks of butter
Calculation:
1 cups of oil * 2.5907 = 2.590667 sticks of butter
End result:
1 cups of oil is equal to 2.590667 sticks of butter
How did we arrive at the 2.59 number? First, multiply the cups by 7.772 to arrive at ounces of oil, then divide by 0.75 (this is the 3/4 oil to butter ratio) to get the substitution amount of butter in ounces, and finally divide by 4 to convert it to sticks of butter. Note that there is a small rounding difference between what you will get in the oil to butter converter above. The practical result is that substituting one cup of vegetable oil requires roughly 2½ sticks of butter.
### Ounces of oil to ounces of butter
Sample task: convert 3 oz of oil to ounces of butter, given that the oil to butter conversion ratio is 1:1.33. Solution:
Formula:
oz of oil * 1.333 = oz of butter
Calculation:
3 oz of oil * 1.3330 = 3.9990 oz of butter
End result:
3 oz of oil is equal to 3.9990 oz of butter
The above means three ounces of oil can be substituted with exactly one 4-ounce stick of butter.
### Cups of vegetable oil to cups of butter
Sample task: convert 2 cups of vegetable oil to cups of butter, knowing that the conversion ratio of oil to butter is 1:4/3 (1:1.33). Solution:
Formula:
cups of oil * 1.333 = cups of butter
Calculation:
0.5 cups of oil * 1.3330 = 0.6665 cups of butter
End result:
0.5 cups of oil is equal to 0.6665 cups of butter
So half a cup of veggie oil can be replaced by three quarters of a cup of butter in most recipes.
### Grams of oil to grams of butter
Sample task: convert 30 grams of oil to grams of butter. Solution:
Formula:
g of oil / 0.75 = g of butter
Calculation:
30 g of oil / 0.75 = 40 g of butter
End result:
30 g of oil is equal to 40 g of butter
or roughly a third of a butter stick.
## Oil to butter conversion table
Oil to butter conversion chart
Vegetable oil Butter
Cups Weight Sticks Weight
1/4 of a cup 2 oz (55 g) 2/3 stick 2½ oz (74 g)
1/3 of a cup 2½ oz (73 g) 1 stick 3½ oz (98 g)
1/2 of a cup 4 oz (113 g) 1⅓ stick 5⅕ oz (147 g)
2/3 of a cup 5⅕ oz (147 g) 1¾ stick 7 oz (196 g)
3/4 of a cup 6 oz (165 g) 2 sticks 7¾ oz (220 g)
1 cup 7.772 oz (220 g) 2½ sticks 10⅓ oz (294 g)
1½ cups 11.657 oz (330 g) 4 sticks 15½ oz (441 g)
2 cups 15.543 oz (330 g) 5⅕ sticks 20¾ oz (587 g)
### How much is a stick of butter?
In the United States, butter is often sold in sticks, each weighing 1/4 pounds or 4 ounces, which is about 113 grams. Due to historical differences in butter cutting and packaging machinery, 4-ounce sticks are mainly produced in two different shapes, with most recipes referring to Elgin-style, a.k.a. Eastern-pack shaped sticks.
See "How much does a stick of butter weigh?" for more on this topic.
#### Cite this converter & page
If you'd like to cite this online converter resource and information as provided on the page, you can use the following citation:
Georgiev G.Z., "Oil to Butter Converter", [online] Available at: https://www.gigacalculator.com/converters/oil-to-butter-converter.php URL [Accessed Date: 29 May, 2023]. | 1,540 | 5,874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-23 | latest | en | 0.947513 |
drowneddesigns.com | 1,558,861,897,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232259015.92/warc/CC-MAIN-20190526085156-20190526111156-00477.warc.gz | 53,453,798 | 3,425 | drowneddesigns.com
Main / Social / Introduction Categorical Data Analysis Agresti Solution Manual
Introduction Categorical Data Analysis Agresti Solution Manual
Name: Introduction Categorical Data Analysis Agresti Solution Manual File size: 790mb Language: English Rating: 10/10 Download
AN INTRODUCTION TO CATEGORICAL DATA ANALYSIS, 2nd ed. Please report any errors in the solutions to Alan Agresti, e-mail [email protected] copyright. Our solution manuals are written by Chegg experts so you can be assured of An Introduction to Categorical Data Analysis Solutions Manual. Get access now with. Get Started. Select your edition Below. by. 2nd Edition. Author: Alan Agresti. Access An Introduction to Categorical Data Analysis 2nd Edition Chapter 2 solutions now. Our solutions are written by Chegg experts so you can be assured of the highest probability solutions manuals / An Introduction to Categorical Data Analysis / 2nd . ISBN ISBN: Alan Agresti Authors.
Page 1 of 1. Ebook Pdf an introduction to categorical data analysis agresti solution manual Verified Book Library. Ebook Pdf an introduction to categorical data. 21 Apr pdf Categorical data analysis alan agresti solution 8 KB 36 to categorical analysis solution manual introduction categorical data. This manual accompanies Agresti's Categorical Data Analysis (). .. can be downloaded from statlib or from the CRAN website (see Introduction π π λ, then discard the solution for the constant λ (called the Lagrange multiplier).
This manual contains solutions and hints to solutions for many of the odd- numbered exercises in Categorical Data Analysis, second edition, by Alan Agresti. This file contains solutions and hints to solutions for some of the exercises in Categorical. Data Analysis, third edition, by Alan Agresti (John Wiley, & Sons, ). 18 Oct - 36 sec - Uploaded by Jo Harrison An Introduction To Categorical Data Analysis Agresti Solution Manual. Jo Harrison. Du er her: ForsidenIntroduction to Categorical Data Analysis ALAN AGRESTI, PhD, is Distinguished Professor Emeritus in the Department of Statistics at of Examples Subject Index Brief Solutions to Some Odd-Numbered Problems Student Solutions Manual for Statistical Methods for the Social Sciences. Agresti, Alan. An introduction to categorical data analysis / Alan Agresti. .. lent free manual prepared by Laura Thompson showing how to use R and S-Plus to.
An Introduction To Categorical Data Analysis Alan Agresti Solution Manual - In this site is not the similar as a solution reference book you buy in a autograph. 16 Feb pendix or specialized manuals while reading the text, as an aid to implement- ing the methods. text, An Introduction to Categorical Data Analysis Wiley, .. denoted by Я, and Я is the solution of a set of likelihood equations. ˆ. ˆŽ. probability G Agresti and Coull ; Blyth and Still drowneddesigns.com: An Introduction to Categorical Data Analysis (): Alan Agresti: Books. Quiz solutions including Quiz 11 are posted Here · Course Outline · Textbook: An introduction to categorical data analysis (2nd ed.) by Alan Agresti.
More: | 667 | 3,096 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-22 | longest | en | 0.725397 |
http://reference.wolfram.com/legacy/v5_2/book/section-1.5.12 | 1,386,411,462,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163053923/warc/CC-MAIN-20131204131733-00074-ip-10-33-133-15.ec2.internal.warc.gz | 200,534,326 | 6,717 | ## 1.5.12 Integral Transforms
LaplaceTransform[expr, t, s] find the Laplace transform of expr InverseLaplaceTransform[expr, s, t] find the inverse Laplace transform of expr
Laplace transforms.
This computes a Laplace transform.
In[1]:= LaplaceTransform[t^3 Exp[a t], t, s]
Out[1]=
Here is the inverse transform.
In[2]:= InverseLaplaceTransform[%, s, t]
Out[2]=
FourierTransform[expr, t, w] find the symbolic Fourier transform of expr InverseFourierTransform[expr, w, t] find the inverse Fourier transform of expr
Fourier transforms.
This computes a Fourier transform.
In[3]:= FourierTransform[t^4 Exp[-t^2], t, w]
Out[3]=
Here is the inverse transform.
In[4]:= InverseFourierTransform[%, w, t]
Out[4]=
Note that in the scientific and technical literature many different conventions are used for defining Fourier transforms. Section 3.8.4 describes the setup in Mathematica.
THIS IS DOCUMENTATION FOR AN OBSOLETE PRODUCT.
SEE THE DOCUMENTATION CENTER FOR THE LATEST INFORMATION. | 270 | 997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2013-48 | latest | en | 0.529888 |
https://blog.csdn.net/Quincuntial/article/details/84897945 | 1,544,741,814,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825112.63/warc/CC-MAIN-20181213215347-20181214000847-00172.warc.gz | 562,252,156 | 34,489 | # Leetcode 63. Unique Paths II
## 2. Solution
• Version 1
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
int m = obstacleGrid.size();
int n = obstacleGrid[0].size();
vector<vector<int>> path(m, vector<int>(n));
path[0][0] = 1;
for(int i = 0; i < m; i++) {
for(int j = 0; j < n; j++) {
if(obstacleGrid[i][j]) {
path[i][j] = 0;
continue;
}
if(i > 0 && j > 0) {
path[i][j] = path[i - 1][j] + path[i][j - 1];
}
else if(i < 1 && j > 0) {
path[i][j] = path[i][j - 1];
}
else if(i > 0 && j < 1) {
path[i][j] = path[i - 1][j];
}
}
}
return path[m - 1][n - 1];
}
};
• Version 2
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
int m = obstacleGrid.size();
int n = obstacleGrid[0].size();
if(!obstacleGrid[0][0]) {
obstacleGrid[0][0] = 1;
}
else {
obstacleGrid[0][0] = 0;
}
for(int i = 0; i < m; i++) {
for(int j = 0; j < n; j++) {
if(i == 0 && j == 0) {
continue;
}
if(obstacleGrid[i][j]) {
obstacleGrid[i][j] = 0;
continue;
}
if(i > 0 && j > 0) {
obstacleGrid[i][j] = obstacleGrid[i - 1][j] + obstacleGrid[i][j - 1];
}
else if(i < 1 && j > 0) {
obstacleGrid[i][j] = obstacleGrid[i][j - 1];
}
else if(i > 0 && j < 1) {
obstacleGrid[i][j] = obstacleGrid[i - 1][j];
}
}
}
return obstacleGrid[m - 1][n - 1];
}
};
SnailTyan
• 擅长领域：
• 深度学习
• PyTorch
• OCR
• Docker
• Caffe | 532 | 1,350 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2018-51 | longest | en | 0.275908 |
https://www.cfd-online.com/Forums/fluent/67211-udf-mass-flux-computing.html | 1,513,369,592,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948579564.61/warc/CC-MAIN-20171215192327-20171215214327-00238.warc.gz | 738,792,716 | 18,893 | # UDF Mass flux computing
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August 7, 2009, 04:41 UDF Mass flux computing #1 New Member Carlo Locci Join Date: Apr 2009 Posts: 8 Rep Power: 10 Hi everybody: I'm trying to write an UDF which calculates the mass flux of an outlet. I need it to calculate a pollutant concentration: for this reason the DEFINE in which my computing is inserted is the DEFINE_ADJUST. my problem is that this udf doesn't return the total mass flux value I get from the REPORT tool. The problem is 2D axisymmetric, so my outlet is seen like a circular area. #include "udf.h" DEFINE_ADJUST(massflux, domain) { int zone_ID = 4; Thread *t = Lookup_Thread(domain,zone_ID); face_t f; float r, x[2]; real summ=0.; real m; real NV_VEC (A); begin_f_loop(f,t) { F_CENTROID(x,f,t); r=x[1]; F_AREA(A,f,t); m=F_U(f,t)*F_R(f,t)*NV_MAG(A)*6.28*r; summ += m; } end_f_loop(f,t) printf("massf: %g\n", summ); } The 6.28 is from 2*pi. It derives because of the area computing in polar coordinates. If you have any idea to carry on the computing in a simpler way, please tell me it. Last thing: I don't get any error while fluent compiles it. It's just that I obtain different values. THANKS
August 11, 2009, 22:41 Please check your formulation #2 New Member Bobby Join Date: Aug 2009 Posts: 7 Rep Power: 10 I believe that your formulation is incorrect for flux. You just add the mass flow rate for each face and then finally divide by pi*Rmax^2. I'm not sure why you are using 2*pi*r*area in the evaluation of m. Instead you need to replace area by dr (i.e cell length in r direction). You will have m_dot = rho*U*2*pi*r*dr. I am assuming that the actual flow velocity is in the U direction i.e flow parallel to the area vector. The other way is to use the F_FLUX(f,t) macro and if needed you can play with it. I'm not sure if this macro accounts for axisymmetric calculations! Good Luck and have fun! Last edited by PawnPace; August 12, 2009 at 01:15.
August 12, 2009, 04:03 Annulus #3 New Member Carlo Locci Join Date: Apr 2009 Posts: 8 Rep Power: 10 Dear PawnPace, let's imagine an outlet of a 2d grid: it is formed by several segments. Now, when you ask to fluent to consider the case as axisymmetric, you obtain a cylinder as you can see in this link: http://jullio.pe.kr/fluent6.1/help/html/ug/node347.htm Now, doing this in the outlet, for each segment you have an Annulus. The area of an Annulus is: (R^2 - r^2)*pi Considering it as infinitesimal you get: ((r+dr)^2-r^2)*pi=(r^2+2rdr+dr^2-r^2)*pi Neglecting the dr^2 you have that the infinitesimal area of the considered Annulus is 2pi*r*dr. Applying all this to our case, you have that r and dr are respectively the y coordinate and the height of the generic segment of the outlet. You have to sum the area of each Annulus multiplied for the density and for the velocity axial component. By this the 2*pi derives. Anyway, I'm going to try your hints and I'll let you know about
August 12, 2009, 13:49 #4 New Member Bobby Join Date: Aug 2009 Posts: 7 Rep Power: 10 I think there was some misunderstanding. Anyway I just wanted to tell you that your formulation of mass flow rate does not match units on right side of your equation. Since, you are using 2d formulation and the area for the face (i think it) is the edge length multiplied by unit depth (not sure). Numerically your equation may be correct but dimensionally it is not. Well I hope my posts were helpful! Regards, PawnPace
August 17, 2009, 11:19 I did it! #5 New Member Carlo Locci Join Date: Apr 2009 Posts: 8 Rep Power: 10 Well, I did it! #include "udf.h" DEFINE_ADJUST(massflux, domain) { int zone_ID = 4; Thread *t = Lookup_Thread(domain,zone_ID); face_t f; real summ=0.; real m; real sumarea=0.; real area; real NV_VEC (A); begin_f_loop(f,t) { F_AREA(A,f,t); m=F_U(f,t)*F_R(f,t)*NV_MAG(A)*6.28; summ += m; area=NV_MAG(A)*6.28; sumarea += area; } end_f_loop(f,t) printf("massf: ,%g,%g\n", summ,sumarea); } Well, this works fine! It makes me get exactly what I wanted both in terms of area and mass flux. Thanks to your suggestion, I thought that multiply an area by a length (in our case r) could be dimensionally wrong. So I decided to multiply only by the area deleting 2*pi*r. After having done it, results were still wrong. Fortunately I've thought to look at what was the ratio between my (wrong) results and the ones I would have wanted to get. Well, that ratio was exactly 6.28! This is the reason the 6.28 lasts. In order to prove if It was right, I've tried it also in others domains and It computes the right area and mass flux in all of them! So I think that the NV_MAG(A) in axisymmetric cases is just the r*dr we spoke above, which is the difference between R^2 and r^2. Thanks you! jyothsna k likes this.
August 17, 2009, 12:22 #6 New Member Bobby Join Date: Aug 2009 Posts: 7 Rep Power: 10 I'm glad that it's resolved! Good Luck and have fun with Fluent! Regards, PawnPace
November 23, 2014, 22:59
thanks a lot! but I use your code but i didn't got the cooperate massflow.
#7
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sam toris
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Originally Posted by Carlo Well, I did it! #include "udf.h" DEFINE_ADJUST(massflux, domain) { int zone_ID = 4; Thread *t = Lookup_Thread(domain,zone_ID); face_t f; real summ=0.; real m; real sumarea=0.; real area; real NV_VEC (A); begin_f_loop(f,t) { F_AREA(A,f,t); m=F_U(f,t)*F_R(f,t)*NV_MAG(A)*6.28; summ += m; area=NV_MAG(A)*6.28; sumarea += area; } end_f_loop(f,t) printf("massf: ,%g,%g\n", summ,sumarea); } Well, this works fine! It makes me get exactly what I wanted both in terms of area and mass flux. Thanks to your suggestion, I thought that multiply an area by a length (in our case r) could be dimensionally wrong. So I decided to multiply only by the area deleting 2*pi*r. After having done it, results were still wrong. Fortunately I've thought to look at what was the ratio between my (wrong) results and the ones I would have wanted to get. Well, that ratio was exactly 6.28! This is the reason the 6.28 lasts. In order to prove if It was right, I've tried it also in others domains and It computes the right area and mass flux in all of them! So I think that the NV_MAG(A) in axisymmetric cases is just the r*dr we spoke above, which is the difference between R^2 and r^2. Thanks you!
I think the original F_FLUX function could bring a similar flux result, but not accurate, hope for your reply!
11-24 11:06
I work it out! I think maybe the unaccuracy was cuased by an unsteady flow.
Tags computing, flux, mass, udf
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All times are GMT -4. The time now is 16:26. | 2,055 | 7,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-51 | latest | en | 0.846906 |
https://slideplayer.com/slide/3620078/ | 1,632,467,582,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057504.60/warc/CC-MAIN-20210924050055-20210924080055-00167.warc.gz | 552,968,516 | 17,910 | # Sec. 11 – 4 Volumes of Prisms & Cylinders Objectives: 1) To find the volume of a prism. 2) To find the volume of a cylinder.
## Presentation on theme: "Sec. 11 – 4 Volumes of Prisms & Cylinders Objectives: 1) To find the volume of a prism. 2) To find the volume of a cylinder."— Presentation transcript:
Sec. 11 – 4 Volumes of Prisms & Cylinders Objectives: 1) To find the volume of a prism. 2) To find the volume of a cylinder.
Volume ►V►V►V►Volume – Is the space that a figure occupies. MMMMeasured in cubic units. ►c►c►c►cm3, in3, m3, ft3 ►T►T►T►Th(10-5) Cavalieri’s Principle IIIIf 2 space figures have the same height & the same cross-sectional area @ every level, then they have the same volume.
I. Finding the volume of a Prism ► Prism – 2 parallel bases, sides (Lats) are rectangles. Height (h) Area of Base (B) V = Bh Height of Prism Area of Base A = bh (Rectangle) A = ½bh (Triangle) A = ½ap (Polygon)
Ex. 1: Finding the Volume of a rectangular prism ► The box shown is 5 units long, 3 units wide, and 4 units high. How many unit cubes will fit in the box? What is the volume of the box?
Ex.1: Find the Volume of the Prism 5in 3in 10in V = Bh = (3in 5in)(10in) = (15in 2 )(10in) = 150in 3 Area of Base B = lw
Ex.2: Find the volume of the following 20m 29m 40m V = Bh = ½bh h = ½(20m)__ (40m) = 210m 2 40m = 8400m 3 Height of the base: a a 2 + b 2 = c 2 a 2 + 20 2 = 29 2 b = 21 Triangle 21
Ex.3: Yet another prism! Find the volume. 8in 10in V = Bh = ½bh h = ½(8in) __ (10in) = (27.7in 2 ) (10in) = 277in 3 4 8 h 60° Sin 60 = h/8.866 = h/8 6.9 = h 6.9
Cavalieri’s Principle If two solids have the same height and the same cross-sectional area at every level, then they have the same volume.
II. Volume of a Cylinder r h V = Bh Volume of right cylinder Height of cylinder Area of base: (Circle) A = r 2
Ex.4: Find the area of the following right cylinder. 16ft 9ft V = Bh = r 2 h = (8ft) 2 (9ft) = 64 ft 2 (9ft) = 576 ft 3 = 1809.6ft 3 Area of a Circle
Ex.5: Find the volume of the following composite figure. Half of a cylinder: V c = Bh = r 2 h = (6in) 2 (4in) = 452in 3 = 452/2 = 226in 3 12in 4in 11in Volume of Prism: V p = Bh = (11)(12)(4) = 528in 3 V T = V c + V p = 226in 3 + 528in 3 = 754in 3
What have we learned?? Volume of a prism or a cylinder: V = Bh Capitol “B” stands for area of the base. Composite Figures: Made up of two separate solids.
Download ppt "Sec. 11 – 4 Volumes of Prisms & Cylinders Objectives: 1) To find the volume of a prism. 2) To find the volume of a cylinder."
Similar presentations | 942 | 2,566 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2021-39 | latest | en | 0.803888 |
https://www.jiskha.com/questions/204153/Jasper-walks-2-km-due-east-He-then-turns-and-walks-3-km-northeast-a-How-far-is | 1,560,823,308,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998600.48/warc/CC-MAIN-20190618003227-20190618025227-00062.warc.gz | 783,064,944 | 5,070 | # Math (Trigonometry)
Jasper walks 2 km due east. He then turns and walks 3 km northeast.
a. How far is Jasper from his starting point? Round your answer to 2 decimal places.
b. What is the direction of Jasper's destination in relation to his starting point? Round your answer to the nearest degree.
1. 👍 0
2. 👎 0
3. 👁 77
1. Did you make a diagram?
I see a triangle with sides 2 and 3 and the contained angle as 135º, let the side opposite that angle be x
I see the cosine law.
x^2 = 2^2 + 3^2 - 2(2)(3)cos 135
= 4 + 9 - 12(-√3/2)
= ....
x = √....
then use the sine law to find the angle at the origin
sin A/3 = sin 135/x
....
1. 👍 0
2. 👎 0
posted by Reiny
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Bill Giordano
Chemistry
Mrs.Rupinski
5/5/10
Specific Heat
Although specific heat measures joules, specific heat needs the capacity, the mass, and the final change in temperature to figure out exactly what the specific is. Specific heat is measured in joules, measures the amount of heat ones temperature can hold. Specific heat’s unit is J/g*C. That unit stands for joules over grams per Celsius. For example copper’s specific heat is 0.385 J/g*C. If you we’re to solve an equation involving specific heat the formula is Q = (Cp) (M) (T). Whereas Cp stands for specific heat, M stands for mass and T stands for change in temperature. The triangle stands for delta as the Greeks had named it long ago. Specific heat is basically heat energy that gets stored in places outdoors or indoors. A copper plate on the bottom of pot on a stove over a flame stores more specific heat than some other elements and metals because H2O particles and atoms are released quicker when heated to create a hotter specific heat.
Specific heat’s greatest partner in role is water. Water is called the universal solvent because it can dissolve so many elements, compounds, and solutions. Water is made up of 70% of the Earth and is in more than 2/3 of the human body. Water is the simplest liquid to be used in specific heat. Water has a high specific heat because hydrogen bonds between the water molecules are weak and temporary. Molecules are constantly breaking apart it takes more energy to heat up water rather than a solid. To find the specific heat of a substance, determine the specific heat capacity of a substance. First drop a substance into a heated container of liquid. Find the mass of the substance then, find the initial and final temperature of the substance. The greater the specific heat capacity of a substance, the more thermal energy is required to increase the temperature of that substance by 1o Celsius.
As liquid crystals and H2O undergo a change, specific heat is to... | 425 | 1,999 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-06 | latest | en | 0.911192 |
https://algebra-class-ecourse.com/question/a-catering-service-offers-7-appetizers-3-main-courses-and-5-desserts-a-customer-is-to-select-4-a-11550554-95/ | 1,632,278,784,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057303.94/warc/CC-MAIN-20210922011746-20210922041746-00292.warc.gz | 140,233,221 | 12,387 | ## A catering service offers 7 appetizers, 3 main courses, and 5 desserts. A customer is to select 4 appetizers, 2 main courses, and 4 desserts
Question
A catering service offers 7 appetizers, 3 main courses, and 5 desserts. A customer is to select 4 appetizers, 2 main courses, and 4 desserts for a banquet. In how many ways can this be done?
0
## Answers ( No )
1. This question is based on combinations
Out of 7 appetizers, 4 are to be selected, this becomes = 7C4
Out of 3 main courses, 2 are to be selected , this is = 3C2
Out of 5 desserts, 4 are to be selected, this is = 5C4
So, total ways become =
7C4 * 3C2 * 5C4
7!/4!(7-4)! * 3!/2!(3-2)! * 5!/4!(5-4)!
= 35*3*5 =525 ways
Hence, this can be done in 525 ways. | 247 | 731 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2021-39 | latest | en | 0.908709 |
https://communities.sas.com/t5/SAS-Programming/Extract-missing-dates-and-additional-dates/m-p/935261 | 1,721,924,419,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763860413.86/warc/CC-MAIN-20240725145050-20240725175050-00466.warc.gz | 153,265,439 | 43,416 | Obsidian | Level 7
## Extract missing dates and additional dates
``````Data sv ;
input cpevent\$1-8 svstdt \$9-18 ;
patient = "1015";
svstdt1 = input(svstdt,date9.);
cards ;
week0 20apr2022
week4 18may2022
week8 20jun2022
week26 19oct2022
week30 22nov2022
week34 14dec2022
week52 18apr2023
week56 17may2023
week60 19jun2023
week78 17oct2023
week82 15nov2023
week86 13dec2023
week104 17apr2024
;
Data qs ;
patient = "1015";
do qsdat = '22apr2022'd to '25apr2022'd,'18may2022'd to '20may2022'd,
'21jun2022'd,'19oct2022'd to '22oct2022'd,'14dec2022'd to '17dec2022'd,
'18apr2023'd to '21apr2023'd ,'17may2023'd to '19may2023'd,'19jun2023'd to '21jun2023'd,
'19oct2023'd to '22oct2023'd,'15nov2023'd to '18nov2023'd,
'13dec2023'd to '16dec2023'd,'18apr2024'd ;
output;
end ;
format qsdat date9.;
run;
``````
I need help in programming based on below conditions :
Created above data based on my real time study, And Requiredoutput dataset also below attached.
(This is emergency requirement for my study)
for the svstdt check the below logics :
1)for day0 check if qsdat = svstdt ,if yes populate the qsdat inday0 column.
keep lt blank
2)for day1 check if qsdat = svstdt+1 ,if yes populate the qsdat in day1 column.
keep lt blank
3)for day2 check if qsdat = svstdt+2 ,if yes populate the qsdat in day2 column.
keep lt blank
4)for day3 check if qsdat = svstdt+3 ,if yes populate the qsdat in day3 column.
keep lt blank
if any columns from day0 ,day1,day2,day3 is blank
then populate flag as "missing"
above defined days (day1,day1,day2,day3)
I am tried below programme everything okay but not getting additional records :
Please any one correct it .
data sv_qs;
if _n_ = 1 then do;
declare hash h(dataset:'qs');
h.defineKey('patient', 'qsdat');
h.defineData('qsdat');
h.defineDone();
end;
set sv;
do i = 0 to 4;
rc = h.find(key:patient, key:svstdt1+i);
if rc = 0 then days[i+1] = svstdt1+i;
/* else days[i+1] =.;*/
end;
rc = h.find(key:patient, key:qsdat);
if rc = 0 and qsdat < svstdt1 or qsdat > svstdt1 + 3 then flag = "extra";
drop rc i;
run;
1 ACCEPTED SOLUTION
Accepted Solutions
Tourmaline | Level 20
## Re: Extract missing dates and additional dates
You could do everything in one data step, but it's easier to code and follow the logic if you save the additional dates in a subsequent step.
``````data SV ;
retain PATIENT '1015';
input CPEVENT :\$8. SVSTDTE :date9. ;
format SVSTDTE date9.;
cards ;
week0 20apr2022
week4 18may2022
week8 20jun2022
week26 19oct2022
week30 22nov2022
week34 14dec2022
week52 18apr2023
week56 17may2023
week60 19jun2023
week78 17oct2023
week82 15nov2023
week86 13dec2023
week104 17apr2024
;
data QS;
retain PATIENT '1015';
do QSDTE = '22apr2022'd to '25apr2022'd, '18may2022'd to '20may2022'd, '21jun2022'd,
'19oct2022'd to '22oct2022'd, '14dec2022'd to '17dec2022'd,
'18apr2023'd to '21apr2023'd, '17may2023'd to '19may2023'd,
'19jun2023'd to '21jun2023'd, '19oct2023'd to '22oct2023'd,
'15nov2023'd to '18nov2023'd, '13dec2023'd to '16dec2023'd, '18apr2024'd ;
output;
end;
format QSDTE date9.;
run;
data FILLDAYS;
%* Define lookup table for QS date;
if _n_ = 1 then do;
if 0 then set QS;
declare hash H(dataset:'QS', ordered: 'y');
h.defineKey( 'PATIENT', 'QSDTE');
h.defineData('PATIENT', 'QSDTE');
h.defineDone();
end;
%* Read SV date and next SV date ;
set SV nobs=NOBS;
if _N_ ne NOBS then set SV(keep=SVSTDTE rename=(SVSTDTE=NEXTDTE) firstobs=2);
else NEXTDTE='31DEC2999'd;
%* Declare DAYn variables;
array DAYS[*] DAY0-DAY3 ;
format DAY0-DAY3 date9. ;
%* Populate DAYn variables;
do I = 0 to 3;
RC = H.find(key:PATIENT, key:SVSTDTE+I);
if RC = 0 then do;
DAYS[I+1] = SVSTDTE+I;
H.remove(key:PATIENT, key:SVSTDTE+I);
end;
end;
%* Set missing flag;
if nmiss(of DAY:)=4 then FLAG = 'missing';
%* Export unused QS dates;
%* Clean up;
drop RC I;
run;
%* Assign additional dates to correct interval;
proc sql;
from FILLDAYS
left join
and ADD1.QSDTE between FILLDAYS.QSDTE and FILLDAYS.NEXTDTE
order by FILLDAYS.PATIENT, FILLDAYS.QSDTE ;
run;
%* Final table, concatenate additional dates ;
data WANT;
by PATIENT QSDTE;
end;
if last.QSDTE then do;
output;
end;
run;
``````
11 REPLIES 11
Tourmaline | Level 20
## Re: Extract missing dates and additional dates
I don't see any DAYn column.
Obsidian | Level 7
## Re: Extract missing dates and additional dates
Obsidian | Level 7
## Re: Extract missing dates and additional dates
Based on the conditions mentioned above and two datasets, I need output like the attached image.
I also pasted my program also (My sample programme)
Tourmaline | Level 20
## Re: Extract missing dates and additional dates
You could do everything in one data step, but it's easier to code and follow the logic if you save the additional dates in a subsequent step.
``````data SV ;
retain PATIENT '1015';
input CPEVENT :\$8. SVSTDTE :date9. ;
format SVSTDTE date9.;
cards ;
week0 20apr2022
week4 18may2022
week8 20jun2022
week26 19oct2022
week30 22nov2022
week34 14dec2022
week52 18apr2023
week56 17may2023
week60 19jun2023
week78 17oct2023
week82 15nov2023
week86 13dec2023
week104 17apr2024
;
data QS;
retain PATIENT '1015';
do QSDTE = '22apr2022'd to '25apr2022'd, '18may2022'd to '20may2022'd, '21jun2022'd,
'19oct2022'd to '22oct2022'd, '14dec2022'd to '17dec2022'd,
'18apr2023'd to '21apr2023'd, '17may2023'd to '19may2023'd,
'19jun2023'd to '21jun2023'd, '19oct2023'd to '22oct2023'd,
'15nov2023'd to '18nov2023'd, '13dec2023'd to '16dec2023'd, '18apr2024'd ;
output;
end;
format QSDTE date9.;
run;
data FILLDAYS;
%* Define lookup table for QS date;
if _n_ = 1 then do;
if 0 then set QS;
declare hash H(dataset:'QS', ordered: 'y');
h.defineKey( 'PATIENT', 'QSDTE');
h.defineData('PATIENT', 'QSDTE');
h.defineDone();
end;
%* Read SV date and next SV date ;
set SV nobs=NOBS;
if _N_ ne NOBS then set SV(keep=SVSTDTE rename=(SVSTDTE=NEXTDTE) firstobs=2);
else NEXTDTE='31DEC2999'd;
%* Declare DAYn variables;
array DAYS[*] DAY0-DAY3 ;
format DAY0-DAY3 date9. ;
%* Populate DAYn variables;
do I = 0 to 3;
RC = H.find(key:PATIENT, key:SVSTDTE+I);
if RC = 0 then do;
DAYS[I+1] = SVSTDTE+I;
H.remove(key:PATIENT, key:SVSTDTE+I);
end;
end;
%* Set missing flag;
if nmiss(of DAY:)=4 then FLAG = 'missing';
%* Export unused QS dates;
%* Clean up;
drop RC I;
run;
%* Assign additional dates to correct interval;
proc sql;
from FILLDAYS
left join
and ADD1.QSDTE between FILLDAYS.QSDTE and FILLDAYS.NEXTDTE
order by FILLDAYS.PATIENT, FILLDAYS.QSDTE ;
run;
%* Final table, concatenate additional dates ;
data WANT;
by PATIENT QSDTE;
end;
if last.QSDTE then do;
output;
end;
run;
``````
Tourmaline | Level 20
## Re: Extract missing dates and additional dates
Why do you post exactly the same question under a different name?
Calcite | Level 5
## extract additional and missing records
``````Data sv ;
input cpevent\$1-8 svstdt \$9-18 ;
patient = "1015";
cards ;
week0 20apr2022
week4 18may2022
week8 20jun2022
week26 19oct2022
week30 22nov2022
week34 14dec2022
week52 18apr2023
week56 17may2023
week60 19jun2023
week78 17oct2023
week82 15nov2023
week86 13dec2023
week104 17apr2024
;
Data qs ;
patient = "1015";
do qsdat = '22apr2022'd to '25apr2022'd,'18may2022'd to '20may2022'd,
'21jun2022'd,'19oct2022'd to '22oct2022'd,'14dec2022'd to '17dec2022'd,
'18apr2023'd to '21apr2023'd ,'17may2023'd to '19may2023'd,'19jun2023'd to '21jun2023'd,
'19oct2023'd to '22oct2023'd,'15nov2023'd to '18nov2023'd,
'13dec2023'd to '16dec2023'd,'18apr2024'd ;
output;
end ;
format qsdat date9.;
run;``````
I need help in programming for below condition. Iam created above data based on my real time project. Required output dataset also attached. Output same as attached Dataset below.
for the svstdt check the below logics :
1)for day0 check if qsdat = svstdt ,if yes populate the qsdat inday0 column.
keep lt blank
2)for day1 check if qsdat = svstdt+1 ,if yes populate the qsdat in day1 column.
keep lt blank
3)for day2 check if qsdat = svstdt+2 ,if yes populate the qsdat in day2 column.
keep lt blank
4)for day3 check if qsdat = svstdt+3 ,if yes populate the qsdat in day3 column.
keep lt blank
if any columns from day0 ,day1,day2,day3 is blank
then populate flag as "missing"
above defined days (day1,day1,day2,day3)
PROC Star
## Re: Extract missing dates and additional dates
@ChrisNZ wrote:
Why do you post exactly the same question under a different name?
That post has now been merged with this one.
--------------------------
The hash OUTPUT method will overwrite a SAS data set, but not append. That can be costly. Consider voting for Add a HASH object method which would append a hash object to an existing SAS data set
Would enabling PROC SORT to simultaneously output multiple datasets be useful? Then vote for
Allow PROC SORT to output multiple datasets
--------------------------
Tourmaline | Level 20
## Re: Extract missing dates and additional dates
I'd have reported the other one as spam rather than merge.
Why 2 different names?
PROC Star
## Re: Extract missing dates and additional dates
@ChrisNZ wrote:
I'd have reported the other one as spam rather than merge.
Why 2 different names?
If there are two different names for this wholly identical post, then it may be that these are classmates asking about a homework task. If so, I think there is a benefit to merging.
--------------------------
The hash OUTPUT method will overwrite a SAS data set, but not append. That can be costly. Consider voting for Add a HASH object method which would append a hash object to an existing SAS data set
Would enabling PROC SORT to simultaneously output multiple datasets be useful? Then vote for
Allow PROC SORT to output multiple datasets
--------------------------
Tourmaline | Level 20
## Re: Extract missing dates and additional dates
The posts are exactly the same, same formatting, same attachment. And the user names too: six digits. Quite unusual.
PROC Star
## Re: Extract missing dates and additional dates
You're right.
--------------------------
The hash OUTPUT method will overwrite a SAS data set, but not append. That can be costly. Consider voting for Add a HASH object method which would append a hash object to an existing SAS data set
Would enabling PROC SORT to simultaneously output multiple datasets be useful? Then vote for
Allow PROC SORT to output multiple datasets
--------------------------
Discussion stats
• 11 replies
• 634 views
• 2 likes
• 4 in conversation | 3,387 | 10,479 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-30 | latest | en | 0.598784 |
https://cubit.sandia.gov/public/15.1/help_manual/WebHelp/geometry/groups/basic_group_operations.htm | 1,553,508,593,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203865.15/warc/CC-MAIN-20190325092147-20190325114147-00453.warc.gz | 458,432,196 | 4,374 | # Basic Group Operations
## Geometry Groups
The command syntax to create or modify a group is:
Group ["name" | <id>] Add <list of topology entities>
For example, the command,
group "exterior" add surface 1 to 2, curve 3 to 5
will create the group named Exterior consisting of the listed topological entities. Any of the commands that can be applied to the "regular" topological entities can also be applied to groups. For example, mesh Exterior , list Exterior , or draw Exterior .
Elements may specified by name as well. For example, the command
group 'interior' add surface with name 'bill' 'john' 'fred'
will add the surfaces named 'bill' 'john' and 'fred' to the group 'interior'.
Wildcards (*) can also be used with names. To add all surfaces with the substring 'bob' in their name, use the command:
group 'interior' add surface with name '*bill*'
A topological entity can be removed from a group using the command:
Group ["name" | <id>] Remove <entity list>
The Xor operation can also be performed on entities in group. Xor means if an entity is already in the group, the command will delete this entity from the group. If it is not in the group, the entity is then added to the group.
Group ["name" | <id>] Xor <entity list>
The Equals operation assigns the group to be exactly the same as the list given. All other existing members of the group will be removed.
Group ["name" | <id>] Equals <entity list>
## Group Booleans
Groups may also be created from existing groups by using boolean operations. Each of these commands will create a new group that contains entities from two existing groups. The intersect command will create a new group that contains elements common to both existing groups. The unite command will contain entities that exist in either group. The subtract command will remove entities that are common to both groups and create a new group from entities that exist in exactly one of the groups.
Group {<'name'>|<id>} Intersect Group <id> with Group <id>
Group {<'name'>|<id>} Unite Group <id> with Group <id>
Group {<'name'>|<id>} Subtract Group <id> from Group <id>
## Mesh Groups
Groups may also contain mesh entities. The commands for adding and removing mesh entities are analogous to those for geometric entities.
Group ["name" | <id>] Add {Hex|Face|Edge|Node <id_list>}
Group ["name" | <id>] Remove {Hex|Face|Edge|Node <id_list>}
Group ["name" | <id>] Xor {Hex|Face|Edge|Node <id_list>}
## Group Copy
Groups may be copied as groups using the group transform commands. Child entities cannot be moved using this command. If a child entity is in the group, its parent entity must be specified as well. In addition, all merge partners must be specified. Only groups containing geometric entities can be copied with these commands. If a geometry entity is meshed, the mesh will be copied as well, unless the [nomesh] option is given. Copied entities can be moved, rotated, reflected, or scaled as well.
Group {<'name'>|<id>} Copy [Move <x> <y> <z>] [nomesh]
Group {<'name'>|<id>} Copy [Move {x|y|z} <distance>...] [nomesh]
Group {<'name'>|<id>} Copy [Move <direction> [distance]] [nomesh]
Group {<'name'>|<id>} Copy [Reflect {x|y|z}] [nomesh]
Group {<'name'>|<id>} Copy [Reflect <x> <y> <z>] [nomesh]
Group {<'name'>|<id>} Copy [Rotate <angle> About {x|y|z}] [nomesh]
Group {<'name'>|<id>} Copy [Rotate <angle> About <x> <y> <z>] [nomesh]
Group {<'name'>|<id>} Copy [Scale <scale> | x <val> y <val> z <val>] [nomesh]
## Group Transformations
Groups may be transformed as groups using the group transform commands. This is especially helpful for transforming groups of free mesh elements, where no geometry exists. The command syntax is shown below.
Group {<'name'>|<id>} [Copy [nomesh]] Move <dx> <dy> <dz>
Group {<'name'>|<id>} [Copy [nomesh]] Move {x|y|z} <distance>...
Group {<'name'>|<id>} [Copy [nomesh]] Reflect {x|y|z}
Group {<'name'>|<id>} [Copy [nomesh]] Reflect <x> <y> <z>
Group {<'name'>|<id>} [Copy [nomesh]] Reflect {x|y|z}
Group {<'name'>|<id>} [Copy [nomesh]] Reflect <x> <y> <z>
Group {<'name'>|<id>} [Copy [nomesh]] Rotate <angle> About {x|y|z}
Group {<'name'>|<id>} [Copy [nomesh]] Rotate <angle> About <x> <y> <z>
Group {<'name'>|<id>} [Copy [nomesh]] Rotate <angle> About Vertex <Vertex-1> [Vertex] <Vertex-2>
Group {<'name'>|<id>} [Copy [nomesh]] Scale <scale> | x <val> y <val> z <val>
The nomesh option applies to the copy part of the command. If the no_mesh option is specify, the mesh will not be copied.
## Deleting Groups
Groups can be deleted with the following command:
Delete Group <id range> [Propagate]
The option propagate will delete the group specified and all of its contained groups recursively.
## Cleaning Out Groups
You can remove all of the entities in a group via the cleanout command:
Group <group_id_range> Cleanout [Geometry|Mesh] [Propagate]
By default all entities will be removed - optionally you can cleanout just geometry or mesh entities. As in delete, the propagate option will cleanout the group specified and all of its contained groups recursively. | 1,358 | 5,092 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2019-13 | latest | en | 0.854208 |
https://www.physicsforums.com/threads/olympic-advantage-high-long-triple-jumps.387682/ | 1,591,250,653,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347439019.86/warc/CC-MAIN-20200604032435-20200604062435-00085.warc.gz | 829,981,760 | 16,302 | Gold Member
## Main Question or Discussion Point
Hello;
I have been wondering this for a while. The Earth is not spherical, so the force of gravity is not the same everywhere on the Earth. Does this mean that it is better to set a world record in high/long/triple jumping in a country with a higher altitude?
Thanks.
Related Other Physics Topics News on Phys.org
i'm pretty much one of the most knowlegeable guys you will ever find anywhere concerning athletics ( not braggadocio - it's likelihood from experience )
gravity per se has little effect on horizontal jumps re: lower g
the effect of gravity is much more significant a factor thru reduced atmospheric pressure with altitude than variations of 9.81 m/s^2 around globe - it's only fractionally less g at mexico city
you can ball-park quantify here :
http://myweb.lmu.edu/jmureika/track/wind/index.html
but to save you labour :
compare 10.00s at your chosen altitude with sea-level ( 0 wind each )
the ratio has to be ^2 to gauge advantage ( higher altitude helps once thru greater runway speed & again thru jump phase )
apply that as correction factor
( can also use it for HJ )
the legend is Bob Beamon & Mexico
that jump was screwed by wind gauge judge ( they swapped nationality judges constantly back then due to politics - the guy for Beamon was clueless ) who rounded 2.0 - 3.0 m/s winds to 2.0 when he should have rounded to 0.1, making it something 2.0 ++ m/s & illegal for record purposes ( but of course, still the win - whoever jumps furthest wins ) - it's likely this part-time judge was thrilled to "make" a WR
Beamon had likely wind likely close to 3m/s
the estimate i got for it 0 altitude/0 wind is ~ 8.60 - 8.65m
Carl Lewis never broke outdoor WR ( had indoors of 8.79m - vastly superior to beamon )
Mike Powell beat Carl in '91 with existing WR of 8.95m to 8.91m ( albeit the runway was illegally hard, offering more than 65% permitted energy return )
Carl was robbed of a legendary jump in '82 by an incompetent judge who foul-judged him despite no mark on the plastiscine
download jump here ( its 8.3 MB download at "The Perfect Jump" - i back fan call - he got extraordinary height for him, when he was a noted "flat" trajectory guy because of his incredible speed )
http://www.arielnet.com/media/
fan discussion here :
http://mb.trackandfieldnews.com/discussion/viewtopic.php?f=5&t=23788&hilit=ariel [Broken]
& check wiki :
http://en.wikipedia.org/wiki/Carl_Lewis
He achieved his 10.00 s clocking the same weekend he leapt 8.61 m twice, and the day he recorded his new low-altitude record 8.76 m at Indianapolis, he had three fouls with his toe barely over the board, two of which seemed to exceed Beamon’s record, the third which several observers said reached 30 ft (about 9.15 m). Some say Lewis should have been credited with setting a world record with that jump, claiming the track officials misinterpreted the rules on fouls.[18]
Last edited by a moderator: | 726 | 2,974 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-24 | longest | en | 0.951973 |
http://planet.racket-lang.org/package-source/cce/dracula.plt/1/7/language/acl2-html-docs/QUANTIFIERS-USING-DEFUN-SK-EXTENDED.html | 1,438,651,791,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042990177.43/warc/CC-MAIN-20150728002310-00161-ip-10-236-191-2.ec2.internal.warc.gz | 195,613,039 | 1,595 | #### QUANTIFIERS-USING-DEFUN-SK-EXTENDED
quantification example with details
```Major Section: QUANTIFIERS
```
See quantifiers-using-defun-sk for the context of this example.
```(in-package "ACL2")
; We prove that if every member A of each of two lists satisfies the
; predicate (P A), then this holds of their append; and, conversely.
; Here is a solution using explicit quantification.
(defstub p (x) t)
(defun-sk forall-p (x)
(forall a (implies (member a x)
(p a))))
; The defun-sk above introduces the following axioms. The idea is that
; (FORALL-P-WITNESS X) picks a counterexample to (forall-p x) if there is one.
#|
(DEFUN FORALL-P (X)
(LET ((A (FORALL-P-WITNESS X)))
(IMPLIES (MEMBER A X) (P A))))
(DEFTHM FORALL-P-NECC
(IMPLIES (NOT (IMPLIES (MEMBER A X) (P A)))
(NOT (FORALL-P X)))
:HINTS (("Goal" :USE FORALL-P-WITNESS)))
|#
; The following lemma seems critical.
(defthm member-append
(iff (member a (append x1 x2))
(or (member a x1) (member a x2))))
; The proof of forall-p-append seems to go out to lunch, so we break into
; directions as shown below.
(defthm forall-p-append-forward
(implies (forall-p (append x1 x2))
(and (forall-p x1) (forall-p x2)))
:hints (("Goal" ; ``should'' disable forall-p-necc, but no need
:use
((:instance forall-p-necc
(x (append x1 x2))
(a (forall-p-witness x1)))
(:instance forall-p-necc
(x (append x1 x2))
(a (forall-p-witness x2)))))))
(defthm forall-p-append-reverse
(implies (and (forall-p x1) (forall-p x2))
(forall-p (append x1 x2)))
:hints (("Goal"
:use
((:instance forall-p-necc
(x x1)
(a (forall-p-witness (append x1 x2))))
(:instance forall-p-necc
(x x2)
(a (forall-p-witness (append x1 x2))))))))
(defthm forall-p-append
(equal (forall-p (append x1 x2))
(and (forall-p x1) (forall-p x2)))
:hints (("Goal" :use (forall-p-append-forward
forall-p-append-reverse))))
``` | 579 | 1,841 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2015-32 | longest | en | 0.84956 |
https://en.academic.ru/dic.nsf/enwiki/6599 | 1,600,819,716,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400208095.31/warc/CC-MAIN-20200922224013-20200923014013-00637.warc.gz | 391,656,281 | 14,557 | # Fresnel equations
Fresnel equations
The Fresnel equations, deduced by Augustin-Jean Fresnel (pronEng|freɪˈnɛl), describe the behaviour of light when moving between media of differing refractive indices. The reflection of light that the equations predict is known as Fresnel reflection.
Explanation
When light moves from a medium of a given refractive index "n"1 into a second medium with refractive index "n"2, both reflection and refraction of the light may occur.
In the diagram on the right, an incident light ray PO strikes at point O the interface between two media of refractive indexes "n"1 and "n"2. Part of the ray is reflected as ray OQ and part refracted as ray OS. The angles that the incident, reflected and refracted rays make to the normal of the interface are given as θi, θr and θt, respectively.The relationship between these angles is given by the law of reflection and Snell's law.
The fraction of the incident power that is reflected from the interface is given by the "reflection coefficient" "R", and the fraction that is refracted is given by the "transmission coefficient" "T". [Hecht (1987), p. 100.] The media are assumed to be "non-magnetic".
The calculations of "R" and "T" depend on polarisation of the incident ray. If the light is polarised with the electric field of the light perpendicular to the plane of the diagram above ("s"-polarised), the reflection coefficient is given by:
: $R_s = left \left[ frac\left\{sin \left( heta_t - heta_i\right)\right\}\left\{sin \left( heta_t + heta_i\right)\right\} ight\right] ^2=left \left[frac\left\{n_1cos\left( heta_i\right)-n_2cos\left( heta_t\right)\right\}\left\{n_1cos\left( heta_i\right)+n_2cos\left( heta_t\right)\right\} ight\right] ^2=left \left[frac\left\{n_1cos\left( heta_i\right)-n_2sqrt\left\{1-left\left(frac\left\{n_1\right\}\left\{n_2\right\} sin heta_i ight\right)^2\left\{n_1cos\left( heta_i\right)+n_2sqrt\left\{1-left\left(frac\left\{n_1\right\}\left\{n_2\right\} sin heta_i ight\right)^2 ight\right] ^2$
where θt can be derived from θi by Snell's law and is simplified using trigonometric identities.
If the incident light is polarised in the plane of the diagram ("p"-polarised), the "R" is given by:
: $R_p = left \left[ frac\left\{ an \left( heta_t - heta_i\right)\right\}\left\{ an \left( heta_t + heta_i\right)\right\} ight\right] ^2=left \left[frac\left\{n_1cos\left( heta_t\right)-n_2cos\left( heta_i\right)\right\}\left\{n_1cos\left( heta_t\right)+n_2cos\left( heta_i\right)\right\} ight\right] ^2=left \left[frac\left\{n_1sqrt\left\{1-left\left(frac\left\{n_1\right\}\left\{n_2\right\} sin heta_i ight\right)^2\right\}-n_2cos\left( heta_i\right)\right\}\left\{n_1sqrt\left\{1-left\left(frac\left\{n_1\right\}\left\{n_2\right\} sin heta_i ight\right)^2\right\}+n_2cos\left( heta_i\right)\right\} ight\right] ^2$
The transmission coefficient in each case is given by "T"s = 1 − "R"s and "T"p = 1 − "R"p. [Hecht (1987), p. 102.]
If the incident light is unpolarised (containing an equal mix of "s"- and "p"-polarisations), the reflection coefficient is "R" = ("R"s + "R"p)/2.
Equations for coefficients corresponding to ratios of the electric field amplitudes of the waves can also be derived, and these are also called "Fresnel equations".
At one particular angle for a given "n"1 and "n"2, the value of "R"p goes to zero and a "p"-polarised incident ray is purely refracted. This angle is known as Brewster's angle, and is around 56° for a glass medium in air or vacuum. Note that this statement is only true when the refractive indexes of both materials are real numbers, as is the case for materials like air and glass. For materials that absorb light, like metals and semiconductors, "n" is complex, and "R"p does not generally go to zero.
When moving from a denser medium into a less dense one (i.e., "n"1 > "n"2), above an incidence angle known as the "critical angle", all light is reflected and "R"s = "R"p = 1. This phenomenon is known as total internal reflection. The critical angle is approximately 41° for glass in air.
When the light is at near-normal incidence to the interface (θi ≈ θt ≈ 0), the reflection and transmission coefficient are given by:
: $R = R_s = R_p = left\left( frac\left\{n_1 - n_2\right\}\left\{n_1 + n_2\right\} ight\right)^2$ : $T = T_s = T_p = 1-R = frac\left\{4 n_1 n_2\right\}\left\{left\left(n_1 + n_2 ight\right)^2\right\}$
For common glass, the reflection coefficient is about 4%. Note that reflection by a window is from the front side as well as the back side, and that some of the light bounces back and forth a number of times between the two sides. The combined reflection coefficient for this case is 2"R"/(1 + "R"), when interference can be neglected.
In reality, when light makes multiple reflections between two parallel surfaces, the multiple beams of light generally interfere with one another, and the surfaces act as a Fabry-Perot interferometer. This effect is responsible for the colours seen in oil films on water, and it is used in optics to make optical coatings that can greatly lower the reflectivity or can be used as an optical filter.
It should be noted that the discussion given here assumes that the permeability μ is equal to the vacuum permeability μ0 in both media. This is approximately true for most dielectric materials, but not for some other types of material. The completely general Fresnel equations are more complicated.
ee also
*Index-matching material
*Fresnel diffraction
*Fresnel integral
*Fresnel lantern
*Fresnel lens
*Fresnel rhomb, Fresnel's apparatus to produce circularly polarized light [http://physics.kenyon.edu/EarlyApparatus/Polarized_Light/Fresnels_Rhomb/Fresnels_Rhomb.html]
*Fresnel zone
*Fresnel zone plate
*Fresnel number
*Fresnel drag
*Specular reflection
References
*cite book | first=Eugene|last=Hecht|year=1987|title=Optics|edition=2nd ed.|publisher=Addison Wesley|id=ISBN 0-201-11609-X
* [http://scienceworld.wolfram.com/physics/FresnelEquations.html Fresnel Equations] – Wolfram
* [http://swiss.csail.mit.edu/~jaffer/FreeSnell/ FreeSnell] – Free software computes the optical properties of multilayer materials
* [http://thinfilm.hansteen.net/ Thinfilm] – Web interface for calculating optical properties of thin films and multilayer materials. (Reflection & transmission coefficients, ellipsometric parameters Psi & Delta)
* [http://www.calctool.org/CALC/phys/optics/reflec_refrac Simple web interface for calculating single-interface reflection and refraction angles and strengths.]
* [http://ReflectionCoefficient.INFO/ ReflectionCoefficient.INFO] – Optical reflection coefficient calculator
Wikimedia Foundation. 2010.
### Look at other dictionaries:
• Fresnel (disambiguation) — Fresnel can refer to physicist Augustin Jean Fresnel, or to the following topics associated with him:*Fresnel equations, describing light reflection and refraction *Huygens Fresnel principle, a description of wave propagation *Fresnel diffraction … Wikipedia
• Fresnel rhomb — A Fresnel rhomb is a prism like device designed in 1817 by Augustin Jean Fresnel for producing circularly polarized light. However, in contrast to a wave plate, the rhomb does not utilise birefringent properties of the material.The rhomb (usually … Wikipedia
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• Augustin-Jean Fresnel — Infobox Scientist name = Augustin Jean Fresnel box width = image width =150px caption = Augustin Jean Fresnel birth date = Birth date|1788|5|10 birth place = Broglie (Eure) death date = death date and age|1827|7|14|1788|5|10 death place =… … Wikipedia
• Maxwell's equations — For thermodynamic relations, see Maxwell relations. Electromagnetism … Wikipedia
• List of equations — This is a list of equations, by Wikipedia page. See also list of equations in classical mechanics, list of relativistic equations, equation solving, theory of equations.Eponymous equations* Arrhenius equation * Bernoulli s equation *… … Wikipedia
• Liste d'équations et formules — Ceci est une Liste des équations et formules par ordre alphabétique. Cette liste contient les équations, les formules, les relations et autres identités, égalités ou inégalités. Sommaire : Haut A B C D E F G H I J K L M N O P Q R S T U V W X … Wikipédia en Français | 2,711 | 9,436 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2020-40 | latest | en | 0.800864 |
http://codingforums.com/other-server-side-languages-issues/177228-math-visual-logic.html?pda=1 | 1,419,081,898,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802769867.110/warc/CC-MAIN-20141217075249-00048-ip-10-231-17-201.ec2.internal.warc.gz | 60,270,296 | 13,869 | ## Math & Visual Logic
Yes, this is not Visual Basic but Visual LOGIC.
OK, so I need to get this done for my Java Programming class.
I can't figure this one out for the life of me! :@
1-3: “As I was going to St. Ives…”: consider the following nursery rhyme:
As I was going to St. Ives, I met a man with seven wives. Every wife and seven sacks, every sack and seven cats, and every cat and seven kittens. Kittens, cats, sacks, and wives, how many were going to St. Ives?
The question at the end is a trick question because only the narrator is going to St. Ives. Write a program to determine the total number of things (including people, animals, and sacks) that were met by the traveler.
An explaination of how to approach this (in Visual Logic) would be awesome.
Basically I have to write a program that calculates the total.
Thanks. | 206 | 841 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2014-52 | latest | en | 0.977325 |
https://www.perlmonks.org/?node_id=86482;replies=1;displaytype=print | 1,723,071,260,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640713269.38/warc/CC-MAIN-20240807205613-20240807235613-00501.warc.gz | 697,922,647 | 2,463 | http://www.perlmonks.org?node_id=86482
eyal_kle has asked for the wisdom of the Perl Monks concerning the following question: (arrays)
How do I create a matrix? is it @@matrix?
Originally posted as a Categorized Question.
Replies are listed 'Best First'.
Re: How do I create a matrix?
by jeffa (Bishop) on Jun 07, 2001 at 12:07 UTC
Heh, close. :)
The trick is to use an array of array references:
```my @matrix = (
[qw(0 0 0 0)],
[qw(0 0 1 0)],
[qw(0 1 0 0)],
[qw(1 0 0 0)],
);
foreach my \$row (@matrix) {
print join(":",@\$row), "\n";
}
# qw(1 0 0 0) is better way of saying (1,0,0,0)
[download]```
Re: How do I create a matrix?
by Beatnik (Parson) on Jun 07, 2001 at 12:33 UTC
You can use PDL::Matrix
```#!/usr/bin/perl -w
use strict;
use PDL::Matrix;
my \$m = pdl [
[1,2,3],
[4,5,6],
[7,8,9]
];
[download]```
Which then also allows you to perform clean, fast math on them.
```my \$m = pdl [
[1,2,3],
[4,5,6],
[7,8,9]
];
my \$n = pdl [
[1,0,0],
[0,1,0],
[0,0,1]
];
my \$r = \$m x \$n;
[download]```
The above appears not to use PDL::Matrix-specific features (which are full of gotchas); better practice would be just to use PDL, and all the rest of the code above will then work as given. Try perldl after installing PDL, the online help is really good. For better matrix support, try also PDL::LinearAlgebra. | 469 | 1,321 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-33 | latest | en | 0.861886 |
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#### Resources tagged with Pythagoras' theorem similar to Eyelids:
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### There are 70 results
Broad Topics > 2D Geometry, Shape and Space > Pythagoras' theorem
### Two Circles
##### Stage: 4 Challenge Level:
Draw two circles, each of radius 1 unit, so that each circle goes through the centre of the other one. What is the area of the overlap?
### Ball Packing
##### Stage: 4 Challenge Level:
If a ball is rolled into the corner of a room how far is its centre from the corner?
### Inscribed in a Circle
##### Stage: 4 Challenge Level:
The area of a square inscribed in a circle with a unit radius is, satisfyingly, 2. What is the area of a regular hexagon inscribed in a circle with a unit radius?
### Semi-detached
##### Stage: 4 Challenge Level:
A square of area 40 square cms is inscribed in a semicircle. Find the area of the square that could be inscribed in a circle of the same radius.
### Six Discs
##### Stage: 4 Challenge Level:
Six circular discs are packed in different-shaped boxes so that the discs touch their neighbours and the sides of the box. Can you put the boxes in order according to the areas of their bases?
### Equilateral Areas
##### Stage: 4 Challenge Level:
ABC and DEF are equilateral triangles of side 3 and 4 respectively. Construct an equilateral triangle whose area is the sum of the area of ABC and DEF.
### Squ-areas
##### Stage: 4 Challenge Level:
Three squares are drawn on the sides of a triangle ABC. Their areas are respectively 18 000, 20 000 and 26 000 square centimetres. If the outer vertices of the squares are joined, three more. . . .
### All Tied Up
##### Stage: 4 Challenge Level:
A ribbon runs around a box so that it makes a complete loop with two parallel pieces of ribbon on the top. How long will the ribbon be?
### Floored
##### Stage: 3 Challenge Level:
A floor is covered by a tessellation of equilateral triangles, each having three equal arcs inside it. What proportion of the area of the tessellation is shaded?
### Round and Round
##### Stage: 4 Challenge Level:
Prove that the shaded area of the semicircle is equal to the area of the inner circle.
### Some(?) of the Parts
##### Stage: 4 Challenge Level:
A circle touches the lines OA, OB and AB where OA and OB are perpendicular. Show that the diameter of the circle is equal to the perimeter of the triangle
### Matter of Scale
##### Stage: 4 Challenge Level:
Prove Pythagoras' Theorem using enlargements and scale factors.
### Napkin
##### Stage: 4 Challenge Level:
A napkin is folded so that a corner coincides with the midpoint of an opposite edge . Investigate the three triangles formed .
### Semi-square
##### Stage: 4 Challenge Level:
What is the ratio of the area of a square inscribed in a semicircle to the area of the square inscribed in the entire circle?
### Compare Areas
##### Stage: 4 Challenge Level:
Which has the greatest area, a circle or a square inscribed in an isosceles, right angle triangle?
### Nicely Similar
##### Stage: 4 Challenge Level:
If the hypotenuse (base) length is 100cm and if an extra line splits the base into 36cm and 64cm parts, what were the side lengths for the original right-angled triangle?
### Holly
##### Stage: 4 Challenge Level:
The ten arcs forming the edges of the "holly leaf" are all arcs of circles of radius 1 cm. Find the length of the perimeter of the holly leaf and the area of its surface.
### Square Pegs
##### Stage: 3 Challenge Level:
Which is a better fit, a square peg in a round hole or a round peg in a square hole?
### Tennis
##### Stage: 3 Challenge Level:
A tennis ball is served from directly above the baseline (assume the ball travels in a straight line). What is the minimum height that the ball can be hit at to ensure it lands in the service area?
### Slippage
##### Stage: 4 Challenge Level:
A ladder 3m long rests against a wall with one end a short distance from its base. Between the wall and the base of a ladder is a garden storage box 1m tall and 1m high. What is the maximum distance. . . .
##### Stage: 4 Challenge Level:
A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground?
### Where to Land
##### Stage: 4 Challenge Level:
Chris is enjoying a swim but needs to get back for lunch. If she can swim at 3 m/s and run at 7m/sec, how far along the bank should she land in order to get back as quickly as possible?
### Under the Ribbon
##### Stage: 4 Challenge Level:
A ribbon is nailed down with a small amount of slack. What is the largest cube that can pass under the ribbon ?
##### Stage: 4 Challenge Level:
The sides of a triangle are 25, 39 and 40 units of length. Find the diameter of the circumscribed circle.
### Grid Lockout
##### Stage: 4 Challenge Level:
What remainders do you get when square numbers are divided by 4?
### Pythagoras
##### Stage: 2 and 3
Pythagoras of Samos was a Greek philosopher who lived from about 580 BC to about 500 BC. Find out about the important developments he made in mathematics, astronomy, and the theory of music.
### Cubic Rotations
##### Stage: 4 Challenge Level:
There are thirteen axes of rotational symmetry of a unit cube. Describe them all. What is the average length of the parts of the axes of symmetry which lie inside the cube?
### Hex
##### Stage: 3 Challenge Level:
Explain how the thirteen pieces making up the regular hexagon shown in the diagram can be re-assembled to form three smaller regular hexagons congruent to each other.
### Rhombus in Rectangle
##### Stage: 4 Challenge Level:
Take any rectangle ABCD such that AB > BC. The point P is on AB and Q is on CD. Show that there is exactly one position of P and Q such that APCQ is a rhombus.
### Get Cross
##### Stage: 4 Challenge Level:
A white cross is placed symmetrically in a red disc with the central square of side length sqrt 2 and the arms of the cross of length 1 unit. What is the area of the disc still showing?
### Tilted Squares
##### Stage: 3 Challenge Level:
It's easy to work out the areas of most squares that we meet, but what if they were tilted?
### Liethagoras' Theorem
##### Stage: 2 and 3
Liethagoras, Pythagoras' cousin (!), was jealous of Pythagoras and came up with his own theorem. Read this article to find out why other mathematicians laughed at him.
### Fitting In
##### Stage: 4 Challenge Level:
The largest square which fits into a circle is ABCD and EFGH is a square with G and H on the line CD and E and F on the circumference of the circle. Show that AB = 5EF. Similarly the largest. . . .
### Pythagorean Triples I
##### Stage: 3 and 4
The first of two articles on Pythagorean Triples which asks how many right angled triangles can you find with the lengths of each side exactly a whole number measurement. Try it!
### Trice
##### Stage: 3 Challenge Level:
ABCDEFGH is a 3 by 3 by 3 cube. Point P is 1/3 along AB (that is AP : PB = 1 : 2), point Q is 1/3 along GH and point R is 1/3 along ED. What is the area of the triangle PQR?
### A Chordingly
##### Stage: 3 Challenge Level:
Find the area of the annulus in terms of the length of the chord which is tangent to the inner circle.
### The Pillar of Chios
##### Stage: 3 Challenge Level:
Semicircles are drawn on the sides of a rectangle ABCD. A circle passing through points ABCD carves out four crescent-shaped regions. Prove that the sum of the areas of the four crescents is equal to. . . .
### Isosceles
##### Stage: 3 Challenge Level:
Prove that a triangle with sides of length 5, 5 and 6 has the same area as a triangle with sides of length 5, 5 and 8. Find other pairs of non-congruent isosceles triangles which have equal areas.
### Rectangular Pyramids
##### Stage: 4 and 5 Challenge Level:
Is the sum of the squares of two opposite sloping edges of a rectangular based pyramid equal to the sum of the squares of the other two sloping edges?
### Medallions
##### Stage: 4 Challenge Level:
Three circular medallions fit in a rectangular box. Can you find the radius of the largest one?
### Pythagorean Triples
##### Stage: 3 Challenge Level:
How many right-angled triangles are there with sides that are all integers less than 100 units?
### Pythagoras Proofs
##### Stage: 4 Challenge Level:
Can you make sense of these three proofs of Pythagoras' Theorem?
### Picturing Pythagorean Triples
##### Stage: 4 and 5
This article discusses how every Pythagorean triple (a, b, c) can be illustrated by a square and an L shape within another square. You are invited to find some triples for yourself.
### Partly Circles
##### Stage: 4 Challenge Level:
What is the same and what is different about these circle questions? What connections can you make?
### Tilting Triangles
##### Stage: 4 Challenge Level:
A right-angled isosceles triangle is rotated about the centre point of a square. What can you say about the area of the part of the square covered by the triangle as it rotates?
### Take a Square
##### Stage: 4 Challenge Level:
Cut off three right angled isosceles triangles to produce a pentagon. With two lines, cut the pentagon into three parts which can be rearranged into another square.
### Pythagorean Triples II
##### Stage: 3 and 4
This is the second article on right-angled triangles whose edge lengths are whole numbers.
### Where Is the Dot?
##### Stage: 4 Challenge Level:
A dot starts at the point (1,0) and turns anticlockwise. Can you estimate the height of the dot after it has turned through 45 degrees? Can you calculate its height?
### Zig Zag
##### Stage: 4 Challenge Level:
Four identical right angled triangles are drawn on the sides of a square. Two face out, two face in. Why do the four vertices marked with dots lie on one line? | 2,402 | 9,917 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2017-47 | latest | en | 0.884252 |
https://community.fabric.microsoft.com/t5/Desktop/Variable-Top-N-depending-on-Column-Conditions/m-p/2591029 | 1,716,739,837,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058956.26/warc/CC-MAIN-20240526135546-20240526165546-00699.warc.gz | 148,912,050 | 120,289 | cancel
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Frequent Visitor
## Variable Top N depending on Column Conditions
Hi Everyone,
I want to create a DAX which gives following result :
So based on the "total" I want to find out "Top 3 brand" for "Category A and B" and "Top 1 Brand" for "Category C".
I am able to show top 3 Brands across all categories but facing difficulty with variable Top selection based on Category.
2 ACCEPTED SOLUTIONS
Frequent Visitor
Thanks Hariharan_R. This measure is absolutely correct. But i have one more doubt for the same.
How to show multiple Categories in this DAX where i need this condition of variable Top N . For eg,
I have Category D and E as well along with C where this filteration of top N is different as compared to Category A and B.
Thanks.
Solution Sage
Hi,
If you have more values then try use disconnected table with the TOPN values otherwise try the below one.
``````Top 3 =
VAR _N =
SWITCH(TRUE(), MIN('Table'[Category])="C",2,MIN('Table'[Category])="D",1,MIN('Table'[Category])="E",1,3)
VAR Top3 =
CALCULATETABLE (
GENERATE (
VALUES ('Table'[Category] ),
TOPN (
_N,
CALCULATETABLE ( VALUES ('Table'[Brand] ) ),
[Sales]
)
),
ALLSELECTED()
)
RETURN
CALCULATE (
1 * ( NOT ISEMPTY ( 'Table' ) ),
KEEPFILTERS ( Top3 )
)``````
Thanks
Hari
If I helped you, click on the Thumbs Up to give Kudos.
6 REPLIES 6
Community Champion
Thanks to the great efforts by MS engineers to simplify syntax of DAX! Most beginners are SUCCESSFULLY MISLED to think that they could easily master DAX; but it turns out that the intricacy of the most frequently used RANKX() is still way beyond their comprehension! DAX is simple, but NOT EASY!
Solution Sage
Hi,
You can use below measure.
``````Top 3 =
VAR _N = IF(MIN('Table'[Category])="C",2,3)
VAR Top3 =
CALCULATETABLE (
GENERATE (
VALUES ('Table'[Category] ),
TOPN (
_N,
CALCULATETABLE ( VALUES ('Table'[Brand] ) ),
[Sales]
)
),
ALLSELECTED()
)
RETURN
CALCULATE (
1 * ( NOT ISEMPTY ( 'Table' ) ),
KEEPFILTERS ( Top3 )
)``````
Sample Data
Thanks
Hari
If I helped you, click on the Thumbs Up to give Kudos.
Frequent Visitor
Thanks Hariharan_R. This measure is absolutely correct. But i have one more doubt for the same.
How to show multiple Categories in this DAX where i need this condition of variable Top N . For eg,
I have Category D and E as well along with C where this filteration of top N is different as compared to Category A and B.
Thanks.
Solution Sage
Hi,
If you have more values then try use disconnected table with the TOPN values otherwise try the below one.
``````Top 3 =
VAR _N =
SWITCH(TRUE(), MIN('Table'[Category])="C",2,MIN('Table'[Category])="D",1,MIN('Table'[Category])="E",1,3)
VAR Top3 =
CALCULATETABLE (
GENERATE (
VALUES ('Table'[Category] ),
TOPN (
_N,
CALCULATETABLE ( VALUES ('Table'[Brand] ) ),
[Sales]
)
),
ALLSELECTED()
)
RETURN
CALCULATE (
1 * ( NOT ISEMPTY ( 'Table' ) ),
KEEPFILTERS ( Top3 )
)``````
Thanks
Hari
If I helped you, click on the Thumbs Up to give Kudos.
Anonymous
Not applicable
your dax function is working fine but i have the categories 6(a,b,c,d,e,f) for each category top 5 brands i want baced on total sales
Solution Sage
Hi @Anonymous
You can remove _N with 5 like below.
``````CALCULATETABLE (
GENERATE (
VALUES ('Table'[Category] ),
TOPN (
5,
CALCULATETABLE ( VALUES ('Table'[Brand] ) ),
[Sales]
)
),
ALLSELECTED()
)
RETURN
CALCULATE (
1 * ( NOT ISEMPTY ( 'Table' ) ),
KEEPFILTERS ( Top3 )
)``````
Thanks
Hari
If I helped you, click on the Thumbs Up to give Kudos.
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# Search results
Found 1240 matches
Density
The density of a material is defined as its mass per unit volume. For a pure substance the density has the same numerical value as its mass concentration. ... more
Runoff equation ( P >Ia)
Surface runoff is the water flow that occurs when the soil is infiltrated to full capacity and excess water from rain. The runoff is depended on the ... more
Logarithmic Mean Size - 1st moment
Calculates the logarithmic mean size (moments method) of the particles’ size distribution of a soil, in phi scale
... more
Nq bearing capacity factor
Karl von Terzaghi was the first to present a comprehensive theory for the evaluation of the ultimate bearing capacity of rough shallow foundations. This ... more
Buoyant force (Archimedes' principle)
Buoyancy is an upward force exerted by a fluid that opposes the weight of an immersed object. Buoyant force equivalent to the weight of the fluid that ... more
Runoff curve number
Surface runoff is the water flow that occurs when the soil is infiltrated to full capacity and excess water from rain. The runoff curve number (also called ... more
Arithmetic Standard Deviation - 2nd moment
Shows how much variation or dispersion from the average exists, on the particles’ size distribution of a soil, in metric scale. Arithmetic mean size (1st ... more
Nγ bearing capacity factor (Terzaghi's theory)
Karl von Terzaghi was the first to present a comprehensive theory for the evaluation of the ultimate bearing capacity of rough shallow foundations. This ... more
Nc bearing capacity factor
Karl von Terzaghi was the first to present a comprehensive theory for the evaluation of the ultimate bearing capacity of rough shallow foundations. This ... more
Evaporation - Penman Equation (Shuttleworth modification)
The Penman equation describes evaporation (E) from an open water surface, and was developed by Howard Penman in 1948. Penman’s equation requires ... more
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Category | 447 | 2,073 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-49 | latest | en | 0.913219 |
https://www.numbersaplenty.com/2011 | 1,656,243,706,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103205617.12/warc/CC-MAIN-20220626101442-20220626131442-00722.warc.gz | 1,000,618,695 | 3,167 | Search a number
2011 is a prime number
BaseRepresentation
bin11111011011
32202111
4133123
531021
613151
75602
oct3733
92674
102011
111569
1211b7
13bb9
14a39
158e1
hex7db
2011 has 2 divisors, whose sum is σ = 2012. Its totient is φ = 2010.
The previous prime is 2003. The next prime is 2017. The reversal of 2011 is 1102.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 2011 - 23 = 2003 is a prime.
It is a plaindrome in base 11.
It is a nialpdrome in base 13.
It is not a weakly prime, because it can be changed into another prime (2017) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1005 + 1006.
It is an arithmetic number, because the mean of its divisors is an integer number (1006).
2011 is a deficient number, since it is larger than the sum of its proper divisors (1).
2011 is an equidigital number, since it uses as much as digits as its factorization.
2011 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 2, while the sum is 4.
The square root of 2011 is about 44.8441746496. The cubic root of 2011 is about 12.6222668331.
Adding to 2011 its reverse (1102), we get a palindrome (3113).
Subtracting from 2011 its reverse (1102), we obtain a palindrome (909).
Multiplying 2011 by its reverse (1102), we get a palindrome (2216122).
It can be divided in two parts, 201 and 1, that added together give a palindrome (202).
The spelling of 2011 in words is "two thousand, eleven", and thus it is an iban number. | 483 | 1,590 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2022-27 | longest | en | 0.908862 |
https://www.accountingexplanation.com/overhead_idle_capacity_variance.htm | 1,723,047,609,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640694594.35/warc/CC-MAIN-20240807143134-20240807173134-00174.warc.gz | 494,998,890 | 6,103 | Home » Standard Costing and Variance Analysis » Overhead Idle Capacity Variance
## Definition and Explanation:
Overhead Idle capacity variance is the difference between the budget allowance based on actual hours worked and actual hours worked multiplied by the standard overhead rate.
Overhead idle capacity variance is calculated when overall or net overhead variance is further analyzed using three variance method. Other two variances that are calculated in three variance method are overhead spending variance and overhead efficiency variance.
## Formula:
Following formula is used for the calculation of this variance:
Idle capacity variance = Budgeted allowance based on actual hours worked - (Actual hours worked × Standard overhead rate)
## Example:
From the following data calculate factory overhead idle capacity variance:
Actual overhead \$7,384 Actual hours worked 3,475 Units produced during the period 850 Standard hours for one unit 4 Standard factory overhead rate: Variable \$1.20 Fixed \$0.80 \$2.00 Normal Capacity in labor hours 4000 hours
### Solution:
Budgeted allowance based on actual hours worked: Fixed expenses budgeted \$3,200 Variable expenses (3,475 actual hours worked × \$1.20 variable overhead rate) 4,170 \$7,370 Actual hours worked at standard rate (3,475 actual hours × \$2.00 standard rate) \$6,950 Idle capacity variance (Unfavorable) \$420 unfav
This variance consists of fixed expense only and can also be computed as follows:
(4,000 normal capacity - 3,475 actual hours) × \$0.80 fixed expenses rate = \$420
This variance indicates the amount of overhead that is either under or over absorbed because actual hours are either less or more than the hours on which the overhead rate was based. In the above example the plant is operated at 86.875% (3,475/4,000) of normal capacity based on actual hours.
## Who is Responsible For Idle Capacity Variance?
This variance is the responsibility of executive management .
## Relevant Articles:
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A D V E R T I S E M E N T | 598 | 2,958 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-33 | latest | en | 0.890326 |
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# Solution - Draw a graph of the line x – 2y = 3. From the graph, find the coordinates of the point when (i) x = – 5 (ii) y = 0. - CBSE Class 10 - Mathematics
ConceptPair of Linear Equations in Two Variables
#### Question
Draw a graph of the line x – 2y = 3. From the graph, find the coordinates of the point when (i) x = – 5 (ii) y = 0.
#### Solution
You need to to view the solution
Is there an error in this question or solution?
#### APPEARS IN
NCERT Mathematics Textbook for Class 10
Chapter 3: Pair of Linear Equations in Two Variables
Q: 0 | Page no. 0
#### Reference Material
Solution for question: Draw a graph of the line x – 2y = 3. From the graph, find the coordinates of the point when (i) x = – 5 (ii) y = 0. concept: Pair of Linear Equations in Two Variables. For the course CBSE
S | 257 | 865 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2018-13 | latest | en | 0.810634 |
theeconomistspoage.net | 1,701,971,420,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100677.45/warc/CC-MAIN-20231207153748-20231207183748-00589.warc.gz | 629,403,096 | 33,722 | Categories :
## Introduction
Resolving delinquent tax debt is both an art and a science, requiring a strategic approach grounded in knowledge and precision. This guide explores the scientific principles behind delinquent tax debt collection relief, providing a structured framework to navigate this complex terrain.
## Understanding the Formulas
### 1. Assessing the Variables
Begin by identifying the variables in your delinquent Tax Debt Attorney in Los Angeles equation. This includes the amount owed, deadlines, and potential legal ramifications. Understanding these variables is crucial for formulating an effective solution.
## Applying Scientific Strategies
### 2. Installment Agreements: The Formula for Gradual Repayment
Apply the installment agreement formula to break down your tax debt into manageable monthly payments. This structured approach prevents immediate financial strain, providing a formula for gradual repayment.
### 3. Offer in Compromise: Calculating the Settling Point
Explore the mathematical precision of an offer in compromise, where you calculate the settling point for your tax debt. This involves negotiating to settle for less than the total amount owed, offering a formula for a fresh financial start.
### 4. Innocent Spouse Relief: Fair Allocation Formula
In cases of joint filing, use the innocent spouse relief formula to ensure a fair allocation of financial responsibility. This legal formula protects individuals not responsible for the tax debt.
## Leveraging Expert Formulas
### 5. Tax Professionals: Equations for Expertise
Engage tax professionals or attorneys as they bring expert equations to the table. Their formulas encompass legal knowledge, negotiation skills, and strategic insights to navigate the complexities of tax debt relief.
## Legal Equations for Resolution
### 6. Appeals Process: The Equation of Challenge
Utilize the appeals process as a formula for challenging assessments. Present evidence and arguments to challenge unfair calculations and reach a fair resolution.
### 7. Compliance and Cooperation: The Trust-building Equation
Demonstrate voluntary compliance and cooperation as a formula for building trust. This equation positively influences negotiations and sets the stage for resolution.
## The Timing Equation
### 8. Swift Resolution: Calculating the Time Factor
Factor in the importance of time in the resolution equation. Swift action is a critical component, influencing the outcome and the effectiveness of relief formulas.
## Conclusion: Achieving Precision in Resolution
The science of delinquent tax debt collection relief involves applying precise formulas and strategic equations. By understanding the variables, applying relief strategies, leveraging expert formulas, navigating legal equations, and factoring in the element of time, taxpayers can achieve precision in resolving delinquent tax debt. Remember, the path to resolution is a calculated journey where each step contributes to the precise outcome of financial relief. | 546 | 3,037 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-50 | longest | en | 0.857459 |
http://www.instructables.com/id/Egor-Stick/ | 1,503,075,725,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104704.64/warc/CC-MAIN-20170818160227-20170818180227-00232.warc.gz | 578,869,327 | 12,830 | ## Introduction: Egor Stick
This is a contraption to finally help mark the location of hidden holes
## Step 1: The Problem
I have this wheelchair and I need to place a flat board on top of the brackets to make a cover or platform.
But how does one know where to place holes in the board to line up with the holes in the brackets?
## Step 2: Reasonable Solution?
You would think that you can just measure the location of the holes . . .
## Step 3: Solution Fail
... But my measuring device seems to have numbers that move when I try to transfer values.
Here you can see that 7.5 inches somehow goes way off of the edge of the board. How's that possible? Geeze. I hate measuring devices.
## Step 4: Actual Solution
So here is my invention to help with this problem: the Egor Stick.
It allows you to mark the board from underneath at the actual location of the holes.
Here's how to make it and use it . . .
## Step 5: Tools, Supplies
1. #2 Pencil, sharpened very fine
2. Drill
3. 1/4 '' bit
4. Lil' tiny saw or Xacto-type knife
5. Paint stirrer stick or similarly narrow board-let (like a shim)
## Step 6: Cut Off Tip of Pencil
Cut off about an inch from the top of the pencil.
NOTE: sharpen the pencil first before sawing.
## Step 7: Pencil Tip
You'll wind up with a pencil stub.
## Step 8: Drill a Hole in the Stick
Drill a hole about an inch from the edge of the stick.
## Step 10: Place Your Board on Your Holes (er Well You Know What I Mean Heh)
Position the board where you need it.
You might want to clamp it down to keep it in place while the hole positions are marked.
## Step 11: Mark Holes With the Stick From Underneath
Slip the stick under the brackets and push the pencil through the holes. Apply slight pressure and move the stick around the bolt hole.
## Step 12: Flip Over the Board
Light registration marks are now on your board.
Figure 1 shows the hole marks, and Figure 2 shows them a little better after circling with a marker.
## Step 13: Drill Holes Based on Registration Marks
Hint: don't use your knee as a backing when doing this.
## Step 14: Place Board Back on and Slip Bolts Through Holes
Here you can see how the bolts go through the board and (surprisingly) line up with the holes in the bracket.
## Step 15: Secure Bolts As Needed
This figure shows the board on the brackets and bolts through the holes.
## Step 17: Tool Complete
Place your new tool among all the other tools where you will never be able to find it again
Toga_Dan (author)2017-07-22
I recently came up with a similar solution for similar problem. but used thread + glue rather than a hole in the stick to secure pencil stub | 656 | 2,659 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2017-34 | longest | en | 0.894866 |
https://numbermatics.com/n/3942/ | 1,639,036,231,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363689.56/warc/CC-MAIN-20211209061259-20211209091259-00384.warc.gz | 490,104,635 | 6,193 | # 3942
## 3,942 is an even composite number composed of three prime numbers multiplied together.
What does the number 3942 look like?
This visualization shows the relationship between its 3 prime factors (large circles) and 16 divisors.
3942 is an even composite number. It is composed of three distinct prime numbers multiplied together. It has a total of sixteen divisors.
## Prime factorization of 3942:
### 2 × 33 × 73
(2 × 3 × 3 × 3 × 73)
See below for interesting mathematical facts about the number 3942 from the Numbermatics database.
### Names of 3942
• Cardinal: 3942 can be written as Three thousand, nine hundred forty-two.
### Scientific notation
• Scientific notation: 3.942 × 103
### Factors of 3942
• Number of distinct prime factors ω(n): 3
• Total number of prime factors Ω(n): 5
• Sum of prime factors: 78
### Divisors of 3942
• Number of divisors d(n): 16
• Complete list of divisors:
• Sum of all divisors σ(n): 8880
• Sum of proper divisors (its aliquot sum) s(n): 4938
• 3942 is an abundant number, because the sum of its proper divisors (4938) is greater than itself. Its abundance is 996
### Bases of 3942
• Binary: 1111011001102
• Hexadecimal: 0xF66
• Base-36: 31I
### Squares and roots of 3942
• 3942 squared (39422) is 15539364
• 3942 cubed (39423) is 61256172888
• The square root of 3942 is 62.7853486095
• The cube root of 3942 is 15.7969122845
### Scales and comparisons
How big is 3942?
• 3,942 seconds is equal to 1 hour, 5 minutes, 42 seconds.
• To count from 1 to 3,942 would take you about five minutes.
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 3942 cubic inches would be around 1.3 feet tall.
### Recreational maths with 3942
• 3942 backwards is 2493
• 3942 is a Harshad number.
• The number of decimal digits it has is: 4
• The sum of 3942's digits is 18
• More coming soon!
## Link to this page
HTML: To link to this page, just copy and paste the link below into your blog, web page or email.
BBCODE: To link to this page in a forum post or comment box, just copy and paste the link code below:
## Cite this page
MLA style:
"Number 3942 - Facts about the integer". Numbermatics.com. 2021. Web. 9 December 2021.
APA style:
Numbermatics. (2021). Number 3942 - Facts about the integer. Retrieved 9 December 2021, from https://numbermatics.com/n/3942/
Chicago style:
Numbermatics. 2021. "Number 3942 - Facts about the integer". https://numbermatics.com/n/3942/
The information we have on file for 3942 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 3942, math, Factors of 3942, curriculum, school, college, exams, university, Prime factorization of 3942, STEM, science, technology, engineering, physics, economics, calculator, three thousand, nine hundred forty-two.
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Some bits of this website may not work unless you switch it on. | 950 | 3,513 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2021-49 | latest | en | 0.877205 |
https://experienceleaguecommunities.adobe.com/t5/adobe-analytics-questions/average-visits-per-hour-of-day/td-p/404190 | 1,723,775,648,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641319057.20/warc/CC-MAIN-20240815235528-20240816025528-00075.warc.gz | 196,214,005 | 48,255 | Expand my Community achievements bar.
SOLVED
## Average visits per hour of day
Level 1
Hi, I'm trying to calculate the average number of visits per hour of day over a given time period.
When I create an Average Visits metric like this:
and then use that to breakdown by the Hour of Day dimension, I would expect it to give me the average per hour of day, however, it gives me only one value: the average of the entire selected time period. Like this:
How do i get the average number of visits that I can break down by hour of day?
1 Accepted Solution
Dear mdep,
Are you looking for the average hourly visits in a day i.e. Visits in a day / 24? Or average hourly visits for the time period selected i.e. Visits for the time period / No. of hours for the time period? If yes, create a calculated metric based on the below.
Once created, you should see the below numbers.
But, not sure about the metric average number of visits per hour of the day, can you explain more?
Thank You, Pratheep Arun Raj B | Xerago | Terryn Winter Analytics
9 Replies
---
Dear mdep,
Are you looking for the average hourly visits in a day i.e. Visits in a day / 24? Or average hourly visits for the time period selected i.e. Visits for the time period / No. of hours for the time period? If yes, create a calculated metric based on the below.
Once created, you should see the below numbers.
But, not sure about the metric average number of visits per hour of the day, can you explain more?
Thank You, Pratheep Arun Raj B | Xerago | Terryn Winter Analytics
Level 1
Hi Pratheep,
Here's what I'm after: I'm trying to find out the most popular hour to visit the site on average. Instead of just summing visits per hour I want to take the average.
Like this:
Date 1AM 2AM 3AM 4AM 5AM 6AM 7AM 4/1/2021 2 1 2 4 6 10 20 4/2/2021 4 0 0 5 4 11 22 4/3/2021 1 1 3 5 6 12 19 Average 2.3 0.7 1.7 4.7 5.3 11.0 20.3
This is just and example, I'd need this for all hours of day (12AM to 12PM) for a certain date range.
So from this I can see that between 1 to 7AM we have on average most number of visits at 7AM.
Any chance you know how to do this in Adobe Analytics?
Manon
Dear Mdep, Still the same metric. But instead of the dimension 'Day', use the dimension 'Hour Of Day'. Thank You, Pratheep Arun Raj B
Level 1
Hi, I'm a little unclear and hoping I can revivie this thread.
I'm attempting to use Hour of Day and Visits to understand on average, how many visits per hour (3am, 4am, etc..) within the designated timeframe.
It's giving me the avg visits per hour for the whole day but not broken down by each hour. The calculated metric just matches the visits for that hour.
Level 1
Hi PratheepArunRaj, that worked! nice for average number of visits per weekday too!
Level 2
Hoping this should solve the problem without creating any additional metrics. Where the Bars are showing the maximum visits in a particular hour from Jan-Feb in this case. Make sure to update the Date range or feel free to use the Rolling date range.
Level 1
Hi @PratheepArunRaj
I have created a calulated metric for Hour of Day But the numbers does not matches to the exact avg . Is this expected?
For example: for 1AM total visits for 40 days is 14717 (avg: 367 ) but the calculated metrics shows 364
Thanks | 893 | 3,285 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-33 | latest | en | 0.867226 |
http://vtmathcoalition.org/talent-search/feb-1997/ | 1,511,341,944,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806543.24/warc/CC-MAIN-20171122084446-20171122104446-00303.warc.gz | 307,891,158 | 2,396 | # Talent Search Mathematics Problems
VERMONT STATE MATHEMATICS COALITION TALENT SEARCH
February 17, 1997
Student Name ________________________________________________________
School ______________________________________________________________
Directions: Solve as many as you can of the problems and list your solutions on this sheet of paper. On separate sheets, in an organized way, show how you solved the problems. You will be awarded full credit for a complete correct answer which is adequately supported by mathematical reasoning. You can receive half credit for correct answers which are the result of guesses, conjectures or incomplete solutions. The decisions of the graders are final. You may earn bonus points for "commendable solutions"- solutions that display creativity, ingenuity and clarity. Your answers and solutions must be postmarked by March 17, 1997 and submitted to:
Tony Trono
Vermont Math Coalition
19 Case Parkway
Burlington, VT 05401.
1. For any positive integer n, show that the positive integer can be expressed in the form where a and b are positive integers.
Your solution should be on the first attached sheet.
2. Tim and Nancy have children Alex and Morgan with Morgan 4 years older than Alex. Tim is 2 years older than Nancy and the sum of ages of the 4 family members is now 96 years. Seven years ago the sum of the ages of all of the family members was 69 years. What is Nancy's present age?
3. Complete the cross-number problem. There are no zero's in the solution.
ACROSS 1. Sum of digits is 13. 3. The digits (in their order)form an arithmetic progression. 5. The digits are all even and their sum is 12. 6. A square. DOWN 1. The sum of the digits is the same as the sum of the digits of 4 down. 2. The sum of the first two digits is the same as the last two digits of 1 down. 3. A cube. 4. The cube of an odd number.
4. The numbers tan(x), cos(x), sec(x) are the first three terms of an arithmetic progression. Find n if the nth term of the progression is cot(x).
5. An equilateral triangle is inscribed in a circle. A circle is inscribed in the triangle as shown in the diagram. Find the ratio a:b, where a and b are lengths of the segments of the chord drawn through the points of tangency. Answer: _____________________________
6. Hugh Colid was good in math, but was a poor speller, sometimes even misspelling his own name. His wife was always teasing him. She asked, "Can't you spell? It's age 'TWO', not 'TO'. All you ever think about is math. You're a square". "Oh yeah", replied Hugh after a moments thought, "if AGE, TO, NOT and TWO are all perfect squares, then do you know what you are? You're a TWO + TO +TOO." His wife now had motivation to figure out what she was being called. What was it that she figured out that he was calling her? | 657 | 2,805 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-47 | latest | en | 0.926944 |
http://www.heidimeijer.nl/api650-coil/portugal_vertical_cy_1057.html | 1,606,909,836,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141708017.73/warc/CC-MAIN-20201202113815-20201202143815-00323.warc.gz | 112,593,591 | 13,165 | • # portugal vertical cylindrical tank oil volume
• Xicheng Science & Technology Building High-tech Development Zone, Zhengzhou, China
• 0086-371-86011881
• [email protected]
• >Online Chating
### Tank Volume Calculator - Oil Tanks
Mar 26, 2015 · The tank size calculator on this page is designed for measuring the capacity of a variety of fuel tanks. Alternatively, you can use this tank volume calculator as a water volume calculator if you need to calculate some specific water volume. The functionality of this Tank Volume CalculatorTotal volume of a cylinder shaped tank is the area, A, of the circular end times the height, h. A = r 2 where r is the radius which is equal to d/2. Therefore: V(tank) = r 2 h The filled volume of a vertical cylinder tank is just a shorter cylinder with the same radius, r, TANK VOLUME CALCULATOR [How to Calculate Tank Cylindrical Oil Tank. Lets say that I have a cylindrical oil tank which measures 7 yards in length and has a round face 5 feet in diameter (the distance across the circular end passing through the central point). I want to calculate the tank volume in cubic feet and work out how much oil
### HOW TO CALCULATE THE VOLUMES OF PARTIALLY FULL
cylindrical tanks, either in horizontal or vertical configuration. Consider, for example, a cylindrical tank with length L and radius R, filling up to a height H. If you want to obtain the volume of the liquid that partially fills the tank, you should indicate if the tank is in horizontal or vertical position.Horizontal Tank Volume Calculations - HagraTools are provided for cylindrical horizontal tanks, and for oval (elliptical) tanks. Horizontal Cylindrical Tank Volume Calculator. Horizontal Oval Tank Volume Calculator. Disclaimer. The calculations on these pages are a purely theoretical exercise! Therefore the outcomes of the calculations on these pages can only be used for indicative , portugal vertical cylindrical tank oil volumeTank Volume Calculator - SA OilCapacity based upon a tank with flat ends (no allowance has been made for dish ends) Calculate. Calculation Results (Approximate) , portugal vertical cylindrical tank oil volume Gallons (UK) 0. 0. Gallons (US) 0. 0. BBL. 0.0. 0.0. Cu.ft. 0. 0. Enter vertical cylindrical tank dimensions: Diameter Height Measurement. Optional: Enter liquid height to work out approximate tank contents , portugal vertical cylindrical tank oil volume
### Tank Volume Calculator - Vertical Cylindrical Tanks - Metric
Vertical cylindrical tank volume calculator diagram: Fill Rate Fill Times @ Litres / Minute. Total Tank Fill Time Current Time to Fill Current Time to Empty If you're cutting blocks, concrete, stone or ANYTHING and there's DUST - DON'T TAKE THE RISK Don't cut it, or cut it wet , portugal vertical cylindrical tank oil volumeTank Volume Calculator - ibec language institute*Program does not calculate liquid volume into upper head on vertical tanks (mixing with liquid in head space is not good practice). Fusion Fluid Equipment provides this tool for reference only. While Fusion Fluid Equipment strives to provide accurate information, no expressed or implied warranty is given by Fusion Fluid Equipment as to the , portugal vertical cylindrical tank oil volumeVERTICAL CYLINDER CALCULATOR - 1728If you want to do calculations for a horizontal cylinder, then go to this link: Horizontal Cylinder Calculator. Example: Inputting tank height = 12, liquid level = 3 and tank diameter = 6, then clicking "Inches" will display the total tank volume in cubic inches and US Gallons and will also show the volume
### volume of oil left in a tank.? | Yahoo Answers
Dec 16, 2008 · volume of oil left in a tank.? A vertical cylindrical oil tank is 2.5 feet across on the inside. The depth of the oil in the tank is 2 feet. If 1 cubic foot of space holds 7.48 gallons, approximately how many gallons of oil are left in the tank? Answer Save. 1 Answer.TANK VOLUME CALCULATOR [How to Calculate Tank Capacity?]Cylindrical Oil Tank. Lets say that I have a cylindrical oil tank which measures 7 yards in length and has a round face 5 feet in diameter (the distance across the circular end passing through the central point). I want to calculate the tank volume in cubic feet and work out how much oil Cylinder, tube tank calculator: surface area, volume , portugal vertical cylindrical tank oil volumeNow it is a simple matter to find the volume of the cylinder tank as it provides estimation of the total and filled volumes of the water tank, oil tank, reservoirs like a tube tank, and others of horizontal or vertical cylindrical shape. Sometimes it is also needed to estimate a surface size of your tank.
### Tank Volume Calculator - Vertical Cylindrical Tanks - Metric
Vertical cylindrical tank volume calculator diagram: Fill Rate Fill Times @ Litres / Minute. Total Tank Fill Time Current Time to Fill Current Time to Empty If you're cutting blocks, concrete, stone or ANYTHING and there's DUST - DON'T TAKE THE RISK Don't cut it, or cut it wet , portugal vertical cylindrical tank oil volumeGreer Tank Calculator | Greer Tank, Welding & Steel1. Click the tab below that represents your tank type 2. Enter your tank measurements length, width etc 3. View your result in the bottom bar of the calculator. If you need assistance with our tank volume calculator or would like to inquire about any of our services, give us a call today on 1-800-725-8108, send us an email or request a , portugal vertical cylindrical tank oil volumeVertical Tank Volume Calculator - AlbertaVertical Tank Volume Calculator The dimensions for diameter, height and depth should be inside dimensions, otherwise the results will be larger than the real volume. Depth to liquid is the dip-stick measurement from the top of the tank to the surface of the liquid.
### Horizontal Cylindrical Tank Volume and Level Calculator
Volume calculation on a partially filled cylindrical tank Some Theory. Using the theory. Use this calculator for computing the volume of partially-filled horizontal cylinder-shaped tanks.With horizontal cylinders, volume changes are not linear and in fact are rather complex as the theory above shows. Fortunately you have this tool to do the work for you.Cylindrical Tank ProblemsThus, the formula for the total gallons of oil in the tank is given by: Now, we can plug in r = 18, x = 10, and l = 48 to get the number of gallons in the tank. We get 47.94 gallons. Here is a graph of the volume function for the oil tank. This graph confirms our estimate of the oil in the tank when the depth is 10".SPREADSHEET MASTER - Tank Level CalculatorCylindrical Horizontal Tank. , portugal vertical cylindrical tank oil volume Your Tank Capacity and the Actual Volume will be calculated ; , portugal vertical cylindrical tank oil volume The tank is in horizontal position (not working for vertical tank) The tank is a cylinder (Not working for oval or other shapes of tank) Both sides of the tank are flat. (Not convex or concave) The tank is not inclined;
### volume of oil left in a tank.? | Yahoo Answers
Dec 16, 2008 · volume of oil left in a tank.? A vertical cylindrical oil tank is 2.5 feet across on the inside. The depth of the oil in the tank is 2 feet. If 1 cubic foot of space holds 7.48 gallons, approximately how many gallons of oil are left in the tank? Answer Save. 1 Answer.Cylindrical Tank CalculatorI n s t r u c t i o n s Use this calculator for computing the volume of partially-filled horizontal cylinders. If you need a vertical cylinder calculator, click here.. Besides calculating volume for any particular depth, this calculator can also produce a "dipstick chart" showing volume across the entire range of tank Tank Charts - Hall Tank Company - Hall Tank CompanyHome > Tank Charts Use this form to generate a chart of tank capacities. Hall Tank does not guarantee the capacity charts accuracy and in no way takes liability for loss due to its content.
### Spill Prevention Control and Countermeasure (SPCC)
Spill Prevention Control and Countermeasure (SPCC) Plan . Single Vertical Cylindrical Tank Inside a Rectangular or Square Dike or Berm EXAMPLE . 3. Determine the percentage of the secondary containment volume, V. SC . to the tank volume, V. Tank (to determine whether the volume of the containment is sufficient to contain the tanks entire , portugal vertical cylindrical tank oil volumeTry Our Tank Volume Calculator To Work Out The Size of , portugal vertical cylindrical tank oil volumeOctanes Tank Volume Calculator makes it really easy to work out the volume of your tank and is totally free for you to use. All you need to do is follow the 4 steps below: 1. Click one of the 3 tabs across the top which represents your tank 2. Select your measurement units 3. Enter your tanks length, width etc 4. Click Calculate.Spill Prevention Control and Countermeasure (SPCC) Spill Prevention Control and Countermeasure (SPCC) Plan . Construct New Secondary Containment EXAMPLE . C. Accounting for the displacements from other horizontal cylindrical tanks to be located in dike or berm with the largest tank 1. For SC. Height (ft), calculate the displacement from additional horizontal cylindrical tanks, Tank 2,
### Elevating The Tank Versus Leaving Dead Volume In Tank , portugal vertical cylindrical tank oil volume
Nov 12, 2013 · Elevating The Tank Versus Leaving Dead Volume In Tank - posted in Industrial Professionals: Hi, We are designing an atmospheric crude storage tank of 3000 m3 volume, 17 m dia with pumping facility. We have to provide NPSH for the crude export pumps ( centrifugal , 100 m3/hr, 50 bar g discharge pressure ) which pump crude out from the tank.Tank Volume - Engineering ToolBoxRelated Topics . Miscellaneous - Engineering related topics like Beaufort Wind Scale, CE-marking, drawing standards and more; Related Documents . Content of Horizontal - or Sloped - Cylindrical Tank and Pipe - Volume of partly filled horizontal or sloped cylindrical tanks and pipes - an online calculator; Content of Pipes and Cylindrical Tanks - Liquid volume in partly filled horizontal tanks , portugal vertical cylindrical tank oil volumeCalculating Tank VolumeCalculating Tank Volume Saving time, increasing accuracy By Dan Jones, Ph.D., P.E. alculating fluid volume in a horizontal or vertical cylindrical or elliptical tank can be complicated, depending on fluid height and the shape of the heads (ends) of a horizontal tank or the bottom of a vertical tank.
### Vertical Tanks - www.specialtytankandwelding
Vertical Round Tanks (Double Wall) Specialty Tank & Welding Double Wall Vertical Cylindrical Tanks are the solution to your aboveground containment needs. All are equipped with extra ports for monitoring of the interstice space. Quick and easy to install. Possess the strength and impermeability of steel.Try Our Tank Volume Calculator To Work Out The Size of , portugal vertical cylindrical tank oil volumeOctanes Tank Volume Calculator makes it really easy to work out the volume of your tank and is totally free for you to use. All you need to do is follow the 4 steps below: 1. Click one of the 3 tabs across the top which represents your tank 2. Select your measurement units 3. Enter your tanks length, width etc 4. Click Calculate.Calculation of Liquid Volume in a Horizontal Cylindrical , portugal vertical cylindrical tank oil volumeCalculation of Liquid Volume in a Horizontal Cylindrical Container This calculator calculates the volume of liquid inside a horizontal cylindrical container at any given height of liquid. The other required dimensions are the diameter and length of the tank.
### Volume of a Cylinder - Web Formulas
The volume of a cylinder is found by multiplying the area of its top or base by its height and is defined as: V = · r 2 · h Example 1: A cylindrical water storage tank has an Blank Worksheet to Calculate Secondary Containment Volume , portugal vertical cylindrical tank oil volumeThis worksheet can be used to calculate the secondary containment volume of a rectangular or square dike or berm for a single vertical cylindrical tank. You may need a PDF reader to view some of the files on this page. See EPAs About PDF page to learn more. Blank Worksheet (PDF) (4 pp, 529 K)Tank calculations - MrExcelDec 19, 2014 · I work in the oil industry and might be able to help you. We use what are called tank strappings for some of our additive tanks. They were built from formulas for cylindrical horizontal or vertical tanks. I pugged in the size of the tanks and a strapping was produced. From there I set up a form using vlookups to the strappings.
### Horizontal Cylindrical Tank Volume Calculator - Imperial
Please help promote this free service - Tell a Friend about this site! Create PDF to print diagrams on this page. Help & Settings Printing Help (new window) Copy all diagrams on this page to bottom of page - Make multiple copies to Print or Compare.Calculator to Find liquid volume for vertically mounted , portugal vertical cylindrical tank oil volumeApr 13, 2017 · Sugar industry or other industries using vertically mounded cylindrical tanks for sealing of vacuum equipment like vacuum condensers and multiple effect evaporator lost body liquid extraction (outlet). Sometimes it has also called mound. In this cylindrical tank sometimes inside of the tank might be separated by vertical plate for easily operation purpose.
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https://hextobinary.com/unit/angularacc/from/radph2/to/arcsecph2 | 1,620,304,939,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988753.97/warc/CC-MAIN-20210506114045-20210506144045-00230.warc.gz | 321,065,529 | 15,081 | Arcsec/Square Hour
###### How many Arcsec/Square Hour are in a Radian/Square Hour?
The answer is one Radian/Square Hour is equal to 206264.81 Arcsec/Square Hour. Feel free to use our online unit conversion calculator to convert the unit from Radian/Square Hour to Arcsec/Square Hour. Just simply enter value 1 in Agate Line and see the result in Arcsec/Square Hour.
###### How to Convert Radian/Square Hour to Arcsec/Square Hour (rad/h2 to arcsec/h2)
By using our Radian/Square Hour to Arcsec/Square Hour conversion tool, you know that one Radian/Square Hour is equivalent to 206264.81 Arcsec/Square Hour. Hence, to convert Radian/Square Hour to Arcsec/Square Hour, we just need to multiply the number by 206264.81. We are going to use very simple Radian/Square Hour to Arcsec/Square Hour conversion formula for that. Pleas see the calculation example given below.
Convert 1 Radian/Square Hour to Arcsec/Square Hour1 Radian/Square Hour = 1 × 206264.81 = 206264.81 Arcsec/Square Hour
###### What is Radian/Square Hour Unit of Measure?
Radian per square hour is a unit of measurement for angular acceleration. By definition, if an object accelerates at one radian per square hour, its angular velocity is increasing by one radian per hour every hour.
###### What is Arcsec/Square Hour Unit of Measure?
Arcsec per square hour is a unit of measurement for angular acceleration. By definition, if an object accelerates at one arcsec per square hour, its angular velocity is increasing by one arcsec per hour every hour.
###### What is the symbol of Arcsec/Square Hour?
The symbol of Arcsec/Square Hour is arcsec/h2. This means you can also write one Arcsec/Square Hour as 1 arcsec/h2.
1206264.81
2412529.61
3618794.42
4825059.22
51031324.03
61237588.84
71443853.64
81650118.45
91856383.26
102062648.06
10020626480.62
1000206264806.25
###### Radian/Square Hour to Other Units Conversion Chart
1 Radian/Square Hour in Degree/Square Second is Equal to0.0000044209706414415
1 Radian/Square Hour in Degree/Square Millisecond is Equal to4.4209706414415e-12
1 Radian/Square Hour in Degree/Square Microsecond is Equal to4.4209706414415e-18
1 Radian/Square Hour in Degree/Square Nanosecond is Equal to4.4209706414415e-24
1 Radian/Square Hour in Degree/Square Minute is Equal to0.01591549430919
1 Radian/Square Hour in Degree/Square Hour is Equal to57.3
1 Radian/Square Hour in Degree/Square Day is Equal to33002.37
1 Radian/Square Hour in Degree/Square Week is Equal to1617116.08
1 Radian/Square Hour in Degree/Square Month is Equal to30574761.15
1 Radian/Square Hour in Degree/Square Year is Equal to4402765604.95
1 Radian/Square Hour in Arcmin/Square Second is Equal to0.00026525823848649
1 Radian/Square Hour in Arcmin/Square Millisecond is Equal to2.6525823848649e-10
1 Radian/Square Hour in Arcmin/Square Microsecond is Equal to2.6525823848649e-16
1 Radian/Square Hour in Arcmin/Square Nanosecond is Equal to2.6525823848649e-22
1 Radian/Square Hour in Arcmin/Square Minute is Equal to0.95492965855137
1 Radian/Square Hour in Arcmin/Square Hour is Equal to3437.75
1 Radian/Square Hour in Arcmin/Square Day is Equal to1980142.14
1 Radian/Square Hour in Arcmin/Square Week is Equal to97026964.86
1 Radian/Square Hour in Arcmin/Square Month is Equal to1834485668.73
1 Radian/Square Hour in Arcmin/Square Year is Equal to264165936297.22
1 Radian/Square Hour in Arcsec/Square Second is Equal to0.01591549430919
1 Radian/Square Hour in Arcsec/Square Millisecond is Equal to1.591549430919e-8
1 Radian/Square Hour in Arcsec/Square Microsecond is Equal to1.591549430919e-14
1 Radian/Square Hour in Arcsec/Square Nanosecond is Equal to1.591549430919e-20
1 Radian/Square Hour in Arcsec/Square Minute is Equal to57.3
1 Radian/Square Hour in Arcsec/Square Hour is Equal to206264.81
1 Radian/Square Hour in Arcsec/Square Day is Equal to118808528.4
1 Radian/Square Hour in Arcsec/Square Week is Equal to5821617891.52
1 Radian/Square Hour in Arcsec/Square Month is Equal to110069140123.84
1 Radian/Square Hour in Arcsec/Square Year is Equal to15849956177833
1 Radian/Square Hour in Sign/Square Second is Equal to1.4736568804805e-7
1 Radian/Square Hour in Sign/Square Millisecond is Equal to1.4736568804805e-13
1 Radian/Square Hour in Sign/Square Microsecond is Equal to1.4736568804805e-19
1 Radian/Square Hour in Sign/Square Nanosecond is Equal to1.4736568804805e-25
1 Radian/Square Hour in Sign/Square Minute is Equal to0.00053051647697298
1 Radian/Square Hour in Sign/Square Hour is Equal to1.91
1 Radian/Square Hour in Sign/Square Day is Equal to1100.08
1 Radian/Square Hour in Sign/Square Week is Equal to53903.87
1 Radian/Square Hour in Sign/Square Month is Equal to1019158.7
1 Radian/Square Hour in Sign/Square Year is Equal to146758853.5
1 Radian/Square Hour in Turn/Square Second is Equal to1.2280474004004e-8
1 Radian/Square Hour in Turn/Square Millisecond is Equal to1.2280474004004e-14
1 Radian/Square Hour in Turn/Square Microsecond is Equal to1.2280474004004e-20
1 Radian/Square Hour in Turn/Square Nanosecond is Equal to1.2280474004004e-26
1 Radian/Square Hour in Turn/Square Minute is Equal to0.000044209706414415
1 Radian/Square Hour in Turn/Square Hour is Equal to0.1591549430919
1 Radian/Square Hour in Turn/Square Day is Equal to91.67
1 Radian/Square Hour in Turn/Square Week is Equal to4491.99
1 Radian/Square Hour in Turn/Square Month is Equal to84929.89
1 Radian/Square Hour in Turn/Square Year is Equal to12229904.46
1 Radian/Square Hour in Circle/Square Second is Equal to1.2280474004004e-8
1 Radian/Square Hour in Circle/Square Millisecond is Equal to1.2280474004004e-14
1 Radian/Square Hour in Circle/Square Microsecond is Equal to1.2280474004004e-20
1 Radian/Square Hour in Circle/Square Nanosecond is Equal to1.2280474004004e-26
1 Radian/Square Hour in Circle/Square Minute is Equal to0.000044209706414415
1 Radian/Square Hour in Circle/Square Hour is Equal to0.1591549430919
1 Radian/Square Hour in Circle/Square Day is Equal to91.67
1 Radian/Square Hour in Circle/Square Week is Equal to4491.99
1 Radian/Square Hour in Circle/Square Month is Equal to84929.89
1 Radian/Square Hour in Circle/Square Year is Equal to12229904.46
1 Radian/Square Hour in Mil/Square Second is Equal to0.000078595033625627
1 Radian/Square Hour in Mil/Square Millisecond is Equal to7.8595033625627e-11
1 Radian/Square Hour in Mil/Square Microsecond is Equal to7.8595033625627e-17
1 Radian/Square Hour in Mil/Square Nanosecond is Equal to7.8595033625627e-23
1 Radian/Square Hour in Mil/Square Minute is Equal to0.28294212105226
1 Radian/Square Hour in Mil/Square Hour is Equal to1018.59
1 Radian/Square Hour in Mil/Square Day is Equal to586708.78
1 Radian/Square Hour in Mil/Square Week is Equal to28748730.33
1 Radian/Square Hour in Mil/Square Month is Equal to543551309.25
1 Radian/Square Hour in Mil/Square Year is Equal to78271388532.51
1 Radian/Square Hour in Revolution/Square Second is Equal to1.2280474004004e-8
1 Radian/Square Hour in Revolution/Square Millisecond is Equal to1.2280474004004e-14
1 Radian/Square Hour in Revolution/Square Microsecond is Equal to1.2280474004004e-20
1 Radian/Square Hour in Revolution/Square Nanosecond is Equal to1.2280474004004e-26
1 Radian/Square Hour in Revolution/Square Minute is Equal to0.000044209706414415
1 Radian/Square Hour in Revolution/Square Hour is Equal to0.1591549430919
1 Radian/Square Hour in Revolution/Square Day is Equal to91.67
1 Radian/Square Hour in Revolution/Square Week is Equal to4491.99
1 Radian/Square Hour in Revolution/Square Month is Equal to84929.89
1 Radian/Square Hour in Revolution/Square Year is Equal to12229904.46 | 2,312 | 7,585 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2021-21 | latest | en | 0.797463 |
https://www.exceltip.com/excel-formula-and-function/excel-array-formulas | 1,606,851,179,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141681209.60/warc/CC-MAIN-20201201170219-20201201200219-00506.warc.gz | 698,148,626 | 14,749 | ### How to Create Array Of Numbers Using Excel INDEX Function
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how to use countif in ms-access
Microsoft Excel, Countif Help!? How can I have multiple validation rules in a SUMIF or COUNTIF function?Can you actually download MS Office with just a key and no discs? You could use Visual Basic to access cell formatting properties but most inbuilt functions that you would type into a cell focus on the cell contents, not the formatting.MS Excel: how to use COUNTIF() dynamically. 0. A: And if the question is about to count those where the amount is > 0 and that you cant use that condition, amount > 0, in the where clause for some reasonQuery similar to COUNTIF in Excel Community Question. Count Distinct Records in MS Access Report Design Community Question. I will start off saying that, yes, I know there is no COUNTIF in Access, and that this question covers a few areas of Access.EDIT: And Im using an older version of Access, so I dont have the option to make Crosstab Queries. Access.This Excel tutorial explains how to use the Excel COUNTIF function with syntax and examples. Description. The Microsoft Excel COUNTIF function counts the number of cells in a range, that meets a given criteria. This article describes how to dynamically change a SQL Passthrough query and a local Access query. A SQL Server View from the AdventureWorks database is used, but a local Access table may be used, too.Using a Linked SQL View in MS Access. This Microsoft Access tutorial walks you through the steps of creating a lookup field in Access 2007.
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# WindandStackOpenArea question
OpenStudio. I want to show you 1) the left pic is the energy simulation of all south-west thermal zones of my building without anything (no hvac, no ventilation, no infilitration) just the building with loads, people, schedules ect. And then the right pic. 2) the building using the "WindandStackOpenArea" measure, setting the minimum indoor T = 22 C, and the maximum indoor T = 26 C. I would have expected that this measure works only during the summer, but in my case temperatures go higher than 26C (the limit I set). Anyway it seems that it works (I don't know why) also during the winter, why? Have a look at the image: in the left case WindandStackOpenArea is not set, and the temperatures in January-February are in the range 20-25 C, but in the right case temperatures in Jan-Feb are LOWER than 22C. Why? I set the minimun indoor T should be 22C, not the maximum. I don't see why WindandStackOpenArea seems that has worked also in these cold months. Then, during the summer it seems that WindandStackOpenArea has no more effect because of the high temperatures, and I can understand it.
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The minimum and maximum indoor temperatures describe when the windows are allowed to open, so when the interior temperatures drop below 22C, the windows close. When temperature go above 26C, the windows also close.
In your model, i'm guessing in the summer your indoor temperatures will at some point go above 26C, since naturally ventilated buildings typically will float this high in most climates. The windows all close at this point, but since you still have internal gains and envelope gains which contribute heat to the building, causing the temperature to continue to rise.
Since the windows are closed and there is no HVAC system, there is no way for the building to lose heat, so heat perpetually increases in the space until the climate has cooled enough to allow the building to cool below 26C and your windows are able to open again.
In the winter, your windows stay open until the building drops below 22C, in which case the windows close. The building still loses heat through the envelope however, so the air temperatures inside fall below the 22C threshold.
I would suggest looking into adding controls for the outdoor temperature as well as in the indoor temperature. Maybe switch your 26C limit to the outdoor temperature limit so that the windows are able to be opened in the summer.
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thank you, very exhaustive. Just one more question: is there a max air velocity that commonly is used for office buildings? something like a limit I can find maybe in ASHRAE? thank you anyway
( 2016-11-24 04:46:20 -0500 )edit
If you're referring to the indoor velocity for use in thermal comfort calculations, ASHRAE standard 55 lays out the criteria in section 5.3.3. Essentially it's as follows:
• if using natural ventilation, max allowable airspeed is 0.8 m/s
• if occupants have user-controllable fans, the maximum airspeed is 1.2 m/s
Anecdotally I've heard that that above 0.6 m/s there are issues with papers getting thrown around.
( 2016-11-28 12:30:10 -0500 )edit
@StefanG no I'm talking about the data to fullfill in the Windstackopenarea measure, that asks the "Max Wind Speed (m/s)", so the outdoor wind. Should I set it as an average wind speed of all the summer season? or should I use the maximum wind speed of the summer season?
( 2016-11-29 05:08:25 -0500 )edit
@poppo92
I'm not sure, but I think the max wind speed is only for closing the windows during high wind events for safety issues. I don't have a good number on hand but I would set as a sufficiently high number to allow natural ventilation to occur most of the times you want it and rely on temperatures to control when the windows open and close.
( 2016-11-29 11:19:10 -0500 )edit | 942 | 3,984 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-38 | latest | en | 0.948054 |
https://devforum.roblox.com/t/how-to-make-a-part-visible-locally-only-if-you-are-close-to-it/723197 | 1,679,969,248,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948756.99/warc/CC-MAIN-20230328011555-20230328041555-00435.warc.gz | 252,105,306 | 5,418 | # How to make a part visible locally only if you are close to it?
Iam trying to create a barrier that can be only visible if you are close to it and if you go far from it goes invisible but I don’t know how I can make that.
You can use magnitude as it returns the length of 2 vectors. Subtract the player’s `HumanoidRootPart` position and the part. Then use magnitude to get the distance and then check if the player is near it or not.
https://developer.roblox.com/en-us/articles/Magnitude
2 Likes
While Silents’ solution would work, if you have a large part, you would need a really large radius in order for the barrier to not turn invisible if you’re near a corner. Doing this would also make the barrier invisible if you’re decently far away, and roughly in line with the center.
A solution to this is to test the distance between the player’s HumanoidRootPart, and the point closest to the player while still inside the barrier.
Implementing this is rather easy, all you need to do is change the player’s position to object space in comparison to the barrier part, and then clamping the player’s position to the size of the cube, and then changing the moved position to be in world space again in comparison to the barrier.
``````function GetClosestPoint(PlayerPos, Barrier) -- Barrier is the barrier object
local RelPoint = Barrier.CFrame:PointToObjectSpace(PlayerPos)
local ClampedPos = Vector3.new(
math.clamp(RelPoint.X, -Barrier.Size.X/2, Barrier.Size.X/2),
math.clamp(RelPoint.Y, -Barrier.Size.Y/2, Barrier.Size.Y/2),
math.clamp(RelPoint.Z, -Barrier.Size.Z/2, Barrier.Size.Z/2)
)
local ClosestPoint = Barrier.CFrame:PointToWorldSpace(ClampedPos)
return ClosestPoint
end
``````
Simply adding this function to your code, and replacing the barrier’s position with the outcome of this function when calculating distance should improve large barriers by a lot. | 438 | 1,873 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-14 | latest | en | 0.822385 |
https://www.eduzip.com/ask/question/evaluate-displaystyle-int-0-1-xleft-1-x-right160-n-dx-580729 | 1,627,172,187,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151531.67/warc/CC-MAIN-20210724223025-20210725013025-00582.warc.gz | 765,208,401 | 8,544 | Mathematics
# Evaluate: $\displaystyle \int _{ 0 }^{ 1 }{ { x\left( 1-x \right) }^{ n } } dx$
##### SOLUTION
$I=\displaystyle \int_{0}^{1}x(1-x)^{n}dx$
$=\displaystyle \int_{0}^{1}(1-x)(1-(1-x))^{n}dx$ [ Using property of definite integral]
$=\displaystyle \int_{0}^{1}(1-x)(x)^{n}dx$
$=\displaystyle \int_{0}^{1}x^{n}-x^{x-1}dx$
$=\displaystyle \left [ \dfrac{x^{n+1}}{n+1} -\dfrac{x^{n+2}}{n+2}\right ]_{0}^{1}$
$=\dfrac{1}{n+1}-\dfrac{1}{n+2}=\dfrac{1}{(n+1)(n+2)}$
Its FREE, you're just one step away
Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105
#### Realted Questions
Q1 Single Correct Medium
$\displaystyle \int_{0}^{1}xe^{x}\ dx=$
• A. $2$
• B. $3$
• C. $4$
• D. $1$
1 Verified Answer | Published on 17th 09, 2020
Q2 Subjective Hard
Evaluate: $\displaystyle \int \dfrac {dx}{\sin^{2}x + 5\sin x \cos x + 2}$
1 Verified Answer | Published on 17th 09, 2020
Q3 Subjective Hard
Evaluate $\displaystyle\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ \cfrac { \sin { x } }{ 1+\cos { x } +\sin { x } } } dx\quad \quad$
1 Verified Answer | Published on 17th 09, 2020
Q4 Single Correct Medium
The value of $\displaystyle\int _{ 0 }^{ \infty }{ \frac { \log { x } }{ { a }^{ 2 }+{ x }^{ 2 } } dx }$
• A. $\dfrac{\pi \log a}{a}$
• B. $\pi \log a$
• C. $0$
• D. $\dfrac{2\pi \log a}{a}$
Let $n \space\epsilon \space N$ & the A.M., G.M., H.M. & the root mean square of $n$ numbers $2n+1, 2n+2, ...,$ up to $n^{th}$ number are $A_{n}$, $G_{n}$, $H_{n}$ and $R_{n}$ respectively. | 651 | 1,536 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2021-31 | latest | en | 0.493897 |
http://photographsbyanjuli.com/which-how-to-buy-a-house/ | 1,679,875,175,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946584.94/warc/CC-MAIN-20230326235016-20230327025016-00381.warc.gz | 36,813,942 | 15,559 | # Which How To Buy A House
Posted on
If you are looking for the answer of which how to buy a house, you’ve got the right page. We have approximately 10 FAQ regarding which how to buy a house. Read it below.
## The Ramos family saved 720,000.00 which is 80% of what
Ask: The Ramos family saved 720,000.00 which is 80% of what they need to buy a house and lot. How much more do they need?
Pa help po with solution po sana
900 more they need.
Solutions:
STEP 1: 720,000,00 = 80% × Y
STEP 2: 720,000,00 = 80 × Y
100
Multiplying boths sides by 100 and dividing both sides of the equation by 80 we will arrive at:
STEP 3: Y = 720 × 100
80
STEP 4: Y = 720 × 100 ÷ 80
STEP 5: Y = 900
There fore the answer is 900
## The Alcantara family saved ₱850,000.00 which is 75% of what
Ask: The Alcantara family saved ₱850,000.00 which is 75% of what they need to buy a house and lot. How much more do they need?
1,062,500
Step-by-step explanation:
850,000 + 25% = 1,062,500
A
HOW MUCH MORE DO THEY NEED
G
P850,000 75%
O
MULTIPLICATION
N
850000×25%=N
S
850000×25%=1,062,500
A
PHP 1,062,500
## Test. General Annuity 1. How much must a man set
1. How much must a man set aside at the end of every six months to raise a fund of Php 1,000,000.00
with interest rate of 5% compounded monthly to purchase his family new house which he plans
1.000.000.000.000.000
Step-by-step explanation:
CARRY ON LEARNING
## the ramos family saved P720,000.00 which is 80% of what
Ask: the ramos family saved P720,000.00 which is 80% of what they need to buy a house and loy. how much more money do they need?
Given: ₱ 720,000.00 and 80%
Solution : ₱720,000.00 x 80% = ₱576,000.00
( 80% is 0.8 )
Explanation: I just multiply the given in the said question.
## Fill in the blank for each of the following question
Ask: Fill in the blank for each of the following question with one of the two words in brackets write your answer down on a piece of paper then check your answer below.
1)________is the name of your dog?(What, How)
2)________orange did you buy?(How many, How far)
3)________did you grow up? (Who, Where)
5)________is the movie?(How cone, How long)
6)________did you study biology?(Which, Why)
7)________handbag is this?(Where, Whose)
9)________did you make that? (How, Who)
10)________away is the office? (How much, How far)
11)________should we send this parcel to?(Why, Whom)
12)________is your father?(How old, How many)
13)________did you talk to?(Whom, Where)
14)________did you jacket cost?(How much, How old)
15)________did you buy this house?(Whose, When)
16)________you went to university?(How far, How come)
1.What is the name of your dog?_____________
2.How many orange did you buy?_____________
3.Where did you grow up?_________
4.Who is your favorite actor? _________
5.How long is the movie? _________
6.Why did you study biology? _________
7.Whose handbag is this? _________
8.Which car did you buy? _________
9.How did you make that? _________
10.How far away is the office? _________
11.Whom should we send this pancel to? __________
12.How old is your father? _________
13.Whom did you talk to? _________
14.How much did your jacket cost? _________
15.When did you buy this house? ________
16.How come you went to university? __________
1 what
2 how many
3 where
4 who
5 how
6 why
7 whose
8 which
9 how
10 how far
11 whom
12 how old
13 whom
14 how much
15 when
16 how come
2 three oranges
3 in manila
4 jennifer lawrence
5 two hours
6 Because I wanted to learn about it.
7 it’s mine
8 the chevrolet
9 i followed a tutorial on Youtube
10 it’s 2 kilometers away
11 send it to my mom
12 54
13 i talked to my teacher
14 900 pesos
15 i bought it last year
16 my parents told me to go to university
## 1. Arthur is trying to decide between two brands of
Ask: 1. Arthur is trying to decide between two brands of paint to use for his house. Brand X costs Php 305.25 for 750 mL while Brand Y costs Php 382.28 for 950 mL. Which of the two brands is a better buy? How will you convince Arthur that such brand is the better buy?
Brand X kaba?? tipid pag may time
## 1 The Donovans need to borrow P30,000 to buy a
Ask: 1 The Donovans need to borrow P30,000 to buy a house Bank A will give them
a 30-year mortgage at 10 %. Bank B will give them a 20-year mortgage at
12 %. Which mortgage should the Donovans assume so that they will pay the
smallest amount of interest? How much will they save?
i sure Donovans will buy a house from bank A because it have a longest discount
Step-by-step explanation:
discount 26.8. choose me brainliest
## 11. Shane and her cousin walked to the convenience store.
Ask: 11. Shane and her cousin walked to the convenience store. Shane bought a pack of chips while her cousin bought a bottle of water Which of the following is question that asks for explicit information?
B. How close is Shane to her cousin?
C. How far is the store to their houses?
D. Why did they buy a pack of chips and a bottle of water?
I’m pretty sure it’s A
Correct me if I’m wrong
## The Ramos family saved #720‚000.00 which is 80% of what
Ask: The Ramos family saved #720‚000.00 which is 80% of what they need to buy house and lot. How much more do they need,
Php 900,000
Solution
Since 720,000 is 80% it means that:
4/5x=720,000 or 4/5(x)
Just transpose 4/5 and you will get:
5/4(720,000)=Php 900,000
## 4. Reggie paid P1.680.00 for the three "Picasso" painting. Hisbrother
Ask: 4. Reggie paid P1.680.00 for the three “Picasso” painting. His
brother Joshua wanted to buy 8 “Picasso” paintings. How
much would Joshua pay
A. P2240
B. P3340
C. P4480
D. P5500
5. Michael wanted to go to his brother’s house which is 24 km
from his house. If he drives his car with a speed of 60 kph,
he will be there in 24 minutes. If he makes his speed slower
to 40 kph, how long would Michael take to reach his
brother’s house?
A. 24 min.
B. 28 min.
C.32 min. D. 36 min. | 1,769 | 5,913 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2023-14 | latest | en | 0.942581 |
http://www.circuitmason.com/work-book/lesson-2-reflections-and-transmissions/lesson-2-section-5 | 1,674,982,387,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499710.49/warc/CC-MAIN-20230129080341-20230129110341-00431.warc.gz | 54,523,364 | 8,684 | ### Section 5: Interesting Scattering Parameter Properties
Scattering Parameters are very useful to RF designers. For one, it turns out that it is easier to measure voltage waves than actual voltages in an RF circuit. When you have pulses and sine waves going through an RF device, with lots of different reflections in the circuit due to mismatches, trying to describe what's going on with a single voltage or current becomes somewhat arbitrary. S-Parameters also have many interesting properties which can help provide insight.
System of S-Parameter Equations
Consider a power divider- many houses have an RF power divider which splits the cable signal to different TVs around the house. Consider specifically a four-way power divider, where port 1 is the input (the cable TV signal goes into here) and ports 2-5 are the output (these signals are sent to TVs around the house). If port 1 is driven, it makes sense that port 2 will have a quarter of the power coming out (as a voltage wave, -6dB less than the input). The actual S-Parameter equations for this system are:
Equation 2.5.1
Now, writing out equations like this can be cumbersome, so we are going to short hand these equations by putting the equivalent variables in matrix formulation. If you aren't familiar with linear algebra, don't worry about the mechanics, just notice that the variables follow the same pattern... and we're going to be talking about patterns. I assure you, these are equivalent sets of equations.
Equation 2.5.2
Reciprocal Devices
A broad class of components, such as our power divider, are known as reciprocal devices. Reciprocal devices have no active devices (no transistors), no magnetic materials (don't worry how they work), and no plasmas. If a device is reciprocal, it has the following property:
Equation 2.5.3
This means the device has symmetry to it. Let's consider the above device where port 1 is driven by the cable signal from outside your house, and port 2 is one of four ports where power is split out to your TV. Equation 2.5.3 says that if S21 is -6dB (25% of the power goes to port 2 from port 1), then S12 is also -6dB. If you put 1W into port 1, then 0.25W will come out of port 2 (assuming everything is matched, no bad reflections). If you put 1W into port 2, then 0.25W will come out of port 1 (assuming everything is matched, no bad reflections). So, S21 = S12 for a reciprocal device.
Figure 2.5.1: Four-way splitter
If we consider a transistor amplifier, the device has gain. The voltage wave coming out of port 2 will be larger than the voltage wave coming out of port 1, for example we described a 13dB gain amplifier above: S21 = 13dB. However, if you connect the device backwards, if you drive power into the output, not much energy will come out of the input: S21 ≠ S12. The S12 is often called the isolation of the amplifier, this is how well what the input is isolated from what happens on the output, and is commonly on the order of -20dB (or only 1% of the power gets through).
Figure 2.5.2: Amplifier with 13dB gain and -20dB isolation | 750 | 3,089 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2023-06 | latest | en | 0.930801 |
https://hackaday.com/tag/transistor/ | 1,534,579,753,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213508.60/warc/CC-MAIN-20180818075745-20180818095745-00533.warc.gz | 680,868,851 | 29,245 | # Circuit VR: A Tale Of Two Transistors
Last time on Circuit VR, we looked at creating a very simple common emitter amplifier, but we didn’t talk about how to select the capacitor values, or much about why we wanted them. We are going to look at that this time, as well as how to use a second transistor in an emitter follower (or common collector) configuration to stiffen the amplifier’s ability to drive an output load.
Several readers wrote to point out that I’d pushed the Ic value a little high for a 2N2222. As it turns out, at least one of the calculations in the comments was a bit high. However, I’ve updated the post at the end to explore what was in the comments, and talk a bit more about how you compute power dissipation with or without LTSpice. If you read that post, you might want to jump back and pick up the update.
## Back to Our Program
As a reminder, the LT Spice circuit we started with appears below. You can download that file and others on GitHub.
## Output Z
Last time, we went over the design equations and even looked at a spreadsheet for figuring out the values. That spreadsheet assumed you wanted to pick a few items including the normal collector current for the device. In some cases, though, your driving design goal will be a certain output impedance. In that case, pick RC accordingly, and go through the same steps but you’ll compute Ic instead of selecting it and skip step 4. You can use this same procedure when the actual load you are driving is the collector resistor, which isn’t uncommon.
It is easy to see that RC is the output impedance if you do a little logic. Remember, this amplifier inverts. So Q1 is as close to off as it is going to get when the input signal is large. Assume Q1 turned all the way off. What would the output circuit look like then? A voltage divider made up of RC and RL. Like any voltage divider, the maximum power in RL is going to occur when RL=RC. If you have more of an engineering mindset, you can think of it as the amplifier’s Thevenin equivalent is a voltage source with RC as the resistor. Or, if you are more graphical, think of a voltage divider with a 10V input and a 100 ohm “top” resistor (R1). If you try values for the bottom resistor (R2) ranging from 1 to 200, it looks like this:
The voltage in R2 keeps going up, but the current goes down. When R2 is 100 ohms, the power maxes out at about 250 mW. This is why you try to match, say, a transmitter with an antenna or speakers with an amplifier.
You might want to control input impedance as well. For the input impedance case, you would have to control the values of Re, R1, and R2 which is quite a bit harder without setting up a lot of simultaneous equations or just iterating. It also will depend on beta, which is notoriously unreliable. If the product of Re and beta is a large number, you can approximate it as R1 and R2 in parallel, and that’s often good enough.
Note that in the above circuit example I just put a large resistor in as the load so it didn’t affect things much. But what if that resistor had been a 16-ohm speaker, perhaps?
## Back to Capacitors
So why are capacitors important? Because the transistor needs a very specific set of DC voltages on its terminals and connecting an input or output to it is going to perturb that. However, we can isolate the circuit from any DC effects using a capacitor on the input and the output. That means we can’t amplify very low-frequency signals well — the capacitors will act like large resistors. But at higher frequencies, it won’t be any problem. You can see that in the simulation where some capacitors guard the inputs and the outputs.
If you want to see the effect in a less distracting way, check out this simulation. Here an input signal is riding a DC level. A voltage divider sets another DC level. With a capacitor between them, the circuit essentially shifts the input to a new DC level, like this:
The reactance of the capacitor, of course, depends on the frequency, according to 1/(2*π*f*C). That means the higher the capacitance, the lower the reactance at a given frequency. In this case, the 100 Hz signal sees the 10 uF capacitor as about 160 ohms of reactance. At 47uF, that drops to about 34 ohms. At 1 kHz, that will divide both of those values by another 10 (16 and 3.4 ohms).
## Gain
The emitter resistor essentially introduces negative feedback which reduces our dependence on beta and makes things generally more stable. However, it also limits gain. If you suppose you have RE as a short-circuit — 0 ohms — you might think you could get infinite gain. But, in fact, you really just get a small internal resistance that is temperature- and current-dependent. At room temperature, though, it is generally just a few ohms at most. It would still increase the gain quite a bit if we could just short the emitter — in theory, up to the beta of the transistor. But without the negative feedback, we get all the other undesirable features we tried to avoid.
However, just as we use capacitors to isolate the input and output, why can’t we use a capacitor to short the signal to ground even if the DC path is through the resistor? As it turns out, you can. Try adding a capacitor across RE and watch the output go higher. Below, you can see the same output with a 47 uF capacitor across RE. Look at the scales. That 0.2V input signal now produces an output of over 5V, peak-to-peak. That’s a gain of about 25, or 5 times the DC gain.
The effect varies on the value of the capacitor and, of course, the frequency. Here’s the output with 10uF, 47uF, and 100 uF capacitors (first graph, below). The second graph shows the effect versus frequency. You typically want the reactance of the capacitor to be about 1/10th of the emitter resistance at the frequency where you will accept a 3dB drop off.
Note that the capacitor works so well, that at some frequencies, we go beyond the allowable gain and clip (see the last graph). Depending on your design goals, you may need to be careful with that.
## Selecting Coupling Capacitors
To know what value to assign the coupling capacitors, you need to know the impedance of the amplifier. That’s fairly easy to estimate, but with LT Spice we can do better. If you look at V2, you know it is putting out 50 mV and you can measure the current drawn from it. Ohms law will tell you that .05 divided by that current must be the resistance V2 “sees.” With C1 set ridiculously high (1F) and V2’s internal resistance set to zero, the circuit draws about 1.75 mA from V2. That’s about 28.6 ohms. So if you know the 3dB frequency you want you simply have to compute the capacitance using the familiar 1/(2*π*f*R) formula. Assume we want 10 kHz as the 3dB point. Since R is 28.6 you need at least 0.6 uF of capacitance. Of course, you can also reverse the formula and determine what your 3 dB point should be given a certain value of capacitor.
Here’s a little WolframAlpha tip. If you try to do the above calculation, you get the answer in scientific notation: 5.56 x 10-7. Sure, you can just shift the decimal point two to the right to get the exponent to -9… or is that to the left? However, you can also just add the words “engineering form” to your query, and you’ll get the answer to the nearest exponent that’s a multiple of 3.
The other problem you’ll often see is that you need to drive a low impedance load which can limit your gain since matching that impedance will prevent you from using a large RC. One answer is to use an output transistor as an emitter follower or common collector amplifier. This is a very simple setup where the input to the base appears practically unchanged on the emitter. So the gain of the stage is nearly 1. That might not seem like a great thing until you realize that the output impedance of such an amplifier is roughly the source impedance divided by beta. Remember, lower output impedance is good because you can drive a wider range of load.
Suppose your RC in the main amplifier is 1600 ohms and you would like to drive a 16-ohm speaker. If the emitter follower beta is 100, the effective impedance seen from the main amp will be 1.6K ohms and the output impedance of the stage will be very low. But because in this case, the emitter resistor is probably the load itself, you won’t want to put a capacitor in the output because it would block the path to ground.
Have a look at this design:
This is very nearly the same amplifier as before, but there’s no coupling capacitor on the output. In addition, the component values changed a bit. When Q1 is turned off, the maximum voltage will go to the load and this will transfer the most power when RL=RC so the output impedance at Q1 is 1600 ohms. This is a poor match for a 16 ohm speaker, but Q2 can get us in the neighborhood in the emitter follower configuration. It is true that beta isn’t reliable, so the match probably won’t be perfect, but it should be good enough for most purposes.
Here’s the output:
Compare that with the output of the original amplifier driving a 16-ohm load. You’ll need to reduce the input drive down to 50mV, but even then the output from the original circuit will be very disappointing.
Of course, Q2 is going to need to be a power transistor. You won’t be able to quite get all 15V on the base of Q2, but you could get close. After the emitter drop, you could have a Ve of about 14V and that’s a little less than 900 mA or around 13 plus watts. Picture a big device with a heat sink. Luckily, the simulation doesn’t care. But, of course, that’s also one of the dangers of simulation is that you can overstress the models and they don’t care.
## The End?
As much as we’ve talked about the common collector amplifier, there’s a lot more to it. What if the collector load is a tuned circuit? Or the emitter bypass? You can construct lots of things including multistage amplifiers using this as a building block.
By the way, you might think that bipolar transistors are old-fashioned compared to FETs, but they do have their uses. Also, all of these amplifier configurations have corresponding FET designs. The ideas are the same but, of course, the design equations are a bit different. FETs operate on voltages and there are other peculiarities. For example, some types of FETs are normally on, so you’ll need a negative bias voltage to get them to turn off. FETs — especially MOSFETs — have very high input impedance which makes input circuits easy to design. However, they also introduce capacitance which can be tricky at higher frequencies. But that’s a topic for a future Circuit VR.
# Circuit VR: Starting an Amplifier Design
Sometimes I wish FETs had become practical before bipolar transistors. A FET is a lot more like a tube and amplifies voltages. Bipolar transistors amplify current and that makes them a bit harder to use. Recently, [Jenny List] did a series on transistor amplifiers including the topic of this Circuit VR, the common emitter amplifier. [Jenny] talked about biasing. I’ll start with biasing too, but in the next installment, I want to talk about how to use capacitors in this design and how to blend two amplifiers together and why you’d want to do that.
But before you can dive into capacitors and cascades, we need a good feel for how to get the transistor biased to start with. As always, there’s good news and bad news. The bad news it that transistors vary quite a bit from device to device. The good news is that we’ll use some design tricks to keep that from being a problem and that will also give us a pretty wide tolerance on component values. The resulting amplifier won’t necessarily be precise, but it will be fine for most uses. As usual, you can find all the design files on GitHub, and we’ll be using the LT Spice simulator.
# How Low Can You Go? Tiny Current Generator
Current limited power supplies are a ubiquitous feature of the bench, and have no doubt helped prevent many calamities and much magic smoke being released from pieces of electronics. But for all their usefulness they are a crude tool that has a current resolution in the range of amps rather than single digit milliamps or microamps.
To address this issue, [Yann Guidon] has produced a precision current source, a device designed to reliably inject tiny currents. And in a refreshing twist, it has an extremely simple circuit in the form of a couple of PNP transistors. It has a range from 20 mA to 5 µA which is set and fine-tuned by a pair of pots, and it has a front-panel ammeter hacked from a surplus pocket multimeter, allowing the current to be monitored. Being powered by its own internal battery (and a separate battery for the ammeter) it is not tied to the same ground as the circuit into which its current is being fed.
[Yann] is a prolific builder whose work has featured here more than once. Take a look at his rubidium reference and his discrete component clocks, for example, and his portable LED flash.
# Biasing That Transistor Part 4: Don’t Forget the FET
Over the recent weeks here at Hackaday, we’ve been taking a look at the humble transistor. In a series whose impetus came from a friend musing upon his students arriving with highly developed knowledge of microcontrollers but little of basic electronic circuitry, we’ve examined the bipolar transistor in all its configurations. It would however be improper to round off the series without also admitting that bipolar transistors are only part of the story. There is another family of transistors which have analogous circuit configurations to their bipolar cousins but work in a completely different way: the Field Effect Transistors, or FETs.
In a way it’s less pertinent to look at FETs in the way we did bipolar transistors, because while they are very interesting devices that power much of what you will do with electronics, you will encounter them as discrete components surprisingly rarely. Every CMOS device you deal with relies on FETs for its operation and every high-quality op-amp you throw a signal at will do so through a FET input, but these FETs are buried inside the chip and you’d be hard-pressed to know they were there if we hadn’t told you. You’d use a FET if you needed a high-impedance audio preamp or a low-noise RF amplifier, and FETs are a good choice for high-current switching applications, but sadly you will probably never have a pile of general-purpose FETs in the way you will their bipolar equivalents.
That said, the FET is a fascinating device. Join us as we take an in-depth look at their operation, and how and where you might use one.
## FET basics
A basic FET has three terminals, a source (the source of electrons), a gate (the control terminal), and a drain (where electrons leave the device). These are analogous to the terminals on a bipolar transistor, in that the source fulfills a similar role to the emitter, the gate to the base, and the drain to the collector. Thus the three basic bipolar transistor circuit configurations have equivalents with a FET; common-emitter becomes common-source, common-base becomes common-gate, and an emitter follower becomes a source follower. It is dangerous to stretch the analogy between bipolar transistors and FETs too far, though, because of their different mode of operation. A closer similarity exists between a FET and a triode tube, if that helps.
The simplest FET for demonstration purposes has a piece of N-type semiconductor with source and drain connections at opposite ends, and a zone of P-type semiconductor deposited in its middle. This is referred to as an N-channel junction FET or JFET, because the channel through which current flows is N-type semiconductor, and because a diode junction exists between gate and channel. There are equivalent P-channel devices, just as there are PNP and NPN bipolar transistors.
Were you to bias an n-channel JFET as you would a bipolar transistor with a positive bias on its gate, the diode between gate and source would conduct, and the transistor would remain a diode with two cathode terminals. If however you give the gate a negative bias compared to the source, the diode becomes reverse-biased, and no current to speak of flows in the gate.
A characteristic of a reverse-biased diode is that it has a depletion zone between anode and cathode, an area in which there are no electrons. This is what causes the diode to no longer conduct, and the size of the depletion zone depends upon the size of the electric field that exists across it. If you’ve ever used a varicap diode, the capacitance between the two sides of this variable-width zone is the property you are exploiting.
In a FET, the depletion zone stretches from the gate region into the channel, and since its size can be adjusted by the gate voltage it can be used to “pinch” the remaining conductive region within the channel. Thus the area through which electrons can flow is controlled by the gate voltage, and thus the current that flows between drain and source is proportional to the gate voltage. We have an amplifier.
In the JFET diagram above, the negative gate bias is represented by a battery. Tube enthusiasts may have encountered equipment that derives negative grid bias from a power supply, and you will find tube power units that include a -150 V rail for this purpose. In general though this is inconvenient in a FET circuit even though the voltage is lower, because of the extra cost of a negative regulator.. Instead the gate is held at a lower potential than the source by careful selection of a source resistor such that the current flowing through it brings the source up above ground, and a gate bias circuit that holds the gate close to ground. The base resistor chain from the bipolar circuit is for this reason often replaced with either a single resistor to ground, or a gate circuit with a very low DC resistance to ground such as an inductor.
## MOSFETs, where the FET becomes more useful
The JFET we have described is the simplest of field-effect devices, but it is not the one you will encounter most frequently. MOSFETs, short for Metal Oxide Semiconductor FETs, have a similar source, gate, and drain, but instead of relying on a depletion zone in a reverse-biased diode, they have a thin layer of insulation. The electric field from the gate acts across this insulation and pinches the conductive region in the channel through repulsion of electrons, with the same effect as it has in the JFET. It is beyond the scope of this piece to go into their mechanisms, but you will encounter two types of MOSFET: depletion mode devices that require the same negative bias as the JFET, and enhancement mode MOSFETS that require a positive bias.
## Why would you use a FET?
So we’ve described the FET, and noted that while its mode of operation is different to that of a bipolar transistor it does a substantially similar job. Why would we use a FET then, what advantages does it offer us? The answer comes from the gate being insulated either by a depletion region in a JFET or by an insulating layer in a MOSFET. A FET is a voltage amplifier rather than a current amplifier, its input impedance is many orders higher than that of a bipolar transistor, and thus you will find FETs used in many applications that require a high impedance small-signal amplifier. The input of a high-performance op-amp will almost certainly be a FET, for example.
The high input impedance has another effect less coupled to small signal work. Where a bipolar transistor requires significant base current to turn itself on, the corresponding FET requires almost none. Thus almost all complex integrated circuit logic devices are FET-based rather than bipolar because of the huge power saving that can be made by not needing to supply the base current demands of many thousands of bipolar transistors.
The same effect influences the choice of FETs for power switching, while a bipolar transistor’s base current is proportional to its collector current and thus it will need a significant driver, by contrast a power MOSFET requires virtually no standing gate current after an initial surge. A MOSFET power switch can thus be built requiring much less in the way of drive electronics and much more efficiently than a corresponding bipolar switch, and makes possible some of the tiny driver boards you might be used to for driving motors in your 3D printer, or your multirotor.
Through the course of this series you should have acquired a solid grounding in basic bipolar transistor principles, and now you should be able to add FETs to that knowledge base. We suggested you buy a bag of 2N3904s to experiment with in one of the previous articles, can we now suggest you do the same with a bag of 2N3819s?
# Biasing That Transistor: The Emitter Follower
We were musing upon the relative paucity of education with respect to the fundamentals of electronic circuitry with discrete semiconductors, so we thought we’d do something about it. So far we’ve taken a look at the basics of transistor biasing through the common emitter amplifier, then introduced a less common configuration, the common base amplifier. There is a third transistor amplifier configuration, as you might expect for a device that has three terminals: the so-called Common Collector amplifier. You might also know this configuration as the Emitter Follower. It’s called a “follower” because it tracks the input voltage, offering increased current capability and significantly lower output impedance.
Just as the common emitter amplifier and common base amplifier each tied those respective transistor terminals to a fixed potential and used the other two terminals as amplifier input and output, so does the common collector circuit. The base forms the input and its bias circuit is identical to that of the common emitter amplifier, but the rest of the circuit differs in that the collector is tied to the positive rail, the emitter forms the output, and there is a load resistor to ground in the emitter circuit.
As with both of the other configurations, the bias is set such that the transistor is turned on and passing a constant current that keeps it in its region of an almost linear relationship between small base current changes and larger collector current changes. With variation of the incoming signal and thus the base current there is a corresponding change in the collector current dictated by the transistor’s gain, and thus an output voltage is generated across the emitter resistor. Unlike the common emitter amplifier this voltage increases or decreases in step with the input voltage, so the emitter follower is not an inverting amplifier.
# Circuit VR: Current Mirrors
Last time we looked at Spice models of a current sink. We didn’t look at some of the problems involved with a simple sink, and for many practical applications, they are perfectly adequate. However, you’ll often see more devices used to improve the characteristics of the current sink or source. In particular, a common design is a current mirror which copies a current from one device to another. Usually, the device that sets the current is in a configuration that makes it very stable while the other device handles the load current.
For example, some transistor parameters vary based on the output voltage which causes small nonlinearities in the output. But if the setting transistor has a fixed voltage across it, that won’t be a problem. The only problem with mirror schemes is that the transistors involved all have to match in key characteristics. For that reason, mirrors are usually better on ICs where the transistors are all more or less the same. You can get discrete transistors that have multiple devices built on a single substrate, but these are not very common.
# Optocouplers: Defending Your Microcontroller, MIDI, and a Hot Tip for Speed
Deep in the heart of your latest project lies a little silicon brain. Much like the brain inside your own bone-plated noggin, your microcontroller needs protection from the outside world from time to time. When it comes to isolating your microcontroller’s sensitive little pins from high voltages, ground loops, or general noise, nothing beats an optocoupler. And while simple on-off control of a device through an optocoupler can be as simple as hooking up an LED, they are not perfect digital devices.
But first a step back. What is an optocoupler anyway? The prototype is an LED and a light-sensitive transistor stuck together in a lightproof case. But there are many choices for the receiver side: photodiodes, BJT phototransistors, MOSFETs, photo-triacs, photo-Darlingtons, and more.
So while implementation details vary, the crux is that your microcontroller turns on an LED, and it’s the light from that LED that activates the other side of the circuit. The only connection between the LED side and the transistor side is non-electrical — light across a small gap — and that provides the rock-solid, one-way isolation. | 5,466 | 24,984 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2018-34 | longest | en | 0.954156 |
http://www.physicsforums.com/showthread.php?s=bef5d019c1aaf77dc009ae18c04bca0e&p=4064776 | 1,368,954,674,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368697232084/warc/CC-MAIN-20130516094032-00087-ip-10-60-113-184.ec2.internal.warc.gz | 648,046,336 | 14,106 | ## Can a magnet's magnetic field perform work on another magnet?
Quote by DaleSpam An opinion unsupported by any good evidence, and one not shared by authors of classical EM textbooks, like Jackson.
Took a look at Jackson 2nd Ed'n, 6.2 'Energy in the Magnetic Field'. The ab-initio assumption is made work is done on any loop current 'to maintain the current constant' against a changing flux through the circuit. OK for real currents flowing through real conducting circuits, but when made a blanket generalization for magnetic media, a strange position given his familiarity with the QM nature of an intrinsic moment. That same position is then translated holus-bolus into 6.8 where Poynting theorem is derived. Can find no account in between where such a classically behaving circuit is contrasted with that of an intrinsic moment. None. Guess that's why you just kept ducking every request of mine to show how E.j type work can be done on an intrinsic moment - Jackson has nothing to say - ipso-facto - neither then can you. Awe and reverence for authority figures. Often works out as good policy, but not always. Myself and others here have given references on the net, even pointing to some respected authority figures here in PF, that dispute Jackson's approach of it seems treating intrinsic magnetic moments as though needing to be maintained against some induced emf as though a classical circulating current. Ho hum. To each their own guru it seems.
Quote by DaleSpam I think that you and I agree on the physics, just not the semantics. So I respect your position and think it is reasonable. This "rubbish" usually goes by the name "Faraday's law of induction".
My "rubbish" comment was ambiguous, so I will restate it to clarify. A moving magnetic field does generate an E field per Faraday Induction Law. That is not "rubbish" at all. The "rubbish" I was referring to was the theory that said induced E field is what exerts the lifting force on the lower magnet.
When the lower magnet is on the floor about to ascend due to lifting force acting upon it, there is initially no motion, hence no induced E field & force. So the force lifting the magnet cannot be induced E field generated force. That theory is pure rubbish. The lift occurs before the induction, so induction & E force cannot account for the lift. Cause must always precede effect, & there are no exceptions or conditions to that rule.
Once again, after the magnet starts moving, an E field is indeed generated per Faraday's Induction Law, no argument there. That's been known since the mid 19th century. I'm glad we generally agree, & I agree that much of this is semantics. To me, the force lifting the magnet is Fm = qvXB. If we expand the system boundary region, another source energized said B field somewhere in history. That source received its energy from another source earlier. The disagreement truly is semantic. At least we agree on that as well. Best regards.
Claude
Mentor
Quote by Q-reeus Took a look at Jackson 2nd Ed'n, 6.2 'Energy in the Magnetic Field'. The ab-initio assumption is made work is done on any loop current 'to maintain the current constant' against a changing flux through the circuit. OK for real currents flowing through real conducting circuits, but when made a blanket generalization for magnetic media, a strange position given his familiarity with the QM nature of an intrinsic moment. That same position is then translated holus-bolus into 6.8 where Poynting theorem is derived. Can find no account in between where such a classically behaving circuit is contrasted with that of an intrinsic moment. None.
You can certainly dispute his editorial choices about what to discuss in his text, but my point is that Jackson (and the authors of all other classical EM textbooks of which I am aware) clearly treats permanent magnets classically. I.e. permanent magnets are part of classical EM.
Quote by Q-reeus Guess that's why you just kept ducking every request of mine to show how E.j type work can be done on an intrinsic moment - Jackson has nothing to say - ipso-facto - neither then can you.
I haven't ducked it. I have consistently answered it.
1) Permanent magnets are described by classical EM (ref Jackson)
2) Poynting's theorem follows from the laws of classical EM (ref derivations)
3) Work done on matter is E.j in all classical EM (ref Lorentz force law)
4) Therefore work done on a permanent magnet is E.j
Quote by Q-reeus Ho hum. To each their own guru it seems.
I don't particularly have a guru; my opinion is just based on the best evidence I have available and subject to revision as new evidence becomes available. It isn't my fault if your argument doesn't measure up by that standard.
Mentor
Quote by cabraham My "rubbish" comment was ambiguous, so I will restate it to clarify... The disagreement truly is semantic. At least we agree on that as well.
Thanks for the clarification. Given our previous discussion, I thought that it was probably just something like that.
Quote by cabraham A moving magnetic field does generate an E field per Faraday Induction Law. That is not "rubbish" at all. The "rubbish" I was referring to was the theory that said induced E field is what exerts the lifting force on the lower magnet. When the lower magnet is on the floor about to ascend due to lifting force acting upon it, there is initially no motion, hence no induced E field & force. So the force lifting the magnet cannot be induced E field generated force. That theory is pure rubbish. The lift occurs before the induction, so induction & E force cannot account for the lift. Cause must always precede effect, & there are no exceptions or conditions to that rule.
I agree with this. The B field does indeed produce the force. I hope I never said otherwise. It is only the work that I was describing, and that is given by E.j. Whether we say that it was B via E.j or E.j via B is less important (IMO) than that we agree on the amount of work done and the forces involved.
Quote by DaleSpam I haven't ducked it. I have consistently answered it. 1) Permanent magnets are described by classical EM (ref Jackson)
As I stated earlier, fully true only insofar as they act as sources of EM fields, or respond to such fields - with the important exclusion of E field curl component (re non-action on intrinsic moments). There is no way a permanent magnet could exist if governed solely by classical EM. Surely you don't dispute that.
2) Poynting's theorem follows from the laws of classical EM (ref derivations)
Agreed.
3) Work done on matter is E.j in all classical EM (ref Lorentz force law)
Agreed. But PM's are not 'all classical'.
4) Therefore work done on a permanent magnet is E.j
The inevitable non sequitur. Only in a strictly formal sense - by pretending Amperian 'currents' are real currents. And even then, there are problems. But for the sake of not distracting things further, I will simply agree to a presumed formal but not real E.j type 'work' being done, whereas in fact all appreciable interactions are magnetic in origin for PM's.
The marathon's not yet over.
Quote by DaleSpam Thanks for the clarification. Given our previous discussion, I thought that it was probably just something like that. I agree with this. The B field does indeed produce the force. I hope I never said otherwise. It is only the work that I was describing, and that is given by E.j. Whether we say that it was B via E.j or E.j via B is less important (IMO) than that we agree on the amount of work done and the forces involved.
Once again our agreement is near complete. But how can B produce force which displaces the mass a distance upward increasing its PE, w/o doing the work? That is illogical. So we agree B produces the Fm Lorentz force that lifts the magnet. That Fm force times height raised is the work done. B did the work. But, E.J is needed to energize B. Once again, it all depends where the system boundary is arbitrarily drawn.
I'm satisfied that we have consensus except for the matter of arbitrarily choosing system boundaries, & minor semantics. I just find it surprising that so many critics agree on B providing the force, but not doing the work. How can the force which lifts not be doing the work? Strange it is indeed. Oh well, no use dwelling on it. Cheers.
Claude
Mentor
Quote by cabraham Once again our agreement is near complete. But how can B produce force which displaces the mass a distance upward increasing its PE, w/o doing the work? That is illogical.
Because B doesn't act on the mass, it acts on the charges which are in motion relative to the mass. The force B exerts on the charges is always perpendicular to their motion, per the Lorentz force law, and therefore does not transfer energy.
Quote by cabraham I'm satisfied that we have consensus except for the matter of arbitrarily choosing system boundaries, & minor semantics. I just find it surprising that so many critics agree on B providing the force, but not doing the work. How can the force which lifts not be doing the work? Strange it is indeed. Oh well, no use dwelling on it.
I agree that the remaining difference is semantic, but hopefully the Lorentz force law and Poynting's theorem help you understand why other people will describe it differently than you do and perhaps help you understand that the description is reasonable, even though it differs from your own.
Mentor
Quote by Q-reeus As I stated earlier, fully true only insofar as they act as sources of EM fields, or respond to such fields - with the important exclusion of E field curl component (re non-action on intrinsic moments).
You did state it earlier, but without evidence.
Quote by Q-reeus There is no way a permanent magnet could exist if governed solely by classical EM. Surely you don't dispute that.
Classical EM doesn't purport to be a theory about the existence of magnets, charges, or fields. It is just a theory about their behavior. Classical EM does, in fact, accurately describe the behavior of permanent magnets, including force, energy, work, etc.
Quote by Q-reeus Only in a strictly formal sense - by pretending Amperian 'currents' are real currents. And even then, there are problems. But for the sake of not distracting things further, I will simply agree to a presumed formal but not real E.j type 'work' being done, whereas in fact all appreciable interactions are magnetic in origin for PM's.
OK, as long as the equations accurately predict the outcome of experiments, that is sufficiently "real" for me. I try to remain agnostic about everything else.
Quote by DaleSpam Because B doesn't act on the mass, it acts on the charges which are in motion relative to the mass. The force B exerts on the charges is always perpendicular to their motion, per the Lorentz force law, and therefore does not transfer energy. I agree that the remaining difference is semantic, but hopefully the Lorentz force law and Poynting's theorem help you understand why other people will describe it differently than you do and perhaps help you understand that the description is reasonable, even though it differs from your own.
Sorry Dale but you can't have it both ways. The E force is also perpendicular to the magnet motion. Not to the charge, but to the direction the mass is moving. Using your definition, E force cannot do work either because E force is not in line with the magnet's motion. I've explained to the point of exhaustion that the work is done on the mass, not the charges. The Fm force (due to B) acts in the right direction along with the motion, but Fe does not. That "Fm is normal to charge motion & thus cannot do work" argument has been demolished beyond sufficiency. We are well beyond that. I can't believe that that argument is still being invoked. I explained that in my 1st post in the 1st thread.
All my posts are consistent with all known laws of physics. Like I said recently, the reason some differ with me is based on where the system boundaries are arbitrarily drawn. If you zoom in close, just looking at Fm & the magnets, then Fm lifts the magnet & does the work. But zoom your lens out to a bigger picture, & B gets energy from another field/source. Zoom out more & another energy source is "doing the work".
Fm acts on the magnet lifting it a finite distance. The Fm direction is along the motion. It does work, not on the charges but on the magnet mass as a whole. The direction is along, not normal. By the definition you cited, that is work being done by Fm (associated Lorentz force per B).
To me it's too easy, & I am at a loss to understand how this can even be argued. We arrived at the right answer long ago. Some just refuse to accept Fm as doing work. It's just a prejudice & nothing more. All attempts to invoke physics laws to prove that E does the work have not withstood scrutiny. BR.
Claude
Mentor
Quote by cabraham Sorry Dale but you can't have it both ways. The E force is also perpendicular to the magnet motion.
The E field is not perpendicular to the current, therefore it does work on the charges.
Quote by cabraham I've explained to the point of exhaustion that the work is done on the mass, not the charges.
This is where we disagree. EM forces only act on charges, not masses. I don't know if this is just another semantic disagreement or not, but if it is not then your position seems problematic to say the least.
Quote by Q-reeus So this is 'agreement'? Ha ha ha ha ha ha ha ha.
As I already gave up on this topic, seeing all this I just have to laugh along with you - but in despair :
Quote by cabraham Once again our agreement is near complete. But how can B produce force which displaces the mass a distance upward increasing its P[otential] E[nergy], w/o doing the work? That is illogical. [..]
Quote by DaleSpam [..] The force B [..] does not transfer energy. [..] I agree that the remaining difference is semantic [..].
Quote by harrylin As I already gave up on this topic, seeing all this I just have to laugh along with you - but in despair :
Ha ha. Harald, we all agree magnets just keep on working regardless of how each see's it. Guess that's all that really matters in the end.
People! Why can't you all agree that the magnetic fields induce the force that does the work. Call it what you want, but the magnetic field generated a force. I still don't understand why we haven't finished this yet. Based on all the laws of Electrodynamics its known as a fact that magnetic fields induce electrical fields and that electrical fields INDUCE magnetic fields. As Claude said before they are of the same coin. Each facing a different side. They are both equal to each other. In what way? In a way that without one of them, the other can't exist or do anything! Peace! Miyze,
Quote by Miyz People! Why can't you all agree that the magnetic fields induce the force that does the work. Call it what you want, but the magnetic field generated a force. I still don't understand why we haven't finished this yet. Based on all the laws of Electrodynamics its known as a fact that magnetic fields induce electrical fields and that electrical fields INDUCE magnetic fields. As Claude said before they are of the same coin. Each facing a different side. They are both equal to each other. In what way? In a way that without one of them, the other can't exist or do anything! Peace! Miyze,
Very true indeed. E & B (or E & H if you like) cannot exist independently if the energy is time varying (power is non-zero). Ampere's Law (AL) tells us that the curl of H equals the displacement current plus the conduction current. The displacement current is dD/dt, & the conduction current is J = σE As we know, D = εE. In a conductor we have conduction current, in an insulator we have displacement current. In either case, E is non-zero if H has a curl. If H is zero, so is E.
Faraday's Law (FL) tells us that the curl of E equals the time rate of change of B. A static B has no E, & an ir-rotational E has no B. But in these cases we've examined, induced E fields are always rotational, hence there has to be a time varying B associated. The B fields enclose a current, hence an E field is present. E forces align with charge carrier velocity.
So whenever one says, this field is "doing the work", another can counter with "no way, the other field is doing it", then produce equations showing how the work cannot be done without the other. They are right, of course, but they don't acknowledge the chicken-egg paradox they are falling victim to. Which comes first is endless.
Fe = qE acts in the wrong direction to lift the magnet, & it cannot be the one doing work. Fm = qvXB is in the right direction lifting the magnet, doing work equal to mgh. Although the Fm is not doing work on charges, it deflects charges w/o changing their KE, & internal bonding forces yank the lattice particles in the direction of the displaced electrons. These forces are E & SN.
I covered this issue in the current loop thread. I was told that my tethering treatise was irrelevant. So I am at a point where it is pointless to continue. I will clarify if asked, but if not asked, then I bid all of you a great & happy weekend, enjoy the pro football games, that is what I will be doing. Cheers.
Claude
Quote by cabraham Very true indeed. E & B (or E & H if you like) cannot exist independently if the energy is time varying (power is non-zero). Ampere's Law (AL) tells us that the curl of H equals the displacement current plus the conduction current. The displacement current is dD/dt, & the conduction current is J = σE As we know, D = εE. In a conductor we have conduction current, in an insulator we have displacement current. In either case, E is non-zero if H has a curl. If H is zero, so is E. Faraday's Law (FL) tells us that the curl of E equals the time rate of change of B. A static B has no E, & an ir-rotational E has no B. But in these cases we've examined, induced E fields are always rotational, hence there has to be a time varying B associated. The B fields enclose a current, hence an E field is present. E forces align with charge carrier velocity. So whenever one says, this field is "doing the work", another can counter with "no way, the other field is doing it", then produce equations showing how the work cannot be done without the other. They are right, of course, but they don't acknowledge the chicken-egg paradox they are falling victim to. Which comes first is endless. Fe = qE acts in the wrong direction to lift the magnet, & it cannot be the one doing work. Fm = qvXB is in the right direction lifting the magnet, doing work equal to mgh. Although the Fm is not doing work on charges, it deflects charges w/o changing their KE, & internal bonding forces yank the lattice particles in the direction of the displaced electrons. These forces are E & SN. I covered this issue in the current loop thread. I was told that my tethering treatise was irrelevant. So I am at a point where it is pointless to continue. I will clarify if asked, but if not asked, then I bid all of you a great & happy weekend, enjoy the pro football games, that is what I will be doing. Cheers. Claude
Good words used in good time.
I probably won't enjoy any games or such! I'll self teach my self about the wonders of or physical world.
Take care Claude! | 4,367 | 19,349 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2013-20 | latest | en | 0.961359 |
https://projekteuler.de/problems/261 | 1,660,621,555,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572220.19/warc/CC-MAIN-20220816030218-20220816060218-00506.warc.gz | 427,182,612 | 3,103 | # Pivotal Square Sums (noch nicht übersetzt)
Problem 261
Let us call a positive integer k a square-pivot, if there is a pair of integers m > 0 and nk, such that the sum of the (m+1) consecutive squares up to k equals the sum of the m consecutive squares from (n+1) on:
(k-m)2 + ... + k2 = (n+1)2 + ... + (n+m)2.
Some small square-pivots are
• 4: 32 + 42 = 52
• 21: 202 + 212 = 292
• 24: 212 + 222 + 232 + 242 = 252 + 262 + 272
• 110: 1082 + 1092 + 1102 = 1332 + 1342
Find the sum of all distinct square-pivots ≤ 1010. | 200 | 523 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2022-33 | latest | en | 0.602598 |
https://www.scirp.org/journal/paperinformation.aspx?paperid=71871 | 1,696,004,642,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510520.98/warc/CC-MAIN-20230929154432-20230929184432-00526.warc.gz | 1,051,860,685 | 32,536 | Optimal Portfolio Strategy with Discounted Stochastic Cash Inflows When the Stock Price Is a Semimartingale
Abstract
This paper discusses optimal portfolio with discounted stochastic cash inflows (SCI). The cash inflows are invested into a market that is characterized by a stock and a cash account. It is assumed that the stock and the cash inflows are stochastic and the stock is modeled by a semi-martingale. The Inflation linked bond and the cash inflows are Geometric. The cash account is deterministic. We do some scientific analyses to see how the discounted stochastic cash inflow is affected by some of the parameters. Under this setting, we develop an optimal portfolio formula and later give some numerical results.
Keywords
Share and Cite:
Baraedi, O. and Offen, E. (2016) Optimal Portfolio Strategy with Discounted Stochastic Cash Inflows When the Stock Price Is a Semimartingale. Journal of Mathematical Finance, 6, 660-684. doi: 10.4236/jmf.2016.64047.
1. Introduction
For example in financial mathematics, the classical model for a stock price is that of a geometric Brownian motion. However, it is argued that this model fails to capture properly the jumps in price changes. A more realistic model should take jumps into account. In the Jump diffusion model, the underlying asset price has jumps super- imposed upon a geometric Brownian motion. The model therefore consists of a noise component generated by the Wiener process, and a jump component. It involves modelling option prices and finding the replicating portfolio. Researchers have increasingly been studying models from economics and from the natural sciences where the underlying randomness contains jumps. According to Nkeki [1] , the wars, decisions of the Federal Reserve, other central banks, and other news can cause the stock price to make a sudden shift. To model this, one would like to represent the stock price by a process that has jumps (Bass [2] ). Liu et al. (2003) [3] solved for the optimal portfolio in a model with stochastic volatility and jumps when the investor can trade the stock and a risk-free asset only. They also found that Liu and Pan (2003) [4] ex- tended this paper to the case of a complete market. Arai [5] considered an incomplete financial market composed of d risky assets and one riskless asset. Branger and Larsen [6] solved the portfolio planning problem of an ambiguity averse investor. They considered both an incomplete market where the investor can trade the stock and the bond only, and a complete market, where he also has access to derivatives. In Guo and Xu (2004) [7] , researchers applied the mean-variance analysis approach to model the portfolio selection problem. They considered a financial market containing assets: risky stocks and one bond. The security returns are assumed to follow a jump-diffusion process. Uncertainty is introduced by Brown motion processes and Poisson processes The general method to solve mean-variance model is the dynamic programming. Dynamic programming technique was firstly introduced by Richard Bellman in the 1950s to deal with calculus of variations and optimal control prob-lems (Weber et al. [8] ). Further developments have been obtained since then by a number of scholars including Florentin (1961, 1962) and Kushner (2006), among others. In Jin and Zhang [9] , researchers solved the optimal dynamic portfolio choice problem in a jump-diffusion model with some realistic constraints on portfolio weights, such as the no-short-selling constraint and the no-borrowing constraint. Beginning with work of Nkeki [1] which involves optimization of the portfolio strategy using discounted stochastic cash inflows, this work explores optimal portfolio strategy using jump diffussion model.
In Nkeki [1] , the stock price is modelled by continuous process which is geometric and but in this work we assume that the stock price process is driven by a sem- imartingale; defined in Shiryaev et al. [10] . The jump diffusion model combines the usual geometric Brownian motion for the diffusion and the general jump process such that the jump amplitudes are normally distributed.
Semimartingales as a tool of modelling stock prices processes has a number of advantages. For example this class contains discrete-time processes, diffusion processes, diffusion processes with jumps, point processes with independent increments and many other processes (Shiryaev [11] ). The class of semimartingales is stable with respect to many transformations: absolutely continuous changes of measure, time changes, localization, changes of filtration and so on as stated in (Sharyaev [11] ). Sto- chastic integration with respect to semimartingales describes the growth of capital in self-financing strategies. In this research, a sufficient maximum principle for the optimal control of jump diffusions is used showing dynamic programming and going applications to financial optimization problem in a market described by such process. For jump diffusions with jumps, a necessary maximum principle was given by Tang and Li, see also Kabanov and Kohlmann (Æksendal and Sulem [12] ). If stochastic control satisfies the maximum principle conditions, then the control is indeed optimal for the stochastic control problem. It is believed that such results involves a useful complicated integro-differential equation (the Hamilton-Jacobi-Bellmann equation) in the jump diffusion case. The investor’s stochastic Cash inflows (CSI) into the cash account, on inflation-linked bond and stock were considered. Most calculations and methods used were influenced by the works of Nkeki [1] , Nkeki [13] Æksendal [14] , Æksendal and Sulem [12] , Klebaner [15] and Cont and Tankov [16] .
2. Model Formulation
Let be a probability space where denotes the “flow of infor- mation” as discussed in the definition. Mathematically the latter means that consists of σ-algebras, i.e. for all. The Brownian motions is a 2-dimensional process on a given filtered probability space, where is the real world probability measure, t is the time period, T is the terminal time period, is the Brownian motion with respect to the “noise” arising from the inflation and is the Brownian motion with respect to the “noise” arising from the stock market.
The dynamics of the cash account with the price is given by:
(1)
where r is the short term interest as defined in Nkeki [1] .
The price of the inflation-linked bond is given by the dynamics:
(2)
where is the volatility of inflation-linked bond, is the market price of inflation risk, is the inflation index at time t and has the dynamics:
where is the expected rate of inflation, which is the difference between nominal interest rate, real interest and is the volatility of inflation index.
Suppose the financial process ( stock return) is given on a filtered probability space. Assume that is of “exponential form”.
(3)
where is a semi-martingale with respect to and.
Using Itô formula for semimartingales (see Appendix) and then differentiating the process we have
(4)
where
(5)
Using random measure if jumps (see [11] )
(6)
hence
(7)
Substituting on Equation (7) into Equation (50) we have
(8)
We know that differential of our stock price can written as
(9)
where and defined as before.
Now comparing Equation (8) with Equation (9), we can now see that when we equate the predictable parts we have
(10)
Equating the continuous parts we get
(11)
and the jump parts give
and hence we let
(12)
From (11) it follows that and hence it follows that
Substituting Equation (12) into Equation (9) we have
(13)
and further simply it to
(14)
where
Using Itó’s formula for jump diffusion
(see Appendix). Hence we define the following
(15)
The market price of the market risk is given by
(16)
where, is the market price of stock market risk. We assume the process which is geometric and with the no arbitrage conditions applied to it obtain the following stochastic differential equation,
(17)
Using Itó’s formula for jump diffusion equation on 17 we have
(18)
where (see Appendix).
is a martingale that is always positive and satisfies.
Now we have the price density given by
(19)
where
(20)
3. The Dynamics of Stochastic Cash Inflows
The dynamics of the stochastic cash inflows with process, is given by
(21)
where is the volatility of the cash inflows and k is the expected growth
rate of the cash inflows. is the volatility arising from inflation and is the volatility arising from the stock market.
Solving for we use Itò’s formula for continuous processes. Let and
(22)
4. The Dynamics of the Wealth Process
If is the wealth process and is the admissible portfolio where is number of units in the cash account, is the number of units in the inflation bond and is the number of units in the stock. In an incomplete market with no arbitrage we have. The dynamics of the wealth process is given by
(23)
where
(24)
(see Appendix). For we have the dynamics of the wealth process as
(25)
For the Poisson jump measure we have the dynamics of the wealth process as
(26)
where is the Poisson measure and is the compensator on the Poisson measure.
5. The Discounted Value of SCI
In this Section, we introduce
Definition 1. The discounted value of the expected future SCI is defined as
(27)
where is the conditional expectation with respect to the Brownian Filtration and is the stochastic discount factor which adjust for nominal interest rate and market price of risks for stock and inflation-linked bond (Nkeki [1] ).
Proposition 1. If is the discounted value of the expected future SCI, then
(28)
Proof. By definition 1, we have that
(29)
Applying change of variable on 30, we have
(30)
starting with
we have
and lastly
We further take note that for we have the discounted value of the SCI as
(31)
The differential form of is given by
(32)
Equation (32) is obtained by differentiating as shown in the proof below
differentiating both sides,
The current discounted cash inflows can be obtained by putting into Equation (28),
(33)
If and we can change the horizon by allowing
our T to go up to i.e.
In case of deterministic case, we have and, so
(34)
and for, and we have
Since is a constant, if we are interested to see how it behaves with respect to we need to take as a function of. Then we can look at the sensitivity analysis of,
Finding partial derivatives of we obtain the followng
Differentiating with respect to T, we have
(35)
Differentiating with respect to, we have
(36)
Differentiating with respect to k, we have
(37)
Differentiating with respect to r, we have
(38)
Differentiating with respect to, we have
(39)
where and and
The following calculations shows how we differentiated with respect to T
differentiating with respect to T
We repeated the following procedure for all other variables.
When we have a deterministic case, differentiating partially we have the fol- lowing
Differentiating with respect to T, we have
(40)
Differentiating with respect to, we have
(41)
Differentiating with respect to r, we have
(42)
Differentiating with respect to, we have
(43)
Differentiating with respect to k, we have
(44)
Table 1 shows the sensitivity of variables. Sensitivity analysis can be incorporated into discounted cash inflows analysis by examining how the discounted cash inflows of each project changes with changes in the inputs used. These could include changes in revenue assumptions, cost assumptions, tax rate assumptions, and discount rates. It also enables management to have contingency plans in place if assumptions are not met. It also shows that the return on the project is sensitive to changes in the projected revenues and costs. Looking at Table 1, one can see that changing a variable can make
Table 1. Simulation of the sensitivity analysis.
an impact on the SCI. An investor must do the sensitivity analysis in order to know changes can be made on the market to improve the results of an investment.
6. The Dynamics of the Value Process
Proposition 2. If is the value process and where is the discounted value of the expected future SCI then the differential form of is given by
(45)
Proof. Differentiating and substituting Equations (32) and (26) on the dif- ferential obtained we have
For, the jump part becomes zero and we obtain
7. Finding Optimal Portfolio
Theorem 3. Let be the worth process whose dynamics is defined by Equation (23), the discounted value of expected future stochastic cash inflow as defined in proportion (1), the value process as defined in proportion (2) and
the utility function and if we assume that, the optimal portfolio is
given by where
and
(46)
The proof is given in Appendix.
From Equation (71), represents the
classical portfolio strategy at time t and represents the inter-
temporal hedging term that offset shock from the SCI at time t.
Some Numerical Values
Figure 1 was obtained by setting, , , , , , , , , and in Equation (70). This figure shows that when, the portfolio value is 0.151 which is equivalent to 15.1% when the value of the wealth is 40,000 and the portfolio value is 0.159 which is equivalent to 15.9% when the value of the wealth is 1,000,000. When, the portfolio value is 0.16 which is equivalent to 16% when the value of the wealth is 40,000
Figure 1. Portfolio value in inflation-linked bond.
and the portfolio value is 0.1604 which is equivalent to 16.04% when the value of the wealth is 1,000,000. This shows that there is a huge increase on the portfolio value from to when the value of the wealth is small and there in less change when the value of the wealth is large.
Figure 2 was obtained by setting, , , , , , , , , and in Equation (71). This figure shows that when, the portfolio value is 0.907 which is equivalent to 90.7% when the value of the wealth is 40,000 and the portfolio value is 0.9019 which is equivalent to 90.19% when the value of the wealth is 1,000,000. When, the portfolio value is 0.9017 which is equivalent to 90.17% when the value of the wealth is 40,000 and the portfolio value is 0.9017 which is equivalent to 90.17% when the value of the wealth is 1,000,000. This shows that there is a huge decrease on the portfolio value from to when the value of the wealth is small and there in less change when the value of the wealth is large.
Figure 3 was obtained by setting, , , , , , , , , and in Equation (72). This figure shows that when, the portfolio value is −0.057 which is equivalent to −5.7% when the value of the wealth is 40,000 and the portfolio value is −0.0613 which is equivalent to −6.13% when the value of the wealth is 1,000,000. When, the portfolio value is −0.0615 which is equivalent to −6.15% when the value of the wealth is 40,000 and the portfolio value is −0.0613 which is equivalent to 6.13% when the value of the wealth is 1,000,000. This shows that there is a huge decrease on the portfolio value from to when the value of the wealth is small and
Figure 2. Portfolio value in stock.
Figure 3. Portfolio value in cash account.
there in less change when the value of the wealth is large.
For, we have a problem because we cannot solve the equation explicitly. we need to come up with a computer program.
8. Conclusion
Semimartingales seems to model financial processes better since the cater for the jumps that occur in the system. The continuous processes may be convenient because one can easily produce results. For example, in our situation we managed to find the portfolio for continuous processes but we couldn’t for the ones with jumps. This work can be extended designing a MATLAB program that will solve the equation for portfolio.
Acknowledgements
We thank the editor and the referee for their comments. We also thank Professor E. Lungu for the guidance he gave us on achieving this. Lastly, we thank the University of Botswana for the resources we used to come up with this paper. Not forgetting the almighty God, the creator.
Appendix
Appendix A
Assume that and. Using Itô formula for sememartingales (see Jacod [?], Protter [?], Shiryaev [11] , Shiryaev [10] ), one obtains
(47)
to find our SDE, assume that and substitute on Equation (47). Simplifying will give the following results
(48)
Differentiating will give;
(49)
Now the differential of the stock process is given by
(50)
where
(51)
then, using Ito’s formula for semimartingales (Protter [?]), we have
(52)
and in differential form, this can be expressed as
(53)
Appendix B
Assuming and substituting it on the formula we get
(54)
Appendix C
Let and
(55)
Appendix D
then
(56)
where
with and
was found by simply dividing by i.e.
Appendix E
Let and define. Then is a sto- chastic process with jumps and
take and substituting on 58 to have
Choosing such that for a given portfolio strategy (not necessarily optimum, we introduce the associated utility
(57)
Substituting, and we now have
Integrating both sides we get
Taking the expectations on both sides we have
For simplicity we have
Where and. Since we know that , we now have
Which gives us
By Equation (57), we have the integral on the right hand side being equals to zero. That is
Differentiating both sides we obtain following partial differential equation with jumps.
Consider the value function
(58)
where J is as in Equation (57) Under technical conditions, the value function V satisfies
(59)
This takes us to the HJB equation;
(60)
where is the second linear operator for jump diffusion. Hence
(61)
Taking our utility function as
(62)
We consider the function of which is
(63)
Differentiating and substitute on (63), we get
(64)
Since is a concave function of, to find its maximum we differentiate (64) with respect to to obtain
(65)
For we can solve for because we have a linear equation below
(66)
and will be given by
(67)
substituting and as defined , we obtain the following
(68)
(69)
where
(70)
and
(71)
We can now see that
(72)
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Conflicts of Interest
The authors declare no conflicts of interest.
[1] Nkeki, C.I. (2013) Optimal Portfolio Strategy with Discounted Stochastic Cash Inflows, Journal of Mathematical Finance, 3, 130-137. http://dx.doi.org/10.4236/jmf.2013.31012 [2] Bass, B.F. (2004) Stochastic Differential Equations with Jumps. Probability Surveys, 1, 1-19. http://dx.doi.org/10.1214/154957804100000015 [3] Liu, J., Pan, J. and Wang, T. (2001) An Equilibrium Model of Rare Event Premia. Working Paper, UCLA, MIT Sloan, and UBC. [4] Liu, J. and Pan, J. (2003) Dynamic Derivative Strategies. Journal of Financial Economics, 69, 401-430. http://dx.doi.org/10.1016/S0304-405X(03)00118-1 [5] Arai, T. (2004) Minimal Martingale Measures for Jump Diffussion Processes. Journal of Applied Probability, 41, 263-270. http://dx.doi.org/10.1017/S0021900200014194 [6] Branger, N. and Larsen, L.S. (2013) Robust Portfolio Choice with Uncertainty about Jump and Diffusion Risk. Journal of Banking and Finance, 37, 5036-5047. http://dx.doi.org/10.1016/j.jbankfin.2013.08.023 [7] Guo, W. and Xu, C. (2004) Optimal Portfolio Selection When Stock Prices Follow an Jump-Diffusion Process. Mathematical Methods of Operations Research, 60, 485-496. http://dx.doi.org/10.1007/s001860400365 [8] Azevedo, N., Pinheiro, D. and Weber, G.W. (2014) Dynamic Programming for a Markov-Switching Jump Diffusion. Journal of Computational and Applied Mathematics, 267, 1-19. http://dx.doi.org/10.1016/j.cam.2014.01.021 [9] Jin, X. and Zhang, K. (2013) Dynamic Optimal Portfolio Choice in a Jump-Diffusion Model with Investment Constraints. Journal of Banking and Finance, 37, 1733-1746. http://dx.doi.org/10.1016/j.jbankfin.2013.01.017 [10] Shiryaev, A.N., Buhlmann, H., Delbaen, F. and Embrechts, P. (1995) No-Arbitrage, Change of Measure and Conditional Esscher Transformations. https://people.math.ethz.ch/~delbaen/ftp/preprints/BDES-CWI.pdf [11] Shiryaev, A. (2003) Essentials of Stochastic Finance: Facts, Models and Theory. World Scientific, Singapore, 3. [12] Oksendal, B. and Sulem, A. (2009) Applied Stochastic Control of Jump Diffusions. 3rd Edition, Springer, Berlin. [13] Nkeki, C.I. (2013) Optimal Investment under Inflation Protection and Optimal Portfolios with Discounted Cash Flows Strategy. Journal of Mathematical Finance, 3, 130-137. http://dx.doi.org/10.4236/jmf.2013.31012 [14] Oksendal, B. (2010) Stochastic Differential Equations: An Introduction with Applications. 6th Edition, Springer, Berlin. [15] Klebaner, F.C. (1998) Introduction to Stochastic Calculus with Applications. Imperial College Press, London. http://dx.doi.org/10.1142/p110 [16] Cont, R. and Tankov, P. (2004) Financial Modelling with Jump Processes. CRC Press, Boca Raton. | 5,097 | 21,264 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2023-40 | latest | en | 0.918308 |
http://us.metamath.org/mpeuni/ax-13.html | 1,632,100,168,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056974.30/warc/CC-MAIN-20210920010331-20210920040331-00485.warc.gz | 62,863,735 | 4,299 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > ax-13 Structured version Visualization version GIF version
Axiom ax-13 2137
Description: Axiom of Quantified Equality. One of the equality and substitution axioms of predicate calculus with equality. An equivalent way to express this axiom that may be easier to understand is (¬ 𝑥 = 𝑦 → (¬ 𝑥 = 𝑧 → (𝑦 = 𝑧 → ∀𝑥𝑦 = 𝑧))) (see ax13b 1914). Recall that in the intended interpretation, our variables are metavariables ranging over the variables of predicate calculus (the object language). In order for the first antecedent ¬ 𝑥 = 𝑦 to hold, 𝑥 and 𝑦 must have different values and thus cannot be the same object-language variable (so they are effectively "distinct variables" even though no \$d is present). Similarly, 𝑥 and 𝑧 cannot be the same object-language variable. Therefore, 𝑥 will not occur in the wff 𝑦 = 𝑧 when the first two antecedents hold, so analogous to ax-5 1793, the conclusion (𝑦 = 𝑧 → ∀𝑥𝑦 = 𝑧) follows. Note that ax-5 1793 cannot prove this because its distinct variable (\$d) requirement is not satisfied directly but only indirectly (outside of Metamath) by the argument above. The original version of this axiom was ax-c9 33068 and was replaced with this shorter ax-13 2137 in December 2015. The old axiom is proved from this one as theorem axc9 2194. The primary purpose of this axiom is to provide a way to introduce the quantifier ∀𝑥 on 𝑦 = 𝑧 even when 𝑥 and 𝑦 are substituted with the same variable. In this case, the first antecedent becomes ¬ 𝑥 = 𝑥 and the axiom still holds. Although this version is shorter, the original version axc9 2194 may be more practical to work with because of the "distinctor" form of its antecedents. A typical application of axc9 2194 is in dvelimh 2228 which converts a distinct variable pair to the distinctor antecedent ¬ ∀𝑥𝑥 = 𝑦. In particular, it is conjectured that it is not possible to prove ax6 2142 from ax6v 1839 without this axiom. This axiom can be weakened if desired by adding distinct variable restrictions on pairs 𝑥, 𝑧 and 𝑦, 𝑧. To show that, we add these restrictions to theorem ax13v 2138 and use only ax13v 2138 for further derivations. Thus, ax13v 2138 should be the only theorem referencing this axiom. Other theorems can reference either ax13v 2138 (preferred) or ax13 2140 (if the stronger form is needed). This axiom scheme is logically redundant (see ax13w 1961) but is used as an auxiliary axiom scheme to achieve scheme completeness (i.e. so that all possible cases of bundling can be proved; see text linked at mmtheorems.html#ax6dgen). It is not known whether this axiom can be derived from the others. (Contributed by NM, 21-Dec-2015.) (New usage is discouraged.)
Assertion
Ref Expression
ax-13 𝑥 = 𝑦 → (𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧))
Detailed syntax breakdown of Axiom ax-13
StepHypRef Expression
1 vx . . . 4 setvar 𝑥
2 vy . . . 4 setvar 𝑦
31, 2weq 1824 . . 3 wff 𝑥 = 𝑦
43wn 3 . 2 wff ¬ 𝑥 = 𝑦
5 vz . . . 4 setvar 𝑧
62, 5weq 1824 . . 3 wff 𝑦 = 𝑧
76, 1wal 1472 . . 3 wff 𝑥 𝑦 = 𝑧
86, 7wi 4 . 2 wff (𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧)
94, 8wi 4 1 wff 𝑥 = 𝑦 → (𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧))
Colors of variables: wff setvar class This axiom is referenced by: ax13v 2138
Copyright terms: Public domain W3C validator | 1,069 | 3,271 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2021-39 | longest | en | 0.879379 |
https://electronics.stackexchange.com/questions/227562/voltage-drop-at-input-of-opamp | 1,696,016,362,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510528.86/warc/CC-MAIN-20230929190403-20230929220403-00286.warc.gz | 259,879,322 | 40,742 | # Voltage drop at input of opamp
simulate this circuit – Schematic created using CircuitLab
At VM1 I theoretically expect about 1mV. When I connect with the opamp I get a voltage drop of 0.5mV at VM1, I get this values in simulation(theoretically). What is the cause of this (something to do with the 10MOhm) and what can I do to prevent this in the practical circuit.
• Welcome to EE.SE A schematic would be useful if you want to get good answers - there is a schematic editor widget available in the edit toolbar. The question also really needs to be edited for readability. Apr 10, 2016 at 16:27
Edited question for non-inverting case, removed circuit + analysis of negative op-amp amplifier.
Yes the problem is most likely the 10MOhm resistor. This happens because while the op-amp assumption is that $$I_+ = I_- = 0$$ this is not really the case, and there is a small input current.
You should decrease the size of both voltage divider resistors while keeping their ratio the same. This will give the required attenuation while also allowing the op-amp terminals to sink some current.
You can look at the datasheet to see how much current the op-amp inverting/non-inverting terminals actually draw. You want this current to make only a small difference compared to the voltage you are amplifying.
I will say that trying to amplify a 500uV signal like this is going to create a lot of noise, and the voltage divider will draw a lot of power. A better way is probably to control the gain with the feedback resistor ratio rather than pre-attenuate it like this, especially for a 1V signal.
• Sadly, this well-written answer is now more or less irrelevant thanks to the original question being updated with a schematic showing that an inverting configuration is not being used.
– pipe
Apr 10, 2016 at 22:42
• The goal of this circuit is to measure currents in de nA range, the source voltage will be around 3V instead of 1V. Is there another option than lowering the resistance values (e.g. another opamp, something with feedback resistors)? Apr 11, 2016 at 10:56
• I found a solution for my problem, by using a JFET opamp the problem does not exist anymore. The reason is that the input impedance is very high (around TOhm) and lower bias current. Apr 11, 2016 at 12:47
• Yes, the '741 op-amp is ancient and no longer used in new designs. Apr 11, 2016 at 19:15 | 574 | 2,371 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-40 | latest | en | 0.929011 |
http://mapleprimes.com/tags/groebner?page=2 | 1,498,318,576,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320264.42/warc/CC-MAIN-20170624152159-20170624172159-00647.warc.gz | 235,777,453 | 34,940 | ## How to find polynomial map?...
how to calculate the polynomial map for a system of polynomials
assume system of polynomial is in terms of a,b,c
how to find polynomial map
(r - something in terms of a,b,c)
(u - something in terms of a,b,c)
(v - something in terms of a,b,c)
## Skew product does not work properly...
Good day everyone,
I want to construct groebner bases over rings of differential Operators.
Thus I used the following:
with(Ore_algebra);
with(Groebner); N := 3;
A := skew_algebra(diff = [D[1], x[1]], diff = [D[2], x[2]], diff = [D[3], x[3]], comm = i, alg_relations = i^2+1); T := MonomialOrder(A, tdeg(D[1], D[2], D[3]));
A["polynomial_indets"]; [returns {D[1], D[2], D[3]}]
A["rational_indets"]; [returns {i, x[1], x[2], x[3]}]
So far everything seems as it should be: The ring i wanted to define here is the third Ring of Differential operators over the field of complex 'rational' functions and maple returns that indeed it will handle the D[i] as monomials and the rest as coefficients for them.
Hover, when i use the Skew product, the following happens:
skew_product(x[1],D[1],A) [returns D[1]x[1]+1}]
skew_product(x[1],D[1],A) [returns D[1]x[1]]
Both is wrong, but maple seems to 'know' this. It used the correct relation to in the first product, the only thing it didn't do was switch D[1] and x[1]. I think maple handles the result of the skew product as if it was a commutative product and always places the D[i] at the left but still 'knows' what the actual result is.
The actual results should have been x[1]D[1]+1 for the first and x[1]*D[1] for the second product.
In the second product, it seems like maple treated x[1]*D[1] as if the Elements were switched already.
What i want though is for maple to correctly display the skew products and return the products so that in every summand the D[i] are at the right side and their coefficients are at the left side (and if possible (i do not know how to do that yet) sort the result of a skew product in a way that displays every different Power product of the D[i] with their coefficients, so that i get (x[1]+x[2])*D[2] and not x[1]*D[2]+x[2]*D[2]).
Can anyone help me here?
## how to eliminate the solution?...
in page 137 of an introduction to groebner bases
how to eliminate the redundant solution (y^2-x*z, 0, -x^2+y*w)
from 3 of them?
eliminate({y,y^2-x*z,-z}, {x, y, z, w});
eliminate({-x,0,y}, {x, y, z});
eliminate({w,-x^2+y*w,-x}, {x, y, z, w});
## 'tord' error in Basis and Normal Form ...
with(Groebner):
T := lexdeg([x,y,z],[e1,e2]);
intermsof1 := y;
intermsof2 := -z;
GB := Basis([e1-intermsof1, e2-intermsof2], 'tord',T);
result := NormalForm(y^2-x*z, GB,'tord', T);
result := NormalForm(y^2-x*z, GB, T);
originally Basis do not have error when without parameter 'tord'
after it has argument error, it has to be added extra parameter tord
NormalForm has the same error too.
i do not understand why it has error, how to solve?
i just want to express y^2-x*z in terms of y and -z
## general case of floats not handled...
g:=Groebner:-Basis([a-2.0*b,b-2], plex);
Groebner:-Reduce(a, g, plex);
Error, (in content/polynom) general case of floats not handled
How to solve this problem simply？
## how to derive poincare series from symmetric funct...
how to derive poincare series from symmetric function in maple
## what is the default characteristic used in hilbert...
without specify the characteristic,
what is the default characteristic used in hilbert series?
## @Markiyan Hirnyk how to calculate for module...
how to calculate for module case
M := [[x*y,y,x],[x^2+x,y+x^2,y],[-y,x,y],[x^2,x,y]];
## example in help file can not succeed to run...
follow Computing non-commutative Groebner bases and Groebner bases for modules
in maple 12
Error, (in Groebner:-Basis) the first argument must be a list or set of polynomials or a PolynomialIdeal
then i find in maple 15 help file is changed from module M := [seq(Vector(subsop(i+1 = 1, [F[i], 0, 0, 0])), i = 1 .. 3)]
to array M := [seq( s^3*F[i] + s^(3-i), i=1..3)];
though it can run, but when apply other example can not run
such as
restart;
with(Groebner):
F := [2*x^2+3*y+z^2, x^2*z^2+z+2*x, x^4*y^7+3*x];
M := [seq( s^3*F[i] + s^(3-i), i=1..3)];
with(Ore_algebra);
A := poly_algebra(x,y,z,s);
T := MonomialOrder(A, lexdeg([s], [x,y,z]), {s});
G := Groebner[Basis](M, T);
Error, (in Groebner:-Basis) the first argument must be a list or set of polynomials or a PolynomialIdeal
G1 := select(proc(a) evalb(degree(a,s)=3) end proc, G);
[seq(Vector([seq(coeff(j,s,3-i), i=0..3)]), j=G1)];
C := Matrix([seq([seq(coeff(j,s,3-i), i=1..3)], j=G1)]);
GB := map(expand, convert(C.Vector(F), list));
Groebner[Basis](F, tdeg(x,y,z));
## how to do term over position (TOP) for leading mon...
follow Computing non-commutative Groebner bases and Groebner bases for modules
f:=[y^2+2*x^2*y, y^2];
f1:=[x+y^2,x];
f2:=[x,y];
g:=[f1,f2];
no result is (0,y^2)
where is wrong?
## What is Groebner?
Maple
What is Groebner? That was asked in different forms several times in MaplePrimes and MathStackExchange (for example, see http://math.stackexchange.com/questions/3550/using-gr?bner-bases-for-solving-polynomial-equations ). In view of this I think the presented post on Groebner basis will be useful. This post consists of two parts: its mathematical background and examples of solutions of polynomial systems by hand and with Maple.
Let us start. Up to Wiki http://en.wikipedia.org/wiki/Gr%C3%B6bner_basis ,Groebner basis computation can be seen as a multivariate, non-linear generalization of both Euclid's algorithm for computing polynomial greatest common divisors, and Gaussian elimination for linear systems. This is implemented in Maple trough the Groebner package.
The simplest introduction to the topic I know is a well-written book of Ivan Arzhantsev (https://zbmath.org/?q=an:05864974) which includes the proofs of all the claimed theorems and the solutions of all the exercises. Here is its digest groebner.pdf done by me (The reader is assumed to be familiar with the ideal notion and ring notion (one may refresh her/his knowledge, looking in http://en.wikipedia.org/wiki/Ideal_%28ring_theory%29)). It should be noted that there is no easy reading about this serious matter.
Referring to the digest as appropriate, we solve the system
S:={a*b = c^2+c, a^2 =a+ b*c, a*c = b^2+b} by hand and with the Groebner package.
For the order a > b > c we construct its ideal
J(S):=<f1 = a*b-c^2-c,f2 = a^2-a-b*c, f3 = a*c-b^2-b>.
The link between f1 and f2 gives
f1*a-f2*b = (-c^2-c)*a + (a + b*c)*b = a*b -a*c + b^2*c -
a*c^2 =f4.
The reduction with f1 produces
f4 ->-a*c^2- a*c + b^2*c + c^2 +c =: f4.
Now the reduction with f3 produces
f4 -> -b^2- b - b*c +c^2 + c =:f4.
The link between f2 and f3 gives:
f2*c - f3*a = a*b +a*b^2 -a*c -b*c^2 = f5.
The reduction with f1 produces
f5 -> -a*c + c*b +c^2 +c =:f5.
The reduction with f3 produces
f5 -> -b^2 -b + c*b +c^2 +c =:f5.
The reduction with f4 produces
f5 -> 2b*c =: f5.
The link between f1 and f3
f1*c - f3*b = b^3 + b^2 -c^3 -c^2=:f6.
The reduction with f4 produces
f6 -> 2b*c + 2b*c^2 -2c^3 -2c^2=:f6.
At last, we reduce f6 by f5, obtaining f6:= -2c^3 -2c^2.
We see the minimal reduced Groebner basis of S consists of
a^2 -a -b*c, -b^2 -b- b*c +c^2 +c, -2c^3 - 2c^2.
Now we find the solution set of the system under consideration. The equation -2c^3 - 2c^2 = 0 implies
c=0, c=0, c=-1. The the equation -b^2 - b - b*c +c^2 + c = 0 gives
b = 0 , b = -1, b = 0, b = -1, b = 0, b = 0 respectively.
At last, knowing b and c, we find a from a^2 -a -b*c = 0.
Hence,
[{a = 0, b = 0, c = 0}, {a = 1, b = 0, c = 0}, {a = 0, b = -1, c = 0}], [{a = 0, b = 0, c = 0}, {a = 1, b = 0, c = 0}, {a = 0, b = -1, c = 0}], [{a = 0, b = 0, c = -1}].
The solution of the system under consideration by the Groebner package is somewhat different because Maple does not find the minimal reduced Groebner basis directly.
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## DegreeReverseLexicographic = T2:=lexdeg([c,b,a],[z...
if DegreeLexicographic is T2:=lexdeg([a,b,c],[x,y,z]);
DegreeReverseLexicographic = T2:=lexdeg([c,b,a],[z,y,x]) ?
## what is the function for degree reverse lex orderi...
how to calculate hlibert series as in maple with Gröbner Bases
would like to know the algorithm and try in another programming language such as F#
i find the algorithm in book Singular introduction to commutative algebra
page 320 and 322
1. is it equal to the hilbert series function in maple?
eq1a := Homogenize(eq1, h);
eq2a := Homogenize(eq2, h);
eq3a := Homogenize(eq3, h);
T3:=lexdeg([a,b,c,h]);
GB := Basis([eq1a,eq2a,eq3a], T3); #a
InterReduce(F, ???);
2. what is the maple function for degree reverse lex ordering ?
## how to eliminate the variable h with resultant aft...
eq1a := Homogenize(eq1, h);
eq2a := Homogenize(eq2, h);
eq3a := Homogenize(eq3, h);
David Cox using Algebraic Geometry page 82 use resultant to eliminate variable h
eq1b := eq1a - x;
eq2b := eq2a - y;
eq3b := eq3a - z;
T2:=lexdeg([a,b,c],[x,y,z]);
GB := Basis([eq1b,eq2b,eq3b], T2);
r1 := resultant(eq1b, eq2b, h);
r2 := resultant(eq1b, eq3b, h);
r1 = r2
page 82 teach how to eliminate, after do question 2, discover r1 and r2 are the same.
how to eliminate the variable h with resultant after homogenize ideal with variable h
## how to find possible system of polynomials which h...
is it possible to find the input ideal from given hilbert series ?
i ask this because maple out of memory when computing all combination of ideals
which book teach how to design hilbert series
i want to research hilbert series and construct similar invariants to classify ideals
i do not know how many memory in order to do this. hope maple can display how much memory need before computation of all combination and display how long will wait for it finish computation.
my home computer can not afford this computation
1 2 3 4 Page 2 of 4
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# Common Questions and Answers about Psych Test Properties
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Explain each of the following in own words. Each should not exceed 6 to 8 doubled spaced lines.
a) Norms- what are they? What should you consider when comparing a client's score with the norms?
b) Age and grade equivalent- norms and their limitations.
c) Raw score, standardized scores, and percentile scores
d) Local versus national norms
e) Correlation
##### Solution Summary
Psychological tests are based on norms. This solution answers common questions about norms, including information on age and grade equivalents, raw scores, standard scores, percentiles, correlation, local norms, and national norms.
##### Solution Preview
Use the following to guide your responses to these questions:
Norms refer to Normative Data that is usually used when discussing psychological and other tests. When a test is normed, usually test publishers seek out psychologists to give new tests to a random sample of people (volunteers usually) so they can see what the average performance on the test is, and analyze the statistical properties of it.
When you are comparing a client's score to norms, it is important to note if the ethnicity and language and other individual factors have been included in the norming sample. For example, if you are giving an IQ test to a person who was born and raised in Alaska, you would like to ensure that the norming sample includes people from Alaska, so that it is a fair comparison.
Raw scores are merely the number correct on a given subtest. Please see the attached page for information on standard scores and percentiles as well as Age and Grade equivalents.
Local norms are merely norms that are based in a certain area. For example if a school district decides they want local norms on a particular test, they give the test to a representative sample in the district, and can then use those norms to compare kids they test with actual children living in the district. National norms are much broader and involve giving the test to a sample that usually matches census data. Then those norms can be used to compare a child's performance to other children their same age or grade, nationwide.
Correlation refers to how one test parallels another, or how the scores on the two tests compare. If a score on one test correlates to another, there is a relationship between the two. Correlation ...
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The topics covered in this course require good knowledge and understanding of calculus, probability theory, inferential statistics (point estimation, likelihood theory, confidence intervals, hypothesis testing), and programming with R. The courses “Mathematics for Statisticians”, “Statistics and Probability” and “Statistical Computing with R” from the MSc in Statistics and Data Science are prerequisites for this course. Students are also expected to have a basic understanding of the linear and logistic regression models.
## Description
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3. Applications of the Monte Carlo method
4. Variance reduction methods
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## Course Objectives
By the end of the course, students are expected to:
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2. be able to recognize when a method can be applied to solve a given problem;
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4. be able to implement the methods and algorithms using R;
5. be able to interpret critically the results yielded by the application of such methods.
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For more information, watch the video or go the the 'help-page' in MyTimetable. Please note: Joint Degree students Leiden/Delft have to merge their two different timetables into one. This video explains how to do this.
## Mode of Instruction
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## Assessment method
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###### FREE Expert Solution
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###### Problem Details
Design a voltaic cell from the following half-reactions:
Cr 2O 7(aq) + 14 H +(aq) + 6 e − → 2 Cr 3+ (aq) + 7 H 2O(l) E o = 1.33
O 2(g) + 4 H +(aq) + 4 e − → H 2O(l) E o = 1.23
In the overall reaction, how many H + ions would be shown and which side would they be on?
A. 18 on the left
B. 16 on the left
C. 10 on the left
D. 10 on the right
E. 12 on the right
What scientific concept do you need to know in order to solve this problem?
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